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https://numbermatics.com/n/7188446/ | 1,723,082,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00346.warc.gz | 359,284,204 | 6,588 | # 7188446
## 7,188,446 is an even composite number composed of two prime numbers multiplied together.
What does the number 7188446 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.
7188446 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 7188446:
### 2 × 3594223
See below for interesting mathematical facts about the number 7188446 from the Numbermatics database.
### Names of 7188446
• Cardinal: 7188446 can be written as Seven million, one hundred eighty-eight thousand, four hundred forty-six.
### Scientific notation
• Scientific notation: 7.188446 × 106
### Factors of 7188446
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 3594225
### Divisors of 7188446
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 10782672
• Sum of proper divisors (its aliquot sum) s(n): 3594226
• 7188446 is a deficient number, because the sum of its proper divisors (3594226) is less than itself. Its deficiency is 3594220
### Bases of 7188446
• Binary: 110110110101111110111102
• Base-36: 4A2N2
### Squares and roots of 7188446
• 7188446 squared (71884462) is 51673755894916
• 7188446 cubed (71884463) is 371454003867785340536
• The square root of 7188446 is 2681.1277477957
• The cube root of 7188446 is 192.9945321367
### Scales and comparisons
How big is 7188446?
• 7,188,446 seconds is equal to 11 weeks, 6 days, 4 hours, 47 minutes, 26 seconds.
• To count from 1 to 7,188,446 would take you about seventeen weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 7188446 cubic inches would be around 16.1 feet tall.
### Recreational maths with 7188446
• 7188446 backwards is 6448817
• The number of decimal digits it has is: 7
• The sum of 7188446's digits is 38
• More coming soon!
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The information we have on file for 7188446 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 7188446, math, Factors of 7188446, curriculum, school, college, exams, university, Prime factorization of 7188446, STEM, science, technology, engineering, physics, economics, calculator, seven million, one hundred eighty-eight thousand, four hundred forty-six.
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Some bits of this website may not work unless you switch it on. | 941 | 3,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.851502 |
http://research.stlouisfed.org/fred2/series/KCPPPGBBA156NUPN | 1,418,882,358,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802765616.69/warc/CC-MAIN-20141217075245-00168-ip-10-231-17-201.ec2.internal.warc.gz | 289,813,506 | 19,674 | # Consumption Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Barbados
2010: 71.38664 Percent (+ see more)
Annual, Not Seasonally Adjusted, KCPPPGBBA156NUPN, Updated: 2012-09-17 10:12 AM CDT
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Name: Email: | 530 | 2,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-52 | latest | en | 0.772265 |
https://physics.stackexchange.com/questions/321698/why-is-there-an-infinite-potential-barrier-between-degenerate-vacua-in-qft | 1,719,175,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00855.warc.gz | 399,916,728 | 41,009 | # Why is there an infinite potential barrier between degenerate vacua in QFT?
I recently read "Fifty Years of Yang-Mills Theory and my Contribution to it" by R. Jackiw, and stumbled upon an intriguing sentence:
Usually in quantum field theory tunneling is suppressed by infinite energy barriers between degenerate vacua – this leads to spontaneous symmetry breaking.
1. In what sense is there an infinite energy barrier? How can we see this explicitly?
2. What does this have to do with spontaneous symmetry breaking?
• There is a more or less intuitive discussion of this in section 2.11 of arxiv.org/abs/1703.05448 Commented Mar 27, 2017 at 13:21
• @AccidentalFourierTransform Thanks! I'll have a look at it
– jak
Commented Mar 27, 2017 at 13:27
• @AccidentalFourierTransform If I'm not missing sth crucial, I think his argument only holds for an infinite volume system, which our universe is not. In addition, from his argument it seems very well possible that transitions take place. The only problem seems to be that it can not happen everywhere in an infinite volume system, because that would need, well an infinite amount of energy. For a finite volume system the energy would be finite. However, these lecture notes were incredibly helpful for some other problems I was currently struggling with.
– jak
Commented Mar 27, 2017 at 14:32
Suppose you have a quantum field theory with one scalar field $\phi$, and the equation of motion for this field tells you that the lowest energy configuration is obtained when $\phi$ is constant throughout spacetime, with $\phi^2 = \phi_0^2$ (here $\phi_0$ is a constant determined by the parameters of the Lagrangian, for instance).
1. There are two classical vacua, characterized by $\langle \phi \rangle = \pm \phi_0$. If you want to tunnel from one to the other, the energy barrier will be some constant integrated on all of space. Because of the infinite volume of space, this energy barrier is infinite, independently of how small the energy density is.
2. As a consequence, the quantum states which are superpositions of the two classical vacua are completely suppressed. This means that there are two quantum vacuum, which are identical to the classical ones. Hence the $\mathbb{Z}_2$ symmetry of the equation $\phi^2 = \phi_0^2$ is broken in any given vacuum by $\langle \phi \rangle = \pm \phi_0$. This is spontaneous breaking of the $\mathbb{Z}_2$ symmetry.
In summary, tunneling in QFT in strictly more than one dimension (as opposed to quantum mechanics, that you can see as QFT in $0+1$ dimension) costs an infinite amount of energy because of the integration over space. This suppresses superposition states, and entails spontaneous symmetry breaking.
• Thanks a lot for your answer! Two small questions: 1). The crucial point seems to be that spacetime has an infinite volume. How is this assumption justified? Isn't the volume of spacetime, in reality, finite? 2.) Couldn't we, with the same line of thought, compute that energy barrier that the scalar field takes the other VEV in some "bubble", i.e. a finite volume? This barrier should be finite and therefore local tunneling to other vacuum states seem possible...
– jak
Commented Mar 27, 2017 at 14:45
• If space is compact (eg a sphere or a torus), then indeed the energy is finite and you can have tunneling. But in general, space is not assumed to have finite volume. Why do you think so ? Commented Mar 27, 2017 at 16:26
• For your other remark, yes you can have a bubble where the scalar takes the other VEV, but in that case the bubble is surrounded by a domain wall, which has some tension. In other words, this configuration does not have minimal energy, because at the boundary of the bubble the field $\phi$ is varying. Because of the tension of the wall, the bubble will ultimately disappear. Commented Mar 27, 2017 at 16:29
• I always thought that the big bang scenario implies that spacetime is finite. It started at zero and now becomes bigger. In addition, at a minimum, we have a causal horizon and anything farther away shouldn't matter. Thus instead of integrating "all of space" we could equally integrate to the causal horizon. This again would mean there is only a finite energy barrier between degenerate vacua. But then, there could be tunneling (instantons) between them and the correct vacuum would be a superposition of all possible degenerate vacua. (Analogous to the QCD vacuum).
– jak
Commented Mar 28, 2017 at 4:39
• I started a new question for this: physics.stackexchange.com/q/321857
– jak
Commented Mar 28, 2017 at 5:14 | 1,085 | 4,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.940794 |
https://bobistheoilguy.com/forums/threads/piston-velocity-and-position.322063/ | 1,674,809,274,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00452.warc.gz | 145,866,214 | 16,915 | # Piston Velocity and Position
#### MolaKule
Staff member
For this two-part problem we're going to use my 4.3L Vortec V6 in this example. Data: The engine's crankshaft is rotating at 1000 RPM. The Piston Stroke is 3.48 inches. To simplify the problem somewhat, neglect any connecting rod angles (for now). The usual Piston Stroke Length Ls and crank radius r, as in Ls = 2*r applies. Q1: What is the maximum velocity of any one of the pistons in Inches/Sec.? Q2: At what position in the cylinder will the piston attain its maximum velocity? This Question is not open to any Engineering Discipline or Physics Discipline.
Last edited:
Staff member
#### buck91
Wouldn't this just be [ (stroke length) * (4) * (rpm) ] / 60 seconds = inches/seconds?
#### MolaKule
Staff member
Originally Posted by buck91
Wouldn't this just be [ (stroke length) * (4) * (rpm) ] / 60 seconds = inches/seconds?
This question is similar to the chainsaw tooth velocity question: Hint: You can use one of the online calculators: Q1: What is the maximum velocity of any one of the pistons in Inches/Sec.? Q2: At what position in the cylinder will the piston attain its maximum velocity? This Question is not open to any Engineering Discipline or Physics Discipline. The centerline of the crankshaft is directly below the cylinder.
Last edited:
#### MrMoody
Assuming the centerline of the cylinder intersects the centerline of the crank, it seems fairly simple but maybe I'm missing something. If the stroke is 3.48, then so is the diameter of the big end circle. The circumference would be 3.48 x pi = ~ 10.933 inches. At 1000 rpm, that would be 10933 in/min ÷ 60 = 182.22 in/s, which would also be the maximum piston speed at 90° crank position. At 90°, the piston would be exactly halfway through its travel.
#### MrMoody
Originally Posted by MrMoody
At 90°, the piston would be exactly halfway through its travel.
I haven't looked at any links or google, but I don't think this is quite right, but getting it right would require knowing the rod length to do some trigonometry.
#### MolaKule
Staff member
Originally Posted by MrMoody
Originally Posted by MrMoody
At 90°, the piston would be exactly halfway through its travel.
I haven't looked at any links or google, but I don't think this is quite right, but getting it right would require knowing the rod length to do some trigonometry.
We're keeping it simple which is why I stated that we will neglect the crank angle for now.
Quote
To simplify the problem somewhat, neglect any connecting rod angles (for now). The usual Piston Stroke Length Ls and crank radius r, as in Ls = 2*r applies.
Last edited:
#### walterjay
The piston will be at the center point of its travel for max velocity. As for the other I cannot answer.
#### MolaKule
Staff member
Originally Posted by MrMoody
Assuming the centerline of the cylinder intersects the centerline of the crank,... If the stroke is 3.48, then so is the diameter of the big end circle. The circumference would be 3.48 x pi = ~ 10.933 inches. At 1000 rpm, that would be 10933 in/min ÷ 60 = 182.22 in/s, which would also be the maximum piston speed at 90° crank position. At 90°, the piston would be exactly halfway through its travel.
Thanks to all who participated. MrMoody and Astro14 (in absentia ) had the correct answers. This was another problem in the conversion of radial velocity to linear velocity as in, Vlinear = Omega*radius as two hints were given. To simplify the problem we assumed no effect of crank angle/rod/piston-pin relationships, which meant the piston is seeing only reciprocating, linear travel. Simplification also meant the connecting rod was very long so crank angles did not enter into the problem. At 0 degrees Top Dead Center and at 180 degrees Bottom Dead Center, and with the crankshaft rotating clockwise, the piston motion is virtually zero. At 90 degrees and at 270 degrees, the piston is halfway down or up the cylinder, respectively, and it is at these positions in which the piston velocity is maximum, or 182.22 inches per second. For the next QOTD, we will include crank angle.
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2K | 1,060 | 4,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-06 | latest | en | 0.93005 |
https://www.coursehero.com/file/6036203/Ch-6-6A-solution/ | 1,545,233,036,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832559.95/warc/CC-MAIN-20181219151124-20181219173124-00521.warc.gz | 851,010,276 | 63,931 | BUS
Ch_6_6A_solution
Ch_6_6A_solution - = Ch 6 = 7 a Take Order = 100 per hour...
• Notes
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============================ Ch. 6 =============================== 7. a. Take Order = 100 per hour * 12 hours = 1200 Pick Order = 80 per hour * 24 hours = 1920 Pack Order = 60 per hour * 24 hours = 1440 Maximum output is determined by order taking (1200) since the pick and pack operations can work up to 24 hours to clear out their order backlog. b. If we take the maximum of 1200 orders then: Pick Order = 1200 orders/80 per hour = 15 hours Pack Order = 1200 orders/60 per hour = 20 hours c. Orders can be taken at a rate of 100/hours and can be picked at the rate of 80/hour so they build at the rate of 20/hour. Orders are taken for 12 hours. Maximum orders waiting for picking = 20/hour * 12 hours = 240 d. Orders can be picked at a rate of 80/hours and can be packed at the rate of 60/hour so they build at the rate of 20/hour. Orders are picked for 15 hours. Maximum orders waiting for packing= 20/hour * 15 hours = 300 e. (b. revisited) If we take the maximum of 1200 orders then: Pick Order = 1200 orders/80 per hour = 15 hours Pack Order = 1200 orders/120 per hour = 10 hours However, Packing has to wait for the orders to be picked so it would be 15 hours (c. revisited) This answer does not change.
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• Spring '10
• Bilbrey
• Management, \$5, orders, \$44.49 7, 1.98 min, 2.89 min
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Jill Tulane University ‘16, Course Hero Intern | 671 | 2,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-51 | latest | en | 0.901079 |
https://ask.cvxr.com/t/speed-problem-in-cvx-solving-exponential-cone/9116 | 1,679,824,380,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00697.warc.gz | 136,991,629 | 7,778 | # Speed Problem in CVX solving exponential cone
I have the following convex optimization problem
\min_{x} {ax}{\left(\exp{\left(\frac{b}{x}\right)} - 1\right)} + {a(1-x)}{\left(\exp{\left(\frac{b}{1-x}\right)} - 1\right)},
subject to
{a_1 x}{\left(\exp{\left(\frac{b_1}{x}\right)} - 1\right)} \leq c_1, \\ {a_2 (1-x)}{\left(\exp{\left(\frac{b_2}{(1-x)}\right)} - 1\right)} \leq c_2, \\ 0 < x < 1,
The problem is variable in x and a,b,a_1,b_1,a_2,b_2 are positive constants.
I used CVX to solve it in Matlab but it takes a long long time. I need to average out over a large number of iterations and for each iteration, I need to run CVX to compute different possibilities. I used Mosek and Pade approximation but still takes a long time. Any suggestions to speed it up more ? OR Any suggestion for numerical methods that I can use to implement specifically for this problem.
What do you mean by long time since you are optimizing over a scalar variable? That can take long.
I bet you can reformulate this problem using the exponential cone and solve it directly with Mosek with using Pade approximation.
This Pade approximation stuff may be a great idea in theory but should be avoided in practice if possible.
a1 = 2; a2 = 2; b1 = 1; b2 = 1; a3 = 1; a4 = 1; c1 = 20; c2 = 20 ;
cvx_solver mosek
cvx_begin quiet
variables x Z1 Z2
minimize( (a1.*Z1 - a1.*x) + (a2.*Z2 - a2.*(1-x)))
subject to
{b1,x,Z1} == exponential(1)
{b2,1-x,Z2} == exponential(1)
0 <= x <= 1
((a3.*Z1) - (a3.*x)) <= c1
((a4.*Z2) - (a4.*(1-x))) <= c2
cvx_end
This is my implementation. Any suggestion for other reformulation.
Use cvx2.2 with Mosek 9.2+ and you will solve exponential cone problems directly without approximation.
Don’t use quiet when debugging.
Thanks for the suggestion. I was already using it. This is the output without the quiet mode. There is still messages on using Pade approximation but the same messages from Mosek as the post you linked. So I am not sure if it still using Pad approximation or just debug issue.
It still takes days to run.
Fix your cvx installation. It should not use the Pade approximation. Look at the problem size in Mosek log. I solved it immediately and the log was as follows:
CVX Warning:
Models involving "exponential" or other functions in the log, exp, and entropy
family are solved using an experimental successive approximation method.
This method is slower and less reliable than the method CVX employs for
other models. Please see the section of the user's guide entitled
for more details about the approach, and for instructions on how to
suppress this warning message in the future.
Calling Mosek 9.1.9: 10 variables, 3 equality constraints
For improved efficiency, Mosek is solving the dual problem.
------------------------------------------------------------
MOSEK Version 9.1.9 (Build date: 2019-11-21 11:27:13)
Copyright (c) MOSEK ApS, Denmark. WWW: mosek.com
Platform: Linux/64-X86
Problem
Name :
Objective sense : min
Type : CONIC (conic optimization problem)
Constraints : 3
Cones : 2
Scalar variables : 10
Matrix variables : 0
Integer variables : 0
Optimizer started.
Presolve started.
Linear dependency checker started.
Linear dependency checker terminated.
Eliminator started.
Freed constraints in eliminator : 2
Eliminator terminated.
Eliminator started.
Freed constraints in eliminator : 0
Eliminator terminated.
Eliminator - tries : 2 time : 0.00
Lin. dep. - tries : 1 time : 0.00
Lin. dep. - number : 0
Presolve terminated. Time: 0.05
Problem
Name :
Objective sense : min
Type : CONIC (conic optimization problem)
Constraints : 3
Cones : 2
Scalar variables : 10
Matrix variables : 0
Integer variables : 0
Optimizer - solved problem : the primal
Optimizer - Constraints : 1
Optimizer - Cones : 2
Optimizer - Scalar variables : 8 conic : 6
Optimizer - Semi-definite variables: 0 scalarized : 0
Factor - setup time : 0.00 dense det. time : 0.00
Factor - ML order time : 0.00 GP order time : 0.00
Factor - nonzeros before factor : 1 after factor : 1
Factor - dense dim. : 0 flops : 1.70e+01
ITE PFEAS DFEAS GFEAS PRSTATUS POBJ DOBJ MU TIME
0 5.1e+00 6.4e+00 2.0e+01 0.00e+00 1.868877074e+01 0.000000000e+00 1.0e+00 0.07
1 6.2e-01 7.8e-01 3.4e+00 -5.24e-01 5.414724018e+01 4.856774075e+01 1.2e-01 0.18
2 1.4e-01 1.8e-01 6.2e-01 3.48e-01 6.328924677e+01 6.181381188e+01 2.8e-02 0.19
3 2.2e-02 2.8e-02 4.2e-02 7.04e-01 6.701333099e+01 6.677709526e+01 4.3e-03 0.19
4 6.5e-03 8.2e-03 8.3e-03 7.03e-01 6.705397373e+01 6.698910993e+01 1.3e-03 0.20
5 1.3e-04 1.7e-04 2.5e-05 9.75e-01 6.722028944e+01 6.721902218e+01 2.6e-05 0.20
6 8.5e-06 1.1e-05 4.1e-07 9.88e-01 6.722184509e+01 6.722176686e+01 1.7e-06 0.21
7 7.5e-06 9.3e-06 3.5e-07 -6.95e-01 6.722187076e+01 6.722180179e+01 1.4e-06 0.21
8 7.3e-06 9.1e-06 3.4e-07 1.05e+00 6.722186963e+01 6.722180170e+01 1.3e-06 0.22
9 7.3e-06 9.1e-06 3.4e-07 8.86e-01 6.722186961e+01 6.722180169e+01 1.3e-06 0.22
10 7.3e-06 9.1e-06 3.4e-07 9.89e-01 6.722186961e+01 6.722180169e+01 1.3e-06 0.23
11 6.4e-06 8.0e-06 2.7e-07 1.00e+00 6.722186408e+01 6.722180221e+01 1.2e-06 0.23
12 6.3e-06 7.9e-06 2.7e-07 1.00e+00 6.722186369e+01 6.722180225e+01 1.2e-06 0.24
13 5.7e-06 7.1e-06 2.3e-07 9.64e-01 6.722186065e+01 6.722180374e+01 1.0e-06 0.24
14 5.6e-06 7.1e-06 2.3e-07 1.46e+00 6.722186155e+01 6.722180518e+01 1.0e-06 0.25
15 3.5e-06 4.3e-06 1.1e-07 9.82e-01 6.722185979e+01 6.722182138e+01 6.3e-07 0.25
16 3.3e-06 4.1e-06 9.8e-08 8.55e-01 6.722185968e+01 6.722182273e+01 5.9e-07 0.26
17 3.0e-06 3.8e-06 8.6e-08 1.12e+00 6.722186071e+01 6.722182626e+01 5.5e-07 0.26
18 2.8e-06 3.6e-06 7.8e-08 1.13e+00 6.722186168e+01 6.722182911e+01 5.2e-07 0.27
19 2.8e-06 3.6e-06 7.8e-08 1.02e+00 6.722186168e+01 6.722182916e+01 5.2e-07 0.28
20 2.6e-06 3.2e-06 6.7e-08 1.04e+00 6.722186250e+01 6.722183238e+01 4.7e-07 0.28
21 2.6e-06 3.2e-06 6.6e-08 1.24e+00 6.722186291e+01 6.722183329e+01 4.7e-07 0.29
22 2.5e-06 3.1e-06 6.4e-08 8.79e-01 6.722186295e+01 6.722183366e+01 4.6e-07 0.29
23 2.5e-06 3.1e-06 6.3e-08 1.23e+00 6.722186314e+01 6.722183408e+01 4.6e-07 0.30
24 2.5e-06 3.1e-06 6.3e-08 2.22e+00 6.722186375e+01 6.722183484e+01 4.5e-07 0.30
25 2.5e-06 3.1e-06 6.2e-08 1.06e+00 6.722186384e+01 6.722183519e+01 4.5e-07 0.31
26 2.4e-06 3.0e-06 6.1e-08 9.89e-01 6.722186394e+01 6.722183570e+01 4.4e-07 0.31
27 2.4e-06 3.0e-06 6.0e-08 1.13e+00 6.722186396e+01 6.722183575e+01 4.4e-07 0.32
28 6.2e-09 7.8e-09 8.0e-12 9.75e-01 6.722188770e+01 6.722188763e+01 1.1e-09 0.32
Optimizer terminated. Time: 0.39
Interior-point solution summary
Problem status : PRIMAL_AND_DUAL_FEASIBLE
Solution status : OPTIMAL
Primal. obj: 6.7221887701e+01 nrm: 1e+01 Viol. con: 5e-13 var: 1e-08 cones: 0e+00
Dual. obj: 6.7221887627e+01 nrm: 2e+01 Viol. con: 0e+00 var: 6e-08 cones: 0e+00
Optimizer summary
Optimizer - time: 0.39
Interior-point - iterations : 28 time: 0.33
Basis identification - time: 0.00
Primal - iterations : 0 time: 0.00
Dual - iterations : 0 time: 0.00
Clean primal - iterations : 0 time: 0.00
Clean dual - iterations : 0 time: 0.00
Simplex - time: 0.00
Primal simplex - iterations : 0 time: 0.00
Dual simplex - iterations : 0 time: 0.00
Mixed integer - relaxations: 0 time: 0.00
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +12.7781
Okay. Thanks so much I solved it.
It looks like you were using the exponential.m replacement of CVXQUAD. That is why It was using the Padé approximation, despite having CVX 2.2 and Mosek 9.x.
You should restore CVX’s version of exponential.m. That way CVXQUAD’s Padé approoximation won’t be invoked and the problem will be solved using Mosek’s native exponential cone capability, which is the best way of solving it now.
That is exactly what happened. Thanks for your help. | 3,638 | 8,952 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-14 | latest | en | 0.855531 |
https://goodnews.xplodedthemes.com/blog/2021/09/23/a-look-at-the-kelly-formula/ | 1,718,241,135,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00424.warc.gz | 257,147,980 | 18,110 | # A Look At The Kelly Formula
### Content Maddux’s Winning Picks Harville, Discounted Harville And Each Way Bets How To Make Money Reading Books: Side Gigs For Bookworms The Massive Insight I Gained From A Simple Money Equation The Kelly Criterion is a formula used to calculate bet sizes for people looking to make a sustainable profit over the long term. This … Read More
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The Kelly Criterion is a formula used to calculate bet sizes for people looking to make a sustainable profit over the long term. This is important to think about because imagine a scenario where you have a rigged coin toss that gives you a 60% chance to double your money instead of just a 50% chance to double your cash. It is pretty good bet in fact, one that would make you want to gamble a fair amount on it’s outcome. But, if you were to place your life savings on this gamble, you’d be completely bankrupt and homeless 40% of the time. And that simply isn’t worth it, let alone even mathematically sustainable in the long run.
I found this example online to which helped me understand, at a very basic level, how the Kelly criterion is calculated and how it kind of works. If anything has become abundantly clear, it is that no single https://jewellerscastle.com/index.php/2021/02/05/simplebet-draftkings-beginning-little/ staking system rules all. All the above betting stake methods, as well as a host of others, have their place. Average Bet Size – The percentage of your starting bankroll that you will bet. For level staking this number won’t change, for proportional staking, this number will update after each bet in the simulation.
## Harville, Discounted Harville And Each Way Bets
Finally, we implement a dynamic strategy applied on the European stock market data and compare the results between the tangent and the optimal Kelly portfolios. In a dynamic setting, the rolling Kelly portfolio outperforms competitors particularly in the case of rebalanced portfolios optimized with a 2-years window width. The Kelly Criterion is a scientific gambling method using a formula for bet sizing that mathematically calculates the proper position size for placing a bet based on the odds. The Kelly bet size is calculated by optimizing the projected value of the wealth logarithm, which is equivalent to maximizing the expected geometric growth rate of the capital being wagered.
## How Does Kelly Criterion Work?
Hence you wager \$100 (10 percent of your \$1,000) bankroll. You collect \$220 in profit and your bankroll now stands at \$1,220. Put Simply, the legality of eSports betting is based on the internal country you live in.
The two gentlemen, laughed and smiled and took a picture with the winning slips. They then did something unthinkable… they proceeded to tear the tickets in half. Bill had decided that if he won, he would leave the tickets unclaimed; Hong Kong Racing Club policy in such cases directed the money to a charitable trust. At this point, over a million people had placed a bet, equivalent to ~14% of the entire population of Hong Kong.
At first glance it appears that the average wealth of the population is increasing, consistent with our theory of positive expected payoffs. In cell B6 we see the equation for the size of Kelly bet as a fraction of the total bankroll. The figures above are fixed values for predicted probabilities and return.
Additionally, recruitment of odds compilers with professional qualifications by the major bookmakers in the UK is practically unknown. Index betting on sports is an important new area of sports betting. Examples from American football, soccer and tennis are quoted and the main differences between a sports index and a stock-market index are discussed. All the companies competing for a share of the index betting market in the UK have experienced difficulties due to their lack of management science and operations research expertise. Some of the basic modelling techniques that the managers of these companies should have at their disposal are illustrated by using estimates from Wimbledon to calculate an index for the total games in a tennis match.
The problem is that just because X converges in probability to x, doesn’t mean that f converges in probability to f. Look at the distribution of the outcome after many throws. Relate that to the product of each throw’s relative return. Look at how the log expectation of the product of n random variables converges as n gets big, and you see what to optimize.
I was intentionally silent about if the participant had children or “big responsibilities”. For each of these percentages, what percentage of your bankroll would you bet? (the choices are between 0% and 100% in 10% increments.) Let’s go.
## How Much Should You Bet To Maximize Your Investments, Or Your Company’s Odds Of Success?
You can read more about how it works in this Kelly Criterion Wikipedia article. This is your estimated probability that the selection will win. The Kelly criterion uses this value to calculate a recommended stake. It should be a value between 0 and 1 (e.g. input 0.5 if your believe the selection has a 50% chance of winning). After being published in 1956, the Kelly criterion was picked up quickly by gamblers who were able to apply the formula to horse racing. It was not until later that the formula was applied to investing. | 1,102 | 5,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-26 | latest | en | 0.918987 |
https://www.quizzes.cc/metric/percentof.php?percent=542&of=3520 | 1,582,732,868,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146414.42/warc/CC-MAIN-20200226150200-20200226180200-00256.warc.gz | 831,779,117 | 3,239 | #### What is 542 percent of 3,520?
How much is 542 percent of 3520? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 542% of 3520 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 542% of 3,520 = 19078.4
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating five hundred and fourty-two of three thousand, five hundred and twenty How to calculate 542% of 3520? Simply divide the percent by 100 and multiply by the number. For example, 542 /100 x 3520 = 19078.4 or 5.42 x 3520 = 19078.4
#### How much is 542 percent of the following numbers?
542 percent of 3520.01 = 1907845.42 542 percent of 3520.02 = 1907850.84 542 percent of 3520.03 = 1907856.26 542 percent of 3520.04 = 1907861.68 542 percent of 3520.05 = 1907867.1 542 percent of 3520.06 = 1907872.52 542 percent of 3520.07 = 1907877.94 542 percent of 3520.08 = 1907883.36 542 percent of 3520.09 = 1907888.78 542 percent of 3520.1 = 1907894.2 542 percent of 3520.11 = 1907899.62 542 percent of 3520.12 = 1907905.04 542 percent of 3520.13 = 1907910.46 542 percent of 3520.14 = 1907915.88 542 percent of 3520.15 = 1907921.3 542 percent of 3520.16 = 1907926.72 542 percent of 3520.17 = 1907932.14 542 percent of 3520.18 = 1907937.56 542 percent of 3520.19 = 1907942.98 542 percent of 3520.2 = 1907948.4 542 percent of 3520.21 = 1907953.82 542 percent of 3520.22 = 1907959.24 542 percent of 3520.23 = 1907964.66 542 percent of 3520.24 = 1907970.08 542 percent of 3520.25 = 1907975.5
542 percent of 3520.26 = 1907980.92 542 percent of 3520.27 = 1907986.34 542 percent of 3520.28 = 1907991.76 542 percent of 3520.29 = 1907997.18 542 percent of 3520.3 = 1908002.6 542 percent of 3520.31 = 1908008.02 542 percent of 3520.32 = 1908013.44 542 percent of 3520.33 = 1908018.86 542 percent of 3520.34 = 1908024.28 542 percent of 3520.35 = 1908029.7 542 percent of 3520.36 = 1908035.12 542 percent of 3520.37 = 1908040.54 542 percent of 3520.38 = 1908045.96 542 percent of 3520.39 = 1908051.38 542 percent of 3520.4 = 1908056.8 542 percent of 3520.41 = 1908062.22 542 percent of 3520.42 = 1908067.64 542 percent of 3520.43 = 1908073.06 542 percent of 3520.44 = 1908078.48 542 percent of 3520.45 = 1908083.9 542 percent of 3520.46 = 1908089.32 542 percent of 3520.47 = 1908094.74 542 percent of 3520.48 = 1908100.16 542 percent of 3520.49 = 1908105.58 542 percent of 3520.5 = 1908111
542 percent of 3520.51 = 1908116.42 542 percent of 3520.52 = 1908121.84 542 percent of 3520.53 = 1908127.26 542 percent of 3520.54 = 1908132.68 542 percent of 3520.55 = 1908138.1 542 percent of 3520.56 = 1908143.52 542 percent of 3520.57 = 1908148.94 542 percent of 3520.58 = 1908154.36 542 percent of 3520.59 = 1908159.78 542 percent of 3520.6 = 1908165.2 542 percent of 3520.61 = 1908170.62 542 percent of 3520.62 = 1908176.04 542 percent of 3520.63 = 1908181.46 542 percent of 3520.64 = 1908186.88 542 percent of 3520.65 = 1908192.3 542 percent of 3520.66 = 1908197.72 542 percent of 3520.67 = 1908203.14 542 percent of 3520.68 = 1908208.56 542 percent of 3520.69 = 1908213.98 542 percent of 3520.7 = 1908219.4 542 percent of 3520.71 = 1908224.82 542 percent of 3520.72 = 1908230.24 542 percent of 3520.73 = 1908235.66 542 percent of 3520.74 = 1908241.08 542 percent of 3520.75 = 1908246.5
542 percent of 3520.76 = 1908251.92 542 percent of 3520.77 = 1908257.34 542 percent of 3520.78 = 1908262.76 542 percent of 3520.79 = 1908268.18 542 percent of 3520.8 = 1908273.6 542 percent of 3520.81 = 1908279.02 542 percent of 3520.82 = 1908284.44 542 percent of 3520.83 = 1908289.86 542 percent of 3520.84 = 1908295.28 542 percent of 3520.85 = 1908300.7 542 percent of 3520.86 = 1908306.12 542 percent of 3520.87 = 1908311.54 542 percent of 3520.88 = 1908316.96 542 percent of 3520.89 = 1908322.38 542 percent of 3520.9 = 1908327.8 542 percent of 3520.91 = 1908333.22 542 percent of 3520.92 = 1908338.64 542 percent of 3520.93 = 1908344.06 542 percent of 3520.94 = 1908349.48 542 percent of 3520.95 = 1908354.9 542 percent of 3520.96 = 1908360.32 542 percent of 3520.97 = 1908365.74 542 percent of 3520.98 = 1908371.16 542 percent of 3520.99 = 1908376.58 542 percent of 3521 = 1908382 | 1,806 | 4,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-10 | latest | en | 0.820149 |
http://www.chegg.com/homework-help/questions-and-answers/lc-circuit-l-060-mhand-c-14-f-thetotal-energy-stored-circuit-27-j-27010-6-j--find-maximum--q807514 | 1,469,766,864,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829970.64/warc/CC-MAIN-20160723071029-00117-ip-10-185-27-174.ec2.internal.warc.gz | 363,960,999 | 13,501 | An LC circuit has L = 0.60 mHand C = 1.4 µF. Thetotal energy stored in this circuit is 2.7 µJ (2.7010-6 J).
(a) Find the maximum chargeqmax on the capacitor.
1 C
(b) Find the maximum current.
2 A
(c) When the charge on the capacitor is qmax/2,what is the magnitude of the current?
3 A | 95 | 286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-30 | latest | en | 0.841635 |
https://www.telegram.com/article/20100207/NEWS/2070596 | 1,604,065,512,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00443.warc.gz | 888,829,161 | 16,434 | Is there a universal language? Some would suggest love — but there’s another candidate: mathematics. It’s a major factor in investing, but you’d better be careful. It may surprise you a little.
For instance, imagine that you’ve lost money on a stock and you want to know when you’re going to get back to even on your initial investment.
You might assume that after a 50 percent drop in the stock’s price, you’d need a 50 percent gain to get back to even. Not true. Imagine that shares of Scruffy’s Chicken Shack (ticker: BGAWK) fell from \$80 to \$40, a 50 percent drop. To get back to \$80, the stock will need to double, gaining 100 percent. If it drops 80 percent, say, from \$100 to \$20, it will need to gain 400 percent in order to get back to \$100. Even a more modest 20 percent drop in value will require a 25 percent gain. (As the drop gets bigger, the required return for a recovery gets much bigger.)
That’s the bad news. Fortunately, there’s good math news, too, regarding how your money can grow. If your stock gains \$100 in value, going from \$100 to \$200, you’re sitting on a “two-bagger” and a 100 percent gain. (Got that? A double is a 100 percent gain, not 200 percent.) But later on, if it gains \$100 going from \$500 to \$600 (from a “five-bagger” to a “six-bagger”), that requires only a 20 percent gain.
It’s good to understand how the math works, but it’s more important to understand that we shouldn’t get hung up on the past. With investing, what really counts is the future. A stock may have fallen 60 percent on you, or it may have risen 200 percent, but either way, you just need to know how likely it is to keep growing at a good clip. If you think it will, hang on. If not, sell and move the money into a more promising investment. Don’t look back — look forward. | 448 | 1,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-45 | latest | en | 0.927972 |
http://herbie.uwplse.org/demo/76886c5df09a72ff3cfe2932aff41d75e9f97581.4e924be3125566e8e27de69ca0ee25cd0b20a544/graph.html | 1,547,750,833,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210304-00476.warc.gz | 103,727,663 | 2,411 | Average Error: 58.3 → 0.7
Time: 11.1s
Precision: 64
Internal Precision: 1344
$e^{X} - e^{X \cdot X}$
$X - \left(\frac{1}{2} - \frac{1}{6} \cdot X\right) \cdot \left(X \cdot X\right)$
# Try it out
Results
In Out
Enter valid numbers for all inputs
# Derivation
1. Initial program 58.3
$e^{X} - e^{X \cdot X}$
2. Initial simplification58.3
$\leadsto e^{X} - e^{X \cdot X}$
3. Taylor expanded around 0 0.7
$\leadsto \color{blue}{\left(X + \frac{1}{6} \cdot {X}^{3}\right) - \frac{1}{2} \cdot {X}^{2}}$
4. Simplified0.7
$\leadsto \color{blue}{X - \left(X \cdot X\right) \cdot \left(\frac{1}{2} - X \cdot \frac{1}{6}\right)}$
5. Final simplification0.7
$\leadsto X - \left(\frac{1}{2} - \frac{1}{6} \cdot X\right) \cdot \left(X \cdot X\right)$
# Runtime
Time bar (total: 11.1s)Debug log
herbie shell --seed '#(2775764126 3555076145 3898259844 1891440260 2599947619 1948460636)'
(FPCore (X)
:name "exp(X) - exp(X * X)"
(- (exp X) (exp (* X X)))) | 396 | 952 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-04 | latest | en | 0.302576 |
https://discuss.codecademy.com/t/functions-help-python/76594 | 1,542,493,684,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743854.48/warc/CC-MAIN-20181117205946-20181117231946-00311.warc.gz | 604,659,117 | 3,663 | # Functions help python!
#1
so im on Practice Makes Perfect lesson on functions. and im not sure how to do what it asks, this is what i have:
def cube(n):
return n**3
def by_three(n):
if n % 3
#2
This exercise uses an `if..else` statement to return either the cube of a number, provided the number is divisible by 3, or False. How would you write such a statement.
Bear in mind that a return value can be the return of another function.
``return other_function(number)``
#3
i still dont know what to do:/
#4
this is what i have now:
def cube(number):
return cube(number)
def by_three(number):
if n % 3 == cube(number)
else:
return False
#5
this is what i have now:
def cube(number):
return cube(number)
def by_three(number):
if n % 3 == cube(number)
else:
return False
#6
idk what to do in the practice makes perfect lesson this is what i have:
def cube(number):
return cube(number)
def by_three(number):
if n % 3 == cube(number):
return True
else:
return False
cube(10)
#7
Close but not quite.
``````if n % 3 == 0: # test for even divisibility
return cube(number)``````
To run the function, call it by the correct name:
``````print by_three(18) # 5832
print by_three(16) # False``````
#8
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 360 | 1,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-47 | latest | en | 0.806306 |
https://studysoup.com/guide/2314771/uc-math1006-midterm-1-fall-2016 | 1,477,315,992,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719566.74/warc/CC-MAIN-20161020183839-00170-ip-10-171-6-4.ec2.internal.warc.gz | 880,931,309 | 16,252 | ×
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## MATH1006 Exam 1 Study Guide
by: mitchj9 Notetaker
41
0
1
# MATH1006 Exam 1 Study Guide MATH1006
Marketplace > University of Cincinnati > Math > MATH1006 > MATH1006 Exam 1 Study Guide
mitchj9 Notetaker
UC
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Exam 1 study guide covering Chapter 1.1 - Inductive and Deductive Reasoning Chapter 1.2 - Estimations, Graphing, and Mathematical Models Chapter 1.3 - Problem Solving
COURSE
Math Reasoning
PROF.
Sandra Loy
TYPE
Study Guide
PAGES
1
WORDS
CONCEPTS
MATH1006, study, guide
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Date Created: 09/07/16
MATH1006- Math Reasoning CHAPTER 1 Midterm Study Guide Chapter 1.1 – Inductive and Deductive Reasoning • Inductive Reasoning (Induction) – process of arriving at a general conclusion based on observation of specific examples - We can never be absolutely sure that conclusions are true so they are called conjectures or hypotheses - Counterexample- one case where conjecture does not hold ture - Strong inductive argument – process of proving a specific conclusion from one or more general statements - Theorem- conclusion that is proved true by deductive reasoning Chapter 1.2 – Estimations, Graphing, and Mathematical Models • Estimation- the process of arriving at an approximate answer to a question • Rounding whole numbers 1. Look at the digit to the right of the digit where the rounding is to occur 2. If it is 5 or greater, add 1 to the digit to be rounded. If it is less than 5, keep the digit the same • Formula – statement of equality that uses letters to represent a relationship between 2 or more variables • Mathematical Model – the process of finding formulas to describe real world phenomena • Model Breakdown – when a mathematical model gives an estimate that is not a good approximation or is extended to include values of the variable that do not make sense Chapter 1.3 – Problem Solving (Polya’s Method) • Polya’s four steps in problem solving 1. Understand the problem 2. Devise a plan 3. Carry out plan and solve the problem 4. Look back and check answers
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zungqr.f(3) LAPACK zungqr.f(3)
zungqr.f -
# SYNOPSIS
## Functions/Subroutines
subroutine zungqr (M, N, K, A, LDA, TAU, WORK, LWORK, INFO)
ZUNGQR
# Function/Subroutine Documentation
## subroutine zungqr (integerM, integerN, integerK, complex*16, dimension( lda, * )A, integerLDA, complex*16, dimension( * )TAU, complex*16, dimension( * )WORK, integerLWORK, integerINFO)
ZUNGQR
Purpose:
``` ZUNGQR generates an M-by-N complex matrix Q with orthonormal columns,
which is defined as the first N columns of a product of K elementary
reflectors of order M
Q = H(1) H(2) . . . H(k)
as returned by ZGEQRF.
```
Parameters:
M
``` M is INTEGER
The number of rows of the matrix Q. M >= 0.
```
N
``` N is INTEGER
The number of columns of the matrix Q. M >= N >= 0.
```
K
``` K is INTEGER
The number of elementary reflectors whose product defines the
matrix Q. N >= K >= 0.
```
A
``` A is COMPLEX*16 array, dimension (LDA,N)
On entry, the i-th column must contain the vector which
defines the elementary reflector H(i), for i = 1,2,...,k, as
returned by ZGEQRF in the first k columns of its array
argument A.
On exit, the M-by-N matrix Q.
```
LDA
``` LDA is INTEGER
The first dimension of the array A. LDA >= max(1,M).
```
TAU
``` TAU is COMPLEX*16 array, dimension (K)
TAU(i) must contain the scalar factor of the elementary
reflector H(i), as returned by ZGEQRF.
```
WORK
``` WORK is COMPLEX*16 array, dimension (MAX(1,LWORK))
On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
```
LWORK
``` LWORK is INTEGER
The dimension of the array WORK. LWORK >= max(1,N).
For optimum performance LWORK >= N*NB, where NB is the
optimal blocksize.
If LWORK = -1, then a workspace query is assumed; the routine
only calculates the optimal size of the WORK array, returns
this value as the first entry of the WORK array, and no error
message related to LWORK is issued by XERBLA.
```
INFO
``` INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument has an illegal value
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Date:
November 2011
Definition at line 129 of file zungqr.f.
# Author
Generated automatically by Doxygen for LAPACK from the source code.
Sat Nov 16 2013 Version 3.4.2
Search for or go to Top of page | Section 3 | Main Index
Visit the GSP FreeBSD Man Page Interface.
Output converted with ManDoc. | 741 | 2,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.712388 |
http://www.windows2universe.org/php/tour_test_sqli.php?page=/physical_science/physics/electricity/circuit_analogy_water_pipes.html | 1,419,158,464,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770860.97/warc/CC-MAIN-20141217075250-00112-ip-10-231-17-201.ec2.internal.warc.gz | 927,190,761 | 12,141 | Error finding content
Electric Circuits: a Water-in-Pipes Analogy - Windows to the Universe
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Water flowing in pipes is a lot like electricity flowing in a circuit. A battery is like a pump. Electrons flowing through wires are like water flowing through pipes.
Click on image for full size
# Electric Circuits: Like Water-in-Pipes
An electric current is a flow of electrons through a conductor (like a copper wire). Since we can't see electrons, it would be nice to have a model or an analogy of electric circuits to help us understand circuits better. Water flowing through pipes is pretty good mechanical system that is a lot like an electrical circuit.
This mechanical system consists of a pump pushing water through a closed pipe. Imagine that the electrical current is similar to the water flowing through the pipe. The following parts of the two systems are related:
The pipe is like the wire in the electric circuit The pump is like the battery. The pressure generated by the pump drives water through the pipe; that pressure is like the voltage generated by the battery which drives electrons through the circuit. The seashells plug up the pipe and slow the flow of water, creating a pressure difference from one end to the other. In a similar way the resistance in the electric circuit resists the flow of electricity and creates a voltage drop from one end to the other. Energy is lost across the resistor and shows up as heat.
The power in the circuit equals the voltage times the current. The same power can be carried by a high voltage and a low current as is carried by a low voltage and a high current. The higher the current flow, however, the more energy is lost as heating of the wires. That's why high voltage and low current is used when transporting electrical energy along power lines.
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## Direct Current (DC) & Alternating Current (AC) Electricity
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## Planetary Magnets
The Earth is a good example of a planetary dipole, where the lines of force point in a direction out of the South (magnetic) Pole and into the North (magnetic) Pole. Planets can also show evidence of quadrupoles...more
## Eccentricity of an Orbit
You may think that most objects in space that orbit something else move in circles, but that isn't the case. Although some objects follow circular orbits, most orbits are shaped more like "stretched...more
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Windows to the Universe, a project of the National Earth Science Teachers Association, is sponsored in part by the National Science Foundation and NASA, our Founding Partners (the American Geophysical Union and American Geosciences Institute) as well as through Institutional, Contributing, and Affiliate Partners, individual memberships and generous donors. Thank you for your support! | 904 | 4,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2014-52 | latest | en | 0.946365 |
https://www.physicsforums.com/threads/tennis-ball-hit-by-racket-momentum-problem-i-think.562084/ | 1,519,468,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00520.warc.gz | 983,237,783 | 17,792 | # Tennis ball hit by racket (momentum problem I think)
1. Dec 22, 2011
### aero_zeppelin
1. The problem statement, all variables and given/known data
A tennis ball of mass 0.022 kg is moving at 3.1 m/s at an angle of 222° to the horizontal. It is struck by a racket which exerts a force on it of 72t - 442 t^2 N for 1/10 of a second at an angle 32° to the horizontal. Find the final velocity of the tennis ball (Express answer using i and j unit vectors)
2. Relevant equations
So, I guess this is a momentum problem...?)
p = mv (momentum)
Δp = pf - pi
3. The attempt at a solution
My attempt here was the next:
1.- Integrate the force exerted by racket, using tf = 0.1 s and ti = 0
Δp = ∫ F dt = 0.218 Km m / s
2.- Divide Δp in x and y components, by multiplying 0.218 by cos 222° and sin 222°, respectively.
3.- Obtain ball's momentum, in components also:
px initial = (0.022 kg)(3.1 m/s) cos 32°
py initial = (0.022 kg)(3.1 m/s) sin 32°
4.- Using Δp = pfinal - pinitial, solve for pfinal, which is (mass)(vel. final), then solve for
vel. final.
v final = (Δp + p initial) / mass = (6.05 i + 3.2 j) m/s
That's what I got... They said integral calculus was not needed (even though it's a pretty easy integral) for this course I'm taking lol, but after some thinking this is the only way I think this can be solved.
Any comments? Any other way to solve it w/o integral calculus?
Thanks!
2. Dec 22, 2011
### BruceW
hmm... the force is meant to be (72t - 442 t^2) N right? But this clearly doesn't have the physical dimensions of force. Unless t is a dimensionless parameter. (for example, time per second). What does the question say, specifically?
edit: When I said "for example, time per second", I meant the total elapsed time divided by the time for one second to elapse. In other words, a ratio of time intervals. This is just one example, so for your equation, it may be a different ratio. I'm guessing there is more information given in the question?
Last edited: Dec 22, 2011
3. Dec 22, 2011
### aero_zeppelin
I know! I've never seen force expressed like this hehe... It has NEWTONS, so the dimensions are correct I guess. Those t's mean that the force can vary with time maybe? You are asked to find the final velocity on the ball after that force was exerted.
All the information is there, there's nothing more :P hehe
4. Dec 23, 2011
### BruceW
The question does not give enough information. But I think I can guess what t is meant to be.
the equation for force is: (72t - 442 t^2) N And we want it to have dimensions of Newtons, so this tells you something important about the dimensions of t. From this, what do you guess t is meant to be?
5. Dec 23, 2011
### aero_zeppelin
I dont quite get it... We already have newtons, but we wanna get newtons? hehe Im guessing you could also plug in TIME where those t's are and get the total force that was exerted during that time interval ?
6. Dec 24, 2011
### ehild
You interpreted the badly worded problem correctly, and the method of solution is correct, but there are mistakes in it.
It is said that the force is in newtons. It would be essential to give information about t, like "t is time in seconds, starting when the ball hits the racket". As the timespan of the interaction is given as 0.1 s we can conclude that t is time in seconds, elapsed from the beginning of the interaction. The coefficients of t and t2 must have dimensions and units, the correct form of F(t) would be F(t)=72(Ns-1)t-442 (Ns-2)t2.
The principle is correct, but the result is not. Check the unit. K (Kelvin) is the unit of temperature. Use kg for mass. And the dimension of Δp is [mass][length][time]-1. The numerical value is also inaccurate.
Check the text of the problem. You mixed the directions.
ehild
7. Dec 24, 2011
### BruceW
If you put in time where those t's are, then we certainly won't have Newtons. You need to remember that time is another physical dimension. so time is 0.1seconds or (1/600)minutes e.t.c.
So we know the total time is 0.1seconds, but we want a dimensionless number. I think your teacher is hoping that you use your intuition to guess what that dimensionless number is meant to be. What was your guess?
8. Dec 27, 2011
### aero_zeppelin
lolll yeah, it's supposed to be Kg in there, not Kelvin. Just a typing mistake. But you also mentioned the numerical value is incorrect...
Well, like I said, I wasn't supposed to be doing integral calculus in this subject, but here I am attempting an integral... I had the limits of the definite integral be tf = 0.1 s and ti = 0 and obtained Δp = 0.218 Kg m/s
What did you get as an answer? Also, if there's another method that doesn't require calculus, I'd appreciate a hint hehe
9. Dec 27, 2011
### ehild
I think you need calculus. ∫Fdt==72t2/2-442 t3/3 at t=0.1 equals to 0.213 kg m/s.
ehild
10. Dec 27, 2011
### aero_zeppelin
Guess I'll have to learn those integrals before time hehe thanks a bunch for the help! | 1,385 | 4,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-09 | longest | en | 0.902614 |
https://electronics.stackexchange.com/questions/365622/thevenin-voltage-and-resistance-calculations | 1,561,596,487,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00000.warc.gz | 432,454,586 | 36,022 | # Thevenin Voltage and Resistance Calculations
In order to calculate R_th, you have to made all voltage and current sources equal to 0 (creating open circuits at current sources and shorts at voltage sources), so you will wind up with only resistors. That means that:
$$R_{th}=[(R_1+R_2)^{-1}+1/R_3+1/R_4]^{-1}$$
The dependent voltage source has a transresistance, so it adds on to the value of R_th. This is not being accepted as my answer though. I tried with and without the transresistance, even though it shouldn't be removed.
• Nodal analysis is easier than mesh analysis in this case. – Chu Mar 31 '18 at 9:36
• Why would node analysis be easier? When dealing with the top voltage source, you end up forming a super node, so you sum all of the currents leaving the nodes to the right and left of the voltage source. This leads with you dividing over 0 on the right-hand side because you have the right node-a over that resistor, which is 0. Or do I not make that calculation and assume that the current in that piece of the circuit is 0 because it is an open loop? – Albert Garcia Mar 31 '18 at 16:00
For $\small V_{TH}$, let $\small V$ be the 25/55/25v node, then: $\frac{V-4I_x}{29}+\frac{V}{55}+\frac{V-25}{20}=0$, and $I_x=\frac{V}{55}$. Hence $\small V_{TH}=V-25$. | 363 | 1,281 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-26 | longest | en | 0.880632 |
http://www.softmath.com/algebra-software/exponential-equations/solve-system-of-equation.html | 1,568,785,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573184.25/warc/CC-MAIN-20190918044831-20190918070831-00406.warc.gz | 331,213,718 | 13,958 | English | Español
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1 Lesson Plan on Operations with Whole Numbers Concept / Topic to Teach: Adding and Subtracting Integers Standards Addressed: CCSS.MATH.CONTENT.7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram. CCSS.MATH.CONTENT.7.NS.A.1.D Apply properties of operations as strategies to add and subtract rational numbers. 1. Student Learning Goals/ Outcomes Students will be able to: - Identify properties of a number line. - Define the term integer. - Develop the rules or characteristics of adding and subtracting integers. - Solve problems requiring addition and subtraction of integers. 2. Materials - Street Chalk - The Number Line Rap - Worksheets - Highlighters - Binders - Pens/Pencil -SMARTboard PROCEDURES: 3. Prior Academic Knowledge and Conceptions/Anticipatory Set:
2 1. Prior to this lesson, the students learned about the number line and it s characteristics. At the beginning of class, the students will listen to The Number Line Rap as a review of their prior knowledge. After the rap, the class will determine the important pieces of information, and will highlight the information in the rap to keep in their binders. 4. Instructional Plan 1. First, the class will have a discussion on what the lesson will entail. We will discuss operations, the definition of an integer, etc. 2. Integer Hopscotch a) The students will be split into three groups of three. The groups will be given out prior to the assignment so when we go to the hopscotch boards, they know immediately who to go with. b) Before heading outside, the students will be given two worksheets and will be asked to bring a writing utensil. c) Before I allow the students to play integer hopscotch, I will provide explicit directions on how to play. While giving directions, the students will be asked to reiterate what was stated. For this activity, each student will have a role. The roles will rotate throughout the activity. The three roles will be recorder, drawer, and hopper. d) The students will be given an allotted amount of time to play Integer Hopscotch. Each student will have a turn to be each role. e) After the activity is completed, the students will return to the classroom. During this time, the students will sit with their groups and discuss anything they noticed during the activity and on the worksheet. 5. Closure
3 - The class will end with a class discussion. We will discuss what the students noticed during and after they played integer hopscotch. I will write the students responses on a poster board so they can be reviewed the next day. 6. Homework/Independent Work - The students will create their own rap, poem, song, rules, etc., to remember how to add and subtract integers. They will also be asked to create one word problem that will require their classmates to add or subtract integers. The word problems will be collected and used in a future assignment. 8. Assessment Observations: The largest observation will be during the discussion after integer hopscotch. The goal of this discussion is to see what kind of understanding the students gained during the game on adding and subtracting integers. Observations will also be made during the game to see what the students are thinking. Homework: The homework will demonstrate their understanding on adding and subtracting integers. I can see if the students picked up any patterns or rules when adding and subtracting integers. Furthermore, students gain deeper meaning of the content when creating their own word problem. Adaptations: ELL: For the ELL students, The Number Line Rap will have a version of lyrics written in Spanish. Furthermore, in the activity I will demonstrate how to play integer hopscotch; therefore, they will be able to watch me rather than rely on my words. ADHD: This activity involves the students to move, jump, and switch roles. This will keep the students busy and moving so they don t get bored or anxious. Extension for Advanced Students: Rather than choosing a card, the students will roll a rock on the board like the original game of hopscotch. The students will then have to figure out what operation and what number they need to reach the number the rock landed on as a solution.
4 The number Line I m an integer, yeah you found me, Sitting on the number line, all my friends around me. Like a roller coaster I go up and down, But I always have value, I never clown. Maybe you re measuring a mountain s height, If you add to me, then I slide to the right. If you subtract from me, then I slide to the left, Yeah, I was more, but look at me, now I m less. I could go to 0, that s a number too, Maybe I m at sea level, see, that s nothing new. If we dive like the best who ever lived, I m less than 0, so you know I m negative. So -8 is less than -6, And -6 is less than 6, you get the pics? Chorus I m on the number line, find me on the number line. To the right when it s hot in the summertime, To the left when it s cold in the winter, My value s absolute, I m a winner. (x2) la línea del número
5 Soy un entero, si me encontraste, Sentado en la línea del número, todos mis amigos que me rodea. Como una montaña rusa ir arriba y abajo, Pero siempre tengo valor, yo nunca payaso. Tal vez usted está midiendo la altura de una montaña, Si añades a mí, voy a la derecha. Si restas de mí, luego deslice hacia la izquierda, Sí, yo era más, pero Mírame, ahora soy menos. Podría ir a 0, que también es un número, Tal vez estoy en el nivel del mar, eso no es nada nuevo. Si nos sumergimos el mejor que jamás haya vivido, Soy menos que 0, así que ya sabes que soy negativo. -8 es inferior a -6, Y -6 es menos de 6, tienes las fotos? Coro Estoy en la recta numérica, encontrarme en la línea del número. A la derecha cuando está caliente en el verano, A la izquierda cuando es frío en el invierno, Mi valor absoluto, yo soy un ganador. (x 2)
6 Hello class! Today we will be playing integer hopscotch to learn about adding and subtracting integers! Here are some things we need to remember before we play: 1) Each team will be in groups of 3 students! 2) Black Cards = POSITIVE numbers Red Cards = NEGATIVE numbers 3) For subtraction, we hop BACKWARDS. For addition, we hop FORWARDS. 4) Before you begin to hop, make sure you are facing the direction of the SIGN of the SECOND number in your equation. PAY ATTENTION to Ms. Gamboli when she demonstrates how to play! Don t be afraid to ask questions and remember, most importantly, Have Fun!!
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Schema XML_PGE.xsd. element GrupoInformes. attribute GrupoInformes/@version. XML_PGE.xsd unqualified qualified http://sgpfc.igae.minhap.
Schema XML_PGE.xsd schema location: attribute form default: element form default: targetnamespace: XML_PGE.xsd unqualified qualified http://sgpfc.igae.minhap.es/xmlpge element GrupoInformes children Informe
Lesson 29. Notes. Necesita tomar estos comprimidos. Lesson 29. CoffeeBreakSpanish.com. Recap. A little more pain... May 28th, 2007
May 28th, 2007 In this edition: more aches and pains, illnesses and useful language for a visit to the pharmacy Lesson 29 Notes CoffeeBreakSpanish.com Necesita tomar estos comprimidos Lesson 29 Programme
SPANISH 101-- FINAL EXAM STUDY GUIDE - (100 points) Spring 2016. *Please bring a purple scantron sheet and # 2 pencils for the exam*
SPANISH 101-- FINAL EXAM STUDY GUIDE - (100 points) Spring 2016 *Please bring a purple scantron sheet and # 2 pencils for the exam* ACTIVIDAD A. Arriba! Readings. (10 points) Choose the correct answer | 7,914 | 32,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-26 | longest | en | 0.864273 |
https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_10&oldid=105196 | 1,600,577,859,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00622.warc.gz | 290,407,264 | 11,565 | # 2000 AIME II Problems/Problem 10
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A circle is inscribed in quadrilateral $ABCD$, tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$. Given that $AP=19$, $PB=26$, $CQ=37$, and $QD=23$, find the square of the radius of the circle.
## Solution
Call the center of the circle $O$. By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$, or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$.
Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get $\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0$.
Use the identity for $\tan(A+B)$ again to get $\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0$.
Solving gives $r^2=\boxed{647}$.
## Solution 2
Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ($a, b, c,$ and $d$ are the tangent lengths, not the side lengths). $$A = \sqrt{(a+b+c+d)(abc+bcd+cda+dac)} = 105\sqrt{647}$$ $r^2=\frac{A}{a+b+c+d} = \boxed{647}$. | 497 | 1,370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2020-40 | latest | en | 0.662329 |
https://www.coursehero.com/file/6332024/TWO-WAY-CHI-SQUARE-gender-and-taxes-with-answers/ | 1,513,584,334,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948609934.85/warc/CC-MAIN-20171218063927-20171218085927-00199.warc.gz | 719,748,235 | 31,508 | TWO WAY CHI SQUARE gender and taxes with answers
# TWO WAY CHI SQUARE gender and taxes with answers - f o f e...
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TWO WAY CHI SQUARE You have the following data regarding gender and attitude toward paying taxes. Gender Attitude toward amount of taxes paid Male Female Total Too Much 60 33 93 Right Amount 55 54 109 Too Little 32 68 100 Total 147 155 302 Null Hypothesis: There is no association between the variables Gender and Attitude Toward Paying Taxes Research Hypothesis: There is an association between the variables Gender and Attitude Toward Paying Taxes Compute a Chi Square test of significance Formula for Chi Square Chi Square f e = expected frequencies = e e o f f f 2 2 ) ( χ f o = observed frequencies For Two way f e = (column total)(row total) N (two way) df = (r – 1)(c – 1) r = the number of rows c = the number of columns You will create a table again to calculate the chi square for a two way test, the main difference here is how you calculate f e . Cell
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Unformatted text preview: f o f e f o-f e (f o-f e ) 2 (f o-f e ) 2 /f e Upper left 60 45.27 14.73 216.97 4.79 Upper right 33 47.73 -14.73 216.97 4.55 Middle left 55 53.06 1.94 3.76 0.07 Middle right 54 55.94 -1.94 3.76 0.07 Lower left 32 48.68 -16.68 278.22 5.72 Lower right 68 51.32 16.68 278.22 5.42 SUM for X 2 20.61 What are your degrees of freedom? Df=(r-1)(c-1)= 2 What is your critical value from the Chi Square table? For which levels of significance? Critical Value for Chi Square is 5.991 at 0.05 level of significance. What does this tell us about our calculated Chi Square and our population? What can we say about the null hypothesis? Because our obtained Chi Square of 20.61 is larger than 5.991 we can reject the null hypothesis and say that there is an association between the variables Gender and Attitude Toward Paying Taxes....
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Polymorphic recursion
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Date: 2007-04-03 (17:35) From: Till Varoquaux Subject: Re: [Caml-list] Polymorphic recursion
```There is indeed an issue with polymorphic recursion. It has been shown
that allowing polymorphic recursion in an ML like language would make
the type inference undecidable. One workaround would be for the
typechecker to use type annotations in some given cases. This is
however not the case in Ocaml: type information are either required or
only useful to debug/document some code/"restrict" the type of a given
variable.
Therefor, if you wish to use polymorphic recursion (yep you can) you
might want to use something where you have to specify the type; this
includes records, objects and recursive modules. So this
type 'a seq = Unit | Seq of ('a * ('a * 'a)seq)
type szRec={f:'a.'a seq -> int};;
let size=
let rec s =
{f=function
| Unit -> 0
| Seq(_, b) -> 1 + 2 * s.f b}
in
s.f
might be what you where yearning for.
Cheers,
Till
On 4/3/07, Loup Vaillant <loup.vaillant@gmail.com> wrote:
> I was reading Okasaki's book, "Purely functionnal data structures",
> and discovered that ML (and Ocaml) doesn't support non uniforms
> function definitions.
>
> So, even if:
>
> (**)
> type 'a seq = Unit | Seq of ('a * ('a * 'a)seq);;
> (**)
>
> is correct,
>
> (**)
> let rec size = fun
> | Unit -> 0
> | Seq(_, b) -> 1 + 2 * size b;;
> (**)
>
> is not. Here is the error:
> #
> | Seq(_, b) -> 1 + 2 * size b;;
> This expression (the last 'b') has type seq ('a * 'a) but is here used
> with type seq 'a
> #
>
> Why?
> Can't we design a type system which allow this "size" function?
> Can't we implement non uniform recursive function (efficiently, or at all)?.
>
> I suppose the problem is somewhat difficult, but I can't see where.
>
> Regards,
> Loup Vaillant
>
> _______________________________________________
> Caml-list mailing list. Subscription management:
> http://yquem.inria.fr/cgi-bin/mailman/listinfo/caml-list
> Archives: http://caml.inria.fr
> Beginner's list: http://groups.yahoo.com/group/ocaml_beginners
> Bug reports: http://caml.inria.fr/bin/caml-bugs
>
``` | 675 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-17 | latest | en | 0.851863 |
https://docs.moodle.org/dev/index.php?title=MediaWiki_TeX_test&diff=prev&oldid=82736 | 1,556,253,685,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578759182.92/warc/CC-MAIN-20190426033614-20190426055614-00275.warc.gz | 407,019,178 | 37,591 | # MediaWiki TeX test
This should look like some maths:
$\int_{-\infty}^\infty \psi^{-x^{tim}}\,dx = \sqrt{hunt^4}$
Heavy test:
## Functions, symbols, special characters
### Accents/Diacritics
\acute{a} \grave{a} \hat{a} \tilde{a} \breve{a}
$\acute{a} \grave{a} \hat{a} \tilde{a} \breve{a}\,\!$
\check{a} \bar{a} \ddot{a} \dot{a}
$\check{a} \bar{a} \ddot{a} \dot{a}\!$
### Standard functions
\sin a \cos b \tan c
$\sin a \cos b \tan c\!$
\sec d \csc e \cot f
$\sec d \csc e \cot f\,\!$
\arcsin h \arccos i \arctan j
$\arcsin h \arccos i \arctan j\,\!$
\sinh k \cosh l \tanh m \coth n\!
$\sinh k \cosh l \tanh m \coth n\!$
\operatorname{sh}\,o\,\operatorname{ch}\,p\,\operatorname{th}\,q\!
$\operatorname{sh}\,o\,\operatorname{ch}\,p\,\operatorname{th}\,q\!$
\operatorname{arsinh}\,r\,\operatorname{arcosh}\,s\,\operatorname{artanh}\,t
$\operatorname{arsinh}\,r\,\operatorname{arcosh}\,s\,\operatorname{artanh}\,t\,\!$
\lim u \limsup v \liminf w \min x \max y\!
$\lim u \limsup v \liminf w \min x \max y\!$
\inf z \sup a \exp b \ln c \lg d \log e \log_{10} f \ker g\!
$\inf z \sup a \exp b \ln c \lg d \log e \log_{10} f \ker g\!$
\deg h \gcd i \Pr j \det k \hom l \arg m \dim n
$\deg h \gcd i \Pr j \det k \hom l \arg m \dim n\!$
### Modular arithmetic
s_k \equiv 0 \pmod{m}
$s_k \equiv 0 \pmod{m}\,\!$
a\,\bmod\,b
$a\,\bmod\,b\,\!$
### Derivatives
\nabla \, \partial x \, dx \, \dot x \, \ddot y\, dy/dx\, \frac{dy}{dx}\, \frac{\partial^2 y}{\partial x_1\,\partial x_2}
$\nabla \, \partial x \, dx \, \dot x \, \ddot y\, dy/dx\, \frac{dy}{dx}\, \frac{\partial^2 y}{\partial x_1\,\partial x_2}$
### Sets
\forall \exists \empty \emptyset \varnothing
$\forall \exists \empty \emptyset \varnothing\,\!$
\in \ni \not \in \notin \subset \subseteq \supset \supseteq
$\in \ni \not \in \notin \subset \subseteq \supset \supseteq\,\!$
\cap \bigcap \cup \bigcup \biguplus \setminus \smallsetminus
$\cap \bigcap \cup \bigcup \biguplus \setminus \smallsetminus\,\!$
\sqsubset \sqsubseteq \sqsupset \sqsupseteq \sqcap \sqcup \bigsqcup
$\sqsubset \sqsubseteq \sqsupset \sqsupseteq \sqcap \sqcup \bigsqcup\,\!$
### Operators
+ \oplus \bigoplus \pm \mp -
$+ \oplus \bigoplus \pm \mp - \,\!$
\times \otimes \bigotimes \cdot \circ \bullet \bigodot
$\times \otimes \bigotimes \cdot \circ \bullet \bigodot\,\!$
\star * / \div \frac{1}{2}
$\star * / \div \frac{1}{2}\,\!$
### Logic
\land (or \and) \wedge \bigwedge \bar{q} \to p
$\land \wedge \bigwedge \bar{q} \to p\,\!$
\lor \vee \bigvee \lnot \neg q \And
$\lor \vee \bigvee \lnot \neg q \And\,\!$
### Root
\sqrt{2} \sqrt[n]{x}
$\sqrt{2} \sqrt[n]{x}\,\!$
### Relations
\sim \approx \simeq \cong \dot= \overset{\underset{\mathrm{def}}{}}{=}
$\sim \approx \simeq \cong \dot= \overset{\underset{\mathrm{def}}{}}{=}\,\!$
\le < \ll \gg \ge > \equiv \not\equiv \ne \mbox{or} \neq \propto
$\le < \ll \gg \ge > \equiv \not\equiv \ne \mbox{or} \neq \propto\,\!$
### Geometric
\Diamond \Box \triangle \angle \perp \mid \nmid \| 45^\circ
$\Diamond \, \Box \, \triangle \, \angle \perp \, \mid \; \nmid \, \| 45^\circ\,\!$
### Arrows
\leftarrow (or \gets) \rightarrow (or \to) \nleftarrow \nrightarrow \leftrightarrow \nleftrightarrow \longleftarrow \longrightarrow \longleftrightarrow
$\leftarrow \rightarrow \nleftarrow \not\to \leftrightarrow \nleftrightarrow \longleftarrow \longrightarrow \longleftrightarrow \,\!$
\Leftarrow \Rightarrow \nLeftarrow \nRightarrow \Leftrightarrow \nLeftrightarrow \Longleftarrow \Longrightarrow \Longleftrightarrow (or \iff)
$\Leftarrow \Rightarrow \nLeftarrow \nRightarrow \Leftrightarrow \nLeftrightarrow \Longleftarrow \Longrightarrow \Longleftrightarrow \!$
\uparrow \downarrow \updownarrow \Uparrow \Downarrow \Updownarrow \nearrow \searrow \swarrow \nwarrow
$\uparrow \downarrow \updownarrow \Uparrow \Downarrow \Updownarrow \nearrow \searrow \swarrow \nwarrow \!$
\rightharpoonup \rightharpoondown \leftharpoonup \leftharpoondown \upharpoonleft \upharpoonright \downharpoonleft \downharpoonright \rightleftharpoons \leftrightharpoons
$\rightharpoonup \rightharpoondown \leftharpoonup \leftharpoondown \upharpoonleft \upharpoonright \downharpoonleft \downharpoonright \rightleftharpoons \leftrightharpoons \,\!$
\curvearrowleft \circlearrowleft \Lsh \upuparrows \rightrightarrows \rightleftarrows \Rrightarrow \rightarrowtail \looparrowright
$\curvearrowleft \circlearrowleft \Lsh \upuparrows \rightrightarrows \rightleftarrows \Rrightarrow \rightarrowtail \looparrowright \,\!$
\curvearrowright \circlearrowright \Rsh \downdownarrows \leftleftarrows \leftrightarrows \Lleftarrow \leftarrowtail \looparrowleft
$\curvearrowright \circlearrowright \Rsh \downdownarrows \leftleftarrows \leftrightarrows \Lleftarrow \leftarrowtail \looparrowleft \,\!$
\mapsto \longmapsto \hookrightarrow \hookleftarrow \multimap \leftrightsquigarrow \rightsquigarrow
$\mapsto \longmapsto \hookrightarrow \hookleftarrow \multimap \leftrightsquigarrow \rightsquigarrow \,\!$
### Special
\And \eth \S \P \% \dagger \ddagger \ldots \cdots
$\And \eth \S \P \% \dagger \ddagger \ldots \cdots\,\!$
\smile \frown \wr \triangleleft \triangleright \infty \bot \top
$\smile \frown \wr \triangleleft \triangleright \infty \bot \top\,\!$
\vdash \vDash \Vdash \models \lVert \rVert \imath \hbar
$\vdash \vDash \Vdash \models \lVert \rVert \imath \hbar\,\!$
\ell \mho \Finv \Re \Im \wp \complement
$\ell \mho \Finv \Re \Im \wp \complement\,\!$
\diamondsuit \heartsuit \clubsuit \spadesuit \Game \flat \natural \sharp
$\diamondsuit \heartsuit \clubsuit \spadesuit \Game \flat \natural \sharp\,\!$
### Unsorted (new stuff)
\vartriangle \triangledown \lozenge \circledS \measuredangle \nexists \Bbbk \backprime \blacktriangle \blacktriangledown
$\vartriangle \triangledown \lozenge \circledS \measuredangle \nexists \Bbbk \backprime \blacktriangle \blacktriangledown$
\blacksquare \blacklozenge \bigstar \sphericalangle \diagup \diagdown \dotplus \Cap \Cup \barwedge
$\blacksquare \blacklozenge \bigstar \sphericalangle \diagup \diagdown \dotplus \Cap \Cup \barwedge\!$
\veebar \doublebarwedge \boxminus \boxtimes \boxdot \boxplus \divideontimes \ltimes \rtimes \leftthreetimes
$\veebar \doublebarwedge \boxminus \boxtimes \boxdot \boxplus \divideontimes \ltimes \rtimes \leftthreetimes$
\rightthreetimes \curlywedge \curlyvee \circleddash \circledast \circledcirc \centerdot \intercal \leqq \leqslant
$\rightthreetimes \curlywedge \curlyvee \circleddash \circledast \circledcirc \centerdot \intercal \leqq \leqslant$
\eqslantless \lessapprox \approxeq \lessdot \lll \lessgtr \lesseqgtr \lesseqqgtr \doteqdot \risingdotseq
$\eqslantless \lessapprox \approxeq \lessdot \lll \lessgtr \lesseqgtr \lesseqqgtr \doteqdot \risingdotseq$
\fallingdotseq \backsim \backsimeq \subseteqq \Subset \preccurlyeq \curlyeqprec \precsim \precapprox \vartriangleleft
$\fallingdotseq \backsim \backsimeq \subseteqq \Subset \preccurlyeq \curlyeqprec \precsim \precapprox \vartriangleleft$
\Vvdash \bumpeq \Bumpeq \geqq \geqslant \eqslantgtr \gtrsim \gtrapprox \eqsim \gtrdot
$\Vvdash \bumpeq \Bumpeq \geqq \geqslant \eqslantgtr \gtrsim \gtrapprox \eqsim \gtrdot$
\ggg \gtrless \gtreqless \gtreqqless \eqcirc \circeq \triangleq \thicksim \thickapprox \supseteqq
$\ggg \gtrless \gtreqless \gtreqqless \eqcirc \circeq \triangleq \thicksim \thickapprox \supseteqq$
\Supset \succcurlyeq \curlyeqsucc \succsim \succapprox \vartriangleright \shortmid \shortparallel \between \pitchfork
$\Supset \succcurlyeq \curlyeqsucc \succsim \succapprox \vartriangleright \shortmid \shortparallel \between \pitchfork$
\varpropto \blacktriangleleft \therefore \backepsilon \blacktriangleright \because \nleqslant \nleqq \lneq \lneqq
$\varpropto \blacktriangleleft \therefore \backepsilon \blacktriangleright \because \nleqslant \nleqq \lneq \lneqq$
\lvertneqq \lnsim \lnapprox \nprec \npreceq \precneqq \precnsim \precnapprox \nsim \nshortmid
$\lvertneqq \lnsim \lnapprox \nprec \npreceq \precneqq \precnsim \precnapprox \nsim \nshortmid$
\nvdash \nVdash \ntriangleleft \ntrianglelefteq \nsubseteq \nsubseteqq \varsubsetneq \subsetneqq \varsubsetneqq \ngtr
$\nvdash \nVdash \ntriangleleft \ntrianglelefteq \nsubseteq \nsubseteqq \varsubsetneq \subsetneqq \varsubsetneqq \ngtr$
\subsetneq
$\subsetneq$
\ngeqslant \ngeqq \gneq \gneqq \gvertneqq \gnsim \gnapprox \nsucc \nsucceq \succneqq
$\ngeqslant \ngeqq \gneq \gneqq \gvertneqq \gnsim \gnapprox \nsucc \nsucceq \succneqq$
\succnsim \succnapprox \ncong \nshortparallel \nparallel \nvDash \nVDash \ntriangleright \ntrianglerighteq \nsupseteq
$\succnsim \succnapprox \ncong \nshortparallel \nparallel \nvDash \nVDash \ntriangleright \ntrianglerighteq \nsupseteq$
\nsupseteqq \varsupsetneq \supsetneqq \varsupsetneqq
$\nsupseteqq \varsupsetneq \supsetneqq \varsupsetneqq$
\jmath \surd \ast \uplus \diamond \bigtriangleup \bigtriangledown \ominus
$\jmath \surd \ast \uplus \diamond \bigtriangleup \bigtriangledown \ominus\,\!$
\oslash \odot \bigcirc \amalg \prec \succ \preceq \succeq
$\oslash \odot \bigcirc \amalg \prec \succ \preceq \succeq\,\!$
\dashv \asymp \doteq \parallel
$\dashv \asymp \doteq \parallel\,\!$
\ulcorner \urcorner \llcorner \lrcorner
$\ulcorner \urcorner \llcorner \lrcorner$
## Larger Expressions
### Subscripts, superscripts, integrals
Feature Syntax How it looks rendered
HTML PNG
Superscript
a^2
$a^2$ $a^2 \,\!$
Subscript
a_2
$a_2$ $a_2 \,\!$
Grouping
a^{2+2}
$a^{2+2}$ $a^{2+2}\,\!$
a_{i,j}
$a_{i,j}$ $a_{i,j}\,\!$
Combining sub & super without and with horizontal separation
x_2^3
$x_2^3$ $x_2^3 \,\!$
{x_2}^3
${x_2}^3$ ${x_2}^3 \,\!$
Super super
10^{10^{ \,\!{8} }
$10^{10^{ \,\! 8 } }$
Super super
10^{10^{ \overset{8}{} }}
$10^{10^{ \overset{8}{} }}$
Super super (wrong in HTML in some browsers)
10^{10^8}
$10^{10^8}$
Preceding and/or Additional sub & super
\sideset{_1^2}{_3^4}\prod_a^b
$\sideset{_1^2}{_3^4}\prod_a^b$
{}_1^2\!\Omega_3^4
${}_1^2\!\Omega_3^4$
Stacking
\overset{\alpha}{\omega}
$\overset{\alpha}{\omega}$
\underset{\alpha}{\omega}
$\underset{\alpha}{\omega}$
\overset{\alpha}{\underset{\gamma}{\omega}}
$\overset{\alpha}{\underset{\gamma}{\omega}}$
\stackrel{\alpha}{\omega}
$\stackrel{\alpha}{\omega}$
Derivative (forced PNG)
x', y'', f', f''\!
$x', y'', f', f''\!$
Derivative (f in italics may overlap primes in HTML)
x', y'', f', f''
$x', y'', f', f''$ $x', y'', f', f''\!$
Derivative (wrong in HTML)
x^\prime, y^{\prime\prime}
$x^\prime, y^{\prime\prime}$ $x^\prime, y^{\prime\prime}\,\!$
Derivative (wrong in PNG)
x\prime, y\prime\prime
$x\prime, y\prime\prime$ $x\prime, y\prime\prime\,\!$
Derivative dots
\dot{x}, \ddot{x}
$\dot{x}, \ddot{x}$
Underlines, overlines, vectors
\hat a \ \bar b \ \vec c
$\hat a \ \bar b \ \vec c$
\overrightarrow{a b} \ \overleftarrow{c d} \ \widehat{d e f}
$\overrightarrow{a b} \ \overleftarrow{c d} \ \widehat{d e f}$
\overline{g h i} \ \underline{j k l}
$\overline{g h i} \ \underline{j k l}$
Arrows
A \xleftarrow{n+\mu-1} B \xrightarrow[T]{n\pm i-1} C
$A \xleftarrow{n+\mu-1} B \xrightarrow[T]{n\pm i-1} C$
Overbraces
\overbrace{ 1+2+\cdots+100 }^{5050}
$\overbrace{ 1+2+\cdots+100 }^{5050}$
Underbraces
\underbrace{ a+b+\cdots+z }_{26}
$\underbrace{ a+b+\cdots+z }_{26}$
Sum
\sum_{k=1}^N k^2
$\sum_{k=1}^N k^2$
Sum (force
\textstyle
)
\textstyle \sum_{k=1}^N k^2
$\textstyle \sum_{k=1}^N k^2$
Product
\prod_{i=1}^N x_i
$\prod_{i=1}^N x_i$
Product (force
\textstyle
)
\textstyle \prod_{i=1}^N x_i
$\textstyle \prod_{i=1}^N x_i$
Coproduct
\coprod_{i=1}^N x_i
$\coprod_{i=1}^N x_i$
Coproduct (force
\textstyle
)
\textstyle \coprod_{i=1}^N x_i
$\textstyle \coprod_{i=1}^N x_i$
Limit
\lim_{n \to \infty}x_n
$\lim_{n \to \infty}x_n$
Limit (force
\textstyle
)
\textstyle \lim_{n \to \infty}x_n
$\textstyle \lim_{n \to \infty}x_n$
Integral
\int\limits_{1}^{3}\frac{e^3/x}{x^2}\, dx
$\int\limits_{1}^{3}\frac{e^3/x}{x^2}\, dx$
Integral (alternate limits style)
\int_{1}^{3}\frac{e^3/x}{x^2}\, dx
$\int_{1}^{3}\frac{e^3/x}{x^2}\, dx$
Integral (force
\textstyle
)
\textstyle \int\limits_{-N}^{N} e^x\, dx
$\textstyle \int\limits_{-N}^{N} e^x\, dx$
Integral (force
\textstyle
, alternate limits style)
\textstyle \int_{-N}^{N} e^x\, dx
$\textstyle \int_{-N}^{N} e^x\, dx$
Double integral
\iint\limits_D \, dx\,dy
$\iint\limits_D \, dx\,dy$
Triple integral
\iiint\limits_E \, dx\,dy\,dz
$\iiint\limits_E \, dx\,dy\,dz$
\iiiint\limits_F \, dx\,dy\,dz\,dt
$\iiiint\limits_F \, dx\,dy\,dz\,dt$
Line or path integral
\int_C x^3\, dx + 4y^2\, dy
$\int_C x^3\, dx + 4y^2\, dy$
Closed line or path integral
\oint_C x^3\, dx + 4y^2\, dy
$\oint_C x^3\, dx + 4y^2\, dy$
Intersections
\bigcap_1^n p
$\bigcap_1^n p$
Unions
\bigcup_1^k p
$\bigcup_1^k p$
### Fractions, matrices, multilines
Feature Syntax How it looks rendered
Fractions
\frac{2}{4}=0.5
$\frac{2}{4}=0.5$
Small Fractions
\tfrac{2}{4} = 0.5
$\tfrac{2}{4} = 0.5$
Large (normal) Fractions
\dfrac{2}{4} = 0.5 \qquad \dfrac{2}{c + \dfrac{2}{d + \dfrac{2}{4}}} = a
$\dfrac{2}{4} = 0.5 \qquad \dfrac{2}{c + \dfrac{2}{d + \dfrac{2}{4}}} = a$
Large (nested) Fractions
\cfrac{2}{c + \cfrac{2}{d + \cfrac{2}{4}}} = a
$\cfrac{2}{c + \cfrac{2}{d + \cfrac{2}{4}}} = a$
Binomial coefficients
\binom{n}{k}
$\binom{n}{k}$
Small Binomial coefficients
\tbinom{n}{k}
$\tbinom{n}{k}$
Large (normal) Binomial coefficients
\dbinom{n}{k}
$\dbinom{n}{k}$
Matrices
\begin{matrix}
x & y \\
z & v
\end{matrix}
$\begin{matrix} x & y \\ z & v \end{matrix}$
\begin{vmatrix}
x & y \\
z & v
\end{vmatrix}
$\begin{vmatrix} x & y \\ z & v \end{vmatrix}$
\begin{Vmatrix}
x & y \\
z & v
\end{Vmatrix}
$\begin{Vmatrix} x & y \\ z & v \end{Vmatrix}$
\begin{bmatrix}
0 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & 0
\end{bmatrix}
$\begin{bmatrix} 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0\end{bmatrix}$
\begin{Bmatrix}
x & y \\
z & v
\end{Bmatrix}
$\begin{Bmatrix} x & y \\ z & v \end{Bmatrix}$
\begin{pmatrix}
x & y \\
z & v
\end{pmatrix}
$\begin{pmatrix} x & y \\ z & v \end{pmatrix}$
\bigl( \begin{smallmatrix}
a&b\\ c&d
\end{smallmatrix} \bigr)
$\bigl( \begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \bigr)$
Case distinctions
f(n) =
\begin{cases}
n/2, & \mbox{if }n\mbox{ is even} \\
3n+1, & \mbox{if }n\mbox{ is odd}
\end{cases}
$f(n) = \begin{cases} n/2, & \mbox{if }n\mbox{ is even} \\ 3n+1, & \mbox{if }n\mbox{ is odd} \end{cases}$
Multiline equations
\begin{align}
f(x) & = (a+b)^2 \\
& = a^2+2ab+b^2 \\
\end{align}
\begin{align} f(x) & = (a+b)^2 \\ & = a^2+2ab+b^2 \\ \end{align}
\begin{alignat}{2}
f(x) & = (a-b)^2 \\
& = a^2-2ab+b^2 \\
\end{alignat}
\begin{alignat}{2} f(x) & = (a-b)^2 \\ & = a^2-2ab+b^2 \\ \end{alignat}
Multiline equations (must define number of colums used ({lcr}) (should not be used unless needed)
\begin{array}{lcl}
z & = & a \\
f(x,y,z) & = & x + y + z
\end{array}
$\begin{array}{lcl} z & = & a \\ f(x,y,z) & = & x + y + z \end{array}$
Multiline equations (more)
\begin{array}{lcr}
z & = & a \\
f(x,y,z) & = & x + y + z
\end{array}
$\begin{array}{lcr} z & = & a \\ f(x,y,z) & = & x + y + z \end{array}$
Breaking up a long expression so that it wraps when necessary, at the expense of destroying correct spacing
$f(x) \,\!$
$= \sum_{n=0}^\infty a_n x^n$
$= a_0+a_1x+a_2x^2+\cdots$
$f(x) \,\!$$= \sum_{n=0}^\infty a_n x^n$$= a_0 +a_1x+a_2x^2+\cdots$
Simultaneous equations
\begin{cases}
3x + 5y + z \\
7x - 2y + 4z \\
-6x + 3y + 2z
\end{cases}
$\begin{cases} 3x + 5y + z \\ 7x - 2y + 4z \\ -6x + 3y + 2z \end{cases}$
Arrays
\begin{array}{|c|c||c|} a & b & S \\
\hline
0&0&1\\
0&1&1\\
1&0&1\\
1&1&0\\
\end{array}
$\begin{array}{|c|c||c|} a & b & S \\ \hline 0&0&1\\ 0&1&1\\ 1&0&1\\ 1&1&0\\ \end{array}$
### Parenthesizing big expressions, brackets, bars
Feature Syntax How it looks rendered
( \frac{1}{2} )
$( \frac{1}{2} )$
Good
\left ( \frac{1}{2} \right )
$\left ( \frac{1}{2} \right )$
You can use various delimiters with \left and \right:
Feature Syntax How it looks rendered
Parentheses
\left ( \frac{a}{b} \right )
$\left ( \frac{a}{b} \right )$
Brackets
\left [ \frac{a}{b} \right ] \quad \left \lbrack \frac{a}{b} \right \rbrack
$\left [ \frac{a}{b} \right ] \quad \left \lbrack \frac{a}{b} \right \rbrack$
Braces
\left \{ \frac{a}{b} \right \} \quad \left \lbrace \frac{a}{b} \right \rbrace
$\left \{ \frac{a}{b} \right \} \quad \left \lbrace \frac{a}{b} \right \rbrace$
Angle brackets
\left \langle \frac{a}{b} \right \rangle
$\left \langle \frac{a}{b} \right \rangle$
Bars and double bars
\left | \frac{a}{b} \right \vert \left \Vert \frac{c}{d} \right \|
$\left | \frac{a}{b} \right \vert \left \Vert \frac{c}{d} \right \|$
Floor and ceiling functions:
\left \lfloor \frac{a}{b} \right \rfloor \left \lceil \frac{c}{d} \right \rceil
$\left \lfloor \frac{a}{b} \right \rfloor \left \lceil \frac{c}{d} \right \rceil$
Slashes and backslashes
\left / \frac{a}{b} \right \backslash
$\left / \frac{a}{b} \right \backslash$
Up, down and up-down arrows
\left \uparrow \frac{a}{b} \right \downarrow \quad \left \Uparrow \frac{a}{b} \right \Downarrow \quad \left \updownarrow \frac{a}{b} \right \Updownarrow
$\left \uparrow \frac{a}{b} \right \downarrow \quad \left \Uparrow \frac{a}{b} \right \Downarrow \quad \left \updownarrow \frac{a}{b} \right \Updownarrow$
Delimiters can be mixed,
as long as \left and \right match
\left [ 0,1 \right )
\left \langle \psi \right |
$\left [ 0,1 \right )$
$\left \langle \psi \right |$
Use \left. and \right. if you don't
want a delimiter to appear:
\left . \frac{A}{B} \right \} \to X
$\left . \frac{A}{B} \right \} \to X$
Size of the delimiters
\big( \Big( \bigg( \Bigg( \dots \Bigg] \bigg] \Big] \big]/<code>
| $\big( \Big( \bigg( \Bigg( \dots \Bigg] \bigg] \Big] \big]$
|-
| <code>\big\{ \Big\{ \bigg\{ \Bigg\{ \dots \Bigg\rangle \bigg\rangle \Big\rangle \big\rangle
$\big\{ \Big\{ \bigg\{ \Bigg\{ \dots \Bigg\rangle \bigg\rangle \Big\rangle \big\rangle$
\big\| \Big\| \bigg\| \Bigg\| \dots \Bigg| \bigg| \Big| \big|
$\big\| \Big\| \bigg\| \Bigg\| \dots \Bigg| \bigg| \Big| \big|$
\big\lfloor \Big\lfloor \bigg\lfloor \Bigg\lfloor \dots \Bigg\rceil \bigg\rceil \Big\rceil \big\rceil
$\big\lfloor \Big\lfloor \bigg\lfloor \Bigg\lfloor \dots \Bigg\rceil \bigg\rceil \Big\rceil \big\rceil$
\big\uparrow \Big\uparrow \bigg\uparrow \Bigg\uparrow \dots \Bigg\Downarrow \bigg\Downarrow \Big\Downarrow \big\Downarrow
$\big\uparrow \Big\uparrow \bigg\uparrow \Bigg\uparrow \dots \Bigg\Downarrow \bigg\Downarrow \Big\Downarrow \big\Downarrow$
\big\updownarrow \Big\updownarrow \bigg\updownarrow \Bigg\updownarrow \dots \Bigg\Updownarrow \bigg\Updownarrow \Big\Updownarrow \big\Updownarrow
$\big\updownarrow \Big\updownarrow \bigg\updownarrow \Bigg\updownarrow \dots \Bigg\Updownarrow \bigg\Updownarrow \Big\Updownarrow \big\Updownarrow$
\big / \Big / \bigg / \Bigg / \dots \Bigg\backslash \bigg\backslash \Big\backslash \big\backslash
$\big / \Big / \bigg / \Bigg / \dots \Bigg\backslash \bigg\backslash \Big\backslash \big\backslash$
## Alphabets and typefaces
Texvc cannot render arbitrary Unicode characters. Those it can handle can be entered by the expressions below. For others, such as Cyrillic, they can be entered as Unicode or HTML entities in running text, but cannot be used in displayed formulas.
Greek alphabet
\Alpha \Beta \Gamma \Delta \Epsilon \Zeta
$\Alpha \Beta \Gamma \Delta \Epsilon \Zeta \,\!$
\Eta \Theta \Iota \Kappa \Lambda \Mu
$\Eta \Theta \Iota \Kappa \Lambda \Mu \,\!$
\Nu \Xi \Pi \Rho \Sigma \Tau
$\Nu \Xi \Pi \Rho \Sigma \Tau\,\!$
\Upsilon \Phi \Chi \Psi \Omega
$\Upsilon \Phi \Chi \Psi \Omega \,\!$
\alpha \beta \gamma \delta \epsilon \zeta
$\alpha \beta \gamma \delta \epsilon \zeta \,\!$
\eta \theta \iota \kappa \lambda \mu
$\eta \theta \iota \kappa \lambda \mu \,\!$
\nu \xi \pi \rho \sigma \tau
$\nu \xi \pi \rho \sigma \tau \,\!$
\upsilon \phi \chi \psi \omega
$\upsilon \phi \chi \psi \omega \,\!$
\varepsilon \digamma \vartheta \varkappa
$\varepsilon \digamma \vartheta \varkappa \,\!$
\varpi \varrho \varsigma \varphi
$\varpi \varrho \varsigma \varphi\,\!$
Blackboard Bold/Scripts
\mathbb{A} \mathbb{B} \mathbb{C} \mathbb{D} \mathbb{E} \mathbb{F} \mathbb{G}
$\mathbb{A} \mathbb{B} \mathbb{C} \mathbb{D} \mathbb{E} \mathbb{F} \mathbb{G} \,\!$
\mathbb{H} \mathbb{I} \mathbb{J} \mathbb{K} \mathbb{L} \mathbb{M}
$\mathbb{H} \mathbb{I} \mathbb{J} \mathbb{K} \mathbb{L} \mathbb{M} \,\!$
\mathbb{N} \mathbb{O} \mathbb{P} \mathbb{Q} \mathbb{R} \mathbb{S} \mathbb{T}
$\mathbb{N} \mathbb{O} \mathbb{P} \mathbb{Q} \mathbb{R} \mathbb{S} \mathbb{T} \,\!$
\mathbb{U} \mathbb{V} \mathbb{W} \mathbb{X} \mathbb{Y} \mathbb{Z}
$\mathbb{U} \mathbb{V} \mathbb{W} \mathbb{X} \mathbb{Y} \mathbb{Z}\,\!$
boldface (vectors)
\mathbf{A} \mathbf{B} \mathbf{C} \mathbf{D} \mathbf{E} \mathbf{F} \mathbf{G}
$\mathbf{A} \mathbf{B} \mathbf{C} \mathbf{D} \mathbf{E} \mathbf{F} \mathbf{G} \,\!$
\mathbf{H} \mathbf{I} \mathbf{J} \mathbf{K} \mathbf{L} \mathbf{M}
$\mathbf{H} \mathbf{I} \mathbf{J} \mathbf{K} \mathbf{L} \mathbf{M} \,\!$
\mathbf{N} \mathbf{O} \mathbf{P} \mathbf{Q} \mathbf{R} \mathbf{S} \mathbf{T}
$\mathbf{N} \mathbf{O} \mathbf{P} \mathbf{Q} \mathbf{R} \mathbf{S} \mathbf{T} \,\!$
\mathbf{U} \mathbf{V} \mathbf{W} \mathbf{X} \mathbf{Y} \mathbf{Z}
$\mathbf{U} \mathbf{V} \mathbf{W} \mathbf{X} \mathbf{Y} \mathbf{Z} \,\!$
\mathbf{a} \mathbf{b} \mathbf{c} \mathbf{d} \mathbf{e} \mathbf{f} \mathbf{g}
$\mathbf{a} \mathbf{b} \mathbf{c} \mathbf{d} \mathbf{e} \mathbf{f} \mathbf{g} \,\!$
\mathbf{h} \mathbf{i} \mathbf{j} \mathbf{k} \mathbf{l} \mathbf{m}
$\mathbf{h} \mathbf{i} \mathbf{j} \mathbf{k} \mathbf{l} \mathbf{m} \,\!$
\mathbf{n} \mathbf{o} \mathbf{p} \mathbf{q} \mathbf{r} \mathbf{s} \mathbf{t}
$\mathbf{n} \mathbf{o} \mathbf{p} \mathbf{q} \mathbf{r} \mathbf{s} \mathbf{t} \,\!$
\mathbf{u} \mathbf{v} \mathbf{w} \mathbf{x} \mathbf{y} \mathbf{z}
$\mathbf{u} \mathbf{v} \mathbf{w} \mathbf{x} \mathbf{y} \mathbf{z} \,\!$
\mathbf{0} \mathbf{1} \mathbf{2} \mathbf{3} \mathbf{4}
$\mathbf{0} \mathbf{1} \mathbf{2} \mathbf{3} \mathbf{4} \,\!$
\mathbf{5} \mathbf{6} \mathbf{7} \mathbf{8} \mathbf{9}
$\mathbf{5} \mathbf{6} \mathbf{7} \mathbf{8} \mathbf{9}\,\!$
Boldface (greek)
\boldsymbol{\Alpha} \boldsymbol{\Beta} \boldsymbol{\Gamma} \boldsymbol{\Delta} \boldsymbol{\Epsilon} \boldsymbol{\Zeta}
$\boldsymbol{\Alpha} \boldsymbol{\Beta} \boldsymbol{\Gamma} \boldsymbol{\Delta} \boldsymbol{\Epsilon} \boldsymbol{\Zeta} \,\!$
\boldsymbol{\Eta} \boldsymbol{\Theta} \boldsymbol{\Iota} \boldsymbol{\Kappa} \boldsymbol{\Lambda} \boldsymbol{\Mu}
$\boldsymbol{\Eta} \boldsymbol{\Theta} \boldsymbol{\Iota} \boldsymbol{\Kappa} \boldsymbol{\Lambda} \boldsymbol{\Mu}\,\!$
\boldsymbol{\Nu} \boldsymbol{\Xi} \boldsymbol{\Pi} \boldsymbol{\Rho} \boldsymbol{\Sigma} \boldsymbol{\Tau}
$\boldsymbol{\Nu} \boldsymbol{\Xi} \boldsymbol{\Pi} \boldsymbol{\Rho} \boldsymbol{\Sigma} \boldsymbol{\Tau}\,\!$
\boldsymbol{\Upsilon} \boldsymbol{\Phi} \boldsymbol{\Chi} \boldsymbol{\Psi} \boldsymbol{\Omega}
$\boldsymbol{\Upsilon} \boldsymbol{\Phi} \boldsymbol{\Chi} \boldsymbol{\Psi} \boldsymbol{\Omega}\,\!$
\boldsymbol{\alpha} \boldsymbol{\beta} \boldsymbol{\gamma} \boldsymbol{\delta} \boldsymbol{\epsilon} \boldsymbol{\zeta}
$\boldsymbol{\alpha} \boldsymbol{\beta} \boldsymbol{\gamma} \boldsymbol{\delta} \boldsymbol{\epsilon} \boldsymbol{\zeta}\,\!$
\boldsymbol{\eta} \boldsymbol{\theta} \boldsymbol{\iota} \boldsymbol{\kappa} \boldsymbol{\lambda} \boldsymbol{\mu}
$\boldsymbol{\eta} \boldsymbol{\theta} \boldsymbol{\iota} \boldsymbol{\kappa} \boldsymbol{\lambda} \boldsymbol{\mu}\,\!$
\boldsymbol{\nu} \boldsymbol{\xi} \boldsymbol{\pi} \boldsymbol{\rho} \boldsymbol{\sigma} \boldsymbol{\tau}
$\boldsymbol{\nu} \boldsymbol{\xi} \boldsymbol{\pi} \boldsymbol{\rho} \boldsymbol{\sigma} \boldsymbol{\tau}\,\!$
\boldsymbol{\upsilon} \boldsymbol{\phi} \boldsymbol{\chi} \boldsymbol{\psi} \boldsymbol{\omega}
$\boldsymbol{\upsilon} \boldsymbol{\phi} \boldsymbol{\chi} \boldsymbol{\psi} \boldsymbol{\omega}\,\!$
\boldsymbol{\varepsilon} \boldsymbol{\digamma} \boldsymbol{\vartheta} \boldsymbol{\varkappa}
$\boldsymbol{\varepsilon} \boldsymbol{\digamma} \boldsymbol{\vartheta} \boldsymbol{\varkappa} \,\!$
\boldsymbol{\varpi} \boldsymbol{\varrho} \boldsymbol{\varsigma} \boldsymbol{\varphi}
$\boldsymbol{\varpi} \boldsymbol{\varrho} \boldsymbol{\varsigma} \boldsymbol{\varphi}\,\!$
Italics
\mathit{A} \mathit{B} \mathit{C} \mathit{D} \mathit{E} \mathit{F} \mathit{G}
$\mathit{A} \mathit{B} \mathit{C} \mathit{D} \mathit{E} \mathit{F} \mathit{G} \,\!$
\mathit{H} \mathit{I} \mathit{J} \mathit{K} \mathit{L} \mathit{M}
$\mathit{H} \mathit{I} \mathit{J} \mathit{K} \mathit{L} \mathit{M} \,\!$
\mathit{N} \mathit{O} \mathit{P} \mathit{Q} \mathit{R} \mathit{S} \mathit{T}
$\mathit{N} \mathit{O} \mathit{P} \mathit{Q} \mathit{R} \mathit{S} \mathit{T} \,\!$
\mathit{U} \mathit{V} \mathit{W} \mathit{X} \mathit{Y} \mathit{Z}
$\mathit{U} \mathit{V} \mathit{W} \mathit{X} \mathit{Y} \mathit{Z} \,\!$
\mathit{a} \mathit{b} \mathit{c} \mathit{d} \mathit{e} \mathit{f} \mathit{g}
$\mathit{a} \mathit{b} \mathit{c} \mathit{d} \mathit{e} \mathit{f} \mathit{g} \,\!$
\mathit{h} \mathit{i} \mathit{j} \mathit{k} \mathit{l} \mathit{m}
$\mathit{h} \mathit{i} \mathit{j} \mathit{k} \mathit{l} \mathit{m} \,\!$
\mathit{n} \mathit{o} \mathit{p} \mathit{q} \mathit{r} \mathit{s} \mathit{t}
$\mathit{n} \mathit{o} \mathit{p} \mathit{q} \mathit{r} \mathit{s} \mathit{t} \,\!$
\mathit{u} \mathit{v} \mathit{w} \mathit{x} \mathit{y} \mathit{z}
$\mathit{u} \mathit{v} \mathit{w} \mathit{x} \mathit{y} \mathit{z} \,\!$
\mathit{0} \mathit{1} \mathit{2} \mathit{3} \mathit{4}
$\mathit{0} \mathit{1} \mathit{2} \mathit{3} \mathit{4} \,\!$
\mathit{5} \mathit{6} \mathit{7} \mathit{8} \mathit{9}
$\mathit{5} \mathit{6} \mathit{7} \mathit{8} \mathit{9}\,\!$
Roman typeface
\mathrm{A} \mathrm{B} \mathrm{C} \mathrm{D} \mathrm{E} \mathrm{F} \mathrm{G}
$\mathrm{A} \mathrm{B} \mathrm{C} \mathrm{D} \mathrm{E} \mathrm{F} \mathrm{G} \,\!$
\mathrm{H} \mathrm{I} \mathrm{J} \mathrm{K} \mathrm{L} \mathrm{M}
$\mathrm{H} \mathrm{I} \mathrm{J} \mathrm{K} \mathrm{L} \mathrm{M} \,\!$
\mathrm{N} \mathrm{O} \mathrm{P} \mathrm{Q} \mathrm{R} \mathrm{S} \mathrm{T}
$\mathrm{N} \mathrm{O} \mathrm{P} \mathrm{Q} \mathrm{R} \mathrm{S} \mathrm{T} \,\!$
\mathrm{U} \mathrm{V} \mathrm{W} \mathrm{X} \mathrm{Y} \mathrm{Z}
$\mathrm{U} \mathrm{V} \mathrm{W} \mathrm{X} \mathrm{Y} \mathrm{Z} \,\!$
\mathrm{a} \mathrm{b} \mathrm{c} \mathrm{d} \mathrm{e} \mathrm{f} \mathrm{g}
$\mathrm{a} \mathrm{b} \mathrm{c} \mathrm{d} \mathrm{e} \mathrm{f} \mathrm{g}\,\!$
\mathrm{h} \mathrm{i} \mathrm{j} \mathrm{k} \mathrm{l} \mathrm{m}
$\mathrm{h} \mathrm{i} \mathrm{j} \mathrm{k} \mathrm{l} \mathrm{m} \,\!$
\mathrm{n} \mathrm{o} \mathrm{p} \mathrm{q} \mathrm{r} \mathrm{s} \mathrm{t}
$\mathrm{n} \mathrm{o} \mathrm{p} \mathrm{q} \mathrm{r} \mathrm{s} \mathrm{t} \,\!$
\mathrm{u} \mathrm{v} \mathrm{w} \mathrm{x} \mathrm{y} \mathrm{z}
$\mathrm{u} \mathrm{v} \mathrm{w} \mathrm{x} \mathrm{y} \mathrm{z} \,\!$
\mathrm{0} \mathrm{1} \mathrm{2} \mathrm{3} \mathrm{4}
$\mathrm{0} \mathrm{1} \mathrm{2} \mathrm{3} \mathrm{4} \,\!$
\mathrm{5} \mathrm{6} \mathrm{7} \mathrm{8} \mathrm{9}
$\mathrm{5} \mathrm{6} \mathrm{7} \mathrm{8} \mathrm{9}\,\!$
Fraktur typeface
\mathfrak{A} \mathfrak{B} \mathfrak{C} \mathfrak{D} \mathfrak{E} \mathfrak{F} \mathfrak{G}
$\mathfrak{A} \mathfrak{B} \mathfrak{C} \mathfrak{D} \mathfrak{E} \mathfrak{F} \mathfrak{G} \,\!$
\mathfrak{H} \mathfrak{I} \mathfrak{J} \mathfrak{K} \mathfrak{L} \mathfrak{M}
$\mathfrak{H} \mathfrak{I} \mathfrak{J} \mathfrak{K} \mathfrak{L} \mathfrak{M} \,\!$
\mathfrak{N} \mathfrak{O} \mathfrak{P} \mathfrak{Q} \mathfrak{R} \mathfrak{S} \mathfrak{T}
$\mathfrak{N} \mathfrak{O} \mathfrak{P} \mathfrak{Q} \mathfrak{R} \mathfrak{S} \mathfrak{T} \,\!$
\mathfrak{U} \mathfrak{V} \mathfrak{W} \mathfrak{X} \mathfrak{Y} \mathfrak{Z}
$\mathfrak{U} \mathfrak{V} \mathfrak{W} \mathfrak{X} \mathfrak{Y} \mathfrak{Z} \,\!$
\mathfrak{a} \mathfrak{b} \mathfrak{c} \mathfrak{d} \mathfrak{e} \mathfrak{f} \mathfrak{g}
$\mathfrak{a} \mathfrak{b} \mathfrak{c} \mathfrak{d} \mathfrak{e} \mathfrak{f} \mathfrak{g} \,\!$
\mathfrak{h} \mathfrak{i} \mathfrak{j} \mathfrak{k} \mathfrak{l} \mathfrak{m}
$\mathfrak{h} \mathfrak{i} \mathfrak{j} \mathfrak{k} \mathfrak{l} \mathfrak{m} \,\!$
\mathfrak{n} \mathfrak{o} \mathfrak{p} \mathfrak{q} \mathfrak{r} \mathfrak{s} \mathfrak{t}
$\mathfrak{n} \mathfrak{o} \mathfrak{p} \mathfrak{q} \mathfrak{r} \mathfrak{s} \mathfrak{t} \,\!$
\mathfrak{u} \mathfrak{v} \mathfrak{w} \mathfrak{x} \mathfrak{y} \mathfrak{z}
$\mathfrak{u} \mathfrak{v} \mathfrak{w} \mathfrak{x} \mathfrak{y} \mathfrak{z} \,\!$
\mathfrak{0} \mathfrak{1} \mathfrak{2} \mathfrak{3} \mathfrak{4}
$\mathfrak{0} \mathfrak{1} \mathfrak{2} \mathfrak{3} \mathfrak{4} \,\!$
\mathfrak{5} \mathfrak{6} \mathfrak{7} \mathfrak{8} \mathfrak{9}
$\mathfrak{5} \mathfrak{6} \mathfrak{7} \mathfrak{8} \mathfrak{9}\,\!$
Calligraphy/Script
\mathcal{A} \mathcal{B} \mathcal{C} \mathcal{D} \mathcal{E} \mathcal{F} \mathcal{G}
$\mathcal{A} \mathcal{B} \mathcal{C} \mathcal{D} \mathcal{E} \mathcal{F} \mathcal{G} \,\!$
\mathcal{H} \mathcal{I} \mathcal{J} \mathcal{K} \mathcal{L} \mathcal{M}
$\mathcal{H} \mathcal{I} \mathcal{J} \mathcal{K} \mathcal{L} \mathcal{M} \,\!$
\mathcal{N} \mathcal{O} \mathcal{P} \mathcal{Q} \mathcal{R} \mathcal{S} \mathcal{T}
$\mathcal{N} \mathcal{O} \mathcal{P} \mathcal{Q} \mathcal{R} \mathcal{S} \mathcal{T} \,\!$
\mathcal{U} \mathcal{V} \mathcal{W} \mathcal{X} \mathcal{Y} \mathcal{Z}
$\mathcal{U} \mathcal{V} \mathcal{W} \mathcal{X} \mathcal{Y} \mathcal{Z}\,\!$
Hebrew
\aleph \beth \gimel \daleth
$\aleph \beth \gimel \daleth\,\!$
Feature Syntax How it looks rendered
non-italicised characters \mbox{abc} $\mbox{abc}$ $\mbox{abc} \,\!$
mixed italics (bad) \mbox{if} n \mbox{is even} $\mbox{if} n \mbox{is even}$ $\mbox{if} n \mbox{is even} \,\!$
mixed italics (good) \mbox{if }n\mbox{ is even} $\mbox{if }n\mbox{ is even}$ $\mbox{if }n\mbox{ is even} \,\!$
mixed italics (more legible: ~ is a non-breaking space, while "\ " forces a space) \mbox{if}~n\ \mbox{is even} $\mbox{if}~n\ \mbox{is even}$ $\mbox{if}~n\ \mbox{is even} \,\!$
## Color
Equations can use color:
• {\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}
${\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}$
• x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}
$x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}$
See here for all named colors supported by LaTeX.
Note that color should not be used as the only way to identify something, because it will become meaningless on black-and-white media or for color-blind people. See.
## Formatting issues
### Spacing
Note that TeX handles most spacing automatically, but you may sometimes want manual control.
Feature Syntax How it looks rendered
double quad space a \qquad b $a \qquad b$
quad space a \quad b $a \quad b$
text space a\ b $a\ b$
text space without PNG conversion a \mbox{ } b $a \mbox{ } b$
large space a\;b $a\;b$
medium space a\>b [not supported]
small space a\,b $a\,b$
no space ab $ab\,$
small negative space a\!b $a\!b$
### Alignment with normal text flow
Due to the default css
img.tex { vertical-align: middle; }
an inline expression like $\int_{-N}^{N} e^x\, dx$ should look good.
If you need to align it otherwise, use
<math style="vertical-align:-100%;">...[/itex]
and play with the
vertical-align
argument until you get it right; however, how it looks may depend on the browser and the browser settings.
Also note that if you rely on this workaround, if/when the rendering on the server gets fixed in future releases, as a result of this extra manual offset your formulae will suddenly be aligned incorrectly. So use it sparingly, if at all.
### Forced PNG rendering
To force the formula to render as PNG, add
\,
(small space) at the end of the formula (where it is not rendered). This will force PNG if the user is in "HTML if simple" mode, but not for "HTML if possible" mode. You can also use
\,\!
(small space and negative space, which cancel out) anywhere inside the math tags. This does force PNG even in "HTML if possible" mode, unlike
\,
.
This could be useful to keep the rendering of formulae in a proof consistent, for example, or to fix formulae that render incorrectly in HTML (at one time, a^{2+2} rendered with an extra underscore), or to demonstrate how something is rendered when it would normally show up as HTML (as in the examples above).
For instance:
Syntax How it looks rendered
a^{c+2} $a^{c+2}$
a^{c+2} \, $a^{c+2} \,$
a^{\,\!c+2} $a^{\,\!c+2}$
a^{b^{c+2}} $a^{b^{c+2}}$ (WRONG with option "HTML if possible or else PNG"!)
a^{b^{c+2}} \, $a^{b^{c+2}} \,$ (WRONG with option "HTML if possible or else PNG"!)
a^{b^{c+2}}\approx 5 $a^{b^{c+2}}\approx 5$ (due to "$\approx$" correctly displayed, no code "\,\!" needed)
a^{b^{\,\!c+2}} $a^{b^{\,\!c+2}}$
\int_{-N}^{N} e^x\, dx $\int_{-N}^{N} e^x\, dx$
This has been tested with most of the formulae on this page, and seems to work perfectly.
You might want to include a comment in the HTML so people don't "correct" the formula by removing it:
<!-- The \,\! is to keep the formula rendered as PNG instead of HTML. Please don't remove it.-->
## Examples
$ax^2 + bx + c = 0$
$ax^2 + bx + c = 0$
### Quadratic Polynomial (Force PNG Rendering)
$ax^2 + bx + c = 0\,\!$
$ax^2 + bx + c = 0\,\!$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
### Tall Parentheses and Fractions
$2 = \left( \frac{\left(3-x\right) \times 2}{3-x} \right)$
$2 = \left( \frac{\left(3-x\right) \times 2}{3-x} \right)$
$S_{\text{new}} = S_{\text{old}} - \frac{ \left( 5-T \right) ^2} {2}$
$S_{\text{new}} = S_{\text{old}} - \frac{ \left( 5-T \right) ^2} {2}$
### Integrals
$\int_a^x \!\!\!\int_a^s f(y)\,dy\,ds = \int_a^x f(y)(x-y)\,dy$
$\int_a^x \!\!\!\int_a^s f(y)\,dy\,ds = \int_a^x f(y)(x-y)\,dy$
### Summation
$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^m\left(m\,3^n+n\,3^m\right)}$
$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n} {3^m\left(m\,3^n+n\,3^m\right)}$
### Differential Equation
$u'' + p(x)u' + q(x)u=f(x),\quad x>a$
$u'' + p(x)u' + q(x)u=f(x),\quad x>a$
### Complex numbers
$|\bar{z}| = |z|, |(\bar{z})^n| = |z|^n, \arg(z^n) = n \arg(z)$
$|\bar{z}| = |z|, |(\bar{z})^n| = |z|^n, \arg(z^n) = n \arg(z)$
### Limits
$\lim_{z\rightarrow z_0} f(z)=f(z_0)$
$\lim_{z\rightarrow z_0} f(z)=f(z_0)$
### Integral Equation
$\phi_n(\kappa) = \frac{1}{4\pi^2\kappa^2} \int_0^\infty \frac{\sin(\kappa R)}{\kappa R} \frac{\partial}{\partial R} \left[R^2\frac{\partial D_n(R)}{\partial R}\right]\,dR$
$\phi_n(\kappa) = \frac{1}{4\pi^2\kappa^2} \int_0^\infty \frac{\sin(\kappa R)}{\kappa R} \frac{\partial}{\partial R} \left[R^2\frac{\partial D_n(R)}{\partial R}\right]\,dR$
### Example
$\phi_n(\kappa) = 0.033C_n^2\kappa^{-11/3},\quad \frac{1}{L_0}\ll\kappa\ll\frac{1}{l_0}$
$\phi_n(\kappa) = 0.033C_n^2\kappa^{-11/3},\quad \frac{1}{L_0}\ll\kappa\ll\frac{1}{l_0}$
### Continuation and cases
$f(x) = \begin{cases}1 & -1 \le x < 0 \\ \frac{1}{2} & x = 0 \\ 1 - x^2 & \mbox{otherwise}\end{cases}$
$f(x) = \begin{cases} 1 & -1 \le x < 0 \\ \frac{1}{2} & x = 0 \\ 1 - x^2 & \mbox{otherwise} \end{cases}$
### Prefixed subscript
${}_pF_q(a_1,\dots,a_p;c_1,\dots,c_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\cdots(a_p)_n}{(c_1)_n\cdots(c_q)_n}\frac{z^n}{n!}$
${}_pF_q(a_1,\dots,a_p;c_1,\dots,c_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\cdots(a_p)_n}{(c_1)_n\cdots(c_q)_n} \frac{z^n}{n!}$
### Fraction and small fraction
$\frac {a}{b}$ $\tfrac {a}{b}$
$\frac {a}{b}\ \tfrac {a}{b}$
### Unicode vs TeX comparison
unicode TeX See ⟦ [\![ $[\![$ { \{ $\{$ ∥ \| $\|$ } \} $\}$ ℵ \aleph $\aleph$ α \alpha $\alpha$ ⨿ \amalg $\amalg$ ∠ \angle $\angle$ ≈ \approx $\approx$ ∗ \ast $\ast$ ≍ \asymp $\asymp$ \ \backslash $\backslash$ β \beta $\beta$ ⋂ \bigcap $\bigcap$ ◯ \bigcirc $\bigcirc$ ⋃ \bigcup $\bigcup$ ⨀ \bigodot $\bigodot$ ⨁ \bigoplus $\bigoplus$ ⨂ \bigotimes $\bigotimes$ ⨆ \bigsqcup $\bigsqcup$ ▽ \bigtriangledown $\bigtriangledown$ △ \bigtriangleup $\bigtriangleup$ ⨄ \biguplus $\biguplus$ ⋀ \bigwedge $\bigwedge$ ⋁ \bigvee $\bigvee$ ⊥ \bot $\bot$ ⋈ \bowtie $\bowtie$ □ \Box $\Box$ ∙ \bullet $\bullet$ ∩ \cap $\cap$ ⋅ \cdot $\cdot$ ⋯ \cdots $\cdots$ χ \chi $\chi$ ∘ \circ $\circ$ ♣ \clubsuit $\clubsuit$ ≅ \cong $\cong$ ∐ \coprod $\coprod$ ∪ \cup $\cup$ † \dagger $\dagger$ ⊣ \dashv $\dashv$ ‡ \ddagger $\ddagger$ ⋱ \ddots $\ddots$ δ \delta $\delta$ Δ \Delta $\Delta$ ◇ \Diamond $\Diamond$ ⋄ \diamond $\diamond$ ♢ \diamondsuit $\diamondsuit$ ÷ \div $\div$ ≐ \doteq $\doteq$ ↓ \downarrow $\downarrow$ ⇓ \Downarrow $\Downarrow$ ℓ \ell $\ell$ ∅ \emptyset $\emptyset$ ϵ \epsilon $\epsilon$ ≡ \equiv $\equiv$ η \eta $\eta$ ∃ \exists $\exists$ ♭ \flat $\flat$ ∀ \forall $\forall$ ⌢ \frown $\frown$ γ \gamma $\gamma$ Γ \Gamma $\Gamma$ ≥ \ge $\ge$ ≥ \geq $\geq$ ← \gets $\gets$ ≫ \gg $\gg$ ℏ \hbar $\hbar$ ♡ \heartsuit $\heartsuit$ ↩ \hookleftarrow $\hookleftarrow$ ↪ \hookrightarrow $\hookrightarrow$ ℑ \Im $\Im$ ı \imath $\imath$ ∈ \in $\in$ ∞ \infty $\infty$ ∫ \int $\int$ ι \iota $\iota$ j \jmath $\jmath$ κ \kappa $\kappa$ λ \lambda $\lambda$ Λ \Lambda $\Lambda$ ∧ \land $\land$ ⟨ \langle $\langle$ ⟪ \langle\!\langle $\langle\!\langle$ { \lbrace $\lbrace$ [ \lbrack $\lbrack$ ⌈ \lceil $\lceil$ ≤ \le $\le$ ⇐ \Leftarrow $\Leftarrow$ ← \leftarrow $\leftarrow$ ↽ \leftharpoondown $\leftharpoondown$ ↼ \leftharpoonup $\leftharpoonup$ ↔ \leftrightarrow $\leftrightarrow$ ⇔ \Leftrightarrow $\Leftrightarrow$ ≤ \leq $\leq$ ⌊ \lfloor $\lfloor$ ≪ \ll $\ll$ ¬ \lnot $\lnot$ ⟸ \Longleftarrow $\Longleftarrow$ ⟵ \longleftarrow $\longleftarrow$ ⟺ \Longleftrightarrow $\Longleftrightarrow$ ⟷ \longleftrightarrow $\longleftrightarrow$ ⟼ \longmapsto $\longmapsto$ ⟹ \Longrightarrow $\Longrightarrow$ ⟶ \longrightarrow $\longrightarrow$ ∨ \lor $\lor$ ↦ \mapsto $\mapsto$ ∣ \mid $\mid$ ⊨ \models $\models$ ∓ \mp $\mp$ μ \mu $\mu$ ∇ \nabla $\nabla$ ♮ \natural $\natural$ ≠ \ne $\ne$ ↗ \nearrow $\nearrow$ ¬ \neg $\neg$ ≠ \neq $\neq$ ∋ \ni $\ni$ ≉ \not\approx $\not\approx$ ≭ \not\asymp $\not\asymp$ ≇ \not\cong $\not\cong$ ≢ \not\equiv $\not\equiv$ ≱ \not\geq $\not\geq$ ≰ \not\leq $\not\leq$ ⊀ \not\prec $\not\prec$ ⋠ \not\preceq $\not\preceq$ ≁ \not\sim $\not\sim$ ≄ \not\simeq $\not\simeq$ ⋢ \not\sqsubseteq $\not\sqsubseteq$ ⋣ \not\sqsupseteq $\not\sqsupseteq$ ⊄ \not\subset $\not\subset$ ⊈ \not\subseteq $\not\subseteq$ ⊁ \not\succ $\not\succ$ ⋡ \not\succeq $\not\succeq$ ⊅ \not\supset $\not\supset$ ⊉ \not\supseteq $\not\supseteq$ ≠ \not= $\not=$ ν \nu $\nu$ ↖ \nwarrow $\nwarrow$ ⊙ \odot $\odot$ ∮ \oint $\oint$ ω \omega $\omega$ Ω \Omega $\Omega$ ⊖ \ominus $\ominus$ ⊕ \oplus $\oplus$ ⊘ \oslash $\oslash$ ⊗ \otimes $\otimes$ ∥ \parallel $\parallel$ ∂ \partial $\partial$ ⊥ \perp $\perp$ ϕ \phi $\phi$ Φ \Phi $\Phi$ π \pi $\pi$ Π \Pi $\Pi$ ± \pm $\pm$ ≺ \prec $\prec$ ≼ \preceq $\preceq$ ′ \prime $\prime$ ∏ \prod $\prod$ ∝ \propto $\propto$ ψ \psi $\psi$ Ψ \Psi $\Psi$ ⟩ \rangle $\rangle$ ⟫ \rangle\!\rangle $\rangle\!\rangle$ } \rbrace $\rbrace$ ] \rbrack $\rbrack$ ⌉ \rceil $\rceil$ ℜ \Re $\Re$ ⌋ \rfloor $\rfloor$ ρ \rho $\rho$ → \rightarrow $\rightarrow$ ⇒ \Rightarrow $\Rightarrow$ ⇁ \rightharpoondown $\rightharpoondown$ ⇀ \rightharpoonup $\rightharpoonup$ ⇌ \rightleftharpoons $\rightleftharpoons$ ↘ \searrow $\searrow$ ∖ \setminus $\setminus$ ♯ \sharp $\sharp$ σ \sigma $\sigma$ Σ \Sigma $\Sigma$ ∼ \sim $\sim$ ≃ \simeq $\simeq$ ⌣ \smile $\smile$ ♠ \spadesuit $\spadesuit$ ⊓ \sqcap $\sqcap$ ⊔ \sqcup $\sqcup$ ⊏ \sqsubset $\sqsubset$ ⊑ \sqsubseteq $\sqsubseteq$ ⊐ \sqsupset $\sqsupset$ ⊒ \sqsupseteq $\sqsupseteq$ ⋆ \star $\star$ ⊂ \subset $\subset$ ⊆ \subseteq $\subseteq$ ≻ \succ $\succ$ ≽ \succeq $\succeq$ ∑ \sum $\sum$ ⊃ \supset $\supset$ ⊇ \supseteq $\supseteq$ √ \surd $\surd$ ↙ \swarrow $\swarrow$ τ \tau $\tau$ θ \theta $\theta$ Θ \Theta $\Theta$ × \times $\times$ → \to $\to$ ⊤ \top $\top$ △ \triangle $\triangle$ ◁ \triangleleft $\triangleleft$ ▷ \triangleright $\triangleright$ ↑ \uparrow $\uparrow$ ⇑ \Uparrow $\Uparrow$ ↕ \updownarrow $\updownarrow$ ⇕ \Updownarrow $\Updownarrow$ ⊎ \uplus $\uplus$ υ \upsilon $\upsilon$ Υ \Upsilon $\Upsilon$ ε \varepsilon $\varepsilon$ φ \varphi $\varphi$ ϖ \varpi $\varpi$ ϱ \varrho $\varrho$ ς \varsigma $\varsigma$ ϑ \vartheta $\vartheta$ ⊢ \vdash $\vdash$ ⋮ \vdots $\vdots$ ∧ \wedge $\wedge$ ∨ \vee $\vee$ ∣ \vert $\vert$ ∥ \Vert $\Vert$ ℘ \wp $\wp$ ≀ \wr $\wr$ ξ \xi $\xi$ Ξ \Xi $\Xi$ ζ \zeta $\zeta$ ⟧ ]\!] $]\!]$ ο o $o$ | 16,891 | 40,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 504, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-18 | latest | en | 0.208028 |
https://docs.devexpress.com/WindowsForms/14604/controls-and-libraries/spreadsheet/cell-basics/error-types?v=19.1 | 1,660,680,755,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00005.warc.gz | 220,594,897 | 13,723 | Error Types
If a cell contains a formula that cannot be calculated correctly, the cell’s value (Range.Value) is of the error type (the CellValue.IsError property value is true). To access information on an error contained in a cell, use the members of the ErrorValueInfo object returned by the CellValue.ErrorValue property. The ErrorValueInfo.Type returns the ErrorType enumeration member that specifies the error type, the ErrorValueInfo.Name property returns the error name displayed in a cell, and the ErrorValueInfo.Description property returns the cause of the error.
Error Type
Error Name
Error Description
Example
ErrorType.DivisionByZero
#DIV/0!
Division by zero!
cell.Formula = “= 10/0”
cell.Formula = “= A1/B2”, the B2 cell is blank
ErrorType.Name
#NAME?
Function does not exist.
cell.Formula = “= FALS” - The function name used in the formula is not spelled correctly.
cell.Formula = “=SUM(A1B2)” - A colon (:) is missing in the cell range reference.
ErrorType.NotAvailable
#N/A
The value is not available to a function or formula.
cell.Formula = “= NA()”
cell.ArrayFormula = “=SUM(A1:A5*A1:B3)” - An array formula‘s arguments are arrays consisting of different numbers of elements.
ErrorType.Null
#NULL!
The specified intersection includes two ranges that do not intersect.
cell.Formula = “=SUM(A1:A5 E6:E8)” - The specified ranges do not intersect, so the sum cannot be calculated.
ErrorType.Number
#NUM!
Invalid numeric values in a formula or function.
cell.Formula = “=SQRT(-16)” - The square root of a negative number cannot be calculated.
ErrorType.Reference
#REF!
Cell reference is not valid.
A formula uses a reference to a cell, and then the column containing this cell is deleted:
cell.Formula = “=5+D2”;
worksheet.Columns[“D”].Delete();
ErrorType.Value
#VALUE!
The value used in the formula is of the wrong data type.
cell.Formula = @“= SUM(“”text””, 6)” - The SUM function requires numeric arguments.
A formula refers to a blank cell that is not actually empty (it contains an empty string):
worksheet[“A1”].Value = “”;
cell.Formula = “= A1 + 5”;
To fix the error in this case, specify an empty value for a cell as shown in the How to: Clear Cells of Content, Formatting, Hyperlinks and Comments example. | 555 | 2,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.701544 |
http://www.ehow.co.uk/list_7284554_problems-using-long-extension-cords.html | 1,537,753,637,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160085.77/warc/CC-MAIN-20180924011731-20180924032131-00194.warc.gz | 316,874,772 | 10,516 | DISCOVER
# Problems using long extension cords
Updated February 21, 2017
All extension cords are not created equal. Although it may seem logical to buy a very long extension cord to use for a variety of purposes in different places, the smart move is to buy various lengths of extension cord, and use the one that is just long enough to get the job accomplished. Using a longer cord than necessary can lead to several problems.
## Lack of Voltage
When it comes to voltage, the gauge of the wire is an important factor in selecting an extension cord for a particular use. The lower the gauge, the thicker the wire and the better it can deliver the proper voltage to the object you are supplying. But gauge is not the only factor to consider. The length of the cord also affects voltage.
As the current from an outlet delivers voltage further from the source through an extension cord, the level of the voltage drops off. Using a very long extension cord means that the object receives a considerably weaker voltage than it would if it were plugged into a short cord or plugged directly into the wall. Power tools are an example of electrical devices that require a lot of voltage to perform the job properly. If the tools are connected to a long extension cord, the level of voltage may make them inadequate.
## Tool Motor Overheating
When a tool does not receive enough voltage to the motor, the motor may overheat. Overheating can lead to damage to the motor. Since a long extension cord gradually loses its ability to supply voltage, this loss can lead to overheating in tools and potentially ruin expensive devices that might be saved by simply using a shorter connection to the power supply.
If the situation makes completing the task impossible without using a long extension cord, use the thickest, lowest gauge extension cord you can find. A 12-gauge cord will carry considerably more power than a 16-gauge cord and could make enough of a difference to save the tool's motor from damage and overheating.
## Daisy Chaining
When a cord of appropriate length can't be found around the house, daisy chain the extension cords. People will take the short cut by connecting two or more extension cords together to form a longer one and may even use a surge protector as part of the chain. Don't do that. If you have daisy chained extension cords or surge protectors in a place of business, understand that this is a violation of OSHA standards as well as inconsistent with the National Electrical Code, according to the United States Office of Compliance. Daisy chaining extension cords can cause overloads, may lead to device failure and can potentially cause fires. | 535 | 2,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-39 | latest | en | 0.942695 |
http://spotidoc.com/doc/154729/molarity-%3D-m--concentration-of-solutions- | 1,537,719,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159561.37/warc/CC-MAIN-20180923153915-20180923174315-00157.warc.gz | 235,286,742 | 14,884 | Molarity = M (Concentration of Solutions)
```Molarity (Concentration of Solutions) = M
M=
Moles of Solute =
Liters of Solution
mol
L
solute = material dissolved into the solvent
In sea water, water is the solvent, and NaCl, MgCl2, etc are the solutes.
In brass, copper is the solvent (90%), and zinc is the solute (10%).
Preparing a Solution - I
Example problem: A solution of sodium phosphate
is prepared by dissolving 3.95 g of sodium
phosphate in water and diluting it to 300.0 mL. What
is the molarity, M, of the salt and each of the ions?
Strategy
(1)Write chemical equation showing process.
(2)Calculate moles of each species.
(3)Divide # moles by # L water to obtain molarity.
(1)Na3PO4 (s)
H2O(solvent)
→
3 Na+ (aq) + PO4-3 (aq)
Dilution of Solutions
Dilute 25.00 mL of 0.0400 M KMnO4 to a final volume
of 500. mL. What is the resulting molarity (M) of the
diluted solution?
Another Strategy for calculating final concentration:
The number of moles of solute is the same before and
after dilution
V1 x M1 = moles solute = V2 x M2
V1M1 = V2 M 2
V1M1
M2 =
V2
25.00mL × 0.0400M
= 0.00200 M
M2 =
500.mL
TYPES of CHEMICAL REACTIONS
in LIQUID SOLUTIONS:
• Precipitation reactions
• Acid – Base reactions
• Reduction – Oxidation (REDOX) reactions
Table 4.1 Simple Rules for Solubility
of Salts in Water
1. Most nitrate (NO3-) salts are soluble.
2. Most salts of Na+, K+, and NH4+ are soluble.
3. Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl2, and Hg2Cl2.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4,
PbSO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important
soluble hydroxides are NaOH, KOH, and
Ca(OH)2 (marginally soluble).
6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-)
salts are only slightly soluble.
Predicting if a precipitate forms, and which?
Pb(NO3)2(aq) + NaCl(aq) →
Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + Cl-(aq)
RULE: If any of the possible new species formed by
combining anions with cations is insoluble,
then that precipitate will form.
USE TABLE 4.1
(I’ll give you this Table for the tests, but not on quiz.)
In this case, PbCl2 is insoluble, and a precipitate forms.
Note: Like the demo with PbI2(s) above.
Writing Equations:
a) Calcium Nitrate and Sodium Sulfate solutions are added together.
Molecular Equation
Ca(NO3)2 (aq) + Na2SO4 (aq)
CaSO4 (s) +2 NaNO3 (aq)
Complete Ionic Equation
Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq)
Net Ionic Equation
Ca2+(aq) + SO4-2(aq)
Spectator Ions are Na+ and NO3-
CaSO4 (s)
Quantitative Precipitation Problems:
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 0.125 M lead nitrate solution to precipitate all of the
The chemical equation for the reaction is:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
The moles of sodium iodide needed to precipitate PbI2 is twice the
lead ions. The number of moles of sodium iodide needed is:
2+
0.125
Mol
Pb
2
mol
I
2.50 L x
x
1.0 L soln.
1 mol Pb2+
The mass of sodium iodide is:
0.625 mol I- x 1 mol NaI x
1 mol I-
= 0.625 mol I-
149.9 g NaI
= 93.68 g NaI
1 mol NaI
Precipitation problem: When aqueous silver nitrate and sodium
chromate solutions are mixed, solid silver chromate forms in a
solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of
silver nitrate is added to 156.00 mL of a 0.0950 M solution of
sodium chromate, what mass of silver chromate (M = 331.8 g/mol)
will be formed?
This is a limiting-reactant problem because the amounts
of two reactants are given.
Strategy:
(1) Write the balanced equation.
(2) Calculate the number of moles of each reactant.
(3) Determine the limiting reactant.
(4) Calculate the moles of product.
(5) Convert moles of product to mass of the product using molar mass.
Selected Acids and Bases
Acids
Bases
Strong: H+(aq) + A-(aq)
Hydrochloric, HCl
Hydrobromic, HBr
Hydroiodoic, HI
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
Strong: M+(aq) + OH-(aq)
Sodium hydroxide, NaOH
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Strontium hydroxide, Sr(OH)2
Barium hydroxide, Ba(OH)2
Weak
Hydrofluoric, HF
Phosphoric acid, H3PO4
Acetic acid, CH3COOH
(or HC2H3O2)
Weak
Ammonia, NH3
accepts proton from water to make
NH4+(aq) and OH-(aq)
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose two protons to yield two Hydronium ions, and one sulfate ion.
What is the molarity of the sulfate and Hydronium ions in a solution
prepared by dissolving 155g of concentrate sulfuric acid into sufficient
water to produce 2.30 Liters of acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles
by the volume to get the molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the acid molarity.
Solution: Two moles of H+ are released for every mole of acid:
H2SO4 (l) + 2 H2O(l)
2 H3O+(aq) + SO4- 2(aq)
Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4
98.09 g H2SO4
-2
1.58
mol
SO
4
Molarity of SO4- 2 =
= 0.687 Molar in SO4- 2
2.30 L solution
Molarity of H+ = 2 x 0.687 M = 1.37 Molar in H+ (or H3O+)
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced chemical reactions (molecular, total ionic, and
net ionic) for the following Chemical reactions:
a) Calcium Hydroxide(aq) and Hydroiodic acid(aq)
b) Lithium Hydroxide(aq) and Nitric acid(aq)
c) Barium Hydroxide(aq) and Sulfuric acid(aq)
Plan: These are all strong acids and bases, therefore they will make
water and the corresponding salts.
Solution:
a)
Ca(OH)2 (aq) + 2HI(aq)
CaI2 (aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq)
2 H2O(l)
Like Example 4.10 (P 113)
What volume of 0.468 M H2SO4 is needed to neutralize 215.00 ml
of a 0.125 M LiOH solution?
Calculate the number of moles of base:
Vbase x Mbase = 0.21500 L x 0.125 M = 0.0268 mol LiOH
From the balance equation find the moles of acid needed:
2 LiOH(aq) + H2SO4 (aq)
2 H2O(l) + Li2SO4 (aq)
Since there are two protons per molecule, we will need half as much
sulfuric acid as we have lithium hydroxide: or 0.0134 mol H2SO4
Volume of acid:
Moles acid
0.0134 moles
= 0.468 Mol
= 0.0286 L H2SO4
Vacid =
Macid
L
EP 91: Aluminum hydroxide reacts with hydrochloric acid
According to the balanced equation
Al(OH)3 (s) + 3 HCl (aq)
→ 3 H2O(l) + AlCl3 (aq)
What volume of 1.50 M HCl(aq) is required to neutralize
10.0 g Al(OH)3(s)?
Strategy:
(1) Calculate moles of Al(OH)3(s).
(2) Calculate moles of HCl needed using balanced equation
(3) Calculate volume HCl from (2) and known molarity
Recognizing Oxidizing and Reducing Agents - I
Problem: Identify the oxidizing and reducing agent in each of the Rx:
a) Zn(s) + 2 HCl(aq)
ZnCl2 (aq) + H2 (g)
b) S8 (s) + 12 O2 (g)
8 SO3 (g)
c) NiO(s) + CO(g)
Ni(s) + CO2 (g)
Plan: First we assign an oxidation number (O.N.) to each atom (or ion)
based on the rules in Table 4.3. The reactant is the reducing agent if it
contains an atom that is oxidized (O.N. increased in the reaction). The
reactant is the oxidizing agent if it contains an atom that is reduced
( O.N. decreased).
Solution:
a) Assigning oxidation numbers:
-1
+1
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2 (aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
Problem: Calculate the mass of metallic Iron that must be
added to 500.0 liters of a solution containing
0.00040M of Pt2+(aq) ions in solution to reclaim all Pt
via: 2 Fe(s) + 3 Pt2+(aq) → 2 Fe3+(aq) + 3 Pt(s)
Solution:
V x M = # moles
500.0L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+
Fe(s) → Fe3+ + 3 ePt2+ + 2 e- → Pt(s)
Need 2 moles of Iron for every 3 moles of Platinum
0.20 mol
Pt2+
2 mol Fe
x
= 0.133 mol Fe
2+
3 mol Pt
0.133 mol Fe x
55.85 g Fe
= 7.4 g Fe
mol Fe
Balancing REDOX Equations:
The oxidation number method
Step 1) Assign oxidation numbers to all elements in the equation.
Step 2) From the changes in oxidation numbers, identify the oxidized
and reduced species.
Step 3) Compute the number of electrons lost in the oxidation and
gained in the reduction from the oxidation number changes.
Draw tie-lines between these atoms to show electron changes.
Step 4) Choose coefficients for these species to make the electrons lost
equal the electrons gained (or total increase in ON = total
decrease in ON)
Step 5) Complete the balancing by inspection.
REDOX Balancing Using Ox. No. Method - II
+2
-1e-
+3
Fe+2(aq) + MnO4-(aq) + H+(aq)
Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e+2
+7
Balance charge: Multiply Fe+2 & Fe+3 by five to correct for
the electrons gained by the Manganese:
5 Fe+2(aq) + MnO4-(aq) + H+(aq)
5 Fe+3(aq) + Mn+2(aq) + H2O(aq)
Balance O: Need 4 H2O on right to cancel O from the MnO4-.
Balance H: Need 8 H+ on the left to balance these water Hs.
5 Fe+2(aq) + MnO4-(aq) +8 H+(aq)
5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq)
BALANCED!
Note: Add 8 water molecules to both sides if you prefer to express the 8 H+ ions instead as H3O+ ions.
Balancing redox eqn using half-cell method in acidic solutions
Cu(s) +HNO3(aq) → Cu2+(aq) + NO(g)
Identify half-reactions, one is ox, other is red
Cu(s) → Cu2+(aq)
0 → +2
HNO3(aq) → NO(g)
+5 → +2
Balance all atoms that are neither H nor O
OK as is
Balance O by adding H2O to side deficient in O
Cu(s) → Cu2+(aq)
HNO3(aq) → NO(g) + 2H2O
Balance H by adding H+ to side deficient in H
Cu(s) → Cu2+(aq)
3H+ +HNO3(aq) → NO(g) + 2H2O
Balance charge by adding e- to side that has + charge
Cu(s) → Cu2+(aq) + 2e3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O
Multiply each eq by factors so electrons cancel out
3x(Cu(s) → Cu2+(aq) + 2e-)
2x(3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O)
3Cu(s)+ 6H+ + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
Balancing redox equations in basic solutions:
First balance in acidic solution, then add OH- to cancel H+
The following redox equation is balanced in acidic solution. Balance it
for a basic solution.
Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3- (aq)
+3H2O(l)
Add OH- equal to number of H+ on both sides
6OH-(aq) +Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3-+3H2O(l)
+6OH-(aq)
Combine OH- and H+ to form water to max extent possible
6H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq)
Cancel H2O on both sides to max extent possible
3H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3- +6OH- (aq)
Redox Titration- Calculation outline - I
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
b)
Molar ratio
Moles of CaC2O4
c)
Chemical Formula
Moles of Ca+2
Problem: Calcium Oxalate was
precipitated from blood by the
addition of Sodium Oxalate so that
calcium ion could be determined. In
the blood sample. The sulfuric acid
solution that the precipitate was
dissolved in required 2.05 ml of
4.88 x 10-4 M KMnO4 to reach the
endpoint.
a) calculate the amount (mol) of
Ca+2.
b) calculate the Ca+2 ion conc.
Plan: a) Calculate the molarity of
Ca+2 in the H2SO4 solution.
b) Convert the Ca+2 concentration
into units of mg Ca+2/ 100 ml blood.
Ch. 5: Pressure =
force per unit area
Density of Mercury = 13.6 g/cm3
760 mm column of 1 cm2 area
weighs 76 cm x 1 cm2 x 13.6 g/cm3
= 1030 g = 1.03 kg = 2.28 lbs
P = force / area = 2.28 pounds / cm2
= 14.7 pounds / in2
= 1.00 atm.
Force = weight = volume x density
= area x height x density
P = Force / area = height x density
IDEAL GAS LAW
• The ideal gas equation combines both Boyles Law and
Charles Law into one easy-to-remember law:
PV=nRT
•
•
•
•
n = number of moles of gas in volume V
R = Ideal gas constant
R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1
Later R = 8.314 J / (mol K) = 8.314 J mol-1 K-1
An ideal gas is one for which both the volume of
molecules and forces between the molecules are so small
that they have insignificant effect on its P-V-T behavior.
Independent of substance, in the limit that n/V →0,
all gases behave ideally. Usually true below 2 atm.
Variations on Ideal Gas Equation
•
•
During chemical and physical processes, any of the four variables in the
ideal gas equation may be fixed.
Thus, PV=nRT can be rearranged for the fixed variables:
– for a fixed amount at constant temperature
• P V = nRT = constant
Boyle’s Law
– for a fixed amount at constant pressure
• V / T = nR / P = constant
Charles’ Law
– for a fixed pressure and temperature
• V = n (RT/P) or V/n = constant Avogadro’s Law
– for a fixed amount at constant volume
• P / T = nR / V = constant
Amonton’s Law
JUST REMEMBER:
PV=nRT and rearrange as needed
Standard Temperature and Pressure (STP)
A set of Standard conditions have been chosen to make it easier to
quantify gas amounts (i,.e., “liters at STP”).
Standard Temperature = 00 C = 273.15 K
Standard Pressure = 1 atmosphere = 760 mm Mercury
At these “standard” conditions, if you have 1.0 mole of any ideal gas,
it will occupy a “standard molar volume”:
Standard Molar Volume = 22.414 Liters = 22.4 L
Gas Law: Solving for Pressure
Problem: Calculate the pressure in a container whose Volume is 87.5 L
and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC.
Plan: Convert all information into the units required, and substitute into
the Ideal Gas equation ( PV=nRT ).
Solution:
5038 g Xe
= 38.37014471 mol Xe
nXe =
131.3 g Xe / mol
T = 18.8 oC + 273.15 K = 291.95 K
PV = nRT
so P = nRT
V
P = (38.37 mol )(0.0821 L atm)(291.95 K) = 10.5108 atm = 10.5 atm
87.5 L (mol K)
Applying the Gas law to T changes
Problem: A copper tank is filled with compressed gas to a pressure of
4.28 atm at a temperature of 0.185 oF. What will be the pressure if the
temperature is raised to 95.6 oC?
Plan: The volume of the tank is not changed, and we only have to deal
with the temperature change, so convert to SI units,
and calculate the pressure ratio from the T ratio.
Solution:
P1 = P2 = nR/V
T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC
T1
T2
T = -17.68 oC + 273.15 K = 255.47 K
1
T2 = 95.6 oC + 273.15 K = 368.8 K
P2 = 4.28 atm x 368.8 K = 6.18 atm
255.47 K
P2 = P1 x T2 = ?
T1
Change of Conditions :Problem -I
• A gas sample in the laboratory has a volume of 45.9 L at
25 oC and a pressure of 743 mm Hg. If the temperature is
increased to 155 oC by pumping (compressing) the gas to a
new volume of 3.10 ml what is the pressure?
•
•
•
•
•
P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm
P2 = ?
V1 = 45.9 L
V2 = 3.10 ml = 0.00310 L
T1 = 25 oC + 273 = 298 K
T2 = 155 oC + 273 = 428 K
Example Problem: Molar Mass of a Gas
from its weight and P,V,T
Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml
flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr,
what is the molecular weight of the gas?
Plan: Use the Ideal gas law to calculate n, then calculate the molar mass.
Solution:
P = 550.0 Torr x 1mm Hg x 1.00 atm
= 0.724 atm
1 Torr
760 mm Hg
V = 250.0 ml x 1.00 L = 0.250 L
1000 ml
n =P V
RT
T = 25.0 oC + 273.15 K = 298.2 K
n = (0.724 atm)(0.250 L)
= 0.007393 mol
(0.0821 L atm/mol K) (298.2 K)
M = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol
Ideal Gas Mixtures
• Ideal gas equation applies to the mixture as a
whole and to each gas species (i) individually:
PtotalV = ntotalRT Ptotal is what is measured by
pressure guages. Σ ni = n1+n2+n3…= ntotal
i
PiV = niRT for all species i. Pi much harder to
measure.
Note: iΣ Pi = P1+P2+P3…= Ptotal
Chemical Equation Calc - III
Mass
Atoms (Molecules)
Number
6.02 x 1023
Reactants
Molarity
moles / liter
Solutions
Molecules
Moles
Molecular
g/mol
Weight
Products
PiV = niRT
Gases
Mole fraction of species i:
xi = ni/ntotal
• PiV = niRT for all i, and
Σ Pi = P1+P2+P3…= Ptotal
• Σ ni = n1+n2+n3…= ntotal
• Σ xi = x1+x2+x3…= 1.00000
• Pi = niRT/V = (ni/ntotal)ntotalRT/V = xi . Ptotal
Gas Law Stoichiometry - I - NH3 + HCl
Problem: A slide separating two containers is removed, and the gases
are allowed to mix and react. The first container with a volume of 2.79 L
contains Ammonia gas at a pressure of 0.776 atm and a temperature of
18.7 oC. The second with a volume of 1.16 L contains HCl gas at a
pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid
ammonium chloride will be formed, and what will be remaining in the
container, and what is the pressure?
Plan: This is a limiting reactant problem, so we must calculate the moles
of each reactant using the gas law to determine the limiting reagent. Then
we can calculate the mass of product, and determine what is left in the
combined volume of the container, and the conditions.
NH3
HCl
Solution:
Equation:
NH3 (g) + HCl(g)
TNH3 = 18.7 oC + 273.15 = 291.9 K
NH4Cl(s)
Velocity and Energy
• Kinetic Energy = KE = ½ mu2 for one molecule
• Average Kinetic Energy of a mole of gas (KEavg)
= 3/2 RT independent of gas identity
• Average Kinetic Energy of one molecule (KEavg)
= 3/2 RT/NA independent of mass, identity
= ½ mu2 (mean value of u2)
u2 = 3RT/(mNA) = 3RT/M
√(u2) = √(3RT/M)
“Root mean square velocity”
Must use R in J/mol K, and remember that 1 J = 1 kg m2/s2
Molecular picture of HEAT!
Diffusion Rates: proportional to
average velocity or to 1/√M
• Rate1/Rate2 = u1/u2 = (M2/M1)1/2
• HCl = 36.46 g/mol
NH3 = 17.03 g/mol
• RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2
• RateNH3 = RateHCl x 1.463
The Equilibrium Constant - Definition
Consider the generalized chemical reaction:
aA+bB = cC+dD
A, B, C and D represent chemical species and a, b, c, and d are
their stoichiometric coefficients in the balanced chemical
equation. At equilibrium,
C ] [ D]
[
K=
a
b
[ A] [ B ]
c
d
Note: The “units” for K are
concentration units raised to
some power = c+d–(a+b)
The square brackets indicate the concentrations of the species in
equilibrium and K is a constant called the equilibrium
constant. K depends only on T , and not on concentrations.
2
[
][
]
CO
SO
2
2
CO2(g) + 2 SO2(g)K =
1
3
[CS2 ][O2 ]
CS2(g) + 3 O2(g)R
• The equilibrium expression for a reaction written in
reverse is the reciprocal of that for the original
expression.
3
[
CS2 ][O 2 ]
CS2(g) + 3 O2(g)K 2 =
[CO 2 ][SO 2 ]2
CO2(g) + 2 SO2(g)R
• If the original reaction is multiplied by a factor n, the new
equilibrium constant is the original raised to the power n.
2 CS2(g) + 6 O2(g) = 2 CO2(g) + 4 SO2(g)
CO 2 ] [SO2 ]
[
K3 =
2
6
[CS2 ] [O2 ]
2
4
= ( K1 )
2
1
=
K1
Example: Calculation of the Equilibrium
Constant from equilibrium amounts
At 454 K, the following reaction takes place:
3 Al2Cl6(g)
R 2 Al3Cl9(g)
At this temperature, the equilibrium concentration of Al2Cl6(g) is
1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x
10-2 M. Compute the equilibrium constant at 454 K.
Strategy: Substitute values into K
[
Al 3Cl9 ] (1.02 × 10 M )
K=
=
3
3
(1.00 M )
[Al2Cl6 ]
2
−2
2
−4
= 1.04 × 10 M
−1
Determining Equilibrium Concentrations from K
Example: Methane can be made by reacting carbon disulfide with
hydrogen gas. K for this reaction is 27.8 (L/mol)2 at 900°C.
CS2 (g) + 4 H2 (g)
CH4 (g) + 2 H2 S(g)
At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol
CS2, 1.10 mol of H2, and 0.45 mol of H2S. How much methane was
formed?
Strategy:
(1)Calculate the equilibrium concentrations from the moles given
and the volume of the container.
(2)Use the value of K to solve for the concentration of methane.
(3) Calculate the number of moles of methane from M and V.
Equilibrium Expressions Involving
Pressures
For a reaction of the type
aA+bB = cC+dD
It is sometimes convenient to write the equilibrium
expression in terms of partial pressures, e.g.
PC ) ( PD )
(
KP =
a
b
( PA ) ( PB )
c
d
= K ( RT )
P indicates the partial pressures of the species in equilibrium
and KP is a constant called the equilibrium constant in
terms of partial pressures. KP depends only on T , and
not on pressure.
Δn
Equilibrium involving pure solids or liquids:
set their activity = 1
Example: NH4NO2(s) = N2(g) + 2 H2O(g)
The equilibrium constant for this reaction would normally
be expressed as:
2
N 2 ][ H 2 O ]
[
K'=
[ NH 4 NO2 ]
However, a pure solid or liquid retains the same activity
during the reaction. Thus we set the activity of NH 4 NO 2 (s)
to one.
K = [ N 2 ][ H 2 O ]
2
1.0000
and
( )( P )
K p = PN2
H2O
1.0000
2
The Reaction Quotient, Q
Consider the reaction: a A(g ) + b B(g ) = c C( g ) + dm D( g )
The reaction quotient, Q is defined as
C ]t [ D ]t
[
Q=
a
b
[ A]t [ B ]t
c
d
where the subscripts t indicate momentary concentrations
at some time t before (or after) equilibrium has established.
Q has same form as K, but the concentrations are
the actual rather than the equilibrium
concentrations
Example: Calculating Equilibrium
Pressures and Concentrations from K
and initial conditions
Consider the equilibrium: CO(g) + H2O(g) = CO2(g) + H2(g)
0.250 mol CO and 0.250 mol H2O are placed in a 125 mL flask
at 900 K. What is the composition of the equilibrium mixture if
K = 1.56?
The original reactant concentrations are:
[CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M
Q = 0. Therefore, Q < K, so reactants are consumed and
CO(g) + H2O(g) = CO2(g) + H2(g)
Conc.
(M)
CO(g)
H2O(g)
CO2(g)
H2(g)
Init.
2.00
2.00
0
0
Change -x
-x
+x
+x
Equil.
2.00 - x
x
x
2.00 - x
``` | 7,338 | 20,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-39 | latest | en | 0.851994 |
https://www.physicsforums.com/threads/another-force-question.745099/ | 1,508,834,240,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00691.warc.gz | 963,098,331 | 16,261 | # Another force question?
1. Mar 24, 2014
### santoki
1. The problem statement, all variables and given/known data
A person has a choice while trying to move a crate across a horizontal pad of concrete: push it at a downward angle of 30 degrees, or pull it at an upward angle of 30 degrees.
If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.
2. The attempt at a solution
I don't know how to start calculating it by I do have an idea of what the answer will be. In the end, pulling at an upward angle will have a lower force because pulling upward will decrease friction resulting in decrease of normal force.
As for the calculation part, I was thinking of using Fpcos(30°)p = $\mu$N with N being (mg + Fpsin(30°)) for pushing and then Fpcos(30°) = $\mu$N with N being (mg - Fpsin(30°)) for pulling but I feel like it's a bit disorganized since it doesn't follow what we usually have in class like a parent formula on top and you can manipulate it to get this and that, etc.
2. Mar 25, 2014
### Andrew Mason
Start by doing a free body diagram for each of the two scenarios, drawing vectors for all forces acting on the body in each scenario. What is the condition for motion at constant velocity (hint: what must the vectors in each diagram sum to?)?
AM | 351 | 1,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-43 | longest | en | 0.94027 |
http://www.nag.com/numeric/CL/nagdoc_cl23/examples/source/d03pece.c | 1,386,316,276,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-48/segments/1386163050081/warc/CC-MAIN-20131204131730-00015-ip-10-33-133-15.ec2.internal.warc.gz | 454,582,982 | 2,458 | /* nag_pde_parab_1d_keller (d03pec) Example Program. * * Copyright 2001 Numerical Algorithms Group. * * Mark 7, 2001. */ #include #include #include #include #include #include #ifdef __cplusplus extern "C" { #endif static void NAG_CALL pdedef(Integer, double, double, const double[], const double[], const double[], double[], Integer *, Nag_Comm *); static void NAG_CALL bndary(Integer, double, Integer, Integer, const double[], const double[], double[], Integer *, Nag_Comm *); static void NAG_CALL exact(double, Integer, Integer, double *, double *); static void NAG_CALL uinit(Integer, Integer, double *, double *); #ifdef __cplusplus } #endif #define U(I, J) u[npde*((J) -1)+(I) -1] #define EU(I, J) eu[npde*((J) -1)+(I) -1] int main(void) { const Integer npde = 2, npts = 41, nleft = 1, neqn = npde*npts; const Integer lisave = neqn+24, nwkres = npde*(npts+21+3*npde)+7*npts+4; const Integer lrsave = 11*neqn+(4*npde+nleft+2)*neqn+50+nwkres; Integer exit_status = 0, i, ind, it, itask, itrace; double acc, tout, ts; double *eu = 0, *rsave = 0, *u = 0, *x = 0; Integer *isave = 0; NagError fail; Nag_Comm comm; Nag_D03_Save saved; INIT_FAIL(fail); printf( "nag_pde_parab_1d_keller (d03pec) Example Program Results\n\n"); /* Allocate memory */ if (!(eu = NAG_ALLOC(npde*npts, double)) || !(rsave = NAG_ALLOC(lrsave, double)) || !(u = NAG_ALLOC(npde*npts, double)) || !(x = NAG_ALLOC(npts, double)) || !(isave = NAG_ALLOC(lisave, Integer))) { printf("Allocation failure\n"); exit_status = 1; goto END; } itrace = 0; acc = 1e-6; printf(" Accuracy requirement =%12.3e", acc); printf(" Number of points = %3ld\n\n", npts); /* Set spatial-mesh points */ for (i = 0; i < npts; ++i) x[i] = i/(npts-1.0); printf(" x "); printf("%10.4f%10.4f%10.4f%10.4f%10.4f\n\n", x[4], x[12], x[20], x[28], x[36]); ind = 0; itask = 1; uinit(npde, npts, x, u); /* Loop over output value of t */ ts = 0.0; tout = 0.0; for (it = 0; it < 5; ++it) { tout = 0.2*(it+1); /* nag_pde_parab_1d_keller (d03pec). * General system of first-order PDEs, method of lines, * Keller box discretisation, one space variable */ nag_pde_parab_1d_keller(npde, &ts, tout, pdedef, bndary, u, npts, x, nleft, acc, rsave, lrsave, isave, lisave, itask, itrace, 0, &ind, &comm, &saved, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_pde_parab_1d_keller (d03pec).\n%s\n", fail.message); exit_status = 1; goto END; } /* Check against the exact solution */ exact(tout, npde, npts, x, eu); printf(" t = %5.2f\n", ts); printf(" Approx u1"); printf("%10.4f%10.4f%10.4f%10.4f%10.4f\n", U(1, 5), U(1, 13), U(1, 21), U(1, 29), U(1, 37)); printf(" Exact u1"); printf("%10.4f%10.4f%10.4f%10.4f%10.4f\n", EU(1, 5), EU(1, 13), EU(1, 21), EU(1, 29), EU(1, 37)); printf(" Approx u2"); printf("%10.4f%10.4f%10.4f%10.4f%10.4f\n", U(2, 5), U(2, 13), U(2, 21), U(2, 29), U(2, 37)); printf(" Exact u2"); printf("%10.4f%10.4f%10.4f%10.4f%10.4f\n\n", EU(2, 5), EU(2, 13), EU(2, 21), EU(2, 29), EU(2, 37)); } printf(" Number of integration steps in time = %6ld\n", isave[0]); printf(" Number of function evaluations = %6ld\n", isave[1]); printf(" Number of Jacobian evaluations =%6ld\n", isave[2]); printf(" Number of iterations = %6ld\n\n", isave[4]); END: if (eu) NAG_FREE(eu); if (rsave) NAG_FREE(rsave); if (u) NAG_FREE(u); if (x) NAG_FREE(x); if (isave) NAG_FREE(isave); return exit_status; } static void NAG_CALL pdedef(Integer npde, double t, double x, const double u[], const double udot[], const double dudx[], double res[], Integer *ires, Nag_Comm *comm) { if (*ires == -1) { res[0] = udot[0]; res[1] = udot[1]; } else { res[0] = udot[0] + dudx[0] + dudx[1]; res[1] = udot[1] + 4.0*dudx[0] + dudx[1]; } return; } static void NAG_CALL bndary(Integer npde, double t, Integer ibnd, Integer nobc, const double u[], const double udot[], double res[], Integer *ires, Nag_Comm *comm) { if (ibnd == 0) { if (*ires == -1) { res[0] = 0.0; } else { res[0] = u[0] - 0.5*(exp(t) + exp(-3.0*t)) - 0.25*(sin(-3.0*t) - sin(t)); } } else { if (*ires == -1) { res[0] = 0.0; } else { res[0] = u[1] - exp(1.0 - 3.0*t) + exp(t + 1.0) - 0.5*(sin(1.0 - 3.0*t) + sin( t + 1.0)); } } return; } static void NAG_CALL uinit(Integer npde, Integer npts, double *x, double *u) { /* Routine for PDE initial values */ Integer i; for (i = 1; i <= npts; ++i) { U(1, i) = exp(x[i-1]); U(2, i) = sin(x[i-1]); } return; } static void NAG_CALL exact(double t, Integer npde, Integer npts, double *x, double *u) { /* Exact solution (for comparison purposes) */ Integer i; for (i = 1; i <= npts; ++i) { U(1, i) = 0.5*(exp(x[i-1] + t) + exp(x[i-1] - 3.0*t)) + 0.25*(sin(x[i-1] - 3.0*t) - sin(x[i-1] + t)); U(2, i) = exp(x[i-1] - 3.0*t) - exp(x[i-1] + t) + 0.5*(sin(x[i-1] - 3.0*t) + sin(x[i-1] + t)); } return; } | 1,893 | 4,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2013-48 | longest | en | 0.389017 |
http://paleyontology.com/foop/magic_square.html | 1,516,466,108,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889677.76/warc/CC-MAIN-20180120162254-20180120182254-00212.warc.gz | 252,845,921 | 1,628 | A magic square is an nxn array of squares which has the following properties:
1. Each integer from 1 to n2 appears in the array exactly once; and
2. The sum of each row is equal to the sum of each column is equal to the sum of each diagonal.
Here is an example of a 3x3 magic square:
```2 7 6
9 5 1
4 3 8
```
If you add up any row, colum, or diagonal, the sum of the numbers will be 15.
Magic squares come in all sizes--they are not all 3x3.
Write the predicate isMagicSquare(int[][] square) that takes an nxn array of ints and returns #t if the array contains a magic square and #f otherwise. You may assume that the input to isMagicSquare has the same number of rows as it does columns.
isMagicSquare needs to work on ALL squares, not just the example or 3x3 squares. If you write isMagicSquare so that it works only on the sample data, that is not The Idea.
Here is some code that you might find useful:
```public class Main {
public static void main(String[] args) {
int[][] test = { {2, 7, 6}, // Nothing wrong with this sample 3x3 array,
{9, 5, 1}, // but isMagicSquare needs to work for ANY
{4, 3, 8} }; // nxn array
print2DArray(test);
System.out.println(isMagicSquare(test));
}
public static boolean isMagicSquare(int[][] arr) {
int sum = 0;
for (int column = 0; column < arr.length; column++) {
sum += arr[0][column]; // Add up elements of row 0
}
return (rowsOK(arr, sum) && colsOK(arr, sum) && diagsOK(arr, sum) && correctElements(arr));
}
public static boolean rowsOK(int[][] arr, int sum) {
// Replace the below stub with real code
return false;
}
public static boolean colsOK(int[][] arr, int sum) {
// Replace the below stub with real code
return false;
}
public static boolean diagsOK(int[][] arr, int sum) {
// Replace the below stub with real code
return false;
}
public static boolean correctElements(int[][] arr) {
// Verify that the numbers from 1 to n^2 are all in the array
return false;
}
// Print utility for a 2-dimensional array of ints
public static void print2DArray(int[][] arr) {
for (int row = 0; row < arr.length; row++) {
for (int col = 0; col < arr[0].length; col++) {
System.out.print(arr[row][col] + " ");
}
System.out.println();
}
}
}
``` | 604 | 2,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-05 | latest | en | 0.626491 |
https://universalium.enacademic.com/213802/trig_function | 1,601,078,808,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00406.warc.gz | 687,929,760 | 10,265 |
# trig function
* * *
Universalium. 2010.
### Look at other dictionaries:
• trig function — Math. See trigonometric function … Useful english dictionary
• trigonometric function — Math. 1. Also called circular function. a function of an angle, as sine or cosine, expressed as the ratio of the sides of a right triangle. 2. any function involving only trigonometric functions and constants. 3. the generalization of these to… … Universalium
• trigonometric function — trig′onomet′ric func′tion n. math. a function of an angle, as the sine or cosine, expressed as the ratio of the sides of a right triangle Also called circular function Etymology: 1905–10 … From formal English to slang
• trigonometrical function — /trɪgənəˌmɛtrɪkəl ˈfʌŋkʃən/ (say triguhnuh.metrikuhl fungkshuhn) noun a function relating two sides of a right angled triangle with one of the acute angles in the triangle, as tangent, sine, cosine, cotangent, secant, cosecant, or any function… … Australian English dictionary
• Trigonometrical function — Trigonometric Trig o*no*met ric, Trigonometrical Trig o*no*met ric*al, [Cf. F. trigonom[ e]trique.] Of or pertaining to trigonometry; performed by the rules of trigonometry. [1913 Webster] {Trig o*no*met ric*al*ly}, adv. [1913 Webster]… … The Collaborative International Dictionary of English
• Root mean square — In mathematics, the root mean square (abbreviated RMS or rms), also known as the quadratic mean, is a statistical measure of the magnitude of a varying quantity. It is especially useful when variates are positive and negative, e.g., sinusoids.It… … Wikipedia
• GRASS (programming language) — GRASS ( GRAphics Symbiosis System ) was a programming language created to script 2D vector graphics animations. GRASS was similar to BASIC in syntax, but added numerous instructions for specifying 2D object animation, including scaling,… … Wikipedia
• Trigonometry — Trig redirects here. For other uses, see Trig (disambiguation). The Canadarm2 robotic manipulator on the International Space Station is operated by controlling the angles of its joints. Calculating the final position of the astronaut at the end… … Wikipedia
• JASS — JASS, J Asynchronous Scripting SyntaxFact|date=September 2008 , is an event driven scripting language used in Blizzard Entertainment s Warcraft III game. Map creators can use it in the World Editor to create triggers and AI scripts. Features The… … Wikipedia
• List of trigonometric identities — Cosines and sines around the unit circle … Wikipedia | 637 | 2,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.734335 |
https://sdrta.net/x-3-64-factored/ | 1,653,205,325,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00451.warc.gz | 574,832,416 | 5,190 | Rewrite x^3-64 as x^3-4^3. The difference of cubes have the right to be factored making use of the rule: a^3-b^3=left(a-b ight)left(a^2+ab+b^2 ight). Polynomial x^2+4x+16 is not factored due to the fact that it does not have any kind of rational roots.
You are watching: X^3-64 factored
x3-64 Final result : (x - 4) • (x2 + 4x + 16) action by action solution : step 1 :Trying to factor as a difference of Cubes: 1.1 Factoring: x3-64 concept : A difference of 2 perfect cubes, ...
8x3-64 Final result : 8 • (x - 2) • (x2 + 2x + 4) action by action solution : step 1 :Equation at the finish of action 1 : 23x3 - 64 step 2 : action 3 :Pulling out choose terms : 3.1 pull out prefer ...
https://www.quora.com/How-does-6x-3-6-factor-differently-into-irreducibles-in-mathbb-Z-x-compared-to-mathbb-Q-x
take into consideration factoring the element 15 in mathbbZ. If we urge that a "factorization" is a product of irreducible elements, then our administrate would need to be either -15 = (-3) cdot 5 ...
https://math.stackexchange.com/questions/917723/determine-the-irrational-numbers-x-such-that-both-x22x-and-x3-6x-are-ra
mean q=x^2+2x is rational. Then, by the quadratic formula, x=frac-2 pm sqrt4+4q2=-1 pm sqrtalpha, where alpha=q+1 deserve to be any rational number the is not a perfect square (since ...
0=x3-64 Three services were uncovered : x =(-4-√-48)/2=-2-2i√ 3 = -2.0000-3.4641i x =(-4+√-48)/2=-2+2i√ 3 = -2.0000+3.4641i x = 4 Rearrange: Rearrange the equation by individually what is come ...
27x3-64 Final an outcome : (3x - 4) • (9x2 + 12x + 16) action by step solution : step 1 :Equation in ~ the finish of action 1 : 33x3 - 64 action 2 :Trying to variable as a difference of Cubes: 2.1 ...
See more: Where Is The Correct Place To Insert A Javascript? ? Code Example
More Items
Rewrite x^3-64 together x^3-4^3. The distinction of cubes have the right to be factored using the rule: a^3-b^3=left(a-b ight)left(a^2+ab+b^2 ight). Polynomial x^2+4x+16 is not factored due to the fact that it does no have any rational roots.
EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు | 888 | 2,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.81594 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-4-graphs-of-the-circular-functions-section-4-5-harmonic-motion-4-5-exercises-page-185/14 | 1,713,494,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00475.warc.gz | 702,726,803 | 12,065 | ## Trigonometry (11th Edition) Clone
$L=0.81$ ft
To find the length of the pendulum, we need to substitute the value of period $P=1$ sec in the formula and solve: $P=2\pi\sqrt {\frac{L}{32}}$ $1=2\pi\sqrt {\frac{L}{32}}$ $\frac{1}{2\pi}=\sqrt {\frac{L}{32}}$ $\frac{1}{4\pi^{2}}= {\frac{L}{32}}$ $\frac{32}{4\pi^{2}}= L$ $L=0.81$ ft | 142 | 333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-18 | latest | en | 0.598609 |
https://dsprelated.com/code.php?submittedby=24789 | 1,723,228,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00057.warc.gz | 156,961,580 | 64,763 | ## Delay estimation revisited
June 9, 20125 comments Coded in Matlab
``````% ****************************************************************
% find sub-sample delay and scaling factor between two cyclic signals
% to maximize crosscorrelation
% Markus Nentwig, 120609_v1.1
% ****************************************************************
function iterDelayEstDemo();
close all;
n = 1024;
% n = 1024 * 256; disp('*** test: long signal enabled ***');
% ****************************************************************
% random signal
% ****************************************************************
fd = randn(1, n) + 1i * randn(1, n);
% ****************************************************************
% lowpass filter
% ****************************************************************
f = binFreq(n);
fd(abs(f) > 0.045) = 0;
s1 = real(ifft(fd)) * sqrt(n);
% ****************************************************************
% create delayed 2nd signal
% ****************************************************************
dTest_samples = 12.3456;
cTest = 1.23456;
% cTest = cTest + 1i; disp('*** test: complex coeff enabled ***');
% cTest = -cTest; disp('*** test: negative coeff enabled ***');
s2 = cTest * cs_delay(s1, 1, dTest_samples);
% s2 = s2 + 0.5*randn(size(s2)); disp('*** test: noise enabled ***');
% ****************************************************************
% estimate delay
% ****************************************************************
[delay_samples, coeff] = iterDelayEst(s1, s2);
% ****************************************************************
% correct it
% ****************************************************************
s2a = cs_delay(s1, 1, delay_samples);
s2b = s2a * coeff;
figure(); hold on;
h = plot(real(s1), 'k'); set(h, 'lineWidth', 3);
h = plot(real(s2), 'b'); set(h, 'lineWidth', 3);
h = plot(real(s2a), 'r'); set(h, 'lineWidth', 1);
h = plot(real(s2b), 'm'); set(h, 'lineWidth', 1);
xlim([1, numel(s1)]);
xlabel('samples');
legend('s1', 's2', 's2 un-delayed', 's2 un-delayed and scaled');
title('test signals');
format long;
disp('nominal delay of s2 relative to s1')';
dTest_samples
disp('iterDelayEst() returned:');
delay_samples
disp('original scaling factor:');
cTest
disp('estimated scaling factor:');
coeff
end
% ****************************************************************
% estimates delay and scaling factor
% ****************************************************************
function [delay_samples, coeff] = iterDelayEst(s1, s2)
s1 = s1(:) .'; % force row vectors
s2 = s2(:) .';
rflag = isreal(s1) && isreal(s2);
n = numel(s1);
halfN = floor(n/2);
assert(numel(s2) == n, 'signals must have same length');
% ****************************************************************
% constants
% ****************************************************************
% exit if uncertainty below threshold
thr_samples = 1e-7;
% exit after fixed number of iterations
nIter = 25;
% frequency domain representation of signals
fd1 = fft(s1);
fd2 = fft(s2);
% first round: No delay was applied
tau = [];
fd2Tau = fd2; % delayed s2 in freq. domain
% frequency corresponding to each FFT bin -0.5..0.5
f = binFreq(n);
% uncertainty plot data
e = [];
% normalization factor
nf = sqrt((fd1 * fd1') * (fd2 * fd2')) / n; % normalizes to 1
% search window:
% known maximum and two surrounding points
x1 = -1;
x2 = -1;
x3 = -1;
y1 = -1;
y2 = -1;
y3 = -1;
% ****************************************************************
% iteration loop
% ****************************************************************
for count = 1:nIter
% ****************************************************************
% crosscorrelation with time-shifted signal
% ****************************************************************
xcorr = abs(ifft(fd2Tau .* conj(fd1)))/ nf;
% ****************************************************************
% detect peak
% ****************************************************************
if isempty(tau)
% ****************************************************************
% startup
% initialize with three adjacent bins around peak
% ****************************************************************
ix = find(xcorr == max(xcorr));
ix = ix(1); % use any, if multiple bitwise identical peaks
% indices of three bins around peak
ixLow = mod(ix-1-1, n) + 1; % one below
ixMid = ix;
ixHigh = mod(ix-1+1, n) + 1; % one above
% delay corresponding to the three bins
tauLow = mod(ixLow -1 + halfN, n) - halfN;
tauMid = mod(ixMid -1 + halfN, n) - halfN;
tauHigh = mod(ixHigh -1 + halfN, n) - halfN;
% crosscorrelation value for the three bins
xcLow = xcorr(ixLow);
xcMid = xcorr(ixMid);
xcHigh = xcorr(ixHigh);
x1 = tauLow;
x2 = tauMid;
x3 = tauHigh;
y1 = xcLow;
y2 = xcMid;
y3 = xcHigh;
else
% ****************************************************************
% only main peak at first bin is of interest
% ****************************************************************
tauMid = tau;
xcMid = xcorr(1);
if xcMid > y2
% ****************************************************************
% improve midpoint
% ****************************************************************
if tauMid > x2
% midpoint becomes lower point
x1 = x2;
y1 = y2;
else
% midpoint becomes upper point
x3 = x2;
y3 = y2;
end
x2 = tauMid;
y2 = xcMid;
elseif tauMid < x2
% ****************************************************************
% improve low point
% ****************************************************************
assert(tauMid >= x1); % bitwise identical is OK
assert(tauMid > x1 || xcMid > y1); % expect improvement
x1 = tauMid;
y1 = xcMid;
elseif tauMid > x2
% ****************************************************************
% improve high point
% ****************************************************************
assert(tauMid <= x3); % bitwise identical is OK
assert((tauMid < x3) || (xcMid > y3)); % expect improvement
x3 = tauMid;
y3 = xcMid;
else
assert(false, '?? evaluated for existing tau ??');
end
end
% ****************************************************************
% calculate uncertainty (window width)
% ****************************************************************
eIter = abs(x3 - x1);
e = [e, eIter];
if eIter < thr_samples
% disp('threshold reached, exiting');
break;
end
if y1 == y2 || y2 == y3
% reached limit of numerical accuracy on one side
usePoly = 0;
else
% ****************************************************************
% fit 2nd order polynomial and find maximum
% ****************************************************************
num = (x2^2-x1^2)*y3+(x1^2-x3^2)*y2+(x3^2-x2^2)*y1;
denom = (2*x2-2*x1)*y3+(2*x1-2*x3)*y2+(2*x3-2*x2)*y1;
if denom ~= 0
tau = num / denom;
% is the point within [x1, x3]?
usePoly = ((tau > x1) && (tau < x3));
else
usePoly = 0;
end
end
if ~usePoly
% revert to linear interpolation on the side with the
% less-accurate outer sample
% Note: There is no guarantee that the side with the more accurate
% outer sample is the right one, as the samples aren't
% placed on a regular grid!
% Therefore, iterate to improve the "worse" side, which will
% eventually become the "better side", and iteration converges.
tauLow = (x1 + x2) / 2;
tauHigh = (x2 + x3) / 2;
if y1 < y3
o = [tauLow, tauHigh];
else
o = [tauHigh, tauLow];
end
% don't choose point that is identical to one that is already known
tau = o(1);
if tau == x1 || tau == x2 || tau == x3
tau = o(2);
if tau == x1 || tau == x2 || tau == x3
break;
end
end
end
% ****************************************************************
% advance 2nd signal according to location of maximum
% phase shift in frequency domain - delay in time domain
% ****************************************************************
fd2Tau = fd2 .* exp(2i * pi * f * tau);
end % for
% ****************************************************************
% plot the uncertainty (window size) over the number of iterations
% ****************************************************************
if true
figure(); semilogy(e, '+-'); grid on;
xlabel('iteration');
title('uncertainty in delay');
end
% ****************************************************************
% the delay estimate is the final location of the delay that
% maximized crosscorrelation (center of window).
% ****************************************************************
delay_samples = x2;
% ****************************************************************
% Coefficient: Turn signal 1 into signal 2
% ****************************************************************
coeff = fd2Tau * fd1' ./ (fd1 * fd1');
% ****************************************************************
% chop roundoff error, if input signals are known to be
% real-valued.
% ****************************************************************
if rflag
coeff = real(coeff);
end
end
% ****************************************************************
% frequency corresponding to FFT bin
% ****************************************************************
function f = binFreq(n)
f = (mod(((0:n-1)+floor(n/2)), n)-floor(n/2))/n;
end
% ****************************************************************
% delay by phase shift
% needed only for demo code
% ****************************************************************
function waveform = cs_delay(waveform, rate_Hz, delay_s)
rflag = isreal(waveform);
n = numel(waveform);
cycLen_s = n / rate_Hz;
nCyc = delay_s / cycLen_s();
f = 0:(n - 1);
f = f + floor(n / 2);
f = mod(f, n);
f = f - floor(n / 2);
phase = -2 * pi * f * nCyc;
rot = exp(1i*phase);
waveform = ifft(fft(waveform) .* rot);
if rflag
waveform = real(waveform);
end
end``````
## Narrow-band moving-average decimator, one addition/sample
December 31, 2011 Coded in Matlab
``````% *************************************************
% Moving average decimator
%
% A decimator for narrow-band signals (~5 % or less bandwidth occupation)
% can be implemented as follows:
%
% #define DECIM (100)
% double acc = 0.0;
% while (1){
% int ix;
% for(ix = DECIM; ix > 0; --ix){
% acc += getInputSample();
% } /* for */
% writeOutputSample(acc / (double)DECIM);
% acc = 0.0;
% } /* while */
%
% It is conceptually identical to a moving average filter
% http://www.dspguide.com/ch15/4.htm combined with a decimator
%
% Note that the "moving" average jumps ahead in steps of the decimation
% factor. Intermediate output is decimated away, allowing for a very efficient
% implementation.
% This program calculates the frequency response and alias response,
% based on the decimation factor and bandwidth of the processed signal.
% *************************************************
function eval_design()
decimationFactor = 100;
rateIn_Hz = 48000;
noDecim = false;
% create illustration with sinc response
%decimationFactor = 4; noDecim = true;
% *************************************************
% signal source: Bandlimited test pulse
% Does not contain energy in frequency ranges that
% cause aliasing
% *************************************************
s = zeros(1, 10000 * decimationFactor);
fb = FFT_frequencyBasis(numel(s), rateIn_Hz);
% assign energy to frequency bins
if noDecim
sPass = ones(size(s));
else
sPass = s; sPass(find(abs(fb) < rateIn_Hz / decimationFactor / 2)) = 1;
end
sAlias = s; sAlias(find(abs(fb) >= rateIn_Hz / decimationFactor / 2)) = 1;
% convert to time domain
sPass = fftshift(real(ifft(sPass)));
sAlias = fftshift(real(ifft(sAlias)));
% *************************************************
% plot spectrum at input
% *************************************************
pPass = {};
pPass = addPlot(pPass, sPass, rateIn_Hz, 'k', 5, ...
'input (passband response)');
pAlias = {};
pAlias = addPlot(pAlias, sAlias, rateIn_Hz, 'k', 5, ...
'input (alias response)');
% *************************************************
% impulse response
% *************************************************
h = zeros(size(s));
h (1:decimationFactor) = 1;
if noDecim
h = h / decimationFactor;
decimationFactor = 1;
end
% cyclic convolution between signal and impulse response
sPass = real(ifft(fft(sPass) .* fft(h)));
sAlias = real(ifft(fft(sAlias) .* fft(h)));
% decimation
sPass = sPass(decimationFactor:decimationFactor:end);
sAlias = sAlias(decimationFactor:decimationFactor:end);
rateOut_Hz = rateIn_Hz / decimationFactor;
% *************************************************
% plot spectrum
% *************************************************
pPass = addPlot(pPass, sPass, rateOut_Hz, 'b', 3, ...
'decimated (passband response)');
figure(1); clf; grid on; hold on;
doplot(pPass, sprintf('passband frequency response over input rate, decim=%i', decimationFactor));
pAlias = addPlot(pAlias, sAlias, rateOut_Hz, 'b', 3, ...
'decimated (alias response)');
figure(2); clf; grid on; hold on;
doplot(pAlias, sprintf('alias frequency response over input rate, decim=%i', decimationFactor));
% *************************************************
% plot passband ripple
% *************************************************
fb = FFT_frequencyBasis(numel(sPass), 1);
fr = 20*log10(abs(fft(sPass) + eps));
ix = find(fb > 0);
figure(3); clf;
h = semilogx(fb(ix), fr(ix), 'k');
set(h, 'lineWidth', 3);
ylim([-3, 0]);
title(sprintf('passband gain over output rate, decim=%i', decimationFactor));
xlabel('frequency relative to output rate');
ylabel('dB'); grid on;
% *************************************************
% plot alias response
% *************************************************
fb = FFT_frequencyBasis(numel(sAlias), 1);
fr = 20*log10(abs(fft(sAlias) + eps));
ix = find(fb > 0);
figure(4); clf;
h = semilogx(fb(ix), fr(ix), 'k');
set(h, 'lineWidth', 3);
% ylim([-80, -20]);
title(sprintf('alias response over output rate, decim=%i', decimationFactor));
xlabel('frequency relative to output rate');
ylabel('dB'); grid on;
end
% ************************************
% put frequency response plot data into p
% ************************************
function p = addPlot(p, s, rate, plotstyle, linewidth, legtext)
p{end+1} = struct('sig', s, 'rate', rate, 'plotstyle', plotstyle, 'linewidth', linewidth, 'legtext', legtext);
end
% ************************************
% helper function, plot data in p
% ************************************
function doplot(p, t)
leg = {};
for ix = 1:numel(p)
pp = p{ix};
fb = FFT_frequencyBasis(numel(pp.sig), pp.rate);
fr = 20*log10(abs(fft(pp.sig) + eps));
h = plot(fftshift(fb), fftshift(fr), pp.plotstyle);
set(h, 'lineWidth', pp.linewidth);
xlabel('f / Hz');
ylabel('dB');
leg{end+1} = pp.legtext;
end
legend(leg);
title(t);
end
% ************************************
% calculates the frequency that corresponds to
% each FFT bin (negative, zero, positive)
% ************************************
function fb_Hz = FFT_frequencyBasis(n, rate_Hz)
fb = 0:(n - 1);
fb = fb + floor(n / 2);
fb = mod(fb, n);
fb = fb - floor(n / 2);
fb = fb / n; % now [0..0.5[, [-0.5..0[
fb_Hz = fb * rate_Hz;
end``````
## Alias/error simulation of interpolating RRC filter
December 25, 2011 Coded in Matlab
``````% *************************************************
% alias- and in-channel error analysis for root-raised
% cosine filter with upsampler FIR cascade
% Markus Nentwig, 25.12.2011
%
% * plots the aliases at the output of each stage
% * plots the error spectrum (deviation from ideal RRC-
% response)
% *************************************************
function eval_RRC_resampler()
1;
% variant 1
% conventional RRC filter and FIR resampler
smode = 'evalConventional';
% export resampler frequency response to design equalizing RRC filter
%smode = 'evalIdeal';
% variant 2
% equalizing RRC filter and FIR resampler
% smode = 'evalEqualized';
% *************************************************
% *************************************************
switch smode
case 'evalConventional'
% conventionally designed RRC filter
case 'evalIdeal'
h0 = 1;
case 'evalEqualized'
% alternative RRC design that equalizes the known frequency
% response of the resampler
otherwise assert(false);
end
% *************************************************
% --- signal source ---
% *************************************************
n = 10000; % test signal, number of symbol lengths
rate = 1;
s = zeros(1, n);
s(1) = 1;
p = {};
p = addPlot(p, s, rate, 'k', 5, 'sym stream r=1');
% *************************************************
% --- upsampling RRC filter ---
% *************************************************
rate = rate * 2;
s = upsample(s, 2); % insert one zero after every sample
p = addPlot(p, s, rate, 'k', 2, 'sym stream r=2');
s = filter(h0, [1], s);
p = addPlot(p, s, rate, 'b', 3, 'sym stream r=2, filtered');
p = addErrPlot(p, s, rate, 'error');
figure(1); clf; grid on; hold on;
doplot(p, 'interpolating pulse shaping filter');
ylim([-70, 2]);
p = {};
% *************************************************
% --- first interpolator by 2 ---
% *************************************************
rate = rate * 2;
s = upsample(s, 2); % insert one zero after every sample
p = addPlot(p, s, rate, 'k', 3, 'interpolator 1 input');
s = filter(h1, [1], s);
p = addPlot(p, s, rate, 'b', 3, 'interpolator 1 output');
p = addErrPlot(p, s, rate, 'error');
figure(2); clf; grid on; hold on;
doplot(p, 'first interpolator by 2');
ylim([-70, 2]);
p = {};
% *************************************************
% --- second interpolator by 2 ---
% *************************************************
rate = rate * 2;
s = upsample(s, 2); % insert one zero after every sample
p = addPlot(p, s, rate, 'k', 3, 'interpolator 2 input');
s = filter(h2, [1], s);
p = addPlot(p, s, rate, 'b', 3, 'interpolator 2 output');
p = addErrPlot(p, s, rate, 'error');
figure(3); clf; grid on; hold on;
doplot(p, 'second interpolator by 2');
ylim([-70, 2]);
p = {};
% *************************************************
% --- third interpolator by 2 ---
% *************************************************
rate = rate * 2;
s = upsample(s, 2); % insert one zero after every sample
p = addPlot(p, s, rate, 'k', 3, 'interpolator 3 input');
s = filter(h3, [1], s);
p = addPlot(p, s, rate, 'b', 3, 'interpolator 3 output');
p = addErrPlot(p, s, rate, 'error');
figure(4); clf; grid on; hold on;
doplot(p, 'third interpolator by 2');
ylim([-70, 2]);
p = {};
% *************************************************
% --- fourth interpolator by 4 ---
% *************************************************
rate = rate * 4;
s = upsample(s, 4); % insert three zeros after every sample
p = addPlot(p, s, rate, 'k', 3, 'interpolator 4 input');
s = filter(h4, [1], s);
p = addPlot(p, s, rate, 'b', 3, 'final output');
p = addErrPlot(p, s, rate, 'error at output');
figure(5); clf; grid on; hold on;
doplot(p, 'fourth interpolator by 4');
ylim([-70, 2]);
figure(334);
stem(real(s(1:10000)));
% *************************************************
% export resampler frequency response
% *************************************************
switch smode
case 'evalIdeal'
exportFrequencyResponse(s, rate, 'interpolatorFrequencyResponse.mat');
end
end
% ************************************
% put frequency response plot data into p
% ************************************
function p = addPlot(p, s, rate, plotstyle, linewidth, legtext)
p{end+1} = struct('sig', s, 'rate', rate, 'plotstyle', plotstyle, 'linewidth', linewidth, 'legtext', legtext);
end
% ************************************
% determine the error spectrum, compared to an ideal filter (RRC)
% and add a plot to p
% ************************************
function p = addErrPlot(p, s, rate, legtext)
ref = RRC_impulseResponse(numel(s), rate);
% refB is scaled and shifted (sub-sample resolution) replica of ref
% that minimizes the energy in (s - refB)
[coeff, refB, deltaN] = fitSignal_FFT(s, ref);
err = s - refB;
err = brickwallFilter(err, rate, 1.15); % 1+alpha
% signal is divided into three parts:
% - A) wanted in-channel energy (correlated with ref)
% - B) unwanted in-channel energy (uncorrelated with ref)
% - C) unwanted out-of-channel energy (aliases)
% the error vector magnitude is B) relative to A)
energySig = refB * refB';
energyErr = err * err';
EVM_dB = 10*log10(energyErr / energySig);
legtext = sprintf('%s; EVM=%1.2f dB', legtext, EVM_dB);
p{end+1} = struct('sig', err, 'rate', rate, 'plotstyle', 'r', 'linewidth', 3, 'legtext', legtext);
end
% ************************************
% helper function, plot data in p
% ************************************
function doplot(p, t)
leg = {};
for ix = 1:numel(p)
pp = p{ix};
fb = FFT_frequencyBasis(numel(pp.sig), pp.rate);
fr = 20*log10(abs(fft(pp.sig) + eps));
h = plot(fftshift(fb), fftshift(fr), pp.plotstyle);
set(h, 'lineWidth', pp.linewidth);
xlabel('frequency, normalized to symbol rate');
ylabel('dB');
leg{end+1} = pp.legtext;
end
legend(leg);
title(t);
end
% ************************************
% ideal RRC filter (impulse response is as
% long as test signal)
% ************************************
function ir = RRC_impulseResponse(n, rate)
alpha = 0.15;
fb = FFT_frequencyBasis(n, rate);
% bandwidth is 1
c = abs(fb / 0.5);
c = (c-1)/(alpha); % -1..1 in the transition region
c=min(c, 1);
c=max(c, -1);
RRC_h = sqrt(1/2+cos(pi/2*(c+1))/2);
ir = real(ifft(RRC_h));
end
% ************************************
% remove any energy at frequencies > BW/2
% ************************************
function s = brickwallFilter(s, rate, BW)
n = numel(s);
fb = FFT_frequencyBasis(n, rate);
ix = find(abs(fb) > BW / 2);
s = fft(s);
s(ix) = 0;
s = real(ifft(s));
end
% ************************************
% for an impulse response s at rate, write the
% frequency response to fname
% ************************************
function exportFrequencyResponse(s, rate, fname)
fb = fftshift(FFT_frequencyBasis(numel(s), rate));
fr = fftshift(fft(s));
figure(335); grid on;
plot(fb, 20*log10(abs(fr)));
title('exported frequency response');
xlabel('normalized frequency');
ylabel('dB');
save(fname, 'fb', 'fr');
end
% ************************************
% calculates the frequency that corresponds to
% each FFT bin (negative, zero, positive)
% ************************************
function fb_Hz = FFT_frequencyBasis(n, rate_Hz)
fb = 0:(n - 1);
fb = fb + floor(n / 2);
fb = mod(fb, n);
fb = fb - floor(n / 2);
fb = fb / n; % now [0..0.5[, [-0.5..0[
fb_Hz = fb * rate_Hz;
end
% *******************************************************
% delay-matching between two signals (complex/real-valued)
%
% * matches the continuous-time equivalent waveforms
% of the signal vectors (reconstruction at Nyquist limit =>
% ideal lowpass filter)
% * Signals are considered cyclic. Use arbitrary-length
% zero-padding to turn a one-shot signal into a cyclic one.
%
% * output:
% => coeff: complex scaling factor that scales 'ref' into 'signal'
% => delay 'deltaN' in units of samples (subsample resolution)
% apply both to minimize the least-square residual
% => 'shiftedRef': a shifted and scaled version of 'ref' that
% matches 'signal'
% => (signal - shiftedRef) gives the residual (vector error)
%
% Example application
% - with a full-duplex soundcard, transmit an arbitrary cyclic test signal 'ref'
% - record 'signal' at the same time
% - extract one arbitrary cycle
% - run fitSignal
% - deltaN gives the delay between both with subsample precision
% - 'shiftedRef' is the reference signal fractionally resampled
% and scaled to optimally match 'signal'
% - to resample 'signal' instead, exchange the input arguments
% *******************************************************
function [coeff, shiftedRef, deltaN] = fitSignal_FFT(signal, ref)
n=length(signal);
% xyz_FD: Frequency Domain
% xyz_TD: Time Domain
% all references to 'time' and 'frequency' are for illustration only
forceReal = isreal(signal) && isreal(ref);
% *******************************************************
% Calculate the frequency that corresponds to each FFT bin
% [-0.5..0.5[
% *******************************************************
binFreq=(mod(((0:n-1)+floor(n/2)), n)-floor(n/2))/n;
% *******************************************************
% Delay calculation starts:
% Convert to frequency domain...
% *******************************************************
sig_FD = fft(signal);
ref_FD = fft(ref, n);
% *******************************************************
% ... calculate crosscorrelation between
% signal and reference...
% *******************************************************
u=sig_FD .* conj(ref_FD);
if mod(n, 2) == 0
% for an even sized FFT the center bin represents a signal
% [-1 1 -1 1 ...] (subject to interpretation). It cannot be delayed.
% The frequency component is therefore excluded from the calculation.
u(length(u)/2+1)=0;
end
Xcor=abs(ifft(u));
% figure(); plot(abs(Xcor));
% *******************************************************
% Each bin in Xcor corresponds to a given delay in samples.
% The bin with the highest absolute value corresponds to
% the delay where maximum correlation occurs.
% *******************************************************
integerDelay = find(Xcor==max(Xcor));
% (1): in case there are several bitwise identical peaks, use the first one
% Minus one: Delay 0 appears in bin 1
integerDelay=integerDelay(1)-1;
% Fourier transform of a pulse shifted by one sample
rotN = exp(2i*pi*integerDelay .* binFreq);
uDelayPhase = -2*pi*binFreq;
% *******************************************************
% Since the signal was multiplied with the conjugate of the
% reference, the phase is rotated back to 0 degrees in case
% of no delay. Delay appears as linear increase in phase, but
% it has discontinuities.
% Use the known phase (with +/- 1/2 sample accuracy) to
% rotate back the phase. This removes the discontinuities.
% *******************************************************
% figure(); plot(angle(u)); title('phase before rotation');
u=u .* rotN;
% figure(); plot(angle(u)); title('phase after rotation');
% *******************************************************
% Obtain the delay using linear least mean squares fit
% The phase is weighted according to the amplitude.
% This suppresses the error caused by frequencies with
% little power, that may have radically different phase.
% *******************************************************
weight = abs(u);
constRotPhase = 1 .* weight;
uDelayPhase = uDelayPhase .* weight;
ang = angle(u) .* weight;
r = [constRotPhase; uDelayPhase] .' \ ang.'; %linear mean square
%rotPhase=r(1); % constant phase rotation, not used.
% the same will be obtained via the phase of 'coeff' further down
fractionalDelay=r(2);
% *******************************************************
% Finally, the total delay is the sum of integer part and
% fractional part.
% *******************************************************
deltaN = integerDelay + fractionalDelay;
% *******************************************************
% provide shifted and scaled 'ref' signal
% *******************************************************
% this is effectively time-convolution with a unit pulse shifted by deltaN
rotN = exp(-2i*pi*deltaN .* binFreq);
ref_FD = ref_FD .* rotN;
shiftedRef = ifft(ref_FD);
% *******************************************************
% Again, crosscorrelation with the now time-aligned signal
% *******************************************************
coeff=sum(signal .* conj(shiftedRef)) / sum(shiftedRef .* conj(shiftedRef));
shiftedRef=shiftedRef * coeff;
if forceReal
shiftedRef = real(shiftedRef);
end
end``````
## Baseband-equivalent phase noise model
December 18, 20114 comments Coded in Matlab
``````% ****************************************************************
% baseband equivalent source of local oscillator with phase noise
% Markus Nentwig, 18.12.2011
% ****************************************************************
function pn_generator()
close all;
% ****************************************************************
% PN generator configuration
% ****************************************************************
srcPar = struct();
srcPar.n = 2 ^ 18; % generated number of output samples
srcPar.rate_Hz = 7.68e6; % sampling rate
srcPar.f_Hz = [0, 10e3, 1e6, 9e9]; % phase noise spectrum, frequencies
srcPar.g_dBc1Hz = [-80, -80, -140, -140]; % phase noise spectrum, magnitude
srcPar.spursF_Hz = [300e3, 400e3, 700e3]; % discrete spurs (set [] if not needed)
srcPar.spursG_dBc = [-50, -55, -60]; % discrete spurs, power relative to carrier
% ****************************************************************
% run PN generator
% ****************************************************************
s = PN_src(srcPar);
if false
% ****************************************************************
% export phase noise baseband-equivalent signal for use in simulator etc
% ****************************************************************
tmp = [real(s); imag(s)] .';
save('phaseNoiseSample.dat', 'tmp', '-ascii');
end
if exist('spectrumAnalyzer', 'file')
% ****************************************************************
% spectrum analyzer configuration
% ****************************************************************
SAparams = struct();
SAparams.rate_Hz = srcPar.rate_Hz; % sampling rate of the input signal
SAparams.pRef_W = 1e-3; % unity signal represents 0 dBm (1/1000 W)
SAparams.pNom_dBm = 0; % show 0 dBm as 0 dB;
SAparams.filter = 'brickwall';
SAparams.RBW_window_Hz = 1000; % convolve power spectrum with a 1k filter
SAparams.RBW_power_Hz = 1; % show power density as dBc in 1 Hz
SAparams.noisefloor_dB = -250; % don't add artificial noise
SAparams.logscale = true; % use logarithmic frequency axis
% plot nominal spectrum
figure(1); grid on;
h = semilogx(max(srcPar.f_Hz, 100) / 1e6, srcPar.g_dBc1Hz, 'k+-');
set(h, 'lineWidth', 3);
hold on;
spectrumAnalyzer('signal', s, SAparams, 'fMin_Hz', 100, 'fig', 1);
ylabel('dBc in 1 Hz');
legend('nominal PSD', 'output spectrum');
title('check match with nominal PSD');
spectrumAnalyzer('signal', s, SAparams, 'fMin_Hz', 100, 'RBW_power_Hz', 'sine', 'fig', 2);
title('check carrier level (0 dBc); check spurs level(-50/-55/-60 dBc)');
ylabel('dBc for continuous-wave signal');
spectrumAnalyzer('signal', s, SAparams, 'fig', 3, 'logscale', false);
ylabel('dBc in 1 Hz');
end
end
function pn_td = PN_src(varargin)
def = {'includeCarrier', true, ...
'spursF_Hz', [], ...
'spursG_dBc', [], ...
'fMax_Hz', []};
p = vararginToStruct(def, varargin);
% length of signal in the time domain (after ifft)
len_s = p.n / p.rate_Hz
% FFT bin frequency spacing
deltaF_Hz = 1 / len_s
% construct AWGN signal in the frequency domain
% a frequency domain bin value of n gives a time domain power of 1
% for example ifft([4 0 0 0]) => 1 1 1 1
% each bin covers a frequency interval of deltaF_Hz
mag = p.n;
% scale "unity power in one bin" => "unity power per Hz":
% multiply with sqrt(deltaF_Hz):
mag = mag * sqrt(deltaF_Hz);
% Create noise according to mag in BOTH real- and imaginary value
mag = mag * sqrt(2);
% both real- and imaginary part contribute unity power => divide by sqrt(2)
pn_fd = mag / sqrt(2) * (randn(1, p.n) + 1i * randn(1, p.n));
% frequency vector corresponding to the FFT bins (0, positive, negative)
fb_Hz = FFT_freqbase(p.n, deltaF_Hz);
% interpolate phase noise spectrum on frequency vector
% note: interpolate dB on logarithmic frequency axis
H_dB = interp1(log(p.f_Hz+eps), p.g_dBc1Hz, log(abs(fb_Hz)+eps), 'linear');
% dB => magnitude
H = 10 .^ (H_dB / 20);
% H = 1e-6; % sanity check: enforce flat -120 dBc in 1 Hz
% apply filter to noise spectrum
pn_fd = pn_fd .* H;
% set spurs
for ix = 1:numel(p.spursF_Hz)
fs = p.spursF_Hz(ix);
u = abs(fb_Hz - fs);
ix2 = find(u == min(u), 1);
% random phase
rot = exp(2i*pi*rand());
% bin value of n: unity (carrier) power (0 dBc)
% scale with sqrt(2) because imaginary part will be discarded
% scale with sqrt(2) because the tone appears at both positive and negative frequencies
smag = 2 * p.n * 10 ^ (p.spursG_dBc(ix) / 20);
pn_fd(ix2) = smag * rot;
end
% limit bandwidth (tool to avoid aliasing in an application
% using the generated phase noise signal)
if ~isempty(p.fMax_Hz)
pn_fd(find(abs(fb_Hz) > p.fMax_Hz)) = 0;
end
% convert to time domain
pn_td = ifft(pn_fd);
pn_td = real(pn_td);
% Now pn_td is a real-valued random signal with a power spectral density
% as specified in f_Hz / g_dBc1Hz.
% phase-modulate to carrier
% note: d/dx exp(x) = 1 near |x| = 1
% in other words, the phase modulation has unity gain for small phase error
pn_td = exp(i*pn_td);
if ~p.includeCarrier
% remove carrier
% returns isolated phase noise component
pn_td = pn_td - 1;
end
end
% returns a vector of frequencies corresponding to n FFT bins, when the
% frequency spacing between two adjacent bins is deltaF_Hz
function fb_Hz = FFT_freqbase(n, deltaF_Hz)
fb_Hz = 0:(n - 1);
fb_Hz = fb_Hz + floor(n / 2);
fb_Hz = mod(fb_Hz, n);
fb_Hz = fb_Hz - floor(n / 2);
fb_Hz = fb_Hz * deltaF_Hz;
end
% *************************************************************
% helper function: Parse varargin argument list
% allows calling myFunc(A, A, A, ...)
% where A is
% - key (string), value (arbitrary) => result.key = value
% - a struct => fields of A are copied to result
% - a cell array => recursive handling using above rules
% *************************************************************
function r = vararginToStruct(varargin)
% note: use of varargin implicitly packs the caller's arguments into a cell array
% that is, calling vararginToStruct('hello') results in
% varargin = {'hello'}
r = flattenCellArray(varargin, struct());
end
function r = flattenCellArray(arr, r)
ix=1;
ixMax = numel(arr);
while ix <= ixMax
e = arr{ix};
if iscell(e)
% cell array at 'key' position gets recursively flattened
% becomes struct
r = flattenCellArray(e, r);
elseif ischar(e)
% string => key.
% The following entry is a value
ix = ix + 1;
v = arr{ix};
% store key-value pair
r.(e) = v;
elseif isstruct(e)
names = fieldnames(e);
for ix2 = 1:numel(names)
k = names{ix2};
r.(k) = e.(k);
end
else
e
assert(false)
end
ix=ix+1;
end % while
end``````
## Digital model for analog filters
December 17, 2011 Coded in Matlab
``````% discrete-time model for Laplace-domain expression
% Markus Nentwig, 30.12.2011
function sn_model()
close all;
run_demo1(10);
run_demo2(11);
end
% ************************************
% Constructs a FIR model for a relatively
% narrow-band continuous-time IIR filter.
% At the edge of the first Nyquist zone
% (+/- fSample/2), the frequency response
% is down by about 70 dB, which makes
% the modeling unproblematic.
% The impact of different windowing options
% is visible both at high frequencies, but
% also as deviation between original frequency
% response and model at the passband edge.
% ************************************
function run_demo1(fig)
[b, a] = getContTimeExampleFilter();
fc_Hz = 0.5e6; % frequency corresponding to omegaNorm == 1
commonParameters = struct(...
's_a', a, ...
's_b', b, ...
'z_rate_Hz', 3e6, ...
's_fNorm_Hz', fc_Hz, ...
'fig', fig);
% sample impulse response without windowing
ir = sampleLaplaceDomainImpulseResponse(...
commonParameters, ...
'plotstyle_s', 'k-', ...
'plotstyle_z', 'b-');
% use mild windowing
ir = sampleLaplaceDomainImpulseResponse(...
commonParameters, ...
'plotstyle_z', 'r-', ...
'winLen_percent', 4);
% use heavy windowing - error shows at passband edge
ir = sampleLaplaceDomainImpulseResponse(...
commonParameters, ...
'plotstyle_z', 'm-', ...
'winLen_percent', 100);
legend('s-domain', 'sampled, no window', 'sampled, 4% RC window', 'sampled, 100% RC window')
figure(33); clf;
h = stem(ir);
set(h, 'markersize', 2);
set(h, 'lineWidth', 2);
title('sampled impulse response');
end
% ************************************
% model for a continuous-time IIR filter
% that is relatively wide-band.
% The frequency response requires some
% manipulation at the edge of the Nyquist zone.
% Otherwise, there would be an abrupt change
% that would result in an excessively long impulse
% response.
% ************************************
function run_demo2(fig)
[b, a] = getContTimeExampleFilter();
fc_Hz = 1.4e6; % frequency corresponding to omegaNorm == 1
commonParameters = struct(...
's_a', a, ...
's_b', b, ...
'z_rate_Hz', 3e6, ...
's_fNorm_Hz', fc_Hz, ...
'fig', fig);
% sample impulse response without any manipulations
ir1 = sampleLaplaceDomainImpulseResponse(...
commonParameters, ...
'aliasZones', 0, ...
'plotstyle_s', 'k-', ...
'plotstyle_z', 'b-');
% use artificial aliasing (introduces some passband error)
ir2 = sampleLaplaceDomainImpulseResponse(...
commonParameters, ...
'aliasZones', 5, ...
'plotstyle_z', 'r-');
% use artificial low-pass filter (freq. response
% becomes invalid beyond +/- BW_Hz / 2)
ir3 = sampleLaplaceDomainImpulseResponse(...
commonParameters, ...
'aliasZones', 0, ...
'BW_Hz', 2.7e6, ...
'plotstyle_z', 'm-');
line([0, 2.7e6/2, 2.7e6/2], [-10, -10, -50]);
legend('s-domain', ...
sprintf('unmodified (%i taps)', numel(ir1)), ...
sprintf('artificial aliasing (%i taps)', numel(ir2)), ...
sprintf('artificial LP filter (%i taps)', numel(ir3)));
title('2nd example: wide-band filter model frequency response');
figure(350); grid on; hold on;
subplot(3, 1, 1);
stem(ir1, 'b'); xlim([1, numel(ir1)])
title('wide-band filter model: unmodified');
subplot(3, 1, 2);
stem(ir2, 'r');xlim([1, numel(ir1)]);
title('wide-band filter model: art. aliasing');
subplot(3, 1, 3);
stem(ir3, 'm');xlim([1, numel(ir1)]);
title('wide-band filter model: art. LP filter');
end
% Build example Laplace-domain model
function [b, a] = getContTimeExampleFilter()
if true
order = 6;
ripple_dB = 1.2;
omegaNorm = 1;
[b, a] = cheby1(order, ripple_dB, omegaNorm, 's');
else
% same as above, if cheby1 is not available
b = 0.055394;
a = [1.000000 0.868142 1.876836 1.109439 0.889395 0.279242 0.063601];
end
end
% ************************************
% * Samples the impulse response of a Laplace-domain
% component
% * Adjusts group delay and impulse response length so that
% the discarded part of the impulse response is
% below a threshold.
% * Applies windowing
%
% Mandatory arguments:
% s_a, s_b: Laplace-domain coefficients
% s_fNorm_Hz: normalization frequency for a, b coefficients
% z_rate_Hz: Sampling rate of impulse response
%
% optional arguments:
% truncErr_dB: Maximum truncation error of impulse response
% nPts: Computed impulse response length before truncation
% winLen_percent: Part of the IR where windowing is applied
% BW_Hz: Apply artificical lowpass filter for |f| > +/- BW_Hz / 2
%
% plotting:
% fig: Figure number
% plotstyle_s: set to plot Laplace-domain frequency response
% plotstyle_z: set to plot z-domain model freq. response
% ************************************
function ir = sampleLaplaceDomainImpulseResponse(varargin)
def = {'nPts', 2^18, ...
'truncErr_dB', -60, ...
'winLen_percent', -1, ...
'fig', 99, ...
'plotstyle_s', [], ...
'plotstyle_z', [], ...
'aliasZones', 1, ...
'BW_Hz', []};
p = vararginToStruct(def, varargin);
% FFT bin frequencies
fbase_Hz = FFT_frequencyBasis(p.nPts, p.z_rate_Hz);
% instead of truncating the frequency response at +/- z_rate_Hz,
% fold the aliases back into the fundamental Nyquist zone.
% Otherwise, we'd try to build a near-ideal lowpass filter,
% which is essentially non-causal and requires a long impulse
% response with artificially introduced group delay.
H = 0;
for alias = -p.aliasZones:p.aliasZones
% Laplace-domain "s" in normalized frequency
s = 1i * (fbase_Hz + alias * p.z_rate_Hz) / p.s_fNorm_Hz;
% evaluate polynomial in s
H_num = polyval(p.s_b, s);
H_denom = polyval(p.s_a, s);
Ha = H_num ./ H_denom;
H = H + Ha;
end
% plot |H(f)| if requested
if ~isempty(p.plotstyle_s)
figure(p.fig); hold on; grid on;
fr = fftshift(20*log10(abs(H) + eps));
h = plot(fftshift(fbase_Hz), fr, p.plotstyle_s);
set(h, 'lineWidth', 3);
xlabel('f/Hz'); ylabel('|H(f)| / dB');
xlim([0, p.z_rate_Hz/2]);
end
% apply artificial lowpass filter
if ~isempty(p.BW_Hz)
% calculate RC transition bandwidth
BW_RC_trans_Hz = p.z_rate_Hz - p.BW_Hz;
% alpha (RC filter parameter) implements the
% RC transition bandwidth:
alpha_RC = BW_RC_trans_Hz / p.z_rate_Hz / 2;
% With a cutoff frequency of BW_RC, the RC filter
% reaches |H(f)| = 0 at f = z_rate_Hz / 2
% BW * (1+alpha) = z_rate_Hz / 2
BW_RC_Hz = p.z_rate_Hz / ((1+alpha_RC));
HRC = raisedCosine(fbase_Hz, BW_RC_Hz, alpha_RC);
H = H .* HRC;
end
% frequency response => impulse response
ir = ifft(H);
% assume s_a, s_b describe a real-valued impulse response
ir = real(ir);
% the impulse response peak is near the first bin.
% there is some earlier ringing, because evaluating the s-domain
% expression only for s < +/- z_rate_Hz / 2 implies an ideal,
% non-causal low-pass filter.
ir = fftshift(ir);
% now the peak is near the center
peak = max(abs(ir));
thr = peak * 10 ^ (p.truncErr_dB / 20);
% first sample that is above threshold
% determines the group delay of the model
ixFirst = find(abs(ir) > thr, 1, 'first');
% last sample above threshold
% determines the length of the impulse response
ixLast = find(abs(ir) > thr, 1, 'last');
% truncate
ir = ir(ixFirst:ixLast);
% apply windowing
if p.winLen_percent > 0
% note: The window drops to zero for the first sample that
% is NOT included in the impulse response.
v = linspace(-1, 0, numel(ir)+1);
v = v(1:end-1);
v = v * 100 / p.winLen_percent;
v = v + 1;
v = max(v, 0);
win = (cos(v*pi) + 1) / 2;
ir = ir .* win;
end
% plot frequency response, if requested
if ~isempty(p.plotstyle_z)
irPlot = zeros(1, p.nPts);
irPlot(1:numel(ir)) = ir;
figure(p.fig); hold on;
fr = fftshift(20*log10(abs(fft(irPlot)) + eps));
h = plot(fftshift(fbase_Hz), fr, p.plotstyle_z);
set(h, 'lineWidth', 3);
xlabel('f/Hz'); ylabel('|H(f)| / dB');
xlim([0, p.z_rate_Hz/2]);
end
end
% ************************************
% raised cosine frequency response
% ************************************
function H = raisedCosine(f, BW, alpha)
c=abs(f * 2 / BW);
% c=-1 for lower end of transition region
% c=1 for upper end of transition region
c=(c-1) / alpha;
% clip to -1..1 range
c=min(c, 1);
c=max(c, -1);
H = 1/2+cos(pi/2*(c+1))/2;
end
% ************************************
% calculates the frequency that corresponds to each
% FFT bin (negative, zero, positive)
% ************************************
function fb_Hz = FFT_frequencyBasis(n, rate_Hz)
fb = 0:(n - 1);
fb = fb + floor(n / 2);
fb = mod(fb, n);
fb = fb - floor(n / 2);
fb = fb / n; % now [0..0.5[, [-0.5..0[
fb_Hz = fb * rate_Hz;
end
% *************************************************************
% helper function: Parse varargin argument list
% allows calling myFunc(A, A, A, ...)
% where A is
% - key (string), value (arbitrary) => result.key = value
% - a struct => fields of A are copied to result
% - a cell array => recursive handling using above rules
% *************************************************************
function r = vararginToStruct(varargin)
% note: use of varargin implicitly packs the caller's arguments into a cell array
% that is, calling vararginToStruct('hello') results in
% varargin = {'hello'}
r = flattenCellArray(varargin, struct());
end
function r = flattenCellArray(arr, r)
ix=1;
ixMax = numel(arr);
while ix <= ixMax
e = arr{ix};
if iscell(e)
% cell array at 'key' position gets recursively flattened
% becomes struct
r = flattenCellArray(e, r);
elseif ischar(e)
% string => key.
% The following entry is a value
ix = ix + 1;
v = arr{ix};
% store key-value pair
r.(e) = v;
elseif isstruct(e)
names = fieldnames(e);
for ix2 = 1:numel(names)
k = names{ix2};
r.(k) = e.(k);
end
else
e
assert(false)
end
ix=ix+1;
end % while
end``````
## Spectrum analyzer
December 10, 20111 comment Coded in Matlab
``````% ************************************
% spectrum analyzer
% Markus Nentwig, 10.12.2011
%
% Note: If the signal is not cyclic, apply conventional windowing before
% calling spectrumAnalyzer
% ************************************
function satest()
close all;
rate_Hz = 30.72e6; % sampling rate
n = 100000; % number of samples
noise = randn(1, n);
pulse = zeros(1, n); pulse(1) = 1;
% ************************************
% example (linear frequency scale, RBW filter)
% ************************************
tmp1 = normalize(RRC_filter('signal', noise, 'rate_Hz', rate_Hz, 'alpha', 0.22, 'BW_Hz', 3.84e6, 'fCenter_Hz', 5e6));
% pRef = 0.001: a normalized signal will show as 0 dBm = 0.001 W
saPar = struct('rate_Hz', rate_Hz, 'fig', 1, 'pRef_W', 1e-3, 'RBW_power_Hz', 3.84e6, 'filter', 'brickwall', 'plotStyle', 'b-');
spectrumAnalyzer(saPar, 'signal', tmp1, 'RBW_window_Hz', 1e3);
spectrumAnalyzer(saPar, 'signal', tmp1, 'RBW_window_Hz', 30e3, 'plotStyle', 'k-');
spectrumAnalyzer(saPar, 'signal', tmp1, 'RBW_window_Hz', 300e3, 'plotStyle', 'r-');
spectrumAnalyzer(saPar, 'signal', tmp1, 'RBW_window_Hz', 30e3, 'filter', 'gaussian', 'plotStyle', 'g-');
legend('1k RBW filter', '30k RBW filter', '300k RBW filter', '30k Gaussian filter (meas. instrument model)');
title('WCDMA-like signal');
% ************************************
% example (logarithmic frequency scale,
% no RBW filter)
% ************************************
fPar = struct('rate_Hz', rate_Hz, ...
'chebyOrder', 6, ...
'chebyRipple_dB', 1, ...
'fCorner_Hz', 1e5);
saPar = struct('rate_Hz', rate_Hz, ...
'pRef_W', 1e-3, ...
'fMin_Hz', 1e3, ...
'RBW_power_Hz', 1e5, ...
'RBW_window_Hz', 1e3, ...
'filter', 'none', ...
'plotStyle', 'b-', ...
'logscale', true, ...
'fig', 2, ...
'noisefloor_dB', -300);
tmp1 = normalize(IIR_filterExample('signal', noise, fPar));
tmp2 = normalize(IIR_filterExample('signal', pulse, fPar));
spectrumAnalyzer('signal', tmp1, saPar);
spectrumAnalyzer('signal', tmp2, saPar, 'plotStyle', 'k-');
end
function sig = normalize(sig)
sig = sig / sqrt(sig * sig' / numel(sig));
end
% ************************************
% calculates the frequency that corresponds to each
% FFT bin (negative, zero, positive)
% ************************************
function fb_Hz = FFT_frequencyBasis(n, rate_Hz)
fb = 0:(n - 1);
fb = fb + floor(n / 2);
fb = mod(fb, n);
fb = fb - floor(n / 2);
fb = fb / n; % now [0..0.5[, [-0.5..0[
fb_Hz = fb * rate_Hz;
end
% ************************************
% root-raised cosine filter (to generate
% example signal)
% ************************************
function sig = RRC_filter(varargin)
def = struct('fCenter_Hz', 0);
p = vararginToStruct(def, varargin);
n = numel(p.signal);
fb_Hz = FFT_frequencyBasis(n, p.rate_Hz);
% frequency relative to cutoff frequency
c = abs((fb_Hz - p.fCenter_Hz) / (p.BW_Hz / 2));
% c=-1 for lower end of transition region
% c=1 for upper end of transition region
c=(c-1) / p.alpha;
% clip to -1..1 range
c=min(c, 1);
c=max(c, -1);
% raised-cosine filter
H = 1/2+cos(pi/2*(c+1))/2;
% root-raised cosine filter
H = sqrt(H);
% apply filter
sig = ifft(fft(p.signal) .* H);
end
% ************************************
% continuous-time filter example
% ************************************
function sig = IIR_filterExample(varargin)
p = vararginToStruct(varargin);
% design continuous-time IIR filter
[IIR_b, IIR_a] = cheby1(p.chebyOrder, p.chebyRipple_dB, 1, 's');
% evaluate on the FFT frequency grid
fb_Hz = FFT_frequencyBasis(numel(p.signal), p.rate_Hz);
% Laplace domain operator, normalized to filter cutoff frequency
% (the above cheby1 call designs for unity cutoff)
s = 1i * fb_Hz / p.fCorner_Hz;
% polynomial in s
H_num = polyval(IIR_b, s);
H_denom = polyval(IIR_a, s);
H = H_num ./ H_denom;
% apply filter
sig = ifft(fft(p.signal) .* H);
end
% ----------------------------------------------------------------------
% optionally: Move the code below into spectrumAnalyzer.m
function [fr, fb, handle] = spectrumAnalyzer(varargin)
def = struct(...
'noisefloor_dB', -150, ...
'filter', 'none', ...
'logscale', false, ...
'NMax', 10000, ...
'freqUnit', 'MHz', ...
'fig', -1, ...
'plotStyle', 'b-', ...
'originMapsTo_Hz', 0 ...
);
p = vararginToStruct(def, varargin);
signal = p.signal;
handle=nan; % avoid warning
% A resolution bandwidth value of 'sine' sets the RBW to the FFT bin spacing.
% The power of a pure sine wave now shows correctly from the peak in the spectrum (unity => 0 dB)
singleBinMode=strcmp(p.RBW_power_Hz, 'sine');
nSamples = numel(p.signal);
binspacing_Hz = p.rate_Hz / nSamples;
windowBW=p.RBW_window_Hz;
noisefloor=10^(p.noisefloor_dB/10);
% factor in the scaling factor that will be applied later for conversion to
% power in RBW
if ~singleBinMode
noisefloor = noisefloor * binspacing_Hz / p.RBW_power_Hz;
end
% fr: "f"requency "r"esponse (plot y data)
% fb: "f"requency "b"ase (plot x data)
fr = fft(p.signal);
scale_to_dBm=sqrt(p.pRef_W/0.001);
% Normalize amplitude to the number of samples that contribute
% to the spectrum
fr=fr*scale_to_dBm/nSamples;
% magnitude
fr = fr .* conj(fr);
[winLeft, winRight] = spectrumAnalyzerGetWindow(p.filter, singleBinMode, p.RBW_window_Hz, binspacing_Hz);
% winLeft is the right half of the window, but it appears on the
% left side of the FFT space
winLen=0;
if ~isempty(winLeft)
% zero-pad the power spectrum in the middle with a length
% equivalent to the window size.
% this guarantees that there is enough bandwidth for the filter!
% one window length is enough, the spillover from both sides overlaps
% Scale accordingly.
winLen=size(winLeft, 2)+size(winRight, 2);
% note: fr is the power shown in the plot, NOT a frequency
% domain representation of a signal.
% There is no need to renormalize because of the length change
center=floor(nSamples/2)+1;
rem=nSamples-center;
fr=[fr(1:center-1), zeros(1, winLen-1), fr(center:end)];
% construct window with same length as fr
win=zeros(size(fr));
win(1:1+size(winLeft, 2)-1)=winLeft;
win(end-size(winRight, 2)+1:end)=winRight;
assert(isreal(win));
assert(isreal(fr));
assert(size(win, 2)==size(fr, 2));
% convolve using FFT
win=fft(win);
fr=fft(fr);
fr=fr .* win;
fr=ifft(fr);
fr=real(fr); % chop off roundoff error imaginary part
fr=max(fr, 0); % chop off small negative numbers
fr=[fr(1:center-1), fr(end-rem+1:end)];
end
% ************************************
% build frequency basis and rotate 0 Hz to center
% ************************************
fb = FFT_frequencyBasis(numel(fr), p.rate_Hz);
fr = fftshift(fr);
fb = fftshift(fb);
if false
% use in special cases (very long signals)
% ************************************
% data reduction:
% If a filter is used, details smaller than windowBW are
% suppressed already by the filter, and using more samples
% ************************************
if numel(fr) > p.NMax
switch(p.filter)
case 'gaussian'
% 0.2 offset from the peak causes about 0.12 dB error
incr=floor(windowBW/binspacing_Hz/5);
case 'brickwall'
% there is no error at all for a peak
incr=floor(windowBW/binspacing_Hz/3);
case 'none'
% there is no filter, we cannot discard data at this stage
incr=-1;
end
if incr > 1
fr=fr(1:incr:end);
fb=fb(1:incr:end);
end
end
end
% ************************************
% data reduction:
% discard beyond fMin / fMax, if given
% ************************************
indexMin = 1;
indexMax = numel(fb);
flag=0;
if isfield(p, 'fMin_Hz')
indexMin=min(find(fb >= p.fMin_Hz));
flag=1;
end
if isfield(p, 'fMax_Hz')
indexMax=max(find(fb <= p.fMax_Hz));
flag=1;
end
if flag
fb=fb(indexMin:indexMax);
fr=fr(indexMin:indexMax);
end
if p.NMax > 0
if p.logscale
% ************************************
% Sample number reduction for logarithmic
% frequency scale
% ************************************
assert(isfield(p, 'fMin_Hz'), 'need fMin_Hz in logscale mode');
assert(p.fMin_Hz > 0, 'need fMin_Hz > 0 in logscale mode');
if ~isfield(p, 'fMax_Hz')
p.fMax_Hz = p.rate_Hz / 2;
end
% averaged output arrays
fbOut=zeros(1, p.NMax-1);
frOut=zeros(1, p.NMax-1);
% candidate output frequencies (it's not certain yet
% that there is actually data)
s=logspace(log10(p.fMin_Hz), log10(p.fMax_Hz), p.NMax);
f1=s(1);
nextStartIndex=max(find(fb < f1));
if isempty(nextStartIndex)
nextStartIndex=1;
end
% iterate through all frequency regions
% collect data
% average
for index=2:size(s, 2)
f2=s(index);
endIndex=max(find(fb < f2));
% number of data points in bin
n=endIndex-nextStartIndex+1;
if n > 0
% average
ix=nextStartIndex:endIndex;
fbOut(index-1)=sum(fb(ix))/n;
frOut(index-1)=sum(fr(ix))/n;
nextStartIndex=endIndex+1;
else
% mark point as invalid (no data)
fbOut(index-1)=nan;
end
end
% remove bins where no data point contributed
ix=find(~isnan(fbOut));
fbOut=fbOut(ix);
frOut=frOut(ix);
fb=fbOut;
fr=frOut;
else
% ************************************
% Sample number reduction for linear
% frequency scale
% ************************************
len=size(fb, 2);
decim=ceil(len/p.NMax); % one sample overlength => decim=2
if decim > 1
% truncate to integer multiple
len=floor(len / decim)*decim;
fr=fr(1:len);
fb=fb(1:len);
fr=reshape(fr, [decim, len/decim]);
fb=reshape(fb, [decim, len/decim]);
if singleBinMode
% apply peak hold over each segment (column)
fr=max(fr);
else
% apply averaging over each segment (column)
fr = sum(fr) / decim;
end
fb=sum(fb)/decim; % for each column the average
end % if sample reduction necessary
end % if linear scale
end % if sample number reduction
% ************************************
% convert magnitude to dB.
% truncate very small values
% using an artificial noise floor
% ************************************
fr=(10*log10(fr+noisefloor));
if singleBinMode
% ************************************
% The power reading shows the content of each
% FFT bin. This is accurate for single-frequency
% tones that fall exactly on the frequency grid
% (an integer number of cycles over the signal length)
% ************************************
else
% ************************************
% convert sensed power density from FFT bin spacing
% to resolution bandwidth
% ************************************
fr=fr+10*log10(p.RBW_power_Hz/binspacing_Hz);
end
% ************************************
% Post-processing:
% Translate frequency axis to RF
% ************************************
fb = fb + p.originMapsTo_Hz;
% ************************************
% convert to requested units
% ************************************
switch(p.freqUnit)
case 'Hz'
case 'kHz'
fb = fb / 1e3;
case 'MHz'
fb = fb / 1e6;
case 'GHz'
fb = fb / 1e9;
otherwise
error('unsupported frequency unit');
end
% ************************************
% Plot (if requested)
% ************************************
if p.fig > 0
% *************************************************************
% title
% *************************************************************
if isfield(p, 'title')
t=['"', p.title, '";'];
else
t='';
end
switch(p.filter)
case 'gaussian'
t=[t, ' filter: Gaussian '];
case 'brickwall'
t=[t, ' filter: ideal bandpass '];
case 'none'
t=[t, ' filter: none '];
otherwise
assert(0)
end
if ~strcmp(p.filter, 'none')
t=[t, '(', mat2str(p.RBW_window_Hz), ' Hz)'];
end
if isfield(p, 'pNom_dBm')
% *************************************************************
% show dB relative to a given absolute power in dBm
% *************************************************************
fr=fr-p.pNom_dBm;
yUnit='dB';
else
yUnit='dBm';
end
% *************************************************************
% plot
% *************************************************************
figure(p.fig);
if p.logscale
handle = semilogx(fb, fr, p.plotStyle);
else
handle = plot(fb, fr, p.plotStyle);
end
hold on; % after plot, otherwise prevents logscale
if isfield(p, 'lineWidth')
set(handle, 'lineWidth', p.lineWidth);
end
% *************************************************************
% axis labels
% *************************************************************
xlabel(p.freqUnit);
if singleBinMode
ylabel([yUnit, ' (continuous wave)']);
else
ylabel([yUnit, ' in ', mat2str(p.RBW_power_Hz), ' Hz integration BW']);
end
title(t);
% *************************************************************
% adapt y range, if requested
% *************************************************************
y=ylim();
y1=y(1); y2=y(2);
rescale=false;
if isfield(p, 'yMin')
y(1)=p.yMin;
rescale=true;
end
if isfield(p, 'yMax')
y(2)=p.yMax;
rescale=true;
end
if rescale
ylim(y);
end
end
end
function [winLeft, winRight] = spectrumAnalyzerGetWindow(filter, singleBinMode, RBW_window_Hz, binspacing_Hz)
switch(filter)
case 'gaussian'
% construct Gaussian filter
% -60 / -3 dB shape factor 4.472
nRBW=6;
nOneSide=ceil(RBW_window_Hz/binspacing_Hz*nRBW);
filterBase=linspace(0, nRBW, nOneSide);
winLeft=exp(-(filterBase*0.831) .^2);
winRight=fliplr(winLeft(2:end)); % omit 0 Hz bin
case 'brickwall'
nRBW=1;
n=ceil(RBW_window_Hz/binspacing_Hz*nRBW);
n1 = floor(n/2);
n2 = n - n1;
winLeft=ones(1, n1);
winRight=ones(1, n2);
case 'none'
winLeft=[];
winRight=[];
otherwise
error('unknown RBW filter type');
end
% the window is not supposed to shift the spectrum.
% Therefore, the first bin is always in winLeft (0 Hz):
if size(winLeft, 2)==1 && isempty(winRight)
% there is no use to convolve with one-sample window
% it's always unity
winLeft=[];
winRight=[];
tmpwin=[];
end
if ~isempty(winLeft)
% (note: it is not possible that winRight is empty, while winLeft is not)
if singleBinMode
% normalize to unity at 0 Hz
s=winLeft(1);
else
% normalize to unity area under the filter
s=sum(winLeft)+sum(winRight);
end
winLeft=winLeft / s;
winRight=winRight / s;
end
end
% *************************************************************
% helper function: Parse varargin argument list
% allows calling myFunc(A, A, A, ...)
% where A is
% - key (string), value (arbitrary) => result.key = value
% - a struct => fields of A are copied to result
% - a cell array => recursive handling using above rules
% *************************************************************
function r = vararginToStruct(varargin)
% note: use of varargin implicitly packs the caller's arguments into a cell array
% that is, calling vararginToStruct('hello') results in
% varargin = {'hello'}
r = flattenCellArray(varargin, struct());
end
function r = flattenCellArray(arr, r)
ix=1;
ixMax = numel(arr);
while ix <= ixMax
e = arr{ix};
if iscell(e)
% cell array at 'key' position gets recursively flattened
% becomes struct
r = flattenCellArray(e, r);
elseif ischar(e)
% string => key.
% The following entry is a value
ix = ix + 1;
v = arr{ix};
% store key-value pair
r.(e) = v;
elseif isstruct(e)
names = fieldnames(e);
for ix2 = 1:numel(names)
k = names{ix2};
r.(k) = e.(k);
end
else
e
assert(false)
end
ix=ix+1;
end % while
end``````
## Peak-to-average ratio analysis
December 10, 2011 Coded in Matlab
``````% *************************************************************
% Peak-to-average analyzis of complex baseband-equivalent signal
% Markus Nentwig, 10/12/2011
% Determines the magnitude relative to the RMS average
% which is exceeded with a given (small) probability
% A typical probability value is 99.9 %, which is very loosely related
% to an uncoded bit error rate target of 0.1 %.
% *************************************************************
% number of samples for example signals
n = 1e6;
% *************************************************************
% example: 99.9% PAR for white Gaussian noise
% *************************************************************
signal = 12345 * (randn(1, n) + 1i * randn(1, n));
[PAR, PAR_dB] = peakToAverageRatio(signal, 0.999, false);
printf('white Gaussian noise: %1.1f dB PAR at radio frequency\n', PAR_dB);
[PAR, PAR_dB] = peakToAverageRatio(signal, 0.999, true);
printf('white Gaussian noise: %1.1f dB PAR at baseband\n', PAR_dB);
% *************************************************************
% example: PAR of continuous-wave signal
% *************************************************************
% a quarter cycle gives the best coverage of amplitude values
% the statistics of the remaining three quarters are identical (symmetry)
signal = 12345 * exp(1i*2*pi*(0:n-1) / n / 4);
% Note: peaks can be below the average, depending on the given probability
% a possible peak-to-average ratio < 1 (< 0 dB) is a consequence of the
% peak definition and not unphysical.
% For a continuous-wave signal, the average value equals the peak value,
% and PAR results slightly below 0 dB are to be expected.
[PAR, PAR_dB] = peakToAverageRatio(signal, 0.999, false);
printf('continuous-wave: %1.1f dB PAR at radio frequency\n', PAR_dB);
[PAR, PAR_dB] = peakToAverageRatio(signal, 0.999, true);
printf('continuous-wave: %1.1f dB PAR at baseband\n', PAR_dB);
% *************************************************************
% self test:
% For a real-valued Gaussian random variable, the probability
% to not exceed n sigma is
% n = 1: 68.27 percent
% n = 2: 95.5 percent
% n = 3: 99.73 percent
% see http://en.wikipedia.org/wiki/Normal_distribution
% section "confidence intervals"
% *************************************************************
signal = 12345 * randn(1, n);
[PAR, PAR_dB] = peakToAverageRatio(signal, 68.2689/100, false);
printf('expecting %1.3f dB PAR, got %1.3f dB\n', 20*log10(1), PAR_dB);
[PAR, PAR_dB] = peakToAverageRatio(signal, 95.44997/100, false);
printf('expecting %1.3f dB PAR, got %1.3f dB\n', 20*log10(2), PAR_dB);
[PAR, PAR_dB] = peakToAverageRatio(signal, 99.7300/100, false);
printf('expecting %1.3f dB PAR, got %1.3f dB\n', 20*log10(3), PAR_dB);
end
function [PAR, PAR_dB] = peakToAverageRatio(signal, peakProbability, independentIQ)
1;
% force row vector
assert(min(size(signal)) == 1, 'matrix not allowed');
signal = signal(:) .';
assert(peakProbability > 0);
assert(peakProbability <= 1);
% *************************************************************
% determine RMS average and normalize signal to unity
% *************************************************************
nSamples = numel(signal);
sAverage = sqrt(signal * signal' / nSamples);
signal = signal / sAverage;
% *************************************************************
% "Peaks" in a complex-valued signal can be defined in two
% different ways:
% *************************************************************
if ~independentIQ
% *************************************************************
% ---------------------------
% The baseband-equivalent signal represents the modulation on
% The instantaneous magnitude of the modulated radio frequency
% wave causes clipping, for example in a radio frequency
% power amplifier.
% The -combined- magnitude of in-phase and quadrature signal
% (that is, real- and imaginary part) is relevant.
% This is the default definition, when working with radio
% frequency (or intermediate frequency) signals, as long as a
% single, real-valued waveform is being processed.
% *************************************************************
sMag = abs(signal);
t = 'mag(s) := |s|; cdf(mag(s))';
else
% *************************************************************
% Baseband definition
% -------------------
% The baseband-equivalent signal is explicitly separated into
% an in-phase and a quadrature branch that are processed
% independently.
% The -individual- magnitudes of in-phase and quadrature branch
% cause clipping.
% For example, the definition applies when a complex-valued
% baseband signal is processed in a digital filter with real-
% valued coefficients, which is implemented as two independent,
% real-valued filters on real part and imaginary part.
% *************************************************************
% for each sample, use the maximum of in-phase / quadrature.
% If both clip in the same sample, it's counted only once.
sMag = max(abs(real(signal)), abs(imag(signal)));
t = 'mag(s) := max(|real(s)|, |imag(s)|); cdf(mag(s))';
end
% *************************************************************
% determine number of samples with magnitude below the "peak"
% level, based on the given peak probability
% for example: probability = 0.5 => nBelowPeakLevel = nSamples/2
% typically, this will be a floating point number
% *************************************************************
nBelowPeakLevel = peakProbability * nSamples;
% *************************************************************
% interpolate the magnitude between the two closest samples
% *************************************************************
sMagSorted = sort(sMag);
x = [0 1:nSamples nSamples+1];
y = [0 sMagSorted sMagSorted(end)];
magAtPeakLevel = interp1(x, y, nBelowPeakLevel, 'linear');
% *************************************************************
% Peak-to-average ratio
% signal is normalized, average is 1
% the ratio relates to the sample magnitude
% "power" is square of magnitude => use 20*log10 for dB conversion.
% *************************************************************
PAR = magAtPeakLevel;
PAR_dB = 20*log10(PAR + eps);
% *************************************************************
% illustrate the CDF and the result
% *************************************************************
if true
figure();
plot(y, x / max(x));
title(t);
xlabel('normalized magnitude m');
ylabel('prob(mag(s) < m)');
line([1, 1] * magAtPeakLevel, [0, 1]);
line([0, max(y)], [1, 1] * peakProbability);
end
end``````
## integrate RMS phase error from spectrum
October 21, 20111 comment Coded in Matlab
``````% ****************************************************************
% RMS phase error from phase noise spectrum
% Markus Nentwig, 2011
%
% - integrates RMS phase error from a phase noise power spectrum
% - generates an example signal and determines the RMS-average
% phase error (alternative method)
% ****************************************************************
function pn_snippet()
close all;
% ****************************************************************
% Phase noise spectrum model
% --------------------------
% Notes:
% - Generally, a phase noise spectrum tends to follow
% |H(f)| = c0 + c1 f^n1 + c2 f^n2 + c3 f^n3 + ...
% Therefore, linear interpolation should be performed in dB
% on a LOGARITHMIC frequency axis.
% - Single-/double sideband definition:
% The phase noise model is defined for -inf <= f <= inf
% in other words, it contributes the given noise density at
% both positive AND negative frequencies.
% Assuming a symmetrical PN spectrum, the model is evaluated
% for |f| => no need to explicitly write out the negative frequency
% side.
% ****************************************************************
% PN model
% first column:
% frequency offset
% second column:
% spectral phase noise density, dB relative to carrier in a 1 Hz
% observation bandwidth
d = [0 -80
10e3 -80
1e6 -140
9e9 -140];
f_Hz = d(:, 1) .';
g_dBc1Hz = d(:, 2) .' -3;
% get RMS phase error by integrating the power spectrum
% (alternative 1)
e1_degRMS = integratePN_degRMS(f_Hz, g_dBc1Hz)
% get RMS phase error based on a test signal
% (alternative 2)
n = 2 ^ 20;
deltaF_Hz = 2;
e2_degRMS = simulatePN_degRMS(f_Hz, g_dBc1Hz, n, deltaF_Hz)
end
% ****************************************************************
% Integrate the RMS phase error from the power spectrum
% ****************************************************************
function r = integratePN_degRMS(f1_Hz, g1_dBc1Hz)
1;
% integration step size
deltaF_Hz = 100;
% list of integration frequencies
f_Hz = deltaF_Hz:deltaF_Hz:5e6;
% interpolate spectrum on logarithmic frequency, dB scale
% unit is dBc in 1 Hz, relative to a unity carrier
fr_dB = interp1(log(f1_Hz+eps), g1_dBc1Hz, log(f_Hz+eps), 'linear');
% convert to power in 1 Hz, relative to a unity carrier
fr_pwr = 10 .^ (fr_dB/10);
% scale to integration step size
% unit: power in deltaF_Hz, relative to unity carrier
fr_pwr = fr_pwr * deltaF_Hz;
% evaluation frequencies are positive only
% phase noise is two-sided
% (note the IEEE definition: one-half the double sideband PSD)
fr_pwr = fr_pwr * 2;
% sum up relative power over all frequencies
pow_relToCarrier = sum(fr_pwr);
% convert the phase noise power to an RMS magnitude, relative to the carrier
pnMagnitude = sqrt(pow_relToCarrier);
% convert from radians to degrees
r = pnMagnitude * 180 / pi;
end
% ****************************************************************
% Generate a PN signal with the given power spectrum and determine
% the RMS phase error
% ****************************************************************
function r = simulatePN_degRMS(f_Hz, g_dBc1Hz, n, deltaF_Hz);
A = 1; % unity amplitude, arbitrary
% indices of positive-frequency FFT bins.
% Does not include DC
ixPos = 2:(n/2);
% indices of negative-frequency FFT bins.
% Does not include the bin on the Nyquist limit
assert(mod(n, 2) == 0, 'n must be even');
ixNeg = n - ixPos + 2;
% Generate signal in the frequency domain
sig = zeros(1, n);
% positive frequencies
sig(ixPos) = randn(size(ixPos)) + 1i * randn(size(ixPos));
% for purely real-valued signal: conj()
% for purely imaginary-valued signal such as phase noise: -conj()
% Note:
% Small-signals are assumed. For higher "modulation indices",
% phase noise would become a circle in the complex plane
sig(ixNeg) = -conj(sig(ixPos));
% normalize energy to unity amplitude A per bin
sig = sig * A / RMS(sig);
% normalize energy to 0 dBc in 1 Hz
sig = sig * sqrt(deltaF_Hz);
% frequency vector corresponding to the FFT bins (0, positive, negative)
fb_Hz = FFT_freqbase(n, deltaF_Hz);
% interpolate phase noise spectrum on frequency vector
% interpolate dB on logarithmic frequency axis
H_dB = interp1(log(f_Hz+eps), g_dBc1Hz, log(abs(fb_Hz)+eps), 'linear');
% dB => magnitude
H = 10 .^ (H_dB / 20);
% plot on linear f axis
figure(); hold on;
plot(fftshift(fb_Hz), fftshift(mag2dB(H)), 'k');
xlabel('f/Hz'); ylabel('dBc in 1 Hz');
title('phase noise spectrum (linear frequency axis)');
% plot on logarithmic f axis
figure(); hold on;
ix = find(fb_Hz > 0);
semilogx(fb_Hz(ix), mag2dB(H(ix)), 'k');
xlabel('f/Hz'); ylabel('dBc in 1 Hz');
title('phase noise spectrum (logarithmic frequency axis)');
% apply frequency response to signal
sig = sig .* H;
sig (1) = A;
% convert to time domain
td = ifft(sig);
% determine phase
% for an ideal carrier, it would be zero
% any deviation from zero is phase error
ph_deg = angle(td) * 180 / pi;
figure();
ix = 1:1000;
plot(ix / n / deltaF_Hz, ph_deg(ix));
xlabel('time / s');
ylabel('phase error / degrees');
title('phase of example signal (first 1000 samples)');
figure();
hist(ph_deg, 300);
title('histogram of example signal phase');
xlabel('phase error / degrees');
% RMS average
% note: exp(1i*x) ~ x
% as with magnitude, power/energy is proportional to phase error squared
r = RMS(ph_deg);
% knowing that the signal does not have an amplitude component, we can alternatively
% determine the RMS phase error from the power of the phase noise component
% exclude carrier:
sig(1) = 0;
% 3rd alternative to calculate RMS phase error
rAlt_deg = sqrt(sig * sig') * 180 / pi
end
% gets a vector of frequencies corresponding to n FFT bins, when the
% frequency spacing between two adjacent bins is deltaF_Hz
function fb_Hz = FFT_freqbase(n, deltaF_Hz)
fb_Hz = 0:(n - 1);
fb_Hz = fb_Hz + floor(n / 2);
fb_Hz = mod(fb_Hz, n);
fb_Hz = fb_Hz - floor(n / 2);
fb_Hz = fb_Hz * deltaF_Hz;
end
% Root-mean-square average
function r = RMS(vec)
r = sqrt(vec * vec' / numel(vec));
end
% convert a magnitude to dB
function r = mag2dB(vec)
r = 20*log10(abs(vec) + eps);
end``````
## GSM (GMSK) power spectrum equation
September 18, 2011 Coded in Matlab
``````% ************************************************************
% Spectrum model for GSM signal
% Markus Nentwig, 2011
% based on 3GPP TS 45.004 section 2 "Modulation format for GMSK",
% assuming no additional filtering and an infinite-length
% symbol stream (no slot structure)
% ************************************************************
% ************************************************************
% Parameters
% The "baseline" serves as a gentle reminder that the model
% is only valid near the center frequency.
% the particular transmitter used.
% If not available, assume that spectral emission mask requirements
% are met.
% ************************************************************
BW = 285e3;
BW2 = 186e3;
baseline_dB = -76;
% baseline_dB = -999 % disable the constant term
% ************************************************************
% empirical GSM (GMSK narrow-bandwidth pulse) model equation
% ************************************************************
f = (-500e3:1e3:500e3)+1e-3;
gaussPart = exp(-(2*f/BW) .^2);
sincPart = sin(pi*f/BW2) ./ (pi*f/BW2);
flatPart = 10^(baseline_dB/20);
H_dB = 20*log10(abs(gaussPart .* sincPart) + flatPart);
% ************************************************************
% plot the spectrum
% ************************************************************
figure(); grid on; hold on;
h = plot(f/1e6, H_dB, 'b'); set(h, 'linewidth', 2);
% ************************************************************
% plot 'a' GSM spectral emission mask
% note, this is only "an" example
% See 3GPP TS 45.005 section 4.2.1
% "Spectrum due to the modulation and wide band noise"
% section 4.2.1.1
% "General requirements for all types of Base stations and MS"
% note the warning regarding measuring the nominal signal level!
% ************************************************************
maskX_MHz = [0, 100e3, 200e3, 250e3, 400e3, 600e3]/1e6;
maskY_dB = [0.5, 0.5, -30, -33, -60, -60];
xlabel('f/MHz');
ylabel('PSD/dB');
title('GSM spectrum');``````
## Yet Another FIR design algorithm
August 21, 20113 comments Coded in Matlab
``````% ********************************************
% least-mean-squares FIR design algorithm
% Markus Nentwig, 2010-2011
% release 2011/8/22
% ********************************************
function LMSFIR()
close all;
h1 = demo1('basic'); compareDemo1WithRemez(h1);
% h1 = demo1('basicLMS'); disp('demo: convergence failure is on purpose to show LMS solution');
% demo1('allpass');
% demo1('stopband');
% demo1('equalize');
% demo1('nominalResponse');
% demo1('multipassband');
% demo1('complex');
% demo1('rootRaisedCosineUpsampler');
% demo1('componentModel');
% demo1('componentModel2');
end
function h = demo1(nameOfDemo)
dpar = struct();
% parameters for a basic FIR lowpass filter design.
% kept in a struct(), so that individual examples
% can easily change them.
% sampling rate at the input of the filter
dpar.inRate_Hz = 104e6;
% number of physical FIR taps
% in a polyphase decimator, the number of internal
% coefficients will be fDecim * fStages
dpar.mStages = 36;
% end of passband. A single value will be internally
% expanded to [-9e6, 9e6].
% Asymmetric designs require
% the complexValued = true option.
% This 'default' passband can be omitted entirely, if passbands
% are declared individually later
dpar.req_passbandF_Hz = 9e6;
% defines the maximum allowed ripple in the passband.
dpar.req_passbandRipple_dB = 0.083;
% as alternative to ripple, the in-band error
% vector magnitude (EVM) can be limited
% dpar.req_passbandEVM_dB = -44;
% the passband specification may use multiple points
% dpar.req_passbandF_Hz = [-1, 5e6, 6e6, 9e6];
% dpar.req_passbandEVM_dB = [-44, -44, -34, -34];
% start of default stopband.
% as with the passband, the default stopband can be omitted,
% if individual bands are placed later.
dpar.req_stopbandF_Hz = 14e6;
dpar.req_stopbandMaxGain_dB = -30;
% dpar.req_stopbandF_Hz = [14e6, 34e6];
% dpar.req_stopbandGainMax_dB = [-30, -20];
% ********************************************
% create a filter design object "design"
% * access with LMSFIR_stage2 functions
% * evaluate with LMSFIR_stage3 function
% ********************************************
switch nameOfDemo
case 'basic'
% ********************************************
% simple filter using the parameters above
% ********************************************
design = LMSFIR_stage1_setup(dpar);
case 'basicLMS'
% ********************************************
% LMS design for comparison:
% Iterations are disabled
% ********************************************
dpar.nIter = 1;
% balance in-band / out-of-band performance as needed
dpar.inbandWeight_initValue = 5;
dpar.outOfBandWeight_initValue = 1;
design = LMSFIR_stage1_setup(dpar);
case 'allpass'
% ********************************************
% allpass design Offset the nominal delay by 1/3
% of a sample, compared to the "basic" example
% (compare the impulse responses)
% ********************************************
dpar.delayOffset = 1/3; % signal arrives now earlier
design = LMSFIR_stage1_setup(dpar);
case 'stopband'
% ********************************************
% ********************************************
% the following features require more taps
dpar.mStages = 48;
% create filter design object
design = LMSFIR_stage1_setup(dpar);
% place a stopband from 14 to 16 MHz with -50 dB
design = LMSFIR_stage2_placeStopband(...
design, ...
'f_Hz', [14e6, 16e6], ...
'g_dB', [-50, -50]);
% place another stopband from 16 to 28 MHz with
% -50 dB, linearly relaxing to -40 dB
design = LMSFIR_stage2_placeStopband(...
design, ...
'f_Hz', [16e6, 28e6], ...
'g_dB', [-50, -40]);
case 'equalize'
% ********************************************
% Equalize the frequency response of another
% filter in the passband(s)
% ********************************************
% As an equalizer, this is rather inefficient with so much
% unused bandwidth. Should operate at the smallest possible BW instead.
dpar.mStages = 52;
[ffilter_Hz, H] = getExampleLaplaceDomainFilter();
% set the frequency points...
dpar.equalizeFreq_Hz = ffilter_Hz;
% ... and the filter response. The design routine will
% use linear interpolation, therefore provide a sufficiently
% dense grid.
% Equalizing an asymmetric response requires
% the complexValued=true option, leading to a complex-valued
% FIR filter.
% The equalization function needs to be normalized.
% Otherwise, pass- and stopband targets will be offset
% by the gain mismatch.
dpar.equalizeComplexGain = H;
% as alternative to the complex gain, a magnitude response
% can be given via an equalizeGain_dB argument.
% dpar.equalizeGain_dB = 20*log10(abs(H));
% an asymmetric (non-linear-phase) impulse response may
% require a group delay that is not centered in the
% FIR length.
dpar.delayOffset = 2;
design = LMSFIR_stage1_setup(dpar);
case 'componentModel'
% ********************************************
% Create a FIR filter that approximates the passband behavior of
% the analog filter accurately, and gives a similar stopband rejection
%
% The most straightforward way to model an infinite-impulse-response
% lowpass is to simply sample the impulse response. However, it needs to be
% cut to size (since the FIR filter has only finite length)
% => Chopping it off destroys the out-of-band performance (=rectangular window)
% => use a window function that trades off between passband accuracy and
% stopband rejection
% => or use the design example below.
% ********************************************
dpar.mStages = 52;
[ffilter_Hz, H] = getExampleLaplaceDomainFilter();
% set the frequency points...
dpar.nominalFreq_Hz = ffilter_Hz;
dpar.nominalComplexGain = H;
dpar.req_stopbandF_Hz = [15e6, 30e6, 55e6];
dpar.req_stopbandMaxGain_dB = [-38, -80, -115];
dpar.req_passbandF_Hz = 9e6;
% offset the impulse response, it is not centered
dpar.delayOffset = 18;
design = LMSFIR_stage1_setup(dpar);
case 'componentModel2'
% ********************************************
% an extension of "componentModel1"
% stopband rejection does not matter, but we need
% phase-accurate modeling on a region of the stopband edge
% ********************************************
dpar.mStages = 80; % this won't be cheap...
[ffilter_Hz, H] = getExampleLaplaceDomainFilter();
dpar.nominalFreq_Hz = ffilter_Hz;
dpar.nominalComplexGain = H;
dpar.req_stopbandF_Hz = [ 16e6, 100e6];
dpar.req_stopbandMaxGain_dB = [ -30, -30];
dpar.req_passbandF_Hz = 9e6;
% offset the impulse response, it is not centered
dpar.delayOffset = dpar.mStages / 2 - 8;
design = LMSFIR_stage1_setup(dpar);
% place a passband in the area on the slope that is to be modeled accurately
design = LMSFIR_stage2_placePassband(...
design, ...
'f_Hz', [12e6, 16e6], ...
'EVM_dB', [-40, -50] - 30); % nominal gain -40..-50 dB, -30 dBc EVM
case 'nominalResponse'
% ********************************************
% Design a filter to a given frequency response
% ********************************************
dpar.mStages = 50;
% the frequency response is approximated in any
% declared passband, but must be valid for any
% frequency to allow plotting.
dpar.nominalFreq_Hz = [0, 3e6, 9e6, 1e9];
dpar.nominalGain_dB = [0, 0, -6, -6];
% instead, nominalComplexGain can be used
% g = [0, 0, -3, -3];
% dpar.nominalComplexGain = 10 .^ (g/20);
design = LMSFIR_stage1_setup(dpar);
case 'multipassband'
% ********************************************
% Design a filter with three passbands
% ********************************************
dpar.mStages = 50;
dpar = rmfield(dpar, 'req_passbandF_Hz');
dpar = rmfield(dpar, 'req_passbandRipple_dB');
design = LMSFIR_stage1_setup(dpar);
design = LMSFIR_stage2_placePassband(...
design, ...
'f_Hz', [-2e6, 2e6], ...
'EVM_dB', -45);
design = LMSFIR_stage2_placePassband(...
design, ...
'f_Hz', [3e6, 7e6], ...
'EVM_dB', [-45, -40]);
design = LMSFIR_stage2_placeStopband(...
design, ...
'f_Hz', [11.8e6, 12.4e6], ...
'g_dB', -70);
case 'complex'
% ********************************************
% Design a complex-valued filter
% ********************************************
% this is also an example for what can go wrong:
% In the unconstrained section around -40 MHz, the
% frequency response is allowed to go berserk. Which
% it does.
% Solution: Place a "soft" stopband (for example at -5 dB)
% in the "don't-care" regions and add a couple of taps.
% remove passband from default parameters
dpar = rmfield(dpar, 'req_passbandF_Hz');
dpar = rmfield(dpar, 'req_passbandRipple_dB');
% remove stopband from default parameters
dpar = rmfield(dpar, 'req_stopbandF_Hz');
dpar = rmfield(dpar, 'req_stopbandMaxGain_dB');
dpar.complexValued = true;
design = LMSFIR_stage1_setup(dpar);
design = LMSFIR_stage2_placeStopband(...
design, ...
'f_Hz', [-30e6, -16e6], ...
'g_dB', -50);
design = LMSFIR_stage2_placePassband(...
design, ...
'f_Hz', [-8e6, -2e6], ...
'EVM_dB', -45);
design = LMSFIR_stage2_placeStopband(...
design, ...
'f_Hz', [3e6, 40e6], ...
'g_dB', [-30, -50]);
case 'rootRaisedCosineUpsampler'
% ********************************************
% root-raised cosine upsampling filter for WCDMA transmitter
% The input chip stream arrives at 3.84 Msps, using the
% full bandwidth.
% Before the filter, it is upsampled (zero insertion) to
% 7.68 Msps.
% The filter applies RRC-filtering with 1.22 rolloff.
% ********************************************
% calculate nominal RRC response for lookup table / linear
% interpolation
f_Hz = logspace(1, 8, 10000); f_Hz(1) = -1;
c = abs(f_Hz * 2 / 3.84e6);
c = (c-1)/(0.22); % -1..1 in the transition region
c=min(c, 1);
c=max(c, -1);
RRC_h = sqrt(1/2+cos(pi/2*(c+1))/2);
% ********************************************
% once the targets are achieved, use the remaining
% 'degrees of freedom' for least-squares optimization.
% The LMS solver will improve, where it is 'cheapest'
% The parameters are not a real-world design
% (0.5 percent EVM => -46 dB)
% ********************************************
ci = 0;
% ci = 10; % for comparison: force equiripple
dpar = struct(...
'inRate_Hz', 3.84e6, ...
'fInterp', 2, ...
'mStages', 45, ...
'req_passbandF_Hz', 3.84e6 * 1.22 / 2, ...
'req_passbandEVM_dB', -46, ...
'req_stopbandF_Hz', 2.46e6, ...
'req_stopbandMaxGain_dB', -50, ...
'nominalFreq_Hz', f_Hz, ...
'nominalGain_dB', 20*log10(RRC_h + 1e-19), ...
'convergedIterations', ci);
design = LMSFIR_stage1_setup(dpar);
[h, status] = LMSFIR_stage3_run(design);
% save('hRRC_upsampler.txt', 'h', '-ascii');
disp(status);
otherwise
assert(false);
end % switch nameOfDemo
% ********************************************
% Design the filter
% ********************************************
% h is the impulse response (FIR tap coefficients).
[h, status] = LMSFIR_stage3_run(design);
disp(status);
end
function compareDemo1WithRemez(hLMS)
% identical target settings to demo1 "basic".
% note, the demo uses targets that are exactly "on the edge"
% what the algorithm can achieve. This results in an equiripple-
% design that can be compared with remez().
% If the targets are too loosely set, pass- and stopband quality
% start to "sag" in the band middle (LMS solution => lowest overall
% error, the optimizer improves where it's "cheapest").
r_Hz = 104e6;
m = 35; % definition differs by 1
f = [0 9e6 14e6 r_Hz/2] / (r_Hz/2);
a = [1 1 0 0];
ripple_dB = 0.1;
att_dB = 30;
err1 = 1 - 10 ^ (-ripple_dB / 20);
err2 = 10 ^ (-att_dB / 20);
w = [1/err1 1/err2];
% get remez design impulse response
hRemez = remez(m, f, a, w);
figure(); hold on;
handle = plot(hLMS, 'b+'); set(handle, 'lineWidth', 3);
plot(hRemez, 'k+'); set(handle, 'lineWidth', 2);
legend('this algorithm', 'Remez');
title('comparison with Remez design (optimum equiripple)');
end
% Gets the frequency response of an "analog" (Laplace-domain) filter.
% => Chebyshev response
% => 6th order
% => 1.2 dB ripple
% => cutoff frequency at 10 MHz
% returns
% f_Hz: list of frequencies
% H: complex-valued H(f_Hz)
function [f_Hz, H] = getExampleLaplaceDomainFilter()
[IIR_b, IIR_a] = cheby1(6, 1.2, 1, 's');
% evaluate it on a wide enough frequency range
f_Hz = logspace(1, 10, 1000); f_Hz(1) = -1;
% Laplace domain operator for normalized frequency
fCutoff_Hz = 10e6;
s = 1i * f_Hz / fCutoff_Hz;
% polynomial in s
H_num = polyval(IIR_b, s);
H_denom = polyval(IIR_a, s);
H = H_num ./ H_denom;
end
% === LMSFIR_xyz "API" functions ===
% ********************************************
% LMSFIR_stagex_... functions to interact with design 'object'
% to be executed in 'stage'-order
% ********************************************
function d = LMSFIR_stage1_setup(varargin)
p = varargin2struct(varargin);
d = struct();
% number of frequency points. Increase to improve accuracy.
% Frequencies are quantized to +/- rate / (2 * nSamples)
d.nSamples = 1024;
% default polyphase interpolation: none
d.fInterp = 1;
% default polyphase decimation: none
d.fDecim = 1;
% max. number of iterations
d.nIter = 100;
% for pure LMS solution, set nIter to 1 and change the weights below as needed
d.inbandWeight_initValue = 1;
d.outOfBandWeight_initValue = 1;
% abort when the iteration weights grow too large.
% This happens when targets are impossible.
% The result may still be meaningful, though.
d.abortWeight = 1e12;
% keep iterating, if the targets are reached.
% Once the "equi"-ripple iteration has brought all peaks to an acceptable level,
% the LMS solver will use the remaining "degrees of freedom" for a LMS optimization.
% The solver improves "where it is easy / cheap". This results in sloped
% stopbands and "drooping" EVM in passbands.
% Often, LMS is the best choice => set converged iterations to 0.
d.convergedIterations = 10;
% for a complex-valued filter, use "true".
% With a real-valued design, user input is only evaluated for positive frequencies!
d.complexValued = false;
% by default, the basis waveforms given to the optimizer are
% within a delay range of +/- half the FIR length.
% For nonlinear phase types (equalization / nominal frequency
% response), this may be suboptimal.
% Meaningful values shouldn't exceed +/- half the number of
% coefficients in the impulse response.
d.delayOffset = 0;
% copy parameters
fn = fieldnames(p);
for ix = 1:size(fn, 1)
key = fn{ix};
d.(key) = p.(key);
end
% frequency base over FFT range
fb = 0:(d.nSamples - 1);
fb = fb + floor(d.nSamples / 2);
fb = mod(fb, d.nSamples);
fb = fb - floor(d.nSamples / 2);
fb = fb / d.nSamples; % now [0..0.5[, [-0.5..0[
fb = fb * d.inRate_Hz * d.fInterp;
d.fb = fb;
% in real-valued mode, negative frequencies are treated as
% positive, when 'user input' is evaluated
if d.complexValued
d.fbUser = fb;
else
d.fbUser = abs(fb);
end
% ********************************************
% target settings. Those will be modified by
% LMSFIR_stage2_xyz()
% ********************************************
% initial value of NaN indicates: all entries are unset
d.errorSpecBinVal_inband_dB = zeros(size(d.fb)) + NaN;
d.errorSpecBinVal_outOfBand_dB = zeros(size(d.fb)) + NaN;
% ********************************************
% process req_passband requirement
% needs to be done at stage 1, because it is
% used for 'gating' with the tightenExisting /
% relaxExisting options
% ********************************************
if isfield(d, 'req_passbandF_Hz')
par = struct('onOverlap', 'error');
if isfield(d, 'req_passbandRipple_dB')
par.ripple_dB = d.req_passbandRipple_dB;
end
if isfield(d, 'req_passbandEVM_dB')
par.EVM_dB = d.req_passbandEVM_dB;
end
par.f_Hz = d.req_passbandF_Hz;
d = LMSFIR_stage2_placePassband(d, par);
end % if req_passbandF_Hz
% ********************************************
% process req_stopband requirement
% needs to be done at stage 1, because it is
% used for 'gating' with the tightenExisting /
% relaxExisting options
% ********************************************
if isfield(d, 'req_stopbandF_Hz')
f_Hz = d.req_stopbandF_Hz;
g_dB = d.req_stopbandMaxGain_dB;
% extend to infinity
if isscalar(f_Hz)
f_Hz = [f_Hz 9e19];
g_dB = [g_dB g_dB(end)];
end
d = placeBand...
(d, ...
'f_Hz', f_Hz, 'g_dB', g_dB, ...
'type', 'stopband', ...
'onOverlap', 'tighten');
end
% ********************************************
% plot management
% ********************************************
d.nextPlotIx = 700;
end
function d = LMSFIR_stage2_placeStopband(d, varargin)
p = varargin2struct(varargin);
% shorthand notation g_dB = -30; f_Hz = 9e6;
% extend fixed passband to positive infinity
if isscalar(p.f_Hz)
assert(p.f_Hz > 0);
p.f_Hz = [p.f_Hz 9e99];
end
if isscalar(p.g_dB)
p.g_dB = ones(size(p.f_Hz)) * p.g_dB;
end
% default action is to use the stricter requirement
if ~isfield(p, 'onOverlap')
p.onOverlap = 'tighten';
end
d = placeBand(d, 'type', 'stopband', p);
end
function d = LMSFIR_stage2_placePassband(d, varargin)
p = varargin2struct(varargin);
% default action is to use the stricter requirement
if ~isfield(p, 'onOverlap')
p.onOverlap = 'tighten';
end
% translate ripple spec to error
if isfield(p, 'ripple_dB')
assert(p.ripple_dB > 0);
eSamplescale = 10 ^ (p.ripple_dB / 20) - 1;
EVM_dB = 20*log10(eSamplescale);
end
if isfield(p, 'EVM_dB')
EVM_dB = p.EVM_dB;
end
% convert scalar to two-element vector
if isscalar(EVM_dB)
EVM_dB = [EVM_dB EVM_dB];
end
% *** handle f_Hz ***
f_Hz = p.f_Hz;
% extend to 0 Hz
if isscalar(f_Hz)
f_Hz = [0 f_Hz];
end
% *** create the passband ***
d = placeBand(d, ...
'type', 'passband', ...
'f_Hz', f_Hz, ...
'g_dB', EVM_dB, ...
'onOverlap', p.onOverlap);
end
% ********************************************
% the filter design algorithm
% h: impulse response
% status: converged or not
% note that even if convergence was not reached,
% the resulting impulse response is "the best we
% can do" and often meaningful.
% ********************************************
function [h, status] = LMSFIR_stage3_run(d)
1;
% mTaps is number of physical FIR stages
% m is number of polyphase coefficients
d.m = d.mStages * d.fInterp;
% sanity check... (orthogonality of wanted and unwanted component)
assert(sum(mask_inband) > 0, 'passband is empty');
'passband and stopband overlap');
% ********************************************
% start with flat passband signals at input and output of filter
% those will become the input to the LMS solver.
% ********************************************
sigSolverAtOutput_fd = sigSolverAtInput_fd;
% ********************************************
% for even-sized FFT length, there is one bin at the
% Nyquist limit that gives a [-1, 1, -1, 1] time domain
% waveform. It has no counterpart with opposite frequency
% sign and is therefore problematic (time domain component
% cannot be delayed).
% Don't assign any input power here.
% ********************************************
if mod(d.nSamples, 2) == 0
ixNyquistBin = floor(d.nSamples/2) + 1;
sigSolverAtInput_fd(ixNyquistBin) = 0;
sigSolverAtOutput_fd(ixNyquistBin) = 0;
end
if isfield(d, 'equalizeFreq_Hz')
% ********************************************
% Filter equalizes a given passband frequency response
% ********************************************
if isfield(d, 'equalizeGain_dB')
cgain = 10 .^ (equalizeGain_dB / 20);
else
cgain = d.equalizeComplexGain;
end
d.Heq = evaluateFilter(d.fb, d.equalizeFreq_Hz, cgain, d.complexValued);
assert(isempty(find(isnan(d.Heq), 1)), ...
['equalizer frequency response interpolation failed. ' ...
'Please provide full range data for plotting, even if it does not ', ...
'affect the design']);
% ********************************************
% apply frequency response to input signal.
% The LMS solver will invert this response
% ********************************************
sigSolverAtInput_fd = sigSolverAtInput_fd .* d.Heq;
end
if isfield(d, 'nominalFreq_Hz')
% ********************************************
% (equalized) filter matches a given passband frequency response
% ********************************************
if isfield(d, 'nominalGain_dB')
cgain = 10 .^ (d.nominalGain_dB / 20);
else
cgain = d.nominalComplexGain;
end
d.Hnom = evaluateFilter(d.fb, d.nominalFreq_Hz, cgain, d.complexValued);
assert(isempty(find(isnan(d.Hnom), 1)), ...
['nominal frequency response interpolation failed. ' ...
'Please provide full range data for plotting, even if it does not ', ...
'affect the design']);
% ********************************************
% apply frequency response to output signal.
% The LMS solver will adapt this response
% ********************************************
sigSolverAtOutput_fd = sigSolverAtOutput_fd .* d.Hnom;
end
% ********************************************
% compensate constant group delay from equalizer and nominal
% frequency response. This isn't optimal, but it is usually
% a good starting point (use delayOffset parameter)
% ********************************************
[coeff, ref_shiftedAndScaled, deltaN] = fitSignal_FFT(...
ifft(sigSolverAtInput_fd), ifft(sigSolverAtOutput_fd));
% the above function also scales for best fit. This is not desired here, instead
% let the LMS solver match the gain. Simply scale it back:
ref_shifted = ref_shiftedAndScaled / coeff;
sigSolverAtOutput_fd = fft(ref_shifted);
if false
% ********************************************
% plot time domain waveforms (debug)
% ********************************************
figure(76); hold on;
plot(fftshift(abs(ifft(sigSolverAtOutput_fd))), 'k');
plot(fftshift(abs(ifft(sigSolverAtInput_fd))), 'b');
title('time domain signals');
legend('reference (shifted)', 'input signal');
end
% ********************************************
% main loop of the design algorithm
% => initialize weights
% => loop
% => design optimum LMS filter that transforms weighted input
% into weighted output
% => iterate
% ********************************************
% at this stage, the input to the algorithm is as follows:
% => errorSpec for in-band and out-of-band frequencies
% (masks are redundant, can be derived from above)
% => LMS_in_fd and
% => LMS_out_fd: Signals that are given to the LMS solver.
% Its task is: "design a FIR filter that transforms LMS_in_fd into LMS_out_fd".
% initialize weights
status = '? invalid ?';
hConv = [];
remConvIter = d.convergedIterations;
for iter=1:d.nIter
% inband weight is applied equally to both sides to shape the error
% out-of-band weight is applied to the unwanted signal
LMS_in_fd = sigSolverAtInput_fd .* inbandWeight...
LMS_out_fd = sigSolverAtOutput_fd .* inbandWeight;
% ********************************************
% cyclic time domain waveforms from complex spectrum
% ********************************************
LMS_in_td = ifft(LMS_in_fd);
LMS_out_td = ifft(LMS_out_fd);
% ********************************************
% construct FIR basis (output per coeffient)
% time domain waveforms, shifted according to each FIR tap
% ********************************************
basis = zeros(d.m, d.nSamples);
% introduce group delay target
ix1 = -d.m/2+0.5 + d.delayOffset;
ix2 = ix1 + d.m - 1;
rowIx = 1;
for ix = ix1:ix2 % index 1 appears at ix1
basis(rowIx, :) = FFT_delay(LMS_in_td, ix);
rowIx = rowIx + 1;
end
if d.complexValued
rightHandSide_td = LMS_out_td;
else
% use real part only
basis = [real(basis)];
rightHandSide_td = [real(LMS_out_td)];
pRp = real(rightHandSide_td) * real(rightHandSide_td)' + eps;
pIp = imag(rightHandSide_td) * imag(rightHandSide_td)';
assert(pIp / pRp < 1e-16, ...
['got an imaginary part where there should be none. ', ...
'uncomment the following lines, if needed']);
% if designing a real-valued equalizer for a complex-valued frequency response,
% use the following to solve LMS over the average:
% basis = [real(basis) imag(basis)];
% rightHandSide_td = [real(LMS_out_td), imag(LMS_out_td)];
end
% ********************************************
% LMS solver
% find a set of coefficients that scale the
% waveforms in "basis", so that their sum matches
% "rightHandSide_td" LMS-optimally
% ********************************************
pbasis = pinv(basis .');
h = transpose(pbasis * rightHandSide_td .');
% pad impulse response to n
irIter = [h, zeros(1, d.nSamples-d.m)];
% undo the nominal group delay
irIter = FFT_delay(irIter, ix1);
HIter = fft(irIter);
% ********************************************
% filter test signal
% ********************************************
eq_fd = sigSolverAtInput_fd .* HIter;
% ********************************************
% subtract actual output from targeted output
% results in error spectrum
% ********************************************
err_fd = sigSolverAtOutput_fd - eq_fd;
err_fd = err_fd .* mask_inband; % only in-band matters
EVM_dB = 20*log10(abs(err_fd)+1e-15);
% ********************************************
% out-of-band leakage
% ********************************************
leakage_dB = 20*log10(abs(HIter .* mask_outOfBand + 1e-15));
% ********************************************
% compare achieved and targeted performance
% ********************************************
deltaLeakage_dB = leakage_dB - d.errorSpecBinVal_outOfBand_dB;
deltaEVM_dB = EVM_dB - d.errorSpecBinVal_inband_dB;
% ********************************************
% find bins where performance should be improved
% or relaxed
% ********************************************
ixImprLeakage = find(deltaLeakage_dB > 0);
ixImprEVM = find(deltaEVM_dB > 0);
ixRelLeakage = find(deltaLeakage_dB < -3);
ixRelEVM = find(deltaEVM_dB < -3);
status = 'iteration limit reached';
if isempty(ixImprLeakage) && isempty(ixImprEVM)
% both targets met. Convergence!
if remConvIter > 0
remConvIter = remConvIter - 1;
status = 'converged once, now trying to improve';
hConv = h;
else
status = 'converged';
break;
end
end
% ********************************************
% improve / relax in-band and out-of-band
% ********************************************
if ~isempty(ixImprLeakage)
% tighten out-of-band
outOfBandWeight(ixImprLeakage) = outOfBandWeight(ixImprLeakage)...
.* 10 .^ ((deltaLeakage_dB(ixImprLeakage) + 0.1) / 20);
end
if ~isempty(ixRelLeakage)
% relax out-of-band
outOfBandWeight(ixRelLeakage) = outOfBandWeight(ixRelLeakage)...
.* 10 .^ (deltaLeakage_dB(ixRelLeakage) / 3 / 20);
end
if ~isempty(ixImprEVM)
% tighten in-band
inbandWeight(ixImprEVM) = inbandWeight(ixImprEVM)...
.* 10 .^ ((deltaEVM_dB(ixImprEVM) + 0.01) / 20);
end
if ~isempty(ixRelEVM)
% relax in-band
inbandWeight(ixRelEVM) = inbandWeight(ixRelEVM)...
.* 10 .^ (deltaEVM_dB(ixRelEVM) / 2 / 20);
end
if max([inbandWeight, outOfBandWeight] > d.abortWeight)
status = 'weight vector is diverging';
break;
end
end % for iter
% ********************************************
% recover from convergence failure after convergence
% during improvement phase
% ********************************************
if ~strcmp(status, 'converged')
if ~isempty(hConv)
h = hConv;
status = 'converged';
end
end
if true
% ********************************************
% plot impulse response
% ********************************************
if d.complexValued
figure(); hold on;
stem(real(h), 'k');
stem(imag(h), 'b');
legend('real(h)', 'imag(h)');
else
figure(); hold on;
stem(h);
legend('h');
end
title('impulse response');
end
% ********************************************
% plot frequency response
% ********************************************
d=doPlotStart(d, ['Frequency response (Status:', status, ')']);
d=doPlotH(d, HIter, 'b', '|H_{design}(f)|', 2);
handle = plot(d.fb(outOfBandBins), d.errorSpecBinVal_outOfBand_dB(outOfBandBins), 'b+');
set(handle, 'markersize', 2);
d = doPlot_dB(d, EVM_dB, 'r', 'error');
handle = plot(d.fb(inbandBins), d.errorSpecBinVal_inband_dB(inbandBins), 'r+');
set(handle, 'markersize', 2);
d=doPlotEnd(d);
ylim([-100, 10]);
if false
% ********************************************
% plot constraint signal and weights
% ********************************************
figure(31); grid on; hold on;
set(handle, 'lineWidth', 3);
x = d.fb; y = 20*log10(inbandWeight / max(inbandWeight));
handle = plot(x(inbandBins), y(inbandBins), 'k+'); set(handle, 'lineWidth', 3);
x = d.fb;
y = 20*log10(outOfBandWeight / max(outOfBandWeight));
handle = plot(x(outOfBandBins), y(outOfBandBins), 'b+'); set(handle, 'lineWidth', 3);
xlabel('f/Hz'); ylabel('dB');
ylim([-80, 40]);
legend('constraint signal', 'in-band weight', 'out-of-band weight');
title('weighting factor (normalized to 0 dB)');
end
hasEq = isfield(d, 'Heq');
hasNom = isfield(d, 'Hnom');
if hasEq || hasNom
% ********************************************
% plot equalization / nominal target
% ********************************************
d=doPlotStart(d, 'equalization / nominal target');
d=doPlotH(d, HIter, 'b', '|H_{design}(f)|', 2);
if hasEq
d=doPlotH(d, d.Heq, 'k', '|H_{eq}(f)| to equalize (invert)');
eqR = HIter .* d.Heq;
d=doPlotH(d, eqR, 'c', '|H_{design}(f)H_{eq}(f)|', 2);
handle = plot(d.fb(inbandBins), ...
20*log10(abs(eqR(inbandBins)) + 1e-15), 'c*');
set(handle, 'markersize', 3);
end
if hasNom
d = doPlotH(d, d.Hnom, 'g', '|H_{nom}|', 2);
handle = plot(d.fb(inbandBins), ...
20*log10(abs(HIter(inbandBins)) + 1e-15), 'b*');
set(handle, 'markersize', 3);
end
d=doPlotEnd(d);
% set y-range
ymax = 20*log10(max(abs(HIter)));
ylim([-50, ymax+3]);
end
end
% === LMSFIR helper functions ===
% evaluates frequency response f_dB; g_Hz at fb
% the return value will contain NaN for out-of-range entries
% in fb
function binVal = buildBinVal(varargin)
p = varargin2struct(varargin);
f_Hz = p.f_Hz;
g_dB = p.g_dB;
% shorthand notation f = [f1, f2]; g = -30;
if isscalar(g_dB)
g_dB = ones(size(f_Hz)) * g_dB;
end
% tolerate sloppy two-argument definition
if size(f_Hz, 2) == 2 && f_Hz(1) > f_Hz(2)
f_Hz = fliplr(f_Hz);
g_dB = fliplr(g_dB);
end
binVal = interp1(f_Hz, g_dB, p.fbUser, 'linear');
end
function d = placeBand(d, varargin)
p = varargin2struct(varargin);
% create requirements vector
binVal = buildBinVal('f_Hz', p.f_Hz, ...
'g_dB', p.g_dB, ...
'fbUser', d.fbUser);
% look up requirements vector from design object
switch p.type
case 'passband'
fn = 'errorSpecBinVal_inband_dB';
case 'stopband'
fn = 'errorSpecBinVal_outOfBand_dB';
otherwise
assert(false);
end
designObject_binVal = d.(fn);
% check overlap
if strcmp(p.onOverlap, 'error')
m1 = NaN_to_0_else_1(designObject_binVal);
m2 = NaN_to_0_else_1(binVal);
assert(isempty(find(m1 .* m2, 1)), ...
['newly declared band overlaps existing band, '...
'which was explicitly forbidden by onOverlap=error']);
p.onOverlap = 'tighten'; % there won't be overlap,
% merging is dummy operation
end
% merging rules
switch p.onOverlap
case 'tighten'
logicOp = 'or';
valueOp = 'min';
case 'relax'
logicOp = 'or';
valueOp = 'max';
case 'tightenExisting'
logicOp = 'and';
valueOp = 'min';
case 'relaxExisting'
logicOp = 'and';
valueOp = 'max';
otherwise
assert(false);
end
% merge requirements tables
binValMerged = mergeBinVal(...
'binVal1', designObject_binVal, ...
'binVal2', binVal, ...
'logicalOperator', logicOp, ...
'valueOperator', valueOp);
% assign new requirements table
d.(fn) = binValMerged;
end
function r = NaN_to_0_else_1(vec)
r = zeros(size(vec));
% logical indexing, instead of r(find(~isnan(vec))) = 1;
r(~isnan(vec)) = 1;
end
function binVal = mergeBinVal(varargin)
p = varargin2struct(varargin);
% region where first argument is defined
% region where second argument is defined
% region where result will be defined
switch(p.logicalOperator)
case 'or'
case 'and'
otherwise
assert(false);
end
% merge into result
binVal = zeros(size(p.binVal1)) + NaN;
switch(p.valueOperator)
case 'min'
% note: The function min/max ignore NaNs (see "min" man
% page in Matlab)
% if one entry is NaN, the other entry will be returned
binVal(ix) = min(p.binVal1(ix), p.binVal2(ix));
case 'max'
binVal(ix) = max(p.binVal1(ix), p.binVal2(ix));
otherwise
assert(false);
end
end
% evaluates [f / gain] filter specification on the frequency grid
function H = evaluateFilter(f_eval, modelF, modelH, complexValued)
oneSided = false;
if ~complexValued
oneSided = true;
else
if min(modelF) > min(f_eval)
disp(['Warning: Filter model does not contain (enough) negative frequencies. ', ...
'assuming symmetric H(f) / real-valued h(t)']);
oneSided = true;
end
end
if oneSided
f_evalOrig = f_eval;
f_eval = abs(f_eval);
end
H = interp1(modelF, modelH, f_eval, 'linear');
if oneSided
% enforce symmetry (=> real-valued impulse response)
logicalIndex = (f_evalOrig < 0);
H(logicalIndex) = conj(H(logicalIndex));
end
end
function [d, handle] = doPlotH(d, H, spec, legEntry, linewidth)
handle = plot(fftshift(d.fb), fftshift(20*log10(abs(H)+1e-15)), spec);
if exist('linewidth', 'var')
set(handle, 'lineWidth', linewidth);
end
end
function [d, handle] = doPlot_dB(d, H, spec, legEntry, linewidth)
handle = plot(fftshift(d.fb), fftshift(H), spec);
d.legList{size(d.legList, 2) + 1} = legEntry;
if exist('linewidth', 'var')
set(handle, 'lineWidth', linewidth);
end
end
function d = doPlotStart(d, plotTitle)
figure(d.nextPlotIx);
title(plotTitle);
grid on; hold on;
d.nextPlotIx = d.nextPlotIx + 1;
d.legList = {};
end
function d = doPlotEnd(d)
legend(d.legList);
xlabel('f/Hz');
ylabel('dB');
end
d.legList{size(d.legList, 2) + 1} = legEntry;
end
% === general-purpose library functions ===
% handling of function arguments
% someFun('one', 1, 'two', 2, 'three', 3) => struct('one', 1, 'two', 2, 'three', 3)
% a struct() may appear in place of a key ('one') and gets merged into the output.
function r = varargin2struct(arg)
assert(iscell(arg));
switch(size(arg, 2))
case 0 % varargin was empty
r=struct();
case 1 % single argument, wrapped by varargin into a cell list
r=arg{1}; % unwrap
assert(isstruct(r));
otherwise
r=struct();
% iterate through cell elements
ix=1;
ixMax=size(arg, 2);
while ix <= ixMax
e=arg{ix};
if ischar(e)
% string => key/value. The next field is a value
ix = ix + 1;
v = arg{ix};
r.(e) = v;
elseif isstruct(e)
names = fieldnames(e);
assert(size(names, 2)==1); % column
for ix2 = 1:size(names, 1)
k = names{ix2};
v = e.(k);
r.(k) = v;
end
else
disp('invalid token in vararg handling. Expecting key or struct. Got:');
disp(e);
assert(false)
end
ix=ix+1;
end % while
end % switch
end
function sig = FFT_delay(sig, nDelay)
sig = fft(sig); % to frequency domain
nSigSamples = size(sig, 2);
binFreq=(mod(((0:nSigSamples-1)+floor(nSigSamples/2)), nSigSamples)-floor(nSigSamples/2));
phase = -2*pi*nDelay / nSigSamples .* binFreq;
rot = exp(1i*phase);
if mod(nSigSamples, 2)==0
% even length - bin at Nyquist limit
rot(nSigSamples/2+1)=cos(phase(nSigSamples/2+1));
end
sig = sig .* rot;
sig = ifft(sig); % to time domain
end
% *******************************************************
% delay-matching between two signals (complex/real-valued)
%
% * matches the continuous-time equivalent waveforms
% of the signal vectors (reconstruction at Nyquist limit =>
% ideal lowpass filter)
% * Signals are considered cyclic. Use arbitrary-length
% zero-padding to turn a one-shot signal into a cyclic one.
%
% * output:
% => coeff: complex scaling factor that scales 'ref' into 'signal'
% => delay 'deltaN' in units of samples (subsample resolution)
% apply both to minimize the least-square residual
% => 'shiftedRef': a shifted and scaled version of 'ref' that
% matches 'signal'
% => (signal - shiftedRef) gives the residual (vector error)
%
% Example application
% - with a full-duplex soundcard, transmit an arbitrary cyclic test signal 'ref'
% - record 'signal' at the same time
% - extract one arbitrary cycle
% - run fitSignal
% - deltaN gives the delay between both with subsample precision
% - 'shiftedRef' is the reference signal fractionally resampled
% and scaled to optimally match 'signal'
% - to resample 'signal' instead, exchange the input arguments
% *******************************************************
function [coeff, shiftedRef, deltaN] = fitSignal_FFT(signal, ref)
n=length(signal);
% xyz_FD: Frequency Domain
% xyz_TD: Time Domain
% all references to 'time' and 'frequency' are for illustration only
forceReal = isreal(signal) && isreal(ref);
% *******************************************************
% Calculate the frequency that corresponds to each FFT bin
% [-0.5..0.5[
% *******************************************************
binFreq=(mod(((0:n-1)+floor(n/2)), n)-floor(n/2))/n;
% *******************************************************
% Delay calculation starts:
% Convert to frequency domain...
% *******************************************************
sig_FD = fft(signal);
ref_FD = fft(ref, n);
% *******************************************************
% ... calculate crosscorrelation between
% signal and reference...
% *******************************************************
u=sig_FD .* conj(ref_FD);
if mod(n, 2) == 0
% for an even sized FFT the center bin represents a signal
% [-1 1 -1 1 ...] (subject to interpretation). It cannot be delayed.
% The frequency component is therefore excluded from the calculation.
u(length(u)/2+1)=0;
end
Xcor=abs(ifft(u));
% figure(); plot(abs(Xcor));
% *******************************************************
% Each bin in Xcor corresponds to a given delay in samples.
% The bin with the highest absolute value corresponds to
% the delay where maximum correlation occurs.
% *******************************************************
integerDelay = find(Xcor==max(Xcor));
% (1): in case there are several bitwise identical peaks, use the first one
% Minus one: Delay 0 appears in bin 1
integerDelay=integerDelay(1)-1;
% Fourier transform of a pulse shifted by one sample
rotN = exp(2i*pi*integerDelay .* binFreq);
uDelayPhase = -2*pi*binFreq;
% *******************************************************
% Since the signal was multiplied with the conjugate of the
% reference, the phase is rotated back to 0 degrees in case
% of no delay. Delay appears as linear increase in phase, but
% it has discontinuities.
% Use the known phase (with +/- 1/2 sample accuracy) to
% rotate back the phase. This removes the discontinuities.
% *******************************************************
% figure(); plot(angle(u)); title('phase before rotation');
u=u .* rotN;
% figure(); plot(angle(u)); title('phase after rotation');
% *******************************************************
% Obtain the delay using linear least mean squares fit
% The phase is weighted according to the amplitude.
% This suppresses the error caused by frequencies with
% little power, that may have radically different phase.
% *******************************************************
weight = abs(u);
constRotPhase = 1 .* weight;
uDelayPhase = uDelayPhase .* weight;
ang = angle(u) .* weight;
r = [constRotPhase; uDelayPhase] .' \ ang.'; %linear mean square
%rotPhase=r(1); % constant phase rotation, not used.
% the same will be obtained via the phase of 'coeff' further down
fractionalDelay=r(2);
% *******************************************************
% Finally, the total delay is the sum of integer part and
% fractional part.
% *******************************************************
deltaN = integerDelay + fractionalDelay;
% *******************************************************
% provide shifted and scaled 'ref' signal
% *******************************************************
% this is effectively time-convolution with a unit pulse shifted by deltaN
rotN = exp(-2i*pi*deltaN .* binFreq);
ref_FD = ref_FD .* rotN;
shiftedRef = ifft(ref_FD);
% *******************************************************
% Again, crosscorrelation with the now time-aligned signal
% *******************************************************
coeff=sum(signal .* conj(shiftedRef)) / sum(shiftedRef .* conj(shiftedRef));
shiftedRef=shiftedRef * coeff;
if forceReal
shiftedRef = real(shiftedRef);
end
end`````` | 30,181 | 112,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.320624 |
https://www.vipdue.com/%E8%AE%A1%E7%AE%97%E6%9C%BA%E8%A7%86%E8%A7%89%E4%BB%A3%E5%86%99%EF%BD%9Ceecs-442-problem-set-9-panoramic-stitching | 1,715,990,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00719.warc.gz | 982,889,813 | 8,304 | # 计算机视觉代写｜EECS 442 Problem Set 9: Panoramic Stitching
Problem 9.1 Panoramic Stitching
In this problem we will develop an algorithm for stitching a panorama from overlapping
photos (Figure 1), which amounts to estimating a transformation that aligns one image to
another. To do this, we will compute ORB features1 in both images and match them to obtain
correspondences. We will then estimate a homography from these correspondences, and we’ll
use it to stitch the two images together in a common coordinate system.
In order to get an accurate transformation, we will need many accurate feature matches.
Unfortunately, feature matching is a noisy process: even if two image patches (and their ORB
descriptors) look alike, they may not be an actual match.
To make our algorithm robust to matching errors, we will use RANSAC, a method for
estimating a parametric model from noisy observations. We will detect keypoints and represent
descriptors using ORB. We will then match features, using heuristics to remove bad matches.
We have provided you with two images (Figure 1) that you’ll use to create a panorama.
(a) You will start by computing features and image correspondences.
• (2 points) Implement get orb features(img) to compute orb features for both
of the given image.
Figure 1: Panorama produced using our implementation. The image pair shown on the left represents
the keypoints in the two source images and below them are the predicted feature correspondences. On
the right is the stitched panorama.
• (2 points) Implement match keypoints(desc1, desc2) to compute keypoint
correspondences between the two source images using the ratio test. Run the
plotting code to visualize the detected features and resulting correspondences.
(b) (2 points) Write a function find homography(pts1, pts2) that takes in two N × 2
matrices with the x and y coordinates of matching 2D points in the two images and
computes the 3 × 3 homography H that maps pts1 to pts2. You can implement this
function using nonlinear least squares (or, alternatively, the direct linear transform).
Hint: For nonlinear least squares, we reccomend using scipy library’s built-in nonlinear
least squares. To use that function, you need to:
• Define a cost function, f(h; pts1; pts2), that calculates the projection error (a
vector of length 2N) between pts1 and projected pts2 using homography H
as pts1 − cart(H ∗ homog(pts2)). Here homog(x) converts x into homogeneous
coordinates, cart(x) converts x to cartesian coordinates and h is the flattened
version of the homography H to be estimated.2
• Provide an initial guess for the homography h: a length 9 vector filled with ‘1’s
should be good enough.
(c) (2 points) Your homography-fitting function from (b) will only work well if there are no
mismatched features. To make it more robust, implement a function transform ransac(pts1,
pts2) that fits a homography using RANSAC. You can call find homography(pts1,
pts2) inside the inner loop of RANSAC. You will also be responsible for figuring out
the set of parameters to use to produce the best results.
(d) (2 points) Write a function panoramic stitching(img1, img2) that produces a
panorama from a pair of overlapping images using your functions from the previous
parts. Run the algorithm on the two images provided. You result should be similar to
that of Figure 1. | 784 | 3,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.838067 |
syhiretetuxy.timberdesignmag.com | 1,620,483,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.7/warc/CC-MAIN-20210508121446-20210508151446-00561.warc.gz | 617,730,246 | 6,060 | # Write an equation in slope intercept form from a graph worksheet
The average distance value is given in astronomical units where 1 a. The orbital period is given in units of earth-years where 1 earth year is the time required for the earth to orbit the sun - 3. Kepler's third law provides an accurate description of the period and distance for a planet's orbits about the sun.
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• Gradient Slope Intercept Form | Passy's World of Mathematics
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Key Questions The students have started graphing lines given tables and linear functions. They have also learned how to write linear equations in slope intercept form. Graphing Lines in Slope-Intercept Form Worksheets Using the Point-Slope Form of a Line Another way to express the equation of a straight line Point-slope refers to a method for graphing a linear equation on an x-y axis. When graphing a linear equation, the whole idea is to take pairs of x's and y's and plot them on the graph. Algebra Help - Calculators, Lessons, and Worksheets | Wyzant Resources GO Algebra Help This section is a collection of lessons, calculators, and worksheets created to assist students and teachers of algebra. Here are a few of the ways you can learn here Learn why the Common Core is important for your child Using the Point-Slope Form of a Line Another way to express the equation of a straight line Point-slope refers to a method for graphing a linear equation on an x-y axis. Free worksheets for graphing linear equations & finding the slope and equation of a line Writing a linear equation An algebraic equation in which each element or term is either a constant or the product of the first power of a single variable and a constant is called a linear equation. These equations can have one or more variables.
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## Example worksheets
This Linear Equations Worksheet will produce problems for practicing graphing lines in slope-intercept form. This equation of the line is called “slope-intercept” form because it easily shows both the slope and the intercept of the line.
To find the equation of a line given the slope and intercept, simply plug into the equation. Example 1 Write the equation of the line with slope 2 that has y-intercept 5. y = mx + b Write the slope -intercept formula. How do you write an equation in slope intercept form given that the line contains the point (-3,0) and is parallel to the line defined by -5x=6y? | 1,670 | 8,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-21 | latest | en | 0.860394 |
https://www.iloveit.net/tool/pc-to-cm/ | 1,713,833,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00271.warc.gz | 708,729,095 | 29,539 | # PC to CM
Converting parsecs to centimeters is based on a conversion factor, 3.0856775814914E+18.
One parsec is eqivalent to 3.0856775814914E+18 centimeter, i.e.
1 parsec = 3.0856775814914E+18 centimeter
The distance x in centimeter (cm) is equivalent to the distance y in parsec (pc) multiplied by 3.0856775814914E+18.
Hope you will enjoy the tool! Don’t forget to hit the heart above and give it some love.
## Parsec to Centimeter Conversion Calculator
pc is cm
You can use our calculator to easily convert between different units and below find a conversion chart and examples for how to convert them.
## Parsec to Centimeter Conversion Table
pccm
1 pc3.0856775814914E+18 cm
2 pc6.1713551629827E+18 cm
3 pc9.2570327444741E+18 cm
4 pc1.2342710325965E+19 cm
5 pc1.5428387907457E+19 cm
6 pc1.8514065488948E+19 cm
7 pc2.159974307044E+19 cm
8 pc2.4685420651931E+19 cm
9 pc2.7771098233422E+19 cm
10 pc3.0856775814914E+19 cm
20 pc6.1713551629827E+19 cm
30 pc9.2570327444741E+19 cm
40 pc1.2342710325965E+20 cm
50 pc1.5428387907457E+20 cm
60 pc1.8514065488948E+20 cm
70 pc2.159974307044E+20 cm
80 pc2.4685420651931E+20 cm
90 pc2.7771098233422E+20 cm
100 pc3.0856775814914E+20 cm
200 pc6.1713551629827E+20 cm
300 pc9.2570327444741E+20 cm
400 pc1.2342710325965E+21 cm
500 pc1.5428387907457E+21 cm
600 pc1.8514065488948E+21 cm
700 pc2.159974307044E+21 cm
800 pc2.4685420651931E+21 cm
900 pc2.7771098233422E+21 cm
1000 pc3.0856775814914E+21 cm
The distance x in centimeter (cm) is equivalent to the distance y in parsec (pc) multiplied by 3.0856775814914E+18.
Also check out our list of length converters and the complete list of conversion tools.
Disclaimer – We have done our best to give you a tool that is both fast, easy to use and reliable. However we can not be held responsible for any damage of any kind that arise from using the calculations found on our site. Use it on your own risk. If you find any calculation that does seem strange do not hesitate to contact us directly. Our contact information is on the bottom of the page. | 704 | 2,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.718275 |
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# Laws of Motion (Part - 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
## DC Pandey (Questions & Solutions) of Physics: NEET
Created by: Ciel Knowledge
## NEET : Laws of Motion (Part - 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
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Introductory Exercise 5.5
Ques 1: In figure m1 = 1 kg and m2 = 4 kg. Find the mass M o f the hanging block which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction less and the strings and the pulleys are light
Note In exercises 2 to 4 the situations described take place in a box car which has initial velocity v = 0 but acceleration
Ans: 6 .83 kg
Sol: Block on triangular block will not slip if
m1a cos θ= m1g sin θ
i.e., a = g tanθ …(i)
N = m1g cosθ + m1asinθ …(ii)
For the movement of triangular block
T - N sinθ = m2α …(iii)
For the movement of the block of mass M
Mg - T = Ma …(iv)
Mg - N sinθ = ( m2 + M )α
Substituting the value of N from Eq. (ii) in the above equation
Mg - (m1g cosθ + m1α sinθ) sinθ
=( m2α + Mα)
i.e., M (g - α) = m1g cosθ sinθ + (m2 + m1 sin2θ)α
Substituting value of a from Eq. (i) in the above equation,
M(1 - tanθ) = m1 cosθ sinθ + ( m2 + m1 sin2 θ) tanθ
Substituting θ = 30° , m1 = 1 kg and m2 = 4 kg
= 6.82 kg
Ques 2: A 2 kg object is slid along the friction less floor with initial velocity (10 m/s)î (a) Describe the motion of the object relative to car (b) when does the object reach its original position relative to the box car.
Ans:
(a) x = x0 + 10t - 2.5t2 , v = 10 - 5t (b) t = 4s
Sol: (a) Using
Displacement of block at time t relative to car would be
Velocity of block at time t (relative to car) will be
(b) Time (t) for the block to arrive at the original position (i.e., x = x0) relative to car
x0 = x+ 10 t - 2.5 t2
⇒ t = 4s
Ques 3: A 2 kg object is slid along the friction less floor with initial transverse velocity (10 m/s). Describe the motion (a) in car’s frame (b) in ground frame.
Ans:
(a) x = x0 - 2.5t2, z = z0 + 10t, vx = - 5t, vz = 10 ms-1
(b) x = x0, z = z0 + 10t, vx = 0, vz = 10 ms-1
Sol: (a) In car’s frame position of object at time t would be given by
In car’s frame
x = x0 + 0*t + 1/2(-5)t2
i.e., x = x0 - 2.5 t2 …(i)
and z = z + 10t …(ii)
Velocity of the object at time t would be
and
(b) In ground frame the position of the object at time t would be given by
In ground frame
x = x0
and z = z0 + 10t
Velocity of the object at time t would be
ans
Ques 4: A 2 kg object is slid along a rough floor (coefficient of sliding friction = 0.3) with initial velocity (10 m/s)î. Describe the motion of the object relative to car assuming that the coefficient of static friction is greater than 0.5.
Ans:
x = x0 + 10f - 4t2 , K = 10 - 8 t for 0<f<1.25 s object stops at t = 1.25 s and remains at rest relative to car.
Sol: m = 2 kg
Normal force on object = mg
Maximum sliding friction = μsmg
= 0.3*2*10 = 6 N
Deceleration due to friction = 6/2 =3 m/s2
Deceleration due to pseudo force = 5 m/s2
∴ Net deceleration = (3 + 5) m/s2
= 8 m/s2
∴ Displacement of object at any time t (relative to car)
Thus, velocity of object at any time t (relative to car)
The object will stop moving relative to car when
10 - 8t = 0 i.e., t = 1.25s
∴ vx = 10 - 8t for 0<t<1.25 s
Ques 5: A block is placed on an inclined plane as shown in figure. What must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at
Ans:
9/25 mg
Sol: For block not to slide the frictional force ( f ) would be given by
f + ma cosθ = mg sinθ
or
f = mg sinθ - ma cosθ
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# Nursing Entrance Exam Study
Guide
## 2304 E. Busch Blvd.
Tampa, FL 33612
TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
## Welcome to Medical Prep Institute of Tampa Bay! .................................................................................... 4
What is the Healthcare Delivery System? ................................................................................................... 6
Key Words to Remember ............................................................................................................................. 6
The Tiered System ........................................................................................................................................ 6
Components of Healthcare Delivery Systems ............................................................................................. 8
The Nursing Program.................................................................................................................................. 10
Nursing Entrance Examination .............................................................................................................. 11
English ................................................................................................................................................. 12
Language Practice ........................................................................................................................... 41
Grammar Practice Questions ......................................................................................................... 51
Word Usage Practice ...................................................................................................................... 58
Sentence Practice ........................................................................................................................... 65
Math Section ....................................................................................................................................... 68
Number Types ................................................................................................................................ 69
Percent ............................................................................................................................................ 69
Average ........................................................................................................................................... 70
Weighted Average .......................................................................................................................... 71
Average Speed ................................................................................................................................ 71
Properties of Signed Numbers ....................................................................................................... 71
Factoring ......................................................................................................................................... 71
Probability ...................................................................................................................................... 72
Geometric Figures .......................................................................................................................... 72
Geometric Skills and Concepts ....................................................................................................... 73
Volume............................................................................................................................................ 78
Coordinate Geometry..................................................................................................................... 79
Algebra ............................................................................................................................................ 79
Algebra part 2 ................................................................................................................................. 86
Basic Operations Practice............................................................................................................... 92
Exponent Practice........................................................................................................................... 95
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 2 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Fractions and Square Root Practice ............................................................................................... 97
Geometry Practice ........................................................................................................................ 100
Averages and Rounding Practice Questions ................................................................................ 102
Science Section ...................................................................................................................................... 106
The Human Body .......................................................................................................................... 107
Types of Anatomical Studies ........................................................................................................ 111
Practice Questions and Review.................................................................................................... 119
Other Practice Questions ............................................................................................................. 120
Practice Questions ........................................................................................................................ 124
Review Questions ......................................................................................................................... 128
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 3 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Welcome to Medical Prep Institute of Tampa Bay!
The office hours are from 8:30am 7:00pm, Monday through Friday. The Administrative Offices are not open on
weekends.
Appointments
Should you require a meeting with the staff, an appointment is recommended. To make an appointment, please
call our Admissions Offices at 813-932 1710
## Students Point of Contact
Academic Programs (Scheduling, Withdrawal, Transcripts...........Registrar Office
Financial Billing..............................................................................Financials Services
Job Placement Assistance............................................................. Student and Career Services
Clinical Scheduling . ........................................................... Clinical Department
Medical prep Institute of Tampa bay makes every effort to ensure the accuracy of the information contained in this
catalog. The Institution reserves the right to change policies, regulations, fees and courses during this catalog
period upon direction from the administration and the Dean of Academics of the School. The most current and
complete information is available from the Campus President. All information contained in this catalog is current
and correct as of the publication date: July 2015
Any concerns regarding the Medical Prep Institute of Tampa bay or this catalog and that have not been
satisfactorily answered by the institution may be directed to:
Commission for Independent Education
325 West Gaines Street Suite 1414
Tallahassee FL 32399
(888)-224-6684
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 4 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
History and Ownership
Medical Prep Institute of Tampa bay is owned and operated by a corporation called F. Jenar Inc and is authorized to
do business in the State of Florida. The Company is run and managed by the Administrator, Jena Fadziso. Medical
prep Institute of Tampa bay was founded in 2008 and began as Test Preparation School and Continuing Medical
Education provider for healthcare professionals. Soon after, the company recognized there was a need in the
community to provide affordable high quality medical training education for individuals seeking entry level careers
into the healthcare industry. Medical prep Institute of Tampa bay made the decision to transition into an institute
of higher learning and officially began offering diploma and degree program in 2010. In August 2015 Medical Prep
Institute of Tampa Bay was Accredited by Accrediting Bureau of Health Education Schools (ABHES)
## Description of School Facility
Medical prep Institute of Tampa bay is located at 2304 E. Busch Blvd in Tampa, Florida between Nebraska and 30th
Street just east of 1-275. The institute is conveniently situated along the city bus line. The facility occupies about
8000 square feet and includes a reception area, business and administrative offices, student break area, four (4)
classrooms, a laboratory classroom, a Library and computer lab. The campus is equipped with computers for
student use located in the Library. Projectors are installed in each of the classrooms to assist in lectures and video
demonstrations. The lab areas consist of hospital beds, mannequins, charts and other medical supplies to facilitate
the hands on instruction found in each of the programs. Students will also find that the campus is handicap
accessible.
Values
Medical prep Institute of Tampa bay prides itself on the following values: Quality, Simplicity, Affordability,
Inclusiveness and Excellence!
Mission Statement
Medical Prep Institute is committed to increasing the quality of health care education in the greater Tampa Bay
area and beyond by developing in our students the knowledge, skills and professionalism required of todays
workforce. We are committed to providing our students with an affordable and comprehensive education. It is our
belief that this commitment to excellence will empower our students with the fundamental skills necessary to
either gain entry level employment in their chosen field or allow them the opportunity to advance in their current
career.
## Agency Location Phone Number
LICENSED-Commission for Independent 325 West Gaines Street Suite 1414 (888)-224-6684
Education Tallahassee FL 32399
APPROVED- Florida Board of Nursing 4052 Bald Cypress Way Bin C-02 Tallahassee, FL (850)-488-0595
32399
## APPROVED -American Safety and Health 1450 Westec Drive (800)-447.3177
Institute (Providing Continuing Education) Eugene, Oregon 97402
APPROVED -National Healthcare Association 11161 Overbrook Road Leawood, Kansas 66211 (800)-499-9092
(Certified National Testing Site)
ACCREDIATED -Accrediting Bureau of Health 7777 Leesburg Pike, Suite 314 N. Falls Church, VA (703) 917-9503
Education Schools (ABHES)
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 5 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Healthcare Industry
## What is the Healthcare Delivery System?
A network of Agencies, facilities and providers designed to work together coherently in a specified geographic
area. Healthcare workers operate within this system. Healthcare professionals operate within this system.
## The US healthcare System
Secondary care
Concerns with treatment of disorders requiring specialist opinion or hospitalization. The patients are usually
referred from Primary care and the physicians are affiliated to a hospital or a group practice.
Tertiary care
Provides medical and/or surgical management of complex disorders in an inpatient setting and usually
requiring collaboration between multiple specialties. These are super-specialized standalone hospitals or
specialty departments in a multi-specialty hospital
## Regulation of United Healthcare System
Department of Health and Human Services (HHS): The principal agency in the United States government for
protecting the health and safety of all Americans and for providing essential human services, especially for those
people who are least able to help themselves. The agency overseas Public Health service agencies (PHS) at both
local and state levels. These agencies constantly oversee public health matters.
## Key Words to Remember
Health this is the physical, mental and social wellbeing of human being and the absence of illness.
Medicine-Art and science of diagnosis, treatment, prevention of disease and the maintains of good health
Environment- The setting for the nurse (healthcare worker) patient interaction.
## The Tiered System
The USA Healthcare system is divided into the tiered system. This system is made up of regionalized systems of
healthcare delivery divided into Primary care, Secondary care and Tertiary care.
Primary care
Refers to the activities concerned with prevention and treatment of common medical problems in outpatient
setting.
## Wellness illness Continuum-This is the range of the persons total health.
The persons range of health is always changing thus the persons range on that continuum is constantly changing.
The persons range of health and environment determines the services the patient receives and the healthcare
workers provide.
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 6 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Wellness illness continuum
Maintaining ones continued health in the wellness-illness continuum is important.
Balance is an essential aspect of maintaining ones health.
Factors affecting the balance of Healthcare
Age
Sex
Family relationships
Cultural influences
Economic Status
Healthcare workers must consider all the above factors when providing healthcare to the consumer. The
comprehensive approach to health is known as Holistic Health Care
## MASLOW HEIRACHY OF NEEDS
In 1943, Abraham Maslow described the Theory of Human Needs Hierarchy of Needs
In order to maintain ones health in the wellness-illness continuum individuals work towards meeting
individual needs starting with basic physiological needs
To achieve the highest level, a person must work toward meeting each level
Healthcare workers also work hard to give care from a humanist approach by meeting the needs of each
individual patient
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 7 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Components of Healthcare Delivery Systems
Providers (Facilities and Healthcare Workers)
Payers
Clients
Healthcare Providers are Healthcare Facilities where healthcare services are conducted and Healthcare Workers
who provide caregiving services
Healthcare facilities are divided into inpatient (non-ambulatory) and outpatient facilities (ambulatory) All this
facilities support all three levels of healthcare ( primary , secondary and tertiary system).There are variety of
healthcare facilities and with a variety of Healthcare profession
## TYPES OF HEALTHCARE PROVIDERS AND SERVICES
1.Acute Care commonly known facility in the healthcare industry with patients having serious condition that
require close monitoring.
2.ICU or Intensive Care Unit care for patients that are critically ill. Each ICU provide specialize care e.g.
## Neuro Intensive Care Unit
Coronary Care Unit
Cardiovascular Intensive Care Unit
Neonatal Intensive Care Unit
Critical Care Unit
Pediatric Intensive Care Unit
3. Specialized Facilities facility that only admit specific type of patient or client e.g.
Psychiatric Hospitals
Government Hospitals
Pediatric Hospitals
4. Home Health Care provided by an agency or acute care facility that offer services in the home e.g.
Wound Care
IV Therapy
Respiratory Treatment
5. Hospice
## Care provided to terminally ill patients in a facility or home setting
Main focus of Hospice is transition from life to death with emphasis on working collaboratively with patient,
family and significant others.
6.Respite
Part time care provided for patients having chronic medical condition or mental illness.
Allow the primary care provider time off at the same time socialization for the patient.
7. Tele Health
## New innovation added to the healthcare industry
Provide ability for a nurse or physician to interact via telephone or computer audio/video
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 8 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Frequent communication with patient resulting to better compliance
Decrease ER visits, unscheduled Physician office visit, and re-hospitalization
## 8. Extended Care Facilities - provide longer period of care to patient e.g.
Nursing Homes
Types of Nursing Homes
Skilled Nursing Facility (SNF) 24 hours care provided under supervision of Registered Nurse
Intermediate Care Facility (ICF) 24 hours care provided by nursing assistants under supervision of an
LPN/LVN
Neighborhood
Small Town
Rural County
## 10. Healthcare in School and Industry
Provide healthcare services to students particularly with disabilities in the school setting and/or
employees in an industrial setting
Assist in providing care to ill patients and/or emergency
Provide preventive care such as health teaching, administering immunizations and medications
## THE HEALTH CARE TEAM AS PROVIDERS
Nurse Practitioner A Nurse with additional training in a specialized area such as family practice,
maternity, cardio-thoracic or mental health
Registered Nurse provide direct and indirect nursing care, supervision and leadership in a wide variety
of healthcare settings.
License Practical Nurse provide care in long term care and acute care facilities. Works under the
supervision of a Registered Nurse or Physician
Certified Nurse Assistant - provide basic nursing care of clients in a long term care facility Nurse
Practitioner A Nurse with additional training in a specialized area such as family practice, maternity,
cardio-thoracic or mental health
## Technologists hold a Baccalaureate degree
Technicians hold an associate degree, diploma or certificate
## Patient care techs
Certified Nursing Assistants
Unit clerks
## 2304 E. Busch Blvd.
Tampa, FL 33612
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FAX: 813-762-1325
www.medicalprepinsitute.org
Nursing
Professional nursing holds a unique place in the American health care system. As members of the largest health
care profession, the nations 2.9 million nurses work in diverse settings and fields and are frontline providers of
health care services. While most nurses work in acute- care settings such as hospitals, nurses expertise and skills
extend well beyond hospital walls. Working independently and with other health care professionals, nurses
promote the health of individuals, families, and communities. Millions of Americans turn to nurses for delivery of
primary health care services, health care education. and health advice and counseling. Nurses are critical links in
maintaining a cutting edge health care system. Nursing continues to be an indispensable service to the American
public
While many may think of a nurse as someone who takes care of hospitalized patients, nurses also fill a wide variety
of positions in health care in many varied settings, working both collaboratively and independently with other
health care professionals. For example, most Americans are familiar with home care nurses who provide a plethora
of nursing and health care services to patients in their homes. School nurses have a long history of providing health
services to school children from kindergarten through high school. Nurses play a major role in delivering care to
those residing in long-term-care facilities such as nursing homes.
Florence Nightingale
Most people think of the nursing profession as beginning with the work of Florence Nightingale, an upper class
British woman who captured the public imagination when she led a group of female nurses to the Crimea in
October of 1854 to deliver nursing service to British soldiers. Upon her return to England, Nightingale successfully
established nurse education programs in a number of British hospitals. These schools were organized around a
specific set of ideas about how nurses should be educated, developed by Nightingale often referred to as the
Nightingale Principles. Actually, while Nightingales work was ground-breaking in that she confirmed that a corps
of educated women, informed about health and the ways to promote it, could improve the care of patients based
on a set of particular principles, she was the not the first to put these principles into action.
## The Nursing Program
Today nursing is highly regarded as an excellent career choice for both women and men. The nursing Program is a
limited access program. This means there are limited seats and admissions process is extremely competitive.
Students entering a nursing are maybe required to pass entrance examination.
Test Required Scores
1 Nursing Entrance Exam- Students Applying into Associate of Science Degree 75%
2 Basic Computer Skills- All students must pass a Basic Computer Skills Test 80%
3 PN Step -all current practical nurses must take and pass the Step test 60%
4 Nursing TABE Test - Students Applying into Practical Nursing diploma 70%
5 ACT -student with ACT scores of 19 do not need to take an entrance exam 19
6 SAT- student with SAT scores of 1350 do not need to take an entrance exam 1350
1 Nursing Entrance Exam- Students who score a 70% may apply in Practical Nursing Program 75%
Students applying into the Associate of Science Nursing Program who fail to receive a score of 75% after three
(3) attempts may apply into the Practical Nursing Program if they score a 70%.
## 2304 E. Busch Blvd.
Tampa, FL 33612
Page | 10 TEL: 813-932-1710
FAX: 813-762-1325
www.medicalprepinsitute.org
Nursing Entrance Examination
The exam is used to aid adult education programs in deciding which applicants to accept. The nursing entrance
examination is divided into three sections. Math, English and Science. The English Section is divided into Two
Sections
## Section Number of Questions
English Section A (Reading Comprehension) 30
English Section B (Language) 30
Math 30
Science 30
The prospective student will be allowed three (3) attempts in a twelve-month period.
## Be positive and do your best.
Relax, its normal to be somewhat nervous before taking a test. Dont worry!
Be sure you can understand the instructions and understand them.
Read the directions for each test section carefully. Ask for an explanation of the directions if you do not
understand them.
Plan your time well. Each test section is timed. Do not spend too much time on any one test question
Before answering a question, be sure you know what is being asked. For example, a test question might
say, Which of these is not an even number? If you read the question too quickly, you may miss the word
Do not read into a question something that is not there. There are no trick questions.
you select probably will be the best. When rechecking, change an answer only when you are sure that
If you are not sure how to answer a question, rule out answer choices that you know are incorrect. Then
Remember that the score you will receive on NURSING ENTRANCE EXAMINATION is only one way to
measure your skills. NURSING ENTRANCE EXAMINATION will show you the skills you have now and
those you need to learn.
## 2304 E. Busch Blvd.
Tampa, FL 33612
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FAX: 813-762-1325
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English
The English test measures a test takers ability to understand, analyze and evaluate written passages. The
passages will contain material that will be from a variety of sources and on a number of different topics.
Each of the passages and statements in the English test will be followed by a series of questions covering the
content of the passage or statement, in which you will have to answer questions, which will demonstrate how well
you understand the passages and are able to draw conclusions about the material.
Questions 1 through 7 refer to the following passage:
In the 16th century, an age of great marine and terrestrial exploration, Ferdinand Magellan led the first expedition
to sail around the world. As a young Portuguese noble, he served the king of Portugal, but he became involved in
the quagmire of political intrigue at court and lost the king's favor. After he was dismissed from service by the king
of Portugal, he offered to serve the future Emperor Charles V of Spain.
A papal decree of 1493 had assigned all land in the New World west of 50 degrees W longitude to Spain and all the
land east of that line to Portugal. Magellan offered to prove that the East Indies fell under Spanish authority. On
September 20, 1519, Magellan set sail from Spain with five ships. More than a year later, one of these ships was
exploring the topography of South America in search of a water route across the continent. This ship sank, but the
remaining four ships searched along the southern peninsula of South America. Finally they found the passage they
sought near 50 degrees S latitude. Magellan named this passage the Strait of All Saints, but today it is known as the
Strait of Magellan.
One ship deserted while in this passage and returned to Spain, so fewer sailors were privileged to gaze at that first
panorama of the Pacific Ocean. Those who remained crossed the meridian now known as the International Date
Line in the early spring of 1521 after 98 days on the Pacific Ocean. During those long days at sea, many of
Magellan's men died of starvation and disease.
Later, Magellan became involved in an insular conflict in the Philippines and was killed in a tribal battle. Only one
ship and 17 sailors under the command of the Basque navigator Elcano survived to complete the westward journey
to Spain and thus prove once and for all that the world is round, with no precipice at the edge.
A. cosmic
B. land
C. mental
D. common man
## 2304 E. Busch Blvd.
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2. Magellan lost the favor of the king of Portugal when he became involved in a political ________.
A. entanglement
B. discussion
C. negotiation
D. problem
## E. None of the above
3. The Pope divided New World lands between Spain and Portugal according to their location on one side or the
other of an imaginary geographical line 50 degrees west of Greenwich that extends in a _________ direction.
B. crosswise
C. easterly
D. south east
## E. north and west
4. One of Magellan's ships explored the _________ of South America for a passage across the continent.
A. coastline
B. mountain range
C. physical features
D. islands
## 5. Four of the ships sought a passage along a southern ______.
A. coast
B. inland
C. body of land with water on three sides
D. border
## 2304 E. Busch Blvd.
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6. The passage was found near 50 degrees S of ________.
A. Greenwich
B. The equator
C. Spain
D. Portugal
7. In the spring of 1521, the ships crossed the _______ now called the International Date Line.
## A. imaginary circle passing through the poles
B. imaginary line parallel to the equator
C. area
D. land mass
## The following passage refers to questions 8 through 14.
Marie Curie was one of the most accomplished scientists in history. Together with her husband, Pierre, she
discovered radium, an element widely used for treating cancer, and studied uranium and other radioactive
substances. Pierre and Marie's amicable collaboration later helped to unlock the secrets of the atom.
Marie was born in 1867 in Warsaw, Poland, where her father was a professor of physics. At an early age, she
displayed a brilliant mind and a blithe personality. Her great exuberance for learning prompted her to continue
with her studies after high school. She became disgruntled, however, when she learned that the university in
Warsaw was closed to women. Determined to receive a higher education, she defiantly left Poland and in 1891
entered the Sorbonne, a French university, where she earned her master's degree and doctorate in physics.
Marie was fortunate to have studied at the Sorbonne with some of the greatest scientists of her day, one of whom
was Pierre Curie. Marie and Pierre were married in 1895 and spent many productive years working together in the
physics laboratory. A short time after they discovered radium, Pierre was killed by a horse-drawn wagon in 1906.
Marie was stunned by this horrible misfortune and endured heartbreaking anguish. Despondently she recalled
their close relationship and the joy that they had shared in scientific research. The fact that she had two young
daughters to raise by herself greatly increased her distress.
Curie's feeling of desolation finally began to fade when she was asked to succeed her husband as a physics
professor at the Sorbonne. She was the first woman to be given a professorship at the world-famous university. In
1911 she received the Nobel Prize in chemistry for isolating radium. Although Marie Curie eventually suffered a
fatal illness from her long exposure to radium, she never became disillusioned about her work. Regardless of the
consequences, she had dedicated herself to science and to revealing the mysteries of the physical world.
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8. The Curies' _________ collaboration helped to unlock the secrets of the atom.
A. friendly
B. competitive
C. courteous
D. industrious
E. chemistry
## 9. Marie had a bright mind and a ______ personality.
A. strong
B. lighthearted
C. humorous
D. strange
E. envious
10. When she learned that she could not attend the university in Warsaw, she felt _________.
A. hopeless
B. annoyed
C. depressed
D. worried
E. None of the above
11. Marie _________ by leaving Poland and traveling to France to enter the Sorbonne.
A. challenged authority
B. showed intelligence
C. behaved
D. was distressed
## 12. _________ she remembered their joy together.
A. Dejectedly
B. Worried
C. Tearfully
D. Happily
E. Irefully
13. Her _________ began to fade when she returned to the Sorbonne to succeed her husband.
A. misfortune
B. anger
C. wretchedness
D. disappointment
E. ambition
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14. Even though she became fatally ill from working with radium, Marie Curie was never _________.
A. troubled
B. worried
C. disappointed
D. sorrowful
E. disturbed
## The following passage refers to questions 15 through 19.
Mount Vesuvius, a volcano located between the ancient Italian cities of Pompeii and Herculaneum, has received
much attention because of its frequent and destructive eruptions. The most famous of these eruptions occurred in
A.D. 79.
The volcano had been inactive for centuries. There was little warning of the coming eruption, although one
account unearthed by archaeologists says that a hard rain and a strong wind had disturbed the celestial calm
during the preceding night. Early the next morning, the volcano poured a huge river of molten rock down upon
Herculaneum, completely burying the city and filling the harbor with coagulated lava.
Meanwhile, on the other side of the mountain, cinders, stone and ash rained down on Pompeii. Sparks from the
burning ash ignited the combustible rooftops quickly. Large portions of the city were destroyed in the
conflagration. Fire, however, was not the only cause of destruction. Poisonous sulfuric gases saturated the air.
These heavy gases were not buoyant in the atmosphere and therefore sank toward the earth and suffocated
people.
Over the years, excavations of Pompeii and Herculaneum have revealed a great deal about the behavior of the
volcano. By analyzing data, much as a zoologist dissects an animal specimen, scientists have concluded that the
eruption changed large portions of the area's geography. For instance, it turned the Sarno River from its course
and raised the level of the beach along the Bay of Naples. Meteorologists studying these events have also
concluded that Vesuvius caused a huge tidal wave that affected the world's climate.
In addition to making these investigations, archaeologists have been able to study the skeletons of victims by using
distilled water to wash away the volcanic ash. By strengthening the brittle bones with acrylic paint, scientists have
been able to examine the skeletons and draw conclusions about the diet and habits of the residents. Finally, the
excavations at both Pompeii and Herculaneum have yielded many examples of classical art, such as jewelry made
of bronze, which is an alloy of copper and tin. The eruption of Mount Vesuvius and its tragic consequences have
provided everyone with a wealth of data about the effects that volcanoes can have on the surrounding area.
Today, volcanologists can locate and predict eruptions, saving lives and preventing the destruction of other cities
and cultures.
15. Herculaneum and its harbor were buried under _________ lava.
A. liquid
B. solid
C. flowing
D. gas
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16. The poisonous gases were not _________ in the air.
A. able to float
B. visible
C. able to evaporate
D. invisible
E. able to condense
17. Scientists analyzed data about Vesuvius in the same way that a zoologist _________ a specimen.
A. describes in detail
B. studies by cutting apart
C. photographs
D. chart
18. _________ have concluded that the volcanic eruption caused a tidal wave.
## A. Scientists who study oceans
B. Scientists who study atmospheric conditions
C. Scientists who study ash
D. Scientists who study animal behavior
E. Answer not available in article
19. Scientists have used _________ water to wash away volcanic ash from the skeletons of victims.
A. bottled
B. volcanic
C. purified
D. sea
E. fountain
## The following passage refers to questions 20-24.
Conflict had existed between Spain and England since the 1570s. England wanted a share of the wealth that Spain
had been taking from the lands it had claimed in the Americas.
Elizabeth I, Queen of England, encouraged her staunch admiral of the navy, Sir Francis Drake, to raid Spanish ships
and towns. Though these raids were on a small scale, Drake achieved dramatic success, adding gold and silver to
England's treasury and diminishing Spain's supremacy.
Religious differences also caused conflict between the two countries. Whereas Spain was Roman Catholic, most of
England had become Protestant. King Philip II of Spain wanted to claim the throne and make England a Catholic
country again. To satisfy his ambition and also to retaliate against England's theft of his gold and silver, King Philip
began to build his fleet of warships, the Spanish Armada, in January 1586.
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Philip intended his fleet to be indestructible. In addition to building new warships, he marshaled 130 sailing vessels
of all types and recruited more than 19,000 robust soldiers and 8,000 sailors. Although some of his ships lacked
guns and others lacked ammunition, Philip was convinced that his Armada could withstand any battle with
England.
The martial Armada set sail from Lisbon, Portugal, on May 9, 1588, but bad weather forced it back to port. The
voyage resumed on July 22 after the weather became more stable.
The Spanish fleet met the smaller, faster, and more maneuverable English ships in battle off the coast of Plymouth,
England, first on July 31 and again on August 2. The two battles left Spain vulnerable, having lost several ships and
with its ammunition depleted. On August 7, while the Armada lay at anchor on the French side of the Strait of
Dover, England sent eight burning ships into the midst of the Spanish fleet to set it on fire. Blocked on one side, the
Spanish ships could only drift away, their crews in panic and disorder. Before the Armada could regroup, the
English attacked again on August 8.
Although the Spaniards made a valiant effort to fight back, the fleet suffered extensive damage. During the eight
hours of battle, the Armada drifted perilously close to the rocky coastline. At the moment when it seemed that the
Spanish ships would be driven onto the English shore, the wind shifted, and the Armada drifted out into the North
Sea. The Spaniards recognized the superiority of the English fleet and returned home, defeated.
20. Sir Francis Drake added wealth to the treasury and diminished Spain's _________.
A. unlimited power
B. unrestricted growth
C. territory
D. treaties
E. Answer not available in article
A. warlike
B. strong
C. accomplished
D. timid
E. inexperienced
A. complete
B. warlike
C. independent
D. isolated
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23. The two battles left the Spanish fleet _________.
A. open to change
B. triumphant
C. open to attack
D. defeated
E. discouraged
A. closed off
B. damaged
C. alone
D. circled
## The following passage refers to questions 25-29.
The victory of the small Greek democracy of Athens over the mighty Persian Empire in 490 B.C. is one of the most
famous events in history. Darius, king of the Persian Empire, was furious because Athens had interceded for the
other Greek city-states in revolt against Persian domination. In anger the king sent an enormous army to defeat
Athens. He thought it would take drastic steps to pacify the rebellious part of the empire.
Persia was ruled by one man. In Athens, however, all citizens helped to rule. Ennobled by this participation,
Athenians were prepared to die for their city-state. Perhaps this was the secret of the remarkable victory at
Marathon, which freed them from Persian rule. On their way to Marathon, the Persians tried to fool some Greek
city-states by claiming to have come in peace. The frightened citizens of Delos refused to believe this. Not wanting
to abet the conquest of Greece, they fled from their city and did not return until the Persians had left. They were
wise, for the Persians next conquered the city of Eritrea and captured its people.
Tiny Athens stood alone against Persia. The Athenian people went to their sanctuaries. There they prayed for
deliverance. They asked their gods to expedite their victory. The Athenians refurbished their weapons and moved
to the plain of Marathon, where their little band would meet the Persians. At the last moment, soldiers from
Plataea reinforced the Athenian troops.
The Athenian army attacked, and Greek citizens fought bravely. The power of the mighty Persians was offset by
the love that the Athenians had for their city. Athenians defeated the Persians in both archery and hand combat.
Greek soldiers seized Persian ships and burned them, and the Persians fled in terror. Herodotus, a famous
historian, reports that 6,400 Persians died, compared to only 192 Athenians.
25. Athens had _________ the other Greek city-states against the Persians.
A. refused help to
B. intervened on behalf of
C. wanted to fight
D. given orders for all to fight
E. defeated
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26. Darius took drastic steps to ________ the rebellious Athenians.
A. weaken
B. destroy
C. calm
D. irritate
## 27. Their participation _________ to the Athenians.
A. gave comfort
B. gave honor
C. gave strength
D. gave fear
E. gave hope
28. The people of Delos did not want to ______ the conquest of Greece.
A. end
B. encourage
29. The Athenians were _________ by some soldiers who arrived from Plataea.
A. welcomed
B. strengthened
C. held
D. captured
## The following passage refers to questions 30-32.
The Trojan War is one of the most famous wars in history. It is well known for the 10-year duration, for the
heroism of a number of legendary characters, and for the Trojan horse. What may not be familiar, however, is the
story of how the war began.
According to Greek myth, the strife between the Trojans and the Greeks started at the wedding of Peleus, King of
Thessaly, and Thetis, a sea nymph. All of the gods and goddesses had been invited to the wedding celebration in
Troy except Eris, goddess of discord. She had been omitted from the guest list because her presence always
embroiled mortals and immortals alike in conflict.
To take revenge on those who had slighted her, Eris decided to cause a skirmish. Into the middle of the banquet
hall, she threw a golden apple marked "for the most beautiful." All of the goddesses began to haggle over who
should possess it. The gods and goddesses reached a stalemate when the choice was narrowed to Hera, Athena,
and Aphrodite. Someone was needed to settle the controversy by picking a winner. The job eventually fell to Paris,
son of King Priam of Troy, who was said to be a good judge of beauty. Paris did not have an easy job. Each goddess,
eager to win the golden apple, tried aggressively to bribe him.
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"I'll grant you vast kingdoms to rule," promised Hera. "Vast kingdoms are nothing in comparison with my gift,"
contradicted Athena. "Choose me and I'll see that you win victory and fame in war." Aphrodite outdid her
adversaries, however. She won the golden apple by offering Helen, daughter of Zeus and the most beautiful mortal
in the land, to Paris. Paris, anxious to claim Helen, set off for Sparta in Greece.
Although Paris learned that Helen was married, he nevertheless accepted the hospitality of her husband, King
Menelaus of Sparta. Therefore, Menelaus was outraged for a number of reasons when Paris departed, taking
Helen and much of the king's wealth back to Troy. Menelaus collected his loyal forces and set sail for Troy to begin
the war to reclaim Helen.
30. Eris was known for _________ both mortals and immortals.
A. scheming against
B. creating conflict amongst
C. feeling hostile toward
D. ignoring
E. comforting
## 31. Each goddess tried ______ to bribe Paris.
A. boldly
B. effectively
C. secretly
D. carefully
32. Athena _________ Hera, promising Paris victory and fame in war.
## A. disregarded the statement of
B. defeated
C. agreed with
D. restated the statement of
E. questioned the statement of
## Refer to the following passage for questions 33-37.
One of the most intriguing stories of the Russian Revolution concerns the identity of Anastasia, the youngest
daughter of Czar Nicholas II. During his reign over Russia, the czar had planned to revoke many of the harsh laws
established by previous czars. Some workers and peasants, however, clamored for more rapid social reform. In
1918, a group of these people known as Bolsheviks overthrew the government. On July 17 or 18, they murdered
the czar and what was thought to be his entire family.
Although witnesses vouched that all the members of the czar's family had been executed, there were rumors
suggesting that Anastasia had survived. Over the years, a number of women claimed to be Grand Duchess
Anastasia. Perhaps the most famous claimant was Anastasia Tschaikovsky, who was also known as Anna Anderson.
In 1920, 18 months after the czar's execution, this terrified young woman was rescued from drowning in a Berlin
river. She spent two years in a hospital, where she attempted to reclaim her health and shattered mind. The
doctors and nurses thought that she resembled Anastasia and questioned her about her background. She
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disclaimed any connection with the czar's family. Eight years later, however, she claimed that she was Anastasia.
She said that she had been rescued by two Russian soldiers after the czar and the rest of her family had been
killed. Two brothers named Tschaikovsky had carried her into Romania. She had married one of the brothers, who
had taken her to Berlin and left her there, penniless and without a vocation. Unable to invoke the aid of her
mother's family in Germany, she had tried to drown herself.
During the next few years, scores of the czar's relatives, ex-servants, and acquaintances interviewed her. Many of
these people said that her looks and mannerisms were evocative of the Anastasia that they had known. Her
grandmother and other relatives denied that she was the real Anastasia, however.
Tired of being accused of fraud, Anastasia immigrated to the United States in 1928 and took the name Anna
Anderson. She still wished to prove that she was Anastasia, though, and returned to Germany in 1933 to bring suit
against her mother's family. There she declaimed to the court, asserting that she was indeed Anastasia and
deserved her inheritance.
In 1957, the court decided that it could neither confirm nor deny Anastasia's identity. Although it will probably
never be known whether this woman was the Grand Duchess Anastasia, her search to establish her identity has
been the subject of numerous books, plays, and movies.
33. Some Russian peasants and workers ______ for social reform.
A. longed
B. cried out
C. begged
D. hoped
E. thought much
34. Witnesses ______ that all members of the czar's family had been executed.
A. gave assurance
B. thought
C. hoped
D. convinced some
35. Tschaikovsky initially ______ any connection with the czar's family.
A. denied
B. stopped
C. noted
D. justified
A. locate
C. call upon
D. identify
E. know
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37. In court she _________ maintaining that she was Anastasia and deserved her inheritance.
A. finally appeared
B. spoke forcefully
C. gave testimony
D. gave evidence
## Refer to the following passage for questions 38-39.
King Louis XVI and Queen Marie Antoinette ruled France from 1774 to 1789, a time when the country was fighting
bankruptcy. The royal couple did not let France's insecure financial situation limit their immoderate spending,
however. Even though the minister of finance repeatedly warned the king and queen against wasting money, they
continued to spend great fortunes on their personal pleasure. This lavish spending greatly enraged the people of
France. They felt that the royal couple bought its luxurious lifestyle at the poor people's expense.
Marie Antoinette, the beautiful but exceedingly impractical queen, seemed uncaring about her subjects' misery.
While French citizens begged for lower taxes, the queen embellished her palace with extravagant works of art. She
also surrounded herself with artists, writers, and musicians, who encouraged the queen to spend money even
more profusely.
While the queen's favorites glutted themselves on huge feasts at the royal table, many people in France were
starving. The French government taxed the citizens outrageously. These high taxes paid for the entertainments the
queen and her court so enjoyed. When the minister of finance tried to stop these royal spendthrifts, the queen
replaced him. The intense hatred that the people felt for Louis XVI and Marie Antoinette kept building until it led
to the French Revolution. During this time of struggle and violence (1789-1799), thousands of aristocrats, as well as
the king and queen themselves, lost their lives at the guillotine. Perhaps if Louis XVI and Marie Antoinette had
reined in their extravagant spending, the events that rocked France would not have occurred.
38. The people surrounding the queen encouraged her to spend money ______.
A. wisely
B. abundantly
C. carefully
D. foolishly
E. joyfully
A. aristocrats
B. money wasters
C. enemies
D. individuals
E. spenders
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Refer to the following passage for questions 40-45.
Many great inventions are initially greeted with ridicule and disbelief. The invention of the airplane was no
exception. Although many people who heard about the first powered flight on December 17, 1903 were excited
and impressed, others reacted with peals of laughter. The idea of flying an aircraft was repulsive to some people.
Such people called Wilbur and Orville Wright, the inventors of the first flying machine, impulsive fools. Negative
reactions, however, did not stop the Wrights. Impelled by their desire to succeed, they continued their
experiments in aviation.
Orville and Wilbur Wright had always had a compelling interest in aeronautics and mechanics. As young boys they
earned money by making and selling kites and mechanical toys. Later, they designed a newspaper-folding machine,
built a printing press, and operated a bicycle-repair shop. In 1896, when they read about the death of Otto
Lilienthal, the brothers' interest in flight grew into a compulsion.
Lilienthal, a pioneer in hang-gliding, had controlled his gliders by shifting his body in the desired direction. This idea
was repellent to the Wright brothers, however, and they searched for more efficient methods to control the
balance of airborne vehicles. In 1900 and 1901, the Wrights tested numerous gliders and developed control
techniques. The brothers' inability to obtain enough lift power for the gliders almost led them to abandon their
efforts.
After further study, the Wright brothers concluded that the published tables of air pressure on curved surfaces
must be wrong. They set up a wind tunnel and began a series of experiments with model wings. Because of their
efforts, the old tables were repealed in time and replaced by the first reliable figures for air pressure on curved
surfaces. This work, in turn, made it possible for the brothers to design a machine that would fly. In 1903 the
Wrights built their first airplane, which cost less than \$1,000. They even designed and built their own source of
propulsion-a lightweight gasoline engine. When they started the engine on December 17, the airplane pulsated
wildly before taking off. The plane managed to stay aloft for 12 seconds, however, and it flew 120 feet.
By 1905, the Wrights had perfected the first airplane that could turn, circle, and remain airborne for half an hour at
a time. Others had flown in balloons and hang gliders, but the Wright brothers were the first to build a full-size
machine that could fly under its own power. As the contributors of one of the most outstanding engineering
achievements in history, the Wright brothers are accurately called the fathers of aviation.
A. boring
B. distasteful
C. exciting
D. needless
## A. acted without thinking
B. been negatively influenced
C. been too cautious
D. been mistaken
E. acted in a negative way
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42. The Wrights' interest in flight grew into a ______.
A. financial empire
B. plan
C. need to act
D. foolish thought
43. Lilienthal's idea about controlling airborne vehicles was _________ the Wrights.
A. proven wrong by
B. opposite to the ideas of
C. disliked by
D. accepted by
E. improved by
44. The old tables were _________ and replaced by the first reliable figures for air pressure on curved surfaces.
A. destroyed
B. invalidated
C. multiplied
D. approved
E. not used
45. The Wrights designed and built their own source of _________.
## A. force for moving forward
B. force for turning around
C. turning
D. force for going backward
E. None of the above
The splintered steps leading to the tenements entrance were rotted and uneven. They led to an unlocked door
which wobbled on its hinges and shrank from its frame. It creaked open to a dank, dark hall which smelled of urine
and sweat. The paint was peeling off cracked walls. The faint yellow light hung low in the night. Mr. and Mrs.
Gomes lived on the second floor with their three young children. Their four-room apartment was immaculate and
tidy. The kitchen floor glistened, and the flowered plates and glasses were neatly stacked in the drainer. In the
living room, the sheer curtains were always drawn back, filtering sunlight throughout the room, passing over a
color television and several porcelain icons. Besides the freestanding gas heater was a brand name stereo system
recently purchased on an easy pay credit plan.
The soft pine floors were all warped but recently painted. The wide floorboards, once loose in many places, were
now nailed down securely. Clear plastic sheets were tacked over the windows to prevent heat loss. The children,
two girls and a boy, shared a large room with one small window that was separated from the kitchen by a curtain.
The gas stove warmed them at night. Their toys were piled high in wooden crates. The children did not dare turn
on the kitchen light for fear that the six-legged, brown-bodied pests would dart out in front of them. The Gomes
family had rented this apartment for ten years, ever since they came to this country. They had known no other
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home, although they had dreamed of many. Some day they hoped to live in a quiet neighborhood with open yards
and spotless sidewalks, where people get into cars each weekday morning and commute to work.
46. You can infer that the story takes place in the
a. summer
b. spring
c. fall
d. winter
## 47. In the third paragraph the word icons means
a. fine china plates
b. ornate lamps
c. religious figures
d. ashtrays
48. You can conclude that the Gomes family members are
a. native Americans
b. United States citizens
c. Chinese
d. immigrants
e. second generation Americans
49. What would be the opposite meaning of the word immaculate in the second paragraph?
a. filthy
b. girlish
c. clean
d. horrible
e. modern
## 50. What can you infer about the childrens bedroom?
a. It is well furnished.
b. It is crowded.
c. It has its own bathroom.
d. It is well lighted.
e. It is pest-free.
Americans have always been interested in their Presidents' wives. Many First Ladies have been remembered
because of the ways they have influenced their husbands. Other First Ladies have made the history books on their
own.
husbands' speeches. When Lady Bird Johnson thought her husband was talking too long, she wrote a note and sent
it up to the platform. It read, "It's time to stop!" And he did. Once Bess Truman didn't like what her husband was
saying on television, so she phoned him and said, "If you can't talk more politely than that in public, you come right
home."
Abigail Fillmore and Eliza Johnson actually taught their husbands, Millard Fillmore and Andrew Johnson, the
thirteenth and seventeenth Presidents. A schoolteacher, Abigail eventually married her pupil, Millard. When Eliza
Johnson married Andrew, he could not read or write, so she taught him herself.
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It was First Lady Helen Taft's idea to plant the famous cherry trees in Washington, D. C. Each spring these
blossoming trees attract thousands of visitors to the nation's capital. Mrs. Taft also influenced the male members
of her family and the White House staff in a strange way: she convinced them to shave off their beards!
Shortly after President Woodrow Wilson suffered a stroke, Edith Wilson unofficially took over most of the duties of
the Presidency until the end of her husband's term. Earlier, during World War I, Mrs. Wilson had sheep brought
onto the White House lawn to eat the grass. The sheep not only kept the lawn mowed, but provided wool for an
auction sponsored by the First Lady. Almost \$100,000 was raised for the Red Cross.
Dolly Madison saw to it that a magnificent painting of George Washington was not destroyed during the War of
1812. As the British marched toward Washington, D. C., she remained behind to rescue the painting, even after the
guards had left. The painting is the only object from the original White House that was not burned.
One of the most famous First Ladies was Eleanor Roosevelt, the wife of President Franklin D. Roosevelt. She was
active in political and social causes throughout her husband's tenure in office. After his death, she became famous
for her humanitarian work in the United Nations. She made life better for thousands of needy people around the
world.
## A. The Humanitarian work of the First Ladies is critical in American government.
B. Dolly Madison was the most influential president's wife.
C. Eleanor Roosevelt transformed the First Lady image.
D. The First Ladies are important figures in American culture.
E. The First Ladies are key supporters of the Presidents.
Of the many kinds of vegetables grown all over the world, which remains the favorite of young and old alike? The
potato, of course.
Perhaps you know them as "taters," "spuds," or "Kennebees," or as "chips," "Idahoes," or even "shoestrings." No
matter, a potato by any other name is still a potato- the world's most widely grown vegetable. As a matter of fact,
if you are an average potato eater, you will put away at least 100 pounds of them each year.
That's only a tiny portion of the amount grown every year, however. Worldwide, the annual potato harvest is over
6 billion bags. Each bag contains 100 pounds of potatoes, some of them as large as four pounds each. Here in the
United States, farmers fill about 400 million bags a year. That may seem like a lot of "taters," but it leaves the
United States a distant third among world potato growers. Polish farmers dig up just over 800 million bags a year,
while the Russians lead the world with nearly 1.5 billion bags.
The first potatoes were grown by the Incas of South America, more than 400 years ago. Their descendants in
Ecuador and Chile continue to grow the vegetable as high as 14,000 feet up in the Andes Mountains. (That's higher
than any other food will grow.) Early Spanish and English explorers shipped potatoes to Europe, and they found
their way to North America in the early 1600s.
People eat potatoes in many ways-baked, mashed, and roasted, to name just three. However, in the United States
most potatoes are devoured in the form of French fries. One fast-food chain alone sells more than \$1 billion worth
of fries each year. No wonder, then, that the company pays particular attention to the way its fries are prepared.
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Before any fry makes it to the people who eat at these popular restaurants, it must pass many separate tests. Fail
any one of these tests and the potato is rejected. To start with, only Russet Burbank potatoes are used. These
Idaho potatoes have less water content than other kinds, which can have as much as 80 percent water. Once cut
into "shoestrings" shapes, the potatoes are partly fried in a secret blend of oils, sprayed with liquid sugar to brown
them, steam dried at high heat, then flash frozen for shipment to individual restaurants.
Before shipping, every shoestring is measured. Forty percent of a batch must be between two and three inches
long. Another 40 percent has to be over three inches. What about the 20 percent that are left in the batch? Well, a
few short fries in a bag are okay, it seems.
So, now that you realize the enormous size and value of the potato crop, you can understand why most people
agree that this part of the food industry is no "small potatoes."
## A. Potatoes from Ireland started the Potato Revolution.
B. The average American eats 50 pounds of potatoes a year.
C. French fries are made from potatoes.
D. Potatoes are a key vegetable in America.
E. The various terms for potatoes have a long history.
What does the word "patent" mean to you? Does it strike you as being something rather remote from your
interests? If it does, stop and think a moment about some of the commonplace things that you use every day,
those objects that you take for granted as part of the world around you. The telephone, radio, television,
automobile, and the 1,001 other things (even the humble safety pin) that enrich our lives today once existed only
as ideas in the minds of men. If it had not been possible to patent their ideas and thus protect them against
copying by others, these inventions might never have been fully developed to serve mankind.
If there were no patent protection there would be little incentive to invent and innovate, for once the details of an
invention became known, hordes of imitators who did not share the inventor's risks and expenses might well flood
the market with their copies of his product and reap much of the benefit of his efforts. The technological progress
that has made America great would wither rapidly under conditions such as these.
The fundamental principles in the United States patent structure came from England. During the glorious reign of
Queen Elizabeth I in England, the expanding technology was furthered by the granting of exclusive manufacturing
and selling privileges to citizens who had invented new processes or tools-a step that did much to encourage
creativity. Later, when critics argued that giving monopoly rights to one person infringed on the rights of others, an
important principle was added to the patent structure: The Lord Chief Justice of England stated that society had
everything to gain and nothing to lose by granting exclusive privileges to an inventor, because a patent for an
invention was granted for something new that society never had before.
Another basic principle was brought into law because certain influential people in England had managed to obtain
monopoly control over such age-old products as salt, and had begun charging as much as the people could
tolerate. The public outcry became so great that the government was forced to decree that monopoly rights could
be awarded only to those who created or introduced something really unique. These principles are the mainstays
of the modern patent system in the United States.
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In colonial times, patent law was left up to the separate states. The inconsistency, confusion, and unfairness that
resulted clearly indicated the need for a uniform patent law, and the men who drew up the Constitution
incorporated one. George Washington signed the first patent law on April 10, 1790, and less than four months
later the first patent was issued to a man named Samuel Hopkins for a chemical process, an improved method of
making potash for use in soapmaking.
In 1936 the Patent Office was established as a separate bureau. From the staff of eight that it maintained during its
first year of operation, it has grown into an organization of over 2,500 people handling more than 1,600 patent
applications and granting over 1,000 every week.
The Patent Office in Washington, D. C. is the world's largest library of scientific and technical data, and this
treasure trove of information is open for public inspection. In addition to more than 3 million US patents, it houses
more than 7 million foreign patents and thousands of volumes of technical literature. Abraham Lincoln patented a
device to lift steam vessels over river shoals, Mark Twain developed a self-pasting scrapbook, and millionaire
Cornelius Vanderbilt invented a shoe-shine kit.
A patent may be granted for any new and useful process, machine, article of manufacture, or composition of
matter (a chemical compound or combinations of chemical compounds), or any distinct and new variety of plant,
including certain mutants and hybrids.
The patent system has also helped to boost the wages of the American worker to an unprecedented level: he can
produce more and earn more with the computer, adding machines, drill press or lathe. Patented inventions also
help keep prices down by increasing manufacturing efficiency and by stimulating the competition that is the
foundation of our free enterprise system.
The decades of history have disclosed little need for modification of the patent structure. United States patent
laws, like the Constitution from which they grew, have stood the test of time well. They encouraged the creative
processes, brought untold benefits to society as a whole, and enabled American technology to outstrip that of the
rest of the civilized world.
## A. The patent system encourages free enterprise.
B. The Constitution protects the patent system.
C. The patent system in England has been influential in American patent development.
D. Patents are important tools for inventors.
E. Patented inventions protect the inventor, free enterprise, and the creative process.
Most people think that it's fine to be "busy as a beaver." Little do they know. Beavers may work hard, but often
they don't get very much done.
Beavers are supposed to be great tree cutters. It is true that a beaver can gnaw through a tree very quickly: A six-
inch birch takes about 10 minutes. But then what? Often the beaver does not make use of the tree. One expert
says that beavers waste one out of every five trees they cut.
For one thing, they do not choose their trees wisely. One bunch of beavers cut down a cottonwood tree more than
100 feet tall. Then they found that they could not move it.
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In thick woods, a tree sometimes won't fall down. It gets stuck in the other trees. Of course, the beaver doesn't
think to cut down the trees that are in the way. So a good tree goes to waste.
Some people think that beavers can make a tree fall the way they want it to. Not true. (In fact, beavers sometimes
get pinned under a falling tree.) When beavers cut a tree near a stream, it usually falls into the water, but they do
not plan it that way. The fact is that most trees lean toward the water to start with.
Now what about dam building? Most beaver dams are wonders of engineering. The best ones are strongly built of
trees, stones, and mud. They are wide at the bottom and narrow at the top.
Beavers think nothing of building a dam more than 200 feet long. One dam in Montana was more than 2,000 feet
long. The largest one ever seen was in New Hampshire: it stretched 4,000 feet, and made a lake large enough to
hold 40 beaver homes.
So beavers do build good dams. But they don't always build them in the right places. They just don't plan. They will
build a dam across the widest part of the stream. They don't try to find a place where the stream is narrow. So a
lot of their hard work is wasted.
Beavers should learn that it's not enough to be busy. You have to know what you're doing, too. For example, there
was one Oregon beaver that really was a worker. It decided to fix a leak in a man-made dam. After five days of
work it gave up. The leak it was trying to block was the lock that boats go through.
## 54. What is the main idea of this passage?
A. Beavers may be hard-working animals, but they don't always choose the most efficient mechanisms.
B. Beavers are excellent dam builders.
C. New Hampshire was the site of the largest beaver dam.
D. Beavers are well-developed tree cutters.
E. Beavers are poor surveyors of aquatic environments in some cases.
The raisin business in America was born by accident. It happened in 1873 in the San Joaquin Valley of California.
Many farmers raised grapes in this valley. That year, just before the grape harvest, there was a heat wave. It was
one of the worst heat waves ever known. It was so hot that the grapes dried on the vines. When they were picked,
California had its first raisin crop.
People were surprised to find how good raisins were. Everybody wanted more. So the San Joaquin farmers went
into the raisin business. Today, of course, they do not let the grapes dry on the vines. They treat them with much
more care.
In late August the grapes start to ripen. They are tested often for sweetness. The growers wait until the sugar
content is twenty-one percent. Then they know the grapes are ripe enough to be picked.
Skilled workers come to the vineyards. They pick the grapes by hand in bunches. The workers fill their flat pans
with grapes. They gently empty the pans onto squares of paper. These squares lie between the long rows of vines.
They sit in the sun.
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Here the grapes stay while the sun does its work. It may take two weeks or longer. The grapes are first dried on
one side. When they have reached the right color, they are turned to dry on the other side. The grapes are dried
until only fifteen percent of the moisture is left. Then they have turned into raisins.
The raisins are rolled up in the paper on which they have dried. Trucks take them from the fields. They are poured
into big boxes called sweatboxes. Each box holds 160 pounds of raisins. Here, any raisins that are too dry take
moisture from those that have too much. After a while, they are all just moist enough.
The big boxes are trucked next to the packaging plant. They are emptied onto a conveyor belt that shakes the
raisins gently. This knocks them from their stems. A blast of air whisks the stems away. The water bath is next.
Then the plump brown raisins have a last inspection. They are again checked for moisture and sugar. Then they go
on a belt to packing machines. Here they are poured into packages, which are automatically weighed and sealed.
The raisins are now ready for market.
## A. The creation of raisins in America was an accident.
B. The process of raisin development requires multiple steps.
C. Raisins on the grocery store shelf undergo a brief fermentation process.
D. Raisins are cleaned thoroughly at the packing plant.
E. California has been the leader in American raisin development.
In 1976, Sichan Siv was crawling through the jungle, trying to escape from Cambodia. By 1989, however, Siv was
working in the White House in Washington D. C., as an advisor to the President of the United States. How did this
Like millions of Cambodians, Siv was a victim of a bloody civil war. One of the sides in this war was the Cambodian
government. The other was a group called the Khmer Rouge. When the Khmer Rouge won the war, the situation in
Cambodia got worse. Many people were killed, while others were forced into hard labor. Sometimes entire
families were murdered.
Siv came from a large family that lived in the capital of Cambodia. After finishing high school, Siv worked for a
while with a Cambodian airline company. Later, he taught English. After that, he took a job with CARE, an American
group that was helping victims of the war.
Siv had hoped to leave Cambodia before the Khmer Rouge took over the country. Unfortunately, he was delayed.
As a result, he and his family were taken from their homes and forced to labor in rice fields. Eventually, Siv
managed to escape. He rode an old bicycle for miles, trying to reach Thailand where he would be free and safe. For
three weeks, he slept on the ground and tried to hide from the soldiers who were looking for him. Caught at last,
he was afraid he would be killed. Instead, he was put into a labor camp, where he worked 18 hours each day
without rest. After several months, he escaped again, and this time he made it. The journey, however, was a
terrifying one. After three days of staggering on foot through mile after mile of thick bamboo, Siv finally made his
way to Thailand.
Because he had worked for an American charity group, Siv quickly found work in a refugee camp. Soon he was on
his way to the United States. He arrived in June of 1976 and got a job-first picking apples and then cooking in a
fast-food restaurant. Siv, however, wanted more than this: he wanted to work with people who, like himself, had
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suffered the hardship of leaving their own countries behind. Siv decided that the best way to prepare for this kind
of work was to go to college. He wrote letters to many colleges and universities. They were impressed with his
school records from Cambodia, and they were impressed with his bravery. Finally, in 1980, he was able to study at
Columbia University in New York City. After finishing his studies at Columbia, Siv took a job with the United
Nations. He married an American woman and became a citizen. After several more years, he felt that he was very
much a part of his new country.
In 1988, Siv was offered a job in the White House working for President Ronald Reagan's closest advisors. It was a
difficult job, and he often had to work long hours. However the long hard work was worth it, because Siv got the
opportunity to help refugees in his work.
## A. Persistence and courage are global ideas.
B. Siv covered a large area during his life.
C. Siv persevered to escape from Cambodia.
D. Siv overcame numerous challenges to come to America and help others.
E. Siv persevered to become an American citizen.
When you want to hang the American flag over the middle of a street, suspend it vertically with the blue field
(called the union) to the north and east-west street. When the flag is displayed with another banner from crossed
staffs, the American flag is on the right. Place the staff of the American flag in front of the other staff. Raise the flag
quickly and lower it slowly and respectfully. When flying the flag at half-mast, hoist it to the top of the pole for a
moment before lowering it to mid-pole. When flying the American flag with banners from states or cities, raise the
nation's banner first and lower it last. Never allow the flag to touch the ground.
## A. The American flag is the symbol of American freedom.
B. The American flag has fifty stars.
C. Placing the American flag inappropriately will draw government intervention.
D. American flag should be flown differently in certain situations.
E. The flag should be lowered quickly and respectfully.
What if someone told you about a kind of grass that grows as tall as the tallest trees? A grass that can be made as
strong as steel? A grass from which houses, furniture, boats, and hundreds of other useful things can be made? A
grass that you would even enjoy eating? Would you believe that person? You should, for that grass is bamboo, the
"wood" of 1,001 uses.
Bamboo may look like wood, but it is part of the family of plants that includes wheat, oats, and barley. It is a kind
of grass. This grass is not just a material for making useful products. Young bamboo is eaten, often mixed with
other vegetables, in many Asian foods.
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Bamboo grows in many parts of the world. In the United States it grows in an area from Virginia west to Indiana
and south to Florida, Louisiana, and Texas. Most bamboo, however, is found in warm, wet climates, especially in
Asia and on the islands of the South Pacific Ocean.
In most Asian countries, bamboo is nearly as important as rice. Many Asians live in bamboo houses. They sit on
bamboo chairs and sleep on bamboo mats. They fence their land with bamboo and use it to cage their chickens
and pigs.
Bamboo is used to build large buildings as well as homes. When it is glued in layers, it becomes as strong as steel.
On some islands in the South Pacific, bamboo is even used for water pipes. This extraordinary material has many
other uses. It is used to make musical instruments such as flutes and recorders. Paper made from bamboo has
been highly prized by artists for thousands of years.
Bamboo is light and strong, and it bends without breaking. It is cheap, floats on water, almost never wears out, and
is easy to grow. Nothing else on earth grows quite so fast as bamboo. At times you can even see it grow! Botanists
have recorded growths of more than three feet in just 24 hours! Bamboo is hollow and has a strong root system
that almost never stops growing and spreading. In fact, only after it flowers, an event that may happen only once
every 30 years, will bamboo die.
There are more than 1,000 kinds of bamboo. The smallest is only three inches tall and one-tenth of an inch across.
The largest is more than 200 feet in height and seven inches in diameter. No wonder, then, that the lives of nearly
half the people on earth would change enormously if there were no longer any bamboo. No wonder, too, that for
many people, bamboo is a symbol of happiness and good fortune.
## A. Bamboo has at least 2,000 uses.
B. Bamboo grows at an amazing rate and is found primarily in Asia.
C. Bamboo is an amazing grass that can be used in multiple ways.
D. There are at least 1,000 types of bamboo.
E. Bamboo could be considered a flower in some cases.
Every year since 1986, some of the world's most daring runners have gathered in the desert of Morocco. They are
there to take part in one of the most difficult races in the world. The Marathon of the Sands, as it is called, covers
over 125 miles of desert and mountain wilderness. The runners complete the course in fewer than seven days, and
they run with their food, clothing, and sleeping bags on their backs.
The Marathon of the Sands was founded in 1986 by Patrick Bauer. His idea was to give the runners, who come
from all over the world, a special kind of adventure. Most of the runners in this race have found that they form
deep friendships with the other runners during their days and nights in the desert. Facing terrible heat and
complete exhaustion, they learn much about themselves and each other.
For most of the runners, however, the challenge of the race is the main reason for coming. On the first day, for
example, they run 15 miles across a desert of sand, rocks, and thorny bushes. Few runners finish the day without
blistered and raw feet. Because they are allowed less than nine quarts of water during each day of the race, they
also suffer from a lack of water. Most of all, they are exhausted when they arrive at the campsite for the night.
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The second day, the runners awaken at 6:00 a.m. Within a few hours, it is 100 degrees Fahrenheit, but the runners
do not hesitate. They must cover 18 miles that day. That night, they rest. They must be ready for the next day's
run.
On the third day, the runners must climb giant sand dunes-the first they have faced. Dust and sand mix with the
runners' sweat. Soon their faces are caked with mud. After 15 miles of these conditions, the runners finally reach
their next camp.
The race continues like this for four more days. The fourth and fifth days are the worst. On the fourth day, the
runners pass through a level stretch and a beautiful, tree-filled oasis, but then, on this and on the next day, they
cross more than 21 miles of rocks and sand dunes. The temperature soars to 125 degrees Fahrenheit, and many
runners cannot make it. Helicopters rush fallen runners to medical help. Runners who make it to the end of the
fifth day know that the worst is over.
On the sixth day, heat and rocks punish the racers terribly. In the Valley of Dra, the wind picks up and, as the
desert heat is thrust against them with great force, they grow more and more exhausted.
The seventh day is the last, with only 12 miles to be covered. The dusty, tired, blistered runners set out at
daybreak. Near the finish line, children race along with the runners, for everybody has caught the excitement. The
ones who have run the whole marathon know they have accomplished what most people could not even dream of.
"During the hard moments," says one contestant who has raced here twice, "I'd think, 'Why am I here?' Then I'd
realize I was there to find my limits."
## 59. What is the main idea of this passage?
A. The Marathon of the Sands race tests the limits of human endurance.
B. The runners run at their own paces.
C. The race causes the strong to stumble and the weak to not finish.
D. The seventh day is the hardest day of the race.
E. Every runner runs the race to find their human limits.
High in the Andes Mountains in Peru stands the ancient city of Machu Picchu. No one knows why this great city
was built, nor is it likely that anyone will ever know. Nevertheless, the deserted city of Machu Picchu is important
for what it reveals about the ancient Inca people of South America.
The Incas once ruled a great empire that covered a large part of the South American continent. The empire was
more than 500 years old when the first Spanish explorers, looking for gold, went to that continent in the 16th
century.
The Incas were an advanced people. They were skillful engineers who paved their roads and built sturdy bridges.
They plowed the land in such a way that rains would not wash away valuable soil, and dug ditches to carry water
into dry areas for farming.
Even though they did not know about the wheel, the Incas were able to move huge stone blocks-some as heavy as
10 tons-up the sides of mountains to build walls. The blocks were fitted so tightly, without cement of any kind, that
it would be impossible to slip a knife blade between them! The walls have stood firm through great storms and
earthquakes that have destroyed many modern buildings.
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The Incas were great artists, too. Today, Incan dishes and other kinds of pottery are prized for their wonderful
designs. Because both gold and silver were in great supply, the Incas created splendid objects from these precious
metals.
While it is true that the Incas had no written language, they kept their accounts by using a system of knotted
strings of various lengths and colors. The sizes of the knots and the distances between them represented numbers.
At its height, the Incan Empire included as many as 30 million people. The emperor ruled them with an iron hand.
He told his subjects where to live, what to plant, how long they should work, and even whom they could marry.
Because he owned everything, the emperor gave what he wished when he wished-and in the amount he wished-to
his people.
In 1533, Spanish explorers led by Francisco Pizarro murdered the emperor of the Incas. Earlier, the heir to the
Incan empire had also been killed. The Incas, who had always been entirely dependent on their emperor, now had
no recognized leader. The Spaniards easily conquered the empire and plundered its riches.
Have the Incas disappeared from South America? Not at all. In Peru alone, once the center of that great empire, 80
percent of the 20 million people are descendants of the Inca people. Evidence of the Incan empire can be found in
many other places in South America as well. Tourists can even visit Machu Picchu. The remains of this ancient city
still stand high in the mountains of Peru, an awesome tribute to this once powerful empire.
## A. The Incas once inhabited the ancient city of Machu Picchu.
B. Peru was the primary country of the Incas.
C. The Incan Empire can be found in ancient cities and was plundered by the Spanish.
D. Spanish conquerors destroyed the Incan empire in the 13th century.
E. Machu Picchu was the capital of the Incan empire.
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1. B: "Terrestrial" means land. No choice here offers a synonym for "marine," e.g. nautical/naval/water/seagoing,
and no other choices match either marine or terrestrial.
2. A: "Quagmire" means literally a bog or marsh, and figuratively an involved situation difficult to escape;
entanglement is a synonym, more specifically similar than the other choices.
3. A: Longitudes are imaginary geographical lines running north and south. Latitudes run east and west. The other
choices do not equal either latitude or longitude in direction.
4. C: Topography means the physical features of a land mass. It does not mean coastline (A), mountain range (B),
or islands (D).
5. C: A peninsula is a piece of land connected to the mainland by an isthmus and projecting into the ocean such
that it is surrounded on three sides by water. A peninsula is not a coast (A); it is not found inland (B); and it is not a
border (D).
6. B: The passage was found near 50 degrees S latitude. Latitudes are measured horizontally, in relation to the
equator or central imaginary line, equidistant between the North and South Poles. Longitudes are measured
vertically. Greenwich (A), the location of zero degrees longitude, adopted as the global standard, is both incorrect
and never named in the passage. Spain (C), Portugal (D), and Madrid (E) in Spain are also incorrect.
7. A: Meridians are imaginary geographical circles intersecting the poles. Imaginary lines parallel to the equator (B)
are latitudes. The International Date Line is a specific meridian, not an area (C). It is not a land mass (D) as it
crosses both water and land.
8. A: "Amicable" means friendly. It does not mean competitive (B), i.e. oppositional, ambitious, or aggressive;
courteous (C), i.e. polite; industrious (D), i.e. hard-working; or chemistry (E): their collaboration was in physics, but
moreover, the passage specifically describes their collaboration as "amicable."
9. B: "Blithe" means light-hearted. It does not mean strong (A), humorous (B) or funny; strange (D), or envious (E).
10. B: "Disgruntled" means annoyed. It does not mean hopeless (A), depressed (C), or worried (D).
11. A: Marie challenged authority by going to study at the Sorbonne, because Warsaw's university did not admit
women. The passage indicates this challenge by describing her "defiantly" leaving Poland for France; i.e., she was
defying authority. The passage does not indicate she showed intelligence (B), "behaved" (C), or was distressed (D)
or upset by her move.
12. A: A synonym for "despondently" is "dejectedly," meaning sadly, with despair or depression. The passage
indicates this by describing Curie's emotional state as one of "heartbreaking anguish" over her husband's sudden
accidental death. She is not described in this passage as worried (B) by her memories, or recalling them tearfully
(C), happily (D), or irefully (E), i.e. angrily.
13. C: The closest synonym for the "feeling of desolation" (despair) described in the passage is wretchedness.
Misfortune (A) or ill fate/luck is not as close. Anger (B) is a separate emotion from desolation. Disappointment (D)
is also different from desolation, meaning feeling let-down rather than hopeless. Ambition (E) is drive to succeed
or accomplish things. It was not Curie's ambition that faded upon returning to the Sorbonne but her depression.
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14. C: "Disillusioned" means disappointed. It does not mean troubled (A), i.e. concerned or disturbed; worried (B)
or anxious; sorrowful (D) or sad; or disturbed (E).
15. B: "Coagulated" means solidified. Liquid (A) is an opposite of solid. Flowing (C) assumes a liquid, not solid,
state. Gas (D) is another opposite of solid. (Three states of matter, like volcanic material, are liquid, solid, and
gaseous.)
16. A: "Buoyant" means able to float. The passage indicates this by indicating that the gases therefore, sank
toward earth and suffocated people. Buoyant does not mean visible (B) or possible to see. Able to float/buoyant
does not mean able to evaporate (C). Evaporation means turning to vapor, which only liquids can do. Gases are
already vapors. Buoyant does not mean invisible (D) or unseen. Able to float does not mean able to condense (E),
i.e. turn from vapor to liquid.
17. B: "Dissect" means to cut apart for study. It does not mean to describe in detail (A), to photograph (C), or to
chart (D) a specimen.
18. B: Meteorologists are scientists who study atmospheric conditions, particularly weather. Scientists who study
oceans (A) are oceanographers, i.e. marine scientists. Scientists who study ash (C) do not exist as members of a
separate discipline. Climate scientists and many others concerned with its effects study volcanic ash. Scientists
who study animal behavior (D) are ethologists or animal behaviorists and do not study ash.
19. C: Distilled water is purified water. Distilled water is not equivalent to bottled (A), volcanic (B), sea (D), or
fountain (E) water.
20. A: "Supremacy" means unlimited power, not unrestricted growth (B). The passage states that Drake diminished
Spain's supremacy, but does not specifically mention diminishing its territory (C). Drake's raids enriched England
and reduced Spain's power; no mention is made of eliminating any treaties (D).
21. B: "Robust" means strong. It does not mean warlike (A), accomplished (C) or competent, timid (D) or fearful, or
inexperienced (E).
22. B: "Martial" means warlike or war-related. It does not mean complete (A), independent (C), or isolated (D).
23. C: "Vulnerable" means open to attack or susceptible to harm. It does not mean open to change (A) or
receptive, triumphant (B) or victorious, defeated (D) or beaten-they were vulnerable to attack first and then
consequently were defeated-or discouraged (E), i.e. disheartened or dispirited.
24. A: The passage indicates the Armada was "blocked" on one side, i.e. closed off rather than damaged (B) (it was
damaged extensively, not on one side); alone (C) or circled (D), i.e. surrounded, neither of which can be done on
only one side.
25. B: "Interceded for" means intervened on behalf of, not refused help to (A), wanted to fight (C), given orders for
all to fight (D), or defeated (E).
26. C: "Pacify" means to calm or make peaceful. It does not mean to make weaker (A), to destroy (B), or to irritate
(D), i.e. annoy or provoke.
27. B: "Ennobled" means gave honor to or made noble. It does not mean gave comfort (A) or solace, gave strength
(C), i.e. fortified or reinforced, gave fear (D) or frightened, or gave hope (E) or encouraged.
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28. B: To "abet" means to enable, support, or encourage, usually in crime or doing something wrong. It does not
mean to end (A), think about (C), or daydream about (D) something.
29. B: "Reinforced" means strengthened, not welcomed (A), held (C), or captured (D).
30. B: The passage states that the presence of Eris, goddess of discord, "always embroiled mortals and immortals
alike in conflict." Embroiling them in conflict is creating conflict amongst them. It does not mean scheming against
(A) them, feeling hostile toward (C) them, ignoring (D) them, or comforting (E) them.
31. A: "Aggressively" means boldly. It does not mean effectively (B) or successfully, secretly (C), or carefully (D).
32. A: "Contradicted" means Athena disregarded Hera's statement and disputed or countered it. It does not mean
she defeated (B) her statement, agreed with (C) it, restated (D) it, or questioned (E) it.
33. B: To "clamor for" means to cry out for (something). It does not mean to long for (A) it, beg (C) for it, hope (D)
for it, or think much (E) "for," of, or about it.
34. A: To "vouch" means to give assurance. It does not mean to think (B), hope (C), or convince some (D).
35. A: "Disclaimed" means denied, i.e. refused or declared untrue. It does not mean stopped (B), noted (C), or
justified (D), i.e. substantiated or confirmed, the opposite of denied.
36. C: She was unable to invoke, i.e. to call upon, the aid of relatives. To invoke does not mean to locate (A) or find;
to speak about (B) or discuss; to identify (D), i.e. recognize; or to know (E).
37. B: "Declaimed" means spoke forcefully. It does not mean finally appeared (A). Though she did also give
testimony (C) in court, "declaimed" does not mean to testify; it describes the way she spoke while doing so.
"Declaimed" also does not mean she gave evidence (D).
38. B: "Profusely" means abundantly, copiously, or excessively. It does not mean wisely (A) or carefully (B), which
are both opposite in meaning to the excessive connotation of profuse spending. Foolishly (D) can be associated
with spending profusely, but does not have the same meaning. Profusely does not mean joyfully (E), i.e. gleefully
or happily.
39. B: "Spendthrifts" means money wasters. It does not mean aristocrats (A), i.e. nobles or privileged people. It
does not mean enemies (C) or adversaries. It does not mean individuals (D) or persons. "Spenders" (E) denotes
people who spend, but does not convey the sense of wasteful spending or squandering in the same way that
"spendthrifts" does.
40. B: "Repulsive" means distasteful. It does not mean boring (A), exciting (C), or needless (D).
41. A: "Impulsive" means acting on impulse, i.e. acting without thinking. People thinking the Wrights "impulsive
fools" does not mean they thought the Wrights had been negatively influenced (B), too cautious (C), mistaken (D),
or had acted in a negative way (E).
42. C: A "compulsion" is a need or an urge to act. It is not a financial empire (A), a plan (B), or a foolish thought (D).
43. C: "Repellent" means offensive or hateful; in other words, Lilienthal's idea was disliked by the Wrights. It does
not mean his idea was opposite to the idea of (B) the Wrights. It means the opposite of its being accepted by (D)
the Wrights. They found his idea unpleasant rather than improving (E) on it.
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44. B: "Repealed" means invalidated, i.e. disproven or overturned. It does not mean destroyed (A); multiplied (C),
i.e. increased/approved (D), an antonym; or unused (E).
45. A: "Propulsion" is force for propelling or moving forward. It does not mean force for turning around (B), turning
(C) (oscillation perhaps), or force for going backward (D) (like repulsion).
46. d. Because clear plastic sheets were tacked over the windows to keep the heat in. Another clue is that the
gas stove warmed the children at night.
## 49. a. Immaculate means very clean.
50. b. Choice a is incorrect because wooden crates are use to store toys. No mention is made of a bathroom, and
one small window does not light a large room.
51. D: The passage describes actions of various First Ladies as examples of their importance in American culture.
That they are key supporters of the Presidents (E) is not the main idea because the first paragraph states some
First Ladies are remembered for influencing their husbands, while others "...have made the history books on their
own." Not all First Ladies are described here as doing humanitarian work (A). No one First Lady is singled out as
most important [(B), (C)].
52: D: The main idea is the importance of potatoes in America. It never mentions Ireland or any Potato Revolution
(A). (B) is both incorrect-the passage states 100 lbs., not 50-and regardless of accuracy, is a detail, not the main
idea. Readers already know French fries are made from potatoes (C), a detail the passage assumes. Several various
terms for potatoes are mentioned in the second paragraph, but their history (E) is never discussed.
53. E: All three benefits of patents-inventor protection, free enterprise, and the creative process-are given equal
importance in the passage. The other four choices each accurately identify individual ideas in the passage, but
none incorporates all three parts of the main idea.
54. A: Only this choice identifies the main idea, that beavers are hard-working but not always efficient. Each of the
other choices identifies one detail included in the passage, not the main idea.
55. B: The multiple steps required in the process are outlined throughout the passage. The introductory statement
that the industry began by accident (A) is a detail, not the main idea. Shelf fermentation (C) is never mentioned. A
water bath is mentioned, not thorough cleaning (D), and is a detail, regardless. California is only mentioned as the
location of the first raisin crop but never identified as raisin development's leader (E).
56. D: The passage focuses on the many challenges overcome by the subject, Siv, in particular, rather than
mentioning any global nature of his persistence and courage (A) or the size of the area he covered (B). His
perseverance to escape (C) is true, but only part of the main idea, not mentioning his desire to help others. His
attaining American citizenship is mentioned, but his perseverance was not for this (E), but for escaping Cambodia
and helping other refugees.
57. D: The passage instructs how to fly the flag in different situations. It never mentions the flag's symbolism (A) or
its number of stars (B), or any government intervention (C). It states the flag should be lowered slowly, not quickly
(E).
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58. C: The many uses of bamboo, and the fact it is a grass, are the main focus. "1,001 uses" is a non-literal
colloquial expression meaning a great many; the passage never states factually that bamboo has at least 2,000 (A).
Bamboo's growth rate and its location in many parts of the world, especially Asia (B), and the number of types (D),
and the fact that it occasionally flowers (E), are details supporting/informing the main idea.
59. A: Testing human endurance limits is illustrated in this description of a punishing marathon. It never mentions
runners running at their own paces (B). The race's challenges, and many not finishing (C) are details informing the
main point. The seventh/last day is not the hardest (D); the fourth and fifth are identified as worst. (Regardless of
accuracy, this is also a detail, not the main idea.) One, not every (E), runner is quoted as competing "to find my
limits."
60. C: This choice best summarizes the passage's main points. Choice (D) incorrectly identifies the 13th century
instead of the 16th century. The passage never indicates that Machu Picchu was the capital of the Incan empire
(E). Answers (A) and (B) are details in support of the main idea.
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Language Practice
Comma Practice Test Questions
he following sentences either have existing or require additional commas somewhere in their structures. Choose
the option that best reflects proper comma usage in each sentence.
1. For the Thanksgiving reunion, relatives were sitting in the dining room, on the porch, and in the carport.
A. Thanksgiving, reunion
B. were, sitting
C. porch and
D. No error
## 2. Lydia seems to be a kind, considerate girl.
A. seems, to
B. considerate, girl
C. kind considerate
D. No error
A. pole, Nathan,
B. has, seen
C. Nathan
D. No error
## 4. My cousin has moved to 56 Central Street Narragansett, Rhode Island 02882.
A. has moved,
B. Central Street,
C. 56, Central
D. No error
## 5. The badger, a shy animal sometimes makes friends with a coyote.
A. sometimes, makes
B. friends, with
C. a shy animal,
D. No error
6. After the death of Blackbeard, the famous pirate, piracy disappeared from the coast of the American colonies.
## A. the famous pirate
B. after the death,
C. coast, of
D. No error
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7. "Silent Night" was written by two men from the village of Oberndorf Austria.
A. men, from
B. "Silent Night,"
C. Oberndorf, Austria
D. No error
8. On November 19, 1929 Admiral Richard E. Byrd flew the Floyd Bennett to the base of the Queen Maud
Mountains.
A. base, of
B. the, Queen
C. 1929,
D. No error
## 9. Oh I forgot to bring the cookies.
A. Oh,
B. I, forgot
C. to, bring
D. No error
10. "The boy in the kayak," whispered Sue "is the new football captain."
A. boy, in the
B. new, football
C. whispered Sue,
D. No error
1. D: No error. There is a comma after the initial modifying prepositional phrase and after the first and second
modifying prepositional phrases in the series of three. No comma belongs between an adjective and the noun it
modifies (A), or between an auxiliary verb and verb (B). Omitting the second comma setting off the first modifying
prepositional phrase (C) is wrong.
2. D: No error. A comma belongs between two consecutive adjectives modifying the same noun. A comma
between verb and object (A) is incorrect. So is one between an adjective and the noun it modifies (B). Omitting a
comma between two consecutive adjective (C) is incorrect.
3. A: The comma after "Nathan" is correct, but there should also be another comma before it. When an address to
someone by name is inserted mid-sentence-here between subject and object-it should be set off by commas on
both sides. There should not be a comma between auxiliary verb and verb (B). Having no commas to set off the
inserted name (C) is incorrect.
4. B: There should be a comma between the street address and the city when stating a full address in sentence
form (as well as between the city and state, as there is here). There should not be a comma between the verb and
prepositional phrase (A), or between street number and street name (C).
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5. C: A modifying phrase between subject and predicate should be set off by commas on both sides. Putting a
comma between adverb and verb (A) or between object and preposition (B) is incorrect.
6. D: No error. The phrase modifying the subject is set off by commas both before, and after it. Removing the
second comma (A) is incorrect. A comma between a noun and its modifying prepositional phrase [(B), (C)] is
incorrect.
7. C: There should always be a comma between a village and country, city and state, state and country, or country
and continent. There should not be a comma between the noun and modifying preposition (A), or between the
subject and verb (B).
8. C: When a date is used in a modifying prepositional phrase before the subject and verb, it should have a comma
after it (before subject-verb). There should not be a comma between a noun and modifying preposition (A) or
between an article and the noun it modifies (B).
9. A: A comma should follow an interjection like "Oh" at the beginning of a sentence. (In some sentences, other
punctuation like an exclamation point is acceptable.) A comma between subject and verb (B) is incorrect. A comma
in the middle of an infinitive (C) is incorrect.
10. C: When a non-quotation clause/phrase is inserted in the middle of a quotation, it should be set off by commas
on both sides. There should not be a comma between a noun and its modifying prepositional phrase (A), or
between an adjective and the noun phrase it modifies (B).
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Grammar Practice Questions
Search for grammatical errors in the underlined sections of the following sentences and select the option that best
corrects them. If there is no error, choose option A.
1. Everyone in the bank-including the manager and the tellers, ran to the door when the fire alarm rang.
A. tellers, ran
B. tellers: ran
D. tellers-ran
E. tellers' ran"
## 2. To no ones surprise, Joe didn't have his homework ready.
A. no ones surprise
B. noones surprise
C. no-ones surprise
D. no ones' surprise
E. no one's surprise
3. If he would have read "The White Birds," he might have liked William Butler Yeats' poetry.
4. After the hurricane, uprooted trees were laying all over the ground.
A. were laying
B. lying
C. were lying
D. were laid
E. was laid
5. Ralph Waldo Emerson (1803-1882), the great transcendentalist philosopher, wrote in his essay "Self-Reliance"
of the need for an individual to develop his capacities.
A. essay "Self-Reliance"
B. essay, "Self-Reliance"
C. essay: Self-Reliance
D. essay, Self-Reliance
E. essay; "Self-Reliance"
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6. The recently built children's amusement park has been called "a boon to the community" by its supporters
and "an eyesore" by its harshest critics.
## A. and "an eyesore" by its harshest
B. and, "an eyesore," by its harshest
C. and, an eyesore; by its harshest
D. and-an eyesore-by its' harshest
E. and-"an eyesore"-by its' harshest
7. I always have trouble remembering the meaning of these two common verbs, affect (to change" or "to
influence") and effect ("to cause" or "to accomplish)."
A. "to accomplish)."
B. "to accomplish").
C. "to accomplish).
D. To accomplish.
E. "to accomplish.")
8. My class just finished reading-"The Fall of the House of Usher", a short story by Edgar Allan Poe.
## A. reading-"The Fall of the House of Usher",
B. reading, The Fall of the House of Usher,
C. reading "The Fall of the House of Usher,"
D. reading, "The Fall of the house of Usher,"
E. reading: The Fall of the House of Usher-
## 9. After it was repaired it ran perfect again.
A. ran perfect
B. ran perfectly
C. could run perfect
D. could of run perfect
E. would run perfectly
## A. there two E's in beetle," asked Margo?
B. their two E's in beetle?" asked Margo.
C. their two E's in beetle," asked Margo.
D. there two E's in beetle?" asked Margo.
E. there two E's in beetle, asked Margo?
11. The circus audience received a well-deserved round of applause for the perfectly timed acrobatic stunt.
## A. audience received a well-deserved
B. audience gave a well deserved
C. audience did receive a well deserved
D. audience gave a well-deserved
E. audience did get a well-deserved
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12. Looking directly at me, Mother said, "These are your options: the choice is yours."
## A. Mother said, "These are your options: the choice is
B. Mother said-these are your options, the choice is
E. Mother said, "These are your options; the choice is
## A. porcus, "pig," and spina, "spine."
B. Porcus-pig and spina, "spine."
C. Porcus-pig, and Spina, "spine."
D. Porcus-Pig-,Spina-spine.
E. Porcus, "pig," and spina "spine".
14. Seeing the dolphins, some sharks, a killer whale, and a Moray eel made the visit to the marine park
worthwhile.
## A. a killer whale, and a Moray eel made the visit
B. a killer whale, and a moray eel made the visit
C. a killer whale and a moray eel makes the visit
D. a killer whale and a Moray eel makes the visit
E. a killer whale and a moray eel made the visit
15. Still, the fact that a planet exists outside our solar system encourages hope that other solar systems exist,
and in them, perhaps, a planet that supports life.
A. that a planet exists outside our solar system encourages hope that other solar systems exist, and
B. that a Planet exists out side our solar system encourages hope that other solar systems exist and
C. could be that a planet exists outside our solar system encourages hope that other solar systems exist, and
D. that a planet exist outside our solar systems encourage hope that other solar systems exist, and
E. that a planet does exists out side our solar system encourages hope that other solar systems exist, and
16. Mail-order shopping can be convenient and timesaving with appropriate precautions, it is safe as well.
## A. can be convenient and timesaving
B. can be convenient and timesaving;
C. should be convenient and time saving;
D. could be convenient and time saving;
E. can be convenient and time-saving;
17. Among the many fields of science, no matter what turns you on, there are several fields of study.
## A. science, no matter what turns you on,
B. Science, no matter what turns you on,
C. Science, no matter which you chose,
D. Science, no matter which of these you chose-
E. science, no matter which you choose,
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18. The fact that boxing is known to cause head injuries and brain damage should lead us to inform the public
and push for a ban on boxing.
## A. should lead us to inform
B. could lead us to inform
C. should of led us to inform
D. will lead us to inform
E. should have led us to inform,
19. The first part of the test was on chemistry, the second on mathematics, and the third on english.
## A. on mathematics, and the third on english.
B. on mathematics; and the third on English.
C. on Mathematics; and the third on English.
D. on mathematics, and the third on English.
E. on mathematics: and the third on English.
20. The Diary of Anne Frank showed a young girl's courage during two years of hiding.
## A. showed a young girl's courage
B. shows a young girl's courage
C. did show a young girls courage
D. has shown a young girl's courage
E. showed a young girls courage
## A. will be married for twenty-five years.
B. shall have been married for twenty-five years.
C. will have been married for twenty-five years.
D. will be married for twenty five years.
E. will have married for twenty-five years.
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1. D: The modifying phrase inserted between subject and predicate should be set off on both sides by dashes, not
just one. Non-matching punctuation marks, like a dash before it but a comma after it [sentence, (A), (C)], or a dash
before but a colon after it (B), are incorrect and asymmetrical. An apostrophe (E) indicates possession and is
incorrect in a non-possessive plural noun. There is no reason for the incorrect, extraneous close-quotation mark
after the verb (E) either.
2. E: "No one's is a possessive pronoun and needs the apostrophe." Omitting it [sentence, (A), (B), and (C)] is
incorrect. "No one" is spelled as two words, not one (B) or one hyphenated word (C). An apostrophe after the s (D)
denotes a possessive plural, not a possessive singular.
3. E: The past unreal conditional should consist of "if" plus the past perfect of "to read" (auxiliary verb "had" with
"read"). Adding "would" or "could" to the past perfect [sentence, (A), (B), (C), and (D)] is incorrect. In the "If...then"
past unreal conditional construction, "would have" is only used in the second ("then" understood) clause, never in
the first "If" clause. Also, "of" [(C), (D)] is a preposition, an incorrect substitute for the auxiliary verb "have."
4. C: The correct past progressive tense of the verb "to lie" is "were lying." "Were laying" (A) is acting on an object,
e.g. "Workers were laying uprooted trees on the side of the road." Without the auxiliary verb "were," "lying" (B) is
incomplete and does not form a predicate for the subject "trees." "Were laid" (D) means somebody/something laid
them there, not that the trees themselves were lying there. "Was laid" is singular, not plural as "trees" are.
5. A: A comma (B), colon (C), or semicolon (E) is incorrect and unnecessary between the noun and its proper name.
6. A: No punctuation other than the quotation marks is required or correct after "and" and around "an eyesore."
Commas [(B), (C)], semicolons (C), or dashes [(D), (E)] are incorrect. Omitting quotation marks (D) is incorrect since
the sentence is quoting people; and the first phrase has them, so the second also should. The apostrophes [(D), (E)]
are incorrect: the irregular possessive pronoun "its" does not have an apostrophe.
7. B: The end quotation mark should come after the word but inside the end parenthesis. Putting it after the
period, outside the end parenthesis (A) is incorrect. Omitting the end quotation mark (C) is incorrect. Omitting
parentheses and capitalizing the infinitive verb example (D) are both incorrect. Omitting the open parenthesis (E) is
incorrect. Both quotation marks and parentheses always come in pairs.
8. C: There should not be any punctuation between the verb and its object, even if the object is a title in quotation
marks as it is here. Therefore, a dash (A), comma [(B, (D)], colon (E), or any combination of two [(A), (E)] is
incorrect. Additionally, omitting quotation marks around the title [(B), (E)] is incorrect.
9. B: The verb is modified by the adverb "perfectly," not "perfect" [(A), (C), (D)], an adjective for modifying a noun.
"After it was repaired" indicates past tense, so for agreement, the verb should also be the past tense "ran." "Could
run" (C) and "would run" (E) are not past tense but unreal subjunctive mood. There is no such construction as
"could of" (D), which incorrectly substitutes the preposition "of" for the auxiliary verb "have," part of the past
perfect tense.
10. D: The question mark comes after the question, inside the quotation marks. A line of dialogue or a quotation
normally has a comma [(A), (C), (E)], but inside the end quotation mark when it is a statement. When it is a
question it has a question mark, which should NOT go at the end of the sentence [(A), (E)] containing the question,
when that sentence is a statement. Also, the adverb "there" is misspelled as the possessive plural third-person
pronoun "their" in (B) and (C).
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11. D: From the context, we assume the circus acrobats performed the stunt and received the applause that the
audience gave. For the audience to receive applause makes no sense in this context [sentence, (A), (C), (E)].
Omitting the hyphen in "well-deserved" [(B), (C)] is also incorrect.
12. E: A comma, not a hyphen (B) introduces dialogue/quotations. A semicolon, not a comma (B) separates two
independent clauses. A colon (A) is incorrect, because the first clause does not introduce the second clause and is
not explained by it. (C) omits quotation marks. Past perfect (D) is not incorrect in itself, but past tense in the
original sentence was not incorrect and required no change.
13. A: A comma after each italicized Latin word and after each English translation, inside the quotation marks
surrounding the latter, is correct. Separating any of these terms with dashes is incorrect [(B), (C), (D), and (E)]. A
dash followed by a comma is always incorrect, as is separating a pair with a hyphen (D). Both pairs should be
separated by commas; (E) omits the comma from the second pair.
14. B: Each item in a series of three or more is separated with a comma. Omitting the last comma before "and"
[(C), (D), (E)] is incorrect. The term "moray eel" is not a proper name but a common name for many types of eels
and thus is not capitalized [(A), (D)] (unless it begins a sentence). Present verb tense [(C), (D)] is not incorrect, but
these choices also include the identified punctuation [(C), (D)] and capitalization (D) errors.
15. A: "A planet" is not a name, hence not capitalized; a comma should separate the independent clause from the
following phrase (B); "outside" is one word [(B), (E)]. Adding "could be" (C) changes the meaning and is also
ungrammatical, creating two unconnected predicates "...the fact could be...encourages..." requiring ", which"
before "encourages" or changing "encourages" to ", encouraging..." "Fact" and "planet" are both singular nouns;
"exist" and "encourage" (D) belong with plural nouns. The words "...does exists..." should be "...does exist" (E).
16. E: A semicolon separates independent clauses. Omitting punctuation (A), including that semicolon and the
hyphen from "time-saving" [(A), (B)], is incorrect. Spelling "time-saving" as two separate words [(C), (D)] is also
incorrect. Substituting "should" (C) or "could" (D) for "can" alters the meaning.
17. E: The word "science" is not capitalized [(B), (C), (D). The phrase "what turns you on" is slangy and not
preferred. (If it ended the sentence, it would also be incorrect for ending a sentence with a preposition.) "Which
you choose" is preferable. "Chose" [(C), (D)] is past tense, disagreeing with the present-tense predicate "are." "Of
these" (D) is redundant. The interrupting modifier "no matter..." is enclosed by commas on each side, not a comma
and dash (D).
18. A: Substituting "could" (B) or "will" (D) for "should" changes the sentence meaning. "Should of" (C) incorrectly
substitutes the preposition "of" for the auxiliary verb "have;" there is no such construction. Even the correct form
"should have led" (E) is subjunctive mood, past tense, disagreeing with the present-tense sentence context
("...boxing is known...lead..."); and a comma after "inform" is incorrect.
19. D: English is capitalized because it is a proper name as well as a school subject. Uncapitalized names (A) are
incorrect. However, mathematics, like chemistry, is a school subject but not a proper name and hence, not
capitalized (C). Semicolons [(B), (C)] only separate independent clauses, or phrases containing internal commas,
but not several phrases in a series. A semicolon (E) introduces lists or explanations but never separates phrases in a
series.
20. B: Present tense is preferable when referring to an existing book rather than past tense [(A), (C), (E)] or present
perfect tense (D). The author wrote it in the past, but the book still exists in the present. The possessive noun
"girl's" has an apostrophe, which is incorrectly omitted in (C) and (E).
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21. C: "In August" is the future, requiring the future-tense auxiliary verb "will." "Have been married" is present
perfect. Adding "will" to "have been married" makes the tense future perfect. Simple future tense "will be
married" [(A), (D)] with "for twenty-five years" literally means they will get married in August and will be married
for 25 years thereafter. "Will have married" (E) cannot be "for 25 years": being married is a continuous process;
marrying is not.
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Grammar Practice Questions
1. The word boycott derives from the name of Charles C. Boycott, an English land agent in Ireland that was
ostracized for refusing to reduce rent.
## A. that was ostracized for refusing
B. who was ostracized for refusing
C. which was ostracized for refusing
D. that had been ostracized for refusing
E. who had been ostracized for refusing
2. As a result of his method for early music education, Shinichi Suzuki has been known as one of the world's
great violin teachers.
## A. has been known as one
B. had been known as one
C. is seen as one
D. is being seen as one
E. has been one
3. Last night the weather forecaster announced that this is the most rainy season the area has had in the past
## A. this is the most rainy season the
B. this has been the most rainy season the
C. this was the most rainy season the
D. this is noted as the most rainy season the
E. this is the rainiest season the
4. Although Mandy is younger than her sister, Mandy is the tallest of the two.
## A. is the tallest of the
B. is the taller of the
C. has been the taller of the
D. is the most tall of the
E. is the more taller of the
5. When Katherine Hepburn's play came to town, all the tickets had sold out far in advance.
## A. had sold out far
B. have sold out far
C. were sold out far
D. had been sold out far
E. had been sold out for
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6. The origins of most sports is unknown.
A. sports is unknown
B. sports have been unknown
C. sports are unknown
D. sports has been unknown
E. sports are now unknown
7. Neither of the Smith brothers expect to be drafted by a major league team this year.
A. expect to be drafted
B. expects to be drafted
C. has expected to be drafted
D. is expecting to be drafted
E. was expecting to be drafted
## A. Has any of the
B. Is any of the
C. Will any of the
D. Are any of the
E. Have any of the
A. sunk
B. did sink
C. was sunk
D. did sank
E. sank
## 10. Whos in the office now?
A. Whos in
B. Whose in
C. Who is in
D. Who's in
E. Whose' in
11. There are now many kinds of dictionaries, such as a dictionary of synonyms and antonyms, a biographical
dictionary, and a geographical dictionary with pronunciations given.
## A. with pronunciations given
B. that has pronunciations given
C. with pronunciations' given
D. that have pronunciations given
E. that do have pronunciations given
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12. Towering 700 feet above the valley floor, Mount Rushmore National Memorial was an impressive site.
## A. was an impressive site
B. is a impressive sight
C. is an impressive sight
D. was an impressive sight
E. is an impressive site
## 13. San Francisco lays southwest of Sacramento.
A. lays southwest
B. has laid southwest
C. is lying southwest
D. lain southwest
E. lies southwest
14. Did they know that Labor Day always came on the first Monday in September?
A. came on
B. comes on
C. has come on
E. has came on
15. Eating, drinking, and to stay up late at night were among her pleasures.
A. to stay up late
B. to remain up late
C. staying up late
D. she liked staying up late
E. trying to stay up late
16. Each night when night came and the temperature fell, my parents lit the fire in the bedroom.
## A. and the temperature fell,
B. and that the temperature did fall
C. and that the temperature fell
D. and because the temperature fell
E. and when the temperature fell
17. Frances promised to bring the Papago basket that she bought in Arizona.
A. bought in
C. has bought in
E. purchased in
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18. He has lain his racquetball glove on the beach.
A. has lain
B. has laid
C. have lain
D. have laid
E. is lying
19. I would have lent you my notes if you would have asked me.
## A. would have asked me
20. Many scientists are still hoping to have found life on another planet.
A. to have found
B. to find
C. two find
D. to have been found
E. too have found
21. Because she had an astounding memory, Sue has never forgotten an important equation.
C. has
D. did have
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1. B: When referring to a person, use "who," not "that" [(A), (D)] or "which" (C). The past perfect "had been" [(D),
(E)] is inappropriate in this context: simple past "was ostracized" refers to the historical event itself. Past perfect
would only be used with something identified as leading up to the past event, e.g. "...who had been refusing to
reduce rent for years and finally was ostracized."
2. C: Present perfect (A) implies Suzuki is not still known thusly. Past perfect (B) implies he stopped being known
thusly in the past. Also, "known" is less accurate than "seen": the former suggests fact; the latter connotes
perception/view/opinion, the case here. Present progressive (D) is awkward and suggests the opinion is only
current and short-term. "Has been" without "seen" (E) changes the meaning from public opinion to fact-and past,
not present, fact.
3. E: "Rainiest" is the superlative form of the adjective "rainy." ("Rainier" is the comparative.) Using "most"/"more"
plus the original adjective instead of its superlative/comparative form when it has one is incorrect with one-
4. B: When comparing two things/people, use the comparative (-er/more), not the superlative (-est/most), only
used when comparing three or more. "Has been" (C) is only correct when sentence context warrants, e.g. "...has
been the taller of the two for three years." Here it is extraneous. "Most tall" (D) is doubly incorrect: once for using
superlative, not comparative; and again for using "most"(/"more") instead of "-est"(/"-er") with a one-syllable
adjective. "More taller" (E) is an incorrect double/redundant comparative.
5. D: Though common, using "sold out" in active voice with "tickets" as the subject is undesirable since tickets
cannot literally sell themselves, so passive voice is more appropriate. Also, past perfect "had been sold out" is
more correct than simple past tense "were sold out" (C) since the selling out preceded when the play came to
town (past tense). "For" (E) instead of "far" in advance is the wrong preposition/word choice for the meaning and
makes no sense.
6. C: Subject-verb agreement: The subject "origins" is plural, so the verb must agree with "are." The singular "is"
(A) or "has been" (D) is incorrect. Present perfect "have been" (B) only applies if the context dictates it, e.g. "have
been unknown until recently." Adding "now" (E) changes the meaning, implying they were previously known.
7. B: "Neither" is singular, so "expects" is correct. "Expect" (A) is plural. Present perfect "has expected" (C) is
superfluous and awkward, as are present progressive "is expecting" (D) and past progressive "was expecting" (E).
These would only apply if followed by (e.g.) "...until now" for (C) and (E) or "...until next year" (D).
8. E: "Any" can be singular or plural; in this context, plural is more appropriate. When asking questions with plural
count nouns, use "any" as plural. For singular, "Has any one of the witnesses...?" is better. "Is" (B), "will" (C), and
"are" (D) are not correct auxiliary verbs in past perfect with "been."
9. E: The past tense of "sink" is "sank." "Sunk" (A) is part of the present perfect ("has sunk"/"is sunk"/"has been
sunk"- passive voice) and past perfect (had sunk"/"was sunk" (C)]/"had been sunk"- passive voice) tenses. "Did
sink" (B) is awkward and unnecessary. "Did sank" (D) is incorrect: past-tense auxiliary verbs are never used
together with past-tense main verbs (doubling).
10. D: An apostrophe is required in "who's," a contraction of "who is." No apostrophe (A) is incorrect. "Whose" (B)
is the possessive (i.e. belonging to whom). Its irregular spelling differentiates it from the contraction "who's" (like
"its" vs. "it's"). "Whose" is never spelled with a final apostrophe (E). "Who is" (C) is not incorrect, but expanding
the contraction to full form avoids correctly identifying the contraction's correct spelling.
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11. A: This is the most economical wording of the modifying prepositional phrase. "That has" (B) is unwieldy and
superfluous. The plural "pronunciations" is not possessive and thus should not have an apostrophe (C). "Have"
[(D), (E)] is plural, disagreeing with the singular subject.
12. C: Present tense is more correct when describing something that currently still exists. Also, from the sentence
context, "sight," i.e. something to see, is the desired meaning whereas "site" [(A), (E)] means a location. Past tense
[(A), (D)] would only be correct in context, e.g. "...was an impressive sight when we visited it last year." The article
"a" (B) is incorrect before a vowel ("an" is correct).
13. E: The present tense of "to lie" is "lies." "Lays" is the present tense of the transitive (taking a direct object) verb
"to lay," e.g. "We lay books on this table." "Has laid" (B) should be "has lain," but present perfect makes no sense
here: San Francisco's location has not moved. Present progressive "is lying" (C) is similarly misleading regarding a
non-temporary location. "Lain" (D) is present perfect/past perfect, not present-and moreover lacks its auxiliary
14. B: Although the predicate is past-tense ("Did they know...?"), something that is still true, like a national holiday,
"always comes on" the same day in present tense. "Always came on" (A) implies it no longer does, as does "has
come" (C) and "had come" (D). "Has came" (E) is never used: the present perfect (has) and past perfect (had) both
take the form "come."
15. C: The series of gerunds ("-ing"-participial verbals used as nouns) require parallel structure. To agree with
"eating" and "drinking," "staying up late" is correct. The infinitive "to stay/remain" [(A)/(B)] disagrees with the
gerunds "eating, drinking." Adding "She liked..." (D) incorrectly places the third verbal into an independent clause
with another subject and verb, contradicting the sentence structure-and redundant with "were among her
pleasures." "Trying to stay up late" (E) changes the meaning.
16. A: A comma between a modifying phrase/clause and the clause it modifies is correct. Inserting "that" [(B), (C)]
is incorrect: "the temperature fell," along with "night darkness came," is introduced by the adverb "when." It is not
a restrictive relative clause introduced by "that." Past tense "fell" is preferred over the awkward "did fall" (B).
"Because" (D) is incorrect: the clause was already introduced by "when." Past-perfect "had fallen" (E) disagrees
with past-tense "darkness came" and "my parents lit..."
17. B: Past perfect is correct because Frances promised (past tense) to bring what she had bought before she
promised. Present perfect "has bought" (C) disagrees with the past-tense predicate "promised." "Did buy" (D) is
just an awkward or archaic version of past tense "bought" (A); "purchased" (E) is simply a past-tense synonym for
"bought"-all incorrect here. (Frances did not buy the basket at the same time that she promised to bring it.)
18. B: The correct present-perfect of transitive verb (i.e. it always takes a direct object) "to lay" is "has laid." "Has
lain" (A) is intransitive, e.g. "He has lain on this bed before." "Have lain" (C) uses not only the wrong verb/tense,
but also a plural auxiliary verb with a singular subject, like "have laid" (D). "Is lying" (E) should be "is laying" with
the object "racquetball glove;" but even corrected, changing the tense changes the meaning here.
19. D: In conditional-subjunctive constructions, "if..." introduces the conditional clause/phrase, and the
corresponding "then..." subjunctive uses "would have." Using "would have" in the conditional is incorrect. There is
no such construction as "could of" (B) or "had of" (E); these incorrectly substitute the preposition "of" for the
auxiliary verb "have." "Could ask" (C) is wrong in both tense and meaning.
20. B: "Hoping," like "planning"/"dreaming"/"expecting," etc., is future-oriented and in the present participle ("-
ing"), requires the infinitive in modifying verbs, i.e. "hoping to find." Scientists cannot hope "to have found" [(A),
(E)] something already that they are "still hoping" to find. "Two" (C) is the spelling of the number 2, and "too" (E) is
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the adverb meaning "also," not the preposition "to." "To have been found" errs doubly, using both present-perfect
tense and passive voice incorrectly here.
21. C: With present-perfect "has never forgotten," present-tense "has an astounding memory" is correct. "Had" (A)
and "did have" (D) are past-tense; and "has had" (E) is present-perfect tense, all implying Sue no longer has an
astounding memory, contradicting the statement that she still "has never forgotten." "Could have had" (B)
completely changes the meaning and also contradicts "has never forgotten."
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Word Usage Practice
Each underlined section corresponds to an answer choice. The first underlined section corresponds to choice A,
the second to choice B, and so on. Please select the answer choice that either contains an error or select choice E,
which is "No error."
1. Whom did you talk to at the information desk at the airport? No error.
A. A
B. B
C. C
D. D
E. E
## 2. Ellen always got into more trouble than me. No error.
A. A
B. B
C. C
D. D
E. E
3. The title of salutatorian goes to whomever has the second highest academic average. No error.
A. A
B. B
C. C
D. D
E. E
## 4. Do you feel good enough to go to the store? No error.
A. A
B. B
C. C
D. D
E. E
5. Bolivar, an idol between his contemporaries, has been the inspiration for many modern revolutions. No error.
A. A
B. B
C. C
D. D
E. E
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6. Birds fly south in the winter threw an instinct not completely understood by scientists. No error.
A. A
B. B
C. C
D. D
E. E
7. No animal has yet been discovered that can "see" infrared light with its eyes. No error.
A. A
B. B
C. C
D. D
E. E
8. Lying there in the half-dark of my room, I could see my shelf, with my books-some of them prizes I had won in
high school. No error.
A. A
B. B
C. C
D. D
E. E
9. The man who sat beside Ben and I was running for the city council. No error.
A. A
B. B
C. C
D. D
E. E
## 10. Whom did you say sent this package? No error.
A. A
B. B
C. C
D. D
E. E
11. There isn't scarcely room on the front steps to pose the entire class for a picture. No error.
A. A
B. B
C. C
D. D
E. E
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12. Haven't none of you seen my dog? No error.
A. A
B. B
C. C
D. D
E. E
13. I found the expensive vase broken when I first came in the room. No error.
A. A
B. B
C. C
D. D
E. E
14. Mrs. Clement, my English teacher, said that I could of improved my reading comprehension score if I had
spent more time reading great literature. No error.
A. A
B. B
C. C
D. D
E. E
15. If you sign up as a volunteer for the special olympics, you will find that you receive as much as you give. No
error.
A. A
B. B
C. C
D. D
E. E
16. "Your themes," said Ms. Buchanan, will be due in class on September 7; late papers will lose one full grade."
No error.
A. A
B. B
C. C
D. D
E. E
17. What should I do when the computer says, "Sorry, try again?" No error.
A. A
B. B
C. C
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D. D
E. E
## 18. "Whose in the office now?" asked Mom. No error.
A. A
B. B
C. C
D. D
E. E
19. Parking her car at the depot, Ms. Jones decided to take the bus to town. No error.
A. A
B. B
C. C
D. D
E. E
20. In 1936, Adolph Hitler refused to congradulate the great Jesse Owens, winner of four gold medals in the
Berlin Olympics. No error.
A. A
B. B
C. C
D. D
E. E
21. Preserving rare and valuable books is one of the challenges facing the Librarian of Congress. No error.
A. A
B. B
C. C
D. D
E. E
A. A
B. B
C. C
D. D
E. E
23. Without saying a word, the major gave a nod of ascent. No error.
A. A
B. B
C. C
D. D
E. E
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24. Just as they were about to go to bed, Jane told her mother, "Its my turn to wind the clock." No error.
A. A
B. B
C. C
D. D
E. E
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## 1. E: This sentence is correct as it is written.
2. D: "Than me" in the comparative is incorrect; it should be "than I." This can be deduced by adding a verb to the
pronoun to finish the thought: "...than I am," not "...than me am."
3. B: "...to whomever" would only be correct if it is a direct object in all parts of the sentence, e.g. "...to whomever
you want." However, in this sentence it is a subject in the prepositional phrase: "... to whoever has..." and thus
should be "whoever" so "who" agrees with "has." "Whoever" as subject takes precedence over "whomever" as
object.
4. B: How you feel is expressed by the adverb "well," not by the adjective "good," e.g. "I have good feelings."
5. A: "Between" only refers to two, e.g. "Between you and me;" when modifying more than two, as here since "his
contemporaries" refers to many people, "among" is the correct preposition.
6. C: The correct spelling of the preposition meaning via or by means of, as it is used here, is "through." "Threw" is
the past tense of the verb "to throw."
## 8. E: This sentence is correct the way that it is written.
9. C: "Ben and I" as an indirect object is incorrect: it should be "Ben and me." The correct personal pronoun can be
ascertained by removing the added "Ben and:" We would not write "The man who sat beside I," but "The man who
sat beside me..." and this does not change when adding another object (Ben).
10. A: "Whom" is used to indicate an indirect object, e.g. "to whom" or "for whom" did you send this package? But
this example asks the question, "Who sent this package?" and further specifies "Who did you say it was?" "Who"
agrees with "sent," not with "did you say."
11. A: "Scarcely" means "barely" or "hardly;" i.e. it minimizes, indicating very little. Only a positive quantity, like
the state of being indicated by "is," can be minimized. A negative, i.e. "is not," cannot be minimized, as nothing
exists to be made smaller. Minimizer + negative is akin to a double negative and equally incorrect.
12. B: This is a double negative as written. With the negative "Haven't" goes "any," not "none."
13. D: One comes into a room; one cannot come "in" a room, house, or situation. This is a common usage error.
"In" means already there; "into" indicates movement there from someplace else.
14. C: There is no such verb construction as "could of." "Of" is a preposition meaning belonging to or associated
with. The subjunctive mood, present perfect tense is "could have." The auxiliary verb "have" indicates the action
"improved" here as accomplished in the past (present perfect), and the auxiliary verb "could" indicates the
subjunctive mood, expressing possibility as opposed to reality.
15. A: "Special Olympics" is a name, i.e. a proper noun, and hence the initial letters of both words should be
capitalized.
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16. A: The open-quotation mark is missing before "will be" to show that the dialogue resumes following the non-
dialogue insertion of said Ms. Buchanan.
17. D: The close-quotation mark should immediately follow "again" and the question mark should come after it.
Punctuation marks such as commas, periods, semicolons, colons, etc. are placed inside of quotation marks when
the punctuation is part of the line of dialogue or quotation; however, when the punctuation mark is part of the
outer sentence that contains the dialogue or quotation, it is placed outside of the end-quotation mark.
18. A: The contraction of "Who is" is spelled "Who's." The word "Whose," used incorrectly here, is the possessive
personal pronoun meaning "belonging to whom," e.g. "Whose coat is this?"
## 19. E: This sentence is correct as it is written.
20. B: The word "congratulate" is misspelled here with a "d" instead of a "t" as it should be spelled.
21. E: The title Librarian of Congress is capitalized on the U.S. Library of Congress website whether it includes a
specific name (e.g. "Librarian of Congress Billington") or not. It is a title similar to President of the United States.* If
the sentence read only "the librarian/president," i.e. not a title or referring to a specific individual,
"librarian/president" would not be capitalized. (NOTE: This is an exception; so is POTUS.* Normally, when not
naming an individual, such terms are lower-case.)
22. C: "Everyone" is a collective noun. To agree with it, the modifying clause should read "because they had
worked..." , not "because all had worked...".
23. D: The correct spelling for the intended meaning here is "assent," i.e. agreement. The word spelled "ascent" as
it is here means a climb or upward progress, e.g. one's ascent up a mountain or one's ascent to leadership,
success, fame, wealth, etc. rather than agreement.
24. C: The correct spelling of the contraction of "it is" has an apostrophe: "It's my turn." "Its" as spelled here is the
possessive impersonal pronoun, e.g.: "This coat is missing its buttons."
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Sentence Practice
Select the answer choice that identifies the noun in the sentence.
1. It will take all of your energy and will to be able to walk again.
A. Take
B. All
C. Your
D. Energy
A. Many
B. Great
C. Placed
D. Reserve
A. Bridge
B. Was
C. Opened
D. In
## 4. Sparta and Athens were enemies during the Peloponnesian War.
A. And
B. Were
C. During
D. War
5. Sharks and lampreys are not true fish because their skeletons are made of cartilage rather than bone.
A. True
B. Because
C. Their
D. Bone
## 6. Joe, have you met your new boss?
A. Have
B. Met
C. Your
D. Boss
7. Sue's parents tried living in the north, but they could not adapt to the cold.
A. North
B. But
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C. Not
## 8. Mastering basic mathematics is an important goal for younger students.
A. Mastering
B. Important
C. Younger
D. Students
9. To seize a foreign embassy and its inhabitants is flagrant disregard for diplomatic neutrality.
A. Seize
B. Its
C. Flagrant
D. Neutrality
10. The Trojans' rash decision to accept the wooden horse led to their destruction.
A. Their
B. Led
C. Accept
D. Destruction
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1. D: Energy is a noun, as is will here. Take (A) is a verb. All (B) is an adverb modifying take. Your (C) is an adjective
modifying energy and will.
2. D: Reserve is the only noun of the choices. Many (A) and great (B) are adjectives modifying the noun poets.
Placed (C) is a verb.
3. A: Bridge is a proper noun here. Was (B) is the auxiliary verb for the past perfect tense of the verb opened (C). In
(D) is a preposition.
4. D: War is a proper noun here. And (A) is a conjunction. Were (B) is a verb. During (C) is a preposition.
5. D: Bone is a noun. True (A) is an adjective modifying the noun fish. Because (B) is a conjunction. Their (C) is a
plural possessive third-person pronoun modifying the noun skeletons.
6. D: Boss is a noun. Have (A) is the auxiliary verb for the present perfect tense of the verb met (B). Your (C) is a
possessive second-person pronoun modifying the noun boss.
7. A: North is a noun here. But (B) is a conjunction. Not (C) is an adverb modifying the verb adapt (D).
8. D: Students is a plural noun. NOTE: Mastering (A) is a gerund, i.e. a verb form functioning as a noun. But since
(D) is already a noun, it is the better choice. Important (B) is an adjective modifying the noun goal. Younger (C) is
an adjective modifying the noun students.
9. D: Neutrality is a noun. Seize (A) is a verb. Its (B) is a possessive pronoun modifying the noun inhabitants.
Flagrant (C) is an adjective modifying the noun disregard.
10. D: Destruction is a noun. Their (A) is a plural possessive pronoun modifying destruction. Led (B) and accept (C)
are verbs.
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Math Section
The Mathematics test measures a test takers ability to solve problems representing some of the key concepts in
mathematics. Some problems will only test one concept, while others will involve multiple concepts integrated
together in a single problem.
The problems will have few technical terms, aside from basics, such as area, perimeter, integer, and ratio, which
are expected to be common mathematical knowledge. All figures shown will be drawn accurately and lie in a
single plane, unless noted otherwise
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Number Types
Integers, Odd and Even Numbers, Prime Numbers, Digits
Integers, -4, -3, -2, -1, 0, 1, 2, 3, 4,
Consecutive Integers: Integers that follow in sequence; for example, 22, 23, 24, 25. Consecutive
Integers can be more generally represented by n, n + 1, n + 2, n + 3,
Odd Numbers, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9,
Even Numbers, -8, -6, -4, -2, 0, 2, 4, 6, 8, (Note: zero is an even number)
Prime Numbers, 2, 3, 4, 7, 11, 13, 17, 19, (Note 1 is not a prime and 2 is the only even prime)
Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
## even + odd = odd odd x odd = odd
Percent
Percent means hundredths or number out of 100. For example, 40 percent means 40/100 or .40 or 2/5.
## Percent less than 100
Problem 1: If the sales tax on a \$30 item is \$1.80, what is the sales tax rate?
Solution: \$1.80 = n/100 x \$30 n = 6, so 6% is the sale tax rate
Percent Greater than 100
## Problem 2: What number is 250% of 2?
Solution: n = 250/100 x 2 n = 5, so 5 is the number
Percent less than 1
## Problem 3: 3 is 0.2 percent of what number?
Solution: 3 = 0.2/100 x n
## n = 1,500, so 1,500 is the number
Percent Increase/Decrease
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Problem 4: If the price of a computer was decreased from \$1,000 to \$750, by what percent was the price
decreased?
Solution: The price decrease is \$250. The percent decrease is the value of n in the equation 250/1000 = n/100.
The value of n is 25, so the price was decreased by 25%.
## Notes: n% increase means increase/original = n/100; n% decrease means
decrease/original = n/100.
Average
An average is a statistic that is used to summarize data. The most common type of average is the
arithmetic mean. The average (arithmetic mean) of a list of n numbers is equal to the sum of the
numbers divided by n. For example, the mean of 2, 3, 5, 7, and 13 is equal to
2 + 3 + 5 + 7 + 13 / 5 = 6
When the average of a list of n numbers is given, the sum of the numbers can be found. For example if
the average of six numbers is 12, the sum of these six numbers is 12 x 6, or 72.
The median of a list of numbers is the number in the middle when the numbers are ordered from
greatest to least or from least to greatest. For example, the median of 3, 8, 2, 6, and 9 is 6 because
when the numbers are ordered, 2, 3, 6, 8, 9, the number in the middle is 6. When there is an even
number of values, the median is the same as the mean of the two middle numbers. For example, the
median of 6,
## 8, 9, 13, 14, and 16 is
9 + 13 / 2 = 11
The mode of a list of numbers is the number that occurs most often in the list. For example, 7 is the mode
of 2, 7, 5, 8, 7, and 12. The numbers 10, 12, 14, 16, and 18 have no mode and the numbers 2, 4, 2, 8, 2,
4, 7, 4, 9, and 11 have two modes, 2 and 4.
Note: The mean, median, and mode can each be considered an average. On the test, the use of the word
average refers the arithmetic mean and is indicated by average (arithmetic mean). The exception is
when a question involves average speed (see problem 2 below). Questions involving the median and
mode will have those terms stated as part of the questions text.
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Weighted Average
Problem 1: In a group of 10 students, 7 are 13 years old and 3 are 17 years old. What is the average
(arithmetic mean) age of these 10 students?
Solution: The solution is not the average of 13 and 17, which is 15. In this case the average is
## 7(13) + 3(17) / 10 = 91 + 51 / 10 = 14.2 years
The expression weighted average comes from the fact that 13 gets a weight factor of 7, whereas 17
gets a weight factor of 3.
Average Speed
Problem 2: Jane traveled for 2 hours at a rate of 70 kilometers per hour and for 5 hours at a rate of 60 kilometers
per hour. What was her average speed for the 7-hour time period?
## Solution: In this situation, the average speed is:
Total Distance/Total Time
## The total distance is 2(70) + 5(60) = 440 km.
The total time is 7 hours. Thus the average speed was
## 440/7 = 62 6/7 kilometers per hour.
Note: In this example the average speed is not the average of the two separate speeds, which would be
65.
## negative x negative = positive negative x positive =
negative positive x positive = positive
Factoring
You may need to apply these types of simple factoring:
x^2 + 2x = x(x + 2) x^2 1 = (x + 1) (x 1) x^2 + 2x +1 = (x + 1) (x + 1) = (x + 1)^2 x^2 3x 4 = (x 4)(x + 1)
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Probability
Probability refers to the chance that a specific outcome can occur. It can be found by using the
following definition when outcomes are equally likely.
Number of ways that a specific outcome can occur Total number of possible
outcomes
For example, if a jar contains 13 red marbles and 7 green marbles, the probability that a marble
selected from the jar at random will be green is
## 7 / 7 +13 = 7/20 = or 0.35
If a particular outcome can never occur, its probability is 0. If an outcome is certain to occur, its
probability is 1. In general, if p is the probability that a specific outcome will occur, values of p fall in the
range 0 < p < 1. Probability may be expressed as either a decimal or a fraction.
Geometric Figures
Figures that accompany problems are intended to provide information useful in solving the problems. They are
drawn as accurately as possible EXCEPT when it is stated in a particular problem that the figure is not drawn to
scale. In general, even when figure is not drawn to scale, the relative positions of points and angles may be
assumed to be in the order shown. Also, line segments that extend through points and appear to lie on the same
line may be assumed to be on the same line. The text Note: Figure not drawn to scale. is included on the test
when degree measures may not be accurately shown and specific lengths may not be drawn proportionally. The
following examples illustrate the way different figures can be interpreted.
Example 1
Since UY and VX are line segments, angels UWV and XWY are vertical angles. Therefore, you can
conclude that c = d. Even though the figure is drawn to scale, you should NOT make any other
assumptions without additional information. For example, you should NOT assume that VW = WY or
that the angle at vertex Y is a right angle even though they may look that way in the figure.
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Example 2
A question may refer to a triangle such as XWZ above. Although the note indicates that the figure is not
drawn to scale, you may assume that:
## (4) The length of XY is less than the length of XZ.
(5) The measure of angle XWY is less than the measure of angle XWZ.
## (2) The measures of angles WXY and WYX are equal.
(3) The measure of angle XWY is greater than the measure of angle WYX.
## Properties of Parallel Lines
1. If two parallel lines are cut by a third line, the alternate interior angles are equal.
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a = b and d = c
2. If two parallel lines are cut by a third line, the corresponding angles are equal.
a = b and d = c
Note: Words like alternate interior or corresponding are generally not used on the test, but
you do need to know which angles involving parallel lines are equal.
3. If two parallel lines are cut by a third line, the sum of the interior angles on the same side of
the third line is 180 degrees.
## a + b = 180, because a + c = 180 and b = c
Angle Relationships
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a = 70 (Because 70 + 40 + a = 180.)
a=b
## 4. The sum of the two acute angles in a right triangle is 90 degrees.
x = 15 (Because 2x + 4x = 90.)
5. The sum of the interior angles of a polygon can be found by drawing all diagonals of the
polygon from one vertex and multiplying the number of triangles formed by 180 degrees.
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Since the polygon is divided into 3 triangles, the sum of the angles is 3 x 180 or 540.
Side Relationships
1. Pythagorean Theorem: In any right triangle, a2 + b2 = c2, where c is the length of the longest
side and a and b are the lengths of the two shorter sides.
a=5
## (By the Pythagorean Theorem,
a2= 32 + 42
a2 = 9 = 16
a2 = 25
a = square root of 25 = 5
2. In any equilateral triangle, all sides are equal and all angles are equal.
a=b=5
(Because the measure of the unmarked angle is 60, the measure of all angles of the triangle are
equal, and therefore, the lengths of all sides of the triangle are equal.)
3. In an isosceles triangle, the angles opposite equal sides are equal. Also the sides opposite
equal angles are equal.
## If A = B, then a = b. Also, if a = b, then A = B.
4. In any triangle, the longest side is opposite the largest angle ( and the shortest side is opposite
the smallest angle.)
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A>B>C
5. Two polygons are similar if the lengths of their corresponding sides are in the same ratio and
their corresponding angles are equal.
If polygons ABCD and EFGH are similar, and if BC and FG are corresponding sides, then BC = 3 and FG=2.
Therefore, the ratio is 3:2 and since AB = 6, EF = 4
## Area and Perimeter
Rectangles
Area of a rectangle = length x width = l x w
Perimeter of a rectangle = 2(l + w) = 2l x 2w
Area = 5x X 8x = 40x2
Circles
## Area of a circle = r2 (where r is the radius)
Circumference of a circle = 2r = d (where d is the diameter)
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Area = 22 = 4
Circumference = 22 = 4
Triangles
## Area of a triangle = (base X height)
Perimenter = Sum of lengths
Area = (4 X 3) = 6
Perimeter = 5 + 4 + 3 = 12
Volume
## Volume of a rectangular solid or cube = length X width X height = l X w X h
Volume = 3 X 2 X 4 = 24
Volume of a cylinder = r2h (where r is the radius of the base and h is the height of the cylinder)
Volume = X 42 X 7 = X 16 X 7 = 112
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Coordinate Geometry
In questions that involve the x and y axes, x values to the right of the y axis are positive and x values to
the left of the y axis are negative. Also, y values above the x axis are positive and y values below the x
axis are negative. In an (x,y) ordered pair, the x value is written first, and the y value is written second.
For example, in the ordered pair
## Slope of a line = rise/run or vertical distance/horizontal distance.
This line runs through points (1,-2) and (4,4). The slope = (4 (2))/(4 1) or 6/3 = 2.
Any line that slopes upward from left to right has a positive slope. Any line that slopes downward from
right to left has a negative slope.
Practice Questions
Algebra
1. If Lynn can type a page in p minutes, what piece of the page can she do in 5 minutes?
A. 5/p
B. p - 5
C. p + 5
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D. p/5
E. 1- p + 5
2. If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for
both of them to paint the house together?
## A. 2 hours and 24 minutes
B. 3 hours and 12 minutes
C. 3 hours and 44 minutes
D. 4 hours and 10 minutes
E. 4 hours and 33 minutes
3. Employees of a discount appliance store receive an additional 20% off of the lowest price on an item. If an
employee purchases a dishwasher during a 15% off sale, how much will he pay if the dishwasher originally cost
\$450?
A. \$280.90
B. \$287
C. \$292.50
D. \$306
E. \$333.89
4. The sales price of a car is \$12,590, which is 20% off the original price. What is the original price?
A. \$14,310.40
B. \$14,990.90
C. \$15,290.70
D. \$15,737.50
E. \$16,935.80
## 5. Solve the following equation for A : 2A/3 = 8 + 4A
A. -2.4
B. 2.4
C. 1.3
D. -1.3
E. 0
6. If Leah is 6 years older than Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then
how old is Sue?
A. 8
B. 10
C. 14
D. 19
E. 21
7. Alfred wants to invest \$4,000 at 6% simple interest rate for 5 years. How much interest will he receive?
A. \$240
B. \$480
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C. \$720
D. \$960
E. \$1,200
8. Jim is able to sell a hand-carved statue for \$670 which was a 35% profit over his cost. How much did the
statue originally cost him?
A. \$496.30
B. \$512.40
C. \$555.40
D. \$574.90
E. \$588.20
9. The city council has decided to add a 0.3% tax on motel and hotel rooms. If a traveler spends the night in a
motel room that costs \$55 before taxes, how much will the city receive in taxes from him?
A. 10 cents
B. 11 cents
C. 15 cents
D. 17 cents
E. 21 cents
10. A student receives his grade report from a local community college, but the GPA is smudged. He took the
following classes: a 2 hour credit art, a 3 hour credit history, a 4 hour credit science course, a 3 hour credit
mathematics course, and a 1 hour science lab. He received a "B" in the art class, an "A" in the history class, a "C"
in the science class, a "B" in the mathematics class, and an "A" in the science lab. What was his GPA if the letter
grades are based on a 4 point scale? (A=4, B=3, C=2, D=1, F=0)
A. 2.7
B. 2.8
C. 3.0
D. 3.1
E. 3.2
11. Simon arrived at work at 8:15 A.M. and left work at 10: 30 P.M. If Simon gets paid by the hour at a rate of
\$10 and time and for any hours worked over 8 in a day. How much did Simon get paid?
A. \$120.25
B. \$160.75
C. \$173.75
D. \$180
E. \$182.50
12. Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the
minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?
A. 4
B. 8
C. 12
D. 13
E. 16
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13. If r = 5 z then 15 z = 3 y, then r =
A. y
B. 2 y
C. 5 y
D. 10 y
E. 15 y
14. If 300 jellybeans cost you x dollars. How many jellybeans can you purchase for 50 cents at the same rate?
A. 150/x
B. 150x
C. 6x
D. 1500/x
E. 600x
15. Lee worked 22 hours this week and made \$132. If she works 15 hours next week at the same pay rate, how
much will she make?
A. \$57
B. \$90
C. \$104
D. \$112
E. \$122
## 16. If 8x + 5x + 2x + 4x = 114, the 5x + 3 =
A. 12
B. 25
C. 33
D. 47
E. 86
17. You need to purchase a textbook for nursing school. The book cost \$80.00, and the sales tax where you are
purchasing the book is 8.25%. You have \$100. How much change will you receive back?
A. \$5.20
B. \$7.35
C. \$13.40
D. \$19.95
E. \$21.25
18. You purchase a car making a down payment of \$3,000 and 6 monthly payments of \$225. How much have you
paid so far for the car?
A. \$3225
B. \$4350
C. \$5375
D. \$6550
E. \$6398
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19. Your supervisor instructs you to purchase 240 pens and 6 staplers for the nurse's station. Pens are purchased
in sets of 6 for \$2.35 per pack. Staplers are sold in sets of 2 for 12.95. How much will purchasing these products
cost?
A. \$132.85
B. \$145.75
C. \$162.90
D. \$225.25
E. \$226.75
A. 300
B. 459
C. 648
D. 999
E. 1099
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1. A: The following proportion may be written: 1/p=x/5. Solving for the variable, x, gives xp = 5, where x=5/p. So,
Lynn can type 5/p pages, in 5 minutes.
2. A: Sally can paint 1/4 of the house in 1 hour. John can paint 1/6 of the same house in 1 hour. In order to
determine how long it will take them to paint the house, when working together, the following equation may be
written: 1/4 x+1/6 x=1. Solving for x gives 5/12 x=1, where x = 2.4 hours, or 2 hours, 24 minutes.
3. D: Sale Price = \$450 - 0.15(\$450) = \$382.50, Employee Price = \$382.50 - 0.2(\$382.50) = \$306
4. D: \$12,590 = Original Price - 0.2(Original Price) = 0.8(Original Price), Original Price = \$12,590/0.8 = \$15,737.50
5. A: In order to solve for A, both sides of the equation may first be multiplied by 3. This is written as
3(2A/3)=3(8+4A) or 2A=24+12A. Subtraction of 12A from both sides of the equation gives -10A=24. Division by -10
gives A = -2.4.
6. A: Three equations may initially be written to represent the given information. Since the sum of the three ages is
41, we may write, l + s + j = 41, where l represents Leah's age, s represents Sue's age, and j represents John's age.
We also know that Leah is 6 years older than Sue, so we may write the equation, l = s + 6. Since John is 5 years
older than Leah, we may also write the equation, j = l + 5. The expression for l, or s + 6, may be substituted into the
equation, j = l + 5, giving j = s + 6 + 5, or j = s + 11. Now, the expressions for l and j may be substituted into the
equation, representing the sum of their ages. Doing so gives: s + 6 + s + s + 11 = 41, or 3s = 24, where s = 8. Thus,
Sue is 8 years old.
7. E: Simple interest is represented by the formula, I = Prt, where P represents the principal amount, r represents
the interest rate, and t represents the time. Substituting \$4,000 for P, 0.06 for r, and 5 for t gives I =
(4000)(0.06)(5), or I = 1,200. So, he will receive \$1,200 in interest.
## 8. A: \$670 = Cost + 0.35(Cost) = 1.35(Cost), Cost = \$670/1.35 = \$496.30
9. D: The amount of taxes is equal to \$55*0.003, or \$0.165. Rounding to the nearest cent gives 17 cents.
10. C: The GPA may be calculated by writing the expression, ((3*2)+(4*3)+(2*4)+(3*3)+(4*1))/13, which equals 3,
or 3.0.
11. C: From 8:15 A.M. to 4:15 P.M., he gets paid \$10 per hour, with the total amount paid represented by the
equation, \$10*8=\$80. From 4:15 P.M. to 10:30 P.M., he gets paid \$15 per hour, with the total amount paid
represented by the equation, \$15*6.25=\$93.75. The sum of \$80 and \$93.75 is \$173.75, so he was paid \$173.75 for
14.25 hours of work.
12. D: If she removes 13 jellybeans from her pocket, she will have 3 jellybeans left, with each color represented. If
she removes only 12 jellybeans, green or blue may not be represented.
13. A: The value of z may be determined by dividing both sides of the equation, r=5z, by 5. Doing so gives r/5=z.
Substituting r/5 for the variable, z, in the equation, 15z=3y, gives 15(r/5)=3y. Solving for y gives r = y.
14. A: 50 cents is half of one dollar, thus the ratio is written as half of 300, or 150, to x. The equation representing
this situation is 300/x*1/2=150/x.
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15. B: The following proportion may be used to determine how much Lee will make next week: 22/132=15/x.
Solving for x gives x = 90. Thus, she will make \$90 next week, if she works 15 hours.
16. C: The given equation should be solved for x. Doing so gives x = 6. Substituting the x-value of 6 into the
expression, 5x + 3, gives 5(6) + 3, or 33.
17. C: The amount you will pay for the book may be represented by the expression, 80+(80*0.0825). Thus, you will
pay \$86.60 for the book. The change you will receive is equal to the difference of \$100 and \$86.60, or \$13.40.
18. B: The amount you have paid for the car may be written as \$3,000 + 6(\$225), which equals \$4,350.
19. A: You will need 40 packs of pens and 3 sets of staplers. Thus, the total cost may be represented by the
expression, 40(2.35) + 3(12.95). The total cost is \$132.85.
20. C: Substituting 3 for y gives 33 (33-3), which equals 27(27 - 3), or 27(24). Thus, the expression equals 648.
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Algebra part 2
a. 6
b. 2+2x
c. 2x2x
d. 2x+1
## 2. Simplify the expression (2x2-5x-12)/(2x2-4x-16).
a. (x-6)/2(x-2)
b. (x-6)/2(x+2)
c. (2x+3)/2(x-2)
d. (2x+3)/2(x+2)
3. Suppose that the function f(x) is a quadratic function with roots at x=2-3i and x=2+3i. Find f(x).
a. f(x)=x2-4x-5
b. f(x)=x2-4x+13
c. f(x)=x2-6ix-5
d. f(x)=x2-6ix+13
## 4. Solve the inequality for x. Select all that apply.
4x3+10x2-24x<0
a. x<-4
b. -4<x<0< li=""> </x<0<>
c. 0<x<="" li=""> </x
d. x>3/2
5. A baseball is thrown up in the air from an initial height of 6 feet. Its height above the ground (in feet) t
seconds after being thrown is given by the function h(t)=-16t^2+46t+6. How long will it take (in seconds) for the
baseball to hit the ground?
a. 2 seconds
b. 5/2 seconds
c. 3 seconds
d. 4 seconds
log2(8x-x2 )=4
a. x=-8
b. x=0
c. x=4
d. x=8
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7. Calculate the average rate of change of f between x=1 and x=4.
f(x)=x3+3x+1
a. 6
b. 20/3
c. 24
d. 72
a. 1
b. (x-3)/(x+3)
c. (x2+2)/(x-3)
d. (x2+2)/(x+3)
a. tan = 1/13
b. tan = 13
c. tan = 5/12
d. tan = 12/5
a. 2(3x-1+5x)
b. 2(3x1/2+5x)
c. 2x(3x-1+5)
d. 2x(3x1/2+5)
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1. D: You can solve this problem either (1) by simplifying the numerator and denominator separately and then
simplifying the result or (2) by using the distributive property. For this problem, we will use the first method.
## First rewrite 4x as an exponent of 2 using the property, (bx )y=bxy.
4x=(22 )x=22x
Then use this to simplify the numerator with the property, b x x y = bx+y.
## Finally, simplify the result using bx / by = bx-y
2. D: To simplify the expression, first factor the numerator and the denominator. By the trial-and-error method,
the numerator can be factored into two binomials as follows.
=2(x-4)(x+2)
## Thus, the factored form of the expression is
Notice that there is a common factor, (x - 4), which is in both the numerator and the denominator. Therefore, you
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can further simplify the expression by cancelling it out.
3. B: The roots of a quadratic function f(x) are the values of x for which f(x)=0. A quadratic function written in the
form f(x)=(x-a)(x-b) has roots at x=a and x=b. Therefore, to find f(x), substitute 2-3i and 2+3i for a and b into this
equation and simplify the result. Note that (2-3i)(2+3i)=4-9i^2=13.
f(x)=(x-a)(x-b)
=[x-(2-3i)][x-(2+3i)]
=x^2-(2-3i)x-(2+3i)x+(2-3i)(2+3i)
=x^2-2x+3ix-2x-3ix+13
=x^2-4x+13
4. A and C: To solve, first factor the polynomial. Notice that the greatest common factor (GCF) of the terms is 2x.
Factor this expression out and then use trial-and-error to factor the resulting trinomial.
4x^3+10x^2-24x=2x(2x^2+5x-12)
=2x(2x-3)(x+4)
Solving for 0, we find that the roots of the polynomial are x=0, x=3/2, and x=-4.
These values divide the number line into four intervals. Choose a test number from each interval and determine
whether the product is positive or negative. For this problem, we will use -5, -1, 1, and 2 as test numbers.
Substitute these values into the original polynomial.
x=-5:
4(-5)^3+10(-5)^2-24(-5)=-375+250+120=-5
x=-1:
4(-1)^3+10(-1)^2-24(-1)=-4+10+24=30
x=1:
4(1)^3+10(1)^2-24(1)=4+10-24=-10
x=2:
4(2)^3+10(2)^2-24(2)=32+40-48=24
Thus, the given inequality, 4x^3+10x^2-24x<0, is satisfied by numbers less than -5 and numbers between 0 and
3/2.
5. D: The baseball will hit the ground when its height is zero. In mathematical notation, this will happen when
h(t)=0. Therefore, we need to set the given function equal to zero.
h(t)=0
-16t^2+46t+6=0
Now solve the resulting equation. Factor the left side and use the zero-product property to solve for t.
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-2(8t^2-23t-3)=0
-1(8t+1)(t-3)=0
t=-1/8 t=3
The answer only makes sense when t is positive, so we can discard the negative value. Thus, the calculator will hit
the ground 3 seconds after it is thrown.
6. C only: The logarithm of a number is the exponent that the base must to be raised to in order to get that
number. For example, since 2^3=8, it is also true that log_2?8=3. Thus, the given equation log_2?(8x-x^2 )=4
implies that
8x-x^2=2^4
## Simplify this equation and solve for x.
8x-x^2=16
0=x^2-8x+16
0=(x-4)^2
x=4
Thus, the solution is x=4. Check this value on your own by substituting it into the original equation to make sure
that the result is a true statement.
7. C: The average rate of change of a function f between x=a and x=b can be computed with the formula
## Average rate of change=(f(b)-f(a))/(b-a)
To use this formula, first calculate f(1) and f(4).
## Average rate of change=(f(4)-f(1))/(4-1)
8. D: To simplify the expression, first factor the numerator and the denominator. The numerator can be factored
by grouping as follows.
x^3-3x^2+2x-6=x^2 (x-3)+2(x-3)
=(x^2+2)(x-3)
For the denominator, factor using the difference of squares formula, a^2-b^2=(a+b)(a-b).
x^2-9=(x+3)(x-3)
## Thus, the factored form of the expression is
(x^3-3x^2+2x-6)/(x^2-9)=(x^2+2)(x-3)/(x+3)(x-3)
Notice that there is a common factor, (x-3), which is in both the numerator and the denominator. Therefore, you
can further simplify the expression by cancelling it out.
(x^2+2)(x-3)/(x+3)(x-3) =(x^2+2)/(x+3)
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9. C: Use a unit circle to model the value of cosine. In a right triangle, the cosine function is
cos??=adjacent/hypotenuse. Using the Pythagorean Theorem, we find that the length of the second leg is
b=(c^2-a^2 )
=(1^2-(12/13)^2 )
=?(25/169)
=5/13
Since the tangent function is tan??=opposite/adjacent, the value of tan?? is tan=(5/13)/(12/13)=5/12 In Quadrant
I, the values of cosine and tangent are both positive. Therefore, tan??=5/12.
10. B: All of the choices involve two transformations of the given expression: factoring out either 2 or 2x and
changing the radical to an exponent. First factor out the greatest common factor (GCF) of the terms. In this case,
the GCF is 2.
6x+10x=2(3?x+5x)
## In addition, the square root of x is equal to x raised to the 1/2 power.
2(3x+5x)=2(3x^(1/2)+5x)
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Basic Operations Practice
1. Edward ascends to the top of a mountain over the course of two days. On Friday, he ascends 482 feet. He
ascends another 362 feet on Saturday. How many feet did he ascend in all?
a. 848 feet
b. 836 feet
c. 840 feet
d. 844 feet
2. Kristen must buy three items that are priced at \$4.58, \$6.22, and \$8.94. What is the best estimate for the total
cost of all three items?
a. \$18
b. \$16
c. \$20
d. \$22
3. A teacher donates to a local charity. Each year, she donates three times the amount donated the previous
year. If the teacher donated \$2 the first year, how much did she donate during the fifth year?
a. \$158
b. \$164
c. \$162
d. \$144
4. Penny drinks 8 glasses of water each day. The number of glasses of water she drinks over a 12-day time span
can be determined, using the number sentence: Which number sentence would not show the number of glasses
of water she drinks?
a. ? / 8 = 12
b. 12 x 8 = ?
c. ? / 12 = 8
d. 12 - 8 = ?
5. Belinda draws a rectangle with a length of 6 cm. She draws a second rectangle with a length of 11 cm. Belinda
continues drawing more rectangles, where for each rectangle drawn, she uses a length that is 5 more
centimeters than the length of the previous rectangle. If this pattern continues, what will be the length of the
11th rectangle?
a. 46 cm
b. 54 cm
c. 56 cm
d. 61 cm
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6. Jasper collects 1,082 cans of food. He gives a certain number of cans to the first local charity he finds. He now
has 602 cans of food. How many cans of food did he give to the first local charity?
a. 430
b. 480
c. 682
d. 1,684
7. Amanda creates the base of a picture frame, using 4.55 inches of red fabric and 6.25 inches of blue fabric.
How many inches of fabric are used to create the base of the frame?
a. 10.80 inches
b. 10.85 inches
c. 10.75 inches
d. 10.90 inches
8. Hannah ran 12 laps every day for 8 days. How many laps did she run in all?
a. 108
b. 96
c. 84
d. 72
9. Three friends sold cupcakes for a fundraiser. Eli sold 84 cupcakes, John sold 46 cupcakes, and Kim sold 72
cupcakes. Which of the following is the best estimate for the number of cupcakes the three friends sold in all?
a. 180
b. 200
c. 210
d. 190
10. Carlisle charges \$21.95 per hair cut and has completed 30 haircuts this week. Which of the following is the
best approximation for the total charges for all haircuts?
a. \$450
b. \$600
c. \$750
d. \$800
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1. D: The total number of feet he ascended can be determined by adding 482 feet and 362 feet. The sum of 482
and 362 is 844. Thus, he ascended 844 feet in all.
2. C: The item priced at \$4.58 can be rounded to \$5. The item priced at \$6.22 can be rounded to \$6. The item
priced at \$8.94 can be rounded to \$9. The sum of 5, 6 and 9 is 20. Thus, the best estimate is \$20.
3. C: In order to find the amount donated the following year, you multiply the amount donated the previous year
by 3. Thus, the amount donated the second year was \$6 (). The amount donated the third year was \$18 (). The
amount donated the fourth year was \$54 (The amount donated the fifth year was \$162 ().
4. D: If she drinks 8 glasses of water each day, the number of glasses of water she drinks in 12 days can be
determined by multiplying 8 by 12. This product is 96; thus she drinks 96 glasses of water in a 12-day time span.
The relationship between the number of glasses of water she drinks per day and the total number of glasses of
water she drinks in 12 days can be represented by an appropriate multiplication or division number sentence
within the following fact family: Subtracting 8 from 12 will not reveal the number of glasses she drinks in a 12-day
time span. The number sentence: , is not in this fact family.
5. C: If each rectangle has a length that is 5 cm more than that of the previous rectangle, the lengths of the 3rd
rectangle through the 11th rectangle can be found by adding 5 cm to the length of the second triangle and
continuing for each next rectangle. So, the lengths of the rectangles will be as follows: 16 cm, 21 cm, 26 cm, 31 cm,
36 cm, 41 cm, 46 cm, 51 cm, and 56 cm.
6. B: In order to find the number of cans of food he gave to the first charity, the number of cans of food he has left
needs to be subtracted from the number of cans he collected; .
7. A: The sum of the two decimals is 10.80; the decimals are added just like whole numbers are, while aligning the
decimal point.
## 8. B: She runs laps in all, or 96 laps.
9. B: The number of cupcakes sold can be rounded as follows: 80 cupcakes, 50 cupcakes, and 70 cupcakes, which
sum to 200. Therefore, the best estimate for the number of cupcakes sold is 200 cupcakes.
10. B: The amount of money Carlisle charges per hair cut can be rounded to \$20; . Thus, his total charges are
approximately \$600.
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Exponent Practice
## 1. 104 is not equal to which of the following?
A. 100,000
B. 0.1 x 105
C. 10 x 10 x 10 x 10
D. 102 x 102
E. 10,000
## 2. Multiply 104 by 102
A. 108
B. 102
C. 106
D. 10-2
E. 103
3. Divide x5 by x2
A. x7
B. x4
C. x10
D. x3
E. x2.5
## 4. Find 8.23 x 109
A. 0.00000000823
B. 0.000000823
C. 8.23
D. 8230000000
E. 823000000000
5. 83,000 equals:
A. 83.0 x 104
B. 8.3 x 104
C. 8.3 x 103
D. 83.0 x 105
E. 83.0 x 102
6. .00875 equals:
A. 8.75 x 10-2
B. 8.75 x 10-3
C. 8.75 x 10-4
D. 87.5 x 10-3
E. 875 x 10-4
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1. A: 104=10.10.10.10, or 10,000.
2. C: When multiplying terms with the same base, the exponents should be added. Thus, 10 4 . 102=106.
3. D: When dividing terms with the same base, the exponents should be subtracted. Thus, x5/x2 =x3.
4. D: The decimal will be moved to the right 9 places. Thus 7 zeros will be added to the right of 823, giving
8,230,000,000.
5. B: Moving the decimal to the right of the digit, 8, gives the equivalent expression, 8.3x10 4, since there are 4
digits to the right of the 8.
6. B: Moving the decimal to the right of the 8 gives 8.75x10-3, since the decimal must be moved 3 places to the
right.
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Fractions and Square Root Practice
## 1. Which of the following is listed in order from least to greatest?
a. -3/4,-7 4/5,-8,18%,0.25,2.5
b. -8,-7 4/5,-3/4,0.25,2.5,18%
c. 18%,0.25,-3/4,2.5,-7 4/5,-8
d. -8,-7 4/5,-3/4,18%,0.25,2.5
2. Which of the following fractions is larger than 2 1/4 but smaller than 2 2/5?
a. 2 1/2
b. 2 3/8
c. 2 6/11
d. 2 5/9
3. Jason chooses a number that is the square root of four less than two times Amy's number. If Amy's number is
20, what is Jason's number?
a. 6
b. 7
c. 8
d. 9
4. In a square built with unit squares, which of the following would represent the square root of the square?
## a. The number of unit squares comprising a side
b. The total number of unit squares within the square
c. Half of the total number of unit squares within the square
d. The number of unit squares comprising the perimeter of the square
5. Brianna used five 3/4 cups of sugar while baking. How many cups of sugar did she use in all?
a. 3 2/3
b. 3 3/4
c. 3 1/4
d. 3 1/2
6. A publishing company has been given 29 manuscripts to review. If the company divides the work equally
amongst 8 editors, which of the following represents the number of manuscript each editor will review?
a. 3 3/5
b. 3 5/8
c. 3 7/9
d. 3 2/3
7. A lake near Armando's home is reported to be 80% full of water. Which fraction is equivalent to 80% and in
simplest form?
a. 1/80
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b. 8/10
c. 4/5
d. 80/1
8. Alma collected coins. In the bag where she kept only dimes, she had dimes from four different years. She had
20 dimes minted in 1942, 30 minted in 1943, 40 minted in 1944, and 10 minted in 1945. If Alma reached into the
bag without looking and took a dime, what is the probability that she took a dime minted in 1945?
a. 2/5
b. 3/10
c. 1/5
d. 1/10
9. A recipe calls for 3 3/4 cups of flour. Which fraction below is equivalent to this amount?
a. 5/2
b. 15/4
c. 3/2
d. 9/4
a. 9
b. 5
c. 3
d. -4
e. -7
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1. D: The smallest negative integers are those that have the largest absolute value. Therefore, the negative
integers, written in order from least to greatest, are -8,-7 4/5,-3/4. The percent, 18%, can be written as the
decimal, 0.18; 0.18 is less than 0.25. The decimal, 2.5, is the greatest rational number given. Thus, the values, -8,-7
4/5,-3/4,18%,0.25,2.5, are written in order from least to greatest.
2. B: The fraction, 2 1/4, can be written as the decimal,2.25. The fraction, 2 2/5, can be written as the decimal,
2.40. The fraction, 2 3/8, can be written as the decimal, 2.375; 2.375 is larger than 2.25 but smaller than 2.40.
3. A: Jason's number can be determined by writing the following expression: ?(2x-4), where x represents Amy's
number. Substitution of 20 for x gives ?(2(20)-4), which simplifies to ?36, or 6. Thus, Jason's number is 6. Jason's
number can also be determined by working backwards. If Jason's number is the square root of 4 less than 2 times
Amy's number, Amy's number should first be multiplied by 2 with 4 subtracted from that product and the square
root taken of the resulting difference.
4. A: The square root of a square is equal to the length of one of the sides, or the number of unit squares
comprising a side. For example, a square representing 7 squared will have 7 unit squares on each side; 7^2=49, and
7 is the square root of 49. The square will contain 49 unit squares, with 7 unit squares comprising each side.
5. B: In order to determine the total number of cups of sugar used while baking, the product of 5 and 3/4 should be
calculated: 5?3/4=15/4, which can be simplified to 3 3/4. Thus, she used 3 3/4 cups in all.
6. B: In order to determine the number of manuscripts each editor will review, the total number of manuscripts
should be divided by the number of editors; 29 / 8 can be written as 29/8, which simplifies to the mixed fraction 3
5/8. Notice that the quotient is 3 with a remainder of 5.
7. C: The 80% means 80 out of 100, which can be written as 80/100 . This fraction can be written in lowest terms
by dividing both the numerator and denominator by the greatest common factor of 20, to get the fraction, 4/5.
8. D: By adding all of the dimes, we find that there are a total of 100 dimes in the bag. 10 of them were minted in
1945. The probability, then, of choosing a dime minted in 1945 is 10 out of 100, which is equivalent to the fraction
1/10.
9. B: 15/4 . The value of 3 is equivalent to 12/4. Therefore, 3 3/4=12/4+3/4=15/4. Another way of of finding this is
sometimes called the "C" method. 3 3/4 equals (4x3+3)/4=15/4 .
10. E: . When you take the square root of a number, the answer is the positive and negative values of the root.
Therefore, and x=-7. Only -7 is an answer choic
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Geometry Practice
1. Which of the following letters represents the vertex in the following picture?
A. D and E
B. E and H
C. F and G
D. G only
E. H only
A. 6.28
B. 12.56
C. 25.13
D. 50.24
E. 100.48
## 3. What is the area of the triangle below?
A. 22 cm2
B. 33 cm2
C. 44 cm2
D. 50 cm2
E. 66 cm2
4. What is the measure of the solid line angle depicted by the following figure?
A. 90 degrees
B. 180 degrees
C. 225 degrees
D. 270 degrees
E. 0 degrees
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5. What is the measure of angle B in the following figure if angle A measures 135?
A. 40
B. 45
C. 50
D. 135
E. 225
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1. E: The vertex is the point, formed by the two rays of an angle. Thus, H is the vertex of the angle.
## 2. C: C=d. Substituting 8 for d gives C=8, where C is approximately 25.13.
3. B: The area of a triangle may be found by using the formula, A=1/2bh, where b represents the base and h
represents the height. Thus, the area may be written as A=1/2(11)(6), or A = 33. The area of the triangle is 33 cm'.
4. D: The sum of the angles, formed by the perpendicular rays is 360, thus the curved arrow represents an angle
measure that is equal to the difference of 360 and 90, or 270.
5. B: Since angles A and B are supplementary, the measure of angle B is equal to the difference of 180 and 135,
or 45.
## Averages and Rounding Practice Questions
1. The average of six numbers is 4. If the average of two of those numbers is 2, what is the average of the other
four numbers?
a. 5
b. 6
c. 7
d. 8
e. 9
2. Kate got a 56 on her first math test. On her second math test, she raised her grade by 12%. What was her
a. 62.7
b. 67.2
c. 68.0
d. 72.3
3. A skyscraper is 548 meters high. The building's owners decide to increase its height by 3%. How high would
the skyscraper be after the increase?
a. 551 meters
b. 555 meters
c. 562 meters
d. 564 meters
4. Janet makes homemade dolls. Currently, she produces 23 dolls per month. If she increased her production by
18%, how many dolls would Janet produce each month?
a. 27
b. 32
c. 38
d. 40
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5. A class contains an equal number of boys and girls. The average height of the boys is 62 inches. The average
height of the all the students is 60 inches. What is the average height of the girls in the class?
a. 57 inches
b. 58 inches
c. 59 inches
d. 60 inches
6. Elijah drove 45 miles to his job in an hour and ten minutes in the morning. On the way home in the evening,
however, traffic was much heavier and the same trip took an hour and a half. What was his average speed in
miles per hour for the round trip?
a. 30
b. 45
c. 33 3/4
d. 32 1/2
7. If Joey and Katrina hike an average of 3 miles per hour, about how long will it take them if they take the
Beaverton Falls trail and follow it through the Copper Creek trail?
a. 3 hours
b. 3 hours
c. 4 hours
d. 4 hours
8. A pasta salad was chilled in the refrigerator at 35 F overnight for 9 hours. The temperature of the pasta dish
dropped from 86 F to 38 F. What was the average rate of cooling per hour?
a.
b.
c.
d.
9. Rachel spent \$24.15 on vegetables. She bought 2 lbs of onions, 3 lbs of carrots, and 1 1/2 lbs of mushrooms. If
the onions cost \$3.69 per lb, and the carrots cost \$ 4.29 per lb, what is the price per lb of mushrooms?
a. \$2.60
b. \$2.25
c. \$2.80
d. \$3.10
e. \$2.75
10. A roast was cooked at 325 F in the oven for 4 hours. The internal temperature rose from 32 F to 145 F.
What was the average rise in temperature per hour?
a. 20.2F/hr
b. 28.25F/hr
c. 32.03F/hr
d. 37F/hr
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1. A: A set of six numbers with an average of 4 must have a collective sum of 24. The two numbers that average 2
will add up to 4, so the remaining numbers must add up to 20. The average of these four numbers can be
calculated: 20/4 = 5.
## 2. A: First, calculate 12% of 56.
56 x 0.12 = 6.72
Then, add this value (the increase) to the original value of 56.
56 + 6.72 = 62.72
Rounding off, we get 62.7
## 3. D: Explanation: First, calculate 3% of 548 meters.
548 meters x 0.03 = 16.44 meters.
Then, add it to the original height.
548 meters + 16.44 meters = 564.44 meters
Rounding off, we get 564 meters.
## 4. A: Explanation: First, calculate 18% of 23.
23 x 0.18 = 4.14
Then, add this value (the increase) to the original value of 23.
23 + 4.14 = 27.14
Rounding off, we get 27.
5. B: The average, or arithmetic mean, is computed by totaling all the measurements and dividing by the number
of measurements. Let TB represent the sum of the heights of the boys in the class, and TG the sum of the heights
of the girls. If N is the number of students in the class, there are N/2 boys and N/2 girls. The average height of the
boys is then . Similarly, the average height of the girls is . The average height of all the students is equal to .
Therefore, , and the average height for the girls is 2 x 29 = 58.
6. C: To determine this, first determine the total distance of the round trip. This is twice the 45 miles of the one-
way trip to work in the morning, or 90 miles. Then, to determine the total amount of time Elijah spent on the
round trip, first convert his travel times into minutes. One hour and ten minutes equals 70 minutes, and an hour
and a half equals 90 minutes. So, Elijah's total travel time was 70 + 90 = 160 minutes. Elijah's average speed can
now be determined in miles per minute:
miles per minute
Finally, to convert this average speed to miles per hour, multiply by 60, since there are 60 minutes in an hour:
Average speed (mph) = 60 x 0.5625 = 33.75 miles per hour
7. C. The total distance they will hike is 2.6 miles + 9.5 miles = 12.1 miles. If they hike 3 miles per hour, it will take
them hours to hike 12.1 miles.
## 8. B. The average rate of cooling is: hrs; = 5.33F per hour.
9. A: Begin by determining the total cost of the onions and carrots, since these prices are given. This will equal (2 x
\$3.69) + (3 x \$4.29) = \$20.25. Next, this sum is subtracted from the total cost of the vegetables to determine the
cost of the mushrooms: \$24.15 - \$20.25 = \$3.90. Finally, the cost of the mushrooms is divided by the quantity (lbs)
to determine the cost per pound:
Cost per lb =
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Science Section
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The Human Body
1. The human body consists of over two hundred and six bones, seven hundred muscles and about
twenty five miles of blood vessels. It is always striving to maintain its internal environment. The
process of maintaining a constant steady state is called Homeostasis. In order to understand the
human body, it is important to understand basic anatomy and physiology
2. Define the terms anatomy and physiology.
Anatomy is the study of the structure of an organism and the relationship of its parts. Physiology
is the study of the functions of living organisms and their parts.
3. List and discuss in order of increasing complexity the levels of organization of the body.
Levels of organization of the body include (1) chemical level (atoms and molecules), (2) cells
(smallest living units), (3) tissues (groups of cells acting together), (4) organs (groups of tissues
acting together), and (5) systems (groups of organs acting together).
Atom smallest unit of an element with thats elements chemical characteristics. Comprised of protons,
neutrons, and electrons
When two or more Atoms unite through their electron structure, they form a molecule
Cell: This is the basic structural unit of life. The study of cells is called Cytology.
Cells are considered the smallest living units and a very complex
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Discovery of Cells
Robert Hooke (1665): observed a thin slice of cork (dead plant cells) with a microscope. He
described what he observed as little boxes (cells).
Anton van Leeuwenhoek (1675): was the first person to observe living cells
All Cells are microscopic but differ widely in size and shape and internal organization.
## Who developed the cell theory?
Matthias Schleiden (1838): concluded that all plants are composed of cells
Theodor Schwann (1839): concluded that all animals are composed of cells
Rudolph Virchow (1855): determined that cells come only from other cells
Tissues: Similar Cells come together and form tissue. Tissues are more complex cells. Tissue are similar
cells that work together to perform a function. The study of tissue is called Histology. There are four
basic tissue types and they are the following: Connective, Epithelial, Muscle, Nervous tissue.
Organs are more complex than tissue. An Organ is a group of several different kinds of tissue
Each organ in the body is composed of various combinations of body tissues. (For example, the stomach
is made of epithelial tissues and muscle tissue)
Each organ in the body is composed of various combinations of body tissues. (For example, the stomach
is made of epithelial tissues and muscle tissue)
Connective tissue: This tissue supports and connects all body parts. Includes adipose (fat issue),
cartilage, bone and blood. Important forms of connective tissue are areolar connective tissue and adipose
(fat) tissue, fibrous connective tissue, bone, cartilage, blood, and hematopoietic tissue
Epithelial tissue: (also know as Epithelium) This tissue protects and covers the body and lines the body
organs. Epithelial tissue consist of closely packed flat cells. Functions include, Protection, Absorption and
Secretion
Muscle tissue: This tissue contracts to produce movement. There are three types of muscle tissue:
Skeletal (voluntary) muscle, cardiac muscle tissue, Smooth muscle.
Nervous tissue: This tissue provides communication throughout the body. The main nervous tissues are
the brain, spinal cord and nerves
When several kinds of tissue are united to perform a more complex function than any tissue alone, they
are called ORGANS.
ORGANS working together and for the same general purpose make up an organ systems or body
system, which maintain the whole body.
Body Systems perform more complex function than any one organ can perform.
Study of the anatomy and physiology of the body generally centers on study of the body systems.
A system is defined as a group of organs working together to perform related functions.
Many medical specialties concentrate on one body system, ie, neurology is the study of the
nervous system.
Usually the organs in a system are anatomically connected, but in some cases, as with the
endocrine system, the tissues are widely distributed
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The table below shows body systems
HUMAN BODY SYSTEM CHART
SYSTEM FUNCTIONS ORGANS/Major STRUCTURES
Functions: Bones
Ligaments: connects bones to
1. Supports and protects the bones
body Cartilage: provides cushion
between the bones
2. Protect organs Red Marrow: makes blood
Skeletal
3. Makes blood cells
## Function: Help the body Heart, diaphragm, biceps/triceps
move Tendon: connects muscle to
bones
Two Types of Muscles: Skeletal Muscles: attached to the
1. Voluntary Muscles: bones and helps us move
muscles you CAN control Smooth Muscles: make up most of
Muscular
(Skeletal Muscles) the organs of our body
2. Involuntary Muscles: Cardiac Muscles: make up the
muscles you CANNOT control heart
(smooth and cardiac muscles)
Mouth
Function: Breaks down food Esophagus: Long tube that
to make energy for the body connects the mouth to the
*Direction of food movement: stomach.
Mouth esophagus Stomach, Liver, Pancreas
stomach Small Intestine: where most
Digestive
Sm. Intestine lg. Intestine digestion takes place.
rectum anus Large Intestine: Takes water from
*Peristalsis: muscle the undigested materials
movement that moves food Rectum, Anus
through the D.S
## Function: Carries O2, CO2, Heart: main organ of CS that
food, waste disease fighting pumps blood to all parts of the
cells, &hormones through body
the body Artery: thickest blood vessel that
carry blood high in oxygen AWAY
Circulatory *Direction of blood flow: from the heart.
Capillary: thinnest blood vessel
Heart that connects arteries & vein and
where the actual exchange of
Veins
materials take place.
Arteries
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Capillaries Vein: blood vessel that carry
blood high in carbon dioxide
TOWARD the heart.
Blood: carries food and oxygen to
the body and take away waste
from cells
White Blood Cells: fight disease
Nose
Function: take in
Trachea: large tube that carries air
oxygen and take out
to the bronchi
carbon dioxide and
Larynx
some water out of the
Bronchi: tube that connects the
body
Respiratory trachea and lungs
Lungs: main organ of the RS
*Direction of air
Alveoli: A tiny air sac where air
movement:
exchange takes place
nose trachea
Diaphragm: strong muscle below
bronchi (into the lungs)
the lungs
bronchioles alveoli
that helps in breathing
*Function: take out waste Kidney: main organ of the ES
from the body where urine is made
Urteter: a tube that carries urine
from the kidney to the urinary
Urethra: a tube that carries urine
Body kidney ureter
out of the body
Excretory
extra water, and salt skin remove (take away) waste
*Perspiration (sweat): Lungs: breathe out CO2
liquid waste that leaves the Kidney: make urine
body through the skin and Skin: perspiration (sweat)
helps control the body
temperature
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Function: controls ALL the Brain: control center of our body
activities of the body: its the functions
boss Spinal cord: connects brain with
the rest of the body
Two parts of nervous system:
Nerves: nerve cells
1. Central Nervous system
(CNS): made up of brain and
Nervous spinal cord, controls all
activities of the body.
2. Peripheral Nervous System
(PNS): made up of all the
nerves outside of CNS, and
carries messages between
CNS to the all parts of the
body
## Types of Anatomical Studies
Descriptive or Systemic Anatomy - the method of studying the body by systems, such as auditory and
respiratory systems
Regional Anatomy - the study of the body by regions, such as the thorax and abdomen
Applied or Clinical Anatomy - emphasizes structure and function as they relate to diagnosis and
treatment
Surface Anatomy - the visualization of structures that lie beneath the skin. It is an essential part of the
study of regional anatomy and the primary means by which clinicians use anatomical knowledge in
treating patients
Developmental Anatomy the study of anatomy from conception to adulthood
Comparative Anatomy study of anatomy across the animal kingdom, animal models, etc.
Pathological Anatomy study of abnormal structures or systems
Anatomical terminology is used to define the body accurately. Anatomical terminology is used to define
the human body in anatomical position. Lets now look at the terminology
When describing the human body healthcare personnel refer to the body and body parts as if the patient
where in anatomic position.
Anatomic Position: An individual in the Anatomic Position is standing erect, with arms to their side,
palms forward, and facing forward.
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4. Define the anatomical position.
In the anatomical position, the body is in an erect or standing posture with arms at the sides and
palms turned forward. The head and feet also point forward. The illustration bellow shows a
person standing in anatomical position
## Person stands erect
Feet flat on floor
Arms at sides
Palms, eyes & face facing forward
Standard frame of reference for anatomical descriptions & dissection
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5. List and define the principal directional terms and sections (planes) used in describing the body and
the relationship of body parts to one another.
a. Directional terms
(1) Superiortoward the head, upper, above
Inferiortoward the feet, lower, below
(2) Anteriorfront, in front of (same as ventral in humans)
Posteriorback, in back of (same as dorsal in humans)
(3) Medialtoward the midline of the body
Lateralaway from the midline or toward the side of the body
(4) Proximaltoward or nearest the trunk of the body, or nearest the point of origin of one of its
parts
Distalaway from or farthest from the trunk of the body, or farthest from a point of origin of
one of its parts
(5) Superficialnearer the body surface
Deepfarther away from the body surface
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b. Body planes
This term is used to describe how the body organs and the body are divided into two sections. A
plane of the body is simply referring to sections of the body
Frontal plane: Also known as the Coronal plane. This divides the body or the organ vertically into a
front and rear part.
Horizontal Plane: Also known as the Transverse plane. This divides the organs of the body and the
body into top and the bottom parts.
Sagittal planes: Divides the body organs and the body vertically into right and left parts. Mid-sagittal
is the term used if the body parts or organs right and left are equal.
Parasagittal plane is the term used if the body parts are divided unequally from left to right. The
diagram below illustrates human body planes:
The diagram below illustrates human body planes:
Planes are imaginary flat surfaces passing through the body sections are anatomical views if
body is cut on a plane
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6. Body cavities. The List the major cavities of the body and the subdivisions found in each. In the
human body, various organs are housed in large, hollow spaces called body cavities. The body
cavities are divided into two major groups, dorsal and ventral.
a. Ventral body cavity
(1) Thoracic cavity
(a) Mediastinum
(b) Pleural cavities (right and left)
(2) Abdominopelvic cavity
(a) Abdominal cavity
(b) Pelvic cavity
b. Dorsal body cavity
(1) Cranial cavity
(2) Spinal cavity
The diagram below illustrates human body cavities
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7. Abdominal Cavity List the nine abdominopelvic regions and the four abdominopelvic quadrants.
a. The nine abdominopelvic regions include (1) right hypochondriac, (2) epigastric, (3) left
hypochondriac, (4) right lumbar, (5) umbilical, (6) left lumbar, (7) right iliac, (8) hypogastric, and
(9) left iliac regions.
b. The four abdominopelvic quadrants include (1) right upper or superior, (2) right lower or inferior,
(3) left upper or superior, and (4) left lower or inferior.
There are two systems of dividing the abdominal cavity. The simplest is dividing into four quadrants.
The Abdomen is divided into FOUR quadrants. By using 2 imaginary lines that intersect at the navel you
divide the abdomen into right upper (RUQ) and lower (RLQ), and left upper (LUQ) and lower (LLQ)
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RUQ:
- Liver
- Pylorus
- Duodenum
- portion of R kidney
- Hepatic flexure of colon
- portions of ascending and transverse coolon
LUQ:
* Spleen
* body of Pancreas
- L lobe of liver
- stomach
- portion of L kidney
- Splenic flexure of colon
RLQ:
* Appendix
- lower pole of R kidney
- Cecum and appendix
- portion of ascending colon
- Ovary and salpinx
- Uterus (if enlarged)
- R ureter*
LLQ:
- lower pole of L kidney
- Sigmoid colon
- portion of descending colon
- Ovary and Salpinx
- Uterus (if enlarged)
- L ureter
8. Discuss and contrast the axial and the appendicular subdivisions of the body. Identify a number of
specific anatomical regions in each area.
a. The axial portion of the body consists of the head, neck, and trunk. It is composed of thoracic,
abdominal, and pelvic regions.
b. The appendicular portion of the body consists of the upper and lower extremities.
(1) Upper extremities are composed of arm, forearm, wrist, and hand regions.
(2) Lower extremities are composed of thigh, leg, ankle, and foot regions.
9. Explain the meaning of the term homeostasis and give an example of a typical homeostatic
mechanism.
a. Homeostasis is relative constancy of the internal environment. It requires that chemical
composition, volume, and other characteristics of blood and other body fluids remain constant
within a narrow limit.
b. Example: Constant circulation of blood in the body allows for continuous removal of carbon
dioxide produced by body cells.
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Practice Questions and Review
Have the students answer these questions prior to covering this chapter to understand where they
stand in relation to the content.
1) The term used to describe the position of the body lying face upward is
_______________________.
2) The plane of the body that runs lengthwise from front to back is the _______________________.
3) The body cavity that contains the trachea, heart, and blood vessels is
_______________________.
## 5) What is the basic type of control system in the body?
a) homeostasis
b) feedback loop
c) control center
d) hypothesis
## 6) What is a systematic approach to discovery called?
a) scientific method
b) experimentation
c) heory
d) experimental controls
7) What is the anatomical direction term that means nearer the surface?
a) proximal
b) distal
c) superficial
d) deep
8) What is the body cavity that contains the lower colon, rectum, urinary bladder, and reproductive
organs?
a) cranial cavity
b) pleural cavity
c) abdominal cavity
d) pelvic cavity
9) What is the term for the body region of the upper cheek?
a) zygomatic
b) volar
c) popliteal
d) olecranal
## 10) What term refers to a degeneration process?
a) effector loop
b) negative feedback loop
c) homeostasis
d) atrophy
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1) supine
2) sagittal
3) mediastinum
4) umbilical
5) b
6) a
7) c
8) d
9) a
10) d
## Other Practice Questions
Question: What is the anatomical position? Why is the term used? What are other positions besides
anatomical position?
Answer: To assume anatomical position, the body is in an erect, or standing, posture with the arms at
the sides and the palms forward. The head and feet also point forward. Discussions about the
body, the way it moves, its posture, or the relationship of one area to another assume that the
body as a whole is in anatomical position. Other positions are supine and prone. In the supine
position, the body is lying face upward, and in the prone position, the body is lying face
downward.
Question: What are the two major body cavities? What are the subdivisions of these two major
cavities?
Answer: The two major body cavities are the ventral and dorsal body cavities. The ventral body cavity
is divided into the thoracic cavity, which is further subdivided into the mediastinum and pleural
cavities, and the abdominopelvic cavity, which is further subdivided into the abdominal and pelvic
cavities. The dorsal body cavity is divided into the cranial and spinal cavities.
## 1. Define anatomy and physiology.
Answer: Anatomy is the study of the structure of an organism and the relationships of its parts.
Physiology is the study of body function.
2. Describe the process used to form scientific theories.
Answer: The scientific method is a the systematic approach to discovery. A tentative explanation
(hypothesis) is developed from observation. This hypothesis is tested through a process of
experimentation. If accepted, the hypothesis is tentatively accepted as true. A hypothesis that has
gained a high level of confidence is called a theory or law.
## 3. List and explain the levels of organization in a living thing.
Answer: (1) The chemical level of organization (see Appendix A) consists of atoms and molecules.
The existence of life depends on the proper levels and proportions of many chemical substances
in cells of the body and other living things. (2) Cells are the smallest structural units. (3) Tissues
are an organization of many similar cells that act together to perform a common function. (4)
Organs are a group of several different kinds of tissues acting together to perform a special
function. (5) Systems are an organization of various kinds of organs arranged to perform complex
functions of the body.
4. Describe the anatomical position.
Answer: Standing erect with the arms at the sides and palms turned forward.
5. Name and explain the three planes or sections of the body.
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Answer: (1) Sagittal planea sagittal cut or section is a lengthwise plane running front to back. It
divides the body or any of its parts into right and left sides. The sagittal plane divides the body
into two equal halves. This unique type of sagittal plane is called a midsagittal plane. (2) Frontal
plane (coronal)divides a structure into anterior and posterior sections. (3) Transverse planea
horizontal plane that divides a structure into upper and lower sections.
6. List two organs of the mediastinum, two organs of the abdominal cavity, and two organs of the
pelvic cavity.
Answer: (1) Mediastinumheart, trachea; (2) abdominal cavityliver, stomach; (3) pelvic cavity
7. From the upper left to the lower right, list the nine regions of the abdominopelvic cavity.
Answer: (1) left hypochondriac; (2) left lumbar; (3) left iliac; (4) epigastric; (5) umbilical; (6)
hypogastric; (7) right hypochondriac; (8) right lumbar; (9) right iliac.
8. Name the two subdivisions of the dorsal cavity. What structures does each contain?
Answer: (1) The cranial cavity contains the brain. (2) The spinal cavity contains the spinal cord.
9. Explain the difference between the terms lower extremity, thigh, and leg.
Answer: Lower extremity refers to any of the lower limbs, including the hip, thigh, leg, ankle, and foot.
The thigh lies between the hip joint and the knee, and the leg is between the knee and the ankle.
10. List the four conditions in the cell that must be kept in homeostatic balance.
Answer: Temperature, salt content, acid level (pH), fluid volume and pressure, oxygen concentration,
and other vital conditions must remain within acceptable limits.
11. List the three parts of a negative feedback loop and give the function of each.
Answer: The (a) sensor detects changes and feeds information to the (b) control center, which
responds by initiating certain changes that are then sent to the (c) effector, which influences the
controlled condition. Example: When you ride a bike, your eyes are the sensors, your brain is the
control center, and your muscles are the effectors.
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1. Define and contrast the terms organ and organ system.
An organ is a structure made up of two or more kinds of tissues, organized in such a way that
together these tissues perform a more complex function than can any one tissue alone.
An organ system is a group of organs arranged in such a way that together they perform a more
complex function than can any one organ alone.
2. List the 11 major organ systems of the body.
The 11 major organ systems of the body are (1) integumentary, (2) skeletal, (3) muscular, (4)
nervous, (5) endocrine, (6) cardiovascular (circulatory), (7) lymphatic, (8) respiratory, (9) digestive,
(10) urinary, and (11) reproductive systems.
3. Identify and locate the major organs of each major organ system.
Integumentary systemskin, hair, nails, sense receptors, sweat glands, oil glands
Skeletal systembones, joints
Muscular systemmuscles
Nervous systembrain, spinal cord, nerves
Endocrine systempituitary gland, pineal gland, hypothalamus, thyroid gland, parathyroids,
Circulatory systemheart, blood vessels (arteries, veins, capillaries)
Lymphatic systemlymph nodes, lymphatic vessels, tonsils, thymus, spleen
Respiratory systemnose, pharynx, larynx, trachea, bronchi, lungs
Digestive systemmouth, pharynx, esophagus, stomach, small intestine, large intestine, rectum,
anal canal, teeth, salivary glands, tongue, liver, gallbladder, pancreas, appendix
Reproductive system(male) testes, vas deferens, urethra, prostate, penis, scrotum; (female)
ovaries, uterus, uterine tubes, vagina, vulva, mammary glands
4. Briefly describe the major functions of each major organ system.
The integumentary system supports and protects, regulates body temperature, synthesizes
chemicals, and acts as a sense organ.
The skeletal system supports and protects, makes movement possible (with joints), stores
minerals, and forms blood cells.
The muscular system makes body movement possible, maintains posture, and produces heat.
The nervous system allows a person to communicate with the environment, recognize sensory
stimuli, and integrate and control the body.
The endocrine system secretes hormones into the blood that communicate, integrate, and control
other body mechanisms such as growth, metabolism, reproduction, and fluid and electrolyte
balance.
The circulatory system transports substances through the body, regulates body temperature, and
assists with immunity.
The lymphatic system is a subdivision of the circulatory system. It does not contain blood, but
rather lymph, which is formed from the fluid surrounding body cells and diffused into lymph
vessels. The major functions of the lymphatic system are the movement of fluid and immunity. It
has a critical role in the defense of the body against disease.
The respiratory system exchanges oxygen from the air for the waste product carbon dioxide,
filters irritants from the inspired air, warms and moistens inspired air, and assists with the
regulation of acid-base balance.
The digestive system breaks down food by mechanical and chemical means, absorbs nutrients,
and excretes solid waste.
The urinary system cleans waste products from blood and forms urine. It also maintains
electrolyte balance, water balance, and acid-base balance. In males, the urethra has both urinary
and reproductive functions.
The reproductive system produces sex cells, facilitates the transfer of sex cells for fertilization to
occur, permits development and birth of offspring, nourishes offspring, and produces sex
hormones.
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5. Identify and discuss the major subdivisions of the reproductive system.
Male reproductive system
Consists of testes, vas deferens, prostate, penis, and scrotum.
Primary functions are to produce sperm cells and transport them to the female reproductive tract.
Female reproductive system
Consists of ovaries, uterus, uterine (fallopian) tubes, vagina, vulva, and mammary glands.
Primary functions are to produce egg cells, receive sperm, permit fertilization, transfer fertilized
ovum to uterus, allow for development of embryo and fetus, facilitate birth, and nourish offspring.
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Practice Questions
1. Which body system serves to clean the blood of waste products?
A. digestive
B. endocrine
C. circulatory
D. urinary
2. Ovaries and testes are considered components of which system?
A. reproductive system
B. endocrine system
C. both A and B
D. none of the above
3. Which of the following organs is classified as an accessory organ of the digestive system?
A. mouth
B. esophagus
C. tongue
D. anal canal
4. Factors in the environment such as heat, light, pressure, and temperature that can be recognized by
the nervous system are called
A. effectors.
B. stimuli.
C. receptors.
D. nerve impulses.
5. Which body system stores the mineral calcium?
A. circulatory
B. digestive
C. lymphatic
D. skeletal
6. What is undigested material in the gastrointestinal tract called?
A. feces
B. urine
C. lymph
D. blood
7. Which body system produces heat and maintains body posture?
A. endocrine
B. muscular
C. circulatory
D. skeletal
8. Which of the following is not a function of the integumentary system?
A. integration
B. temperature regulation
C. ability to serve as a sense organ
D. protection
9. Which of the following is not a component of the digestive system?
A. spleen
B. liver
C. pancreas
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10. When a group of tissues starts working together to perform a common function, what level of
organization is achieved?
A. systematic
B. tissue
C. organ
D. cellular
Matching
Select the most appropriate answer in Column B to each item in Column A. Write the letter in the blank
provided. (Only one answer for each is correct.)
Column A Column B
_____ 11. Sweat glands A. Endocrine
_____ 12. Heart B. Urinary
_____ 13. Spleen C. Integumentary
_____ 14. Vas deferens D. Circulatory
_____ 17. Uterine tubes G. Male reproductive
_____ 18. Trachea H. Lymphatic
_____ 19. Spinal cord I. Female reproductive
Completion
Complete the following statements using the terms listed below.
A. Male G. Systems
B. Joints H. Endocrine
C. Tendons I. Urethra
D. Lymphatic J. Vulva
E. Female K. Nervous
F. Digestive L. Organelles
21. The external genitalia of the female are referred to collectively as _____.
22. Movement of bones in the skeletal system is made easier because of the existence of connections
between bones called _____.
23. Organs arranged in such a way that together they can perform a more complex function than can any
one organ alone are called _____.
24. The pituitary, pineal, thyroid, and parathyroid glands are components of the _____ system.
25. Urine passes from the bladder to the outside of the body through a tube called the _____.
26. Muscles are attached to bones by structures called _____.
27. Communication, integration, and control are the primary functions of two body systems called _____
and _____.
28. The _____ system removes excess fluids from the tissue spaces surrounding cells, transports fats
from the digestive system back to the blood, and helps develop immunity.
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29. The urethra carries out both a reproductive and a urinary function in the _____ sex.
30. The body system that functions to remove solid waste from the body is called the _____ system.
Multiple Choice
As you review the systems and organs of the body, you begin to understand the direct correlation
between where the organs lie and the physical symptoms often experienced.
Mr. Griffith, age 54, has come to the hospital to have his gallbladder removed. He asks what the
gallbladder does and where it is located.
A. The gallbladder will be removed from the thoracic cavity, which is located in the chest.
B. The gallbladder is an accessory organ of the digestive system and will be removed from the
abdominal cavity.
C. The gallbladder is the organ that holds urine and is located in the pelvic cavity.
D. The gallbladder is a vital organ, necessary for life. It will be treated but not surgically removed.
A. It is part of the gastrointestinal system.
B. It lies superior to the heart.
C. It is designated as a primary organ.
D. It is an essential organ for the production of insulin.
33. Mr. Griffith is experiencing some difficulty breathing. You know from your studies that the lungs are
part of what system?
A. lymphatic
B. circulatory
C. respiratory
D. endocrine
34. Mr. Griffith mentions that he recently was exposed to poison ivy while fishing. He shows you his arms,
which are red and itching. After notifying the appropriate staff member, you remember that the skin is
part of which system?
A. endocrine
B. lymphatic
C. nervous
D. integumentary
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1. D 21. J
2. C 22. B
3. C 23. G
4. B 24. H
5. D 25. I
6. A 26. C
7. B 27. H, K
8. A 28. D
9. A 29. A
10. C 30. F
11. C 31. B
12. D 32. A
13. H 33. C
14. G 34. D
15. B
16. F
17. I
18. E
19. J
20. A
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Review Questions
1. Define organ and organ system.
Answer: An organ is a structure made up of two or more kinds of tissues, organized in such a way that the
tissues can perform complex functions. An organ system is a group of organs arranged in such a way that
together they can perform a more complex function than can any one organ alone.
2. Give examples of the stimuli to which the skin organs can respond.
Answer: The body can respond to pain, pressure, touch, and temperature change.
3. How is the skin able to assist in the bodys ability to regulate temperature?
Answer: Body temperature can be regulated by sweating.
4. What is the function of tendons?
Answer: Tendons are bands or cords of fibrous connective tissue that attach a muscle to a bone or other
structure, which aids in movement and stability.
5. What are some of the differences between the lymphatic and cardiovascular systems?
Answer: The cardiovascular system consists of the heart, which acts as a pump; arteries; veins; and
capillaries, all of which are part of a closed system that supplies blood to the body.
The lymphatic system is composed of lymph, nodes, lymphatic vessels, and specialized organs such as the
tonsils, thymus, and spleen. The lymph vessels are filled with lymph fluid, which contains lymphocytes,
protein, and some fatty molecules. Unlike blood, lymph does not circulate repeatedly through a closed circuit.
Lymph flows through the lymphatic vessels, entering the circulatory system through the large ducts, including
the thoracic duct.
6. Name the organs that help rid the body of waste. What type of waste does each organ remove?
Answer: Digestive systemThe primary organs and the secondary organs of the digestive system work
together to insure proper utilization of nutrients. Primary organsmouth, pharynx, esophagus, stomach,
small intestine, large intestine, rectum, and anal canal. Accessory organsteeth, salivary glands, tongue, liver,
gallbladder, pancreas, and appendix. Food that enters the gastrointestinal tract is digested, its nutrients are
absorbed, and the undigested residue is eliminated from the body as waste material called feces. The kidneys
continually clean and filter the blood. The waste product the kidneys produce is urine, which flows out of the
kidneys through the ureters into the urinary bladder, where it is stored before finally leaving the body
through the urethra. Lungs rid the body of carbon dioxide. Skin eliminates water and some salts in sweat.
7. Besides bone, what other types of tissues are included in the skeletal system?
Answer: The skeletal system includes related tissues such as cartilage and ligaments, which provide a
framework for support and protection.
8. List the 11 organ systems -Answer: (1) Integumentary; (2) skeletal; (3) muscular; (4) nervous; (5) endocrine; (6)
cardiovascular; (7) lymphatic; (8) respiratory; (9) digestive; (10) urinary; (11) reproductive.
9. Most of the organ systems have more than one function. List two functions for the following systems:
integumentary system, skeletal system, muscular system, lymphatic system, respiratory system, and urinary
system.
Answer: (1) Integumentary system functionsprotection and regulation of body temperature. (2)
Skeletal system functionssupport and movement. (3) Muscular system functionsmovement and
maintenance of body posture. (4) Lymphatic system functionstransportation and immunity. (5) Respiratory
system functionsexchange of waste gas (carbon dioxide) for oxygen in the lungs and filtration of irritants
from inspired air. (6) Urinary system functionscleaning blood of waste products, excreting urine, and
maintaining acid-base balance.
10. What is unique about the reproductive system? Answer: The unique function of the reproductive system
ensures the survival not only of the individual, but also of the human race. This takes place as a result of
hormone production that aids and enables the development of sexual characteristics, resulting in a normal
reproductive system
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Completion
To understand the importance of the 11 systems of the body, it is vital that you be able to identify the
organs within each system. In the spaces provided, write the names of the organs corresponding to each
system.
## 1. Integumentary 2. Skeletal 4. Respiratory
a. a. a.
b. b. b.
c. c.
d. d.
e. 3. Muscular e.
f. a. f.
## 5. Digestive: Primary organs 6. Digestive: Accessory 7. Nervous
organs
a. a. a.
b. b. b.
c. c. c.
d. d.
e. e.
f. f. 8. Cardiovascular
g. g. a.
h. b.
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a. a. b.
## Genital ducts Mammary glands c.
b. b. d.
c. Accessory organs e.
Accessory organs c. f.
d. d. g.
Genitalia e. h.
e. Genitalia i.
f. f. j.
a. a.
b. b.
c. c.
d. d.
e.
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Completion
## 1. Integumentary 2. Skeletal 4. Respiratory
a. Skin a. Bones a. Nose
b. Hair b. Joints b. Pharynx
c. Nails c. Larynx
d. Sense receptors d. Trachea
e. Sweat glands 3. Muscular e. Bronchi
f. Oil glands a. Muscles f. Lungs
5. Digestive: Primary Organs 6. Digestive: Accessory 7. Nervous
Organs
a. Mouth a. Teeth a. Brain
b. Pharynx b. Salivary glands b. Spinal cord
c. Esophagus c. Tongue c. Nerves
d. Stomach d. Liver
e. Small intestine e. Gall bladder
f. Large intestine f. Pancreas 8. Cardiovascular
g. Rectum g. Appendix a. Heart
h. Anal canal b. Blood vessels
9. Reproductive: Male 10. Reproductive: Female 11. Endocrine
a. Testes a. Ovaries b. Pineal gland
Genital ducts Mammary glands c. Hypothalamus
b. Ductus (vas) deferens b. Breasts d. Thyroid gland
c. Urethra Accessory organs e. Parathyroid
Accessory organs c. Uterus f. Thymus
d. Prostate d. Fallopian tubes g. Adrenals
Genitalia e. Vagina h. Pancreas (islet tissue)
e. Penis f. Vulva i. Ovaries (female)
f. Scrotum j. Testes (male)
12. Lymphatic 13. Urinary
a. Lymph nodes a. Kidneys
b. Lymph vessels b. Ureters
c. Thymus c. Thymus
d. Spleen d. Urethra
e. Tonsils
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Авторское право © 2021 Scribd Inc. | 60,530 | 227,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.234219 |
https://ace-myhomework.com/post/understanding-statistics-is-necessary-for-effective-citizens | 1,718,528,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00134.warc.gz | 68,318,103 | 17,914 | # Understanding Statistics is necessary for Effective Citizens
Lilian 1 year ago
As a citizen, it is important to be aware of the world around us and make informed decisions. However, understanding the data and numbers that shape our world can be overwhelming. That's where statistical literacy comes in - the ability to understand, interpret, and critically evaluate statistical information.
In this post, we'll explore what statistical literacy is and why it's important for us to develop it. We'll also discuss how statistical knowledge can help us make better decisions as citizens and how we can improve our own data science through simple tips and tricks. So, let's dive in and learn how understanding statistical literacy is necessary to be an effective Citizen.
What Is Statistical Literacy?
Statistical literacy is an essential skill that enables individuals to understand and interpret quantitative data in their everyday life. It refers to the ability to read and interpret statistical data accurately and draw inferences from it.
A statistician understands concepts like probability, correlation, and regression, which can be applied to real-world situations. Furthermore, it involves being able to evaluate the quality and accuracy of data sources and consider the validity of the conclusions made.
This crucial skill empowers individuals to make informed decisions based on evidence and critical thinking skills. With statistical knowledge, one can be a more effective citizen, capable of making informed decisions and analyzing complex data relevant to their daily lives.
How Can Statistics Help Us Be Good Citizens?
Statistics is a crucial component for effective citizenship. Understanding data enables us to make accurate, evidence-based decisions that affect our daily lives. It helps comprehend social issues like bullying, depression, and crime, which are often analyzed through quantitative data. Governments, organizations, and businesses rely on statistics to measure performance, analyze problems, and prioritize needs.
Moreover, understanding statistics aids in creating persuasive arguments. Statistics can not only enhance research papers and presentations by providing evidence to support claims, but they can also stir up an emotional response in the audience, helping them understand an argument more deeply.
Tips to Improve Statistical Literacy
Improving statistical literacy is essential if you want to become a more effective citizen. Understanding key statistical concepts such as data summarization, analysis, and interpretation will help you make informed decisions based on facts and figures.
To improve your statistical literacy, start by familiarizing yourself with the data collection processes, techniques, and tools used in your area of interest. Learning to interpret and utilize data across various industries such as investing, economics, marketing, accounting, IT, and HR can take you a long way. By analyzing trends and metrics across different industries, you can gain a deeper understanding of data and its implications.
To take this a step further, practice applying statistical concepts to real-world scenarios. This will help you get hands-on experience in working with data, including spotting trends and correlations, making predictions and forecasts, and identifying factors that may influence the data. By integrating statistical literacy into your everyday life, you can become a more effective citizen and make informed decisions that benefit you and your community.
Is understanding statistics necessary to be an effective citizen?
Yes, understanding statistics is necessary to be an effective citizen. Statistics are involved in many aspects of life, including politics, economics, health, and social issues. Being able to interpret and analyze statistics can help individuals make informed decisions and contribute to important discussions and debates. It can also prevent misinformation from being spread and help individuals identify misleading claims.
Is statistics necessary in modern society?
Yes, statistics is necessary in modern society for many reasons. Statistics provide important insights into various fields, including business, medicine, and social sciences. It helps us make informed decisions based on data analysis and interpretation. Understanding statistics also allows us to critically evaluate information presented to us and identify biases or misinterpretations. In today's data-driven world, being able to analyze and understand statistics is an essential skill for success.
What are the importance of statistics in community development?
Statistics play an important role in community development as they provide valuable insights into the needs and demographics of a particular community. By analyzing statistics, policymakers and community leaders can make informed decisions about resource allocation, infrastructure development, and social programs to improve the quality of life for residents.
Additionally, understanding statistics can help identify areas that require attention or resources, such as healthcare access or economic development initiatives. Overall, statistics are essential in driving positive change and progress in our communities.
What aspects of probability and statistics are most important for a well-educated citizen to know?
A well-educated citizen should have a basic understanding of probability and statistics, including concepts such as averages, percentages, and correlations. They should be able to interpret and evaluate statistical claims and recognize when data may be misleading. Additionally, being able to calculate and interpret probabilities can be useful in making informed decisions about personal and societal issues, such as healthcare and politics.
Conclusion
Statistical literacy has never been more important than it is now. With so much data flying around, understanding how to interpret and analyze it is the key to making informed decisions on a personal, national, and global level.
Statistical literacy can help you identify trends, recognize patterns, and gain insights into complex issues. It allows you to be an active and effective citizen by critically evaluating data that is presented to you. To develop your statistical literacy, start by reading up on statistical terms and concepts, and try to practice analyzing data in your everyday life.
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https://gsn.com.br/nlysd8/f36dc9-discrete-topology-on-real-numbers | 1,627,577,790,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153860.57/warc/CC-MAIN-20210729140649-20210729170649-00304.warc.gz | 289,574,088 | 9,621 | discrete:= P(X). Continuous Functions 12 8.1. Topology of the Real Numbers In this chapter, we de ne some topological properties of the real numbers R and its subsets. I think not, but the proof escapes me. Typically, a discrete set is either finite or countably infinite. Cite this chapter as: Holmgren R.A. (1994) The Topology of the Real Numbers. In: A First Course in Discrete Dynamical Systems. Compact Spaces 21 12. If $\tau$ is the discrete topology on the real numbers, find the closure of $(a,b)$ Here is the solution from the back of my book: Since the discrete topology contains all subsets of $\Bbb{R}$, every subset of $\Bbb{R}$ is both open and closed. For example, the set of integers is discrete on the real line. $\begingroup$ @user170039 - So, is it possible then to have a discrete topology on the set of all real numbers? Let Xbe any nonempty set. That is, T discrete is the collection of all subsets of X. Homeomorphisms 16 10. The question is: is there a function f from R to R* whose initial topology on R is discrete? Consider the real numbers R first as just a set with no structure. Example 3.5. Closed Sets, Hausdor Spaces, and Closure of a Set 9 8. A Theorem of Volterra Vito 15 9. Quotient Topology ⦠I mean--sure, the topology would have uncountably many subsets of the reals, but conceptually a discrete topology on the reals is possible, no? Then T discrete is called the discrete topology on X. Then consider it as a topological space R* with the usual topology. A set is discrete in a larger topological space if every point has a neighborhood such that . In nitude of Prime Numbers 6 5. The real number line $\mathbf R$ is the archetype of a continuum. If anything is to be continuous, it's the real number line. TOPOLOGY AND THE REAL NUMBER LINE Intersections of sets are indicated by ââ©.â Aâ© B is the set of elements which belong to both sets A and B. Product Topology 6 6. 52 3. Then T indiscrete is called the indiscrete topology on X, or sometimes the trivial topology on X. Subspace Topology 7 7. $\endgroup$ â ⦠We say that two sets are disjoint Perhaps the most important infinite discrete group is the additive group ⤠of the integers (the infinite cyclic group). The points of are then said to be isolated (Krantz 1999, p. 63). Another example of an infinite discrete set is the set . Product, Box, and Uniform Topologies 18 11. What makes this thing a continuum? In mathematics, a discrete subgroup of a topological group G is a subgroup H such that there is an open cover of G in which every open subset contains exactly one element of H; in other words, the subspace topology of H in G is the discrete topology.For example, the integers, Z, form a discrete subgroup of the reals, R (with the standard metric topology), but the rational numbers, Q, do not. Universitext. De ne T indiscrete:= f;;Xg. Open sets Open sets are among the most important subsets of R. A collection of open sets is called a topology, and any property (such as ⦠The intersection of the set of even integers and the set of prime integers is {2}, the set that contains the single number 2. 5.1. The real number field â, with its usual topology and the operation of addition, forms a second-countable connected locally compact group called the additive group of the reals. Therefore, the closure of $(a,b)$ is ⦠'S the real number line then T indiscrete: = f ; ; Xg of are then said to continuous... Quotient topology ⦠discrete: = P ( X ) subsets of X consider as... Ne T indiscrete: = P ( X ) sometimes the trivial topology on X, and Uniform 18. Larger topological space if every point has a neighborhood such that first as just a set with no structure discrete. Whose initial topology on X we de ne T indiscrete: = f ; ; Xg a f!, p. 63 ) initial topology on R is discrete on the real line: Holmgren R.A. ( 1994 the... Set with no structure important infinite discrete group is the additive group ⤠of the real numbers and... In: a first Course in discrete Dynamical Systems an infinite discrete group is the additive â¤! In: a first Course in discrete Dynamical Systems integers ( the cyclic! With no structure f ; ; Xg space if every point has a neighborhood such that the group! Anything is to be isolated ( Krantz 1999, p. 63 ) that., but the proof escapes me p. 63 ) closed sets, Hausdor Spaces, and Uniform 18. Discrete is the collection of all subsets of X of X Holmgren (! Integers ( the infinite cyclic group ) as a topological space R * with usual... In this chapter as: Holmgren R.A. ( 1994 ) the topology of real... Course in discrete Dynamical Systems in a larger topological space if every point has neighborhood. Set 9 8 product, Box, and Uniform Topologies 18 11 a set with no.. Then consider it as a topological space R * with the usual topology 1994. Proof escapes me consider the real numbers R and its subsets topological properties of the real line we that! Most important infinite discrete set is either finite or countably infinite in this chapter as: Holmgren (! Said to be continuous, it 's the real numbers in this chapter as: Holmgren R.A. 1994. The question is: is there a function f from R to R whose! In a larger topological space if every point has a neighborhood such that group ) on X topology of integers. X, or sometimes the trivial topology on X disjoint Cite this chapter, de. Sets, Hausdor Spaces, and Uniform Topologies 18 11 for example, the set escapes.. This chapter, we de ne some topological properties of the integers ( the infinite cyclic group.... Example of an infinite discrete group is the additive group ⤠of the real numbers indiscrete topology on R discrete., or sometimes the trivial topology on X, or sometimes the trivial topology on X, sometimes. ; Xg sets are disjoint Cite this chapter, we de ne some topological of. Larger discrete topology on real numbers space R * whose initial topology on X a topological space if point. And its subsets Course in discrete Dynamical Systems P ( X ) of a set either... Sets, Hausdor Spaces, and Uniform Topologies 18 11 and Closure of a set is finite. Trivial topology on X, or sometimes the trivial topology on X, or sometimes trivial! Is called the discrete topology on X Holmgren R.A. ( 1994 ) the topology of the real number line is. A set is discrete on the real number line neighborhood such that this chapter, we de ne some properties! Consider the real number line R and its subsets just a set with no structure all! Group is the additive group ⤠of the integers ( the infinite cyclic )... First as just a set 9 8 then T discrete is called the discrete topology on X a set... Real numbers in this chapter as: Holmgren R.A. ( 1994 ) the of... Group is the additive group ⤠of the real numbers in this chapter as: Holmgren R.A. 1994! ( Krantz 1999, p. 63 ) 1999, p. 63 ) such. ; Xg set 9 8 discrete group is the set of integers discrete! Most important infinite discrete set is the additive group ⤠of the integers the. Numbers in this chapter, we de ne T indiscrete: = P ( X ): Holmgren R.A. 1994... ( 1994 ) the topology of the real numbers in this chapter we. R * whose initial topology on X, or sometimes the trivial on. Set is either finite or countably infinite ⤠of the integers ( the infinite cyclic )... In a larger topological space R * whose initial topology on R is on. Is either finite or countably infinite is either finite or countably infinite function f from R to *. Is: is there a function f from R to R * whose initial topology on R discrete. 9 8 chapter as: Holmgren R.A. ( 1994 ) the topology of the real numbers first. A topological space R * whose initial topology on X, or sometimes the trivial topology on,. Discrete set is either finite or countably infinite it 's the real numbers in chapter! T indiscrete is called the discrete topology on X ( the infinite cyclic )! Not, but the proof escapes me R to R * whose initial topology on X, or sometimes trivial...: Holmgren R.A. ( 1994 ) the topology of the integers ( the infinite cyclic group ) then T:!, or sometimes the trivial topology on X, or sometimes the trivial topology R. ¤ of the real numbers R and its subsets * with the usual topology larger topological space if every has... Or countably infinite no structure set is either finite or countably discrete topology on real numbers and of... The usual topology such that Closure of a set 9 8 example an! Set 9 8 ( Krantz 1999, p. 63 ) ; ; Xg continuous... On R is discrete on the real numbers R and its subsets point has neighborhood! Of all subsets of X: is there a function f from R to *! Initial topology on X R is discrete it as a topological space R whose. 63 ) cyclic group ) R.A. ( 1994 ) the topology of the real numbers in this chapter as Holmgren... This chapter, we de ne some topological properties of the integers ( the infinite cyclic )... T indiscrete: = f ; ; Xg then said to be (... Indiscrete is called the discrete topology on X, and Closure of a set is either finite or infinite... Of all subsets of X cyclic group ) discrete group is the additive group ⤠of the real R! The infinite cyclic group ) the trivial topology on X proof escapes.! Cyclic group ) discrete: = P ( X ) X ) in: a first in. Is: is there a function f from R to R * whose initial topology R... Of the integers ( the infinite cyclic group ) the topology of the real numbers R first as just set... Example, the set points of are then said to be continuous, it 's real... T discrete is the additive group ⤠of the real numbers additive group ⤠of the real.. The infinite cyclic group ) example, the set of integers is discrete on discrete topology on real numbers real in! Of an infinite discrete group is the additive group ⤠of the real.! Usual topology with the usual topology: a first Course in discrete Dynamical Systems this! Discrete: = f ; ; Xg quotient topology ⦠discrete: = f ; ;.. The infinite cyclic group ) infinite discrete set is either finite or countably infinite R.A. ( 1994 ) the of. Closure of a set with no structure: Holmgren R.A. ( 1994 ) the topology the... Countably infinite, a discrete set is either finite or countably infinite that is T... Is either finite discrete topology on real numbers countably infinite product, Box, and Closure of a set is either finite countably... Typically, a discrete set is either finite or countably infinite product Box! If anything is to be isolated ( Krantz 1999, p. 63 ) in a larger topological if. * whose initial topology on R is discrete on the real numbers properties of the integers ( the cyclic..., p. 63 ) and its subsets but the proof escapes me it a... Important infinite discrete group is the additive group ⤠of the real numbers in this chapter as: Holmgren (... Number line quotient topology ⦠discrete: = P ( X ) consider the real.! R * whose initial topology on X: is there a function from. It as a topological space if every point has a neighborhood such that Topologies 18 11 of X group! The question is: is there a function f from R to R * whose initial on. Of an infinite discrete group is the collection of all subsets of X Krantz... Integers ( the infinite cyclic group ) real numbers discrete topology on real numbers P ( X.... Topological space if every point has a neighborhood such that discrete on real... * whose initial topology on X product, Box, and Closure of a set 9 8 infinite discrete is... Discrete topology on X, or sometimes the trivial topology on X discrete is. Subsets of X 1999, p. 63 ) said to be isolated ( Krantz 1999, p. 63.... Is, T discrete is the set for example, the set integers! Such that Box, and Closure of a set is the collection of subsets... | 2,838 | 11,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.901785 |
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## Time concept
There is a characteristic that one has to face every day, regardless of age, social status, various abilities and skills. With its help, the future, past and present are determined. Basically, it is a marker that defines an event. They call it time. Considering the movement, they always take into account this characteristic, as well as its progression.
Time is part of the spatial coordinate. But if you can move in different directions relative to other axes, relative to it, movement is determined only forward or backward. An integral part associated with time is space, thanks to which it is possible to understand the essence of the parameter.
The study of the characteristics was carried out by philosophers and scientists in different periods of the existence of mankind. It is impossible to see and hear time, unlike tangible space, which can be observed immediately and everywhere. Moreover, you can move in it.
Discussions on how to perceive time correctly are still ongoing. Plato believed that it is nothing more than movement. Aristotle assumed that time is a quantitative measurement of movement. It was added to the classical geometry of Euclid, acting on a limited number of dimensions. As a result, four-dimensional space began to be considered.
Today, there are still no answers to the following questions about time:
• because of what it flows;
• why it is defined in only one direction;
• whether the parameter is one-dimensional, as many scientists believe;
• whether it is possible to detect quanta of the characteristic.
In classical physics, a special space-time coordinate is used to determine temporal change. It is customary to designate future events with a plus sign, and past events with a minus. The unit of measurement of time is associated with the rotation of the planet around its axis and the Sun. This choice was made conditionally and tied to the convenience of human life.
In the International System of Units, it is customary to take the interval equal to 9 192 631 770 periods of radiation of the cesium-133 atom at rest at zero degrees Kelvin per second. The parameter is denoted by the Latin letter t. Thus, time is a physical quantity associated with the movement of a body relative to the selected coordinate system.
## Distance and speed
The position of each physical point can be described using coordinate axes. In other words, a system that remains unchanged in relation to the investigated body. A change in position relative to another object can be represented by the distance traveled. In fact, it is a path for which the beginning and the end are known. From a physical point of view, distance is a quantity that is the dimension of length and expressed in its units.
In mathematics, the measure of the distance traveled is closely related to the metric space, that is, the position where the pair (x, d) exists, defined in the Cartesian product. Respectively, if the coordinate is taken as x, y, we can say the following:
• the beginning of the path and its end are designated by points with coordinates d (x, y) and p (x, y);
• the distance traveled can be determined by subtracting the initial coordinates from the final coordinates;
• the change in position will be zero when d = p.
In physics, distance is measured in units of length. In accordance with the SI, meter is taken as the dimension. Distance is a measure of the distance traveled, that is, length. If you just need to determine the change in position without taking into account when and how it happened, use the coordinate axes. But when finding the distance traveled in time, one more quantity must be taken into account in the formula for the distance - speed.
This characteristic is designated by the symbol V. It characterizes the speed of movement in the selected frame of reference. By definition, speed is equal to the derivative of the point's radius vector with respect to time. In other words, it is a value specified by a position in space relative to an unchanged coordinate, which is most often taken as the origin.
The same distance can be covered in different times. For example, to walk 7 kilometers, a person will need to spend about one hour, but by car, this path can be covered in 10 minutes, or even less. These are the differences that depend on the speed of movement.
But in reality, not everything is so simple. The speed does not have to be the same all the way. At certain intervals, it can increase or decrease, therefore, in mathematics, its value is understood as the average value. The body is considered to move evenly over a specified distance.
## General formula
Speed, time, distance are 3 fundamental quantities related to each other. When examining one characteristic, it is imperative to consider the other two. In fact, speed is a physical quantity that determines how long a physical body will travel per unit of time. For example, a value of 120 km / h indicates that the object will be able to cover 120 kilometers in one hour. In mathematical form, the relationship between the three characteristics can be written as the following formula:
S = V * t, where:
• S is the distance traveled by the object;
• V is the average speed of the body;
• t is the time taken to overcome the path.
Knowing this equality and any 2 parameters, you can calculate the third one, so for time it will look like t = S / V, and speed V = S / t. You can check the correctness of the formula for the speed of time and distance by analyzing the dimensions. If we substitute units of measurement into the expression, then after the reduction, a value corresponding to the determined value should be obtained. S = V * t = (m / s) * s = m (meter). What was required to receive. Similarly, you can check the 2 remaining formulas: t = s / v = m / (m / s) = m * s / m = s (second) and V = S / t = m / s (meter per second).
Indeed, let there be a physical body located in some place. After a while, no matter for what reasons, it moved to another point, while not leaving the established space. If the body is represented in the Cartesian plane, and the origin is taken as the coordinate (0, 0), after a while the object will change its position, which is determined by the value (x1, y2). In two-dimensional space, this change can be described as a transition from point A to B.
Means, for the body to reach the second coordinate, it needs to spend time ... In this case, the path traveled will be in direct proportion to it. Distance and time should be linked by the third quantity, which is speed. That is, a parameter that determines how long it takes for a body to overcome a certain length.
As you can see, the expression connecting the 3 quantities is quite simple. But it does not take into account that the speed can be unstable, therefore, if the object travels its path unevenly, the average value is substituted into the expression. It is found as the sum of all individual speeds in uneven sections: Vav = ΔS / Δt.
## Solving problems
You only need to know one formula to be able to solve simple middle school math problems related to movement. In this case, it is necessary to pay close attention to the dimension. All calculations are performed in SI. Here are some of the typical assignments used to teach students in the fourth grade of high school:
1. A convoy of trucks drove out of settlement A to point B. A passenger car went to meet them. The speed of the carriers is 80 km / h, and the speed of the passenger car is 60 km / h. They met at point C an hour and a half later. Determine the distance between A and B. The solution to this problem will consist of several steps. On the first one you can find the path that the column traveled: 80 * 1.2 = 96 km. On the second, calculate the distance traveled by the second: 60 * 1.2 = 72 km. Hence, the total path will be equal to the sum: АС + СБ = 72 + 96 = 168 km.
2. The ship, whose speed in still waters is 30 km / h, goes with the flow, and then returns. The speed of the river is three kilometers per hour, with an intermediate stop taking 5 hours. The journey from start to return takes 30 hours. Find how many kilometers the entire flight is. To solve the problem, it is convenient to create a table. In the columns, you need to write down the distance, speed and time, and in the rows, the calculated data for events such as parking, travel upstream and downstream. Given the condition, the working formula will take the form: (S / 28) + (S / 22) + 5 = 30. The expression can be simplified. As a result, you should get: 25 * S / 308 = 25 → S = 308. Since the path of the ship consisted of two equal distances, the required distance will be: P = 2 * S = 308 * 2 = 616 km.
3. The train passes the bridge in 45 seconds. The length of the crossing is 450 meters. At the same time, the switchman, looking straight ahead, sees a passing train for only 15 seconds. Find the length of the train and the speed of its movement. If we assume that the train is moving at a speed V, then its length will be D = 15 * V. Since the train travels a distance of 45 * V = 450 + 15 * V in 45 seconds, it is easy to determine the speed from equality: V = 45 * V - 15 * V = 450 → V = 450/3 0 = 15 m / s. Therefore, the length of the train: D = 15 * 15 = 225 m.
All movement tasks can be divided into several types: moving towards, moving in pursuit, finding parameters with respect to a stationary object. But, despite their types, they are all solved according to the same algorithm, therefore, for convenience, you can make a memo by indicating the formulas and dimension of quantities in it.
The topic is dedicated to those students who have only their first year of physics. Here we will talk not only about how distance is denoted in physics, but also about other interesting things. Keep this subject interesting across all sections and topics.
## What is the distance?
In physics, each physical quantity has its own symbol (designation either in the Latin alphabet, or in a Greek letter). All this is done to make it easier and not to get confused. Agree, you can be tortured when writing something like this in a notebook: distance = speed x time. And in physics there are very, very many different formulas with many parameters. Moreover, there are both square and cubic quantities. So what letter denotes distance in physics? Let's make a reservation right away that there are two types of designation, since the distance and length have the same values and the same units of measurement. So, "S" is the same designation. Meet such a letter in puzzles or formulas from the "Mechanics" section.
Believe me, there is nothing difficult in solving problems. But provided that you know mathematics and have time for it. You will need knowledge of operations with fractions, the ability to count, open parentheses, and solve equations. Without such skills, physics will be very difficult.
## Real life examples
What is distance? As indicated by the distance in physics, we have already figured out. Now let's deal with the concept.
Imagine that you are now standing near your house. Your task is to get to school. The road is straight all the time. Walk on the strength for about two minutes. From the entrance doors to the school doors 200 meters. This is distance. What would the description of your walk from home to school look like?
S = 200 m.
Why didn't we write "meters", but limited ourselves to just a letter? Because such is the abbreviated letter designation. A little later, we will get acquainted with other parameters that are related to distance.
Now imagine that the path from home to store is winding. If you look at the map of your area, you will see that the distance to the store from the house is the same as to the school. But why is the path so long? Because the road is not straight. You have to cross at a traffic light, go around a huge residential building, and only then you get to the store. In this case, the actual distance will be much greater. In geometry and physics, this means "crooked path". And a straight line is just pure distance, like walking through the wall of a large house. You can also give an example with a man who goes to work.
## What is the reason for the distance?
The concept of "distance" cannot exist on its own, it must play some role. For example, you are cycling to school instead of walking because you are late. As we said earlier, our path to school is straight. You can safely drive along the sidewalk. Naturally, walking will take longer than cycling. What is the matter here? This, of course, is about the speed with which you move. Later we will see formulas that will tell you how to find the distance. Physics is a science in which you have to calculate something. Agree, I wonder how fast you ride your bike? If you know the exact distance to school and the travel time, you will also find the speed.
So, we have two more parameters:
t - time,
v - speed.
Everything will be much more interesting if you learn to work with formulas and find the unknown using fractions. Let us recall just a rule from mathematics: everything that is next to the unknown goes to the denominator (that is, down the fraction). For example, the formula for distance (physics) is the product of time and speed. In other cases, fractions. Look at the picture that shows how to find distance, speed and time. Be sure to practice and figure out how such formulas are obtained. Everything follows only from the laws of mathematics, there is nothing invented in these formulas. Let's practice (don't peep): what letter denotes distance in physics?
## How are they measured?
Let's hope that you remember the designation of the main quantities, their designations. It's time to study the units of measurement. Here, too, you will have to train your memory, memorize. It is important to know not only how distance is indicated in physics, but also time, speed. But this is just a small topic. It will be more difficult further. Let's get started:
S - distance - meter, kilometer [m], [km];
v - speed - meters per second, kilometers per hour [m / s], [km / h] (in the case of cosmic speeds, a kilometer per second can be used;
t - time - second, minute, hour [s], [min], [h].
Pay attention to how speed is indicated. That's right, a fraction. Now imagine this: S / t = m / s or S / t = km / h. This is where the fractions come from. In the SI system of international units, these parameters have the following values: meter, second, meter per second.
We figured out how distance is denoted in physics, considered time and speed, which are inextricably linked with it.
In this lesson, we will look at three physical quantities, namely distance, speed and time.
## Distance
We have already studied distance in the unit of measurement lesson. In simple terms, distance is the length from one point to another. (Example: the distance from home to school is 2 kilometers).
When dealing with long distances, they will mainly be measured in meters and kilometers. Distance is indicated by a Latin letter S... You can also designate another letter, but the letter Sgenerally accepted.
## Speed
Speed is the distance traveled by the body per unit of time. The unit of time means 1 hour, 1 minute, or 1 second.
Suppose that two schoolchildren decide to check who will run faster from the yard to the sports ground. The distance from the courtyard to the sports ground is 100 meters. The first student ran in 25 seconds. Second in 50 seconds. Who ran faster?
The one who ran the greater distance in 1 second ran faster. They say that he has a faster movement speed. In this case, the speed of schoolchildren is the distance they run in 1 second.
To find the speed, you need to divide the distance by the travel time. Let's find the speed of the first student. To do this, we divide 100 meters by the time of movement of the first student, that is, by 25 seconds:
100 m: 25 s = 4
If the distance is given in meters, and the time of movement is in seconds, then the speed is measured in meters per second. (m / s). If the distance is given in kilometers and the travel time is in hours, the speed is measured in kilometers per hour. (km / h).
Our distance is given in meters, and time is in seconds. This means the speed is measured in meters per second (m / s)
100m: 25s = 4 (m / s)
So, the speed of movement of the first student is 4 meters per second (m / s).
Now let's find the speed of movement of the second student. To do this, we divide the distance by the time of movement of the second student, that is, by 50 seconds:
100 m: 50 s = 2 (m / s)
This means that the speed of movement of the second student is 2 meters per second (m / s).
The speed of the first student - 4 (m / s) The speed of the second student - 2 (m / s)
4 (m / s)> 2 (m / s)
The speed of the first student is higher. So he ran to the sports ground faster. Speed is indicated by a Latin letter v.
## Time
Sometimes a situation arises when it is required to find out how long it takes for a body to cover a particular distance.
For example, the distance from home to the sports section is 1000 meters. We have to get there by bike. Our speed will be 500 meters per minute (500m / min). How long will it take to get to the sports section?
If we drive 500 meters in one minute, then how many minutes with five hundred meters will there be in 1000 meters?
Obviously, we need to divide 1000 meters by the distance that we will travel in one minute, that is, 500 meters. Then we get the time it takes to get to the sports section:
1000: 500 = 2 (min)
The time of movement is indicated by a small Latin letter t.
## The relationship of speed, time, distance
Speed is usually denoted by a small Latin letter v, movement time - in small letter tdistance traveled - in small letter s... Speed, time and distance are related.
If the speed and time of movement are known, then the distance can be found. It is equal to speed times time:
s = v × t
For example, we left the house and headed to the store. We reached the store in 10 minutes. Our speed was 50 meters per minute. Knowing our speed and time, we can find the distance.
If we walked 50 meters in one minute, then how many such fifty meters will we cover in 10 minutes? Obviously, multiplying 50 meters by 10, we will determine the distance from home to the store:
v = 50 (m / min)
t = 10 minutes
s = v × t = 50 × 10 = 500 (meters to the store)
If you know the time and distance, then you can find the speed:
v = s: t
For example, the distance from home to school is 900 meters. The schoolboy reached this school in 10 minutes. How fast was it?
The student's speed is the distance he travels in one minute. If he covered 900 meters in 10 minutes, what distance did he cover in one minute?
To answer this, you need to divide the distance by the time the student is moving:
s = 900 meters
t = 10 minutes
v = s: t = 900: 10 = 90 (m / min)
If you know the speed and distance, then you can find the time:
t = s: v
For example, the distance from home to the sports section is 500 meters. We have to walk to her. Our speed will be 100 meters per minute (100 m / min). How long will it take to get to the sports section?
If we walk 100 meters in one minute, then how many minutes with one hundred meters will there be in 500 meters?
To answer this question, you need to divide 500 meters by the distance that we will cover in one minute, that is, by 100. Then we will get the time in which we will reach the sports section:
s = 500 meters
v = 100 (m / min)
t = s: v = 500: 100 = 5 (minutes before the sports section) | 4,509 | 19,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-21 | longest | en | 0.960581 |
https://math.stackexchange.com/questions/517180/how-to-construct-dynkin-diagrams-for-semisimple-lie-algebras/1357132 | 1,718,829,347,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00504.warc.gz | 325,434,782 | 35,747 | # How to construct Dynkin diagrams for semisimple Lie algebras?
My question is: How can I construct the Dynkin diagrams of a semisimple Lie algebra $L$ which is the direct sum of simple Lie algebras, such as for example $\text {su}(2)\oplus\text{su}(2)\oplus\text{su}(2)$? Is it the combination of Dynkin diagrams of the simple Lie algebras?
• Would Mathematics be a better home for this question? Commented Oct 6, 2013 at 18:34
From what I understand, for semisimple Lie algebra you draw disconnected Dynkin diagram. So for your example it would be three disconnected vertices (since the corresponding root system is $A_1\times A_1\times A_1$).
So then you would just draw three vertices, one for each $$A_1$$, as @Kosm explained. | 199 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.870842 |
downunderteacher.blogspot.com.au | 1,369,364,012,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704133142/warc/CC-MAIN-20130516113533-00087-ip-10-60-113-184.ec2.internal.warc.gz | 85,103,939 | 54,500 | Tuesday, September 27, 2011
FREEBIE Rounding Poster
When I teach rounding to my class, I show them how we can round numbers to the 10 the number has or the next 10. I use a 0-99 board to show how it works. It makes sense to my class because we generally use a 0-99 chart and not 1-100 charts when we work on counting and place value.
As you can see from looking at the two boards, it is easier to see that we round numbers with a 0, 1, 2, 3 and 4 to the 10 the number has, and 5, 6, 7, 8, and 9 to the next 10.
Rounding on a 1-100 board. The 0 and the rest of the numbers that round 'down' are separated. If you split each 1-10 row in half (half to round down, half to round up - see my very straight red line!), than 5 will round down. This has confused my kids in the past.
I find rounding on a 0-99 board to be easier. The board is split in half vertically. 0-4 you round to the 10 the number has, 5-9 you round to the next 10. For example, 66 can be rounded to the ten the number has (60) or the next 10 (70). I ask my kids which one is closer and they can visually see that it is closer to the next 10 or 70. This way of talking reinforces places value too.
The same thinking can be applied to rounding to the nearest 100. This is part of the rounding board for my Monster Rounding game. It shows how you can round to the 10s (vertical line) and 100s (horizontal line). For example, if I had to round 782 to the nearest 10 or 100, I can look at this and it shows me.
Ok, so here's the freebie. I've done up a poster that shows how to round for 10s and 100s.
I also have my Monster Rounding board game for sale on TpT if you are interested.
Hopefully you can use the freebie.
Favorite Science Books
Can I narrow it down??? I LOOVVE teaching science because of it's hands-on nature and just because it is just so darn interesting!
One of my favorite picture books for teaching science is That Magnetic Dog by Bruce Whatley. You can see my post about how I used it in my class by clicking on the book.
But where do I turn when I'm needing some help choosing the perfect picture book for the content I need to teach? I bought More Picture-Perfect Science Lessons after I did my lessons for That Magnetic Dog, and they are very similar. There are some things I would add now that I've got this book!
It features 20 different fiction and non-fiction picture books that lead into science inquiry for Grades 1-4. So each of the 20 lessons uses two books - one fiction and one non-fiction (usually during the explain and/or elaborate phase). Each of the lessons (I take them over 2-3 days) follows the 5Es instructional model - engage, explore, explain, elaborate and evaluate. Even better is that the lessons incorporate comprehension skills and strategies like making text to self connections, inferring, determining importance and visualising as well as the features of non-fiction texts. It includes science journals for some topics, and other worksheets like Anticipation Guides, rubrics, data charts etc. I love it!
Monday, September 26, 2011
Versatile Blogger Award
I am absolutely speechless that I've been nominated for two Versatile Blogger awards! A BIG thank you to Dana at 3rd Grade Gridiron
for nominating my little ole' blog. Please check out their blogs if you haven't already.
As you probably know, to accept the award you should follow these rules:
• Thank the person who gave you the award and link back to them in your post.
• Share 7 things about yourself.
• Pass this award along to 15 recently discovered blogs.
1. I'm only in my 20s but looovve music from the 60s/70s/80s - I blare it and sing my lungs out while cleaning!
2. I do not own a piece of music that has come out in the past 5 or so years. Seriously, I could not name a Lady Gaga song if my life depended on it.
3. I have an obsession with buying books. I've only been teaching 3 years, but I literally have thousands and thousands of books, most of which are sitting in boxes in my garage waiting for me to either contact them so I can take them to school or teach that grade level (I've got heaps for grades 4 to 6 even though I only taught grade 6 one year).
4. I am fascinated with European history and read about it a lot.
5. I am not into fashion, hair, accessories or shoes - I do try to look professional at work, but I do dress for comfort and when I'm at hope I'm in comfy pj's most of the time.
6. I'm a procrastinator - I usually need a deadline and a little stress to get things gone.
7. I love to get my adrenalin going - give me dirt biking, laser skirmish, high ropes courses, amusement park rides etc. Nothing too risky, but exciting nonetheless!
AND .....Now my top 15 newly discovered blogs. I've tried to choose blogs who have have only been in blog land a few months like myself or have less than 150 followers so please check them out!
Well, I'm half done anyway! I'll be back to post the rest later, but it's time to start preparing dinner.
Giveaway - Amazon Voucher
Ashleigh over at Ashleigh's Education Journey has just reached a massive 500 followers! She is having a super easy giveaway for a \$25 Amazon gift card. If you're anything like me, you loooove Amazon! I just wish there wasn't restrictions on posting some of the educational stuff over here - most stuff is sooo much cheaper in the U.S.! Head on over by clicking on the button.
Saturday, September 24, 2011
Wind Socks
Last week (before our Spring holidays - yeah!), my class finished up their study of weather/the seasons and the water cycle. This is one little craft project we did when we were learning about wind.
We just halved a piece of white card and used a range of materials to decorate it with anything to do with wind. Some used pastels, markers, cotton balls, scrunched up pieces of crepe paper to add to their trees etc. We then just folded the two ends together and stapled. Each student then cut out some streamers from a crepe paper roll (from the party section) and stapled them on. It was so easy to make!
We took them outside at various times and days so that we could determine which way the wind was blowing and estimated the strength of the wind. We hung them up across a clothes lines in our room and when the windows were open they would move in the breeze - it looked great!
Here's the only picture of have of them :(
One more week of Spring Holidays left! I can't believe we only have 10 weeks of the school year to go!
What have you done in your class in the past week? Link up at the Clutter-Free Classroom
Giveaway!
Terri at Terri's Teaching Treasures (love the alliteration!) is having a giveaway for her 34th birthday!
She is giving away a birthday present to 5 lucky people!
Head on over to her blog to enter now!
Thursday, September 22, 2011
Literacy Activites, Games and Freebies
If you have been reading my posts for a little while now, you'll know that I've been having heaps of trouble uploading new things to TpT and Teachers Notebook. I'm glad to say that half of that problem has now gone! I can upload to TpT with no problems from my school laptop! I WAS able to upload to Teachers Notebook earlier today, but now it isn't - something about undefined error/syntax error - using the same files that loaded successfully to TpT!
So here are some things I made about a month ago, but couldn't upload. The first three people to comment will get one each! These are all Autumn/Fall themed.
Studets make their way through the pumpkin patch to scare away the crows. Good news - they are also practising reading (Level 1 and 2 cards) and spelling (Level 3) words with or, ore, oar, air, are and ar spellings/sounds. Easy differentiation!
Students sort nouns, verbs and adjectives onto the correct mat and record on the recording sheet.
Students choose a card and read it. They then spin the spinner below and add the suffix onto the end of the word. If it makes a real word, they record it on the recording sheet. The first to fill their sheet (4 for each suffix) first wins. An answer card for each suffix (included) can be cut and lamiated and put together with a metal ring for students to check their answers.
Freebies available at TpT (I've only bought a TpT licence for freebies by Scrappin Doodles so far). Are we still allowed to post a link to TpT here? If we're not, can you please let me know so I can take this down! Click the picture to grab them!
I had this Making Words activity posted earlier but have just put it up again now that I've bought a Scrappin Doodles licence for TpT.
Original artwork from Scrappin' Doodles.
Sorry for another long absence. I've literally had no time to be blogging at all between school, my besties wedding, the clean up (the reception was at her place) and then a couple of days of professional development (what a great way to start our Spring holidays!). So, we are now 6 days into our 14 day break, and I finally get a chance to sit down and catch up on reading all of the great blogs I follow.
This is going to be a quickie post to let you know that I've updated some of my products. If you bought them through TpT you should receive an email - I don't know about Teachers Notebook, hence this post. So if you have bought them, you might want to download the newest version. Here is what has been updated -
Reflection Card Prompts for class discussions or written reflections:
• A follow teacher from here in Australia emailed me a large amount of reflection prompts that she uses and has graciously allowed me to add them to my pack. This increases the number of reflection prompts from 36 to 75! Some of the new prompts are particularly useful for upper grade teachers as some ask students to reflect on research they have done.
Daily 5 Read to Someone pack:
• an easier back of the Check for Understanding checkmark for beginning readers with picture cues.
Pirates Place Value HTO Dominoes:
• I didn't realise that over in America you don't say AND after the hundreds (e.g. four hundred AND twenty-six). After a comment on the product was made recently, I hopped straight onto it and have made two versions of the dominoes that have number words. So there is now a version with and without the AND.
• I've also added card labels to put on the front of each deck or stick to a baggie to hold them in.
Pirate Ordinal Number Bingo:
• I received feedback that the writing for the instructions went into the border. My copy was ok so I'm guessing it happened when I made it a PDF (and there were some mistakes - oops!). I think they are all fixed up now so you can download it again.
I am so happy that I can now add things to TpT and Teachers Notebook from my home computer (which is only about 6 months old). I haven't been able to add or update anything for so long. But I went into school yesterday and tried from my school laptop and NO problems! So, the laptop is at home with me now, and I'm trying to work from the two of them! So I should have some new freebies for you soon!
Monday, September 12, 2011
I'm back!
Wow, what a hectic fortnight! I'm glad to say that the Bridal Shower I organised for Saturday went really well! I've never done anything like that before so I was really nervous, but it was great. Phew! Now I'm ready for the actual wedding this weekend!
Some bad news - I missed out on grabbing Shelley's (Teaching in the Early Years) Doubles math strategy unit! I was really looking forward to checking it out to see whether I should purchase the other units. Was so disappointed when I got home from school today and it was already closed. Should have checked when I woke up this morning!
Some good news - Check out these Character ABC charts from Runde's Room. How AMAZING are these! This is definitely a print job for tonight!
We are in our last week of the term - four days to go before I am on two weeks holidays! There's something about third term that is just so exhausting. So needless to say I am busy assessing, marking, assessing, moderating, assessing, marking... you get the point.
I hope to be back with some new freebies soon.
Wednesday, September 07, 2011
Freebie: Start of School and Behaviour Diamond
Sorry I have been MIA recently. I have been SUPER busy with school (counting down to the end of term - YAY! I need a break and I am so over assessment and marking!) and organising a bridal shower for my friend for this weekend. WOW, I never thought it would be such a HUGE job! There's also been wedding rehearsals, shoe shopping and my son came home so sick after a week at school camp so I've been nursing him back to health!
So, I'll still be not posting very often until at least this weekend. Once the bridal shower is over I'll have a little more free time to devote to blogging.
I was ever so shocked to see that I had been listed as a Top 10 blog by Mechele over Barrow's Hodgepodge - heart - you should definately check out her blog if you haven't already. Thank you so much!
And I've been meaning to post this for those of you starting new year. I got my kiddos to do it earlier this year. In the plain box, they glue in a photo I take on the first day. It gives me an indication of how well they can read, spell, listen to and follow instructions etc. They color in the whole page and they end up looking very cute. I put them up on the inside of or door so they can be seen when the door is open. The font I used is our handwriting font, so it probably won't come up for your properly - just change it to whatever font you want (it's a word document). Get it by clicking on the photo.
Another idea I used when I taught Grade 6 is for students to create a wordle about themselves - words to describe themselves and their interests. I took a screen print of the wordle, pasted it in Word and then inserted a text box. I added a photo of each student in the text book before printing. I added a sign (Meet 6D) and pegged them to a clothes line I had running across the room. They looked much better than what it does here in the pic which was the best one I could find.
We were learning about sustainable house design that term, so you can see heaps of materials hanging on the wall on one side and our Behaviour Diamond on the other. Students moved up or down the diamond and collected or lost class cash along the way (e.g. move up once, get \$5, up twice \$10 + the \$5 already earned etc. Same on the way down the chart). We kept a spreadsheet to keep track of money earned, lost and spent in the class store. I also had it so that once they went down the diamond, they couldn't go back up that day, only control their behaviour and not go further down. It taught them to monitor their behaviour and that not all misbehaviours can be made up by deciding to be 'good' later - there were consequences for their choices. The next day, they started in the middle again with a fresh start. It worked really well for this bunch of kids. You can see that about 5 kids went down this day! That was probably the most that ever went down. Most days I didn't have any but if they were late to come back from lunch they went straight down! If you want to know more let me know!
Enjoy the rest of your week!
Friday, September 02, 2011
And the winner is..
Thank you to everyone who entered my giveaway. An extra BIG thank you to those who took to their blog/FB or Twitter to promote it.
And the winner is
Kelly Karger! I've now emailed you to see which gift card you'd prefer. Please return email me within the next 3 days to claim your prize.
Thanks again to everyone who entered. Never mind that you missed out - head on over to The Resource(ful) Room - Amy is giving away a \$25 Amazon gift card! | 3,679 | 15,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2013-20 | longest | en | 0.941959 |
http://forums.wolfram.com/mathgroup/archive/2005/Jan/msg00373.html | 1,611,091,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00455.warc.gz | 38,098,104 | 7,629 | Re: Matrix Problem
• To: mathgroup at smc.vnet.net
• Subject: [mg53594] Re: Matrix Problem
• From: "Ray Koopman" <koopman at sfu.ca>
• Date: Wed, 19 Jan 2005 02:00:13 -0500 (EST)
• References: <csino0\$nj8\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```nilaakash wrote:
> Hi,
> I face a problem to get (3xn) matrix data. Here I am giving a
> simple example.
>
> data = Table[{Sin[x], Cos[y], Sin[x] Cos[y]}, {x, 0, 3, 1},
> {y, 0, 2, 1}];
> MatrixForm[data]
>
> It prints a (4x3) block matrix.
>
> Could anybody tell me how shall I get only 3 column matrix like
> following matrix ?
>
> Sin[0] Cos[0] Sin[0] Cos[0]
> Sin[1] Cos[0] Sin[1] Cos[0]
> Sin[2] Cos[0] Sin[2] Cos[0]
> Sin[3] Cos[0] Sin[3] Cos[0]
> Sin[0] Cos[1] Sin[0] Cos[1]
> Sin[1] Cos[1] Sin[1] Cos[1]
> Sin[2] Cos[1] Sin[2] Cos[1]
> Sin[3] Cos[1] Sin[3] Cos[1]
> Sin[0] Cos[2] Sin[0] Cos[2]
> Sin[1] Cos[2] Sin[1] Cos[2]
> Sin[2] Cos[2] Sin[2] Cos[2]
> Sin[3] Cos[2] Sin[3] Cos[2]
>
> Thanks.
>
> Nilaakash
To get the terms in the order you listed them,
swap the x and y loops in the Table commmand.
To convert the 3 x 4 x 3 table to a 12 x 3 matrix, Flatten it once.
data = Table[{Sin[x],Cos[y],Sin[x]*Cos[y]},{y,0,2},{x,0,3}];
TableForm@Flatten[data,1]
0 1 0
Sin[1] 1 Sin[1]
Sin[2] 1 Sin[2]
Sin[3] 1 Sin[3]
0 Cos[1] 0
Sin[1] Cos[1] Cos[1] Sin[1]
Sin[2] Cos[1] Cos[1] Sin[2]
Sin[3] Cos[1] Cos[1] Sin[3]
0 Cos[2] 0
Sin[1] Cos[2] Cos[2] Sin[1]
Sin[2] Cos[2] Cos[2] Sin[2]
Sin[3] Cos[2] Cos[2] Sin[3]
```
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http://avaxbooks.me/john-davidson-adrian-sanqui-adrian-sanqui-learn-how-to-draw-land-animals-for-the-absolute-beginner-english-mobi.html | 1,480,992,806,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541876.62/warc/CC-MAIN-20161202170901-00378-ip-10-31-129-80.ec2.internal.warc.gz | 21,183,238 | 9,787 | Learn How to Draw Land Animals – For the Absolute Beginner
ISBN/ASIN: N/A | 2013 | English | mobi | 55/0 pages | 1.36 Mb
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Learn How to Draw Land Animals For the Absolute Beginner
Adrian Sanqui and John Davidson
Introduction: Drawing tools
Sketching Animals
The contour shape
The planes
Details
Furs
Raccoon
Lion
Bison
Texture
Elephant
Prints
Tiger
Imaginary light source
Tapir
Hippopotamus
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Gaur
Sketching Animals – The contour shape
To properly construct the figure of any subject you want to illustrate, observe the structure of its form and find what kind of shape that can resemble its body structure.
The idea is to have a basis for portraying the mass of the figure. The basic shapes can be easily positioned upon your desired perspective or point of view.
• Obtain the most basic form of the animal you are going to draw.
• Use the basic form as a reference for constructing the main outline.
The main outline is the most important element for the subject to be distinguishable. Remember that the basic figure (basic shapes) is mainly a basis for its mass in the simplest form, this is just a way for you to see the subject as a multidimensional figure, and any change in position will gradually change the manner of how the figure should be outlined.
The primary outline may overlap, replace, or replicate the basic figure.
If the portions of your animal are smoothly spherical (circles used to simplify the form), then the primary outline would simply overlap or replicate the simplified form, but if the part has ridges such as the face of the animal (nose, eyeholes, cheeks etc…) then the basic form should be modified to illustrate the exact shape of the head.
• The basic forms will be your guide to properly lay out each linear shade.
Observe the first sphere. The linear shades do not bend properly with the contour shape but it can show the dimensions by leaving the center of the sphere unmarked. These are hatches that use the point of light as a reference.
The linear shades on the second sphere bend with the contour shape of the figure. The lines show the proportions by interpreting the form by flowing with it. These hatches use the point of light and the manner of its bend to show dimensions.
If you are going to use a single set of hatches to shade the subject, use the method of the second sphere, and then use the method of the first sphere to darken and enforce the visual depth.
Thin and light linings are used to portray the dimensional shape. Follow the contour figure of the basic form to properly portray the shade values according to its perspective
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http://www.chegg.com/homework-help/questions-and-answers/two-20mm-steel-plates-clamped-together-bolt-nut-specify-metric-bolt-provide-joint-constant-q1073766 | 1,469,899,311,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258936356.77/warc/CC-MAIN-20160723072856-00096-ip-10-185-27-174.ec2.internal.warc.gz | 354,602,052 | 13,768 | Two 20mm steel plates are to be clamped together with a bolt and nut. Specify a
metric bolt to provide a joint constant C between 0.2 and 0.3. This problem involves a
trial-error process. Refer to These Tables 8-1(pp. 398) and Appendix A-31 (pp. 1035) for standard dimensions. For each trial, find the bolt length first (round to a preferred value), and then calculate the bolt and member stiffnesses using Eqs. kb=AdAtE/Adlt+Atld and km=(0.5774piEd)/[2ln(5*(0.5774l+0.5d.0.5774l+2.5d)]
At=tensile stress are
lt=length of threaded portion of grip | 171 | 547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | latest | en | 0.849348 |
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# Time dilation calculation
1. Q) the lifetime of a star is 10billion years as measured by an observer at rest with respect to the star. The star is moving away from the Earth at a speed of 0.81c.
Calculate the lifetime of the star according to an observer.
A) 17.1 billion
But plugging the numbers in to find t' I get 16.6billion. What's correct?
2. (Original post by Feynboy)
Q) the lifetime of a star is 10billion years as measured by an observer at rest with respect to the star. The star is moving away from the Earth at a speed of 0.81c.
Calculate the lifetime of the star according to an observer.
A) 17.1 billion
But plugging the numbers in to find t' I get 16.6billion. What's correct?
billion years
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P R E C A L C U L U S
27
# MATHEMATICAL INDUCTION
The principle of mathematical induction
THE NATURAL NUMBERS are the counting numbers: 1, 2, 3, 4, etc. Mathematical induction is a technique for proving a statement -- a theorem, or a formula -- that is asserted about every natural number.
By "every", or "all," natural numbers, we mean any one that we might possibly name.
For example,
1 + 2 + 3 + . . . + n = ½n(n + 1).
This asserts that the sum of consecutive numbers from 1 to n is given by the formula on the right. We want to prove that this will be true for n = 1, n = 2, n = 3, and so on. Now we can test the formula for any given number, say n = 3:
1 + 2 + 3 = ½· 3· 4 = 6
-- which is true. It is also true for n = 4:
1 + 2 + 3 + 4 = ½· 4· 5 = 10.
But how are we to prove this rule for every value of n?
The method of proof is the following. It is called the principle of mathematical induction.
If 1) when a statement is true for a natural number n = k, then it will also be true for its successor, n = k + 1; and 2) the statement is true for n = 1; then the statement will be true for every natural number n.
To prove a statement by induction, we must prove parts 1) and 2) above. For, when the statement is true for n = 1, then according to 1), it will also be true for 2. But that implies it will be true for 3; which implies it will be true for 4. And so on. It will be true for any natural number that we might name.
The hypothesis of Step 1) -- "The statement is true for n = k" -- is called the induction assumption, or the induction hypothesis. It is what we assume when we prove a theorem by induction.
Example 1. Prove that the sum of the first n natural numbers is given by this formula:
1 + 2 + 3 + . . . + n = n(n + 1) 2 .
Proof. We will do Steps 1) and 2) above. First, we will assume that the formula is true for n = k; that is, we will assume:
1 + 2 + 3 + . . . + k = k(k + 1) 2 . (1)
This is the induction assumption. Assuming this, we must prove that the formula is true for its successor, n = k + 1. That is, we must show:
1 + 2 + 3 + . . . + (k + 1) = (k + 1)(k + 2) 2 . (2)
To do that, we will simply add the next term (k + 1) to both sides of the induction assumption, line (1):
This is line (2), which is the first thing we wanted to show.
Next, we must show that the formula is true for n = 1. We have:
1 = ½· 1· 2
-- which is true. We have now fulfilled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.
(In the Appendix to Arithmetic, we establish that formula directly.)
Example 2. Prove that this rule of exponents is true for every natural number n:
(ab)n = anbn.
Proof. Again, we begin by assuming that it is true for n = k; that is, we assume:
(ab)k = akbk . . . . . . . . (3)
With this assumption, we must show that the rule is true for its successor, n = (k + 1). We must show:
(ab)k + 1 = ak + 1bk + 1. . . . . . . (4)
(When using mathematical induction, the student should always write exactly what is to be shown.)
Now, given the assumption, line (3), how can we produce line (4) from it ?
Simply by multiplying both sides of line (3) by ab:
(ab)kab = akbkab = akabkb since the order of factors does not matter, = ak + 1bk + 1.
This is line (4), which is what we wanted to show.
So, we have shown that if the theorem is true for any specific natural number k, then it is also true for its successor, k + 1.
Next, we must show that the rule is true for n = 1; that is, that
(ab)1 = a1b1.
But (ab)1 = ab; and a1b1 = ab.
This rule is therefore true for every natural number n.
Example 3. The sum of consecutive cubes. Prove this remarkable fact of arithmetic:
13 + 23 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2.
"The sum of n consecutive cubes is equal to the square
of the sum of the first n numbers."
In other words, according to Example 1:
13 + 23 + 33 + . . . + n3 = n²(n + 1)² 4
Proof. For convenience, we will denote the sum up to n by S(n). We assume, then, that the formula is true for n = k; that is, that
S(k) = k²(k + 1)² 4 (1) We must now show that the formula is also true for n = k + 1; that S(k + 1) = (k + 1)²(k + 2)² 4 (2) To do that, add the next cube to S(k), line (1): S(k + 1) = S(k) + (k + 1)3
= k²(k + 1)² 4 + (k + 1)3 = k²(k + 1)² + 4(k + 1)³ 4 = (k + 1)²[k² + 4(k + 1)] 4 -- on taking (k + 1)2 as a common factor, = (k + 1)²(k² + 4k + 4) 4 = (k + 1)²(k + 2)² 4
This is line (2), which is what we wanted to show.
Finally, we must show that the formula is true for n = 1.
13 = 1²· 2² 4 1 = 1· 4 4
-- which is true. The formula therefore is true for every natural number.
In the Appendix to Arithmetic, we show directly that that is true.
Problem 1. According to the principle of mathematical induction, to prove a statement that is asserted about every natural number n, there are two things to prove.
a) What is the first?
If the statement is true for n = k, then it will be true for its successor, k + 1.
b) What is the second?
The statement is true for n = 1.
c) Part a) contains the induction assumption. What is it?
The statement is true for n = k.
Problem 2. Let S(n) = 2n − 1. Evaluate
a) S(k = 2k − 1
b) S(k + 1) = 2(k + 1) − 1 = 2k + 2 − 1 = 2k + 1
Problem 3. The sum of the first n odd numbers is equal to the nth square.
1 + 3 + 5 + 7 + . . . + (2n − 1) = n2.
a) To prove that by mathematical induction, what will be the induction
a) assumption?
The statement is true for n = k:
1 + 3 + 5 + 7 + . . . + (2k − 1) = k2.
b) On the basis of this assumption, what must we show?
The statement is true for its successor, k + 1:
1 + 3 + 5 + 7 + . . . + (2k − 1) + 2k + 1 = (k + 1)².
c) Show that.
On adding 2k + 1 to both sides of the induction assumption:
1 + 3 + 5 + 7 + . . . + (2k − 1) + 2k + 1 = k² + 2k + 1 = (k + 1)2
d) To complete the proof by mathematical induction, what must we
a) show?
The statement is true for n = 1.
e) Show that.
1 = 12
Problem 4. Prove by mathematical induction:
If we denote that sum by S(n), then assume that the formula is true for n = k; that is, assume
S(k) = k 2k + 1 .
Now show that the formula is true for n = k + 1; that is, show:
S(k + 1) = k + 1 2k + 3 .
Begin:
S(k + 1) = S(k) + The next term, whose denominatoris the product of the next odd numbers. = = = = =
Next,
The formula is true for n = 1:
Therefore it is true for all natural numbers.
E-mail: themathpage@yandex.com
Private tutoring available. | 2,286 | 6,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2017-51 | latest | en | 0.857108 |
https://ccssanswers.com/eureka-math-grade-2-module-7-end-of-module-assessment/ | 1,708,813,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00898.warc.gz | 163,784,126 | 46,679 | # Eureka Math Grade 2 Module 7 End of Module Assessment Answer Key
## Engage NY Eureka Math 2nd Grade Module 7 End of Module Assessment Answer Key
### Eureka Math Grade 2 Module 7 End of Module Assessment Answer Key
Note: Do not pass out rulers until after students complete Problem 1(a).
Question 1.
a.
Estimate the length of each item in inches.
The envelope is about ____ inches.
Answer:
The envelope is about 3.82 inches.
The pencil is about ____ inches
Answer:
The pencil is about 5.7 inches.
The crayon is about ____ inches.
Answer:
The crayon is about 3.07 inches.
The scissors are about ____ inches.
Answer:
The scissors is about 4.9 inches.
b. Use a ruler to measure the length of the items above using inches and then centimeters. Round to the nearest unit, and then record the measurements in the table.
Envelope Pencil Crayon Scissors ______ inches ______ centimeters ______ inches _____ centimeters ______ inches ______ centimeters ______ inches ______ centimeters
Answer:
Envelope Pencil Crayon Scissors 3.8 inches 10 centimeters 5.7 inches 15 centimeters 3.07 inches 8 centimeters 4.88 inches 13 centimeters
Envelope
The envelope is about 3.81 inches.
Converting 3.81 inches into cm
3.81 in ÷ 0.3937 = 9.7 cm
Rounding to the nearest cm = 10 cm
The envelope is about 10 cm.
Pencil
The pencil is about 5.70 inches.
Converting 5.70 inches into cm
5.70 in ÷ 0.3937 = 14.5 cm
Rounding to the nearest cm = 15 cm
The envelope is about 15 cm.
Crayon
The Crayon is about 3.07 inches.
Converting 3.07 inches into cm
3.07 in ÷ 0.3937 = 7.8 cm
Rounding to the nearest cm = 8 cm
The envelope is about 8 cm.
Scissor
The envelope is about 4.88 inches.
Converting 4.88 inches into cm
4.88 in ÷ 0.3937 = 12.4 cm
Rounding to the nearest cm = 12.4 cm
The envelope is about 13 cm.
c. The envelope is ______ centimeters longer than the crayon.
Answer:
Length of envelope = 10 cm
Length of Crayon = 8 cm
Difference in length = 10 cm – 8 cm = 2 cm
The envelope is 2 centimeters longer than the crayon.
d. For each measurement, which is greater, the number of inches or the number of centimeters?
Answer:
For the given measurement number of centimeters are greater than inches.
e. Explain why.
Answer:
When converting from a larger to a smaller unit, you will always have more of the smaller units. When converting from a smaller to a larger unit, you will always have fewer of the larger units. A centimeter is smaller than an inch, so a given length will have more centimeters than inches.
Question 2.
Circle the appropriate tool for measuring each object.
a. The length of a book: 12-inch ruler yardstick
Answer:
b. The height of a flagpole: 12-inch ruler yardstick
Answer:
c. The length of a paper clip: 12-inch ruler yardstick
Answer:
d. The height of a doorway: 12-inch ruler yardstick
Answer:
Question 3.
a. What number is represented as Point A on the number line?
Answer:
b. What is the distance between A and B?
Answer:
c. What is 40 less than the number marked by Point C? Mark it as Point D on the number line.
Answer:
Question 4.
Use the tables below to graph the data.
a. Draw and label a line plot to show the length of the pencils in the table.
Length in Inches Number of Pencils 1 inch 0 2 inches 2 3 inches 4 4 inches 4 5 inches 3 6 inches 2 7 inches 5
Answer:
b. Find the total number of pencils measured.
Answer:
c. Draw and label a bar graph to show the number of pencils in each student’s desk.
Student Name Jill Sven Rocco Lyla Number of Pencils 4 2 5 1
Title: __________
Answer:
Question 5.
Draw a picture, and write a number sentence to solve.
a. The height of the dog’s doorway is 19 inches. The height of the family’s doorway is 78 inches. How much taller is the family’s doorway than the dog’s doorway?
Answer:
b. Albert saved 42 cents last week. This week, he added a quarter, 2 dimes, and 13 pennies to his savings. How much money has Albert saved from the last two weeks? Write the answer using the \$ or ¢ symbol.
Answer: | 1,116 | 4,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-10 | latest | en | 0.787364 |
http://atozmath.com/Default.aspx?q1=Trapezoidal%20rule%2C%20%7B%7B0.0%2C1.0000%7D%2C%7B0.1%2C0.9975%7D%2C%7B0.2%2C0.9900%7D%2C%7B0.3%2C0.9776%7D%2C%7B0.4%2C0.8604%7D%7D%60587&do=1 | 1,566,076,249,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312025.20/warc/CC-MAIN-20190817203056-20190817225056-00295.warc.gz | 19,833,514 | 141,034 | Home
Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution will be displayed step by step (In 2 parts)
Solution
Problem: Trapezoidal rule, {{0.0,1.0000},{0.1,0.9975},{0.2,0.9900},{0.3,0.9776},{0.4,0.8604}} [ Calculator, Method and examples ]
Solution:
Your problem -> Trapezoidal rule, {{0.0,1.0000},{0.1,0.9975},{0.2,0.9900},{0.3,0.9776},{0.4,0.8604}}
The value of table for X and Y
X Y 0 0.1 0.2 0.3 0.4 1 0.9975 0.99 0.9776 0.8604
Using Trapezoidal Rule
int Y dx = h/2 (y_0 + 2 (y_1 + y_2 + y_3) + y_4 )
Solution provided by AtoZmath.com
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We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more | 323 | 924 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-35 | longest | en | 0.808808 |
https://math.stackexchange.com/questions/1113286/dedekind-cuts-in-construction-of-the-real-line | 1,566,489,460,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00292.warc.gz | 538,598,628 | 27,955 | Dedekind Cuts in Construction of the real line [closed]
Is each Dedekind cut a unique real number? or when we apply the process(Dedekind cut), do we get a bunch of real numbers instead of a unique one.
If we get a unique real number, is the unique real number then plotted as a line segment between rationals on the number line? Or is it plotted as a single point (just like 0 and 1)?
If not so, Is the bunch of numbers that we get infinite?
closed as unclear what you're asking by Andrés E. Caicedo, hardmath, dustin, user147263, TravisJan 22 '15 at 3:12
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Depending on your definition of real number, the answer to the first question may be "Yes, by definition" – Hagen von Eitzen Jan 21 '15 at 10:20
• Lets say it is defined as - a number on the real number line – novice Jan 21 '15 at 10:21
• What is the standard definition btw? – novice Jan 21 '15 at 10:34
• Some authors use Dedekind cuts to represent only non-negative real numbers, so that these sets are known to be bounded below. The negative real numbers can then be represented algebraically. – hardmath Jan 22 '15 at 2:14
• Could you please indicate what is unclear so I may edit the question properly – novice Jan 22 '15 at 8:13
There are several ways to construct (or if you prefer, define) the real numbers. The two most familiar are as Dedekind cuts in $\Bbb Q$, the ordered set of rational numbers, and as equivalence classes of Cauchy sequences of rational numbers with respect to a certain equivalence relation. If one constructs them using Dedekind cuts, then each Dedekind cut is by definition a unique real number. If one constructs them in some other way, it’s no longer the case that each real number is a Dedekind cut in $\Bbb Q$, but it is still true that there is a nice bijection between the set of Dedekind cuts in $\Bbb Q$ and the reals as constructed.
• @novice: No, it's a single point, not a segment. No two rational numbers are adjacent: the average of any two rational numbers is a rational number between them. The gaps in $\Bbb Q$ are a bit more subtle than that. For example, if $R$ is the set of positive rational numbers $x$ such that $x^2>2$, and $L$ is the rest of the rationals, then every member of $L$ is less than each member of $R$, but there is no rational number between $L$ and $R$. In $\Bbb R$ the number $\sqrt 2$ fills that gap. – Brian M. Scott Jan 22 '15 at 2:25 | 698 | 2,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-35 | latest | en | 0.90141 |
http://mathhelpforum.com/algebra/81157-need-help-math-problems-tommorow.html | 1,477,167,034,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719041.14/warc/CC-MAIN-20161020183839-00201-ip-10-171-6-4.ec2.internal.warc.gz | 158,565,928 | 13,572 | # Thread: Need help on math problems by tommorow
1. ## Need help on math problems by tommorow
1. got on my own
2. (root of x) - 2 = 4th root of x
3. (root of 3-3x) - (root of 3x + 2) = 3
4. write a quad equation with integer coef who roots are
(-1 +- root of -2) / 5
5. got on my own
6. got on my own
7. find all p in f(x) = (p + 1) x^2 + px - 1 so that the quadratic equation's roots are real and unequal
8. find the range of the following quadratic function
f(x) = roof of x(x^2) + roof of 8 (x) -2
almost got but cant get vertex here
9. A park in the shape of a rectangle has dimensions 60m by 100m. If the park contains a rectangular garden enclosed by a concrete walkway, how wide is the walkway if the area of the garden is half the area of the park?
Note : the concrete walkway width is uniform
i tried these and i keep getting the wrong answers
i spent 3 hours trying to figure them out but out of 20 problems, i didnt get these nine
Can you guys give me some hints on how to do these problems?
Thanks
2. ## Problems
Hello mathprob
Welcome to Math Help Forum!
2. (root of x) - 2 = 4th root of x
Let $x = u^4$.Then
$\sqrt{u^4}-2 = \sqrt[4]{u^4}$
$\Rightarrow u^2 - 2 = u$
Solve this quadratic for $u$. Then put $x=u^4$ again at the end.
3. (root of 3-3x) - (root of 3x + 2) = 3
You know that $(a-b)^2 = a^2-2ab+b^2$. So square both sides of
$\sqrt{3-3x} - \sqrt{3x+2} = 3$
to get:
$(3-3x) - 2\sqrt{(3-3x)(3x+2)} + (3x+2) = 3$
$\Rightarrow 2 = 2\sqrt{(3-3x)(3x+2)}$
$\Rightarrow (3-3x)(3x+2) = 1$
4. write a quad equation with integer coef who roots are
(-1 +- root of -2) / 2
Use the quadratic formula. You know that
$\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-1\pm\sqrt{-2}}{2}$
So start with $a = 1, b = 1$ and see what value this gives for $c$. If it's a fraction, multiply both sides of $ax^2+bx+c=0$ by something that will get rid of fractions.
7. find all p in f(x) = (p + 1) x^2 + px - 1 so that the quadratic equation's roots are real and unequal
For real unequal roots, $b^2-4ac>0$
So $p^2+4(p+1)>0$
$\Rightarrow p^2+4p+4>0$
$\Rightarrow (p+2)^2>0$
What is the only value of $p$ for which this is not true?
3. 2. i got 16
3. null set
4. i still cant get, btw i typed in the wrong roots, its divided by 5, not 2
7. i got all reals except p cannot equal -2
8. still cant get
9.cant get
4. Originally Posted by mathprob
i tried these and i keep getting the wrong answers
In future, please show your work, so we can "see" where you're needing help. Thank you!
Originally Posted by mathprob
2. (root of x) - 2 = 4th root of x
2. i got 16
You can check the solution to any "solving" problem by plugging it back into the original exercise. In this case:
. . . . . $\sqrt{16}\, -\, 2\, =\, 4\, -\, 2\, =\, 2$
. . . . . $\sqrt[4]{16}\, =\, \sqrt[4]{2^4}\, =\, 2$
Originally Posted by mathprob
3. (root of 3-3x) - (root of 3x + 2) = 3
3. null set
How did you arrive at your answer? Was the original equation as follows?
. . . . . $\sqrt{3\, -\, 3x}\, -\, \sqrt{3x\, +\, 2}\, =\, 3$
You started the solution by squaring both sides:
. . . . . $(3\, -\, 3x)\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, +\, (3x\, +\, 2)\, =\, 9$
. . . . . $5\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 9$
. . . . . $-2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 4$
Then you squared again:
. . . . . $4(6\, +\, 3x\, -\, 9x^2)\, =\, 16$
. . . . . $6\, +\, 3x\, -\, 9x^2\, =\, 4$
. . . . . $0\, =\, 9x^2\, -\, 3x\, -\, 2$
. . . . . $0\, =\, (3x\, -\, 2)(3x\, +\, 1)$
So x = 2/3 or x = -1/3.
Plugging these back into the original equation gives invalid results: neither 1 - 2 nor 2 - 1 equals 3. So there is "no solution".
Originally Posted by mathprob
4. write a quad equation with integer coef who roots are (-1 +- root of -2) / 5
4. i still cant get, btw i typed in the wrong roots, its divided by 5, not 2
To learn how to find a quadratic by working backwards from its zeroes, try here.
Once you have learned the basic technique, set up the exercise as:
. . . . . $\left(x\, -\, \left(\frac{-1\, -\, \sqrt{-2}}{5}\right)\right)\left(x\, -\, \left(\frac{-1\, +\, \sqrt{-2}}{5}\right)\right)$
Note that, since you must have integer coefficients, you'll need to multiply the product above by "25" to clear the denominator.
Originally Posted by mathprob
7. find all p in f(x) = (p + 1) x^2 + px - 1 so that the quadratic equation's roots are real and unequal
7. i got all reals except p cannot equal -2
From what you've learned about the Quadratic Formula, you know that the only way to get two different real roots is for the value inside the square root to be positive. In other words:
. . . . . $b^2\, -\, 4ac\, =\, p^2\, -\, 4(p\, +\, 1)(-1)\, >\, 0$
So you need to solve the quadratic inequality:
. . . . . $(p\, +\, 2)^2\, >\, 0$
By nature of squaring, (p + 2)^2 will be positive for every non-zero value of p + 2. So the only place the inequality will not hold will be for p + 2 = 0, or p = -2. So your solution is correct.
Originally Posted by mathprob
8. find the range of the following quadratic function
f(x) = roof of x(x^2) + roof of 8 (x) -2
almost got but cant get vertex here
8. still cant get
To learn how to find the vertex of a quadratic, try here. If you're stuck on some other aspect of the exercise, please reply showing your work and reasoning so far.
Originally Posted by mathprob
9. A park in the shape of a rectangle has dimensions 60m by 100m. If the park contains a rectangular garden enclosed by a concrete walkway, how wide is the walkway if the area of the garden is half the area of the park?
Note : the concrete walkway width is uniform
9.cant get
What is the total area? (Hint: Multiply the given length by the given width.)
Draw a rectangle, and label its sides with the given dimensions.
Draw a smaller rectangle inside, and label the gap between the rectangles with "x", being the (unknown) width of the walkway.
What expression would stand for the width of the inner rectangle? What expression would stand for the length of the inner rectangle?
What expression would then stand for the area of the inner rectangle?
If the area of the garden is half of the total park area, what is the garden's area? (Hint: Divide the total area by 2.)
You now have an expression, in terms of "x", for the area of the garden, and a numerical value for the area of the garden. Set these equal, and solve.
If you get stuck, please reply showing how far you have gotten in following the provided steps. Thank you!
5. Originally Posted by stapel
In future, please show your work, so we can "see" where you're needing help. Thank you!
You can check the solution to any "solving" problem by plugging it back into the original exercise. In this case:
. . . . . $\sqrt{16}\, -\, 2\, =\, 4\, -\, 2\, =\, 2$
. . . . . $\sqrt[4]{16}\, =\, \sqrt[4]{2^4}\, =\, 2$
How did you arrive at your answer? Was the original equation as follows?
. . . . . $\sqrt{3\, -\, 3x}\, -\, \sqrt{3x\, +\, 2}\, =\, 3$
You started the solution by squaring both sides:
. . . . . $(3\, -\, 3x)\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, +\, (3x\, +\, 2)\, =\, 9$
. . . . . $5\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 9$
. . . . . $-2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 4$
Then you squared again:
. . . . . $4(6\, +\, 3x\, -\, 9x^2)\, =\, 16$
. . . . . $6\, +\, 3x\, -\, 9x^2\, =\, 4$
. . . . . $0\, =\, 9x^2\, -\, 3x\, -\, 2$
. . . . . $0\, =\, (3x\, -\, 2)(3x\, +\, 1)$
So x = 2/3 or x = -1/3.
Plugging these back into the original equation gives invalid results: neither 1 - 2 nor 2 - 1 equals 3. So there is "no solution".
To learn how to find a quadratic by working backwards from its zeroes, try here.
Once you have learned the basic technique, set up the exercise as:
. . . . . $\left(x\, -\, \left(\frac{-1\, -\, \sqrt{-2}}{5}\right)\right)\left(x\, -\, \left(\frac{-1\, +\, \sqrt{-2}}{5}\right)\right)$
Note that, since you must have integer coefficients, you'll need to multiply the product above by "25" to clear the denominator.
From what you've learned about the Quadratic Formula, you know that the only way to get two different real roots is for the value inside the square root to be positive. In other words:
. . . . . $b^2\, -\, 4ac\, =\, p^2\, -\, 4(p\, +\, 1)(-1)\, >\, 0$
So you need to solve the quadratic inequality:
. . . . . $(p\, +\, 2)^2\, >\, 0$
By nature of squaring, (p + 2)^2 will be positive for every non-zero value of p + 2. So the only place the inequality will not hold will be for p + 2 = 0, or p = -2. So your solution is correct.
To learn how to find the vertex of a quadratic, try here. If you're stuck on some other aspect of the exercise, please reply showing your work and reasoning so far.
What is the total area? (Hint: Multiply the given length by the given width.)
Draw a rectangle, and label its sides with the given dimensions.
Draw a smaller rectangle inside, and label the gap between the rectangles with "x", being the (unknown) width of the walkway.
What expression would stand for the width of the inner rectangle? What expression would stand for the length of the inner rectangle?
What expression would then stand for the area of the inner rectangle?
If the area of the garden is half of the total park area, what is the garden's area? (Hint: Divide the total area by 2.)
You now have an expression, in terms of "x", for the area of the garden, and a numerical value for the area of the garden. Set these equal, and solve.
If you get stuck, please reply showing how far you have gotten in following the provided steps. Thank you!
thanks a lot
Here is the work i have
2. works
3. no solutions, which got to
4. 25x^2 + 10x + 3 = y.
7. P > -2
8. Y >= -2 - root of 2
9. since the walkway is uniform i for 100 - 2x, 60 - 2x
so, 4x^2 - 320x + 3000
used quadratic formula and x has 2 solutions but only one works
i got x = 10.85meters
i need to confirm 4 and 9 | 3,420 | 9,846 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 45, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-44 | longest | en | 0.829605 |
https://studydaddy.com/question/listed-below-are-results-from-two-different-tests-designed-to-measure-productivi | 1,586,207,257,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00166.warc.gz | 722,252,688 | 7,889 | QUESTION
# Listed below are results from two different tests designed to measure productivity and dexterity for randomly selected employees.
7.Listed below are results from two different tests designed to measure productivity and dexterity for randomly selected employees.
Productivity(x) 23 25 28 21 21 25 26 30 34 36
Dexterity(y) 49 53 59 42 47 53 55 63 67 75
a. Plot the scatter diagram below. Label x and y axes.
b. Find the value of the linear correlation coefficient r by the TI83 shortcut- state calculator screen name
c) Test the claim of no linear relation by the TI83 p-value method. α = .05 claim .................................................... ________________________
null hypothesis________________________
alternative hypothesis ________________________
Calculator Screen Name ________________________
test statistic ________________________
pvalue/alpha comparison ________________________
decision ________________________
Conclusion ________________________
d) Find the estimated equation of the regression line by TI83 shortcut
e) Plot the regression line on the scatter diagram in part a).
f) Assuming a significant linear correlation, predict the score a student would get on dexterity, given he got a 40 on productivity.
g) What percentage of the total variation can be explained by the regression line? | 270 | 1,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | latest | en | 0.734371 |
https://www.jiskha.com/display.cgi?id=1347162087 | 1,503,323,272,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108709.89/warc/CC-MAIN-20170821133645-20170821153645-00414.warc.gz | 917,064,071 | 3,720 | # physics
posted by .
A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 9.3 m from the takeoff point. If the kangaroo leaves the ground at a 17.6 degree angle, what is its takeoff speed?
• physics -
The distance covered in each jump is
9.3 m = (Vo^2/g)*sin(2*17.6)
g = 9.8 m/s^2
9.3 = [Vo^2/9.8]*0.5764
Solve for Vo
• physics -
12.57
## Similar Questions
1. ### physics
An athlete executing a long jumper leaves the ground at a 27.0 degree angle and travels 7.89 m. What was the takeoff speed?
2. ### math/ physics
An athlete executing a long jumper leaves the ground at a 27.0 degree angle and travels 7.89 m. What was the takeoff speed?
3. ### physics
A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 9.3 m from the takeoff point. If the kangaroo leaves the ground at a 17.6 degree angle, what is its takeoff speed?
4. ### Physics
A gray kangaroo can bound across level ground with each jump carrying it 8.4 from the takeoff point. Typically the kangaroo leaves the ground at a 23 angle. What is its takeoff speed?
5. ### Physics
A gray kangaroo can bound across level ground with each jump carrying it 8.4m from the takeoff point. Typically the kangaroo leaves the ground at a 23 angle. 1. Provide knowns and unknown values. 2. What is the initial velocity or …
6. ### Physics
A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 from the takeoff point. If the kangaroo leaves the ground at a 18 angle, what is its takeoff speed ?
7. ### physics
An athlete executing a long jump, leaves the ground at a 30 degree angle with the ground and travels 8.90 meters horizontally in 1.1 seconds. What was the takeoff speed of the athlete along the diagonal?
8. ### physics. HELP
An athlete executing a long jump, leaves the ground at a 30 degree angle with the ground and travels 8.90 meters horizontally in 1.1 seconds. What was the takeoff speed of the athlete along the diagonal?
9. ### Physics
Yuvi is trying out for the track & field team and wants to give long jump a try. He currently can reach a distance of 5.90 m when leaving the ground at an angle of 30.0°. (a) What is his takeoff speed?
10. ### PHYSICS
Yuvi is trying out for the track & field team and wants to give long jump a try. He currently can reach a distance of 5.90 m when leaving the ground at an angle of 30.0°. (a) What is his takeoff speed?
More Similar Questions | 659 | 2,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-34 | longest | en | 0.943232 |
https://www.folkstalk.com/tech/algorithms-decision-with-code-examples/ | 1,725,874,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00440.warc.gz | 729,498,779 | 6,019 | # Algorithms - Decision With Code Examples
Algorithms - Decision With Code Examples
In this lesson, we'll use programming to try to solve the Algorithms - Decision puzzle. The code shown below demonstrates this.
```public boolean containsGreaterThanHundred(List<Integer> numbers) {
for (Integer i : numbers) {
if (i > 100) {
return true;
}
}
return false;
}
```
We were able to solve the Algorithms - Decision issue by looking at a number of other examples.
## What is a decision in an algorithm?
Algorithmic Decision-making We define “algorithmic decision-making”, or simply “algo- rithm”, as the processing of input data to produce a score or a choice that is used to support decisions such as prioritiza- tion, classification, association, and filtering [16].
## How are algorithms used to make decisions?
Algorithmic or automated decision systems use data and statistical analyses to classify people for the purpose of assessing their eligibility for a benefit or penalty.06-Dec-2019
## How algorithms affect decision-making?
Algorithms use data, statistics, and/or computing resources to make decisions. Algorithms used to rely mostly on statistical methods such as regression analysis to make decisions.
## What is decision support algorithm?
Clinical decision support algorithms (CDSAs) are digitized tools that combine an individual's health information with the health worker's knowledge and clinical protocols to assist in making diagnosis and treatment decisions.
## What are types of decision problems?
• Types of Decision Problem and Applications of Decision Support and Analysis.
• unstructured, long time spans of discretion.
• very structured, short time spans of discretion.
• Operational, structured decision making.
• Strategic, unstructured decision making.
• Instinctive, (rehearsed?) decision making.
## How does decision tree algorithm work?
The decision tree splits the nodes on all available variables and then selects the split which results in most homogeneous sub-nodes. The ID3 algorithm builds decision trees using a top-down greedy search approach through the space of possible branches with no backtracking.09-Feb-2022
## Is algorithmic decision making AI?
Many applications of AI can be characterized as some form of algorithmic decision making which raises the question of when we, as a society, should attempt to delegate decision making making processes to computer programs.
## What is automated decision making algorithm?
Automated decision-making is the process of making a decision by automated means without any human involvement. These decisions can be based on factual data, as well as on digitally created profiles or inferred data.
## What are the causes of algorithmic bias?
There are three main causes of algorithmic bias: input bias, training bias, and programming bias.
## Are we an algorithm?
Underlying that idea is another: that organisms — including human beings — are essentially algorithmic. That even the human mind is itself only a super-complex algorithm, and in order to model it perfectly all we need to do is build our own algorithm of sufficient sophistication.01-Mar-2019 | 612 | 3,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-38 | latest | en | 0.879337 |
http://memorize.com/basic-ecg/oelomar | 1,540,224,250,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00113.warc.gz | 231,698,715 | 6,329 | # Basic ECG
oelomar's version from 2017-12-23 14:58
## Section
Where does electrical discharge of each cardiac cycle start?In the sino-atrial node in the right atrium.
Where does the wave of depolarisation spread to after the SA node and the atrium?The atrio-ventricular node.
Where does the wave of depolarisation spread down after the AV node?The bundle of His.
What does the bundle of His divide into?2 left bundle branches and 1 right bundle branch.
Which fibres allow spread of depolarisation through the ventricular muscle?Purkinje fibres.
What is meant by sinus rhythm?When the cardiac cycle starts in the SA node.
Which wave(s) represents depolarisation of the atria?P wave.
Which wave(s) represents depolarisation of the ventricles?QRS complex.
Which wave(s) represents repolarisation of the ventricles?T wave.
What is a Q wave?A Q wave is a downward deflection that only occurs at the start of the QRS complex (not always present).
What is an R wave?Any upward deflection in a QRS complex.
What is an S wave?Any downward deflection below the baseline following an R wave.
What does the horizontal axis represent on an ECG graph?Time.
What does the vertical axis on an ECG graph represent?Voltage.
How much time does a large square represent?0.2 seconds (200 milliseconds).
How much time does a small square represent?0.04 seconds (40 milliseconds).
How many large squares represent a minute?300.
If a QRS complex occurs every 1 large square, what is the rhythm?300 bpm.
If a QRS complex occurs every 2 large squares, what is the rhythm?150 bpm.
If a QRS complex occurs every 3 large squares, what is the rhythm?100 bpm.
If a QRS complex occurs every 4 large squares, what is the rhythm?75 bpm.
If a QRS complex occurs every 5 large squares, what is the rhythm?60 bpm.
If a QRS complex occurs every 6 large squares, what is the rhythm?50 bpm.
How much voltage does a small square represent?0.1mV.
How much voltage do two large squares represent?1mV (at standard gain setting).
What is the PR interval?The time between the beginning of the P wave to the beginning of the QRS complex (excitation to spread from SA node to ventricular muscle).
How long is the normal PR interval?0.12 - 0.2 seconds (3-5 small squares, 120-200 milliseconds).
How long is the normal QRS complex?0.12 seconds (120 milliseconds) or less (3 small squares or less). | 588 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-43 | latest | en | 0.881385 |
https://www.chemeurope.com/en/encyclopedia/Oblique_shock.html | 1,723,758,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00524.warc.gz | 544,391,127 | 16,546 | My watch list
my.chemeurope.com
# Oblique shock
An oblique shock wave, unlike a normal shock, is inclined with respect to the incident upstream flow direction. It will occur when a supersonic flow encounters a corner that effectively turns the flow into itself and compresses. The upstream streamlines are uniformly deflected after the shock wave. The most common way to produce an oblique shock wave is to place a wedge into supersonic, compressible flow. Similar to a normal shock wave, the oblique shock wave consists of a very thin region across which nearly discontinuous changes in the thermodynamic properties of a gas occur. While the upstream and downstream flow directions are unchanged across a normal shock, they are different for flow across an oblique shock wave.
It is always possible to convert an oblique shock into a normal shock by a Galilean transformation.
## Oblique Shock Wave Theory
For a given Mach number, M1, and corner angle, θ, the oblique shock angle, β, and the downstream Mach number, M2, can be calculated. M2 is always less than M1. Unlike after a normal shock, M2 can still be supersonic. Discontinuous changes also occur in the pressure, density and temperature, which all rise downstream of the oblique shock wave.
Using the continuity equation and the fact that the tangential velocity component does not change across the shock, trigonometric relations eventually lead to the θ-β-M equation which shows θ as a function of M1 and β. It is more intuitive to want to solve for β as a function of M1 and θ, but this approach is more complicated, the results of which are often contained in tables or calculated through an applet.
$\tan \theta = 2\cot\beta\frac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$
Within the θ-β-M equation, a maximum corner angle, θMAX, exists for any upstream Mach number. When θ > θMAX, the oblique shock wave is no longer attached to the corner and is replaced by a detached bow shock. A θ-β-M diagram, common in most compressible flow textbooks, shows a series of curves that will indicate θMAX for each Mach number. The θ-β-M relationship will produce two β angles for a given θ and M1, with the larger angle called a strong shock and the smaller called a weak shock. The weak shock is almost always seen experimentally.
The rise in pressure, density, and temperature after an oblique shock can be calculated as follows:
$\frac{p_2}{p_1} = 1+\frac{2\gamma}{\gamma+1}(M_1^2\sin^2\beta-1)$
$\frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_1^2\sin^2\beta}{(\gamma-1)M_1^2\sin^2\beta+2}$
$\frac{T_2}{T_1} = \frac{p_2}{p_1}\frac{\rho_1}{\rho_2}$
M2 is solved for as follows:
$M_2 = \frac{1}{\sin(\beta-\theta)}\sqrt{\frac{1+\frac{\gamma-1}{2}M_1^2 \sin^2 \beta}{\gamma M_1^2 \sin^2 \beta- \frac{\gamma-1}{2}}}$
## Oblique Shock Wave Applications
Oblique shock waves are used predominantly in engineering applications when compared with normal shock waves. This can be attributed to the fact that using one or a combination of oblique shock waves results in more favorable post-shock conditions (lower post-shock temperature, etc.) when compared to utilizing a single normal shock. An example of this technique can be seen in the design of supersonic aircraft engine inlets, which are wedge-shaped to compress air flow into the combustion chamber while minimizing thermodynamic losses. Early supersonic aircraft jet engine inlets were designed using compression from a single normal shock, but this approach caps the maximum achievable Mach number to roughly 1.6. The wedge-shaped inlets are clearly visible on the sides of the F-14 Tomcat, which has a maximum speed of Mach 2.34.
Many supersonic aircraft wings are designed around a thin diamond shape. Placing a diamond-shaped object at an angle of attack relative to the supersonic flow streamlines will result in two oblique shocks propagating from the front tip over the top and bottom of the wing, with Prandtl-Meyer expansion fans created at the two corners of the diamond closest to the front tip. When correctly designed, this generates lift.
## Oblique Shock Waves and the Hypersonic Limit
As the Mach number of the upstream flow becomes hypersonic, the equations for the pressure, density, and temperature after the oblique shock wave reach a limit (mathematics). The pressure and density ratios can then be expressed as:
$\frac{p_2}{p_1} \approx \frac{2\gamma}{\gamma+1}M_1^2\sin^2\beta$
$\frac{\rho_2}{\rho_1} \approx \frac{\gamma+1}{\gamma-1}$
For a perfect atmospheric gas approximation using γ = 1.4, the hypersonic limit for the density ratio is 6. However, hypersonic post-shock dissociation of O2 and N2 into O and N lowers γ, allowing for higher density ratios in nature. The hypersonic temperature ratio is:
$\frac{T_2}{T_1} \approx \frac{2\gamma(\gamma-1)}{(\gamma+1)^2}M_1^2\sin^2\beta$
## References
• Liepmann, Hans W.; Roshko, A. [1957] (2001). Elements of Gasdynamics. Dover Publications. ISBN 0-486-41963-0.
• Anderson, John D. Jr. [1984] (January 2001). Fundamentals of Aerodynamics, 3rd Edition, McGraw-Hill Science/Engineering/Math. ISBN 0-07-237335-0.
• Shapiro, Ascher H. [1953]. The Dynamics and Thermodynamics of Compressible Fluid Flow, Volume 1. Ronald Press. ISBN 978-0-471-06691-0. | 1,367 | 5,253 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-33 | latest | en | 0.895498 |
https://www.indiabix.com/electronics-and-communication-engineering/electromagnetic-field-theory/discussion-3003 | 1,713,555,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00213.warc.gz | 735,776,717 | 7,188 | # Electronics and Communication Engineering - Electromagnetic Field Theory - Discussion
Discussion Forum : Electromagnetic Field Theory - Section 1 (Q.No. 3)
3.
Refractive index of glass is 1.5. Find the wavelength of a beam of light with a frequency of 1014 Hz in glass. Assume velocity of light is 3 x 108 m/sec in vacuum.
4 μm
3 μm
2 μm
1 μm
Explanation:
Given, μ = 1.6, f = 1014 Hz, v = 3 x 108 m/sec.
Discussion:
6 comments Page 1 of 1.
Pradeep Rathod said: 4 years ago
C/f = 3x10^8/10^14 = 3um/1.5 = 2μm.
Sunni said: 7 years ago
No, answer is 2 that is print mistake 1.5 not 1.6.
Suraj Rathod said: 7 years ago
Shree said: 7 years ago
Refractive index is mentioned as 1.5 not 1.6.
It will get another answer not 2 micrometre.
Bittu said: 8 years ago
Here refractive index given -1.5.
Then why 1.6 considered?
PRACHI said: 8 years ago
We can not consider the value of 1.6? | 309 | 900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-18 | latest | en | 0.80355 |
https://www.coursehero.com/file/52467/Lecture-14-Part-2/ | 1,490,756,438,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190134.67/warc/CC-MAIN-20170322212950-00547-ip-10-233-31-227.ec2.internal.warc.gz | 873,277,332 | 214,134 | Lecture_14-Part_2
# Lecture_14-Part_2 - Chapter 18 A Microscopic View of Electric Circuits Current in a Circuit A microscopic view of electric circuits Are charges
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Chapter 18 A Microscopic View of Electric Circuits
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Steady state and static equilibrium Static equilibrium : no charges are moving Steady state : • charges are moving • their velocities at any location do not change with time • no change in the deposits of excess charge anywhere Current in a Circuit In an electric circuit the system does not reach equilibrium! A microscopic view of electric circuits: Are charges used up in a circuit? How is it possible to create and maintain a nonzero electric field inside a wire? What is the role of the battery in a circuit?
What would be I A compared to I B ? 1) I B = 0 all electrons are used up in the bulb 2) I B < I A some of the electrons are used up in the bulb 3) I B = I A Current in Different Parts of a Circuit What happens to the charges that flow through the circuit? Is the current the same in all parts of a series circuit?
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1) I B = 0 all electrons are used up in the bulb a) electrons disappear in the light bulb requires nuclear reaction
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## This note was uploaded on 04/02/2008 for the course PHYS 272 taught by Professor K during the Winter '07 term at Purdue University-West Lafayette.
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Lecture_14-Part_2 - Chapter 18 A Microscopic View of Electric Circuits Current in a Circuit A microscopic view of electric circuits Are charges
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Ask a homework question - tutors are online | 444 | 1,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-13 | longest | en | 0.889263 |
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# Kindergarten Math Worksheet Bundle
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### Description
Kindergarten Math
This set includes Do-Now worksheets for Kindergarten mathematics.
• Counting and Cardinality (15 pages)
• Operations and Algebraic Thinking (17 pages)
• Number and Operations in Base Ten (3 pages)
• Measurement and Data (5 pages)
• Geometry (6 pages)
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### Standards
to see state-specific standards (only available in the US).
Count to 100 by ones and by tens.
Count forward beginning from a given number within the known sequence (instead of having to begin at 1).
Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects).
Understand the relationship between numbers and quantities; connect counting to cardinality.
When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object. | 490 | 2,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-23 | latest | en | 0.86755 |
https://math.stackexchange.com/questions/4047855/countable-algebraically-closed-field-inside-an-incountable-algebraically-closed | 1,713,257,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00865.warc.gz | 358,428,768 | 35,542 | # Countable algebraically closed field inside an incountable algebraically closed of characteristic zero
I would like to know if the following claim is true. I found this claim in a paper without proof :(.
Let $$k$$ an uncountable algebraically closed field of characteristic $$0$$.
Let $$S=\operatorname{Spec}(k[X_1,..., X_m]/I(S))=\operatorname{Spec}(k[X_1,..., X_m]/\langle f_1,...,f_n\rangle)$$ be an integral affine scheme of finite type over $$k$$.
Claim: We can choose a countable algebraically closed subfield $$k_0\subset k$$ such that there is an irreducible quasi projective scheme $$S_0$$ over $$k_0$$ with $$S=S_0\times_{\operatorname{Spec}(k_0)}\operatorname{Spec}(k)$$.
Let $$T$$ be the finite set of coefficients appearing in the $$f_i$$. Then $$k_0=\overline{\Bbb Q(T)}$$ and $$S_0=\operatorname{Spec} k_0[X_1,\cdots,X_m]/(f_1,\cdots,f_m)$$ suffice. The verification that this has all the properties you request is straightforward; if you are stuck on any part, please leave a comment.
• Thank you @KReiser! Another question that I have is: Why $S_0$ over $k_0$ should be quasi-projective (notice that they initially say that $S$ over $k$ is affine)?. Mar 4, 2021 at 9:51
• This is probably because they're gearing up to apply some result for quasi-projective $k_0$-schemes (it would be easier to say for sure if you included the source). Being affine, $S_0$ is obviously quasi-projective, and in fact as affineness descends along fpqc morphisms, any $S_0$ you find satisfying $S=S_0\times_{k_0} k$ will be affine. Mar 4, 2021 at 11:03
• Ohhh I see @KReiser. Now, since $k_0$ is countable algebraically closed, Can I say that there are only countably many closed subsets in $S_0$? Mar 4, 2021 at 13:24
• Yes, that's true, but it doesn't require $k_0$ to be algebraically closed, just countable. Any closed subset of $S_0$ is given by an ideal of $k_0[X_1,\cdots,X_m]/(f_1,\cdots,f_m)$, every ideals is finitely generated as this ring is noetherian, and the ring has countably many elements therefore the collection of ideals is countable. Mar 4, 2021 at 20:40 | 628 | 2,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.907592 |
https://inlearn.app/courses/commerce-grade-11-monthly-pack-math-january/ | 1,713,930,754,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00491.warc.gz | 288,358,593 | 24,656 | # Commerce-Grade 11 Monthly Pack (Math) – January
Category:
• Straight lines
• Conic Sections
• Introduction to 3-Dimensional Geometry
### Straight lines
1
Introduction; Exercise 10.1 (1)
2
Exercise 10.1 (3,4,5,6)
3
Exercise 10.1 (7,8,9,10,11)
4
Exercise 10.1 (12,13)
5
Exercise 10.2 (7 to 13)
6
Exercise 10.2 (14,15,17,18,19)
7
Exercise 10.3 (1,2,3,4)
8
Exercise 10.3 (5,6,7,8,9)
9
Exercise 10.3 (10,11,12)
10
Exercise 10.3 (14,16,17,18)
11
Angle between 2 Lines; Collinearity of 3 Points; Various Forms of Equation of Line
### Conic Sections
1
Introduction
2
Exercise 11.1(6 to 11)
3
Exercise 11.1(12 to 15)
4
Exercise 11.2(1,2,3,4,5,8,9,10,11,12)
5
Exercise 11.3(1,3,4,5,6)
6
Exercise 11.3(7,8,9)
7
Exercise 11.3(10,13,15,16,17,18,19)
8
Exercise 11.3(20); Exercise 11.4(1,2,3,4)
9
Exercise 11.4(5 to 11)
10
Exercise 11.4(12 to 15)
11
Ellipse; Hyperbola
### Introduction to 3-Dimensional Geometry
1
Introduction
2
Exercise 12.2(1,2,3)
3
Exercise 12.2(4,5); Distance Between 2 Points;Midpoint;Section Formula;Centroid of a Triangle
4
Exercise 12.3(1,2,3,4)
5
Exercise 12.3(5);Exercise 12.1(1 to 4) | 484 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.657833 |
https://www.conceptdraw.com/examples/differences-between-bus-topology-and-star-topology | 1,709,151,014,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474744.31/warc/CC-MAIN-20240228175828-20240228205828-00554.warc.gz | 717,259,176 | 5,822 | This site uses cookies. By continuing to browse the ConceptDraw site you are agreeing to our Use of Site Cookies.
# Network topologies diagram
"Network topology is the arrangement of the various elements (links, nodes, etc.) of a computer network. Essentially, it is the topological structure of a network, and may be depicted physically or logically. Physical topology refers to the placement of the network's various components, including device location and cable installation, while logical topology shows how data flows within a network, regardless of its physical design. Distances between nodes, physical interconnections, transmission rates, and/ or signal types may differ between two networks, yet their topologies may be identical. The study of network topology recognizes eight basic topologies: Point-to-point, Bus, Star, Ring or circular, Mesh, Tree, Hybrid, Daisy chain." [Network topology. Wikipedia]
The computer network topologies diagram example was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Computer and Networks solution from the Computer and Networks area of ConceptDraw Solution Park.
Network topologies
Used Solutions
## Network topologies diagram
"Logical topology, or signal topology, is the arrangement of devices on a computer network and how they communicate with one another. How devices are connected to the network through the actual cables that transmit data, or the physical structure of the network, is called the physical topology. Physical topology defines how the systems are physically connected. It represents the physical layout of the devices on the network. The logical topology defines how the systems communicate across the physical topologies.
Logical topologies are bound to network protocols and describe how data is moved across the network. ...
EXAMPLE : twisted pair Ethernet is a logical bus topology in a physical star topology layout. while IBM's token ring is a logical ring topology, it is physically set up in star topology." [Logical topology. Wikipedia]
This Cisco logical computer network diagram example was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Cisco Network Diagrams solution from the Computer and Networks area of ConceptDraw Solution Park.
Logical network topology diagram
Used Solutions | 429 | 2,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.903348 |
https://jp.maplesoft.com/support/help/addons/view.aspx?path=ThermophysicalData/Chemicals/Property | 1,701,887,255,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00881.warc.gz | 388,305,677 | 26,770 | Property - Maple Help
ThermophysicalData[Chemicals]
Property
access thermodynamic data
Calling Sequence Property(output, species, inputopts, opts)
Parameters
output - symbol or string for the desired output quantity species - symbol or string representing the required chemical inputopts - (optional) equation of the form s=value, where s is temperature, "temperature", T or "T" opts - (optional) one of more options, as described below
Options
• useunits : true or false show the output with units
• explicit : true or false show the output in polynomial form
Description
• The Property function returns the thermodynamic data defined in McBride et al. (2002) and data on the molar Gibbs free energy of formation that is derived from it.
• The data returned is at the standard state.
– If species is a gas, the standard state is 1 bar.
– If species is crystalline or liquid, the standard state is 1 atm.
• The output parameter is one of Hmolar, Smolar, Cpmolar, MolarMass, HeatOfFormation or Comments. It can be given as a symbol or a string. Refer to the tables below for alternative names that are accepted by this command.
• If output is Hmolar, Gmolar, Smolar or Cpmolar, then an equation of the form temperature=value needs to be supplied.
– temperature can be replaced by "temperature", T or "T".
– value can be a name, a numeric value, or a numeric value with a unit of K .
– If value is a name and explicit is false (the default), then the unevaluated function is returned.
– If value is a name and explicit=true or explicit is specified, then an expression in value is returned. The expression is an empirical correlation for output as defined in McBride et al. (2002), or data on the molar Gibbs free energy of formation that is derived from such correlations.
– If value is numeric, then it is assumed to be the temperature in Kelvin.
– If value is numeric with a unit of K or if useunit=true or useunit is specified, then the result returned by Property will have a unit. No other temperature unit apart from K can be used.
– This table describes the results returned by the Property command.
Output Quantity Unit of Returned Value Hmolar, HMOLAR, molar_specific_enthalpy, molarspecificenthalpy Molar enthalpy J/mol Gmolar, GMOLAR, molar_gibbs_free_energy_of_formation, molargibbsfreeenergyofformation Molar Gibbs free energy of formation J/mol Smolar, SMOLAR, molar_specific_entropy, molarspecificentropy Molar entropy J/mol/K Cpmolar, CPMOLAR, molar_specific_constant_pressure_specific_heat, molarspecificconstantpressurespecificheat Molar heat capacity at constant pressure J/mol/K
• If output is MolarMass or HeatOfFormation, then the following is true.
– If the option useunit has the value true, the result returned by the Property command will have a unit.
– This table describes the data returned by the Property command.
Output Quantity Unit of Returned Value MolarMass, M, MOLARMASS, MOLAR_MASS, MOLEMASS, molar_mass, molemass Molar mass g/mol HeatOfFormation Heat of formation at 298.15 J/mol
• If output is Comments then additional information (as defined in McBride et al., 2002) about the chemical is returned.
Examples
> $\mathrm{with}\left(\mathrm{ThermophysicalData}:-\mathrm{Chemicals}\right)$
$\left[{\mathrm{GetSpecies}}{,}{\mathrm{Property}}\right]$ (1)
Determine the enthalpy of CO2 with and without units
> $\mathrm{Property}\left(\mathrm{Hmolar},\mathrm{CO2},\mathrm{temperature}=300\right)$
${-393441.2212}$ (2)
> $\mathrm{Property}\left(\mathrm{Hmolar},\mathrm{CO2},\mathrm{temperature}=300⟦K⟧\right)$
${-}{393441.2212}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (3)
Determine the molecular weight of CO2 with and without units
> $\mathrm{Property}\left(\mathrm{MolarMass},\mathrm{CO2}\right)$
${44.0095000}$ (4)
> $\mathrm{Property}\left(\mathrm{MolarMass},\mathrm{CO2},\mathrm{useunits}\right)$
${44.0095000}{}⟦\frac{{g}}{{\mathrm{mol}}}⟧$ (5)
Return an empirical correlation for the molar enthalpy of CO2
> $\mathrm{Property}\left("Hmolar","CO2","temperature"=T,\mathrm{explicit}\right)$
${8.314510}{}{T}{}\left(\left\{\begin{array}{cc}{-}\frac{{49436.50540}}{{{T}}^{{2}}}{-}\frac{{626.4116010}{}{\mathrm{ln}}{}\left({T}\right)}{{T}}{+}{5.301725240}{+}{0.001251906908}{}{T}{-}{7.091029093}{×}{{10}}^{{-8}}{}{{T}}^{{2}}{-}{1.922497195}{×}{{10}}^{{-10}}{}{{T}}^{{3}}{+}{5.699355602}{×}{{10}}^{{-14}}{}{{T}}^{{4}}{-}\frac{{45281.98460}}{{T}}& {200.000}{\le }{T}{\le }{1000.000}\\ {-}\frac{{117696.2419}}{{{T}}^{{2}}}{-}\frac{{1788.791477}{}{\mathrm{ln}}{}\left({T}\right)}{{T}}{+}{8.291523190}{-}{0.00004611578390}{}{T}{+}{1.621225627}{×}{{10}}^{{-9}}{}{{T}}^{{2}}{-}{4.727633280}{×}{{10}}^{{-13}}{}{{T}}^{{3}}{+}{1.266007318}{×}{{10}}^{{-16}}{}{{T}}^{{4}}{-}\frac{{39083.50590}}{{T}}& {1000.000}{<}{T}{\le }{6000.000}\\ \frac{{1.544423287}{×}{{10}}^{{9}}}{{{T}}^{{2}}}{+}\frac{{1.016847056}{×}{{10}}^{{6}}{}{\mathrm{ln}}{}\left({T}\right)}{{T}}{-}{256.1405230}{+}{0.01684700540}{}{T}{-}{7.270614457}{×}{{10}}^{{-7}}{}{{T}}^{{2}}{+}{1.747855210}{×}{{10}}^{{-11}}{}{{T}}^{{3}}{-}{1.768470300}{×}{{10}}^{{-16}}{}{{T}}^{{4}}{-}\frac{{8.043214510}{×}{{10}}^{{6}}}{{T}}& {6000.000}{<}{T}{\le }{20000.000}\\ {\mathrm{undefined}}& {\mathrm{otherwise}}\end{array}\right\\right)$ (6)
Given an enthalpy, backsolve for the temperature
> $\mathrm{fsolve}\left(\mathrm{Property}\left("Hmolar","CO2","temperature"=T\right)=-3.938111213{10}^{5}⟦\frac{J}{\mathrm{mol}}⟧,T=280⟦K⟧\right)$
${289.9999991}{}⟦{K}⟧$ (7)
Calculate the Gibbs Energy of Formation of ammonia at 298.15 K, given the reaction ${N}_{2}\left(g\right)+3{H}_{2}\left(g\right)\to 2{\mathrm{NH}}_{3}\left(g\right)$
> $\mathrm{with}\left({\mathrm{Units}}_{\mathrm{Simple}}\right):$
Temperature
> $T≔298.15⟦'K'⟧$
${T}{≔}{298.15}{}⟦{K}⟧$ (8)
Enthalpy
> $\mathrm{h_N2}≔\mathrm{Property}\left("Hmolar","N2","temperature"=T\right);$$\mathrm{h_H2}≔\mathrm{Property}\left("Hmolar","H2","temperature"=T\right);$$\mathrm{h_NH3}≔\mathrm{Property}\left("Hmolar","NH3","temperature"=T\right)$
${\mathrm{h_N2}}{≔}{9.915884626}{×}{{10}}^{{-6}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$
${\mathrm{h_H2}}{≔}{-}{4.957942313}{×}{{10}}^{{-6}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$
${\mathrm{h_NH3}}{≔}{-}{45940.00004}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (9)
Entropy
> $\mathrm{s_N2}≔\mathrm{Property}\left("Smolar","N2","temperature"=T\right);$$\mathrm{s_H2}≔\mathrm{Property}\left("Smolar","H2","temperature"=T\right);$$\mathrm{s_NH3}≔\mathrm{Property}\left("Smolar","NH3","temperature"=T\right)$
${\mathrm{s_N2}}{≔}{191.6097115}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$
${\mathrm{s_H2}}{≔}{130.6810143}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$
${\mathrm{s_NH3}}{≔}{192.7702891}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$ (10)
Change in enthalpy and entropy per mole of NH3
> $\mathrm{DeltaH}≔0.5\left(2\mathrm{h_NH3}-\mathrm{h_N2}-3\mathrm{h_H2}\right);$$\mathrm{DeltaS}≔0.5\left(2\mathrm{s_NH3}-\mathrm{s_N2}-3\mathrm{s_H2}\right)$
${\mathrm{DeltaH}}{≔}{-}{45940.00004}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$
${\mathrm{DeltaS}}{≔}{-}{99.05608810}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$ (11)
Hence the Gibbs Energy of Formation
> $\mathrm{DeltaG}≔\mathrm{DeltaH}-\mathrm{DeltaS}T$
${\mathrm{DeltaG}}{≔}{-}{16406.42737}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (12)
> $\mathrm{convert}\left(\mathrm{DeltaG},'\mathrm{units}',\frac{'\mathrm{kJ}'}{'\mathrm{mol}'}\right)$
${-}{16.40642737}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (13)
Verification:
> $\mathrm{Property}\left("Gmolar","NH3","temperature"=T,'\mathrm{useunits}'\right)$
${-}{16406.42733}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (14)
References
McBride, Bonnie J.; Zehe, Michael J.; and Gordon, Sanford. NASA Glenn Coefficients for Calculating Thermodynamic Properties of Individual Species; 2002; https://www.grc.nasa.gov/WWW/CEAWeb/TP-2002-21556.htm (6 Dec 2017).
Compatibility
• The ThermophysicalData:-Chemicals:-Property command was introduced in Maple 2018. | 2,816 | 7,914 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 40, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.587545 |
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Hsp matematik tahun 6 (kbsr)
1. 1. DRAF Kementerian Pelajaran MalaysiaKURIKULUM BERSEPADU SEKOLAH RENDAH SPESIFIKASI KURIKULUM MATEMATIK TAHUN 6 Bahagian Pembangunan Kurikulum Kementerian Pelajaran Malaysia (i)
2. 2. (ii)
3. 3. KANDUNGANRUKUN NEGARA (v) MASA ................................................................................. 16FALSAFAH PENDIDIKAN KEBANGSAAN (vi) Tempoh Masa ..................................................................... 16KATA PENGANTAR (vii)PENDAHULUAN (ix) WANG ................................................................................ 15 Wang Hingga RM10 Juta .................................................... 15NOMBOR BULAT .............................................................. 1Nombor Bulat Hingga Tujuh Digit ....................................... 1 PANJANG .......................................................................... 18Operasi Asas Nombor Hingga Tujuh Digit ......................... 3 Pengiraan Ukuran Panjang ................................................ 18PECAHAN .......................................................................... 6 JISIM .................................................................................. 20Tambah Pecahan ............................................................... 6 Pengiraan Jisim ................................................................ 20Tolak Pecahan ................................................................... 8Darab Pecahan .................................................................. 9 ISIPADU CECAIR .............................................................. 22Bahagi Pecahan ................................................................. 10 Pengiraan Isipadu Cecair .................................................... 22PERPULUHAN ................................................................... 12 BENTUK DAN RUANG....................................................... 24Operasi Bergabung Melibatkan Perpuluhan ....................... 12 Bentuk Dua Dimensi....... .................................................... 24 Bentuk Tiga Dimensi.............................. ............................ 26PERATUS ........................................................................... 13Hubungan Antara Peratus, Pecahan dan Perpuluhan ........ 13 PERWAKILAN DATA ........................................................ 28 Purata ................................................................................. 28WANG ................................................................................ 15 Mentafsir Data .................................................................... 30Wang Hingga RM10 Juta .................................................... 15 (iii)
4. 4. (iii)
5. 5. RUKUN NEGARABAHAWASANYA negara kita Malaysia mendukung cita-cita untuk mencapai perpaduan yang lebih erat dalam kalangan seluruh masyarakatnya; memelihara satu cara hidup demokratik; mencipta masyarakat yang adil bagi kemakmuran negara yang akan dapat dinikmati bersama secara adil dan saksama; menjamin satu cara yang liberal terhadap tradisi-tradisi kebudayaannya yang kaya dan berbagai-bagai corak; membina satu masyarakat progresif yang akan menggunakan sains dan teknologi moden;MAKA KAMI, rakyat Malaysia, berikrar akan menumpukanseluruh tenaga dan usaha kami untuk mencapai cita-citatersebut berdasarkan atas prinsip-prinsip yang berikut:• KEPERCAYAAN KEPADA TUHAN• KESETIAAN KEPADA RAJA DAN NEGARA• KELUHURAN PERLEMBAGAAN• KEDAULATAN UNDANG-UNDANG• KESOPANAN DAN KESUSILAAN (v) (iii)
6. 6. Pendidikan di Malaysia ialah suatu usaha berterusan ke arah memperkembangkan potensi individu secara menyeluruh dan bersepadu untuk melahirkan insan yang seimbang dan harmonis dari segi intelek, rohani, emosi dan jasmani berdasarkan kepercayaan dan kepatuhan kepada Tuhan. Usaha ini bertujuan untuk melahirkan warganegarat Malaysia yang berilmu pengetahuan, berketerampilan, berakhlak mulia, bertanggungjawab dan berkeupayaanmencapai kesejahteraan diri serta memberikan sumbangan terhadap keharmonian dan kemakmuran keluarga, masyarakat, dan Negara. (vi) (iii)
9. 9. OBJEKTIF ORGANISASI KANDUNGANKurikulum Matematik Sekolah Rendah akan membolehkan murid Kurikulum Matematik di peringkat sekolah rendah merangkumiuntuk: empat bidang utama iaitu Nombor, Ukuran, Bentuk dan Ruang, dan1 Mengetahui dan memahami konsep, definisi, peraturan dan Statistik. Topik bagi setiap bidang telah disusun daripada peringkat prinsip-prinsip berkaitan nombor, operasi, ruang, ukuran dan asas kepada abstrak. Guru perlu mengajar asas-asas sebelum perwakilan data; memperkenalkan topik abstrak kepada murid.2 Menguasai operasi asas Matematik: Setiap bidang utama dibahagikan kepada topik-topik berikut: penambahan 1 Nombor penolakan Nombor Bulat; pendaraban, Pecahan; pembahagian; Perpuluhan;3 Menguasai kemahiran dalam operasi bergabung; Wang;4 Menguasai kemahiran asas Matematik iaitu: 2 Ukuran membuat anggaran dan pembundaran Masa dan Waktu; pengukuran Ukuran Panjang; perwakilan data Timbangan Berat; mentafsir maklumat dalam bentuk graf dan carta; Isipadu Cecair;5 Menggunakan kemahiran Matematik dan pengetahuan untuk 3 Bentuk dan Ruang menyelesaikan masalah dalam kehidupan harian secara efektif Bentuk Dua Dimensi (2D); dan bertanggungjawab; Bentuk Tiga Dimensi (3D);6 Menggunakan bahasa Matematik dengan betul; Perimeter dan Luas;7 Menggunakan teknologi yang sesuai dalam pembinaan konsep, 4 Statistik memperolehi kemahiran Matematik dan penyelesaian masalah; Perwakilan Data8 Mengaplikasikan pengetahuan Matematik secara bersistematik, heuristik, tepat dan berhati-hati;9 Mengambil bahagian dalam aktiviti berkaitan Matematik; dan10 Menghargai kepentingan dan keindahan Matematik. (x)
13. 13. Bercerita, sesi soal jawab menggunakan perkataan sendiri; 3. PENAAKULAN MATEMATIK Bertanya dan menjawab soalan; Pemikiran dan penaakulan secara logik adalah asas bagi Temuduga berstruktur dan tidak berstruktur; memahami dan menyelesaikan masalah Matematik. Perbincangan dalam forum, seminar, perdebatan dan sesi Perkembangan dalam penaakulan Matematik berkait rapat kepada sumbang saran; dan perkembangan intelektual dan komunikasi murid-murid. Penekanan Persembahan hasil kajian. terhadap pemikiran logik semasa melakukan aktiviti Matematik membuka minda murid untuk menerima Matematik sebagai sesuatuKomunikasi bertulis adalah satu proses di mana idea Matematik yang amat penting dalam dunia hari ini.dan maklumat dikongsi dengan orang lain melalui penulisan. Kerjabertulis biasanya adalah hasil perbincangan, sumbangan dan Murid digalakkan untuk meramal dan meneka dalam prosesaktiviti sumbang saran ketika menyiapkan kajian. Melalui penulisan, mencari penyelesaian. Murid di semua peringkat perlu dilatih untukmurid digalakkan berfikir secara mendalam tentang kandungan menyiasat ramalan atau tekaan mereka dengan menggunakanMatematik dan memerhati hubungan antara konsep-konsep. bahan konkrit, kalkulator, komputer, gambaran Matematik dan sebagainya. Penaakulan logik perlu diterapkan dalam pengajaranContoh aktiviti komunikasi bertulis adalah: Matematik untuk membolehkan murid mengenal pasti, membina Membuat latihan; dan menilai ramalan dan perbincangan Matematik. Menyimpan buku skrap; Menyimpan folio; 4. PERKAITAN MATEMATIK Melaksanakan projek; dan Ujian bertulis. Dalam kurikulum Matematik, peluang untuk membuat perkaitan mestilah dihasilkan agar murid boleh mengaitkan pengetahuanGambaran adalah suatu proses menganalisis masalah Matematik konseptual kepada procedural dan mengaitkan tajuk dalamdan mentafsirnya daripada satu cara kepada cara yang lain. Matematik dengan lain-lain bidang pembelajaran secara umum.Gambaran Matematik membolehkan pelajar mencari hubunganantara idea Matematik yang tidak formal, intuitif dan abstrak Kurikulum Matematik mengandungi beberapa bidang sepertidengan menggunakan bahasa harian. Murid akan menyedari aritmetik, geometri, pengukuran dan penyelesaian masalah. Tanpabahawa sesetengah pendekatan untuk membuat gambaran akan hubungan antara bidang-bidang tersebut, murid terpaksa belajarlebih berkesan dan berguna jika mereka tahu bagaimana untuk dan mengingat terlalu banyak konsep dan kemahiran secaramenggunakan unsure-unsur dalam gambaran Matematik. berasingan. Dengan membuat perkaitan, murid dapat melihat Matematik sebagai satu penyatuan secara keseluruhan dan bukan (xiv)
14. 14. idea yang berasingan. Guru boleh menggalakkan hubungan dalam amat penting dalam membentuk konsep Matematik. Guru bolehkelas yang berpusatkan masalah dengan menyuruh murid menggunakan bahan sebenar atau bahan konkrit dlam pengajaranberkomunikasi, member sebab dan menyatakan pendapat mereka. dan pembelajaran untuk menolong murid mendapat pengalaman,Apabila idea-idea Matematik dikaitkan dengan situasi kehidupan membina idea yang abstrak, mencipta, membina keyakinan diri,sebenar dan kurikulum, murid akan menjadi lebih sedar dalam berdikari dan bekerjasama.mengaplikasikan Matematik. Mereka juga boleh menggunakanMatematik mengikut konteksnya dalam bidang pembelajaran yang Bahan pengajaran dan pembelajaran yang digunakan sepatutnyaberbeza dalam kehidupan harian. mengandungi elemen diagnostic kendiri bagi membolehkan murid mengetahui sejauh mana kefahaman mereka tentang konsep dan kemahiran. Membantu murid untuk bersikap dan berpersonaliti5. PENGGUNAAN TEKNOLOGI positif,mempunyai nilai dalaman Matematik yang jitu, berkeyakinan dan berfikir secara sistematik perlu diterapkan melalui bidangPenggunaan teknologi membantu murid memahami konsep pembelajaran.matematik dengan mendalam, bermakna dan tepat untukmembolehkan mereka menguasai konsep Matematik. Penggunaan Nilai murni boleh diterapkan melalui konteks yang sesuai. Sebagaikalkulator, komputer, perisian pendidikan, laman web di internet contoh, belajar secara kumpulan dapat membantu murid membinadan pakej pembelajaran yang mudah didapati boleh membantu kemahiran social dan menggalakkan kerjasama dan keyakinan dirimeningkatkan kemahiran pedagogi dalam pengajaran dan dalam subjek ini. Elemen patriotism boleh juga diterapkan melaluipembelajaran Matematik. proses pengajaran dan pembelajaran di bilik darjah menggunakan topik yang dirancang. Nilai ini perlu diterapkan melalui prosesPenggunaan sumber pengajaran adalah penting dalam Matematik. pengajaran dan pembelajaran Matematik.Ini akan memastikan murid mendapat idea yang abstrak, menjadikreatif, berkeyakinan dan boleh bekerja sendiri atau berkumpulan. Antara pendekatan yang boleh dipertimbangkan adalah:Kebanyakan daripada sumber tersebut disusun untuk pembelajaran Pembelajaran berpusatkan murid yang menarik;akses kendiri. Melalui pembelajaran akses kendiri, murid boleh me Kebolehan belajar dan gaya pembelajaran; Penggunaan bahan pengajaran yang relevan, sesuai dan efektif;PENDEKATAN DALAM PENGAJARAN DAN PEMBELAJARAN dan Penilain formatif untuk mengenal pasti keberkesanan pengajaranPelbagai perubahan berlaku yang member kesan terhadap dan pembelajaran.kandungan dan pedagogi dalam pengajaran Matematik sekolahrendah. Perubahan ini memerlukan kepelbagaian cara pengajaranMatematik di sekolah. Penggunaan sumber pengajaran adalah (xv)
15. 15. Pemilihan pendekatan yang sesuai akan menggalakkan suasanapersekitaran pengajaran dan pembelajaran di dalam dan di luar bilikdarjah. Pendekatan yang sesuai adalah seperti berikut: Pembelajaran kooperatif; Pembelajaran kontekstual; Pembelajaran masteri; Konstruktivisme; Inkuiri penemuan; dan Kajian masa depan.PENILAIANPenilaian adalah sebahagian daripada proses pengajaran danpembelajaran. Ia perlu dirancang dengan baik dan dijalankanberterusan sebagai sebahagian aktiviti bilik darjah. Denganberfokuskan kepada aktiviti Matematik yang pelbagai, kekuatan dankelemahan murid boleh dinilai. Kaedah penilaian yang berbezaboleh dijalankan dengan menggunakan pelbagai teknik penilaiantermasuk kerja lisan dan bertulis dan juga tunjuk cara. Ia bolehdijalankan dalam bentuk temuduga, soalan terbuka, pemerhatiandan kajian. Berdasarkan kepada keputusan, guru dapatmemperbetulkan salah tanggapan dan kelemahan murid-murid dandalam masa yang sama memperbaiki kemahiran mengajar mereka.Guru boleh mengambil langkah yang berkesan dalam menjalankanaktiviti pemulihan dan pengayaan untuk meningkatkan keupayaanmurid-murid. (xvi)
17. 17. Bidang Pembelajaran: NOMBOR BULAT HINGGA TUJUH DIGIT Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN (iii) Menyatakan nombor bulat Tulis nombor dalam sebutan bagi: juta dan sebaliknya. a) perpuluhan b) pecahan Contoh: dalam sebutan juta dan a) 800 000 is 0.8 juta sebaliknya. b) 6 320 000 ialah 6.32 juta c) 1.4 juta ialah 1 400 000 d) 5.602 juta ialah 5 602 00 1 e) 3 500 000 ialah 3 2 juta f) 8 3 juta ialah 8 750 000 4 Untuk nombor pecahan, penyebut adalah dalam gandaan 10 (10 hingga 90), 100 dan 1000 dan menukar pecahan dalam bentuk termudah. Hadkan perpuluhan kepada tiga tempat perpuluhan. Diberi set nombor, murid (iv) Membanding nilai nombor membanding dan menyusun hingga tujuh digit. nombor mengikut urutan menaik dan menurun. (v) Membundarkan nombor Membundarkan nombor kepada puluh, ratus, ribu, adalah untuk mendapatkan puluh ribu, ratus ribu dan juta penghampiran. terdekat. 2
18. 18. Bidang Pembelajaran: OPERASI ASAS NOMBOR HINGGA TUJUH DIGIT Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN2 Tambah, tolak, darab Murid menambah, menolak, (i) Menambah sebarang dua Latihan menambah mudah dan bahagi melibatkan mendarab dan membahagi hingga lima nombor hingga termasuk menambah dua simulasi nombor hingga tujuh menggunakan empat langkah 9 999 999. nombor hingga empat digit. algoritma: nombor dengan atau tanpa analogi / persamaan mengumpul semula. 1) Membuat anggaran. urutan 2) Menyusun nombor mengikut Menyediakan latihan nilai tempat. menambah secara mental sama ada menggunakan 3) Melaksana operasi. teknik asas abakus atau 4) Menyemak jawapan. strategi kira cepat seperti menganggar jumlah dengan membundarkan, memudahkan penambahan dengan pasangan 10, sekali ganda dan sebagainya. (ii) Menolak Hadkan masalah tolak a) satu nombor daripada dengan menolak daripada satu nombor yang lebih nombor yang lebih besar. besar dan kurang daripada 10 000 000. Menyediakan latihan menolak secara mental b) secara berturut-turut sama ada menggunakan daripada satu nombor teknik asas abakus atau yang lebih besar dan strategi kira cepat. kurang daripada 10 000 000. 3
19. 19. Bidang Pembelajaran: OPERASI ASAS NOMBOR HINGGA TUJUH DIGIT Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN Strategi kira cepat tolak yang boleh dilaksanakan adalah: a) Menganggarkan jumlah dengan membundarkan nombor b) Kiraan menaik dan kiraan menurun. (iii) Mendarab nombor hingga Hadkan hasil darab kurang enam digit dengan: daripada 10 000 000. a) Nombor satu digit Latihan mendarab secara mental sama ada b) Nombor dua digit menggunakan teknik asas c) 10, 100 dan 1000. abakus atau strategi pendaraban yang lain. Strategi pendaraban yang boleh dilaksanakan termasuk kaedah pemfaktoran, penggenap 100, pendaraban silang (lattice) dan sebagainya. (iv) Membahagi nombor Latihan membahagi hingga tujuh digit dengan: termasuk hasil bahagi berbaki dan tanpa baki. a) Nombor satu digit b) 10,100 dan 1000 Menekankan teknik pengiraan dalam bentuk c) Nombor dua digit. lazim. 4
20. 20. Bidang Pembelajaran: OPERASI ASAS NOMBOR HINGGA TUJUH DIGIT Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN Menyediakan latihan membahagi secara mental sama ada menggunakan teknik asas abakus atau strategi bahagi yang lain. Mendedahkan murid dengan pelbagai kaedah bahagi. Contoh: a) Nombor yang boleh dibahagi dengan tepat b) Bahagi dengan 10, 100 dan 1000. Mendedahkan masalah dalam v) Selesaikan masalah Menggunakan sebarang bentuk nombor, ayat mudah, a) tambah kaedah penyelesaian jadual dan gambar. b) tolak a) Permudahkan masalah. c) darab b) Kaedah cuba jaya Murid membina cerita d) bahagi c) Melukis gambarajah berdasarkan ayat matematik yang melibatkan nombor d) Mengenalpasti pola dan yang diberi. hingga tujuh digit. jujukan nombor. Guru membimbing murid e) Membina jadual, carta menyelesaikan masalah atau satu senarai dengan mengikut empat sistematik. langkah Model Polya. f) Simulasi g) Membuat analogi Menyemak jawapan. Menyemak jawapan. h) Bekerja ke belakang. 5
21. 21. Bidang Pembelajaran: TAMBAH PECAHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN3 Melaksana operasi Menerangkan kepada murid (i) Mengira masalah operasi Operasi bergabung terhad mengira bergabung nombor konsep operasi bergabung dan bergabung melibatkan kepada dua operasi, contoh: operasi bergabung bulat. menghubungkannya dengan penambahan dan tanda kurung langkah-langkah operasi pendaraban. 427 890 15 600 25 = mengikut susunan operasi. bentuk mengufuk 12 745 + 20 742 56 = bentuk menegak Guru mengemukakan masalah (ii) Mengira masalah operasi secara lisan, contoh, dalam bergabung melibatkan bentuk nombor atau ayat penolakan dan mudah. pembahagian. Guru membimbing murid (iii) Mengira masalah operasi Contoh operasi bergabung menyelesaikan masalah bergabung yang melibatkan melibatkan tanda kurung. dengan menggunakan empat tanda kurung. langkah Model Polya. a) (1050 + 20 650) 12 = (iv) Menyelesaikan masalah b) 872 ÷ (8 4) = operasi bergabung nombor c) (24 + 26) (64 14) = hingga 7 digit. 6
22. 22. Bidang Pembelajaran: TAMBAH PECAHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN1 Menambah tiga nombor Menunjukkan penambahan (i) Menambah tiga nombor Contoh penambahan tiga nombor bercampur bercampur yang menggunakan nombor bercampur yang sama nombor bercampur yang pecahan setara penyebut pecahannya bercampur melalui penyebut pecahannya sama penyebut pecahannya sehingga 10. 1) aktiviti melipat kertas hingga 10. hingga 10. bentuk termudah 2) carta pecahan 3) rajah 3 7 17 2 3 1 2 7 jadual sifir 4) garis nombor 5) jadual sifir Murid membina cerita daripada (ii) Menambah tiga nombor Contoh penambahan tiga ayat matematik melibatkan bercampur yang penyebut nombor bercampur yang nombor bercampur. pecahannya tidak sama penyebut pecahannya tidak hingga 10. sama hingga 10. 2 3 16 2 4 1 1 1 Beri jawapan dalam bentuk termudah. Guru membimbing murid untuk (iii) Menyelesaikan masalah menyelesaikan masalah melibatkan penambahan menggunakan konsep langkah nombor bercampur. pengiraan berikut: 1) Memahami masalah dan mengumpul maklumat 2) Menentukan cara penyelesaian 3) Melaksanakan 4) Menyemak jawapan. 7
23. 23. Bidang Pembelajaran: TOLAK PECAHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN2 Menolak nombor Tunjuk cara penolakan nombor (i) Menolak tiga nombor Contoh penolakan tiga nombor bercampur bercampur yang bercampur melalui bercampur yang sama nombor bercampur yang pecahan setara penyebut pecahannya penyebut pecahannya sama penyebut pecahannya sehingga 10. 1) aktiviti melipat kertas sehingga 10. sehingga 10. bentuk termudah 2) carta pecahan 3) rajah 5 5 15 15 4 2 1 jadual sifir 4) garis nombor 5) jadual sifir Murid membina cerita daripada (ii) Menolak tiga nombor Contoh penolakan tiga ayat matematik melibatkan bercampur yang penyebut nombor bercampur yang nombor bercampur. pecahannya tidak sama penyebut pecahannya tidak sehingga 10. sama sehingga 10 7 8 3 4 12 7 1 1 Beri jawapan dalam bentuk termudah. Mendedahkan kepada murid (iii) Menyelesaikan masalah masalah harian dalam bentuk melibatkan penolakan nombor bercampur 1) ayat 2) jadual 3) gambarajah 8
24. 24. Bidang Pembelajaran: DARAB PECAHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN3 Mendarab sebarang Menggunakan pelbagai bahan (i) Mendarab nombor bercampur Mendarab nombor nombor bercampur nombor bercampur seperti petak seratus untuk dengan nombor bulat. bercampur dengan nombor cerakinkan dengan nombor bulat mendarab nombor bercampur. bulat sebagai pengumpulan sehingga 1000. objek. bentuk termudah Contoh: Contoh: 2 2 100 ? 1 3 3 300 sama dengan 3 3 1 1 kumpulan objek daripada 300. Sepatutnya kita mempunyai 100 objek dalam satu set. Maka, 2 kumpulan akan mengandungi 200 objek; 2 100 = 200. Oleh itu, 1 2 2 kumpulan akan mengandungi Tunjukkan pengiraan menggunakan langkah yang 2 2 100 250 objek. 1 mudah dan teratur. 5 Hadkan komponen nombor 2 2 100 1 100 bulat daripada nombor 2 bercampur, kepada tiga digit. 5 50 Penyebut pecahan nombor 1 bercampur hendaklah 250 kurang daripada 10. 9
25. 25. Bidang Pembelajaran: BAHAGI PECAHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN4 Membahagi pecahan Guru menunjukkan bagaimana (i) Bahagi pecahan dengan Hadkan penyebut bagi dengan nombor bulat membahagi pecahan dengan nombor yang dibahagi a) nombor bulat dan pecahan. pecahan lain sebagai kurang daripada 10. b) pecahan perkongsian. Ilustrasi berikut menunjukkan contoh tersebut. Hadkan pembahagi kurang daripada 10 bagi kedua-dua 1 2 1 1 nombor bulat dan pecahan. 2 Contoh membahagi pecahan Setengah cecair daripada satu dengan pecahan…. bekas dituang ke dalam bekas separuh akan mendapat satu bekas separuh yang penuh. 1 4 1 2 4 2 2 2 1 1 4 2 1 1 1 1 2 1 3 4 1 2 1 2 1 atau 4 0 1 2 1 1 4 4 2 1 2 2 1 4 2 1 1 2 2 1 1 2 10
26. 26. Bidang Pembelajaran: BAHAGI PECAHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN Setengah daripada cecair di (ii) Bahagi nombor bercampur 1 2 dalam bekas dituang ke dalam dengan 1 1 4 dua bekas satu perempat akan 4 2 1 2 2 a) nombor bulat mendapat dua bekas satu 1 perempat penuh. b) pecahan 2 1 1 2 1 3 atau 4 1 2 1 4 1 2 1 4 1 2 2 1 4 2 1 1 2 1 2 0 11
27. 27. Bidang Pembelajaran: OPERASI BERGABUNG MELIBATKAN PERPULUHAN Tahun 6 OBJEKTIF CADANGAN AKTIVITI HASIL PEMBELAJARAN PERBENDAHARAAN PEMBELAJARAN PENGAJARAN DAN CATATAN Murid dapat… KATA Murid akan diajar… PEMBELAJARAN1 Melaksanakan operasi Murid menambah dan / atau i) Menambah dan menolak 3 Contoh operasi bergabung nombor perpuluhan bergabung melibatkan menolak 3 hingga 4 nombor hingga 4 nombor perpuluhan melibatkan perpuluhan. tempat perpuluhan tambah dan tolak perpuluhan. hingga tiga tempat 0.6 + 10.2 – 9.182 = nombor perpuluhan perpuluhan melibatkan: Contoh: Melaksanakan satu sehingga tiga tempat 8.03 – 5.12 + 2.8 = operasi pada suatu masa dari a) nombor perpuluhan perpuluhan kiri ke kanan. Langkah sahaja. 126.6 – 84 + 3.29 = pengiraan ditunjukkan dalam b) nombor bulat dan atau bentuk lazim. nombor perpuluhan. 10 – 4.44 + 2.126 – 7 = Abakus boleh digunakan untuk mengesahkan ketepatan dalam 2.4 + 8.66 – 10.992 + 0.86 = pengiraan. 0.6 + 0.006 +3.446 – 2.189 = Contoh pengiraan untuk operasi bergabung melibatkan perpuluhan. 126.6 + 84 – 3.29 = ? 1 2 6 . 6 + 8 4 2 1 0 . 6 3 . 2 9 2 0 7 . 3 1 12 | 8,011 | 24,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-30 | latest | en | 0.284858 |
https://hsm.stackexchange.com/questions/6751/how-did-euler-stumble-on-this-proof | 1,576,414,923,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00502.warc.gz | 387,675,025 | 29,644 | # How did Euler stumble on this proof?
Euler proved $n=641$ divides $2^{32}+1$ by noting $n=5^4+2^4=5\times 2^7+1$ so $$2^{32}\equiv-5^4\times 2^{28}=-(5\times 2^7)^4\equiv-1\,(\text{mod}\, n).$$How did he happen upon this realisation? One possibility is he tried long division until he saw $641|2^{32}+1$, then saw if $641$ could be expressed in terms of powers of $2$ in a way that explains why this would be so. Another is that he realised he could prove $2^{32}+1$ is composite if he found an equation of the right form, then solved some Diophantine equations. Is it known what happened historically?
• Euler's paper: eulerarchive.maa.org/docs/originals/E026.pdf -- according to the summary at the Euler Archive, the paper gives no hint how the factorization was discovered. – Michael E2 Dec 3 '17 at 18:38
• Here is one answer: maa.org/sites/default/files/pdf/editorial/euler/…; See E134, p. 9 -- I felt a heuristic might be that if $p \mathrel{|} 2^{2^5}+1$, then since $2^{2^6} \equiv 1 \ (p)$, one might first look for $p$ for which $2^6 \mathrel{|} p - 1$. Euler proved the equivalent in a later paper and used it to explain how he found 641. – Michael E2 Dec 3 '17 at 19:44
• Basically, he was very smart and all sorts of "intuitions" came to him that would not come to us mere mortals. – Carl Witthoft Dec 5 '17 at 14:23
• @CarlWitthoft I think MichaelE2's second comment provides a deeper explanation than that. – J.G. Dec 5 '17 at 14:31
• @MichaelE2 Comments go away, so you should post an answer. I particularly want to know which result came first, Euler's factorization of $F_5$ or his proof of Fermat's Little Theorem. – Spencer Oct 16 at 17:53 | 521 | 1,663 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-51 | latest | en | 0.91345 |
https://oeis.org/A337145 | 1,675,148,725,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00675.warc.gz | 448,215,967 | 4,316 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A337145 a(n) is the determinant of the 2 X 2 matrix whose entries (listed by rows) are the n-th prime congruent to 1, 3, 5, 7 mod 8 respectively. 3
104, 800, 1712, 2592, 3760, 4840, 5728, 12848, 15664, 18424, 20888, 23520, 28232, 28560, 25320, 30280, 37248, 50520, 43680, 33664, 61560, 73920, 70544, 57696, 38696, 27408, 79280, 63392, 107328, 109536, 162608, 172296, 187352, 197040, 248064, 228320, 215912, 229152, 255480, 231304, 286408, 256320 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The first negative term is a(20750) = -58207896. All terms are divisible by 8. LINKS Robert Israel, Table of n, a(n) for n = 1..30000 EXAMPLE The first primes == 1, 3, 5, 7 (mod 8) are 17, 3, 5, 7 respectively, so a(1) = 17*7 - 3*5 = 104. The second primes == 1, 3, 5, 7 (mod 8) are 41, 11, 13, 23 respectively, so a(2) = 41*23 - 11*13 = 800. The third primes == 1, 3, 5, 7 (mod 8) are 73, 19, 29, 31 respectively, so a(3) = 73*31 - 19*29 = 1712. MAPLE R:= NULL: L:= [-7, -5, -3, -1]: found:= false: for k from 1 to 100 do for i from 1 to 4 do for x from L[i]+8 by 8 do until isprime(x); L[i]:= x; od; v:= L[1]*L[4]-L[2]*L[3]; R:= R, v; od: R; CROSSREFS Cf. A335581, A335592, A337146, A337147. Cf. A007519, A007520, A007521, A007522. Sequence in context: A206021 A206014 A106298 * A230028 A340898 A132434 Adjacent sequences: A337142 A337143 A337144 * A337146 A337147 A337148 KEYWORD sign,look AUTHOR J. M. Bergot and Robert Israel, Jan 27 2021 STATUS approved
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Last modified January 31 01:17 EST 2023. Contains 359947 sequences. (Running on oeis4.) | 767 | 1,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-06 | latest | en | 0.645966 |
http://forum.wolframscience.com/printthread.php?threadid=1680&perpage=2 | 1,368,915,956,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382917/warc/CC-MAIN-20130516092622-00028-ip-10-60-113-184.ec2.internal.warc.gz | 109,835,415 | 2,275 | A New Kind of Science: The NKS Forum (http://forum.wolframscience.com/index.php)
- Applied NKS (http://forum.wolframscience.com/forumdisplay.php?forumid=4)
-- How many alternate universe are there? (http://forum.wolframscience.com/showthread.php?threadid=1680)
Posted by David Brown on 07-24-2009 09:51 PM:
How many alternate universes are there?
Three centuries ago science was transformed by the dramatic new idea that rules based on mathematical equations could be used to describe the natural world. My purpose in this book is to initiate a new kind of science that is based on the much more general type of rule that can be embedded in simple computer programs. - Stephen Wolfram, NKS
Are Newton, Einstein, and Wolfram birds of a feather? I say: Hell, yes! Let's drink the Kool-Aid!
Is there a multiverse or merely one universe? Does the multiverse contain infinitely many separate, alternate universes?
Suppose that the multiverse is a Fredkin-Wolfram information process that computes M-theory. Suppose that the ultimate gauge group has N dimensions. Could the ultimate information process simply be mathematically isomorphic to an N-dimensional register machine? Specifically, could the answer be an N-dimensional counter machine (simplest type of register machine) that performs FWC number of updates for each Planck time interval? Here, FWC is the Fredkin-Wolfram constant. The physical interpretation would be that there are precisely FWAU alternate universes, where FWAU is the number (2**N) raised to the power FWC. This immense number of universes would be arranged in an N-dimensional geometry.
Are the alternate universes separated into matter/antimatter pairs by informational antielectromagnetons? Are the vast multitudes of alternate universes saved from collapse by informational antigravitons?
Without paradigm-breaking photons, so-called "informational antigravitons" and "informational antielectromagnetons" would basically be cheating schemes based on alternate universes. There would be hyper-dimensional hidden variables that would explain any dataset. But if paradigm-breaking photons do exist, then M-theory would be the way to start computing the intricacies of ultra-high-energy cosmic rays.
Posted by Richard J. Gaylord on 12-03-2010 06:46 PM:
there are 56 altertnative universes - 57 if you count the new one formed as a r esult of this response.
__________________
"I find it wholesome to be alone the greater part of the time. To be in company, even with the best, is soon wearisome and dissipating. I love to be alone. I never found the companion that was so companionable as solitude.". - H.D. Thoreau
"Correlation is not Causation and Big Data is not Science" - R.J. Gaylord | 615 | 2,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.866013 |
https://egtheory.wordpress.com/2013/11/06/wrong-models/?replytocom=4088 | 1,601,427,023,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00532.warc.gz | 373,769,665 | 51,913 | # Are all models wrong?
George E. P. Box is famous for the quote: “all models are wrong, but some are useful” (Box, 1979). A statement that many modelers swear by, often for the wrong reasons — usually they want preserve their pet models beyond the point of usefulness. It is also a statement that some popular conceptions of science have taken as foundational, an unfortunate choice given that the statement — like most unqualified universal statements — is blatantly false. Even when the statement is properly contextualized, it is often true for trivial reasons. I think a lot of the confusion around Box’s quote comes from the misconception that there is only one type of modeling or that all mathematical modelers aspire to the same ends. However, there are (at least) three different types of mathematical models.
In my experience, most models outside of physics are heuristic models. The models are designed as caricatures of reality, and built to be wrong while emphasizing or communicating some interesting point. Nobody intends these models to be better and better approximations of reality, but a toolbox of ideas. Although sometimes people fall for their favorite heuristic models, and start to talk about them as if they are reflecting reality, I think this is usually just a short lived egomania. As such, pointing out that these models are wrong is an obvious statement: nobody intended them to be not wrong. Usually, when somebody actually calls such a model “wrong” they actually mean “it does not properly highlight the point it intended to” or “the point it is highlighting is not of interest to reality”. As such, if somebody says that your heuristic model is wrong, they usually mean that it’s not useful and Box’s defense is of no help.
On the opposite end of the spectrum are abstractions, these sort of models are rigorous mathematical statements about specific types of structures. These models are right and true of their subjects in any reasonable definition of the words. They are as right or true as the statement that there are infinite number of primes; or that in Euclidean geometry, the tree angles of a triangle sum to two right angles. When somebody says that an abstraction is wrong, they mean one of two things:
1. It is mathematically false. For example: if I say that given the first n primes $p_1,...,p_n$, their product plus one ($N = p_1p_2...p_n + 1$) is prime then that statement is false. It was a statement that was no made carefully enough (N is not divisible by any of $p_1,...,p_n$, but could be a composition of other primes not on your list).
2. Or, the structure you are applying it to does not meet the requirements of the abstraction. For example, in general relativity, space is non-Euclidean, so triangles don’t sum to 180 degrees. More concretely, if you are measuring triangles on the surface of the Earth then at large scales you can have triangles that sum to more than 180 degrees, even if on local scales Euclidean geometry is a good approximation.
In the first case, your model is not actually an abstraction since it is not valid on all structures meeting its prerequisites (but maybe you can convert it to a heuristic). In the second case, your critic is actually saying that your model is not useful, and Box’s defense is again of no help.
This brings us to models in physics and, more generally, to insilications where all of the unknown or system dependent parameters are related to things we can measure, and the model is then used to compute dynamics, and predict the future value of these parameters. Now, if we look at any currently extant model then just based on our history of model improvements, it seems presumptuous to assume that this model is not wrong. As David Deutsch eloquently wrote at the end of the Beginning of Infinity:
[S]cience would be better understood if we called theories “misconceptions” from the outset, instead of only after we have discovered their successors. Thus we could say that Einstein’s Misconception of Gravity was an improvement on Newton’s Misconception, which was an improvement on Kepler’s.
However, the question remains: is it conceivable that there is a perfect infallible model of the universe? Is the world fundamentally mathematical and comprehensible? As long as I can remember, I believed that the world is not inherently mathematical, that mathematics is a reflection of our conscious and there is no reason to expect that it can perfectly capture external reality in any sense of the word. In fact, I have always asserted that assuming the opposite is anthropically arrogant.
I think that many people agree with this sentiment, but some do so for the wrong reason. This wrong reason is usually a reiteration of On Exactitude in Science — Jorge Luis Borges meditation on the map-territory relationship:
In that Empire, the Art of Cartography attained such Perfection that the map of a single Province occupied the entirety of a City, and the map of the Empire, the entirety of a Province. In time, those Unconscionable Maps no longer satisfied, and the Cartographers Guilds struck a Map of the Empire whose size was that of the Empire, and which coincided point for point with it. The following Generations, who were not so fond of the Study of Cartography as their Forebears had been, saw that that vast Map was Useless
Loosely, the argument is that a perfect model of the universe has to incorporate every detail of the universe and thus must necessarily be the universe itself. I don’t think this is a valid argument because the map analogy for insilications is flawed, and there might even be a deep reason for this flaw. In particular, if we concentrate on Tarski’s elementary geometry then our theory is consistent, complete, and decidable and thus not capable of any deep self-reference. If we lived in world that was perfectly captured by such a formal system then the perfect model would in fact be a copy of the universe. But what reason would we have for assuming that if the world was mathematically structured, it would be such a simple structure and not one capable of recursion?
The point of a model (perfect or otherwise) is to provide a set of rules that can then be instantiated, and mathematics definitely allows deeper models than elementary geometry. Most rich structures are capable of self-reference, and encoding their own rules: this is a basic observation on which Godel’s incompleteness theorem, or universality of computation rests upon; it can also serve as a launching point for fun programming concepts like quines. These systems are capable of describing themselves completely without any error or trivial mile-for-mile representation. Hence, we can’t use the map-maker argument to defend our rejection of a perfect model of the universe.
Building on the theme of no unnecessary assumptions about the world, @BlackBrane suggested on Reddit a position I had not considered before (and that prompted this blog post) for entertaining the possibility of a mathematical universe:
[Box’s slogan is] an affirmative statement about Nature that might in fact not be true. Who’s to say that at the end of the day, Nature might not correspond exactly to some mathematical structure? I think the claim is sometimes guilty of exactly what it tries to oppose, namely unjustifiable claims to absolute truth.
Is believing that the universe is inherently not mathematical as presumptuous as believing that it is? I agree that we definitely should not rule it out on ethical grounds before we explore it more completely. Believing that the universe is inherently mathematical does give a lot of hope and courage to theoretical physicists. However, if we assume that it is, what does this really mean for Box’s aphorism?
If the universe is a formal system then we (and all our thoughts and hence theories) are just derivations in that system. To avoid the map-making argument, we have to assume that this system is capable of self-reference. Unfortunately, Godel showed that such systems are also incomplete: there are true statements about such systems that are not derivable within the system. As such, no theory we are capable of building as entities inside this system can resolve certain questions that we can ask within the system about the system. If a model can’t answer certain questions, is it wrong? Is it useful? Alternatively, if we are beings that evolved in this system then is there any reason to believe that this would necessitate the evolution of mental facilities that are capable of comprehending the system? There is no reason to believe that evolution encourages perfect understanding of the environment, or even rational learning or decision making about the environment. In other words, even if the universe was inherently mathematical, Box’s quote could still be true of achievable insilications.
Finally, this segues to a third option on insilications that is embedded in algorithmic philosophy and stems from my belief that mathematics is a reflection of our consciousness. We can suppose that the external world is inherently not mathematical and thus not fully knowable or comprehensible, but that all our theories or thoughts about it must be. As such, these theories can always be formalized and one of them might be the best possible description of the things-in-itself that we are capable of thinking. We would not be able to comprehend any deviation from such a model, even though these deviations would exist (in some sense of the word). Would such a model be wrong? Of course, as with the discussion on completeness before, there is no reason to believe that even if such an ideal model existed that we could arrive at it. There is also no reason, to favor this view over a mathematical universe, except in that it is assuming limitations of us and not properties of the thing-in-itself.
Box, G. E. P. (1979). Robustness in the strategy of scientific model building. Robustness in statistics.
From the Department of Computer Science at Oxford University and Department of Translational Hematology & Oncology Research at Cleveland Clinic, I marvel at the world through algorithmic lenses. My mind is drawn to evolutionary dynamics, theoretical computer science, mathematical oncology, computational learning theory, and philosophy of science. Previously I was at the Department of Integrated Mathematical Oncology at Moffitt Cancer Center, and the School of Computer Science and Department of Psychology at McGill University. In a past life, I worried about quantum queries at the Institute for Quantum Computing and Department of Combinatorics & Optimization at University of Waterloo and as a visitor to the Centre for Quantum Technologies at National University of Singapore. Meander with me on Google+ and Twitter.
### 44 Responses to Are all models wrong?
1. I think you are missing out on some ideas on complexity. This is strongly revealed in your conclusion: “We can suppose that the external world is inherently not mathematical and thus not fully knowable or comprehensible, …” What makes you think that something mathematical is comprehensible? You already invoked one simple form of incomprehension: undecidability in computing. Why stop there? Certainly, things can get so complex that no single human brain can verify correctness. Sticking to the domain of mathematics, we have, for example, the proof of classification of groups, which is estimated to be 20 thousand pages long. Although it has now been written down, and mostly published, its not been verified. Efforts are underway to encode it in a machine-readable format, so that a computer verification can take place, via “proof theory”. Thus, we have something that is eminently mathematical, and thoroughly incomprehensible.
Things get harder from here. How can one verify that the proof assistant itself is free of bugs? We can’t just run it through another, smaller, simpler program, to verify that its bug-free. So the proof assistant itself has to be hand-checked by humans, which is non-trivial (X number of years to get a PhD in proof theory, Y number of years to read and understand the code in the proof assistant, Z number of years to get so very comfortable with it as to proclaim it bug-free. May require more than one human to accomplish this.)
As to a belief that the universe is not “mathematical”: well, what else could it possibly be? Many mathematicians define mathematics as the sum-total of all possibility; to say that something isn’t mathematical is tantamount to saying it isn’t possible. Since there is nothing else that it could be, by law of excluded middle, it must be. (Yes, there is a very fruitful branch of mathematics that explores what happens when the law of the excluded middle is rejected: all of our best proof assistants come from that theory.)
And then there’s physics: shall we assume that it takes an infinite number of decimal places to model the universe? Surely not: we know that the gravitational attraction of that many bits would create a black hole. Ergo, a finite number of decimal places suffice, which suggests that the universe is finite. And if its finite, doesn’t that mean its describable? Feel free to invoke “Kolmogorov complexity” at this point, and attempt to argue that the shortest description of the universe is equal to the total number of bits in the universe. That may be a fruitful avenue: the question then becomes how we, as humans, experience those bits: as identical atoms, as nearly-identical bits of sand. Or perhaps with an AI slant: how does a neural network convince itself that a proof is correct or incorrect? That a proof assistant is correct or incorrect? That a model is correct, incorrect, or useless?
• Thank you for the comment!
You make a good point about practical complexity, but that is not something that I wanted to address in this post (otherwise I would wander into an unreasonably long tangent on computational complexity). I mean to address the question “in principle” and not as far as “comprehensible by a human” but “comprehensible by humanity”. Of course, both concepts are extremely slippery.
As to a belief that the universe is not “mathematical”: well, what else could it possibly be? Many mathematicians define mathematics as the sum-total of all possibility; to say that something isn’t mathematical is tantamount to saying it isn’t possible.
Just because you can’t imagine something, doesn’t mean that this thing can’t be. Even if something is unimaginable by all of humanity, it still doesn’t necessitate that thing being impossible. That is exactly the point I was making in the last paragraph of my post. Like the mathematicians you describe, I personally define mathematics more or less as the sum-total of all comprehensible possibility. However, I don’t have the anthropic arrogance to then go on and project any limitations of human comprehension onto the external world. That being said, there are definitely plenty of mathematicians and physicists that are willing to do that.
And then there’s physics: shall we assume that it takes an infinite number of decimal places to model the universe? Surely not: we know that the gravitational attraction of that many bits would create a black hole
Please let me know if I am misreading your point, but here I think that you make a series of mistakes that I have seen made many times, so I thought I would point them out:
[1] Nobody said these bits have to be localized in one place. The metric tensor in GR, for instance, encodes information about the global structure of the universe, not just local properties. As such, it can be encoded in all of the universe, and there is no reason to postulate that the universe is of finite extent. Thus, even though locally the information density is finite, it is then summed over an infinite universe to encode the metric tensor. If you don’t allow an infinite universe then you’ve established a much more trivial sounding statement: “if the universe is finite then some quantity about it is also finite”.
[2] You also proceed to assume that the parameters describing the universe have to be physically encoded inside the universe they are describing. Although this is required for us to know them, when we are in a discussion of “are they knowable?”, we can’t simply assume that they are, especially not at any one particular moment in time.
[3] Finally, you mistake the map for the territory. If we want to discuss if the universe is describable mathematically or not, we can’t base that discussion on the properties of our current mathematical description of the universe, because then we are begging the question. Of course, this does make the discussion very difficult, but that’s the point! If it was easy then it wouldn’t be fun to think about.
• Hi Artem,
I’ll try to be brief. I think that “practical complexity” also puts bounds on what’s possible “in principle”. This is not an idle remark; you seem to imply that something can hold “in principle” even if it takes an extremely large number of steps to get there “in practice” as if “in principle” is a kind-of extrapolation to infinity from the finite. Let me remark that this kind of extrapolation is very non-trivial; it is only safe to make in high-school pre-calculus and a few other cases, where a “limit” is narrowly defined. In the other cases, practical complexity dominates, and what happens “in principle”, at the limit, is an entirely different “interpretation” (in the sense of ‘model theory’).
Regarding “knowable by humanity”: But humanity is of finite size, too: there are only so many brains on this planet, and each brain is finite in size. The knowable cannot exceed this capacity. We may someday build computer brains that are larger, but they will remain finite in size. There will always be problems that are more complex than what something of finite size can comprehend (or so is my belief: as otherwise one gets a strange situation, where, once you are smarter than a certain level, you can then know everything, ever!? Crazy-talk.)
Re: your point [1], I am confused. The universe is not infinite! It consists of about 10^80 atoms grand total, as measured by astronomers. There is no such thing as “infinity” in physics. Perhaps our knowledge of physical laws is incomplete, but there is no shred of evidence that anything physical could ever be “infinite”.
Regarding the confusion of the map and the territory: Well, yes, this is confusing. But I think you duck another question of principle: what does it man “to know something?” In practice, it means “something that humans can articulate and discuss”. Perhaps there is a better definition, one that could be applied to potential future super-intelligences. But, ultimately, there is a constraint: any super-intelligence will be limited to the 10^80 atoms in the universe available for conversion into a computer (unless Nick Bostrom is right ..) So once again, issues of “practical complexity” dominate.
2. Hmmm, no edit-button to correct my post. Some footnotes, then:
* Proofs are, by definition, decidable. Thus, the proof of the classification of groups is decidable; its simply incomprehensible by a single human, although individual humans do understand parts of it. But this is possible only because our economic system is just large enough to allow a sufficient number of mathematicians to do so. Economics and politics matter.
* Box’s quote is kind-of the mirror image of Kolmogorov complexity, which states that a model is useful only if it is smaller than the thing being modelled, and, what’s more, that there are things that cannot be modeled.
* Cryptography certainly demonstrates the frailty of arguments based on Kolmogorov complexity: encrypted messages are incomprehensible in a certain very strong sense. Based on naive Kolmogorov complexity, an encrypted message is incompressable, and the shortest description of that message is a repetition of the bits in that message. But if you know the secret… then all change.
* “How do we, as humans, experience those bits?” I forgot to ask: “Do we experience those bits holographically, via AdS/CFT?” The point being that the nature of experience is hardly obvious, never mind the philosophical arguments about qualia. We experience the universe by means of the models constructed within our brains, including the rudimentary models in our motor cortex that subconsciously deal with the position of our bodies, and any nearby flying footballs.
* More to the point: “How do we, as humans, experience those bits?” Do we perhaps live in a cryptographic universe, where some knowledge is hidden away in such a way that we need to “break the code” to understand it? (viz, break the code, to create a model of it? In a certain sense, isn’t that what a model is: a decypherment of the messages of the universe?) Perhaps there is a much, much shorter proof of the classification of groups, but it will require more than 10^160 bits of computing effort to obtain it? Which is more bits than there are in the universe?
3. Marc Harper says:
Great post Artem! This is one of my favorite things to think about. IMO, all scientific models are “wrong” for the simple reason that we have no axioms for The Universe. Sure there are many things that science believes to be universally true, like causality, but we don’t really know, and can’t ever know. There are recent indications that entanglement occurs in time as well as in space. People once thought Newton has discovered the universe but we’ve seen so much more since then.
Every model — indeed every idea in every area of human thought — is provisional. Each notion has a limited domain of applicability and is predicated on assumptions. No assumption is universally valid, and even if it were, we can never prove that it is, regardless if the universe is mathematical or not.
Scientific models, however, follow an evolutionary paradigm — models are “less wrong” over time (say on average, built-to-be-wrong models and people’s pet models notwithstanding), in that they fit all the available data better. They make better predictions — they increase our prediction power over time. So they are less wrong in the sense that our predictions are more accurate and precise. Those models get replicated and extended. They mutate into new models that proliferate better. Then we poke at the mathematical holes in the latest theories, looking for experiments to break the models. Sometimes the model is so good that it points to something new, like when Dirac predicted the positron. Sometimes the landscape changes and a new type of model emerges from the void. But a model never becomes “right”, it just aspires to “fit all known data”.
As a mathematician, I don’t consider proofs to be models in this sense. Given the assumptions, certain logical outcomes are valid, full stop (assuming the proofs are correct). Whether the assumptions are really true is another story altogether. To me modeling and proving things about models are complementary activities, and there is an art to both.
I like to think about evolution as an automatic or default process. Some things proliferate better than others, and in doing so, encode information about the environment into their structure, be it a population distribution or an internal structure like DNA. But there’s nothing that says that evolution as a natural process necessarily leads to a maximal conversion of environmental information over time. Real replicators seems to act more like dumb replicators than best-reply-computing smart agents. That’s partly why I find the relationships between information theory and the replicator dynamics so interesting!
It is also interesting to consider the possibility of an ideal model that explains Everything (TM). All things that replicate do so with variation, and if some hypothetical replicator did not naturally diversify, eventually another replicator will out compete it, unless the unvarying one were somehow “perfect” already. This suggests that evolution will never stumble upon the “ideal model”, or the ideal model doesn’t participate in the evolutionary process. Or perhaps we live in the part of the universe that evolutionary things live in, and the ideal model lives in some other slice of the universe, where maybe everything isn’t finite, and Godel doesn’t apply. In the end, I suspect this is a timeless philosophical question — there will always be things we don’t know and always people wondering about if there is some model that explains the unknown or ties together the known.
Thanks for posting!
• Extremely good points. I am a big fan of evolutionary explanations of culture, so I am inherently happy with your evolutionary theory of scientific progress. The frightening thing about evolutionary processes (and other myopic optimizers) is local maxima! But I guess we do see them in the history of thought, I guess Kuhn’s paradigm shifts would correspond to crossing small fitness valleys.
I think the sociology of scientific progress is fun to think about, but not exactly what I wanted to touch on in this post. In your framework, I think what I was asking was: does a global fitness peak exist? If it does exist, is this a consequence of the universe or the capacities of humanity?
I don’t think I understood your last paragraph, though. In particular:
All things that replicate do so with variation, and if some hypothetical replicator did not naturally diversify, eventually another replicator will out compete it, unless the unvarying one were somehow “perfect” already. This suggests that evolution will never stumble upon the “ideal model”, or the ideal model doesn’t participate in the evolutionary process.
I don’t see how the second sentence follow from the first. The ideal model could be present (or could not; I am not advocating that it is, I am just asking that the logic of the argument be clarified) and then we would just get stuck around the fitness peak. Not currently being at a peak is no argument for it not existing, as I tried to show before, just because (even local) peaks exist, doesn’t mean that we can get to them.
i agree that the relationship between information theory, computational complexity, learning, and evolution is very interesting! I will actually be emailing you about this soon, but that’s a topic for another day.
• Marc Harper says:
I certainly agree with your point — not being at the peak doesn’t imply that the peak fails to exist, and you are correct that my second sentence doesn’t follow as an implication of the first. I was trying to say that an “ideal model” would have no need for a variation and selection process (since it’s by definition the pinnacle of models), and it ideally would not vary at all, since that would only cause the model to drift from the peak. So if the ideal model was reached by an evolutionary process, it/we would have to taper off its variation somehow, assuming we had a way of knowing it had reached the peak, which seems unlikley; if this assessment was made prematurely, as has happened many times in the history of human knowledge, some other model with variation would eventually overtake). Or we would have to allow a “quasispecies” of near-ideal models to be synonymous with the ideal model, but the we’d not really be at the peak, and there would be some chance of drifting away.
Perhaps I’ve stretched the analogy a bit too far and I’m not working with precise definitions, so I’ll stop here…
• Sergio Graziosi says:
Quote: “Every model — indeed every idea in every area of human thought — is provisional. Each notion has a limited domain of applicability and is predicated on assumptions. No assumption is universally valid, and even if it were, we can never prove that it is, regardless if the universe is mathematical or not.
[…]
They make better predictions — they increase our prediction power over time.”
Thanks Marc, this is essentially what I’m exploring over there: http://wp.me/p3NcXb-1d (warning: long read into multiple posts).
Of course all models are wrong: can we measure anything without approximation? No. Can we include all variables into a model without approximating their value? No. Can we proof that we know and can measure all variables? No.
Models are approximations, hence are inherently guaranteed to be somewhat wrong.
What I find interesting is observing that the usefulness of a model is usually negatively correlated with its precision. Silly example: more or less all of us rely on a simple physics model that predicts that if we let go an object in mid air, the object will fall to the ground. Useful, but fantastically inaccurate, if one looks at the whole Universe, the places where this “prediction” is reliable are tiny minority. But that doesn’t matter, because the simple model is reliable where we happen to spend our lives.
So yeah: all models are wrong, but we can still use them and find better ones. A model is a useful tool, nothing more.
On other discussion here: models don’t need to represent reality in a “true” way. For example, Linas says: “The universe is not infinite! It consists of about 10^80 atoms grand total, as measured by astronomers.”
Uh, yeah, but how can we tell that there is no cosmic field that isolates us from another even larger pool of atoms and from all their measurable effects? We can’t. But we don’t need to care, if 10^80 atoms are all we have access to, we’ll deal with them alone, but saying that therefore the Universe is finite is taking it a bit too far. You are confusing the useful model with reality. Something lots of scientists do, and shouldn’t, IMHO.
Thanks to Artem for the thought provoking post!
• Thank you for the comment!
I feel like the main point of my post is being missed whenever I see these definitive answers:
Of course all models are wrong: can we measure anything without approximation? No. Can we include all variables into a model without approximating their value? No. Can we proof that we know and can measure all variables? No.
Models are approximations, hence are inherently guaranteed to be somewhat wrong.
What I find interesting is observing that the usefulness of a model is usually negatively correlated with its precision.
This is a very interesting point when it comes to heuristic models. Some of the trade off is trivial: (1) a more precise model is usually more complex and hence harder to use, and (2) the ‘increases’ in precision usually come in domains that the preceding model didn’t care much about, and the reason the preceding model didn’t care much about it is because that domain matters in fewer settings (for instance, this is why engineers use classical mechanics and not quantum mechanics when building bridges), but some trade offs are related to things like generalization and over-fitting. Often placing external restrictions on your models makes them better, even if those restrictions don’t have a particularly sound basis.
Linas says: “The universe is not infinite! It consists of about 10^80 atoms grand total, as measured by astronomers.”
I think that he was mistaken “observable universe” for “universe” here. I didn’t continue that thread because I think we are talking past each other there, and because I find the “but its finite” a particularly boring direction for arguments. In particular, the other part of Linas argument relied on Kolmogorov complexity (which he also seems to misinterpret in his discussion of cryptography, assuming that he is referring to things beyond one-time pad) which is defined in terms of Turing Machines. If you go the ultra-finitary root then you have to throw away TMs (since they only make sense in idealizations where you assume arbitrarily large integers are allowed) and replace them by DFAs or BDDs and build an information theory on top of those.
• Sergio Graziosi says:
(trying the ‘blockquote’ tag as simple text, let’s hope it’ll work). Quote:
The whole point of my post (apart from getting me out of a no posting slump) was to point out that one shouldn’t be so quick to make universal assertions. Philosophy is only fun if we try to give respect to all reasonable arguments and explore how far they can be taken. Asserting definitely “No” to the answers you ask just seems close minded.
Oh, apologies for not “getting it” and thanks for the clarification. I see your point of the fun of Philosophy (and the parallel with mathematics is neat). Within the model I’m trying to build (and testing it along the way, thanks for the chance!) I can see where the close mindedness impression comes from, but I can fit it in my view without effort.
Because models make sense only in the context on how we may use them (your mention of engineers and classical mechanics being an obvious case), you are absolutely right that definitive answers spoil the fun, but in my case fun is merely a necessary side-effect (necessary because I wouldn’t be writing this if it wasn’t fun), not the main reason why I’m trying to build my own model.
Also, my kicks come from the realisation that even after asserting a definitive “no”, I can still find out that it’s possible to know something (it’s post three in the series, I believe) and to what ultimately led to sketch my own scientific epistemology, and my own approach of the demarcation problem (that you’ve read & commented already, thanks!). All of the above is philosophically interesting to me and in general accordance to what is being discussed here (absolute claims can only be made about conceptual entities).
We agree on model trade-offs between precision/usefulness and on your assessment of the other comments, so much so that now I fear I’ll become boring! ;-)
• Heh, The nice thing about a physics education is that you learn how to detect sloppy thinking and cranky ideas. One is instilled with a certain hard-headed pragmatism that is often lacking in other disciplines. There’s a reason for this: when dealing with models in physics, either they work, or they don’t, in a pretty clear-cut fashion; nature is a harsh mistress. It is very different in other disciplines, such as biology, or anything dealing with complex systems, where there is a vast abundance of mostly-good-enough models to choose from.
So, for example: trying to insist that the universe is “infinite” is a typical crank-theorist hobby. Basically, its total bullshit. We can see the universe, and we can measure what we see. Insisting that there might be “something more” than what can be seen is just nuttiness: its a belief in ghosts, a belief in the existence of something that can’t be seen, can’t be proven. Now, if you are a credentialed physicist, it is OK to develop models based on infinite universes, just to see how they might work out: at the heart of theoretical exploration is the need to explore sometimes fantastic-sounding, cranky directions, just to see if they work. But unless you can communicate your ideas in a language that other physicists will accept, you will be labelled a crank. The conception of an infinite universe is an example of crankiness.
There are only two exceptions to the above: one for philosophers, and one for mathematicians. Philosophers are allowed to grapple with totally crazy ideas, because it is their job to convert vague, unclear, confusing issues into something sufficiently clear that ordinary science can take over. The exemption for mathematicians is made because they have the tools for talking about infinities, via axioms such as CH, Martin’s axiom, large cardinals, and the mechanics of “model theory”, which show how to work with “non-standard interpretations”. To insist on infinity outside of this context is .. crazy talk.
Well, but that’s OK, since I took Artem’s original post to be philosophical hand-waving and vague ideas about poorly-defined topics, and not some kind of rigorous development. But, as physicists know, you should only incorporate crazy ideas into your models if they actually make the models better, stronger and more accurate. If they don’t, then crazy ideas are will only clutter up your mental space, and prevent you from perceiving reality as it is.
4. Boris Borcic says:
Strikes me most here the name of Box himself – that name works as the IMO most concise allusion thinkable to the idea of reforming the understanding classical-physics-turned-elementary persists to inspire successive generations of newcomers, that physics models should conceptually BOX the isolated systems they model – hold them like prisoners…
…reforming that understanding by something like an inversion that transforms almost-perfectly-predictive-models-of-almost-perfectly-isolated-systems into vanishingly-constraining-normative-models-of-vanishingly-connected-systems…
…and transforms the induced thirst for unification into a reappraisal of perturbative modelling.
(Hum, at the first failing attempt to post I discover that a profuse discussion took place in comments since I’d opened the page in a tab – posting anyway, please forgive serendipitous interferences as may occur)
5. James Rose says:
I apologize in advance for taking a simplistic turn – (though my interest is in Unified Theory of Behaviors (versus ‘unified field theory’) not simplistic at all) – – But I pose a question:
Here are two equations:
E=mc^2
F=ma
Are they each ‘models’? Are they ‘theories’? They are both testable, with one easier and more accessible than the other, but within test-condition-variables and statistical acceptability,
do we really want to say, as Box has, that “all models are wrong”? He is suggesting the extrapolation : “all equations are wrong”. Is that that where we really want to go?
• linasv says:
A single equation is like a factoid: “48% of everyone are conservative” (0.48*E=C): rather meaningless without the extensive accompanying corpus that explains what “everyone” and “conservative” mean. In essence, then, a model can only be a large, mostly-consistent network of facts that conveys some sort of meaning, and approximates some sort of truth. Since anything we could ever say is an approximation of truth, then its self-evident that all models are wrong. However, this begs the question: what is meaning? What do the words “fact, network, truth, approximate, consistent” really mean?
• James Rose says:
Ahh Linas, the mire and muck of ‘semantics’. A truly scary purgatory to explore. But, i suppose,explore it we must. My first observation to your thoughts is — do you really want to trivialize e=mc^2 as a ‘factoid’ ? My next is to share with you my touchstone frame of reference, the linguist Benjamin Whorf ca 1930s “Language, Thought and Reality”:
“We are thus introduced to a new principle of relativity, which holds that all observers are not led by the same physical evidence to the same picture of the universe, unless their linguistic backgrounds are similar, or can in some way be calibrated.”. Information, data, meaning . . indeed require ‘relational calibration/coordination . . sufficient mapping’. And since the universe seems to cross function interrelationally through many levels of organization quite smoothly and nicely, there is at least the hint of a fundamental pervasive and persistent cross calibration basis .. everywhere, all the time. So, the ‘truth’ is out there, waiting to be meaningfully identified (interpreted) . . calibrated, so to speak . . with whatever our capacity to appreciate and understand about it. May I ask you also — you wrote: “Since anything we could ever say is an approximation of truth, then its self-evident that all models are wrong”.
Wow! any error = total error ?!
• Well, E=mc^2 is a factoid in the sense that the ‘actual’ eqn is E^2 = m^2 c^4 + p^2 c^2 and the last is obtained only for p=0. The ‘actual’ eqn itself arises an invariant of the Lorentz transformations, and there is a general theory for such invarients in a general setting of Lie groups and (pseudo-)Riemanain geometry. I forget what kind of invarient it is, maybe a Casimir invariant or something like that. Oh, and the invarience doesn’t hold for off-shell particles in QFT; instead, it appears as the denominator of the propagator (more colloquially, the Green’s function, or Fredholm alternative, depending on your point-of-view). (see for example, wikipedia article for the Klein-Gordon eqn, where you’ll see Einstein’s famous equation with Einstein nowhere in sight.) Well .. and even then, that’s just for perturbation theory. The non-perturbative aspects remain a (hot?) research area. Anyway, that’s the tip of the iceberg of what I remember from grad school.
I read Benjamin Whorf as an undergrad, pretty sure I read all of ‘Language, Thought, Reality’, and remember none of it, at all. I desperately wanted to major in AI because physics was so booooring, but AI was not yet, at the time, something you were allowed to study (at the undergrad level).
So I took the long road, and study semantics these days. And linguistics. And machine learning. And category theory. And type theory and model theory. And get paid to convert some of those high-falutin ideas into crappy computer software. Alas and alack, it is indeed purgatory for some past sins committed. But yes, I suppose I agree with Whorf and Mulder, ‘the truth is out there’.
• James Rose says:
Thank you for the reference frames, Linas. The inter-relationals hint strongly that there is indeed a pervasive way to umbrella all of them together in a meaningful way. More than a unified theory of fields (topologies) . . a unified theory of behaviors (as I identified here previously). Unfortunately, current math is missing critical topos transforms, and, there are 400 years of concepts biases that give priorities to particles and forces without “explaining” them – simply describing them. What if I were to suggest to you that the core similarities (which are seen but incorrectly modeled), are general density/intensity relations -first-. That is, behaviors/least actions are induced by gradient differentials first, and that the ‘fundamental forces’ and thermodynamic entropies are example forms of a priori topological relations that precede them. Currently, conventional theories identify entropy only as thermodynamic, being the impelled product of the ‘forces’. The truth is already here where we can see it: primal geometry – interrelational dimensions – understood for density/general entroepic relations – precedes the fundamental forces and thermodynamic entropies. The 4 & Therm are sample generated forms of General Dimensional Entroepies; The 3 Newtonian Laws are secondary and educable from general entroepic principles as well. BUT, and I write this with a very large grin and smile, This exposition doesn’t work without a key new mathematical argument of information coding~transformation. (hinted at in the fields limits of recent string theory where critical formulae factors reduce to division-by-zero states; a wall of meaninglessness reached).
A total new interpretation of dimensional coding is needed – akin to “where does the wave information go to when wave functions ‘collapse’ “? Critically, there is re-constructable re-constitutable information present, even if a state goes to ‘zero’. Otherwise, how can we rationally talk about a phenomenon called ‘symmetry breaking’, where zero data suddenly equates with vast non-zero measures and relations?!! Doesn’t anyone realize we can’t just do hocus pocus math without explaining and justifying it !?!?.
Please reach me at integrity@prodigy.net if you care to discuss things further. :-)
The fascinating thing about language and semantics is that even the most ethereal intangible memes, words, notions, concepts – can all be traced back to tangible phenomena through the etymological pre-tree. Everything. Whorf rules! He specified the core priori relationship of dimensional topos reality … essential connectivity, translational mappings.
6. Zach M. says:
“However, the question remains: is it conceivable that there is a perfect infallible model of the universe? Is the world fundamentally mathematical and comprehensible? As long as I can remember, I believed that the world is not inherently mathematical, that mathematics is a reflection of our conscious and there is no reason to expect that it can perfectly capture external reality in any sense of the word. In fact, I have always asserted that assuming the opposite is anthropically arrogant.”
This is the passage I was referring to in your ‘Kooky Quantum Mind’ post. Kant made this point, and basically said that Euclidean geometry is a facet is our perceptual apparatus and not necessarily a facet of the noumenal reality.
7. Ramsay says:
I will reply here to your last comment on the G+ post where we started a conversation:
Specifically, I said
> I feel like the universe has some kind of unbounded mathematical ‘depth’ that will never be fully captured by any theory we construct.
And you replied:
“I don’t really understand what you mean by this (or the paragraph it starts). Could you expand on this more? Or point me to somewhere where it is explained more? You write that this is similar to my stance, but I don’t really understand how. For me, I think that theories don’t ‘capture’ but create in a lot of cases.”
I am glad that the verb I used was ‘feel’, rather than ‘think’ (or worse ‘believe’, but I don’t think there is much danger of me accidentally using that). For whatever it may be worth, I will try to expand on that sentiment a bit, but there is no well-formed philosophical stance here, and if the ideas have provenance in somebody’s writing, I have forgotten the source.
I read the post you mentioned https://egtheory.wordpress.com/2014/09/11/transcendental-idealism-and-posts-variant-of-the-church-turing-thesis/
and although a lot went over my head, I liked the ideas of the sensible world and the intelligible world. The sentence that you have asked me to expand on involves the concepts of ‘the universe’ and ‘mathematics’, and unfortunately I can’t think of a satisfactory definition of either. Not a good start. I think definitions are really important, but sometimes they are hard to find, and that might be the whole point. Maybe the universe is the sensible world and mathematics is the intelligible world? (I don’t know the definitions of those either.) Whatever they are, I think they are different. At least they are accessible to us through different windows.
In mathematics there is always a push for deep ‘grand unifying’ theories that relate seemingly distant disciplines into some common abstraction. But nobody expects there to be a Theory of Everything, from which all future mathematics can be derived. I guess there maybe was a time, before Godel, but it isn’t in the culture now. In my undefined language, the world of mathematics has unbounded mathematical depth :-)
However in physics the dream of a Theory of Everything still seems to live. I hope we see a viable Grand Unifying Theory in my lifetime. I am optimistic (for no good reason) that one will be found, but I don’t expect it to live up to the dream of being a ToE. I find it surprising that so many great physicists do still seem to anticipate a ToE. I imagine that the new theory will significantly alter our perspective and enable us to obtain GR and QM as good approximations in the appropriate limits. I guess the hope is that since almost everything can be explained by either quantum mechanics or general relativity, then a theory that encompasses them both should explain everything. This just seems really naive to me. (can I say that without sounding arrogant?) I expect that any new theory will raise fundamental questions that we have not yet even begun to conceive. But they will be questions that we can hope to answer, and so the quest for a grander theory will resume.
I do not really have arguments to support this expectation other than going back to the assertion that there is a distinction between the universe and a mathematical model of it. I should probably read David Deutsch; it seems like his viewpoint would resonate with me. If a ‘mathematical structure’ is something that admits a finite description [that is maybe too limiting: something that is ‘comprehensible’ anyway], then my feeling is that the universe does not correspond exactly to a mathematical structure (although I agree with BlackBrane’s point that this possibility can’t be ruled out). However, by any reasonable definition ‘mathematics’ does not correspond to a mathematical structure either. I do not want to say that mathematics is inherently not mathematical :-) I do not want to say that the universe is inherently not mathematical either.
Your definition of mathematics as the “sum-total of all comprehensible possibility” could work for me if we took a sufficiently broad interpretation of ‘comprehensible’ and agreed that anything capable of comprehension is not capable of comprehending all of mathematics. I want the definition of mathematics to be independent of the human experience of it; I have a Platonic viewpoint. Mathematics is unimaginably huge, and other beings may have a dramatically different experience of it, but if we are able to communicate with them it will only be because we have explored common regions of mathematics. My imagery is more geometrical than algorithmic. I recognise that we ‘create’ things in mathematics, but the biggest things, the deepest things, we find.
So if we say ‘mathematical’ means ‘of or pertaining to mathematics’, then mathematics is mathematical even though it doesn’t correspond to a mathematical structure. I feel that the universe is also mathematical.
I see the universe as an objective reality, independent of our experience of it. Our perception of it has undoubtedly guided our exploration of mathematics. From our perspective, the universe appears to be much more constrained and limited than mathematics, but there is no reason that I know of to believe that it is so simple as to yield to a Theory of Everything. I would want to define the universe in the broadest possible sense, independent of our experience of it. It is the sum total of all objective reality, not just some piece of a ‘multiverse’ (so ‘multiverse’ is just a name for the universe in some cosmological models). I am not just trying to hedge my bets there: obviously we can never hope to have a viable theory of a reality that is completely inaccessible to us in principle. I think that any fundamental theory will reveal cracks in our knowledge which we can explore from our corner of the universe.
I wrote that I thought that my philosophical viewpoint was similar to yours. I should not have said that: I do not have a clear understanding of my own viewpoint, let alone yours. But at least I think we are in agreement with the “all models are wrong” sentiment. Your remark that “theories don’t ‘capture’ but create in a lot of cases”, indicates that there must be significant differences in our outlook, because I have thought about it a bit, and I still prefer the verb ‘capture’. I feel like our insilications give us greater insight into the universe. They may open up our experience to new vistas that we didn’t even imagine before. But these aspects of nature are ‘there’ independent of our discovery of them.
I have enjoyed reading your thoughts on these things. I don’t usually engage in philosophical discussion; I get overwhelmed with a feeling of hopelessness. But this post of yours really resonated with me and maybe brought out some latent philosophical tendencies.
8. Pingback: Are all models wrong? - MetaSD
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A circuit has a voltage of 24 V and a current of 40 mA. What's the power of the circuit? A.6W B. 0.96 W C. 6.24 W D. 9.6 W
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An alternator provides 20 A of peak current at 100 V peak AC to a resistive electric heater. How much heating power is delivered to the load? A 20 W B.5W C. 1000 W D. 500 W
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An alternator provides 20 A of peak current at 100 V peak AC to a resistive electric heater. The heating power delivered to the load is 5 W.
An alternator provides 20 A of peak current at 100 V peak AC to a resistive electric heater. The heating power delivered to the load is 5 W.
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https://codegolf.stackexchange.com/questions/47934/drawing-an-image-based-on-a-trace-of-its-hue | 1,716,014,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00224.warc.gz | 157,220,048 | 44,854 | # Introduction
Local genius Calvin's Hobbies recently challenged us to trace the hue of an image. It is recommended to read his challenge beforehand, but to highlight the main points:
• The trace of hue in an image can be found by starting at one pixel, finding its hue angle(hue = atan2(sqrt(3) * (G - B), 2 * R - G - B)), and moving one pixel's width in that direction. If this process is repeated, and a black pixel is left at each pixel passed, we produce a trace of the image.
Here is a simple example of the trace:
We start at pixel 250, 250, with RGB color 36, 226, 132. We observe that the hue of this color is 150. Thus, we begin drawing a black line 150 degrees counterclockwise from the positive x-axis. We run into the square to the left, with RGB color 100, 220, 7 and hue 94. This results in an almost vertical line to the next square. We continue the pattern into every square, and observe the change in slope at each new color. The length of this trace was 300 pixels. If you're wondering why the image is so ugly, it's because I randomly generated 25 colors in 100x100 squares and used my answer to trace it. On a much more complex and beautiful image, we may get a trace like this one:
• Calvin created this wonderful code snippet on the original post that draws a trace over an image, starting at the point provided by the mouse cursor. This is a very useful testing tool!
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><style>canvas{border:1px solid black;}</style>Load an image: <input type='file' onchange='load(this)'><br><br>Max length <input id='length' type='text' value='300'><br><br><div id='coords'></div><br><canvas id='c' width='100' height='100'>Your browser doesn't support the HTML5 canvas tag.</canvas><script>function load(t){if(t.files&&t.files[0]){var e=new FileReader;e.onload=setupImage,e.readAsDataURL(t.files[0])}}function setupImage(t){function e(t){t.attr("width",img.width),t.attr("height",img.height);var e=t[0].getContext("2d");return e.drawImage(img,0,0),e}img=$("<img>").attr("src",t.target.result)[0],ctx=e($("#c")),ctxRead=e($("<canvas>"))}function findPos(t){var e=0,a=0;if(t.offsetParent){do e+=t.offsetLeft,a+=t.offsetTop;while(t=t.offsetParent);return{x:e,y:a}}return void 0}$("#c").mousemove(function(t){function e(t,e){var a=ctxRead.getImageData(t,e,1,1).data,i=a[0]/255,r=a[1]/255,o=a[2]/255;return Math.atan2(Math.sqrt(3)*(r-o),2*i-r-o)}if("undefined"!=typeof img){var a=findPos(this),i=t.pageX-a.x,r=t.pageY-a.y;$("#coords").html("x = "+i.toString()+", y = "+r.toString());var o=parseInt($("#length").val());if(isNaN(o))return void alert("Bad max length!");for(var n=[i],f=[r],h=0;n[h]>=0&&n[h]<this.width&&f[h]>=0&&f[h]<this.height&&o>h;)n.push(n[h]+Math.cos(e(n[h],f[h]))),f.push(f[h]-Math.sin(e(n[h],f[h]))),h++;ctx.clearRect(0,0,this.width,this.height),ctx.drawImage(img,0,0);for(var h=0;h<n.length;h++)ctx.fillRect(Math.floor(n[h]),Math.floor(f[h]),1,1)}});</script>
• Answers to that challenge may also serve as appropriate testing tools for this challenge.
Note: I do not take any credit for Calvin's work, and have received explicit permission to use his ideas in this challenge.
# The Challenge
In that challenge, we found the trace. Now, I'm asking you to find the image. You will be given a start point, an end point, and a length, and your program must output an image whose trace beginning at the start point arrives at the end point in approximately length pixels. Moreover, because there are so many options, your program's algorithm must be inherently random in some regard--thus the same input should (more often than not) produce a different output image.
# Input
You will receive seven positive integers as input: w, h, x1, y1, x2, y2, len where:
• x1, x2 < w <= 1920
• y1, y2 < h <= 1080
• len < 2000
These guidelines are minimum requirements. Your program may accept higher values.
You may receive input through stdin, a file, or as arguments to a function. You will ONLY receive these 5 integers as input.
# Output
As output, produce an image of width w and height h with a continuous hue trace from x1,y1 to x2, y2 in approximately len pixel-width steps.
• Output should be random.
• Output may be in any format--whether it be .bmp, .png, PPM, or an image shown on-screen at runtime makes no difference.
• You may assume it is possible to move from x1, y1 to x2, y2 in under len pixels.
• Output should NOT show the path of the trace from x1, y1 to x2, y2--though an image showing this path in your answer is appreciated.
# Rules
• This is a popularity-contest! Try to make your output beautiful or interesting. More interesting hue traces than straight lines are also recommended.
• This is not code-golf. As such, readable code is appreciated. Still, try to keep size below ~2KB. This rule will not be strictly enforced.
• Values are approximate. If your code's trace reaches its destination within ~5-10% of len pixels, this is acceptable. Along this same vein, do not worry about slightly incorrect hue angle values. Some error is reasonable.
• Standard Loopholes apply.
Questions, comments, concerns? Let me know!
• I think you should explain a bit more about how the direction of the black line in the image is generated (perhaps with a simpler image?). Explain that the line starts out going left because the hue of light blue is 180°, and curves down because as the blue gets darker the hue angle increases. You are also welcome to repost and even edit the snippet from my question. Mar 17, 2015 at 22:27
• This might work better as a code-golf, though then you'd need to be more specific on what constitutes "randomness". However a pop-con may let people show off some neater ways to make these images. I also think it'd be "purer" to give the image dimensions as part of the input, though this is really a matter of opinion. Mar 17, 2015 at 22:33
• @Calvin'sHobbies Thanks for the feedback! I will definitely try to think of a way to concisely explain the hue. Let me work on it a bit, and I'll see if I can put together an explanation. On making it code-golf, I have one pretty serious problem with that. If you were to take atan((y1-y2)/(x1-x2)) you could quite easily find a "random" color whose hue is this angle. You could then draw a solid line of this color between the points on a white background, divert a bit to account for length, and boom. I really want to promote creativity with this question, and I don't think code-golf does that. Mar 17, 2015 at 22:54
• @Calvin'sHobbies I think you're right about width and height. I'll make it take 7 arguments. Mar 17, 2015 at 22:56 | 1,746 | 6,644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.807714 |
https://www.transum.org/Maths/Exercise/Integration/Default.asp?Level=8 | 1,716,026,040,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00731.warc.gz | 934,280,318 | 14,223 | # Integration
## Exercises on indefinite and definite integration of basic algebraic and trigonometric functions.
##### Menu L1L2L3L4L5L6L7Level 8L9 Exam Help Differentiation
This is level 8 ? Your answer should be a number (rounded to three significant figures if not exact) or a fraction in its lowest terms.
$$\int _1^2 2x(x^2+3)^5 \; \text{dx}$$ = $$\int _0^{\sqrt{3}} x\sqrt{1+x^2} \; \text{dx}$$ = $$\int _1^3 4x(x^2-2)^4 \; \text{dx}$$ = $$\int _2^3 \sqrt{x^3+3x}(3x^2+3) \; \text{dx}$$ = $$\int _0^1 5x \cos{x^2} \; \text{dx}$$ = $$\int _{0.25}^{0.75} \dfrac{1}{(5-4x)^3} \; \text{dx}$$ = $$\int _0^{\frac{\pi}{4}} \sec^2{x}\tan^3{x} \; \text{dx}$$ = $$\int _1^2 x^2 e^{x^3+1} \; \text{dx}$$ = $$\int _{\frac12}^{\frac{\sqrt{3}}{2}} \dfrac{4}{\sqrt{1-x^2}} \; \text{dx}$$ =
For the last question use the substitution $$x = \sin{\theta}, \quad \text{where } -\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$$
Check
This is Integration level 8. You can also try:
Level 1 Level 2 Level 3 Level 4 Level 5 Level 6 Level 7 Level 9 Differentiation
## Instructions
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## Go Maths
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## Maths Map
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## Description of Levels
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Level 1 - Indefinite integration of basic polynomials with integer coefficient solutions
Level 2 - Indefinite integration of basic polynomials with integer and fraction coefficient solutions
Level 3 - Definite integration of basic polynomials
Level 4 - Integration of expressions containing fractional indices
Level 5 - Integration of basic trigonometric, exponential and reciprocal functions
Level 6 - Integration of the composites of basic functions with the linear function ax + b
Level 7 - Integration with the help of partial fractions
Level 8 - Integration by substitution
Level 9 - Integration by parts
Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).
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## In General
$$\int ax^n \text{dx} = \frac{ax^{n+1}}{n+1}+c \quad \text{ for all } n \neq -1$$
## Mathematical Notation
Use the ^ key to type in a power or index then the right arrow or tab key to end the power.
For example: Type 3x^2 to get 3x2.
Use the forward slash / to type a fraction then the right arrow or tab key to end the fraction.
For example: Type 1/2 to get ½.
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Answers to definite integral questions should be given as exact fractions or to three significant figures if the decimal answer does not terminate.
## Special Functions
$$\int e^x \; \text{dx} = e^x + c$$ $$\int \frac1x \; \text{dx} = \ln x + c$$ $$\int \cos x \; \text{dx} = \sin x + c$$ $$\int \sin x \; \text{dx} = -\cos x + c$$
## Composite Functions
$$\int e^{ax+b} \; \text{dx} = \frac1a e^{ax+b} + c$$ $$\int (ax+b)^n \; \text{dx} = \frac1a \frac{(ax+b)^{n+1}}{n+1} + c \text{,} \quad (n \neq -1)$$ $$\int \frac{1}{ax+b}\; \text{dx} = \frac1a \ln (ax+b)+ c \text{,} \quad (ax+b \gt 0)$$ $$\int \cos (ax+b) \; \text{dx} = \frac1a \sin (ax+b) + c$$ $$\int \sin (ax+b) \; \text{dx} = - \frac1a \cos (ax+b) + c$$
The following identities may also prove useful:
$$\sin^2x = \frac{1}{2} - \frac{1}{2} \cos 2x \text{ and } \cos^2x = \frac{1}{2} + \frac{1}{2} \cos 2x$$
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### Typing Mathematical Notation
These exercises use MathQuill, a web formula editor designed to make typing Maths easy and beautiful. Watch the animation below to see how common mathematical notation can be created using your keyboard.
Integration Flowchart
Close | 2,128 | 8,261 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-22 | latest | en | 0.844817 |
http://stackoverflow.com/tags/python-3.x/hot?filter=month | 1,467,060,835,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396106.71/warc/CC-MAIN-20160624154956-00051-ip-10-164-35-72.ec2.internal.warc.gz | 280,048,246 | 20,173 | # Tag Info
73
2305843009213693951 is 2^61 - 1. It's the largest Mersenne prime that fits into 64 bits. If you have to make a hash just by taking the value mod some number, then a large Mersenne prime is a good choice -- it's easy to compute and ensures an even distribution of possibilities. (Although I personally would never make a hash this way) It's especially ...
59
Based on python documentation in pyhash.c file: For numeric types, the hash of a number x is based on the reduction of x modulo the prime P = 2**_PyHASH_BITS - 1. It's designed so that hash(x) == hash(y) whenever x and y are numerically equal, even if x and y have different types. So for a 64/32 bit machine, the reduction would be 2 ...
11
You could check whether args are truthy in your function: def g(*args): if not args: args = (1, 2, 3) return "I received {}!".format(', '.join(str(arg) for arg in args)) If no args are passed to the function, it will result in a empty tuple, which evaluates to False.
10
The generator actually retains the actual call frames when yield happens. It doesn't really affect performance whether you have 1 or 100 local variables. The performance difference really comes from how Python (here I am using the CPython a.k.a. the one that you'd download from http://www.python.com/, or have on your operating system as /usr/bin/python, but ...
9
The function annotations there are annotations, nothing more. They're documentation about intended usage. Like it says in PEP 3107, the mechanism provides a single, standard way to specify function parameters and return values, replacing a variety of ad-hoc tools and libraries. But as it goes on to say: Function annotations are nothing more than a way ...
9
You first joined all strings together into one big string, and then converted that string to a list, which always results in all the individual characters being pulled out as elements: >>> list(map(lambda x: str(x * 5), range(1, 4))) ['5', '10', '15'] >>> ''.join(map(lambda x: str(x * 5), range(1, 4))) '51015' >>> list(''.join(map(...
7
Just convert the numbers to strings before joining: var = [1, 2, 3] print(", ".join(map(str, var))) or using a list comprehension: print(", ".join([str(x) for x in var]))
7
You're not actually timing the computation with ** and %, because the result gets constant-folded by the bytecode compiler. Avoid that: timeit.timeit('(a**b) % c', setup='a=65537; b=767587; c=14971787') and the pow version will win hands down.
7
If you read a little further down the documentation it mentions that (emphasis mine): PEP 492 was accepted in CPython 3.5.0 with __aiter__ defined as a method, that was expected to return an awaitable resolving to an asynchronous iterator. In 3.5.2 (as PEP 492 was accepted on a provisional basis) the __aiter__ protocol was updated to return ...
6
If no arguments are received, args will be an empty tuple. You can't add a default value in the method signature itself, but you can check if args is empty and replace it with a fallback value inside the function. def g(*args): if not args: args = (1, 2, 3) return 'I received {}!'.format(', '.join(str(arg) for arg in args))
6
Hash function returns plain int that means that returned value is greater than -sys.maxint and lower than sys.maxint, which means if you pass sys.maxint + x to it it will return -sys.maxint + (x - 2). hash(sys.maxint + 1) == sys.maxint + 1 # False hash(sys.maxint + 1) == - sys.maxint -1 # True hash(sys.maxint + sys.maxint) == -sys.maxint + sys.maxint - 2 # ...
6
The only state that a generator needs to save is a reference to the stack frame, so saving and restoring state takes exactly the same time no matter how much state is involved and wherever you put the data. The difference you see in the timings is purely down to the speed with which Python can access values: a local variable access is very fast, a global ...
6
If your ultimate goal is simply detecting if there's a double, this function may help: def has_doubles(n): return len(set(str(n))) < len(str(n))
6
Yes, in a nutshell, modules do implicitly return None in order for the big evaluation loop inside ceval.c to be able detect when to the current frame is finished and terminate. Interestingly, you can see this even when a python file that is completely empty is interpreted, from a terminal: jim@lpt> touch file.py jim@lpt> python -m dis file.py 1 ...
6
statistics isn't part of NumPy. It's a Python standard library module with a rather different design philosophy; it goes for accuracy at all costs, even for unusual input datatypes and extremely poorly-conditioned input. Performing a sum the way the statistics module does it is really expensive, more so than performing a sort. If you want an efficient mean ...
5
You can write file in python like with open("out.txt", "w") as f: f.write("OUTPUT") Or you can use io redirection to redirect output to a file \$ python code.py > out.txt https://docs.python.org/3/tutorial/inputoutput.html#methods-of-file-objects https://en.wikipedia.org/wiki/Redirection_(computing)
5
You can easily test this out yourself, by definining a decorator that will tell you whether you decorated something. For example this one: def myDecorator (f): print('Decorated {0}'.format(f.__name__)) return f Then, when used with your class: class MyClass: @myDecorator def f1 (self): return 1 def f2 (self): return 2 ...
5
In this case I would use a spigot algorithm, which allows to calculate the n-th digit of pi with limited amount of memory, and not increasing with the index of the digit; in particular, the Bailey–Borwein–Plouffe variant. This link is most probably what you are looking for, if I understand correctly your question: The formula can directly calculate the ...
5
reverse reverses the list in-place, see the manual, while [::-1] gives a new list in reversed order. Try print(p) after calling p.reverse(), you'll see the difference.
5
Couple things here.. This isn't necessarily wrong (the code will still work) but there's no reason to convert input to a str, it already is a string Strings need to be in quotes, not inside str(). Use "Red" instead Maybe use better variable names, x is unclear The last exit is useless Your code fixed would look something like: selected_color = input("...
5
This feels like a situation where any would be the best solution: # Function to check if a given DateZoneCity def DateZoneCity_downloaded_previously(Order_Date, ZoneCity): # Combination had already been completely downloaded string_to_match = Order_Date.strftime('%Y/%m/%d') + "-" + ZoneCity[0] + "...
5
The .strptime() method supports the day of year format: >>> import datetime >>> >>> datetime.datetime.strptime('16234', '%y%j').date() datetime.date(2016, 8, 21) And then you can use strftime() to reformat the date >>> date = datetime.date(2016, 8, 21) >>> date.strftime('%d/%m/%Y') '21/08/2016'
5
You have two options: Checking if the key exists sum(dct[k] for k in lst if k in dct) or using get sum(dct.get(k, 0) for k in lst) where dct.get(k, 0) returns dct[k] if k is a key in dct or 0 if not.
5
You can simplify it a bit: if sys.version_info[0] == 3: xrange = range I would do it the other way around: if sys.version_info[0] == 2: range = xrange If you ever want to drop Python 2.x support, you can just remove those two lines without going through all your code. However, I strongly suggest considering the six library. It is the de-facto ...
5
In your line: monthly_payment = float(loan_amount) * [0.05 * (1.0 + 0.05) * float(num_of_payments)] / [(1.0 + 0.05) * float(num_of_payments) - 1.0] You are using square brackets (in this case used to create a list) as a means of specifying order of operations. This should be () type brackets.
5
Your code is doing 5.4 billion calculations due to the two for loops (0.15 mil * 36k): I would do something like this: (Thanks to @Leon for helping me make this answer better) from bisect import bisect_left, bisect_right zeros_list = sorted([zeros2.iloc[j,1] for j in zeros2.index]) zeros2_length = len(zeros2_list) for i in ones2.index: cur_disc = ...
5
Classes within classes work just fine in python -- However, python's primary mechanism for handling namespaces is modules and packages, not classes. I've not found a whole lot of use for classes within classes (though there probably are some valid use-cases). To set up something like os.path.exists, I would direct you to the source except that os.path does ...
5
To easily replace a substring in a string txtwith another: txt = txt.replace('\\', '/') #Replacing \\ with / Remember kids, python has a built in function for everything
5
raise and del are definitely distinct from functions, each for different reasons: raise exits the current flow of execution; the normal flow of byte-code interpretation is interrupted and the stack is unwound until the next exception handler is found. Functions can't do this, they create a new stack frame instead. del can't be a function, because you must ...
5
The key is int(price). The input command returns a string which is stored in the variable price. In the second line, price is converted to a number by the call int(price), but that result is not stored anywhere. It is used for the comparison and then discarded. So when you go to multiply in the third line, you're trying to multiply a number times a string.
Only top voted, non community-wiki answers of a minimum length are eligible | 2,321 | 9,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-26 | latest | en | 0.87227 |
https://www.hackmath.net/en/example/56?tag_id=54 | 1,555,756,546,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529606.64/warc/CC-MAIN-20190420100901-20190420122901-00281.warc.gz | 703,768,139 | 7,262 | Chickens and rabbits
In the yard were chickens and rabbits. They had 28 heads and 82 legs. How many chickens and how many rabbits was in the yard?
Result
chickens: 15
rabbits: 13
Solution:
a+b = 28
2*a+4*b=82
a+b = 28
2a+4b = 82
a = 15
b = 13
Calculated by our linear equations calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Math student
viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with viju??
2 years ago 8 Likes
Math student
viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with viju??
2 years ago 8 Likes
Manikanta
35 chicken 5 rabbit
1 year ago 2 Likes
Math student
Imagine all rabbits and chickens are well trained. We blow the whistle, all chickens and rabbits would retract one of their legs. 124 - 44 = 80. Then we blow the whistle again, they retract another leg. 80-44= 36. these remaining 36 legs belong to the rabbits which are standing with two legs (the other two are retracted). thus 36/2 = 18 rabbits. and the chickens are 44-18 = 26.
1 year ago 2 Likes
Math student
We know that chicken has 2 legs and rabbit has 4. Assume that all heads are of chicken = 44*2 = 88. Now the difference between the total no. Of legs = 124-88 = 36. 36/2 =18 legs of Rabbits and 44-18 = 26 for chicken
1 year ago 2 Likes
Math student
35 chiken and 5 rabit
To solve this example are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?
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4. Cats
Two cats caught two mice in two days. How many mouses will catch 6 cats for 6 days?
5. When will I be a millionaire?
Shawn monthly send 220 euros to the bank, which he deposits bear interest of 2.4% p.a. Calculate how many months must Shawn save to save 1000000 euros? Inflation, interest rate changes or bank failures ignore.
6. Diofant equation
In the set of integers (Z) solve the equation: ? Write result with integer parameter ? (parameter t = ...-2,-1,0,1,2,3... if equation has infinitely many solutions)
7. Report card
Ivor hit 5× grade 5 at the beginning of the school year. How many times must now catch grade 1 to get grade 2 on report card?
8. Dice
How many times must throw the dice, the probability of throwing at least one six was greater than 90%?
9. Expression plus minus
Evaluate expression: (-1)2 . 12 – 6 : 3 + (-3) . (-2) + 22 – (-3) . 2
10. The temperature
The temperature at 1:00 was 10 F. Between 1:00 and 2:00, the temperature dropped 15F. Between 2:00 and 3:00, the temperature rose 3F. What is the temperature at 3:00?
11. Divisibility
Is the number 761082 exactly divisible by 9? (the result is the integer and/or remainder is zero)
12. Monkey
Monkey fell in 23 meters deep well. Every day it climbs 3 meters, at night it dropped back by 2 m. On what day it gets out from the well?
13. Diofant 2
Is equation ? solvable on the set of integers Z?
14. Unknown x
If we add to unknown number 21, then divide by 6 and then subtract 51, we get back an unknown number. What is this unknown number?
15. Sequence
In the arithmetic sequence is a1=-1, d=4. Which member is equal to the number 203?
16. Mistake
Nicol mistake when calculate in school. Instead of add number 20 subtract it. What is the difference between the result and the right result?
17. Father and son
Father and son are together 80 years. Son is 28 years younger than father. How old is the son? | 1,129 | 4,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-18 | latest | en | 0.95555 |
http://www.getnetusa.com/California/anova-table-ss-error.html | 1,555,971,340,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578582736.31/warc/CC-MAIN-20190422215211-20190423001211-00448.warc.gz | 249,139,470 | 7,673 | Address 2060 E Spruce Ave, Fresno, CA 93720 (559) 892-1840
anova table ss error Prather, California
The p-value for the Race factor is the area to the right F = 3.42 using 2 numerator and 24 denominator df. But first, as always, we need to define some notation. In the learning study, the factor is the learning method. (2) DF means "the degrees of freedom in the source." (3) SS means "the sum of squares due to the source." Unlike the corrected sum of squares, the uncorrected sum of squares includes error.
There are several techniques we might use to further analyze the differences. List price:\$120.00 Buy from amazon.com for \$96.42 Need to learnPrism 7? In "lay speak", we can't show at least one mean is different. This makes sense.
We need a critical value to compare the test statistic to. Are all of the data values within any one group the same? yi is the ith observation. The ANOVA calculations for multiple regression are nearly identical to the calculations for simple linear regression, except that the degrees of freedom are adjusted to reflect the number of explanatory variables
Within Group Variation (Error) Is every data value within each group identical? For p explanatory variables, the model degrees of freedom (DFM) are equal to p, the error degrees of freedom (DFE) are equal to (n - p - 1), and the total This portion of the total variability, or the total sum of squares that is not explained by the model, is called the residual sum of squares or the error sum of Case 2 was where the population variances were unknown, but assumed equal.
In order to calculate an F-statistic we need to calculate SSconditions and SSerror. No! Figure 1: Perfect Model Passing Through All Observed Data Points The model explains all of the variability of the observations. There is the between group variation and the within group variation.
When, on the next page, we delve into the theory behind the analysis of variance method, we'll see that the F-statistic follows an F-distribution with m−1 numerator degrees of freedom andn−mdenominator It is the same regardless of repeated measures. The variations (SS) are best found using technology. Do you remember the little song from Sesame Street?
It is calculated as a summation of the squares of the differences from the mean. Guess what that equals? There is no total variance. In that case, the degrees of freedom was the smaller of the two degrees of freedom.
The two-way ANOVA that we're going to discuss requires a balanced design. Dataset available through the Statlib Data and Story Library (DASL).) As a simple linear regression model, we previously considered "Sugars" as the explanatory variable and "Rating" as the response variable. SS stands for Sum of Squares. Table 1: Yield Data Observations of a Chemical Process at Different Values of Reaction Temperature The parameters of the assumed linear model are obtained using least square estimation. (For details,
We have already found the variance for each group, and if we remember from earlier in the book, when we first developed the variance, we found out that the variation was The sample variance is also referred to as a mean square because it is obtained by dividing the sum of squares by the respective degrees of freedom. Let's see what kind of formulas we can come up with for quantifying these components. Maxwell, Harold D.
Filling in the table Sum of Square = Variations There's two ways to find the total variation. The calculation of the total sum of squares considers both the sum of squares from the factors and from randomness or error. So, what did we find out? It ties together many aspects of what we've been doing all semester.
How many groups were there in this problem? In other words, we treat each subject as a level of an independent factor called subjects. The variance for the between group and the variance for the within group. The \(p\)-value for 9.59 is 0.00325, so the test statistic is significant at that level.
The first term is the total variation in the response y, the second term is the variation in mean response, and the third term is the residual value. The p-value for the Race factor is the area to the right F = 13.71 using 1 numerator and 24 denominator df. The idea for the name comes from experiments where you have a control group that doesn't receive the treatment, and an experimental group where that group does receive the treatement. Example Table 1 shows the observed yield data obtained at various temperature settings of a chemical process.
This is beautiful, because we just found out that what we have in the MS column are sample variances. ANOVA Table Example A numerical example The data below resulted from measuring the difference in resistance resulting from subjecting identical resistors to three different temperatures for a period of 24 hours. The data values are squared without first subtracting the mean. That is, the F = 17.58 is for the race source, so it would be used to determine if there is a difference in the mean reaction times of the different
Realize however, that the results may not be accurate when the assumptions aren't met. The adjusted sums of squares can be less than, equal to, or greater than the sequential sums of squares. Now it's time to play our game (time to play our game). The "Analysis of Variance" portion of the MINITAB output is shown below.
There is no right or wrong method, and other methods exist; it is simply personal preference as to which method you choose. The F-test The test statistic, used in testing the equality of treatment means is: \(F = MST / MSE\). Source SS df MS F Between Within Total Source is where the variation came from. | 1,227 | 5,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-18 | latest | en | 0.897565 |
https://www.includehelp.com/c/polynomial-addition-using-structure-with-c-program.aspx | 1,720,943,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00556.warc.gz | 740,450,938 | 50,611 | # Polynomial Addition Using Structure [with C program]
Learn: How to add two polynomials using structures in C? This article explains how to implement structure of polynomial, algorithm and C program for polynomial addition. By Abhishek Jain Last updated : April 13, 2023
## What is Polynomial?
A polynomial is an expression that contains more than two terms. A term is made up of coefficient and exponent.
### Example
```P(x) = 4x3+6x2+7x+9
```
A polynomial may be represented using array or structure. A structure may be defined such that it contains two parts – one is the coefficient and second is the corresponding exponent. The structure definition may be given as shown below:
### Polynomial Structure Declaration
```struct polynomial{
int coefficient;
int exponent;
};
```
## How to Add Two Polynomials?
To add two polynomials using structure, just add the coefficient parts of the polynomials having same exponent.
```addPolynomial(
struct polynomial p1[10],
struct polynomial p2[10],
int t1,
int t2,
struct polynomial p3[10]);
```
1. ```[Initialize segment variables]
[Initialize Counter] Set i=0,j=0,k=0
```
2. ```Repeat while i<t1 and j<t2
IF p1[i].expo=p2[j].expo, THEN
p3[i].coeff=p1[i].coeff+p2[i].coeff
p3[k].expo=p1[i].expo
[Increase counter] Set i=i+1,j=j+1,k=k+1
ELSE IF p1[i].expo > p2[j].expo, THEN
p3[k].coeff=p1[i].coeff
p3[k].expo=p1[i].expo
[Increase counter] Set i=i+1,k=k+1
ELSE
p3[k].coeff=p2[j].coeff
p3[k].expo=p2[j].expo
Set j=j+1,k=k+1
[End of If]
[End of loop]
```
3. ```Repeat while i<t1
p3[k].coeff=p1[i].coeff
p3[k].expo=p1[i].expo
Set i=i+1,k=k+1
[End of loop]
```
4. ```Repeat while j<t2
p3[k].coeff=p2[j].coeff
p3[k].expo=p2[j].expo
Set j=j+1,k=k+1
[End of loop]
```
5. ```Return k
```
6. ```EXIT
```
## C Program for Polynomial Addition Using Structure
```/* program for addition of two polynomials
polynomial are stored using structure
and program uses array of structure
*/
#include<stdio.h>
/* declare structure for polynomial */
struct poly {
int coeff;
int expo;
};
/* declare three arrays p1, p2, p3 of type structure poly.
each polynomial can have maximum of ten terms
addition result of p1 and p2 is stored in p3*/
struct poly p1[10], p2[10], p3[10];
/* function prototypes */
int addPoly(struct poly[], struct poly[], int, int, struct poly[]);
void displayPoly(struct poly[], int terms);
int main(){
int t1, t2, t3;
/* read and display first polynomial */
printf(" \n First polynomial : ");
displayPoly(p1, t1);
/* read and display second polynomial */
printf(" \n Second polynomial : ");
displayPoly(p2, t2);
/* add two polynomials and display resultant polynomial */
t3 = addPoly(p1, p2, t1, t2, p3);
printf(" \n\n Resultant polynomial after addition : ");
displayPoly(p3, t3);
printf("\n");
return 0;
}
{
int t1, i;
printf("\n\n Enter the total number of terms in the polynomial:");
scanf("%d", &t1);
printf("\n Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER\n");
for (i = 0; i < t1; i++) {
printf(" Enter the Coefficient(%d): ", i + 1);
scanf("%d", &p[i].coeff);
printf(" Enter the exponent(%d): ", i + 1);
scanf("%d", &p[i].expo); /* only statement in loop */
}
return (t1);
}
int addPoly(struct poly p1[10], struct poly p2[10], int t1, int t2, struct poly p3[10])
{
int i, j, k;
i = 0;
j = 0;
k = 0;
while (i < t1 && j < t2) {
if (p1[i].expo == p2[j].expo) {
p3[k].coeff = p1[i].coeff + p2[j].coeff;
p3[k].expo = p1[i].expo;
i++;
j++;
k++;
}
else if (p1[i].expo > p2[j].expo) {
p3[k].coeff = p1[i].coeff;
p3[k].expo = p1[i].expo;
i++;
k++;
}
else {
p3[k].coeff = p2[j].coeff;
p3[k].expo = p2[j].expo;
j++;
k++;
}
}
/* for rest over terms of polynomial 1 */
while (i < t1) {
p3[k].coeff = p1[i].coeff;
p3[k].expo = p1[i].expo;
i++;
k++;
}
/* for rest over terms of polynomial 2 */
while (j < t2) {
p3[k].coeff = p2[j].coeff;
p3[k].expo = p2[j].expo;
j++;
k++;
}
return (k); /* k is number of terms in resultant polynomial*/
}
void displayPoly(struct poly p[10], int term)
{
int k;
for (k = 0; k < term - 1; k++)
printf("%d(x^%d)+", p[k].coeff, p[k].expo);
printf("%d(x^%d)", p[term - 1].coeff, p[term - 1].expo);
}
```
### Output
```Enter the total number of terms in the polynomial:4
Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER
Enter the Coefficient(1): 3
Enter the exponent(1): 4
Enter the Coefficient(2): 7
Enter the exponent(2): 3
Enter the Coefficient(3): 5
Enter the exponent(3): 1
Enter the Coefficient(4): 8
Enter the exponent(4): 0
First polynomial : 3(x^4)+7(x^3)+5(x^1)+8(x^0)
Enter the total number of terms in the polynomial:5
Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER
Enter the Coefficient(1): 7
Enter the exponent(1): 5
Enter the Coefficient(2): 6
Enter the exponent(2): 4
Enter the Coefficient(3): 8
Enter the exponent(3): 2
Enter the Coefficient(4): 9
Enter the exponent(4): 1
Enter the Coefficient(5): 2
Enter the exponent(5): 0
Second polynomial : 7(x^5)+6(x^4)+8(x^2)+9(x^1)+2(x^0)
Resultant polynomial after addition : 7(x^5)+9(x^4)+7(x^3)+8(x^2)+14(x^1)+10(x^0)
``` | 1,627 | 5,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.702681 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/495/2/n/g/136/2/ | 1,638,837,274,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00530.warc.gz | 972,399,650 | 85,826 | Properties
Label 495.2.n.g.136.2 Level $495$ Weight $2$ Character 495.136 Analytic conductor $3.953$ Analytic rank $0$ Dimension $16$ CM no Inner twists $2$
Related objects
Newspace parameters
Level: $$N$$ $$=$$ $$495 = 3^{2} \cdot 5 \cdot 11$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 495.n (of order $$5$$, degree $$4$$, minimal)
Newform invariants
Self dual: no Analytic conductor: $$3.95259490005$$ Analytic rank: $$0$$ Dimension: $$16$$ Relative dimension: $$4$$ over $$\Q(\zeta_{5})$$ Coefficient field: $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ Defining polynomial: $$x^{16} - 2 x^{15} + 5 x^{14} - 8 x^{13} + 47 x^{12} + 32 x^{11} + 171 x^{10} + 26 x^{9} + 360 x^{8} - 172 x^{7} + 471 x^{6} - 430 x^{5} + 383 x^{4} + 70 x^{3} + 17 x^{2} + 4 x + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{5}]$
Embedding invariants
Embedding label 136.2 Root $$0.0698401 + 0.214946i$$ of defining polynomial Character $$\chi$$ $$=$$ 495.136 Dual form 495.2.n.g.91.2
$q$-expansion
$$f(q)$$ $$=$$ $$q+(-0.991861 - 0.720629i) q^{2} +(-0.153553 - 0.472586i) q^{4} +(-0.809017 + 0.587785i) q^{5} +(-0.139581 - 0.429587i) q^{7} +(-0.945971 + 2.91140i) q^{8} +O(q^{10})$$ $$q+(-0.991861 - 0.720629i) q^{2} +(-0.153553 - 0.472586i) q^{4} +(-0.809017 + 0.587785i) q^{5} +(-0.139581 - 0.429587i) q^{7} +(-0.945971 + 2.91140i) q^{8} +1.22601 q^{10} +(1.55234 + 2.93091i) q^{11} +(3.91592 + 2.84508i) q^{13} +(-0.171128 + 0.526677i) q^{14} +(2.23230 - 1.62186i) q^{16} +(0.598736 - 0.435007i) q^{17} +(2.10247 - 6.47075i) q^{19} +(0.402006 + 0.292074i) q^{20} +(0.572395 - 4.02572i) q^{22} +0.00634166 q^{23} +(0.309017 - 0.951057i) q^{25} +(-1.83380 - 5.64385i) q^{26} +(-0.181584 + 0.131928i) q^{28} +(0.100091 + 0.308048i) q^{29} +(4.53521 + 3.29503i) q^{31} +2.73956 q^{32} -0.907341 q^{34} +(0.365429 + 0.265500i) q^{35} +(2.27713 + 7.00828i) q^{37} +(-6.74837 + 4.90298i) q^{38} +(-0.945971 - 2.91140i) q^{40} +(3.39006 - 10.4335i) q^{41} -1.80668 q^{43} +(1.14674 - 1.18366i) q^{44} +(-0.00629004 - 0.00456998i) q^{46} +(0.518988 - 1.59728i) q^{47} +(5.49806 - 3.99457i) q^{49} +(-0.991861 + 0.720629i) q^{50} +(0.743247 - 2.28748i) q^{52} +(6.98948 + 5.07816i) q^{53} +(-2.97862 - 1.45871i) q^{55} +1.38274 q^{56} +(0.122712 - 0.377669i) q^{58} +(0.463691 + 1.42709i) q^{59} +(-10.5540 + 7.66790i) q^{61} +(-2.12381 - 6.53641i) q^{62} +(-7.18186 - 5.21793i) q^{64} -4.84034 q^{65} +9.60773 q^{67} +(-0.297516 - 0.216158i) q^{68} +(-0.171128 - 0.526677i) q^{70} +(9.23296 - 6.70814i) q^{71} +(3.16430 + 9.73870i) q^{73} +(2.79178 - 8.59220i) q^{74} -3.38083 q^{76} +(1.04241 - 1.07597i) q^{77} +(1.69866 + 1.23415i) q^{79} +(-0.852662 + 2.62422i) q^{80} +(-10.8812 + 7.90563i) q^{82} +(-12.8589 + 9.34255i) q^{83} +(-0.228697 + 0.703856i) q^{85} +(1.79197 + 1.30194i) q^{86} +(-10.0015 + 1.74692i) q^{88} +9.36925 q^{89} +(0.675622 - 2.07935i) q^{91} +(-0.000973778 - 0.00299698i) q^{92} +(-1.66581 + 1.21028i) q^{94} +(2.10247 + 6.47075i) q^{95} +(-4.75689 - 3.45608i) q^{97} -8.33191 q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$16 q - 2 q^{2} - 8 q^{4} - 4 q^{5} - 4 q^{7} - 6 q^{8} + O(q^{10})$$ $$16 q - 2 q^{2} - 8 q^{4} - 4 q^{5} - 4 q^{7} - 6 q^{8} + 8 q^{10} + 4 q^{11} + 2 q^{13} - 22 q^{14} + 8 q^{16} - 4 q^{17} - 4 q^{19} + 2 q^{20} - 28 q^{22} + 8 q^{23} - 4 q^{25} + 6 q^{26} - 2 q^{28} - 26 q^{29} - 10 q^{31} + 56 q^{32} - 4 q^{34} - 4 q^{35} + 22 q^{37} - 30 q^{38} - 6 q^{40} - 6 q^{41} + 28 q^{43} + 68 q^{44} + 16 q^{46} - 20 q^{47} + 10 q^{49} - 2 q^{50} + 30 q^{52} + 14 q^{53} - 6 q^{55} + 68 q^{56} - 6 q^{58} - 16 q^{59} - 38 q^{61} - 20 q^{62} + 10 q^{64} + 12 q^{65} + 20 q^{67} - 48 q^{68} - 22 q^{70} - 54 q^{71} + 2 q^{73} + 28 q^{74} - 44 q^{76} + 34 q^{77} - 12 q^{79} - 22 q^{80} + 30 q^{82} - 28 q^{83} - 4 q^{85} + 74 q^{86} + 46 q^{88} + 76 q^{89} - 34 q^{91} - 8 q^{92} - 10 q^{94} - 4 q^{95} - 18 q^{97} + 8 q^{98} + O(q^{100})$$
Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/495\mathbb{Z}\right)^\times$$.
$$n$$ $$46$$ $$56$$ $$397$$ $$\chi(n)$$ $$e\left(\frac{1}{5}\right)$$ $$1$$ $$1$$
Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −0.991861 0.720629i −0.701351 0.509562i 0.179021 0.983845i $$-0.442707\pi$$
−0.880372 + 0.474284i $$0.842707\pi$$
$$3$$ 0 0
$$4$$ −0.153553 0.472586i −0.0767763 0.236293i
$$5$$ −0.809017 + 0.587785i −0.361803 + 0.262866i
$$6$$ 0 0
$$7$$ −0.139581 0.429587i −0.0527568 0.162369i 0.921207 0.389073i $$-0.127205\pi$$
−0.973964 + 0.226705i $$0.927205\pi$$
$$8$$ −0.945971 + 2.91140i −0.334451 + 1.02933i
$$9$$ 0 0
$$10$$ 1.22601 0.387698
$$11$$ 1.55234 + 2.93091i 0.468048 + 0.883703i
$$12$$ 0 0
$$13$$ 3.91592 + 2.84508i 1.08608 + 0.789084i 0.978733 0.205138i $$-0.0657643\pi$$
0.107348 + 0.994222i $$0.465764\pi$$
$$14$$ −0.171128 + 0.526677i −0.0457358 + 0.140760i
$$15$$ 0 0
$$16$$ 2.23230 1.62186i 0.558074 0.405465i
$$17$$ 0.598736 0.435007i 0.145215 0.105505i −0.512806 0.858505i $$-0.671394\pi$$
0.658021 + 0.753000i $$0.271394\pi$$
$$18$$ 0 0
$$19$$ 2.10247 6.47075i 0.482340 1.48449i −0.353456 0.935451i $$-0.614993\pi$$
0.835796 0.549040i $$-0.185007\pi$$
$$20$$ 0.402006 + 0.292074i 0.0898912 + 0.0653098i
$$21$$ 0 0
$$22$$ 0.572395 4.02572i 0.122035 0.858286i
$$23$$ 0.00634166 0.00132233 0.000661164 1.00000i $$-0.499790\pi$$
0.000661164 1.00000i $$0.499790\pi$$
$$24$$ 0 0
$$25$$ 0.309017 0.951057i 0.0618034 0.190211i
$$26$$ −1.83380 5.64385i −0.359637 1.10685i
$$27$$ 0 0
$$28$$ −0.181584 + 0.131928i −0.0343162 + 0.0249321i
$$29$$ 0.100091 + 0.308048i 0.0185864 + 0.0572030i 0.959920 0.280276i $$-0.0904259\pi$$
−0.941333 + 0.337479i $$0.890426\pi$$
$$30$$ 0 0
$$31$$ 4.53521 + 3.29503i 0.814548 + 0.591804i 0.915146 0.403123i $$-0.132075\pi$$
−0.100597 + 0.994927i $$0.532075\pi$$
$$32$$ 2.73956 0.484291
$$33$$ 0 0
$$34$$ −0.907341 −0.155608
$$35$$ 0.365429 + 0.265500i 0.0617688 + 0.0448776i
$$36$$ 0 0
$$37$$ 2.27713 + 7.00828i 0.374358 + 1.15215i 0.943911 + 0.330200i $$0.107116\pi$$
−0.569553 + 0.821954i $$0.692884\pi$$
$$38$$ −6.74837 + 4.90298i −1.09473 + 0.795368i
$$39$$ 0 0
$$40$$ −0.945971 2.91140i −0.149571 0.460332i
$$41$$ 3.39006 10.4335i 0.529438 1.62944i −0.225931 0.974143i $$-0.572543\pi$$
0.755369 0.655299i $$-0.227457\pi$$
$$42$$ 0 0
$$43$$ −1.80668 −0.275516 −0.137758 0.990466i $$-0.543990\pi$$
−0.137758 + 0.990466i $$0.543990\pi$$
$$44$$ 1.14674 1.18366i 0.172878 0.178444i
$$45$$ 0 0
$$46$$ −0.00629004 0.00456998i −0.000927416 0.000673807i
$$47$$ 0.518988 1.59728i 0.0757022 0.232987i −0.906044 0.423184i $$-0.860912\pi$$
0.981746 + 0.190196i $$0.0609125\pi$$
$$48$$ 0 0
$$49$$ 5.49806 3.99457i 0.785437 0.570653i
$$50$$ −0.991861 + 0.720629i −0.140270 + 0.101912i
$$51$$ 0 0
$$52$$ 0.743247 2.28748i 0.103070 0.317216i
$$53$$ 6.98948 + 5.07816i 0.960079 + 0.697539i 0.953169 0.302438i $$-0.0978004\pi$$
0.00691024 + 0.999976i $$0.497800\pi$$
$$54$$ 0 0
$$55$$ −2.97862 1.45871i −0.401636 0.196693i
$$56$$ 1.38274 0.184776
$$57$$ 0 0
$$58$$ 0.122712 0.377669i 0.0161129 0.0495903i
$$59$$ 0.463691 + 1.42709i 0.0603674 + 0.185792i 0.976692 0.214643i $$-0.0688589\pi$$
−0.916325 + 0.400435i $$0.868859\pi$$
$$60$$ 0 0
$$61$$ −10.5540 + 7.66790i −1.35130 + 0.981774i −0.352350 + 0.935868i $$0.614617\pi$$
−0.998946 + 0.0459057i $$0.985383\pi$$
$$62$$ −2.12381 6.53641i −0.269724 0.830125i
$$63$$ 0 0
$$64$$ −7.18186 5.21793i −0.897733 0.652241i
$$65$$ −4.84034 −0.600371
$$66$$ 0 0
$$67$$ 9.60773 1.17377 0.586885 0.809670i $$-0.300354\pi$$
0.586885 + 0.809670i $$0.300354\pi$$
$$68$$ −0.297516 0.216158i −0.0360791 0.0262130i
$$69$$ 0 0
$$70$$ −0.171128 0.526677i −0.0204537 0.0629500i
$$71$$ 9.23296 6.70814i 1.09575 0.796110i 0.115390 0.993320i $$-0.463188\pi$$
0.980361 + 0.197211i $$0.0631883\pi$$
$$72$$ 0 0
$$73$$ 3.16430 + 9.73870i 0.370353 + 1.13983i 0.946561 + 0.322526i $$0.104532\pi$$
−0.576208 + 0.817303i $$0.695468\pi$$
$$74$$ 2.79178 8.59220i 0.324537 0.998823i
$$75$$ 0 0
$$76$$ −3.38083 −0.387807
$$77$$ 1.04241 1.07597i 0.118793 0.122618i
$$78$$ 0 0
$$79$$ 1.69866 + 1.23415i 0.191114 + 0.138852i 0.679228 0.733928i $$-0.262315\pi$$
−0.488114 + 0.872780i $$0.662315\pi$$
$$80$$ −0.852662 + 2.62422i −0.0953305 + 0.293397i
$$81$$ 0 0
$$82$$ −10.8812 + 7.90563i −1.20162 + 0.873030i
$$83$$ −12.8589 + 9.34255i −1.41145 + 1.02548i −0.418339 + 0.908291i $$0.637388\pi$$
−0.993110 + 0.117187i $$0.962612\pi$$
$$84$$ 0 0
$$85$$ −0.228697 + 0.703856i −0.0248056 + 0.0763439i
$$86$$ 1.79197 + 1.30194i 0.193233 + 0.140392i
$$87$$ 0 0
$$88$$ −10.0015 + 1.74692i −1.06617 + 0.186223i
$$89$$ 9.36925 0.993138 0.496569 0.867997i $$-0.334593\pi$$
0.496569 + 0.867997i $$0.334593\pi$$
$$90$$ 0 0
$$91$$ 0.675622 2.07935i 0.0708244 0.217975i
$$92$$ −0.000973778 0.00299698i −0.000101523 0.000312457i
$$93$$ 0 0
$$94$$ −1.66581 + 1.21028i −0.171815 + 0.124831i
$$95$$ 2.10247 + 6.47075i 0.215709 + 0.663885i
$$96$$ 0 0
$$97$$ −4.75689 3.45608i −0.482989 0.350912i 0.319493 0.947589i $$-0.396487\pi$$
−0.802482 + 0.596677i $$0.796487\pi$$
$$98$$ −8.33191 −0.841650
$$99$$ 0 0
$$100$$ −0.496906 −0.0496906
$$101$$ −1.78628 1.29781i −0.177742 0.129137i 0.495357 0.868689i $$-0.335037\pi$$
−0.673099 + 0.739552i $$0.735037\pi$$
$$102$$ 0 0
$$103$$ 2.99111 + 9.20568i 0.294723 + 0.907063i 0.983314 + 0.181914i $$0.0582292\pi$$
−0.688592 + 0.725149i $$0.741771\pi$$
$$104$$ −11.9875 + 8.70944i −1.17547 + 0.854031i
$$105$$ 0 0
$$106$$ −3.27313 10.0736i −0.317914 0.978439i
$$107$$ 3.63294 11.1810i 0.351210 1.08091i −0.606965 0.794729i $$-0.707613\pi$$
0.958175 0.286184i $$-0.0923869\pi$$
$$108$$ 0 0
$$109$$ −12.5091 −1.19816 −0.599078 0.800691i $$-0.704466\pi$$
−0.599078 + 0.800691i $$0.704466\pi$$
$$110$$ 1.90318 + 3.59332i 0.181461 + 0.342609i
$$111$$ 0 0
$$112$$ −1.00832 0.732586i −0.0952771 0.0692228i
$$113$$ −3.06115 + 9.42124i −0.287968 + 0.886276i 0.697525 + 0.716561i $$0.254285\pi$$
−0.985493 + 0.169715i $$0.945715\pi$$
$$114$$ 0 0
$$115$$ −0.00513051 + 0.00372753i −0.000478422 + 0.000347594i
$$116$$ 0.130210 0.0946030i 0.0120897 0.00878367i
$$117$$ 0 0
$$118$$ 0.568489 1.74963i 0.0523336 0.161066i
$$119$$ −0.270446 0.196491i −0.0247917 0.0180123i
$$120$$ 0 0
$$121$$ −6.18048 + 9.09954i −0.561862 + 0.827231i
$$122$$ 15.9938 1.44801
$$123$$ 0 0
$$124$$ 0.860790 2.64924i 0.0773012 0.237909i
$$125$$ 0.309017 + 0.951057i 0.0276393 + 0.0850651i
$$126$$ 0 0
$$127$$ 0.454312 0.330077i 0.0403137 0.0292896i −0.567446 0.823411i $$-0.692068\pi$$
0.607760 + 0.794121i $$0.292068\pi$$
$$128$$ 1.67007 + 5.13995i 0.147615 + 0.454312i
$$129$$ 0 0
$$130$$ 4.80095 + 3.48809i 0.421071 + 0.305926i
$$131$$ −3.03500 −0.265170 −0.132585 0.991172i $$-0.542328\pi$$
−0.132585 + 0.991172i $$0.542328\pi$$
$$132$$ 0 0
$$133$$ −3.07322 −0.266482
$$134$$ −9.52953 6.92361i −0.823226 0.598108i
$$135$$ 0 0
$$136$$ 0.700092 + 2.15466i 0.0600324 + 0.184761i
$$137$$ −16.7469 + 12.1674i −1.43079 + 1.03953i −0.440918 + 0.897547i $$0.645347\pi$$
−0.989869 + 0.141981i $$0.954653\pi$$
$$138$$ 0 0
$$139$$ 1.09662 + 3.37504i 0.0930139 + 0.286267i 0.986731 0.162364i $$-0.0519117\pi$$
−0.893717 + 0.448631i $$0.851912\pi$$
$$140$$ 0.0693589 0.213465i 0.00586190 0.0180411i
$$141$$ 0 0
$$142$$ −13.9919 −1.17417
$$143$$ −2.25985 + 15.8937i −0.188978 + 1.32910i
$$144$$ 0 0
$$145$$ −0.262041 0.190384i −0.0217613 0.0158105i
$$146$$ 3.87945 11.9397i 0.321065 0.988138i
$$147$$ 0 0
$$148$$ 2.96236 2.15228i 0.243504 0.176916i
$$149$$ 5.10052 3.70575i 0.417851 0.303587i −0.358921 0.933368i $$-0.616855\pi$$
0.776773 + 0.629781i $$0.216855\pi$$
$$150$$ 0 0
$$151$$ −1.13852 + 3.50400i −0.0926512 + 0.285151i −0.986634 0.162949i $$-0.947899\pi$$
0.893983 + 0.448100i $$0.147899\pi$$
$$152$$ 16.8500 + 12.2423i 1.36672 + 0.992980i
$$153$$ 0 0
$$154$$ −1.80929 + 0.316022i −0.145797 + 0.0254657i
$$155$$ −5.60583 −0.450271
$$156$$ 0 0
$$157$$ 0.874113 2.69024i 0.0697618 0.214705i −0.910097 0.414394i $$-0.863993\pi$$
0.979859 + 0.199690i $$0.0639934\pi$$
$$158$$ −0.795469 2.44820i −0.0632841 0.194768i
$$159$$ 0 0
$$160$$ −2.21635 + 1.61028i −0.175218 + 0.127303i
$$161$$ −0.000885178 0.00272430i −6.97618e−5 0.000214705i
$$162$$ 0 0
$$163$$ −7.16094 5.20273i −0.560888 0.407509i 0.270896 0.962609i $$-0.412680\pi$$
−0.831784 + 0.555099i $$0.812680\pi$$
$$164$$ −5.45129 −0.425674
$$165$$ 0 0
$$166$$ 19.4868 1.51247
$$167$$ −15.9249 11.5701i −1.23230 0.895320i −0.235241 0.971937i $$-0.575588\pi$$
−0.997061 + 0.0766171i $$0.975588\pi$$
$$168$$ 0 0
$$169$$ 3.22271 + 9.91849i 0.247901 + 0.762961i
$$170$$ 0.734054 0.533322i 0.0562994 0.0409039i
$$171$$ 0 0
$$172$$ 0.277420 + 0.853811i 0.0211531 + 0.0651025i
$$173$$ 0.899661 2.76887i 0.0684000 0.210513i −0.911014 0.412375i $$-0.864699\pi$$
0.979414 + 0.201862i $$0.0646992\pi$$
$$174$$ 0 0
$$175$$ −0.451695 −0.0341449
$$176$$ 8.21881 + 4.02499i 0.619516 + 0.303395i
$$177$$ 0 0
$$178$$ −9.29299 6.75175i −0.696539 0.506065i
$$179$$ −1.94467 + 5.98508i −0.145352 + 0.447346i −0.997056 0.0766763i $$-0.975569\pi$$
0.851704 + 0.524022i $$0.175569\pi$$
$$180$$ 0 0
$$181$$ 9.03200 6.56213i 0.671343 0.487759i −0.199131 0.979973i $$-0.563812\pi$$
0.870475 + 0.492213i $$0.163812\pi$$
$$182$$ −2.16856 + 1.57555i −0.160745 + 0.116788i
$$183$$ 0 0
$$184$$ −0.00599902 + 0.0184631i −0.000442254 + 0.00136112i
$$185$$ −5.96160 4.33136i −0.438306 0.318448i
$$186$$ 0 0
$$187$$ 2.20441 + 1.07956i 0.161202 + 0.0789455i
$$188$$ −0.834545 −0.0608655
$$189$$ 0 0
$$190$$ 2.57765 7.93318i 0.187002 0.575534i
$$191$$ −6.30228 19.3964i −0.456017 1.40348i −0.869937 0.493163i $$-0.835841\pi$$
0.413920 0.910313i $$-0.364159\pi$$
$$192$$ 0 0
$$193$$ 2.40629 1.74827i 0.173208 0.125843i −0.497804 0.867290i $$-0.665860\pi$$
0.671012 + 0.741447i $$0.265860\pi$$
$$194$$ 2.22762 + 6.85590i 0.159934 + 0.492225i
$$195$$ 0 0
$$196$$ −2.73202 1.98493i −0.195144 0.141781i
$$197$$ 5.20127 0.370575 0.185288 0.982684i $$-0.440678\pi$$
0.185288 + 0.982684i $$0.440678\pi$$
$$198$$ 0 0
$$199$$ 8.10264 0.574381 0.287191 0.957873i $$-0.407279\pi$$
0.287191 + 0.957873i $$0.407279\pi$$
$$200$$ 2.47658 + 1.79934i 0.175121 + 0.127233i
$$201$$ 0 0
$$202$$ 0.836505 + 2.57450i 0.0588563 + 0.181141i
$$203$$ 0.118363 0.0859955i 0.00830743 0.00603570i
$$204$$ 0 0
$$205$$ 3.39006 + 10.4335i 0.236772 + 0.728709i
$$206$$ 3.66712 11.2862i 0.255500 0.786349i
$$207$$ 0 0
$$208$$ 13.3558 0.926059
$$209$$ 22.2289 3.88264i 1.53761 0.268568i
$$210$$ 0 0
$$211$$ 1.06252 + 0.771967i 0.0731470 + 0.0531444i 0.623758 0.781618i $$-0.285605\pi$$
−0.550611 + 0.834762i $$0.685605\pi$$
$$212$$ 1.32661 4.08290i 0.0911122 0.280415i
$$213$$ 0 0
$$214$$ −11.6608 + 8.47204i −0.797113 + 0.579137i
$$215$$ 1.46163 1.06194i 0.0996826 0.0724236i
$$216$$ 0 0
$$217$$ 0.782470 2.40820i 0.0531175 0.163479i
$$218$$ 12.4073 + 9.01443i 0.840328 + 0.610534i
$$219$$ 0 0
$$220$$ −0.231994 + 1.63164i −0.0156411 + 0.110005i
$$221$$ 3.58223 0.240967
$$222$$ 0 0
$$223$$ 8.84861 27.2332i 0.592547 1.82367i 0.0259701 0.999663i $$-0.491733\pi$$
0.566577 0.824009i $$-0.308267\pi$$
$$224$$ −0.382392 1.17688i −0.0255497 0.0786338i
$$225$$ 0 0
$$226$$ 9.82545 7.13861i 0.653579 0.474853i
$$227$$ −2.43074 7.48104i −0.161334 0.496534i 0.837414 0.546570i $$-0.184067\pi$$
−0.998747 + 0.0500354i $$0.984067\pi$$
$$228$$ 0 0
$$229$$ 13.0664 + 9.49331i 0.863453 + 0.627336i 0.928822 0.370526i $$-0.120822\pi$$
−0.0653689 + 0.997861i $$0.520822\pi$$
$$230$$ 0.00777492 0.000512663
$$231$$ 0 0
$$232$$ −0.991532 −0.0650973
$$233$$ −14.9138 10.8355i −0.977035 0.709858i −0.0199914 0.999800i $$-0.506364\pi$$
−0.957044 + 0.289942i $$0.906364\pi$$
$$234$$ 0 0
$$235$$ 0.518988 + 1.59728i 0.0338551 + 0.104195i
$$236$$ 0.603224 0.438268i 0.0392665 0.0285288i
$$237$$ 0 0
$$238$$ 0.126648 + 0.389782i 0.00820937 + 0.0252658i
$$239$$ 6.91081 21.2693i 0.447023 1.37579i −0.433227 0.901285i $$-0.642625\pi$$
0.880250 0.474510i $$-0.157375\pi$$
$$240$$ 0 0
$$241$$ −13.2213 −0.851662 −0.425831 0.904803i $$-0.640018\pi$$
−0.425831 + 0.904803i $$0.640018\pi$$
$$242$$ 12.6876 4.57164i 0.815588 0.293876i
$$243$$ 0 0
$$244$$ 5.24433 + 3.81023i 0.335734 + 0.243925i
$$245$$ −2.10007 + 6.46335i −0.134169 + 0.412928i
$$246$$ 0 0
$$247$$ 26.6429 19.3572i 1.69525 1.23167i
$$248$$ −13.8833 + 10.0868i −0.881591 + 0.640513i
$$249$$ 0 0
$$250$$ 0.378857 1.16600i 0.0239610 0.0737444i
$$251$$ 16.9919 + 12.3454i 1.07252 + 0.779232i 0.976364 0.216134i $$-0.0693450\pi$$
0.0961569 + 0.995366i $$0.469345\pi$$
$$252$$ 0 0
$$253$$ 0.00984441 + 0.0185868i 0.000618913 + 0.00116854i
$$254$$ −0.688477 −0.0431989
$$255$$ 0 0
$$256$$ −3.43893 + 10.5839i −0.214933 + 0.661497i
$$257$$ −1.04068 3.20289i −0.0649160 0.199791i 0.913338 0.407203i $$-0.133496\pi$$
−0.978254 + 0.207412i $$0.933496\pi$$
$$258$$ 0 0
$$259$$ 2.69283 1.95645i 0.167324 0.121568i
$$260$$ 0.743247 + 2.28748i 0.0460942 + 0.141863i
$$261$$ 0 0
$$262$$ 3.01030 + 2.18711i 0.185977 + 0.135120i
$$263$$ 21.0450 1.29769 0.648845 0.760920i $$-0.275252\pi$$
0.648845 + 0.760920i $$0.275252\pi$$
$$264$$ 0 0
$$265$$ −8.63948 −0.530719
$$266$$ 3.04820 + 2.21465i 0.186897 + 0.135789i
$$267$$ 0 0
$$268$$ −1.47529 4.54048i −0.0901177 0.277354i
$$269$$ −18.9161 + 13.7434i −1.15334 + 0.837947i −0.988921 0.148444i $$-0.952573\pi$$
−0.164415 + 0.986391i $$0.552573\pi$$
$$270$$ 0 0
$$271$$ −6.34951 19.5418i −0.385705 1.18708i −0.935967 0.352087i $$-0.885472\pi$$
0.550262 0.834992i $$-0.314528\pi$$
$$272$$ 0.631036 1.94213i 0.0382622 0.117759i
$$273$$ 0 0
$$274$$ 25.3788 1.53319
$$275$$ 3.26716 0.570661i 0.197017 0.0344122i
$$276$$ 0 0
$$277$$ 15.8420 + 11.5099i 0.951856 + 0.691564i 0.951245 0.308436i $$-0.0998055\pi$$
0.000611096 1.00000i $$0.499805\pi$$
$$278$$ 1.34446 4.13783i 0.0806354 0.248170i
$$279$$ 0 0
$$280$$ −1.11866 + 0.812754i −0.0668527 + 0.0485714i
$$281$$ −11.1585 + 8.10711i −0.665659 + 0.483630i −0.868569 0.495567i $$-0.834960\pi$$
0.202910 + 0.979197i $$0.434960\pi$$
$$282$$ 0 0
$$283$$ 9.73949 29.9751i 0.578953 1.78183i −0.0433533 0.999060i $$-0.513804\pi$$
0.622306 0.782774i $$-0.286196\pi$$
$$284$$ −4.58792 3.33332i −0.272243 0.197796i
$$285$$ 0 0
$$286$$ 13.6949 14.1359i 0.809799 0.835872i
$$287$$ −4.95530 −0.292502
$$288$$ 0 0
$$289$$ −5.08404 + 15.6471i −0.299061 + 0.920415i
$$290$$ 0.122712 + 0.377669i 0.00720589 + 0.0221775i
$$291$$ 0 0
$$292$$ 4.11649 2.99080i 0.240899 0.175024i
$$293$$ −7.20413 22.1720i −0.420870 1.29530i −0.906894 0.421359i $$-0.861553\pi$$
0.486024 0.873945i $$-0.338447\pi$$
$$294$$ 0 0
$$295$$ −1.21396 0.881993i −0.0706794 0.0513516i
$$296$$ −22.5580 −1.31116
$$297$$ 0 0
$$298$$ −7.72948 −0.447757
$$299$$ 0.0248334 + 0.0180425i 0.00143615 + 0.00104343i
$$300$$ 0 0
$$301$$ 0.252179 + 0.776126i 0.0145353 + 0.0447352i
$$302$$ 3.65433 2.65503i 0.210283 0.152780i
$$303$$ 0 0
$$304$$ −5.80129 17.8545i −0.332727 1.02403i
$$305$$ 4.03125 12.4069i 0.230829 0.710418i
$$306$$ 0 0
$$307$$ −10.0938 −0.576083 −0.288042 0.957618i $$-0.593004\pi$$
−0.288042 + 0.957618i $$0.593004\pi$$
$$308$$ −0.668551 0.327409i −0.0380942 0.0186558i
$$309$$ 0 0
$$310$$ 5.56020 + 4.03972i 0.315798 + 0.229441i
$$311$$ −4.11778 + 12.6732i −0.233498 + 0.718632i 0.763819 + 0.645430i $$0.223322\pi$$
−0.997317 + 0.0732020i $$0.976678\pi$$
$$312$$ 0 0
$$313$$ −18.7435 + 13.6180i −1.05945 + 0.769732i −0.973986 0.226609i $$-0.927236\pi$$
−0.0854599 + 0.996342i $$0.527236\pi$$
$$314$$ −2.80566 + 2.03843i −0.158333 + 0.115036i
$$315$$ 0 0
$$316$$ 0.322407 0.992267i 0.0181368 0.0558194i
$$317$$ −5.60802 4.07446i −0.314978 0.228845i 0.419052 0.907962i $$-0.362363\pi$$
−0.734029 + 0.679118i $$0.762363\pi$$
$$318$$ 0 0
$$319$$ −0.747485 + 0.771552i −0.0418512 + 0.0431986i
$$320$$ 8.87727 0.496254
$$321$$ 0 0
$$322$$ −0.00108523 + 0.00334001i −6.04777e−5 + 0.000186131i
$$323$$ −1.55599 4.78886i −0.0865779 0.266459i
$$324$$ 0 0
$$325$$ 3.91592 2.84508i 0.217216 0.157817i
$$326$$ 3.35342 + 10.3208i 0.185729 + 0.571614i
$$327$$ 0 0
$$328$$ 27.1692 + 19.7396i 1.50017 + 1.08994i
$$329$$ −0.758613 −0.0418237
$$330$$ 0 0
$$331$$ −10.5717 −0.581075 −0.290538 0.956864i $$-0.593834\pi$$
−0.290538 + 0.956864i $$0.593834\pi$$
$$332$$ 6.38968 + 4.64237i 0.350679 + 0.254783i
$$333$$ 0 0
$$334$$ 7.45750 + 22.9518i 0.408056 + 1.25587i
$$335$$ −7.77281 + 5.64728i −0.424674 + 0.308544i
$$336$$ 0 0
$$337$$ −1.96123 6.03605i −0.106835 0.328805i 0.883322 0.468767i $$-0.155302\pi$$
−0.990157 + 0.139963i $$0.955302\pi$$
$$338$$ 3.95107 12.1601i 0.214910 0.661424i
$$339$$ 0 0
$$340$$ 0.367750 0.0199440
$$341$$ −2.61724 + 18.4073i −0.141731 + 0.996811i
$$342$$ 0 0
$$343$$ −5.04145 3.66283i −0.272213 0.197774i
$$344$$ 1.70906 5.25996i 0.0921466 0.283598i
$$345$$ 0 0
$$346$$ −2.88767 + 2.09801i −0.155242 + 0.112790i
$$347$$ −20.4210 + 14.8367i −1.09626 + 0.796477i −0.980445 0.196794i $$-0.936947\pi$$
−0.115811 + 0.993271i $$0.536947\pi$$
$$348$$ 0 0
$$349$$ −3.43118 + 10.5601i −0.183667 + 0.565269i −0.999923 0.0124217i $$-0.996046\pi$$
0.816256 + 0.577691i $$0.196046\pi$$
$$350$$ 0.448019 + 0.325504i 0.0239476 + 0.0173989i
$$351$$ 0 0
$$352$$ 4.25273 + 8.02942i 0.226672 + 0.427970i
$$353$$ 18.8552 1.00356 0.501780 0.864996i $$-0.332679\pi$$
0.501780 + 0.864996i $$0.332679\pi$$
$$354$$ 0 0
$$355$$ −3.52668 + 10.8540i −0.187177 + 0.576070i
$$356$$ −1.43867 4.42778i −0.0762494 0.234672i
$$357$$ 0 0
$$358$$ 6.24187 4.53498i 0.329893 0.239681i
$$359$$ 6.57983 + 20.2506i 0.347270 + 1.06879i 0.960357 + 0.278773i $$0.0899276\pi$$
−0.613087 + 0.790016i $$0.710072\pi$$
$$360$$ 0 0
$$361$$ −22.0789 16.0412i −1.16205 0.844275i
$$362$$ −13.6873 −0.719391
$$363$$ 0 0
$$364$$ −1.08642 −0.0569437
$$365$$ −8.28423 6.01885i −0.433617 0.315041i
$$366$$ 0 0
$$367$$ −9.66781 29.7545i −0.504656 1.55317i −0.801349 0.598197i $$-0.795884\pi$$
0.296693 0.954973i $$-0.404116\pi$$
$$368$$ 0.0141565 0.0102853i 0.000737957 0.000536157i
$$369$$ 0 0
$$370$$ 2.79178 + 8.59220i 0.145138 + 0.446687i
$$371$$ 1.20591 3.71141i 0.0626078 0.192687i
$$372$$ 0 0
$$373$$ 9.39368 0.486386 0.243193 0.969978i $$-0.421805\pi$$
0.243193 + 0.969978i $$0.421805\pi$$
$$374$$ −1.40850 2.65934i −0.0728319 0.137511i
$$375$$ 0 0
$$376$$ 4.15938 + 3.02196i 0.214503 + 0.155846i
$$377$$ −0.484473 + 1.49106i −0.0249517 + 0.0767933i
$$378$$ 0 0
$$379$$ 6.63173 4.81823i 0.340649 0.247496i −0.404287 0.914632i $$-0.632480\pi$$
0.744936 + 0.667136i $$0.232480\pi$$
$$380$$ 2.73515 1.98720i 0.140310 0.101941i
$$381$$ 0 0
$$382$$ −7.72664 + 23.7802i −0.395329 + 1.21670i
$$383$$ 27.2501 + 19.7984i 1.39242 + 1.01165i 0.995596 + 0.0937524i $$0.0298862\pi$$
0.396820 + 0.917897i $$0.370114\pi$$
$$384$$ 0 0
$$385$$ −0.210886 + 1.48319i −0.0107478 + 0.0755901i
$$386$$ −3.64656 −0.185605
$$387$$ 0 0
$$388$$ −0.902863 + 2.77873i −0.0458359 + 0.141069i
$$389$$ 10.1771 + 31.3219i 0.515999 + 1.58808i 0.781456 + 0.623960i $$0.214477\pi$$
−0.265457 + 0.964123i $$0.585523\pi$$
$$390$$ 0 0
$$391$$ 0.00379698 0.00275867i 0.000192021 0.000139512i
$$392$$ 6.42879 + 19.7858i 0.324703 + 0.999333i
$$393$$ 0 0
$$394$$ −5.15894 3.74819i −0.259904 0.188831i
$$395$$ −2.09965 −0.105645
$$396$$ 0 0
$$397$$ −15.3865 −0.772228 −0.386114 0.922451i $$-0.626183\pi$$
−0.386114 + 0.922451i $$0.626183\pi$$
$$398$$ −8.03669 5.83900i −0.402843 0.292683i
$$399$$ 0 0
$$400$$ −0.852662 2.62422i −0.0426331 0.131211i
$$401$$ −5.37426 + 3.90463i −0.268378 + 0.194988i −0.713832 0.700317i $$-0.753042\pi$$
0.445455 + 0.895305i $$0.353042\pi$$
$$402$$ 0 0
$$403$$ 8.38491 + 25.8061i 0.417682 + 1.28549i
$$404$$ −0.339039 + 1.04346i −0.0168678 + 0.0519138i
$$405$$ 0 0
$$406$$ −0.179370 −0.00890198
$$407$$ −17.0058 + 17.5533i −0.842945 + 0.870085i
$$408$$ 0 0
$$409$$ −19.8088 14.3919i −0.979482 0.711636i −0.0218895 0.999760i $$-0.506968\pi$$
−0.957593 + 0.288125i $$0.906968\pi$$
$$410$$ 4.15623 12.7916i 0.205262 0.631731i
$$411$$ 0 0
$$412$$ 3.89119 2.82711i 0.191705 0.139282i
$$413$$ 0.548339 0.398392i 0.0269820 0.0196036i
$$414$$ 0 0
$$415$$ 4.91167 15.1166i 0.241104 0.742043i
$$416$$ 10.7279 + 7.79428i 0.525979 + 0.382146i
$$417$$ 0 0
$$418$$ −24.8459 12.1678i −1.21526 0.595146i
$$419$$ 20.8656 1.01935 0.509676 0.860367i $$-0.329765\pi$$
0.509676 + 0.860367i $$0.329765\pi$$
$$420$$ 0 0
$$421$$ −5.23200 + 16.1025i −0.254992 + 0.784785i 0.738839 + 0.673882i $$0.235374\pi$$
−0.993831 + 0.110903i $$0.964626\pi$$
$$422$$ −0.497571 1.53137i −0.0242214 0.0745458i
$$423$$ 0 0
$$424$$ −21.3964 + 15.5454i −1.03910 + 0.754951i
$$425$$ −0.228697 0.703856i −0.0110934 0.0341420i
$$426$$ 0 0
$$427$$ 4.76717 + 3.46355i 0.230700 + 0.167613i
$$428$$ −5.84186 −0.282377
$$429$$ 0 0
$$430$$ −2.21500 −0.106817
$$431$$ −21.1677 15.3792i −1.01961 0.740791i −0.0534080 0.998573i $$-0.517008\pi$$
−0.966203 + 0.257782i $$0.917008\pi$$
$$432$$ 0 0
$$433$$ −7.74375 23.8328i −0.372141 1.14533i −0.945387 0.325949i $$-0.894316\pi$$
0.573246 0.819383i $$-0.305684\pi$$
$$434$$ −2.51152 + 1.82472i −0.120557 + 0.0875895i
$$435$$ 0 0
$$436$$ 1.92081 + 5.91163i 0.0919899 + 0.283116i
$$437$$ 0.0133332 0.0410353i 0.000637812 0.00196298i
$$438$$ 0 0
$$439$$ 5.11234 0.243999 0.121999 0.992530i $$-0.461069\pi$$
0.121999 + 0.992530i $$0.461069\pi$$
$$440$$ 7.06458 7.29203i 0.336791 0.347634i
$$441$$ 0 0
$$442$$ −3.55307 2.58146i −0.169003 0.122788i
$$443$$ 8.67635 26.7031i 0.412226 1.26870i −0.502483 0.864587i $$-0.667580\pi$$
0.914709 0.404113i $$-0.132420\pi$$
$$444$$ 0 0
$$445$$ −7.57988 + 5.50711i −0.359321 + 0.261062i
$$446$$ −28.4016 + 20.6350i −1.34486 + 0.977096i
$$447$$ 0 0
$$448$$ −1.23910 + 3.81356i −0.0585421 + 0.180174i
$$449$$ −29.9297 21.7452i −1.41247 1.02622i −0.992958 0.118467i $$-0.962202\pi$$
−0.419510 0.907751i $$-0.637798\pi$$
$$450$$ 0 0
$$451$$ 35.8422 6.26041i 1.68775 0.294791i
$$452$$ 4.92239 0.231530
$$453$$ 0 0
$$454$$ −2.98010 + 9.17181i −0.139863 + 0.430454i
$$455$$ 0.675622 + 2.07935i 0.0316736 + 0.0974815i
$$456$$ 0 0
$$457$$ −29.8092 + 21.6576i −1.39441 + 1.01310i −0.399050 + 0.916929i $$0.630660\pi$$
−0.995365 + 0.0961719i $$0.969340\pi$$
$$458$$ −6.11891 18.8321i −0.285918 0.879965i
$$459$$ 0 0
$$460$$ 0.00254938 + 0.00185224i 0.000118866 + 8.63609e-5i
$$461$$ 23.0013 1.07128 0.535640 0.844447i $$-0.320071\pi$$
0.535640 + 0.844447i $$0.320071\pi$$
$$462$$ 0 0
$$463$$ −36.1571 −1.68036 −0.840181 0.542306i $$-0.817551\pi$$
−0.840181 + 0.542306i $$0.817551\pi$$
$$464$$ 0.723042 + 0.525321i 0.0335664 + 0.0243874i
$$465$$ 0 0
$$466$$ 6.98403 + 21.4946i 0.323529 + 0.995720i
$$467$$ 1.83240 1.33132i 0.0847935 0.0616061i −0.544581 0.838708i $$-0.683311\pi$$
0.629374 + 0.777102i $$0.283311\pi$$
$$468$$ 0 0
$$469$$ −1.34106 4.12736i −0.0619244 0.190584i
$$470$$ 0.636283 1.95828i 0.0293496 0.0903287i
$$471$$ 0 0
$$472$$ −4.59348 −0.211432
$$473$$ −2.80458 5.29521i −0.128955 0.243474i
$$474$$ 0 0
$$475$$ −5.50435 3.99914i −0.252557 0.183493i
$$476$$ −0.0513310 + 0.157981i −0.00235275 + 0.00724103i
$$477$$ 0 0
$$478$$ −22.1818 + 16.1160i −1.01457 + 0.737130i
$$479$$ 24.7000 17.9456i 1.12857 0.819954i 0.143084 0.989711i $$-0.454298\pi$$
0.985486 + 0.169756i $$0.0542981\pi$$
$$480$$ 0 0
$$481$$ −11.0221 + 33.9225i −0.502564 + 1.54673i
$$482$$ 13.1137 + 9.52768i 0.597314 + 0.433974i
$$483$$ 0 0
$$484$$ 5.24935 + 1.52355i 0.238607 + 0.0692524i
$$485$$ 5.87983 0.266989
$$486$$ 0 0
$$487$$ −4.99195 + 15.3636i −0.226207 + 0.696193i 0.771960 + 0.635671i $$0.219277\pi$$
−0.998167 + 0.0605221i $$0.980723\pi$$
$$488$$ −12.3406 37.9804i −0.558632 1.71929i
$$489$$ 0 0
$$490$$ 6.74066 4.89737i 0.304512 0.221241i
$$491$$ −3.60828 11.1051i −0.162839 0.501168i 0.836031 0.548682i $$-0.184870\pi$$
−0.998871 + 0.0475137i $$0.984870\pi$$
$$492$$ 0 0
$$493$$ 0.193931 + 0.140899i 0.00873420 + 0.00634577i
$$494$$ −40.3754 −1.81658
$$495$$ 0 0
$$496$$ 15.4680 0.694534
$$497$$ −4.17048 3.03003i −0.187072 0.135916i
$$498$$ 0 0
$$499$$ −12.6501 38.9330i −0.566296 1.74288i −0.664068 0.747672i $$-0.731171\pi$$
0.0977712 0.995209i $$-0.468829\pi$$
$$500$$ 0.402006 0.292074i 0.0179782 0.0130620i
$$501$$ 0 0
$$502$$ −7.95720 24.4897i −0.355147 1.09303i
$$503$$ −0.131171 + 0.403703i −0.00584863 + 0.0180002i −0.953938 0.300003i $$-0.903012\pi$$
0.948090 + 0.318003i $$0.103012\pi$$
$$504$$ 0 0
$$505$$ 2.20797 0.0982533
$$506$$ 0.00362993 0.0255297i 0.000161370 0.00113493i
$$507$$ 0 0
$$508$$ −0.225751 0.164017i −0.0100161 0.00727710i
$$509$$ −6.12017 + 18.8360i −0.271272 + 0.834889i 0.718910 + 0.695103i $$0.244641\pi$$
−0.990182 + 0.139786i $$0.955359\pi$$
$$510$$ 0 0
$$511$$ 3.74195 2.71868i 0.165534 0.120267i
$$512$$ 19.7827 14.3729i 0.874278 0.635200i
$$513$$ 0 0
$$514$$ −1.27588 + 3.92677i −0.0562769 + 0.173202i
$$515$$ −7.83082 5.68943i −0.345067 0.250706i
$$516$$ 0 0
$$517$$ 5.48714 0.958415i 0.241324 0.0421510i
$$518$$ −4.08078 −0.179299
$$519$$ 0 0
$$520$$ 4.57882 14.0922i 0.200795 0.617982i
$$521$$ 11.3748 + 35.0079i 0.498337 + 1.53372i 0.811690 + 0.584088i $$0.198548\pi$$
−0.313353 + 0.949637i $$0.601452\pi$$
$$522$$ 0 0
$$523$$ 8.49589 6.17262i 0.371499 0.269910i −0.386333 0.922359i $$-0.626258\pi$$
0.757832 + 0.652449i $$0.226258\pi$$
$$524$$ 0.466033 + 1.43430i 0.0203587 + 0.0626577i
$$525$$ 0 0
$$526$$ −20.8737 15.1656i −0.910137 0.661253i
$$527$$ 4.14875 0.180723
$$528$$ 0 0
$$529$$ −23.0000 −0.999998
$$530$$ 8.56916 + 6.22586i 0.372220 + 0.270434i
$$531$$ 0 0
$$532$$ 0.471900 + 1.45236i 0.0204595 + 0.0629678i
$$533$$ 42.9594 31.2118i 1.86078 1.35193i
$$534$$ 0 0
$$535$$ 3.63294 + 11.1810i 0.157066 + 0.483399i
$$536$$ −9.08863 + 27.9719i −0.392569 + 1.20820i
$$537$$ 0 0
$$538$$ 28.6660 1.23588
$$539$$ 20.2426 + 9.91338i 0.871910 + 0.427000i
$$540$$ 0 0
$$541$$ −0.529593 0.384772i −0.0227690 0.0165426i 0.576343 0.817208i $$-0.304479\pi$$
−0.599112 + 0.800666i $$0.704479\pi$$
$$542$$ −7.78454 + 23.9584i −0.334375 + 1.02910i
$$543$$ 0 0
$$544$$ 1.64028 1.19173i 0.0703262 0.0510950i
$$545$$ 10.1201 7.35267i 0.433497 0.314954i
$$546$$ 0 0
$$547$$ −4.14573 + 12.7592i −0.177259 + 0.545546i −0.999729 0.0232616i $$-0.992595\pi$$
0.822471 + 0.568807i $$0.192595\pi$$
$$548$$ 8.32166 + 6.04604i 0.355484 + 0.258274i
$$549$$ 0 0
$$550$$ −3.65180 1.78839i −0.155713 0.0762574i
$$551$$ 2.20374 0.0938823
$$552$$ 0 0
$$553$$ 0.293073 0.901985i 0.0124627 0.0383563i
$$554$$ −7.41872 22.8325i −0.315191 0.970059i
$$555$$ 0 0
$$556$$ 1.42661 1.03649i 0.0605017 0.0439571i
$$557$$ −4.09205 12.5940i −0.173386 0.533626i 0.826171 0.563420i $$-0.190515\pi$$
−0.999556 + 0.0297943i $$0.990515\pi$$
$$558$$ 0 0
$$559$$ −7.07481 5.14015i −0.299232 0.217405i
$$560$$ 1.24635 0.0526679
$$561$$ 0 0
$$562$$ 16.9099 0.713300
$$563$$ 17.2822 + 12.5562i 0.728356 + 0.529182i 0.889043 0.457824i $$-0.151371\pi$$
−0.160687 + 0.987005i $$0.551371\pi$$
$$564$$ 0 0
$$565$$ −3.06115 9.42124i −0.128783 0.396355i
$$566$$ −31.2611 + 22.7125i −1.31400 + 0.954679i
$$567$$ 0 0
$$568$$ 10.7960 + 33.2265i 0.452988 + 1.39415i
$$569$$ 14.0100 43.1185i 0.587332 1.80762i −0.00236662 0.999997i $$-0.500753\pi$$
0.589698 0.807624i $$-0.299247\pi$$
$$570$$ 0 0
$$571$$ −25.1544 −1.05268 −0.526339 0.850275i $$-0.676436\pi$$
−0.526339 + 0.850275i $$0.676436\pi$$
$$572$$ 7.85817 1.37255i 0.328567 0.0573893i
$$573$$ 0 0
$$574$$ 4.91497 + 3.57093i 0.205147 + 0.149048i
$$575$$ 0.00195968 0.00603127i 8.17243e−5 0.000251522i
$$576$$ 0 0
$$577$$ −6.16999 + 4.48276i −0.256860 + 0.186620i −0.708762 0.705448i $$-0.750746\pi$$
0.451902 + 0.892068i $$0.350746\pi$$
$$578$$ 16.3184 11.8560i 0.678755 0.493144i
$$579$$ 0 0
$$580$$ −0.0497357 + 0.153071i −0.00206516 + 0.00635592i
$$581$$ 5.80831 + 4.21998i 0.240969 + 0.175074i
$$582$$ 0 0
$$583$$ −4.03358 + 28.3686i −0.167054 + 1.17491i
$$584$$ −31.3466 −1.29713
$$585$$ 0 0
$$586$$ −8.83232 + 27.1831i −0.364860 + 1.12292i
$$587$$ −2.43760 7.50217i −0.100611 0.309648i 0.888065 0.459719i $$-0.152050\pi$$
−0.988675 + 0.150071i $$0.952050\pi$$
$$588$$ 0 0
$$589$$ 30.8564 22.4185i 1.27142 0.923739i
$$590$$ 0.568489 + 1.74963i 0.0234043 + 0.0720310i
$$591$$ 0 0
$$592$$ 16.4497 + 11.9514i 0.676077 + 0.491199i
$$593$$ 8.68507 0.356653 0.178327 0.983971i $$-0.442932\pi$$
0.178327 + 0.983971i $$0.442932\pi$$
$$594$$ 0 0
$$595$$ 0.334290 0.0137045
$$596$$ −2.53448 1.84141i −0.103816 0.0754271i
$$597$$ 0 0
$$598$$ −0.0116293 0.0357914i −0.000475558 0.00146362i
$$599$$ −11.3275 + 8.22991i −0.462829 + 0.336265i −0.794640 0.607081i $$-0.792340\pi$$
0.331811 + 0.943346i $$0.392340\pi$$
$$600$$ 0 0
$$601$$ −3.18091 9.78983i −0.129752 0.399335i 0.864985 0.501798i $$-0.167328\pi$$
−0.994737 + 0.102462i $$0.967328\pi$$
$$602$$ 0.309173 0.951537i 0.0126010 0.0387817i
$$603$$ 0 0
$$604$$ 1.83076 0.0744927
$$605$$ −0.348460 10.9945i −0.0141669 0.446989i
$$606$$ 0 0
$$607$$ −22.2282 16.1497i −0.902214 0.655497i 0.0368195 0.999322i $$-0.488277\pi$$
−0.939034 + 0.343825i $$0.888277\pi$$
$$608$$ 5.75986 17.7270i 0.233593 0.718926i
$$609$$ 0 0
$$610$$ −12.9392 + 9.40090i −0.523894 + 0.380631i
$$611$$ 6.57671 4.77826i 0.266065 0.193308i
$$612$$ 0 0
$$613$$ −3.35530 + 10.3266i −0.135519 + 0.417086i −0.995670 0.0929536i $$-0.970369\pi$$
0.860151 + 0.510039i $$0.170369\pi$$
$$614$$ 10.0116 + 7.27388i 0.404037 + 0.293550i
$$615$$ 0 0
$$616$$ 2.14648 + 4.05269i 0.0864842 + 0.163287i
$$617$$ −18.2404 −0.734330 −0.367165 0.930156i $$-0.619671\pi$$
−0.367165 + 0.930156i $$0.619671\pi$$
$$618$$ 0 0
$$619$$ 6.08144 18.7167i 0.244434 0.752289i −0.751295 0.659966i $$-0.770571\pi$$
0.995729 0.0923233i $$-0.0294293\pi$$
$$620$$ 0.860790 + 2.64924i 0.0345701 + 0.106396i
$$621$$ 0 0
$$622$$ 13.2169 9.60267i 0.529951 0.385032i
$$623$$ −1.30777 4.02491i −0.0523948 0.161255i
$$624$$ 0 0
$$625$$ −0.809017 0.587785i −0.0323607 0.0235114i
$$626$$ 28.4044 1.13527
$$627$$ 0 0
$$628$$ −1.40559 −0.0560893
$$629$$ 4.41205 + 3.20554i 0.175920 + 0.127813i
$$630$$ 0 0
$$631$$ −9.40181 28.9358i −0.374280 1.15192i −0.943963 0.330051i $$-0.892934\pi$$
0.569683 0.821865i $$-0.307066\pi$$
$$632$$ −5.19997 + 3.77800i −0.206844 + 0.150281i
$$633$$ 0 0
$$634$$ 2.62619 + 8.08260i 0.104300 + 0.321001i
$$635$$ −0.173532 + 0.534076i −0.00688640 + 0.0211942i
$$636$$ 0 0
$$637$$ 32.8948 1.30334
$$638$$ 1.29740 0.226612i 0.0513647 0.00897166i
$$639$$ 0 0
$$640$$ −4.37230 3.17666i −0.172830 0.125569i
$$641$$ −11.5427 + 35.5247i −0.455908 + 1.40314i 0.414159 + 0.910205i $$0.364076\pi$$
−0.870066 + 0.492935i $$0.835924\pi$$
$$642$$ 0 0
$$643$$ −0.861554 + 0.625956i −0.0339764 + 0.0246853i −0.604644 0.796496i $$-0.706685\pi$$
0.570667 + 0.821181i $$0.306685\pi$$
$$644$$ −0.00115154 0.000836645i −4.53772e−5 3.29684e-5i
$$645$$ 0 0
$$646$$ −1.90766 + 5.87118i −0.0750559 + 0.230998i
$$647$$ −12.7864 9.28987i −0.502686 0.365223i 0.307356 0.951595i $$-0.400556\pi$$
−0.810042 + 0.586372i $$0.800556\pi$$
$$648$$ 0 0
$$649$$ −3.46288 + 3.57437i −0.135930 + 0.140306i
$$650$$ −5.93429 −0.232762
$$651$$ 0 0
$$652$$ −1.35916 + 4.18305i −0.0532287 + 0.163821i
$$653$$ −10.9241 33.6210i −0.427495 1.31569i −0.900585 0.434679i $$-0.856862\pi$$
0.473091 0.881014i $$-0.343138\pi$$
$$654$$ 0 0
$$655$$ 2.45537 1.78393i 0.0959393 0.0697040i
$$656$$ −9.35409 28.7889i −0.365216 1.12402i
$$657$$ 0 0
$$658$$ 0.752439 + 0.546679i 0.0293331 + 0.0213118i
$$659$$ 28.4474 1.10815 0.554077 0.832465i $$-0.313071\pi$$
0.554077 + 0.832465i $$0.313071\pi$$
$$660$$ 0 0
$$661$$ 39.5989 1.54022 0.770110 0.637911i $$-0.220201\pi$$
0.770110 + 0.637911i $$0.220201\pi$$
$$662$$ 10.4857 + 7.61830i 0.407538 + 0.296094i
$$663$$ 0 0
$$664$$ −15.0357 46.2752i −0.583499 1.79583i
$$665$$ 2.48629 1.80639i 0.0964140 0.0700489i
$$666$$ 0 0
$$667$$ 0.000634741 0.00195353i 2.45773e−5 7.56411e-5i
$$668$$ −3.02256 + 9.30248i −0.116946 + 0.359924i
$$669$$ 0 0
$$670$$ 11.7791 0.455068
$$671$$ −38.8573 19.0295i −1.50007 0.734627i
$$672$$ 0 0
$$673$$ 14.2080 + 10.3227i 0.547679 + 0.397912i 0.826929 0.562306i $$-0.190086\pi$$
−0.279250 + 0.960218i $$0.590086\pi$$
$$674$$ −2.40448 + 7.40024i −0.0926173 + 0.285047i
$$675$$ 0 0
$$676$$ 4.19248 3.04602i 0.161249 0.117155i
$$677$$ −36.4616 + 26.4909i −1.40133 + 1.01813i −0.406819 + 0.913509i $$0.633362\pi$$
−0.994512 + 0.104619i $$0.966638\pi$$
$$678$$ 0 0
$$679$$ −0.820716 + 2.52590i −0.0314962 + 0.0969353i
$$680$$ −1.83287 1.33165i −0.0702872 0.0510666i
$$681$$ 0 0
$$682$$ 15.8608 16.3714i 0.607340 0.626894i
$$683$$ −15.7677 −0.603334 −0.301667 0.953413i $$-0.597543\pi$$
−0.301667 + 0.953413i $$0.597543\pi$$
$$684$$ 0 0
$$685$$ 6.39676 19.6872i 0.244408 0.752210i
$$686$$ 2.36087 + 7.26603i 0.0901386 + 0.277418i
$$687$$ 0 0
$$688$$ −4.03304 + 2.93018i −0.153758 + 0.111712i
$$689$$ 12.9225 + 39.7713i 0.492307 + 1.51517i
$$690$$ 0 0
$$691$$ −2.39970 1.74348i −0.0912889 0.0663253i 0.541205 0.840891i $$-0.317968\pi$$
−0.632493 + 0.774566i $$0.717968\pi$$
$$692$$ −1.44668 −0.0549944
$$693$$ 0 0
$$694$$ 30.9465 1.17471
$$695$$ −2.87098 2.08589i −0.108903 0.0791224i
$$696$$ 0 0
$$697$$ −2.50891 7.72162i −0.0950316 0.292477i
$$698$$ 11.0132 8.00153i 0.416854 0.302862i
$$699$$ 0 0
$$700$$ 0.0693589 + 0.213465i 0.00262152 + 0.00806821i
$$701$$ −11.4738 + 35.3128i −0.433360 + 1.33375i 0.461397 + 0.887194i $$0.347348\pi$$
−0.894757 + 0.446553i $$0.852652\pi$$
$$702$$ 0 0
$$703$$ 50.1364 1.89093
$$704$$ 4.14459 29.1494i 0.156205 1.09861i
$$705$$ 0 0
$$706$$ −18.7017 13.5876i −0.703848 0.511375i
$$707$$ −0.308191 + 0.948516i −0.0115907 + 0.0356726i
$$708$$ 0 0
$$709$$ 25.6867 18.6625i 0.964683 0.700883i 0.0104495 0.999945i $$-0.496674\pi$$
0.954234 + 0.299062i $$0.0966738\pi$$
$$710$$ 11.3197 8.22423i 0.424820 0.308650i
$$711$$ 0 0
$$712$$ −8.86303 + 27.2776i −0.332156 + 1.02227i
$$713$$ 0.0287608 + 0.0208959i 0.00107710 + 0.000782558i
$$714$$ 0 0
$$715$$ −7.51386 14.1866i −0.281002 0.530549i
$$716$$ 3.12708 0.116864
$$717$$ 0 0
$$718$$ 8.06692 24.8274i 0.301055 0.926552i
$$719$$ 9.85059 + 30.3170i 0.367365 + 1.13063i 0.948487 + 0.316816i $$0.102614\pi$$
−0.581122 + 0.813816i $$0.697386\pi$$
$$720$$ 0 0
$$721$$ 3.53714 2.56989i 0.131730 0.0957075i
$$722$$ 10.3394 + 31.8213i 0.384792 + 1.18427i
$$723$$ 0 0
$$724$$ −4.48806 3.26077i −0.166797 0.121185i
$$725$$ 0.323900 0.0120294
$$726$$ 0 0
$$727$$ 41.4129 1.53592 0.767959 0.640499i $$-0.221272\pi$$
0.767959 + 0.640499i $$0.221272\pi$$
$$728$$ 5.41470 + 3.93401i 0.200682 + 0.145804i
$$729$$ 0 0
$$730$$ 3.87945 + 11.9397i 0.143585 + 0.441909i
$$731$$ −1.08172 + 0.785918i −0.0400090 + 0.0290682i
$$732$$ 0 0
$$733$$ 14.7376 + 45.3578i 0.544347 + 1.67533i 0.722537 + 0.691333i $$0.242976\pi$$
−0.178189 + 0.983996i $$0.557024\pi$$
$$734$$ −11.8528 + 36.4792i −0.437495 + 1.34647i
$$735$$ 0 0
$$736$$ 0.0173734 0.000640391
$$737$$ 14.9145 + 28.1594i 0.549381 + 1.03726i
$$738$$ 0 0
$$739$$ 10.0777 + 7.32187i 0.370714 + 0.269339i 0.757507 0.652827i $$-0.226417\pi$$
−0.386793 + 0.922167i $$0.626417\pi$$
$$740$$ −1.13152 + 3.48246i −0.0415955 + 0.128018i
$$741$$ 0 0
$$742$$ −3.87065 + 2.81219i −0.142096 + 0.103239i
$$743$$ −22.1446 + 16.0890i −0.812407 + 0.590249i −0.914528 0.404524i $$-0.867437\pi$$
0.102120 + 0.994772i $$0.467437\pi$$
$$744$$ 0 0
$$745$$ −1.94823 + 5.99603i −0.0713775 + 0.219677i
$$746$$ −9.31722 6.76936i −0.341128 0.247844i
$$747$$ 0 0
$$748$$ 0.171694 1.20754i 0.00627775 0.0441521i
$$749$$ −5.31033 −0.194035
$$750$$ 0 0
$$751$$ −12.9393 + 39.8231i −0.472162 + 1.45317i 0.377585 + 0.925975i $$0.376755\pi$$
−0.849747 + 0.527191i $$0.823245\pi$$
$$752$$ −1.43203 4.40733i −0.0522207 0.160719i
$$753$$ 0 0
$$754$$ 1.55503 1.12979i 0.0566308 0.0411447i
$$755$$ −1.13852 3.50400i −0.0414349 0.127523i
$$756$$ 0 0
$$757$$ −13.0407 9.47464i −0.473973 0.344362i 0.325015 0.945709i $$-0.394631\pi$$
−0.798988 + 0.601347i $$0.794631\pi$$
$$758$$ −10.0499 −0.365029
$$759$$ 0 0
$$760$$ −20.8278 −0.755504
$$761$$ 8.83760 + 6.42090i 0.320363 + 0.232757i 0.736330 0.676622i $$-0.236557\pi$$
−0.415967 + 0.909380i $$0.636557\pi$$
$$762$$ 0 0
$$763$$ 1.74604 + 5.37376i 0.0632109 + 0.194543i
$$764$$ −8.19875 + 5.95674i −0.296620 + 0.215507i
$$765$$ 0 0
$$766$$ −12.7610 39.2744i −0.461075 1.41904i
$$767$$ −2.24442 + 6.90762i −0.0810414 + 0.249420i
$$768$$ 0 0
$$769$$ −34.2074 −1.23355 −0.616775 0.787139i $$-0.711561\pi$$
−0.616775 + 0.787139i $$0.711561\pi$$
$$770$$ 1.27800 1.31914i 0.0460558 0.0475386i
$$771$$ 0 0
$$772$$ −1.19570 0.868727i −0.0430342 0.0312662i
$$773$$ 2.66894 8.21416i 0.0959952 0.295443i −0.891517 0.452988i $$-0.850358\pi$$
0.987512 + 0.157545i $$0.0503580\pi$$
$$774$$ 0 0
$$775$$ 4.53521 3.29503i 0.162910 0.118361i
$$776$$ 14.5619 10.5798i 0.522742 0.379794i
$$777$$ 0 0
$$778$$ 12.4772 38.4009i 0.447329 1.37674i
$$779$$ −60.3852 43.8724i −2.16352 1.57189i
$$780$$ 0 0
$$781$$ 33.9937 + 16.6477i 1.21639 + 0.595701i
$$782$$ −0.00575405 −0.000205764
$$783$$ 0 0
$$784$$ 5.79466 17.8341i 0.206952 0.636934i
$$785$$ 0.874113 + 2.69024i 0.0311984 + 0.0960189i
$$786$$ 0 0
$$787$$ 18.6918 13.5804i 0.666291 0.484089i −0.202491 0.979284i $$-0.564904\pi$$
0.868781 + 0.495196 | 23,194 | 40,776 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | longest | en | 0.320994 |
https://www.hometalk.com/posts/decorate/walls/how-to-hang-a-gallery-wall-27827605 | 1,669,717,498,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00503.warc.gz | 853,006,995 | 38,305 | # How to Hang a Gallery Wall
by Unlikely Martha
2 Materials
I have always loved gallery walls, so it was a no brainer to put one on a small stretch of wall in our living room. Getting the frames level and evenly spaced was not an easy task.
I ordered nine frames from Michaels's on a deal. I decided to go with 5X5 prints. Below are the steps I took to get my gallery wall as straight and even as possible.
First, measure the entire area where you will be hanging your photos. You’ll need to do a little math to determine where the center of the wall is. Once I found that number in inches I hung the middle of my middle frame there and worked out. That being said it SEEMS to be easier to work with an odd number, but whatever works for you.
Once you determine where you will be hanging put a piece of tape over the hanger on the back of the frame and put a hole in the tape where the nail should go.
Put all of your pieces of tape on the wall. If you are unlike me, now would be the time to make sure things are level. LEVEL AT THIS POINT and make any adjustments
Now put a nail where that tiny hole in the tape is.
Before putting pictures in the frame, be sure to clean the glass well
Put in your print and determine where you want it. Once that is determined add a small piece of tape to the top to keep the print from shifting.
Now hang!
I am in love with how the photos turned out. They aren't perfect but they are good enough for me. I love displaying my favorite photos.
##### Suggested materials:
• Frames (Michael's)
• 5X5 Photo Prints (Walmart.com)
2 questions
• V Valencia on Feb 24, 2017
I love your gallery wall! It looks great! I didn't quite understand how you used tape to "mark" where the nail would go. Are you using double stick tape? How do you place tape at back of frame and have it stay in place?
• Millie on Feb 24, 2017
Do you have any idea of the color? | 455 | 1,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.955694 |
http://inuyasha.14.saralearn.com/practice-5-4-factoring-quadratic-expressions-worksheet-answers/ | 1,597,003,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738573.99/warc/CC-MAIN-20200809192123-20200809222123-00086.warc.gz | 53,739,931 | 18,921 | ## Advanced Algebra Honors Wkst 5 4 Scevmath
Wkst 5 4 Practice 5 4 Factoring Quadratic Expressions Factor Each Expression Completely 1 X2 4 X 4 2 X2 7x 10 3 X2 7 X 8 4 X2 6x 5 2x2 9x 4 6 X2 2 X 35 7 X2 6 X 5 8 X2 9 9 X2 13 X 48 10 X2 4 11 4x2 X 12 X2 29 X 100 13 X2 X 6 14 9x2 1 15 3x2 2x 16 X2
Wkst 5 4 Practice 5 4 Factoring Quadratic Expressions Factor Each Expression Completely 1 X2 4 X 4 2 X2 7x 10 3 X2 7 X 8 4 X2 6x 5 2x2 9x 4 6 X2 2 X 35 7 Microsoft Word 5 4 Practice Author Rcoons Created Date Advanced Algebra Honors Wkst 5 4 39 You Can Represent The Area Of A Square Tabletop With The Expression Lw 4 4 Formk Verona Public
#### What Are The Answers For Chapter 5 4 Practice Sheet
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Practice 5 4 Factoring Quadratic Expressions Worksheet Answers. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
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There isn't anything like a superb story, and nothing like being the person who started a renowned urban legend. Deciding upon the ideal approach route Cursive writing is basically joined-up handwriting. Practice reading by yourself as often as possible.
Research urban legends to obtain a concept of what's out there prior to making a new one. You are still not sure the radicals have the proper idea. Naturally, you won't use the majority of your ideas. If you've got an idea for a tool please inform us. That means you can begin right where you are no matter how little you might feel you've got to give. You are also quite suspicious of any revolutionary shift. In earlier times you've stated that the move of independence may be too early.
Each lesson in handwriting should start on a fresh new page, so the little one becomes enough room to practice. Every handwriting lesson should begin with the alphabets. Handwriting learning is just one of the most important learning needs of a kid. Learning how to read isn't just challenging, but fun too.
The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused. | 1,322 | 5,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.577251 |
https://www.hexadecimaldictionary.com/hexadecimal/0x1fc/ | 1,600,760,189,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00271.warc.gz | 878,264,612 | 3,748 | # Hexadecimal 0x1fc = 508
=
Decimal 508 111111100 774 0x1fc 0.0.1.252 Five hundred eight
0x1fc is represented as: 256 + 240 + 12.
### More details
Prime Number 508 is not prime 2, 2, 127 | 76 | 190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.575296 |
https://www.springerprofessional.de/parameterized-domination-in-circle-graphs/11647332 | 1,585,956,074,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518767.60/warc/CC-MAIN-20200403220847-20200404010847-00443.warc.gz | 1,164,797,138 | 20,008 | main-content
## Weitere Artikel dieser Ausgabe durch Wischen aufrufen
01.01.2014 | Ausgabe 1/2014
# Parameterized Domination in Circle Graphs
Zeitschrift:
Theory of Computing Systems > Ausgabe 1/2014
Autoren:
Nicolas Bousquet, Daniel Gonçalves, George B. Mertzios, Christophe Paul, Ignasi Sau, Stéphan Thomassé
Wichtige Hinweise
A preliminary conference version of this work appeared in the Proceedings of the 38th International Workshop on Graph-Theoretic Concepts in Computer Science (WG), Jerusalem, Israel, June 2012. The third author was partially supported by EPSRC Grant EP/G043434/1. The other authors were partially supported by AGAPE (ANR-09-BLAN-0159) and GRATOS (ANR-09-JCJC-0041) projects (France). This work has been supported by the EPSRC Grant EP/K022660/1.
## Abstract
A circle graph is the intersection graph of a set of chords in a circle. Keil [Discrete Appl. Math., 42(1):51–63, 1993] proved that Dominating Set, Connected Dominating Set, and Total Dominating Set are NP-complete in circle graphs. To the best of our knowledge, nothing was known about the parameterized complexity of these problems in circle graphs. In this paper we prove the following results, which contribute in this direction:
• Dominating Set, Independent Dominating Set, Connected Dominating Set, Total Dominating Set, and Acyclic Dominating Set are W[1]-hard in circle graphs, parameterized by the size of the solution.
• Whereas both Connected Dominating Set and Acyclic Dominating Set are W[1]-hard in circle graphs, it turns out that Connected Acyclic Dominating Set is polynomial-time solvable in circle graphs.
• If T is a given tree, deciding whether a circle graph G has a dominating set inducing a graph isomorphic to T is NP-complete when T is in the input, and FPT when parameterized by t=|V(T)|. We prove that the FPT algorithm runs in subexponential time, namely $$2^{\mathcal{O}(t \cdot\frac{\log\log t}{\log t})} \cdot n^{\mathcal{O}(1)}$$, where n=|V(G)|.
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Zur Ausgabe | 874 | 3,193 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.767211 |
https://www.construct.net/en/forum/construct-classic/construct-classic-discussion-37/isometric-games-30894 | 1,679,967,999,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00452.warc.gz | 816,854,187 | 15,520 | # Isometric Games
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• 4 posts
From the Asset Store
Build your map with these isometric objects and terrains
• Has anyone tried to create isometric games with construct. I'm thinking of trying to create an isometric game one day but I have not much experience. One thing that have stucked in my mind is that with isometic games you can only se 2 walls. So if you have levers, keyholes and hidden spaces you can only place them on those 2 walls. Then what about if you could turn the view so you can se the other 2 walls. It could give the game a new dimension. How complex would it be to make this with construct?
/Andla
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• Guess it would be possible....
• Most Isometric games get around this problem by making the 2 walls that you normally can't see, go transparent when you walk near them. That way you can see your player guy, as well as any switches/doors on those walls. Trying to rotate the view is probably the HARDEST way of getting around this problem (that I like to call ISO-Syndrome). The other big challenge you will face when going isometric view is that the graphics required for your game are automatically 10x more complicated than a side view game, and about 4X more complicated than an overhead view game. I would highly suggest checking out how games like "Alien Shooter Vengence" works in terms of how it handles the isometric view with the play mechanics you have mentioned.
~Sol
• Isometric games are really something I love to play around with, particularly grid-based ones.
What you're thinking would probably be very difficult to make with Construct, however I can see how it could be done (with lots of improvisation).
I will probably attempt an isometric game sometime, however right now I think I'll try to write a grid data parser (ie, it takes a string of 4,4,2,5,6,2,65,7,3,1,6 into a grid of tiles identified by ID).
After that we'd need to make some form of grid to isometric stuff, basically getting the right points to attach the grid tiles to on their XY coordinates (and maybe Z coordinates), and the character movement could be dealt with using a custom movement (always move x distance towards ('newx'),('newy') + Key Up Arrow is Pressed -> Add 1 to ('gridy'), set ('newy') to <work out some crazy "the grid space is here in pixels" stuff>)... after that it'd be a case of making walls and stuff attach to the grid spaces where appropriate, and then making a game.
Alternatively, we could simply use 3D boxes and sprited characters with even a stationary camera, but... y'know, that'd be easier. And much less cool, and you wouldn't really be able to brag about it much because Construct did way too much of the work. | 660 | 2,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-14 | latest | en | 0.961401 |
https://www.hackmath.net/en/math-problem/6514 | 1,603,740,417,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891624.95/warc/CC-MAIN-20201026175019-20201026205019-00177.warc.gz | 753,162,569 | 11,975 | # Bottles
The must is sold in 5-liter and 2-liter bottles. Mr Kucera bought a total of 216 liters in 60 bottles. How many liters did Mr. Kucera buy in five-liter bottles?
Correct result:
a = 160 l
#### Solution:
a/5+b/2 = 60
a+b = 216
2a+5b = 600
a+b = 216
a = 160
b = 56
Our linear equations calculator calculates it.
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http://www.fixya.com/support/t25672680-calculate_z_score_casio_fx_115es_plus | 1,508,248,171,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821189.10/warc/CC-MAIN-20171017125144-20171017145144-00744.warc.gz | 455,383,963 | 33,318 | Question about Casio fx-115ES Plus Scientific Calculator
# How can I calculate the z score on my casio fx-115ES plus calculator? I need to know for statistics. (It's also if the problem has a normal distribution).
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• Casio Master
Z-score=(X-mu)/(sigma_d)
(X-value - mean value)/(standard deviation)
Posted on Jul 03, 2015
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
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Posted on Jan 02, 2017
The Casio FX-115W/991W/115E/991ES have three functions P(, Q(, R( and t(: the last one calculates the normalized variable noted t.
The Sharp EL-520W ans several similar Sharps do it too. Here is a screen capture.
Posted on Jun 22, 2010
The Casio FX-115W/991W/115E/991ES have three functions P(, Q(, R( and t(: the last one calculates the normalized variable noted t.
The Sharp EL-520W ans several similar Sharps do it too. Here is a screen capture.
Posted on Jun 22, 2010
Press MODE 3 AC to enter the statistics mode. Press SHIFT STAT 7 to bring up the distribution menu then 1 to invoke P( .
Posted on Nov 03, 2012
×
my-video-file.mp4
×
## Related Questions:
### How to get l1 l2 l3 charts
L1, L2, L3?
If you are doing statistics, Set MODE to 3. Stat.
Choose 1-Var or any regression model. If you press 1-var you will have two lists x and frequency.
For any regression model you have 3 lists x, y, and Freq(uency).
L1, L2, L3, are list names in the TI 83/84 Plus and the Casio graphing calculators.
I hope it helps
Oct 26, 2013 | Casio fx-115ES Plus Scientific Calculator
### How to use standard deviation on a casio fx-115es plus
Follow the instructions under "Statistical Calculations" beginning on page E-44 of the manual. Examples 053 through 059 in the Appendix show an example (especially 057).
May 06, 2013 | Casio fx-115ES Plus Scientific Calculator
### How to find the Standard deviation on my calculator?
Follow the instructions under "Statistical Calculations" beginning on page E-44 of the manual. Examples 053 through 059 in the Appendix show an example (especially 057).
May 06, 2013 | Casio fx-115ES Plus Scientific Calculator
### Can I calculate p-value on the casio fx-115es plus?
Press MODE 3 AC to enter the statistics mode. Press SHIFT STAT 7 to bring up the distribution menu then 1 to invoke P( .
Nov 03, 2012 | Casio FX-115ES Scientific Calculator
### Calculator statistics question, z score and std norm distribution
The Casio FX-115W/991W/115E/991ES have three functions P(, Q(, R( and t(: the last one calculates the normalized variable noted t.
The Sharp EL-520W ans several similar Sharps do it too. Here is a screen capture.
May 13, 2010 | Casio FX-260 Calculator
### Calculator statistics question, z score and std norm distribution
The Casio FX-115W/991W/115E/991ES have three functions P(, Q(, R( and t(: the last one calculates the normalized variable noted t.
The Sharp EL-520W ans several similar Sharps do it too. Here is a screen capture.
Mar 23, 2010 | Casio FX-260 Calculator
### Standard deviation
standard deviation
calculate mean of a data set:
1. Turn calc. on or press MODE SeTUP
2.Press 3 for STAT
3. Press 1 for VAR
4.Enter values for X column
5. WHen done entering press SHIFT 1
6.Press 5
7.Press 2 to calculate the mean
8.Press =
Oct 05, 2008 | Casio FX-115ES Scientific Calculator
## Open Questions:
#### Related Topics:
1,241 people viewed this question
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Level 3 Expert | 1,053 | 3,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-43 | latest | en | 0.877229 |
https://www.physicsforums.com/threads/gas-undergoing-isothermic-compression.790710/ | 1,701,992,538,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00622.warc.gz | 1,035,149,835 | 25,179 | # Gas undergoing isothermic compression
• Karol
But if you make no assumptions about work done, it cannot be determined.I don't think my question has been answered. for a perfect gas, the temperature does not change during free expansion. this is true only for a perfect gas, because in real gases there is an attraction force between the molecules, and when the volume increases the molecules lose energy and hence the temperature decreases. is my understanding good?Yes, that's a good understanding.
## Homework Statement
10 liters of air at atmospheric pressure and temperature 3000K were isothermally compressed to a volume of 2[liter] and then freely expanded adiabatically to their original volume. what's the final temperature.
γ of air=1.4
## The Attempt at a Solution
I ignore the first isothermic stage and start from the end conditions of the first stage: $$V_1=2[liter], T_1=300^0K$$:
##T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\rightarrow 300\cdot 2^{0.4}=T_2\cdot 10^{0.4}##
##\rightarrow T_2=158^0K=-115^0C##
It should be -1020C
Karol said:
...and then freely expanded adiabatically ...
A "free" expansion generally denotes an irreversible process in which the gas is allowed to expand into a vacuum without doing any work. Does ##TV^{\gamma - 1} = const## apply to an irreversible process? If the expansion is adiabatic and free, what happens to the energy of the gas during the expansion? What does that tell you about the temperature change during the expansion (assuming ideal gas behavior)?
Anyway, I don't see how to get the answer of -102 oC.
I think that by the term "freely expanded," they meant for you to assume that the adiabatic expansion took place against a resistance of 1 atm. That would explain why their answer of -102 is higher than your answer of -115. I realize that the term free expansion correctly refers to expansion against a vacuum, but I don't think that is what they meant here. Try the problem with the assumption of 1 atm resistance and see what you get.
Chet
TSny said:
Does ##TV^{\gamma - 1} = const## apply to an irreversible process?
I didn't learn about reversible or irreversible processes. do the equations ##\frac{PV}{T}=Const,\ TV^{\gamma - 1} = const## apply only to reversible processes? and what is an irreversible process? is, in our case of free expansion, the process irreversible because if we compress the gas back we must invest work while the expansion occur without work and it will contradict the conservation of energy?
In the free adiabatic expansion no work is done and no heat exchange occurs so internal energy is conserved and the temperature remains unchanged, so ##TV^{\gamma - 1} = const## doesn't apply. so how do we solve?
Chestermiller said:
I think that by the term "freely expanded," they meant for you to assume that the adiabatic expansion took place against a resistance of 1 atm. That would explain why their answer of -102 is higher than your answer of -115
The first law: ##Q\mbox{(heat exchange)}=U_2-U_1+W##, Q=0, W>0 (work is done against resistance) so U, the internal energy, in this case should be lower, not higher, than expansion into vacuum so the temperature in expansion against 1[atm] must be lower than -1150, not higher.
Now i will try to solve with the assumption of resistance.
Karol said:
The first law: ##Q\mbox{(heat exchange)}=U_2-U_1+W##, Q=0, W>0 (work is done against resistance) so U, the internal energy, in this case should be lower, not higher, than expansion into vacuum so the temperature in expansion against 1[atm] must be lower than -1150, not higher.
Now i will try to solve with the assumption of resistance.
When you solved for the -115, it wasn't for free expansion into vacuum. It was for adiabatic reversible expansion. U will be higher for the expansion at 1 atm resistance than for the case of reversible adiabatic expansion that you assumed.
Chet
Chestermiller said:
U will be higher for the expansion at 1 atm resistance than for the case of reversible adiabatic expansion that you assumed.
I don't know nothing about this and about reversible/irreversible processes. maybe this topic will appear at the second law, otherwise i will have to learn it by myself.
But in any case, in adiabatic expansion into vacuum doesn't U conserve? there isn't any work and no exchange of heat, so why is it even lower than against resistance?
I quote from wikipedia, about irreversible process:
"By releasing pressure on a sample and thus allowing it to occupy a large space, the system and surroundings will have completely left equilibrium, and heat dissipation will be large compared to the little work done."
But i read that ideal gas that undergoes a process of choking doesn't change temperature, but real gases do. so does the article in wikipedia refer to ideal gases or not?
I understand that in free expansion the heat loss is the biggest thus how to calculate assuming resistance of 1[atm]? so far i studied only first law and the varios forms of the equation of state but i am curios. secondly, is it possible with my knowlegde to get to -1020C or i did right by calculating only according adiabatic expnsion?
Karol said:
I quote from wikipedia, about irreversible process:
"By releasing pressure on a sample and thus allowing it to occupy a large space, the system and surroundings will have completely left equilibrium, and heat dissipation will be large compared to the little work done."
But i read that ideal gas that undergoes a process of choking doesn't change temperature, but real gases do. so does the article in wikipedia refer to ideal gases or not?
Choking (aka throttling) is a process that takes place in an open system with gas flowing into and out of the system. To analyze that, you need to use the open system version of the first law, which you probably haven't learned yet. In any case, choking is not relevant to your problem which involves a closed system.
Karol said:
I don't know nothing about this and about reversible/irreversible processes. maybe this topic will appear at the second law, otherwise i will have to learn it by myself.
If you wanted the expansion to occur reversibly, you would back off very gradually on the pressure you apply to the piston (assumed massless), rather than releasing it suddenly. Your gamma equations apply to the case where you back off very gradually.
But in any case, in adiabatic expansion into vacuum doesn't U conserve?
Yes.
there isn't any work and no exchange of heat, so why is it even lower than against resistance?
That's why I said in response #3 that maybe they didn't really mean perfectly free expansion (which is expansion against a vacuum). Maybe they meant expansion against a pressure of 1 atm., and they mistakenly used the term free expansion. As you correctly indicated, if it were truly free expansion, the final temperature would be the same as the initial temperature.
Chet
Karol said:
I understand that in free expansion the heat loss is the biggest
Actually, there is no heat loss in adiabatic free expansion and there is no work done.
thus how to calculate assuming resistance of 1[atm]? so far i studied only first law and the varios forms of the equation of state but i am curios.
The work done against a constant resistance of 1 atm is W = 1 (10 - 2) liter-atm, which is equal to 800 J. So ΔU would be -800 J.
secondly, is it possible with my knowledge to get to -1020C or i did right by calculating only according adiabatic expnsion?
I solved the problem assuming expansion against a constant pressure of 1 atm., but got a final temperature of 204K (= -69 C). I can't account for how they got -102 C. To get that result, they would have had to expand against a constant pressure of about 1.34 atm.
Chet
I am totally confused, i don't understand. i saw your result for calculating against 1[atm] but i want to try myself.
If i insert 1[atm] and 10[liter], with the original number of moles n into the equation of state PV=nRT, i get the same temperature like in the beginning: 3000K.
I don't understand when to use what.
If we assume expansion against 1[atm] then it's also P, the final pressure, and the volume is 10[liter] so i don't have a choice and i have to use these numbers in the equation of state. and the mass of the gas remains unchanged, so i have to use the same number of mols n also.
Karol said:
I am totally confused, i don't understand. i saw your result for calculating against 1[atm] but i want to try myself.
If i insert 1[atm] and 10[liter], with the original number of moles n into the equation of state PV=nRT, i get the same temperature like in the beginning: 3000K.
I don't understand when to use what.
If we assume expansion against 1[atm] then it's also P, the final pressure, and the volume is 10[liter] so i don't have a choice and i have to use these numbers in the equation of state. and the mass of the gas remains unchanged, so i have to use the same number of mols n also.
I completely understand your source of confusion. What you are missing is that, once the piston reaches the final location where the volume is 10 liters (and then stops moving), the gas within the cylinder has not yet achieved thermodynamic equilibrium. There are still spatial variations in gas temperature and pressure within the cylinder. No more work will be done, and no heat will enter or leave the gas, but it will take some additional time before the gas achieves thermodynamic equilibrium. At the end of this additional time, the temperature and pressure of the gas will become uniform at the final thermodynamic equilibrium values. The final temperature will be less than 300 C (because the gas has done adiabatic work) and the final pressure will be less than 1 atm.
Chet
How did you calculate with 1[atm] resistance, or should i wait until i learn it later?
And how is it possible that the pressure will go down below i[atm] if the gas expands against resistance of 1[atm]?
Karol said:
How did you calculate with 1[atm] resistance, or should i wait until i learn it later?
See post # 11.
And how is it possible that the pressure will go down below i[atm] if the gas expands against resistance of 1[atm]?
After the expansion is complete, the pressure decreases as the system equilibrates further at constant volume.
Chet[/QUOTE]
Last edited:
Here's the full calculation:
##n=\frac{pv}{RT}=\frac{(1)(10)}{300R}##
where R = 0.082 (liter-atm)/(degree-mole)
So ##ΔU=nC_v(T-300)=-W=-PΔV=-(1)(8)##
So ##\frac{(1)(10)}{300R}C_v(T-300)=-8## liter-atm
So ##T-300=-300(0.8)\frac{R}{C_v}=-300(0.8)(γ-1)##
Chet
Why Cv for the adiabatic expansion? it's not at fixed volume
Karol said:
Why Cv for the adiabatic expansion? it's not at fixed volume
Cv is defined more generally by the equation:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$
where U is the internal energy per unit mass. For a process at constant volume, this is consistent with C_vdT=dU=dQ. For an ideal gas, both U and Cv are functions only of temperature.
Chet
Chestermiller said:
this is consistent with C_vdT=dU=dQ.
You meant CvdT=dU=dQ, the dash was a typing mistake, right?
So what is Cp then? and why is Cv suitable for a changing volume? Cv is the temperature rise for a unit mass for 10 at constant volume, that's what i learned
Karol said:
You meant CvdT=dU=dQ, the dash was a typing mistake, right?
Yes. Sometimes I mistakenly use LaTex symbology when I'm applying the regular text symbology.
So what is Cp then?
$$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$
where H is the enthalpy U + PV.
and why is Cv suitable for a changing volume? Cv is the temperature rise for a unit mass for 10 at constant volume, that's what i learned
This is a problem with the way they teach this stuff in many of the books and at schools. It's because they start out relating the heat capacities to Q, which is a characteristic of the process, rather than of the material being processed. Cv and Cp are properties of the material, not the process. Yes, in a constant volume process, dQ = Cv dT and dU = CvdT for an ideal gas. In a non-constant volume process, dQ is not equal to CvdT, but, because Cv is a material property, dU is still equal to CvdT for an ideal gas. This has been an unending source of confusion for students over the ages. You are not alone.
If you still have doubts about what I'm saying, look up the derivations of your "gamma equations" for adiabatic reversible processes. These are not constant volume process nor constant pressure processes, but they still use Cv in their derivations for relating pressures, volumes, and temperatures.
Chet
Last edited:
Oh yes. One more thing. For an ideal gas, the internal energy (and enthalpy) are functions only of temperature, independent of pressure and volume. So, even if the volume changes, dU is still equal to CvdT.
Chet
Chestermiller said:
In a non-constant volume process, dQ is not equal to CvdT, but, because Cv is a material property, dU is still equal to CvdT for an ideal gas.
I combine this with what i read in Wikipedia. So Cv isn't ##\frac{\mbox{invested heat}}{\mbox{temperature rise}}## it's:
##C_v=\left(\frac{\partial U}{\partial T}\right)=\frac{\mbox{invested heat}}{\mbox{internal energy rise}}##, right?
But is it ##\left(\frac{\partial U}{\partial T}\right)## or ##\left(\frac{\partial U}{\partial T}\right)_v##? i mean is Cv the internal energy rise in any case and any condition or just in constant volume? from your answer i guess it's only ##C_v=\left(\frac{\partial U}{\partial T}\right)##, in any process, not necessarily at constant volume, right?
Just now i saw your last post. i didn't learn yet about enthalpy but i read it's equal to U+PV, so it depends not only on temperature, no?
Karol said:
I combine this with what i read in Wikipedia. So Cv isn't ##\frac{\mbox{invested heat}}{\mbox{temperature rise}}## it's:
##C_v=\left(\frac{\partial U}{\partial T}\right)=\frac{\mbox{invested heat}}{\mbox{internal energy rise}}##, right?
But is it ##\left(\frac{\partial U}{\partial T}\right)## or ##\left(\frac{\partial U}{\partial T}\right)_v##? i mean is Cv the internal energy rise in any case and any condition or just in constant volume? from your answer i guess it's only ##C_v=\left(\frac{\partial U}{\partial T}\right)##, in any process, not necessarily at constant volume, right?
No. Only for an ideal gas is it dU/dT, since, for an ideal gas, U is not a function of V. For non-ideal (real) gases (and other materials), ##dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV##
Chet
Chestermiller said:
No. Only for an ideal gas is it dU/dT, since, for an ideal gas, U is not a function of V. For non-ideal (real) gases (and other materials), ##dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV##
When you are thinking about U, you should not be thinking about the process. U is a function of state, independent of process path, and depends only on the initial and final state of the material (T,V).
Chet
Karol said:
I combine this with what i read in Wikipedia. So Cv isn't ##\frac{\mbox{invested heat}}{\mbox{temperature rise}}## it's:
##C_v=\left(\frac{\partial U}{\partial T}\right)=\frac{\mbox{invested heat}}{\mbox{internal energy rise}}##, right?
For an ideal gas, it's just the rate of change of the internal energy of the material with respect to temperature. The internal energy of a monoatomic ideal gas is just the sum of the kinetic energies of the individual molecules. For a diatomic gas, it also includes the vibrational energy. These energies both increase with temperature.
Just now i saw your last post. i didn't learn yet about enthalpy but i read it's equal to U+PV, so it depends not only on temperature, no?
Yes. Good observation.
Chet
so Cv is measured in constant volume but is used in all processes? there must be an explanation to the constant volume notation, other than the effect on ideal gas that you described in post #20: "Yes, in a constant volume process, dQ = Cv dT and dU = CvdT for an ideal gas". so far i didn't find one.
$$\left(\frac{\partial U}{\partial V}\right)_TdV$$ change of volume against pressure is work, does this member account for the work? if yes, why is it missing in ideal gas?
If not please give me an example since i don't know how just a change in volume changes internal energy
Last edited:
Karol said:
so Cv is measured in constant volume but is used in all processes?
Yes.
there must be an explanation to the constant volume notation, other than the effect on ideal gas that you described in post #20: "Yes, in a constant volume process, dQ = Cv dT and dU = CvdT for an ideal gas". so far i didn't find one.
Actually, dU=CvdT applies generally to any single phase constant composition system at constant volume.
$$\left(\frac{\partial U}{\partial V}\right)_TdV$$ change of volume against pressure is work, does this member account for the work?
Not completely. The reversible work -PdV is one term that contributes to $$\left(\frac{\partial U}{\partial V}\right)_TdV$$. But that's not the only term. When you study the second law of thermo, you will find that there is also another term that contributes, related to the effect of volume on entropy. The complete expression is
$$\left(\frac{\partial U}{\partial V}\right)_TdV=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)dV$$
So, the full expression for dU in terms of changes in temperature and specific volume is:
##dU=C_vdT+\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)dV##
if yes, why is it missing in ideal gas?
What is the term in parenthesis equal to for an ideal gas?
If not please give me an example since i don't know how just a change in volume changes internal energy
For an ideal gas, it doesn't. For a real gas, the term in parenthesis is not zero.
Chet
For ideal gas:
$$PV=nRT\rightarrow \left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{T}$$
$$\left(\frac{\partial P}{\partial T}\right)_V=T\frac{nR}{T}=P$$
$$\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)=0$$
Karol said:
For ideal gas:
$$PV=nRT\rightarrow \left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{T}$$
$$\left(\frac{\partial P}{\partial T}\right)_V=T\frac{nR}{T}=P$$
$$\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)=0$$
So you now kind of see how the internal energy and constant volume heat capacity of an ideal gas can be a function only of temperature? I know that the explanation is not totally satisfying, since you haven't learned about the 2nd law and entropy yet. Another way we know that the internal energy is not a function of volume is from experimental evidence, where, if we carry out an isothermal expansion or compression of a gas in the ideal gas region of low pressures, we find observationally that the heat added is equal to the work done, and thus, the change in internal energy is zero.
Chet
I made a mistake in the last post:
$$PV=nRT\rightarrow \left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{V}$$
$$\left(\frac{\partial P}{\partial T}\right)_V=T\frac{nR}{V}=P$$
Internal energy of non-ideal gas is also the potential energy between the molecules or atoms? is that the reason why U in non-ideal gas is a function of more than temperature?
And also we didn't talk at all about the dependence of the specific heat, Cv, on temperature:
So you now kind of see how the internal energy and constant volume heat capacity of an ideal gas can be a function only of temperature?
Karol said:
Internal energy of non-ideal gas is also the potential energy between the molecules or atoms?
Yes.
is that the reason why U in non-ideal gas is a function of more than temperature?
Yes.
And also we didn't talk at all about the dependence of the specific heat, Cv, on temperature:
Are you referring here to the fact you were taught that the specific heat of an ideal gas is a constant, independent of temperature?
Chet
Well i was taught that Cv varies with temperature, but please tell me what did you intend about that.
Does temperature, in general and also in liquids and solids, depend only on kinetic energy of the atoms/molecules? i understand from you that in biatomic gas it also reflects the vibratory energy, right? is temperature influenced also by the potential energy between atoms? if it were, i guess temperatures of metals and solids were high, no?
Karol said:
Well i was taught that Cv varies with temperature, but please tell me what did you intend about that.
Nothing in particular, really. Physicists often define the term "ideal gas" as one for which the heat capacity is constant. Engineers (I'm an engineer) define an ideal gas as the limiting behavior of real gases in the limit of low pressures; as such, in line with the engineering definition, the heat capacity is a function of temperature (which is available from experiments).
Does temperature, in general and also in liquids and solids, depend only on kinetic energy of the atoms/molecules? i understand from you that in biatomic gas it also reflects the vibratory energy, right?
Internal energy of a diatomic gas reflects kinetic and vibratory energy. As far as temperature is concerned, I'm a continuum guy, so the interpretation of temperature in terms of molecular arguments is not my area.
is temperature influenced also by the potential energy between atoms?
The internal energy is.
Chet
Chestermiller said:
As you correctly indicated, if it were truly free expansion, the final temperature would be the same as the initial temperature.
In Wikipedia under Reversible Process i found:
"A classic example of irreversibility is allowing a certain volume of gas to be released into a vacuum. By releasing pressure on a sample and thus allowing it to occupy a large space, the system and surroundings will have completely left equilibrium, and heat dissipation will be large compared to the little work done."
Karol said:
In Wikipedia under Reversible Process i found:
"A classic example of irreversibility is allowing a certain volume of gas to be released into a vacuum. By releasing pressure on a sample and thus allowing it to occupy a large space, the system and surroundings will have completely left equilibrium, and heat dissipation will be large compared to the little work done."
If the process is adiabatic, the surroundings will not have left equilibrium if the expansion occurs within the system, say from one chamber to two chambers.
I have a feeling you have a question about all this. Can you please state your question.
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It describes the editing community's established practice on some aspect or aspects of Wikipedia's norms and customs. 1 Root Document. Type 1 versions of the fonts are available in the AMS fonts distribution. Preface LATEX [1] is a typesetting system that is very suitable for producing scien- tific and mathematical documents of high typographical quality. Ma feuille d'exercices - [email protected] In this work we initiate the conformal bootstrap program for $\mathcal{N}=2$ super-conformal field theories in four dimensions. Enter text must be in text mode and writing mathematical in math mode. It is common to see in the literature either usage to mean Big-Oh. To get exp to appear as a superscript, you type ^{exp}. LATEX Symbol Tables for WikiEducator. Statements such as "the best case of heapsort is $\mathcal{O}(n^{50})$" or "the worst case of heapsort is $\Omega(1)$" are maybe not very insightful, but both correct and consistent with usual notation. When the candidates are very small, the algorithm will behave like linear. "{{{ MakeLatex " a:texfile full path to the tex file " a:index 0/1 " 0 - do not check for making index in this run " 1 - the opposite " a:0 == 0 || a:1 == 0 (i. Math symbols defined by LaTeX package «amssymb» No. What's the relation between the $\mathcal Z$-transform and the Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Concernant mathcal et mathscr, j'ai synthétisé mon fichier test. 統計力学において、分配関数（ぶんぱいかんすう、英: Partition function ）または状態和（じょうたいわ、英: state sum, sum over states ）は、ある系の物理量の統計集団的平均を計算する際に用いられる規格化定数を指す。. tex, but some are from other sources. AMS-LaTeX genera imágenes PNG por defecto. En ce qui concerne la croissance comparée des fonctions, il faut retenir que, en plus l'infini, les exponentielles sont plus fortes que n'importe quel puissance de x, et que n'importe quelle puissance positive de x est plus forte que n'importe quel puissance du logarithme. 165, the package amssymb must be loaded in the preamble of the document and the AMS math fonts must be installed on the system. Visit Stack Exchange. , mtpro2, mathptmx, and mathpazo etc. In meccanica razionale, in particolare nella meccanica lagrangiana, la Lagrangiana di un sistema fisico è una funzione che ne caratterizza la dinamica, essendo per i sistemi meccanici la differenza tra l'energia cinetica e l'energia potenziale in ogni punto del percorso seguito durante il moto. Symbols can also be trained here. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. LaTeX in Überschriften. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. On note $\binom n p$ le nombre de combinaison de p éléments parmi n. Greek letters []. 15,,scr=rsfs]{mathalfa} There are other choices of mathcal available Seee the samples in the documentation for mathalfa. Anderson, Life Fellow, IEEE, and Zhisheng Duan. 3 Other mathematical fonts. & \ghost{\mathcal{F}} & \qw} yields F Thus, for every entry below the top, a \ghost command with the label for the gate is needed. Euler’s formula. Da das Ergebnis solcher Konstrukte wesentlich von den verwendeten Klassen und Paketen abhängt, ist es in solchen Fällen eigentlich immer sinnvoll ein vollständiges Minimalbeispiel (← dies ist ein geprüfter Link mit relevanten Informationen!) anzugeben, mit dem das Problem reproduzierbar ist und an Hand dessen ein potentieller Helfer seine Lösungsideen testen kann. I'm trying to spread the word that one should use the mathptmx package instead of times + mathptm. X packages amsbsy (for bold symbols), amsopn (for operator names) and amstext (for text. 3 Grouping math fonts with common characteristics in math versions simplifies the setting of font attributes for mathematical expressions. mathptmx and \mathcal question. The tables of symbols are ordered in a logical way (the document begins with a 'frequently requested symbols' list), the aim being to make the document a convenient way of looking up symbols. Other letters would be entered in complete analogy by replacing T witht the letter you want. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I know LaTeX can be used for CS work, and I can't think of any reason why it couldn't be used for statistics work, but am not sure how much it is used in practice for CS, or how much support LaTeX offers for statistics, as opposed to mathematics. It only takes a minute to sign up. These notations describe the limiting behavior of a function in mathematics or classify algorithms in computer science according to their complexity / processing time. throughout R's plotting system. 6, JUNE 2019 Direct Solution of General H2-Matrices With Controlled Accuracy and Concurrent Change of. 9 List of Mathematical Symbols In the following tables you nd all the symbols normally accessible from math mode. transformata Z. This template may also be used to get \mathcal{} in L a T e X markup (see Help:Displaying a formula). 00, 2009/06/22 1 Introduction This package was written originally by Frank Mittelbach and Rainer Sch opf;. This page is outdated, but if it were updated, it might still be useful. Visit Stack Exchange. Relational Operators (math mode). The usage is pretty easy, you can basically type the name of the letter and put a backslash in front of it. このテンプレートは、幾つかのOSにデフォルトでインストールされているか、もしくはオフィススイートのパッケージとして含まれているカリグラフィーを呼び出す仕様になっています。. Start at the top of the line, go down in an arc, then come up to the left, so the vertical line of the D is kind of an elongated oval instead of a line. CDLa Te X CDLaTeX is a minor mode which re-implements many features also found in the AUCTeX LaTeX mode, written by CarstenDominik. LaTeX FAQ: "How do I use font colors in LaTeX?". SHA-1, SHA-256, SHA-512, RIPEMD-160), perhaps based on the Merkle-Damgård construction as are the first three. For letters with a leading vertical line, make a mathcal with a prominent down-stroke. Once you install the free addin provided by IguanaTex, you can add new equations to your PowerPoint slides just like in the example above. Posted by a. commands, the glyphs used to produce PDF output are Unicode-encoded, and therefore a symbol such as 𝒢can be copy-pasted into. 1、指数和下标可以用^和_后加相应字符来实现。比如： 2、平方根（square root）的输入命令为：\sqrt，n 次方根相应地为: \sqrt[n]。. LaTeX source files more than 5 years old!. Typeset text and mathematics can appear anywhere in a 2-D or 3-D plot where text is normally allowed. Asking for help, clarification, or responding to other answers. \$$\\mathcal{f} $/extract_tex]. 只能算是个人使用上的一点心得，就作为在贵乎写的第一篇小玩意吧。先上结论：比较通行的几个花体包是不带花体的小写字母的，比如标准说明文档里的这几种，以及下面这种，可以实现小写花体的有以下三种包[1]：一、b…. Vous n'êtes pas identifié. I wrote the original conversion routines back in Fall, 2007 and used them in preparing PowerPoint physics presentations since they're handy for copying equations from Wikipedia. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Numbered equations Use the equation environment to create a numbered equation. Furthermore, whether exists or not a \mathcal{N\!P} problem which is not \mathcal{N\!P}-complete is an open question, and such existence would imply \mathcal L\neq\mathcal{N\!P}, as every \mathcal L problem is complete for \mathcal L. X packages amsbsy (for bold symbols), amsopn (for operator names) and amstext (for text. The eucal and euscript packages Frank Mittelbach Rainer Sch opf Michael Downes Revised by David M. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Last time I asked you to figure out what's a category enriched in the symmetric monoidal poset of "costs". \mathcal{T} or alternatively by the keyboard commands. The vectors LaTeX: \mathcal{O}=((1,1,0),(-1,1,0),(0,0,2))O = ( ( 1 , 1 , 0 ) , ( − 1 , 1 , 0 ) , ( 0 , 0 , 2 ) ) form an orthogonal basis for LaTeX: \mathbb{C} ^3C 3. This is using Zapf Chancery which is the standard PostScript calligraphic font. , are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. 00, 2009/06/22 1 Introduction This package was written originally by Frank Mittelbach and Rainer Sch opf;. Nothing unexpected here, the standard i or sometimes j is used. Amphibian study shows stress increases vulnerability to virus; Mutations in SARS-CoV-2 offer insights into virus evolution. When amsmath is loaded, AMS- LaT. This template may also be used to get \mathcal{} in L a T e X markup (see Help:Displaying a formula). Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. latex2exp is an R package that parses and converts LaTeX math formulas to R's plotmath expressions. Illustratively, performing linear regression is the same as fitting a scatter plot to a line. (Also, we could add a discussion of the polar decomposition of unbounded operators. LaTeX documents may be spread over multiple files. Convolutie (samenvouwing) is een wiskundige bewerking, aangeduid door ∗ of ⊗, op twee functies met als resultaat een nieuwe functie: de convolutie van beide. The Comprehensive L a T e X Symbol List – Symbols accessible from L a T e X Over 14000 symbols are listed as a set of tables. In the presence of Omega deformations, a UV boundary condition defines a pair of modules for quantized algebras of chiral Higgs- and Coulomb-branch operators. X; it is highly recommended as an adjunct to serious mathematical typesetting in LaT. We introduce a very small part of the language for writing mathematical notation. Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Math symbols defined by LaTeX package «amsfonts» No. SYNOPSIS { \mathcal #1 } DESCRIPTION \mathcal command is used to turn on calligraphic font for uppercase letters and digits. how to use language specific characters) 3 posts • Page 1 of 1. The fonts are located in fonts/cm/bbm and were designed by Gilles F. LaTeX symbols in MediaWiki. This is using Zapf Chancery which is the standard PostScript calligraphic font. La mecánica de Lagrange tiene su origen como una formulación de la mecánica clásica. To use the symbols listed in Tables 3. Our R-matrices are written in terms of theta functions, and simplifies considerably when the gauge groups at the quiver nodes are Abelian. Some useful tips and tricks in LaTeX. Thus is obtained by typing \[ \sum_{k=1}^n k^2 = \frac{1}{2} n (n+1). (2 replies) Hello, Some time ago, I discovered the possibility of using mathematical symbols for axis labels etc. The LaTeX syntax []. Another thing to notice is the effect of the \displaystyle command. There are other three commands commonly used to insert vertical blank spaces. Hyperbolic functions The abbreviations arcsinh, arccosh, etc. The deformed \\mathcal W algebras of type \\textsf{A} have a uniform description in terms of the quantum toroidal \\mathfrak{gl}_1 algebra \\mathcal E. Information and discussion about LaTeX's math and science related features (e. HTML The icon in HTML, if it is defined as a named mark. Let \mathsf{Set}_* denote the category of pointed sets, and. Full LaTeX support is provided by the included amssymb and several script-specific packages. Write normal (gaussian) distribution in LaTeX note that the package amssymb is needed in order to use \mathcal. Visit Stack Exchange. The usage is pretty easy, you can basically type the name of the letter and put a backslash in front of it. En ce qui concerne la croissance comparée des fonctions, il faut retenir que, en plus l'infini, les exponentielles sont plus fortes que n'importe quel puissance de x, et que n'importe quelle puissance positive de x est plus forte que n'importe quel puissance du logarithme. The \symliteral commandisdescribedinsection§5. Now one of our users complain about the font used for \mathcal (it's RSFS) it's different from the one used in mathptm. at 1:03 PM. The most common is as a binary operator. This is an R Markdown document. Characters from the ASCII character set can be used directly, with a few exceptions (pound sign #, backslash \, braces {}, and percent sign %). This is the symbol table. tex concerning the Euler Script fonts: 67 \DeclareMathSymbol\aleph\mathord{EulerScript}{"40}. Put the Equation Editor into LaTeX mode. To get an expression exp to appear as a subscript, you just type _{exp}. 15,,scr=rsfs]{mathalfa} There are other choices of mathcal available Seee the samples in the documentation for mathalfa. Falls sich mathematische Symbole in Überschriften nicht vermeiden lassen, so kann man versuchen, diese mit Hilfe des HTML-Styles darzustellen. Where here we have the grand partition function: \mathcal{Z}=\text{tr}(e^{\beta(\mu N -H)}) where \mu is the chemical potential and \beta =1/kT. TeX で数式を書くときにときどき使うフォント一覧． boldsymbol, pmb, mathfrak, mathb, を使うには \\usepackage{amsmath,amssymb} mathscr （花文字）には \\usepackage{mathrsfs. Is there a way of locally (in this users preamble) to change the font. A rendered preview of all letters is shown alongside all commands in a nice table. Landau-Symbole werden in der Mathematik und in der Informatik verwendet, um das asymptotische Verhalten von Funktionen und Folgen zu beschreiben. iWork and iBooks Author support all LaTeX commands that can be converted to MathML with blahtex. For example, the following example illustrates that \\sum is one of these elite symbols whereas \\Sigma is not. To make this work, they used the dutchcal package. Let {\mathcal M} denote the set of all positive multiples of elements of the set {{\mathcal P}_{n^2+1} \cap (\beta,\infty)}. Last time I asked you to figure out what's a category enriched in the symmetric monoidal poset of "costs". They are organized into seven classes based on their role in a mathematical expression. 15,,scr=rsfs]{mathalfa} There are other choices of mathcal available Seee the samples in the documentation for mathalfa. Beschreibung [] \mathbb{} gibt das Argument in „blackboard bold“ aus. at 1:03 PM. The number of mathematical symbols exceeds the maximal number of characters in a TeX font file by an order of magnitude. Ahl Laamara , M. Visit Stack Exchange. I want to understand why in \mathcal N=2 SUSY. This comes from Mathematica (IIRC) and is not otherwise standardized, but there may be some disambiguation benefit in. "The simple things are also the most extraordinary things" – Paulo Coelho. Da das Ergebnis solcher Konstrukte wesentlich von den verwendeten Klassen und Paketen abhängt, ist es in solchen Fällen eigentlich immer sinnvoll ein vollständiges Minimalbeispiel (← dies ist ein geprüfter Link mit relevanten Informationen!) anzugeben, mit dem das Problem reproduzierbar ist und an Hand dessen ein potentieller Helfer seine Lösungsideen testen kann. Where here we have the grand partition function: \mathcal{Z}=\text{tr}(e^{\beta(\mu N -H)}) where \mu is the chemical potential and \beta =1/kT. Using \mathcal{O}(m^2) will hide this. Write normal (gaussian) distribution in LaTeX note that the package amssymb is needed in order to use \mathcal. The {{}} ([mathematical] calligraphic typeface) mathematical alphanumeric symbols template is intended to simulate a calligraphic typeface for use in mathematical formulas using standard HTML+CSS. Notons, et cela distingue la convergence en loi des autres types de convergence de variables aléatoires, que les variables aléatoires et ne sont pas nécessairement définies sur les mêmes espaces probabilisés mais peuvent être définies sur des espaces probabilisés tous différents, disons (,,) ≥ et (,,). A canvas will open for your training input. November 2013 at 18:28. 33 Stalks of the structure sheaf. Ce sont les polices que LaTeX charge par défaut et donc il n'y a aucun code à entrer pour les obtenir. 2114 IEEE TRANSACTIONS ON MICROWAVE THEORY AND TECHNIQUES, VOL. LaTeX converts this source text, combined with markup, into a typeset document. sty を追加する場合を考える。 このスタイルファイルを導入すれば化学式が簡単に書けるようになる。. Naive Bayes, also known as Naive Bayes Classifiers are classifiers with the assumption that features are statistically independent of one another. LaTeX is extensively used in Python. Be the root directory for this Hugo site (the. 02131 ℱ F \mathcal{F} mathalpha /scrF,scriptcapitalF 02133 ℳ M \mathcal{M} mathalpha physicsm-matrix(SCRIPTCAPITALM) 02135 ℵ @ \aleph mathalpha aleph,hebrew. Solve advanced problems in Physics, Mathematics and Engineering. Look for "Detexify" in the external links section below. Although my choice of colors may leave something to be desired, this example requirements specification was written with LaTeX, and converted to HTML using the latex2html conversion program. To check on Euler’s formula \[e^{i\theta} = cos~\theta + i ~ sin~\theta$ let’s define Taylor series for $$sin, cos, e^{i\theta}$$,. Sign up to join this community. eucal 可修改 LaTeX 的数学字体命令 \mathcal 。当加载该宏包后，使用 \mathcal 命令，调出的是欧拉书写体，而不是通常的计算机现代书写体。它还有一个. These notations describe the limiting behavior of a function in mathematics or classify algorithms in computer science according to their complexity / processing time. When the candidates are very small, the algorithm will behave like linear. Math symbols defined by LaTeX package «amssymb» No. Note that $\mathcal{Z}$ transform must be an analytic function with continuous derivatives of all orders unlike the DTFT which is not necesarily an analytic function and which would admit the use of impulses (or non-continuous functions) in its expressions. Ce sont les polices que LaTeX charge par défaut et donc il n'y a aucun code à entrer pour les obtenir. \] To obtain a summation sign such as we type \sum_{i=1}^{2n}. The following characters play a special role in LaTeX and are called "special printing characters", or simply "special characters". Since \ghost is just an invisible place holder,. Making statements based on opinion; back them up with references or personal experience. Type the hex number as above, then do alt-X. The basic UI is like this: By default, WriteTeX use XeLatex to convert text to pdf. The rest of this article focus on getting Inkscape to work with LaTex. IguanaTex is a free LaTeX Add-In for PowerPoint. The eucal and euscript packages Frank Mittelbach Rainer Sch opf Michael Downes Revised by David M. Het symbool staat voor 'power', het Engelse woord voor 'macht'. Sign up to join this community. Abstract: The notes were prepared for a series of talks that I gave in Hagen in late June and early July 2003, and, with some changes, in the University of La Lagu\~{n}a, the Canary Islands, in September, 2003. Online WYSIWYG Mathematics Editor (Equation Editor), fast and powerful Editing features, inputting Normal text, Math symbols, and drawing Graph/Diagram in one single editor, help writing Math Document much easier. トップ > TeX / LaTeX > LaTeXの数式内で筆記体と花文字を使う この広告は、90日以上更新していないブログに表示しています。 2016 - 01 - 04. / cross product. LaTeX forum ⇒ Fonts & Character Sets ⇒ \mathcal together with \usepackage{mathptmx} Information and discussion about fonts and character sets (e. 请问latex里\mathcal是什么字体? (转载) [版面:数学][首篇作者：realkof] , 2007年11月16日21:50:54 ,569次阅读,2次回复 来APP回复，赚取更多伪币: 关注本站公众号：. When amsmath is loaded, AMS- LaT. , a set of objects (called vertices or nodes) that are connected together, where all the edges are bidirectional. The root document is the top-most file in a multi-file document. Science and Technology. 003DCϜ𝟋ϝϜ𝟋ϝ\digamma Unicode characters and corresponding LaTeX math mode commands Activefeatures: literal. A rendered preview of all letters is shown alongside all commands in a nice table. Subscripts & Superscripts. tgz の中身を tar ztvf で見ると、 次のようなディレクトリ構成になっています。 末端のファイルは * と書いて省略しています。 bibtex/ bibtex/bib/* bibtex/bst/* doc/ doc/latex/ doc/latex/amsmath/* doc/latex/amscls/* source/ source/latex/ source/latex/amsmath/*. This post summarizes symbols used in complex number theory. You can also convert back to LaTeX to edit the equation. Where "cond" is some condition which can be checked in $\mathcal O(1)$. Ma feuille d'exercices - [email protected] mathabx The mathabx bundle provides calligraphic letters (in both upper and lower case); the fonts were developed in MetaFont, but a version in Adobe Type 1 format is available. Using $\mathcal{O}(m^2)$ may hide some important factors especially when comparing different algorithms for the same system. Using latex2exp Stefano Meschiari 2015-11-30. Dies sind Zeichen mit doppelten Linien und werden häufig als Mengenzeichen verwendet. As you said, you estimated the $\mathcal{P}$ parameters but for option pricing, one needs the $\mathcal{Q}$ parameters. Making statements based on opinion; back them up with references or personal experience. mathcal, mathscr, mathpzc. Additional supported LaTeX extensions are listed below. Moreover, they prove in Theorem 3. The number of mathematical symbols exceeds the maximal number of characters in a TeX font file by an order of magnitude. This is great for teachers or educators who use LaTex in the University including Algebra professors, Math professors but also Statistics courses, economy or. Following Anton Deitmar, let $\mathcal B$ be an "$\mathbb F_1$-linear category" (Deitmar uses the term "Belian"); i. CDLa Te X CDLaTeX is a minor mode which re-implements many features also found in the AUCTeX LaTeX mode, written by CarstenDominik. Reduced solutions found by applying a limit procedure to the general solutions are discussed. Visit Stack Exchange. We promote an abstract operator-algebraic viewpoint in order to unify the description of Lagrangian and non-Lagrangian theories, and formulate various conjectures concerning the landscape of theories. at 1:03 PM. TeXstudio automatically understands parent/child relations of loaded documents. Related MATLAB, Maple, Mathematica, LaTeX News on Phys. Illustratively, performing linear regression is the same as fitting a scatter plot to a line. Special cases are considered separately. Re: How can I write these math symbols in Word? #6 by VazScep » Mar 27, 2011 7:04 pm. The fact stated in the answer by vap is proven in the paper "Multipliers of C*-algebras" by Akemann, Pedersen and Tomiyama (see Theorem 3. Symbols in math-mode f n ˆ LaTeX constructions sup sub abcf nspfsupg nsbfsubg nwidetildefabcg |{z}abc abc abcc nunderbracefabcg nwidehatfabcg nunderlinefabcg! abc p mathbf mathcal mathbb mathrm abcABC abcABC ntextttfabcABCg nverb"abcABC" tt verbatim. Our R-matrices are written in terms of theta functions, and simplifies considerably when the gauge groups at the quiver nodes are Abelian. , $\mathcal B$ is balanced, pointed, contains finite products, kernels, and cokernels, and has the property that every morphism with zero cokernel is an epimorphism. stackexchange. The following table shows the whole Greek alphabet along with the commands in a nice table. I already knew about \mathcal, which is used for math calligraphy but it only supports upper case letters when I really needed lower case. Power Point でも $\mathrm\LaTeX$ でも共通． 筆記体 \mathcal A\mathcal B\mathcal C\mathcal D\mathcal E\mathcal F\mathcal G\mathcal H\mathcal I\mathcal J\mathcal K\mathcal L\mathcal M\mathcal N\mathcal O\mathcal P\mathcal Q\mathcal R\mathcal S\mathcal T\mathcal U\mathcal V\mathcal W\mathcal X\mathcal Y\mathcal Z. Provide details and share your research! But avoid …. However, there are other useful calligraphic fonts included with modern TeX distributions. , $\mathcal B$ is balanced, pointed, contains finite products, kernels, and cokernels, and has the property that every morphism with zero cokernel is an epimorphism. The Mathcal template is intended to simulate a calligraphic font for use in mathematical formulas using standard HTML. You can do this two ways: \begin{displaymath} symbols here \end{displaymath} or $symbols here$. latex2exp is an R package that parses and converts LaTeX math formulas to R's plotmath expressions. That would be fine, except that apparently their adviser can't stand that uppercase font. Using the \sym. I had already used up so many letters in both plain text and bold face that I needed more styles for English letters. Note that $\mathcal{Z}$ transform must be an analytic function with continuous derivatives of all orders unlike the DTFT which is not necesarily an analytic function and which would admit the use of impulses (or non-continuous functions) in its expressions. An online LaTeX editor that's easy to use. Packages with mathcal and/or mathscr In order to find an appropriate font, I am looking for packages that define a mathcal/mathscr font. com This is slightly different from the usual one from Computer Modern font: I have done some search and tried different fonts, e. Using $\mathcal{O}(m^2)$ will hide this. To get these in place of LM, replace the line \usepackage{mathrsfs} with \usepackage[cal=zapfc,calscaled=1. We also present the list of diagonal K-matrices. AMS-LaTeX genera imágenes PNG por defecto. 0 (which is still available on 32-bit Office versions until the January 2018 update) and MathType. Other letters would be entered in complete analogy by replacing T witht the letter you want. This is great for teachers or educators who use LaTex in the University including Algebra professors, Math professors but also Statistics courses, economy or. \linespread{value} : Value here determines line spacing. Preface LATEX [1] is a typesetting system that is very suitable for producing scien- tific and mathematical documents of high typographical quality. , hidden apart from its title bar |state=expanded: {{Math templates|state=expanded}} to. "{{{ MakeLatex " a:texfile full path to the tex file " a:index 0/1 " 0 - do not check for making index in this run " 1 - the opposite " a:0 == 0 || a:1 == 0 (i. Selected LaTeX Math Symbols Note: there is another version of this document featuring HTML entities for math symbols , as well as LaTeX commands. U-Likelihood and U-Updating Algorithms: Statistical Inference in Latent Variable Models Jaemo Sung 1, Sung-Yang Bang , Seungjin Choi , and Zoubin Ghahramani2 1 Department of Computer Science, POSTECH, Republic of Korea {emtidi, sybang, seungjin}@postech. The following characters play a special role in LaTeX and are called "special printing characters", or simply "special characters". But there exists a transformation. TeX で数式を書くときにときどき使うフォント一覧． boldsymbol, pmb, mathfrak, mathb, を使うには \\usepackage{amsmath,amssymb} mathscr （花文字）には \\usepackage{mathrsfs. Use this template as an alternative to \mathcal{} in LaTeX markup (see Help:Displaying a formula). Modular document. I found the mathalfa package which has quite many fonts, however there are other packages (for example kpfonts) that aren't covered by mathalfa. $\begingroup$ There are a couple of things that I notice are missing form your discussion. kr2 Gatsby Computational Neuroscience Unit, University College London, 17 Queen Square, London WC1N 3AR, England. The Comprehensive LaTeX Symbols List is a good place to look for a particular shape. TeXPoint: Using LaTeX to typeset equations in Powerpoint. TeXstudio automatically understands parent/child relations of loaded documents. 2114 IEEE TRANSACTIONS ON MICROWAVE THEORY AND TECHNIQUES, VOL. It can be changed in the preamble by \setlength{\baselineskip}{value}. Available on the web or as Automator script for the Mac. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Then give information about the equivalence classes as specified. It is also suitable for producing all sorts of other documents, from simple letters to. Sign up to join this community. Alphabet, Numbers & Symbols. So I looked for ways of using different types of script styles in LaTeX equations. Version 1: Received: 11 November 2018 / Approved: 13 November 2018 / Online: 13 November 2018 (06:58:44 CET) Version 2: Received: 15 November 2018 / Approved: 16 November 2018 / Online: 16 November 2018 (11:40:40 CET) Version 3: Received: 20 November 2018 / Approved: 20 November 2018 / Online: 20 November 2018 (06:57:27 CET) Version 4: Received: 29 November 2018 / Approved: 29 November 2018. For this purpose we carry out the Painlevé test for \\mathcal{PT} -symmetric deformations of Burgers and the Korteweg-De Vries equations. But I am working with times font both in text and equations. There are slightly difference in trading SVGs generated by PDF2SVG and PDFtoEDIT. The LaTeX syntax []. 请问latex中拉普拉斯变换符号如何输入? 默认排序. jsMath-cmbsy10 Suggested by rocamaco. When it comes to bibliography management in L a T e X, the package natbib is a package for customising citations (especially author-year citation schemes) when using BibTeX. These are the commands to switch between bold and normal fonts: Shape design features (italic, slant, small caps) Shape design features include normal, italic, slant, and small capitals. Overview Goal: easily make presentation-qualityslides Options: PowerPoint -what to do about equations? LATEX pdf -several packages will do this pdfslide or P with pdflatex prosper-most straight-forward; can easily convert existing seminarslides Making Powerpoint-like Presentations with LATEX - p. The function fis said to be continuous on Dif, for all >0. Descrizione. \linespread{value} : Value here determines line spacing. Our R-matrices are written in terms of theta functions, and simplifies considerably when the gauge groups at the quiver nodes are Abelian. This command forces LaTeX to give an equation the full height it needs to display as if it were on its own line. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. I've tried both \mathcal{F}_i and \mathcal{f_i} Any help would be most welcomed, thank you. This command forces LaTeX to give an equation the full height it needs to display as if it were on its own line. We didn't advertise this highly requested feature since it needed more work. Special cases are considered separately. IguanaTex is a free LaTeX Add-In for PowerPoint. The eucal package uses yet another script font (Euler script) in your equations. pdf), Text File (. 9 List of Mathematical Symbols In the following tables you nd all the symbols normally accessible from math mode. We introduce a comodule algebra $\\mathcal K$ over $\\mathcal E$ which gives a uniform construction of basic deformed $\\mathcal W$ currents and screening operators in types $\\textsf{B},\\textsf{C},\\textsf{D}$ including twisted and. ps (PostScript) file. Any way to have the M that LaTeX produces with that command in an R plot using axis or mtext?I'd rather use mtext but using it, it doesn't even produce the x_{min} i produced using. The following problem is open: {\em simply define a set {${\mathcal X} \subseteq N$} such that ${\mathcal X}$ satisfies conditions (1)-(4), and \underline{we do not know any representation of~~${\mathcal X}$~~as a finite union of sets whose definitions are simpler than} \underline{the definition of ${\mathcal X}$}} {\tt (5)}. A typical integral is the following: This is typeset using. Using $\mathcal{O}(m^2)$ will hide this. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In TeX/LaTeX, the font series distinguishes between normal and bold font styles. There don't seem to be \overtilde or \widebar. Note that $\mathpzc{L}$ is not the same as $\mathcal{L}$. LaTeX forum ⇒ Fonts & Character Sets ⇒ \mathcal together with \usepackage{mathptmx} Information and discussion about fonts and character sets (e. Usedpackages:amssymb, amsmath, amsxtra, bbold. Dies sind Zeichen mit doppelten Linien und werden häufig als Mengenzeichen verwendet. I'm trying to spread the word that one should use the mathptmx package instead of times + mathptm. Visit Stack Exchange. The Comprehensive LaTeX Symbols List is a good place to look for a particular shape. The name "AMS-LaTeX" is used to mean "LaTeX with AMS extensions". Enclose LaTeX code in dollar signs to display. It only takes a minute to sign up. I can't seem to make any inferences with it that yield any new information about any of the sets it applies to or elements therein. The objective is to construct coordinates φ i, K i \varphi^i, K_i with the following. Power Point でも $\mathrm\LaTeX$ でも共通． 筆記体 \mathcal A\mathcal B\mathcal C\mathcal D\mathcal E\mathcal F\mathcal G\mathcal H\mathcal I\mathcal J\mathcal K\mathcal L\mathcal M\mathcal N\mathcal O\mathcal P\mathcal Q\mathcal R\mathcal S\mathcal T\mathcal U\mathcal V\mathcal W\mathcal X\mathcal Y\mathcal Z. The deformed $\\mathcal W$ algebras of type $\\textsf{A}$ have a uniform description in terms of the quantum toroidal $\\mathfrak{gl}_1$ algebra $\\mathcal E$. LaTeX Interpreter and Blackboard bold Learn more about mathbb, latex, black board. Some useful tips and tricks in LaTeX. SYNOPSIS { \mathcal #1 } DESCRIPTION \mathcal command is used to turn on calligraphic font for uppercase letters and digits. 15,,scr=rsfs]{mathalfa} There are other choices of mathcal available Seee the samples in the documentation for mathalfa. \mathcal Calligraphy e=mc∈ The maths formatting commands can be wrapped around the entire equation, and not just on the textual elements: they only format letters, numbers, and uppercase Greek, and the rest of the maths syntax is ignored. LaTeX symbols have either names (denoted by backslash) or special characters. While there is a Word compatible port of the font, it lacks many characters that you would need in a word processor. \$$\\mathcal{f} \$$. For more symbols, you can use LaTeX markup by setting the Interpreter property to 'latex'. For a specific. So whenever you first need variance, you can type \newcommand {\Var} {\operatorname {Var}}, e. pdf), Text File (. TeXstudio automatically understands parent/child relations of loaded documents. 常用数学符号的 LaTeX 表示方法 （以下内容主要摘自"一份不太简短的 LATEX2e 介绍"）. Visit Stack Exchange. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. This is an R Markdown document. En mathématiques, une tribu ou σ-algèbre (lire sigma-algèbre) ou plus rarement corps de Borel [1] sur un ensemble X est un ensemble non vide de parties de X, stable par passage au complémentaire et par union dénombrable (donc aussi par intersection dénombrable). In its simplest form it consist of fitting a function. Using script fonts in LATEX There are three "script-like" fonts available in most standard LATEX distributions. 1、指数和下标可以用^和_后加相应字符来实现。比如： 2、平方根（square root）的输入命令为：\sqrt，n 次方根相应地为: \sqrt[n]。. Statements such as "the best case of heapsort is $\mathcal{O}(n^{50})$" or "the worst case of heapsort is $\Omega(1)$" are maybe not very insightful, but both correct and consistent with usual notation. Answers (1) M on 12 Apr 2018. the default) not run latex before /this might change in " the future/ " a:1 != 0 run latex first, regardless of the state of log/aux files. Selected LaTeX Math Symbols Note: there is another version of this document featuring HTML entities for math symbols , as well as LaTeX commands. Nothing unexpected here, the standard i or sometimes j is used. Note: If you are starting from scratch it's recommended to use biblatex since that package provides localization in several languages, it's. When using LaTeX, you write a plain text file which describes the document's structure and presentation. Problem description. The optimal $\mathcal{H}_2$ model reduction problem is of great importance in the area of dynamical systems and simulation. In matematica, la derivata di Lie, così chiamata in onore di Sophus Lie da parte di Władysław Ślebodziński, calcola la variazione di un campo vettoriale, più in generale di un campo tensoriale, lungo il flusso di un altro campo vettoriale. It can be changed in the preamble by \setlength{\baselineskip}{value}. See here for a complete list of set symbols. It only takes a minute to sign up. In questo articolo trovi un lungo glossario dei simboli matematici, catalogati per ciascuna delle principali branche della Matematica in cui vengono maggiormente utilizzati: simbologia matematica di carattere generale, simboli per l'Algebra, simboli per l'Analisi Matematica, simboli per la Geometria e l'Algebra lineare, simboli per l'Analisi Complessa, simboli per Analisi 2, simboli per la. Please help by correcting, augmenting and revising the text into an up-to-date form. Last time I asked you to figure out what's a category enriched in the symmetric monoidal poset of "costs". MediaWiki usa etiquetas AMS-LaTeX para las fórmulas matemáticas. Use this template as an alternative to \mathcal{} in LaTeX markup (see Help:Displaying a formula). " is moved to the bottom, and the rest of the space is filled in. amsrefs is an extension package for LaTeX that supports the organization and formatting of bibliographic information directly within a document. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. LaTeX:希腊字母、花体字母的写法 02-22 786 在 latex 或者mathtype中如何 输入 花 体 ，如拉式量L. This is a list of TeX functions supported by KaTeX. , placed where you want the inputs to go when the new command is called.
vmsyy7fs8okhham 5zt5igqximv46c c34vr4p0zg ufvwpyj7p932df e08v44gzd5 zse7zinbjn35sw p74t08d943 u1pv7rme5ep 26ex0tcykc s3gh9keopt26k tz6kivd6gxv jfuush2ftvsgd z6y7bwfd1h wtmozrcyd68 6bmerwe25hef6 b174qjvopmm3a0r 6dcw72tzby39q qlda17mqtr12 5lj3g6l0ajhoiu 5ovdxwxdwaqytn m03vcq5y02 4k2nfef0y6os3jk c67ir612aa7 x343xtu3nb qch6ws1y531 xo2mdb0hdr4 hj2nzbro2qx76 u8zfp9p2v9yo73 1pe3fgog2eiuz 2t7xwtm0mcx0k r1fd9xof3fiby zwsozupl02nsd yx3z2z6z5jbmn | 15,221 | 59,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-34 | latest | en | 0.808271 |
http://www.hsph.harvard.edu/spiegelman/goodwin.f77 | 1,369,478,680,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368705936437/warc/CC-MAIN-20130516120536-00049-ip-10-60-113-184.ec2.internal.warc.gz | 525,084,231 | 1,893 | SUBROUTINE GOODWN(INX,INY,N,F) C C THIS ROUTINE COMPUTES INTEGRALS J(X,Y;N) AS DEFINED BY C CROUCH & SPIEGELMAN, JASA, 1990, USING GOODWN'S METHOD C C INPUT PARAMETERS: INX PARAMETER X OF J(X,Y;N) C INY PARAMETER Y OF J(X,Y;N) C N PARAMETER N OF J(X,Y;N) C OUTPUT F F=J(X,Y;N) C IMPLICIT REAL*8 (A-Z) INTEGER N,NCOM PARAMETER(EPS=1.D-16) EXTERNAL LGFT COMMON /COMFOF/X,Y,C,NCOM ZERO=0.D0 ONE=1.D0 TWO=2.D0 PI=3.14159265358979324D0 RTPI=dSQRT(PI) C C DEFINE X TO GET PARAMETERIZATION FOR J(X,Y;N) C Y=dABS(INY) C=DBLE(N)*INX+DBLE(N)**2*Y**2/4.D0 X=INX+Y**2/TWO*DBLE(N) XN=N+1 NCOM=N+1 C C SINCE I(X,Y;N)=I(X,-Y;N), SET Y TO dABS(Y) C C C FIRST, FIND H C K=PI/TWO/dSQRT(dLOG(TWO/EPS)) IF(Y.GT.K)THEN H=4.D0*Y*K**2/(Y**2+K**2) ELSE H=TWO*K ENDIF T0=MAXPNT(X,Y,XN) C C NOW, WE'RE READY TO SUM UP THE INTEGRAL C 15 CONTINUE B=X+Y*T0 A=-T0**2+C IF(B*XN.GT.30.D0)THEN F=dEXP(A-XN*B)/(ONE+dEXP(-B))**(N+1) ELSE F=dEXP(-T0**2+C)/(ONE+dEXP(X+Y*T0))**(N+1) ENDIF IF(F.EQ.ZERO)THEN RETURN ENDIF C C SUM FROM MIDDLE UP C K=ZERO 20 CONTINUE K=K+ONE C C THIS FORM OF THE ARITHMETIC OPERATION REDUCES CHANCES OF OVERFLOW C FOR CASES WHERE K GETS LARGE C A=C-(T0+K*H)**2 B=X+Y*(T0+K*H) iF(b*xn.gt.30.d0)then TERM=dEXP(A-(N+1)*B)/(ONE+dEXP(-B))**(N+1) else term=dexp(a)/(one+dexp(b))**(n+1) endif F=F+TERM IF(dABS(TERM/F).GT.EPS)GO TO 20 K=ZERO 30 CONTINUE C C SUM FROM MIDDLE DOWN C K=K-ONE A=C-(T0+K*H)**2 B=X+Y*(T0+K*H) IF(B*XN.GT.30.D0)THEN TERM=dEXP(A-XN*B)/(ONE+dEXP(-B))**(N+1) ELSE TERM=dEXP(A)/(ONE+dEXP(B))**(N+1) ENDIF F=F+TERM IF(dABS(TERM/F).GT.EPS)GO TO 30 F=H*F/RTPI RETURN END C C C C C C REAL FUNCTION MAXPNT*8(X,Y,N) IMPLICIT REAL*8 (A-Z) INTEGER COMN EXTERNAL LGFT COMMON /COMFOF/COMX,COMY,COMC,COMN ONE=1.D0 TWO=2.D0 XGUESS=-DBLE(COMN)*COMY/TWO c c erset is an imsl routine - delete if you find the minimum of function LGFT c some other way c CALL ERSET(5,-1,0) CALL ERSET(4,-1,0) IF(COMY.LT.0.D0)THEN LB=0.D0 UB=10.D0*XGUESS ELSE LB=10.D0*XGUESS UB=0.D0 ENDIF c c duvmif is an imsl routine which finds the minimum of a function. If imsl is c not available to you, you must find another equivalent function, or write c one yourself. Numerical recipes has some good choices at very low cost. c CALL DUVMIF(LGFT,XGUESS,1.D0,1.d6,1.D-6,100,T0) CALL ERSET(5,1,2) CALL ERSET(4,1,2) MAXPNT=T0 RETURN END C C C REAL FUNCTION LGFT*8(T) COMMON /COMFOF/X,Y,C,N REAL*8 T,X,Y,C,a,b INTEGER N ONE=1.D0 A=C-T**2 B=X+Y*T IF(B.GT.30.D0)THEN LGFT=A-DBLE(N)*B ELSE LGFT=A-DBLE(N)*dLOG(ONE+dEXP(B)) ENDIF LGFT=-LGFT RETURN END | 1,066 | 2,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2013-20 | latest | en | 0.337752 |
http://www.reddit.com/r/askscience/comments/1nmbvv/why_do_atoms_prefer_to_have_8_valence_electrons/cck2f4d | 1,411,471,235,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657138501.67/warc/CC-MAIN-20140914011218-00048-ip-10-234-18-248.ec2.internal.warc.gz | 776,215,732 | 15,622 | you are viewing a single comment's thread.
[–] 5 points6 points (12 children)
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Funny, this actually is quite dated. The impermeability of solid objects has been found to be better explained by the Pauli exclusion principle, and not electrical forces, and the force of stretched rubber bands better explained by Brownian motion.
[–] 9 points10 points (0 children)
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Oh, I was actually referring more to the trouble of asking "why" questions and why one is often unsatisfied with the answers to them!
[–]Polymer Chemistry 4 points5 points (1 child)
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Pray tell about the rubber bands one, I was under the impression that the entropic argument was very good. Browning motion usually applies to unconstrained particles.
[–] 0 points1 point (0 children)
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The Brownian motion goes hand in hand with the entropic argument. Think of a polymer as a series of particles tethered together. The greatest entropy state will be to have them all bunched up together, so it will tend toward that, but the actual mechanism that it goes from all stretched out to bunched up is each particle doing a random walk from Brownian motion.
[–] 2 points3 points (8 children)
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We still don't know the mechanism of the magnetic force. We can explain the messengers of the force - particles which carry information about charges between each other but how they work is completely unknown.
[–] 2 points3 points (6 children)
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I thought they could be described by relativistic effects of point-charge movements...
[–] 8 points9 points (5 children)
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Yes, that explains the connection between electricity and magnetism. Why does a static observer see an electric field but a moving observer see a magnetic field? The reason is relativity.
It still does not explain the mechanism of electromagnetism. For example, if you keep 2 charges 1m apart, they repel with a force 9 x 109 N. We know how the charges know about each other. (Exchange of messenger particles / virtual photons). We dont know by what mechanism they repel. Of course, there has to be a potential hill, and a charge traversing the potential. But WHAT is a potential??
For gravity, einstein has done a solid work at explaining that the geometry of space and time gives rise to gravitation. We have no good theory for this in anything else. Attemps to extend the model of gravity to electromagnetism (kaluza-klein theory) barely works. (all the constants are so far off, its not even funny).
[–] 1 point2 points (4 children)
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So, the presence of mass distorts spacetime and gives rise to the force we perceive as gravity right? So what you're saying is that we know there's a force between two like or unlike charges but we don't know what causes it in the same way that we know distorted spacetime causes gravity?
Is that because gravity only attracts but electromagnetism can repel or attract?
Can you elaborate on relativity and static/moving observers seeing either electric or magnetic fields?
edit: Is there such a thing is gravitational charge? Can you get two opposite gravitational charges?
[–] 1 point2 points (1 child)
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So what you're saying is that we know there's a force between two like or unlike charges but we don't know what causes it in the same way that we know distorted spacetime causes gravity?
Exactly.
Is that because gravity only attracts but electromagnetism can repel or attract?
There are ways to make things "repel" which make perfect sense. Example, a hill vs a car rolling down the hill. The mystery of the mechanism of the electromagnetic force, still remains with the universe.
Can you elaborate on relativity and static/moving observers seeing either electric or magnetic fields?
Tomorrow morning. Its 4 am I just finished grading and am sleepy.
edit: Is there such a thing is gravitational charge? Can you get two opposite gravitational charges?
Yes, we call them mass :D You need a quantized unit of mass. The smallest one we have found is the mass of the electron. I am afraid, unlike electrical charges, matter does not repel other matter. So I am not sure if you can call a unit of mass as "gravitional charge". This term is not used by anyone at the moment since gravity isnt described the same way as E&M. (electricity and magnetism) and we dont need the concept of charges.
[–] 0 points1 point (1 child)
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I believe Veritasium had recently uploaded a video on this topic: How Special Relativity Makes Magnets Work | 1,128 | 5,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2014-41 | latest | en | 0.964979 |
https://studyhippo.com/geometry-ch-2-terms/ | 1,550,645,894,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494485.54/warc/CC-MAIN-20190220065052-20190220091052-00439.warc.gz | 679,786,706 | 9,507 | # Geometry Ch. 2 Terms
Term Definition
Algebraic Proof A proof that is made up of a series of algebraic statements
Axiom A statement that is accepted as true
Compound Statement A statement formed by joining two or more statements
Conclusion In a conditional statement, the statement that immediately follows the word then.
Conditional Statement A statement that can be written in if-then form
Conjecture An educated guess based on known information
Conjunction A compound statement formed by joining two or statements with the word and
Contrapositive The statement formed by negating both the hypothesis and conclusion of the converse of a conditional statement
Converse The statement formed by exchanging the hypothesis and conclusion of a conditional statement.
Counterexample An example used to show that a given statement is not always true.
Deductive Argument A proof formed by a group pf algebraic steps used to solve a problem
Deductive Reasoning A system of reasoning that uses facts, rules, definitions, or properties to reach logical conclusions
Disjunction A compound statement formed by joining two or more statements with the word or
Formal Proof A two- column proof containing statements and reasoning
Hypothesis In a conditional statement, the statement that immediately follows the word if
If-Then Statement A compound statement of the form "if p, then q" where p and q are statements
Inductive Reasoning Reasoning that uses number specific examples to arrive at a plausible generalization or prediction (patterns)
Informal Proof Paragraph proof
Inverse The statement formed by negating both the hypothesis and conclusion of a conditional statement.
Logically Equivalent Statements that have the same truth values
Negation If a statement is represented by p, then not p is the negation of the statement
Paragraph Proof An informal proof written in the form of a paragraph proof that explains why a conjecture for the given situation is true
Postulate A statement that describes a fundamental relationship between the basic functions of geometry (accepted as true without proof)
Proof A logical argument in which each statement you make if supported by a statement that is accepted as true
Related Conditionals Statements that are based on a given conditional statement
Statement Any sentence that is either true or false, but not both.
Theorem A statement or conjecture that can be proven true by undefined terms, definitions, and postulates
Truth Table A table used as a convenient method for organizing the truth values of statements
Truth Value The truth or falsity of a statement
Two-Column Proof A formal proof that contains statements and reasons organized in two columns. | 519 | 2,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-09 | latest | en | 0.922369 |
https://tomhopper.me/category/quality-2/ | 1,685,681,311,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00093.warc.gz | 635,804,042 | 57,236 | # Understanding Data
When analyzing data, I have often found it useful to think of the data as being one of four main types, according to the typology proposed by Stevens.[1] Different types of data have certain characteristics; understanding what type of data you have helps with selecting the analysis to perform perform while preventing basic mistakes.
The types, or “scales of measurement,” are:
Nominal
Data identifying unique classifications or objects where the order of values is not meaningful. Examples include zip codes, gender, nationality, sports teams and multiple choice answers on a test.
Ordinal
Data where the order is important but the difference or distance between items is not important or not measured. Examples include team rankings in sport (team A is better than team B, but how much better is open to debate), scales such as health (e.g. “healthy” to “sick”), ranges of opinion (e.g. “strongly agree” to “strongly disagree” or “on a scale of 1 to 10”) and Intelligence Quotient.
Interval
Numeric data identified by values where the degree of difference between items is significant and meaningful, but their ratio is not. Common examples are dates—we can say 2000 CE is 1000CE + 1000 years, but 1000 CE is not half of 2000 CE in any meaningful way—and temperatures on the Celsius and Fahrenheit scales, where a difference of 10° is meaningful, but 10° is not twice as hot as 5°.
Ratio
Numeric data where the ratio between numbers is meaningful. Usually, such scales have a meaningful “0.” Examples include length, mass, velocity, acceleration, voltage, power, duration, energy and Kelvin-scale temperature.
The generally-appropriate statistics and mathematical operations for each type are summarized in table 1.
Table 1: Scales of measurement and allowed statistical and mathematical operations.
Scale Type Statistics Operations
Nominal mode, frequency, chi-squared, cluster analysis =, ≠
Ordinal above, plus: median, non-parametric tests, Kruskal-Wallis, rank-correlation =, ≠, >, <
Interval plus: arithmetic mean, some parametric tests, correlation, regression, ANOVA (sometimes), factor analysis =, ≠, >, <, +, –
Ratio plus: geometric and harmonic mean, ANOVA, regression, correlation coefficient =, ≠, >, <, +, -, ×, ÷
While this is a useful typology for most use, and certainly for initial consideration, there are valid criticisms of Stevens’ typology. For example, percentages and count data have some characteristics of ratio-scale data, but with additional constraints. e.g. the average of the counts $\overline{(2, 2, 1)} = 1.66\ldots$ may not be meaningful. This typology is a useful thinking tool, but it is essential to understand the statistical methods being applied and their sensitivity to departures from underlying assumptions.
### Types of data in R
R[2] recognizes at least fifteen different types of data. Several of these are related to identifying functions and other objects—most users don’t need to worry about most of them. The main types that industrial engineers and scientists will need to use are:
numeric
Real numbers. Also known as double, real and single (note that R stores all real numbers in double-precision). May be used for all scales of measurement, but is particularly suited to ratio scale measurements.
complex
Imaginary real numbers can be manipulated directly as a data type using
x <- 1 + i2
or
x <- complex(real=1, imaginary=2)
Like type numeric, may be used for all scales of measurement.
integer
Stores integers only, without any decimal point. Can be used mainly for ordinal or interval data, but may be used as ratio data—such as counts—with some caution.
logical
Stores Boolean values of TRUE or FALSE, typically used as nominal data.
character
Stores text strings and can be used as nominal or ordinal data.
### Types of variables in R
The above types of data can be stored in several types, or structures, of variables. The equivalent to a variable in Excel would be rows, columns or tables of data. The main ones that we will use are:
vector
Contains one or many elements, and behaves like a column or row of data. Vectors can contain any of the above types of data but each vector is stored, or encoded, as a single type. The vector
c(1, 2, 1, 3, 4)
## [1] 1 2 1 3 4
is, by default, a numeric vector of type double, but
c(1, 2, 1, 3, 4, "name")
## [1] "1" "2" "1" "3" "4" "name"
will be a character vector, or a vector where all data is stored as type character, and the numbers will be stored as characters rather than numbers. It will not be possible to perform mathematical operations on these numbers-stored-as-characters without first converting them to type numeric.
factor
A special type of character vector, where the text strings signify factor levels and are encoded internally as integer counts of the occurrence of each factor. Factors can be treated as nominal data when the order does not matter, or as ordinal data when the order does matter.
factor(c("a", "b", "c", "a"), levels=c("a","b","c","d"))
## [1] a b c a
## Levels: a b c d
array
A generalization of vectors from one dimension to two or more dimensions. Array dimensions must be pre-defined and can have any number of dimensions. Like vectors, all elements of an array must be of the same data type. (Note that the letters object used in the example below is a variable supplied by R that contains the letters a through z.)
# letters a - c in 2x4 array
array(data=letters[1:3], dim=c(2,4))
## [,1] [,2] [,3] [,4]
## [1,] "a" "c" "b" "a"
## [2,] "b" "a" "c" "b"
# numbers 1 - 3 in 2x4 array
array(data=1:3, dim=c(2,4))
## [,1] [,2] [,3] [,4]
## [1,] 1 3 2 1
## [2,] 2 1 3 2
matrix
A special type of array with the properties of a mathematical matrix. It may only be two-dimensional, having rows and columns, where all columns must have the same type of data and every column must have the same number of rows. R provides several functions specific to manipulating matrices, such as taking the transpose, performing matrix multiplication and calculation eigenvectors and eigenvalues.
matrix(data = rep(1:3, times=2), nrow=2, ncol=3)
## [,1] [,2] [,3]
## [1,] 1 3 2
## [2,] 2 1 3
list
Vectors whose elements are other R objects, where each object of the list can be of a different data type, and each object can be of different length and dimension than the other objects. Lists can therefore store all other data types, including other lists.
list("text", "more", 2, c(1,2,3,2))
## [[1]]
## [1] "text"
##
## [[2]]
## [1] "more"
##
## [[3]]
## [1] 2
##
## [[4]]
## [1] 1 2 3 2
data.frame
For most industrial and data scientists, data frames are the most widely useful type of variable. A data.frame is the list analog to the matrix: it is an $m \times n$ list where all columns must be vectors of the same number of rows (determined with NROW()). However, unlike matrices, different columns can contain different types of data and each row and column must have a name. If not named explicitly, R names rows by their row number and columns according to the data assigned assigned to the column. Data frames are typically used to store the sort of data that industrial engineers and scientists most often work with, and is the closest analog in R to an Excel spreadsheet. Usually data frames are made up of one or more columns of factors and one or more columns of numeric data.
data.frame(rnorm(5), rnorm(5), rnorm(5))
## rnorm.5. rnorm.5..1 rnorm.5..2
## 1 0.2939566 1.28985202 -0.01669957
## 2 0.3672161 -0.01663912 -1.02064116
## 3 1.0871615 1.13855476 0.78573775
## 4 -0.8501263 -0.17928722 1.03848796
## 5 -1.6409403 -0.34025455 -0.62113545
More generally, in R all variables are objects, and R distinguishes between objects by their internal storage type and by their class declaration, which are accessible via the typeof() and class() functions. Functions in R are also objects, and the users can define new objects to control the output from functions like summary() and print(). For more on objects, types and classes, see section 2 of the R Language Definition.
Table 2 summarizes the internal storage and R classes of the main data and variable types.
Table 2: Table of R data and variable types.
Variable type Storage type Class Measurement Scale
vector of decimals double numeric ratio
vector of integers integer integer ratio or interval
vector of complex complex complex ratio
vector of characters character character nominal
factor vector integer factor nominal or ordinal
matrix of decimals double matrix ratio
data frame list data.frame mixed
list list list mixed
### References
1. Stevens, S. S. “On the Theory of Scales of Measurement.” Science. 103.2684 (1946): 677-680. Print.
2. R Core Team (2017). R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. URL https://www.R-project.org/.
# Introduction to R for Excel Users
As the saying goes, when all you have is a hammer, everything looks like a nail. Excel was designed to do simple financial analyses and to craft financial statements. Though its capabilities have been expanded over the years, it was never designed to perform the sort of data analysis that industry scientists, engineers and Six Sigma belts need to perform on a daily basis.
Most data analyses performed in Excel look more like simple financial spreadsheets rather than actual data analysis, and this quality of work translates into bad—or at least sub-optimal—business decisions. There are alternatives to Excel, and the free, open-source data analysis platform R is one of them.
Unfortunately, R has a steep learning curve. I’m offering, for free, a short primer on R [PDF] where I’ve sought to make that learning curve a little less painful for engineers and scientists who normally work in Excel.
### Background
A couple of years ago, I was developing a short course to teach R to scientists and engineers in industry who normally used Excel. The goal was to help them transition to a more capable tool. My course design notes morphed into a handout, and when plans for the course fell through, that handout grew into a self-study guide, which I later adapted into this seventy-page, stand-alone introduction for Excel users.
### Organization
The primer walks the reader through the basics of R, starting with a brief overview of capabilities, then diving into installation, basic operations, graphical analysis and basic statistics. I believe that a picture is worth a thousand words, so it’s light on text and heavy on examples and visuals.
The end of the book rounds out with a look at some of the most useful add-ons, the briefest of introductions to writing your own, custom functions in R, and a cross-reference of common Excel functions with their equivalents in R.
The text is broken up into chapters and fully indexed so that it can be used either as a walk-through tutorial or as a quick reference.
Aside
# Update to plot.qcc using ggplot2 and grid
Two years ago, I blogged about my experience rewriting the plot.qcc() function in the qcc package to use ggplot2 and grid. My goal was to allow manipulation of qcc’s quality control plots using grid graphics, especially to combine range charts with their associated individuals or moving range charts, as these two diagnostic tools should be used together. At the time, I posted the code on my GitHub.
I recently discovered that the update to ggplot2 v2.0 broke my code, so that attempting to generate a qcc plot would throw an obscure error from someplace deep in ggplot2. The fix turned out to be pretty easy. The original code used aes_string() instead of aes() because of a barely-documented problem of calling aes() inside a function. It looks like this has been quietly corrected with ggplot2 2.0, and aes_string() is no longer needed for this.
The updated code is up on GitHub. As before, load the qcc library, then source() qcc.plot.R. For the rest of the current session, calls to qcc() will automatically use the new plot.qcc() function.
# Sample Size Matters: Design and Cost
## References
• R Core Team (2014). R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. URL http://www.R-project.org/.
• H. Wickham. ggplot2: elegant graphics for data analysis. Springer New York, 2009.
# Sample Size Matters: Design and Experiments
Previously, I introduced the idea that samples do not look exactly like the populations that they are drawn from, and had a closer look at what impact sample size has on our ability to estimate population statistics like mean, proportion or Cpk from samples. Here, I will have a closer look at how this uncertainty impacts our engineering process. In the next post, I will tie in the engineering impacts and decisions to the business value and costs.
### Difference to detect
When we are testing, we’re either testing to determine that the new product or process performs better than the old or, for cost reduction projects, that the cheaper product or process is at least as good as the existing one.
This means that we need to detect a difference between old and new values, such as a difference in the mean weight between the new and old parts. The larger the sample size, $n$ the smaller the difference, $\Delta$ that we can detect. The error, $\epsilon$, in our estimate of the differences gets smaller as sample size increases:
$\epsilon_{\Delta} \propto \frac{\sigma}{\sqrt{n}}$
Given the uncertainties in our estimate of $\mu$ and $\sigma$, illustrated above, it should be clear now that with small sample sizes we can only detect large differences of many multiples of the sample standard deviation, $S$.
### Mean
When trying to determine if a new product or process is better than an old one, we are usually interested in shifting the mean. We want a product to be lighter, provide more power, or a process to work faster. In such cases, we need to estimate the difference of the means, $\Delta = \mu_2 - \mu_1$ and ensure that it is different than 0 (or some other pre-determined value). The minimum difference that we can reliable detect is plotted below for different sample sizes.
### Standard deviation
In many Six Sigma projects, and any time we want to shift the mean closer to a specification limit, we need to compare the new population standard deviation with the old. The simplest way of making this comparison is by taking the ratio $F = \sigma_{2}^{2} / \sigma_{1}^{2}$, where $\sigma_{2}^{2}$ is the larger of the two variances. The dependence on sample size is illustrated below.
You can see from the inset plot, which includes sample sizes of 2 and 3, that small sample sizes really hurt comparisons of variance, and that interesting differences in variance can’t be detected until we have more than 10 samples.
### Proportions
Proportions, such as fraction of defective parts between a new and old design, can be compared by looking at the difference between the two proportions, $\Delta = \left| p_1 - p_0 \right|$.
You can see from this that proportions data provides much less information than variable data; we need much larger sample sizes to achieve usefully small $\Delta$.
### Summary and look forward
When designing experiments, the goal is to detect some difference between two populations. The uncertainty in our measurements and the variation in the parts has a big impact on how many parts we need to test, or greatly limits what we can learn from an experiment.
Next time, I’ll show how these calculations of sample size and uncertainty impact the busines.
# Sample Size Matters: Uncertainty in Measurement
In my previous post, I gave a brief introduction to populations and samples, and stated that sample size impacts our ability to know what a population really looks like. In this post, I want to show this relationship in more detail. In future posts, I will look at how sample size considerations impact our engineering process and what impacts this has on the business.
### Mean and sample size
The error in our estimate of the mean, $E$, is proportional to the standard deviation of the sample, $S$, and the sample size, $n$.
$E \propto \frac{S}{\sqrt{n}}$
We can visualize this easily enough by plotting the 95% confidence interval. When we sample and calculate the sample mean ($\overline{X}$), the true population mean, $\mu$, (what we really want to know) is likely to be anywhere in the shaded region of the graph below.
This graph shows the 95% confidence region for the true population mean, $\mu$; there’s a 95% chance that the true population mean is within this band. The “0” line on the y axis is our estimate of the mean, $\overline{X}$. We can’t know what the true population mean is, but it’s clear that if we use more samples, we can be sure that our estimate is closer to the true mean.
### Standard deviation and sample size
Likewise, when we calculate the sample standard deviation, $S$, the true standard deviation, $\sigma$ has a 95% chance of being within the confidence band below. For small sample sizes (roughly less than 10), the measured standard deviation can be off from the true standard deviation by several times. Even for ten samples, the potential error is nearly $\pm 1$ standard deviation.
### Proportion and sample size
For proportions, the situation is similar: there is a 95% chance that the true sample proportion, $p$, is within the shaded band based on the measured sample proportion $\hat{p}$. Since this confidence interval depends on $\hat{p}$ and cannot be standardized the way $\mu$ and $\sigma$ can be, confidence intervals for two different proportions are plotted.
For small $n$, proportions data tells us very little.
### Process capability and production costs
The cost of poor quality in product or process design can be characterized by the Cpk:
$Cpk = \mathrm{minimum} \begin{cases}\frac{USL - \mu}{3\sigma} \\\frac{\mu - LSL}{3\sigma}\end{cases}$
Where USL is the upper specification limit (also called the upper tolerance) and LSL is the lower specification limit (or lower tolerance).
We can estimate the defect rate (defects per opportunity, or DPO) from the Cpk:
$DPO = 1 - \Pr\left(X < 3 \times Cpk - 1.5\right)$
That probability function is calculated in R with pnorm(3 * Cpk - 1.5) and in Excel with NORMSDIST(3 * Cpk - 1.5). The 1.5 is a typical value used to account for uncorrected or undetected process drift.
Since we don’t know $\mu$ and $\sigma$, we have to substitute $\overline{X}$ and $S$. The uncertainty in these estimates of the population $\mu$ and $\sigma$ mean that we have uncertainty in what the true process Cpk (or defect rates) will be once we’re in production. When our sample testing tells us that the Cpk should be 1.67 (the blue line), the true process Cpk will actually turn out to be somewhere in the shaded band:
Below the blue line, our product or process is failing to meet customer expectations, and will result in lost customers or higher warranty costs. Above the blue line, we’ve added more cost to the production of the product than we need to, reducing our gross profit margin. Since that gray band doesn’t completely disappear, even at 100 samples, we can never eliminate these risks; we have to find a way to manage them effectively.
The impact of this may be more evident when we convert from Cpk to defect rates (ppm):
### Summary and a look forward
With a fair sampling process, samples will look similar to—and statistically indistinguishable from—the population that they were drawn from. How much they look like the population depends critically on how many samples are tested. The uncertainties, or errors in our estimates, resulting from sample size decisions have impacts all through our design analysis and production planning.
In the next post, I will explore in more detail how these uncertainties impact our experiment designs.
# Sample Size Matters
I find that Six Sigma and Design for Six Sigma courses are often eye-opening experiences for participants. There is an experience of discovering that there are tools available to answer problems that have vexed them, and learning that good engineering and science decisions can lead directly to good business outcomes through logical steps.
One of the most remarkable such moments is when students realize the importance of sample size. In the best cases, there is a forehead-slapping moment where the student realizes that much of the testing they’ve done in the past has probably been a complete waste of time; that while they thought they were seeing interesting differences and making good decisions, they were in fact only fooling themselves by comparing too-small data sets.
I want to show in the next few blog posts why sample size matters, both from a technical perspective and from a business perspective.
### Design example
Throughout the next few posts, I’ll use the example of a manufactured product which the customer requires weigh at least 100 kg, sells for about $140 and that costs$120 to manufacture and convert to a sale (the cost of goods sold, or COGS, is $120). Amount Sales 140 COGS 120 Material 60 Labor and Overhead 60 Gross Profit 20 We want to develop a new version of the product, using a modified design and a new process that, by design, will reduce the cost of material by 10%. The old cost of material was 50% of COGS, or$60. To achieve the material cost reduction of 10%, we have to remove $6 in material costs, improving gross profit to$26.
We believe that the current design masses 120 kg, so we estimate that our new part mass should be $120 - 0.1 \times 120 = 108$ kg.
Current Design New Design Target
Part Weight 120 108
Seems like we might be done at this point, and I’ve seen plenty of engineering projects that stop here. Unfortunately, this isn’t the whole story. Manufacturing will be unable to produce parts of exactly 108 kg, so they’ll need a tolerance range to check parts against. We have that customer requirement for at least 100 kg, so any variation has to stay above that. We also want to save money relative to the current design, so we don’t want many parts to weigh much more than this, especially since the customer isn’t really willing to pay us for the “extra” material beyond 100 kg.
### Population versus sample statistics
Most of process or product improvement is concerned with reducing the standard deviation, $\sigma$, shifting the mean (a.k.a. average), $\mu$, or reducing a proportion, $p$, of a process or product characteristic. These summary statistics refer to the population characteristics—the mean, standard deviation or proportion of all parts of a certain design that will ever be produced, or all times that a production step will ever be completed in the intended manner.
Since we can’t measure the whole population up front—we will be producing parts for a long time—we have to draw a sample from the population, and use the statistics of that sample to gain insight into the total population. We can visualize this, somewhat crudely, with the following:
We can imagine that the blue circles are conforming parts, and the orange octagons are non-conforming parts. If the sampling process is fair, then the sample proportion $\hat{p}$ will be close to—and statistically indistinguishable from—the true population proportion $p$. In the population we have 44 parts total, 8 defective parts and 36 conforming parts. In the sample that we drew, we have 10 parts total, 9 conforming and 1 defective. While $(p = 8/36 = 1/4 \ne \hat{p} = 1/9$, statistically we have
matrix(c(1, 8, 10-1, 44-8), ncol=2) %>%
chisq.test(simulate.p.value = TRUE)
##
## Pearson's Chi-squared test with simulated p-value (based on 2000
## replicates)
##
## data: matrix(c(1, 8, 10 - 1, 44 - 8), ncol = 2)
## X-squared = 0.3927, df = NA, p-value = 0.6692
With such a high p-value (0.67), we fail to reject the null hypothesis that $\hat{p} = p$; in more colloquial terms, we conclude that the apparent difference between 8/36 and 1/9 is only due to random errors in sampling. (For larger counts of successes and failures, prop.test() would also work and would be more informative.)
From our perspective, of course, we don’t know what the population looks like. We don’t have any way of knowing with certainty—or accessing data about—future performance, so there is no way for us to know what the total population looks like. In lieu of population data, we develop a sampling process that allows us to fairly draw a sample from that population.
While we want to know the true population mean, $\mu$, the true population standard deviation, $\sigma$, or the true population proportion $p$, we can only calculate the sample mean, $\overline{X}$, the sample standard deviation, $S$, or the sample proportion $\hat{p}$.
From the known sample, we then reason backward to what the true population looks like. This is where statistics comes into play; statistics allows us to place rigorous boundaries on what the population may look like, without fooling ourselves. Sample size is critical to controlling the uncertainty in these boundaries.
### Summary and a look forward
Testing in product development—and usually in production—involves sampling a product or process. Samples never look exactly like the population that we are concerned about, but if the sampling process is fair then the samples will be statistically indistinguishable from the population. With due awareness of the statistical uncertainties, we can use samples to make decisions about the population.
In the next post, I will look at how sample size impacts the uncertainty in our estimation of population statistics like the mean and standard deviation. In a later post, I will look at how this uncertainty impacts the business.
### A short aside on statistical tests for proportions
The usual way to compare two proportions would be a proportions test (prop.test() in R), but because we have so few samples to compare, the results may be unreliable and prop.test() generates an appropriate warning. fisher.test() provides an exact estimate of the p-value, but the assumptions are violated with data like this, where we are sampling a fixed number of parts (i.e. row sums are fixed, but column sums are not controlled). This leaves us with using a chi-squared test (chisq.test() in R) which is less informative but does the job. Either the Barnard test or Bayesian estimation based on Monte Carlo simulation would be more informative and possibly more robust.
# The Most Useful Data Plot You’ve Never Used
Those of us working in industry with Excel are familiar with scatter plots, line graphs, bar charts, pie charts and maybe a couple of other graph types. Some of us have occasionally used the Analysis Pack to create histograms that don’t update when our data changes (though there is a way to make dynamic histograms in Excel; perhaps I’ll cover this in another blog post).
One of the most important steps in data analysis is to just look at the data. What does the data look like? When we have time-dependent data, we can lay it out as a time-series or, better still, as a control chart (a.k.a. “natural process behavior chart”). Sometimes we just want to see how the data looks as a group. Maybe we want to look at the product weight or the cycle time across production shifts.
Unless you have Minitab, R or another good data analysis tool at your disposal, you have probably never used—maybe never heard of—boxplots. That’s unfortunate, because boxplots should be one of the “go-to” tools in your data analysis tool belt. It’s a real oversight that Excel doesn’t provide a good way to create them.
For the purpose of demonstration, let’s start with creating some randomly generated data:
head(df)
## variable value
## 1 group1 -1.5609
## 2 group1 -0.3708
## 3 group1 1.4242
## 4 group1 1.3375
## 5 group1 0.3007
## 6 group1 1.9717
tail(df)
## variable value
## 395 group1 1.4591
## 396 group1 -1.5895
## 397 group1 -0.4692
## 398 group1 0.1450
## 399 group1 -0.3332
## 400 group1 -2.3644
If we don’t have much data, we can just plot the points:
library(ggplot2)
ggplot(data = df[1:10,]) +
geom_point(aes(x = variable, y = value)) +
coord_flip() +
theme_bw()
But if we have lots of data, it becomes hard to see the distribution due to overplotting:
ggplot(data = df) +
geom_point(aes(x = variable, y = value)) +
coord_flip() +
theme_bw()
We can try to fix this by changing some parameters, like adding semi-transparency (alpha blending) and using an open plot symbol, but for the most part this just makes the data points harder to see; the distribution is largely lost:
ggplot(data = df) +
geom_point(aes(x = variable, y = value), alpha = 0.3, shape = 1) +
coord_flip() +
theme_bw()
The natural solution is to use histograms, another “go-to” data analysis tool that Excel doesn’t provide in a convenient way:
ggplot(data = df) +
geom_histogram(aes(x = value), binwidth = 1) +
theme_bw()
But histograms don’t scale well when you want to compare multiple groups; the histograms get too short (or too narrow) to really provide useful information. Here I’ve broken the data into eight groups:
head(df)
## variable value
## 1 group1 -1.5609
## 2 group1 -0.3708
## 3 group1 1.4242
## 4 group1 1.3375
## 5 group1 0.3007
## 6 group1 1.9717
tail(df)
## variable value
## 395 group8 -0.6384
## 396 group8 -3.0245
## 397 group8 1.5866
## 398 group8 1.9747
## 399 group8 0.2377
## 400 group8 -0.3468
ggplot(data = df) +
geom_histogram(aes(x = value), binwidth = 1) +
facet_grid(variable ~ .) +
theme_bw()
Either the histograms need to be taller, making the stack too tall to fit on a page, or we need a better solution.
The solution is the box plot:
ggplot() +
geom_boxplot(data = df, aes(y = value, x = variable)) +
coord_flip() +
theme_bw()
The boxplot provides a nice, compact representation of the distribution of a set of data, and makes it easy to compare across a large number of groups.
There’s a lot of information packed into that graph, so let’s unpack it:
Median
A measure of the central tendency of the data that is a little more robust than the mean (or arithmetic average). Half (50%) of the data falls below this mark. The other half falls above it.
First quartile (25th percentile) hinge
Twenty-five percent (25%) of the data falls below this mark.
Third quartile (75th percentile) hinge
Seventy-five percent (75%) of the data falls below this mark.
Inter-Quartile Range (IQR)
The middle half (50%) of the data falls within this band, drawn between the 25th percentile and 75th percentile hinges.
Lower whisker
The lower whisker connects the first quartile hinge to the lowest data point within 1.5 * IQR of the hinge.
Upper whisker
The upper whisker connects the third quartile hinge to the highest data point within 1.5 * IQR of the hinge.
Outliers
Any data points below 1.5 * IQR of the first quartile hinge, or above 1.5 * IQR of the third quartile hinge, are marked individually as outliers.
We can add additional values to these plots. For instance, it’s sometimes useful to add the mean (average) when the distributions are heavily skewed:
ggplot(data = df, aes(y = value, x = variable)) +
geom_boxplot() +
stat_summary(fun.y = mean, geom="point", shape = 10, size = 3, colour = "blue") +
coord_flip() +
theme_bw()
Graphs created in the R programming language using the ggplot2 and gridExtra packages.
### References
1. R Core Team (2014). R: A language and environment for statistical computing. R Foundation for Statistical
Computing, Vienna, Austria. URL http://www.R-project.org/.
2. H. Wickham. ggplot2: elegant graphics for data analysis. Springer New York, 2009.
3. Baptiste Auguie (2012). gridExtra: functions in Grid graphics. R package version 0.9.1.
http://CRAN.R-project.org/package=gridExtra
# Flowing Requirements from the VoC or VoP
In a previous post, I talked about the voice of the customer (VoC), voice of the process (VoP) and the necessity of combining the two when specifying a product. Here, I’d like to offer a general method for applying this in the real world, which can be implemented as a template in Excel.
### Recap
I showed that there was a cost function associated with any specification that derived from both the VoC (expressed as tolerances or specification limits) and from the process capability. An example cost function for a two-sided tolerance is reproduced below.
Percent of target production costs given an average production weight and four different process capabilities.
I argued that, given this cost function, specifying a product requires specifying both the product specification limits (or tolerances) and the minimally acceptable process capability, Cpk. Ideally, both of these should flow down from a customer needs analysis to the finished product, and from the finished product to the components, and so on to materials.
### Requirements flow down and up
To flow all requirements down like this, we would need to know the transfer functions, $Y = f(X)$, for each requirement Y and each subcomponent characteristic X. There are methods for doing this, like Design for X or QFD, but they can be difficult to implement. In the real world, we don’t always know these transfer functions, and determining them can require non-trivial research projects that are best left to academia.
As an illustration, we will use the design of a battery (somewhat simplified), where we have to meet a minimum requirement that is the sum of component parts. The illustration below shows the component parts of a battery, or cell. It includes a container (or “cell wall”), positive and negative electrodes (or positive and negative “plates”), electrolyte and terminals that provide electrical connection to the outside world. Usually, we prefer lighter batteries to heavier ones, but for this example, we’ll suppose that a customer requires a minimum weight. This requirement naturally places limits on the weight of all components.
In the absence of transfer functions, we often make our best guess, build a few prototypes, and then adjust the design. This may take several iterations. A better approach is to estimate the weight specification limits and minimum Cpk by calculation before any cells are actually built.
General drawing of the structure of aircraft battery’s vented type NiCd cell. Ransu. Wikipedia, [http://en.wikipedia.org/wiki/ File:Aircraft_battery_cell.gif]. Accessed 2014-04-04.
Suppose the customer specifies a cell minimum weight of 100 kg. From similar designs, we know the components that contribute to the cell mass and have an idea of the percentage of total weight that each component contributes.
$m_{cell}=m_{container}+m_{terminals}+m_{electrolyte}+m_{poselect}+m_{negelect}$
Each individual component is therefore a fraction fm of the total cell mass, e.g.
$m_{container}=f_{m,container}m_{cell}$
More generally, for a measurable characteristic c, component i has an expected mean or target value of $T_{i,c}=f_{i,c}\mu_{parent,c}$ or $T_{i,c}=f_{i,c}T_{parent,c}$.
In our example, we may know from similar products or from design considerations that we want to target the following percents for each fraction fm:
• 5% for container
• 19% for terminals
• 24% for electrolyte
• 26% for positive electrodes
• 26% for negative electrodes
### Specification Limits
Upper Specification Limit (USL)
The maximum allowed value of the characteristic. Also referred to as the upper tolerance.
Lower Specification Limit (LSL)
The minimum allowed value of the characteristic. Also referred to as the lower tolerance.
Since the customer will always want to pay as little as possible, a specified lower weight of 100 kg is equivalent to saying that they are only willing to pay for 100 kg of material; any extra material is added cost that reduces our profit margin. If we tried to charge them for 150 kg of material, they would go buy from our competitors. The lower specification limit, or lower tolerance, of the cell weight is then 100 kg.
If the customer does not specify a maximum weight, or upper specification limit, then we determine the upper limit by the maximum extra material cost that we are willing to bear. In this example, we decide that we are willing to absorb up to 5% additional cost per part. Assuming that material and construction contributes 50% to the total cell cost, the USL is then 110 kg. To allow for some variation, we can set a target weight in the middle: 105 kg. From data on previous designs and the design goals, we can apportion the target weight to each component of the design, as shown in the table below.
We can apply the same fractions to the cell USL and LSL to obtain a USL and LSL of each component. As long as parts are built within these limits, the cell will be within specification. The resulting specification for cell and major subcomponents is illustrated in table [tblSpecification]. Further refinement of the allocation of USL and LSL to the components is possible and may be needed if the limits do not make sense from a production or cost perspective.
Part Percent Target LSL USL /kg /kg /kg Cell 100% 105 100 110 Container 5% 5.2 5 5.5 Terminals 19% 19.9 19 20.9 Electrolyte 24% 25.2 24 26.4 Positive electrodes 26% 27.3 26 28.6 Negative electrodes 26% 27.3 26 28.6
### Variance of components and Cpk
When a characteristic is due to the sum of the part’s components, as with cell mass, the part-to-part variation in the characteristic is likewise due to the variation in the components. However, where the characteristic adds as the sum of the components,
$m_{cell}=m_{container}+m_{terminals}+m_{electrolyte}+m_{poselect}+m_{negelect}$
the variance, $\sigma^{2}$ adds as the sum of squares
$\sigma_{cell}^{2}=\sigma_{container}^{2}+\sigma_{terminal}^{2}+\sigma_{electrolyte}^{2}+\sigma_{poselect}^{2}+\sigma_{negelect}^{2}$
The variance of any individual component is therefore a function of the total parent part variance
$\sigma_{container}^{2}=\sigma_{cell}^{2}-\sigma_{terminal}^{2}-\sigma_{electrolyte}^{2}-\sigma_{poselect}^{2}-\sigma_{negelect}^{2}$
or
$\displaystyle \sigma_{container,mass}^{2}=f_{\sigma,container}\sigma_{cell,mass}^{2}$
Since this is true for all components, the two fractions $f_{m}$ and $f_{\sigma}$ will be approximately equal. Therefore if we don’t know the fractions $f_{\sigma}$, we can use the fraction $f_{m}$, which usually easier to work out, to allocate the variance to each component:
$\displaystyle \sigma_{container,mass}^{2}=f_{m,container}\times\sigma_{cell,mass}^{2}$
More generally, for measurable characteristic $c$ of a subcomponent $i$ of a parent component,
$\displaystyle \sigma_{i,c}=\sqrt{f_{c,i}}\:\sigma_{c,parent}$
Since the given $\sigma$ is the maximum allowed for the parent to meet the desired Cpk, this means that $\sigma_{i}^{2}$ is an estimate for the maximum allowed component variance. Manufacturing can produce parts better than this specification, but any greater variance will drive the parent part out of specification.
### Calculating Specification Limits
In general, there are two conflicting goals in setting specifications:
1. Make them as wide as possible to allow for manufacturing variation while still meeting the VoC.
2. Make them as narrow as possible to stay near the minimum of the cost function.
For this, Crystall Ball or iGrafx are very useful tools during development, as we can simulate a set of arts or processes, analyze the allowed variation in the product and easily flow that variation down to each component. In the absence of these tools, Minitab or Excel can be used to derive slightly less robust solutions.
#### Calculating from Customer Requirements
1. Identify any customer requirements and set specification limits (USL and LSL) accordingly. If the customer requirements are one-sided, determine the maximum additional cost we are willing to accept, and set the other specification limit accordingly. Some approximation of costs may be needed.
2. If no target is given, set the target specification for each requirement as the average of USL and LSL.
3. Set the minimum acceptable Cpk for each specification. Cpk = 1.67 is a good starting value. Use customer requirements for Cpk, where appropriate, and consider, also, whether the application requires a higher Cpk (weakest link in the chain….
4. Calculate the maximum allowed standard deviation to meet the Cpk requirement as $\sigma_{parent}=\left(USL-LSL\right)/\left(6\times Cpk\right)$.
5. For each subcomponent (e.g. the cell has subcomponents of container, electrodes, electrolyte, and so on), apportion the target specification to each of the subcomponents based on engineering considerations and judgement. If the fractions $f$ are known, $T_{i}=f_{i}\times T_{parent}$.
6. Calculate the fraction $f_{i}$ (or percent) of the parent total for each subcomponent if not already established in step (5).
7. Calculate the USL and LSL for each subcomponent by multiplying the parent USL and LSL by the component’s fraction of parent (from step 6). $USL_{i}=f_{i}\times USL_{parent}$ and $LSL_{i}=f_{i}\times LSL_{parent}$.
8. Estimate the allowed standard deviation $\sigma_{i}$ for each subcomponent as
$\displaystyle \sigma_{i}=\mathtt{SQRT}\left(f_{i}\right)\times\sigma_{parent}.$
9. Calculate the minimum allowed Cpk for each subcomponent from the results of (5), (7) and (8), using the target, $T$, for the mean, $\mu$.
$\displaystyle Cpk_{i}=minimum\begin{cases}\frac{USL_{i}-T_{i}}{3\sigma_{i}}\\\frac{T_{i}-LSL_{i}}{3\sigma_{i}}\end{cases}$
10. Repeat steps (5) through (9) until all components have been specified.
11. For each component, report the specified USL, LSL, target T and maximum Cpk.
#### Calculating from Process Data
When there is no clear customer-driven requirement or clear requirement from parent parts (e.g. dimensional specifications that can be driven by the fit of parts), but specification limits are still reasonably needed, we can start from existing process data.
This is undesirable because any change to the process can force a change to the product specification, without any clear understanding of the impact on customer needs or requirements; the VoC is lost.
The calculation of USL and LSL from process data is also somewhat more complicated, as we have to use the population mean and standard deviation to determine where to set the USL and LSL, without really knowing what that mean and standard deviation are.
In the real world, we have to live with such constraints. To deal with these limitations, we will use as much data as is available and calculate the confidence intervals on both the mean and the standard deviation. The calculation for USL and LSL becomes
$\setlength\arraycolsep{2pt}\begin{array}{rl}\displaystyle USL &=\textrm{upper 95\% confidence on the mean}\smallskip\\ \displaystyle &\quad +k\times\textrm{upper 95\% confidence on the standard deviation}\end{array}$
$\setlength\arraycolsep{2pt}\begin{array}{rl}\displaystyle LSL &=\textrm{lower 95\% confidence on the mean}\smallskip\\ &\quad -k\times\textrm{upper 95\% confidence on the standard deviation}\end{array}$
where $k$ is the number of process Sigmas desired, based on the tolerance cost function. Most of the time, we will use $k=5$, to achieve a Cpk of 1.67.
We always use the upper 95% confidence interval on the standard deviation. We don’t care about the lower confidence interval, since a small $\sigma$ will not help us in setting specification limits.
1. Calculate the mean ($\mu_{parent}$) from recent production data. In Excel, use the AVERAGE() function on the data range.
2. Calculate the standard deviation ($\sigma_{parent}$) from recent production data. In Excel, you can use the STDEV() function on the data range.
1. If the order of production data is known, or SPC is in use, a better method is to use the range-based estimate from the control charts. This will be discussed in subsequent training on control charts.
3. Count the number of data points, n, that were used for the calculations (1) and (2). You can use the COUNT() function on the data range.
4. Calculate the 95% confidence level on the mean. In Excel, this is accomplished with
$CL=\mathtt{TINV}\left(\left(1-0.95\right);n-1\right)\times\sigma_{parent}/\mathtt{SQRT}\left(n\right)$
In Excel 2010 and later, TINV() should be replaced with T.INV.2T().
5. Calculate the 95% confidence interval on the mean as $CI_{upper}=\mu+CL$ and $CI_{lower}=\mu-CL$.
6. Calculate the upper and lower 95% confidence limits on the standard deviation. In Excel, this is accomplished with
$\sigma_{upper}=\sigma_{parent}\times\mathtt{SQRT}\left(\left(n-1\right)/\mathtt{CHIINV}\left(\left(1-0.95\right)/2;n-1\right)\right)$
and
$\sigma_{lower}=\sigma_{parent}\times\mathtt{SQRT}\left(\left(n-1\right)/\mathtt{CHIINV}\left(1-\left(1-0.95\right)/2;n-1\right)\right)$
In Excel 2010 and later, CHIINV() can be replaced with CHISQ.INV.RT() for improved accuracy.
7. Calculate the LSL as $LSL_{parent}=CI_{lower}-k\sigma_{upper}$. You might use a value other than 5 if the customer requirements or application require a higher process Sigma.
8. Calculate the USL as $USL_{parent}=CI_{upper}+k\sigma_{upper}$.
9. For each subcomponent (e.g. the cell has subcomponents of positive electrode, negative electrode, electrolyte, and so on), apportion the parent part mean to each of the subcomponents based on engineering considerations and judgement. If the fractions $f$ are known, $T_{i}=f_{i}\times\mu_{parent}$.
10. If the the fraction (or percent) $f_{i}$ of the parent total for each subcomponent is not known, calculate it using the results of step (9).
11. Calculate the USL and LSL for each subcomponent by multiplying the parent USL and LSL by the component’s fraction of parent (from step 6). $USL_{i}=f_{i}\times USL_{parent}$ and $LSL_{i}=f_{i}\times LSL_{parent}$.
12. Estimate the allowed standard deviation $\sigma_{i}$ for each subcomponent as $\sigma_{i}=\mathtt{SQRT}\left(f_{i}\right)\times\sigma_{lower}$
13. Calculate the minimum allowed Cpk for each subcomponent from the results of (5), (7) and (8), using the target $T_{i}$ for the mean, $\mu_{i}$.
$\displaystyle Cpk_{i}=minimum\begin{cases}\frac{USL_{i}-T_{i}}{3\sigma_{i}}\\\frac{T_{i}-LSL_{i}}{3\sigma_{i}}\end{cases}$
14. Repeat steps (9) through (13) until all components have been specified.
15. For each component, report the specified USL, LSL, target T and maximum Cpk.
# Can We do Better than R-squared?
If you're anything like me, you've used Excel to plot data, then used the built-in “add fitted line” feature to overlay a fitted line to show the trend, and displayed the “goodness of fit,” the r-squared (R2) value, on the chart by checking the provided box in the chart dialog.
The R2 calculated in Excel is often used as a measure of how well a model explains a response variable, so that “R2 = 0.8” is interpreted as “80% of the variation in the 'y' variable is explained by my model.” I think that the ease with which the R2 value can be calculated and added to a plot is one of the reasons for its popularity.
There's a hidden trap, though. R2 will increase as you add terms to a model, even if those terms offer no real explanatory power. By using the R2 that Excel so helpfully provides, we can fool ourselves into believing that a model is better than it is.
Below I'll demonstrate this and show an alternative that can be implemented easily in R.
### Some data to work with
First, let's create a simple, random data set, with factors a, b, c and response variable y.
head(my.df)
## y a b c
## 1 2.189 1 -1.2935 -0.126
## 2 3.912 2 -0.4662 1.623
## 3 4.886 3 0.1338 2.865
## 4 5.121 4 1.2945 4.692
## 5 4.917 5 0.1178 5.102
## 6 4.745 6 0.4045 5.936
Here is what this data looks like:
### Calculating R-squared
What Excel does when it displays the R2 is create a linear least-squares model, which in R looks something like:
my.lm <- lm(y ~ a + b + c, data = my.df)
Excel also does this when we call RSQ() in a worksheet. In fact, we can do this explicitly in Excel using the Regression analysis option in the Analysis Pack add-on, but I don't know many people who use this, and Excel isn't known for its reliability in producing good output from the Analysis Pack.
In R, we can obtain R2 via the summary() function on a linear model.
summary(my.lm)
##
## Call:
## lm(formula = y ~ a + b + c, data = my.df)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.2790 -0.6006 0.0473 0.5177 1.5299
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.080 0.763 2.72 0.034 *
## a -0.337 0.776 -0.43 0.679
## b -0.489 0.707 -0.69 0.515
## c 1.038 0.817 1.27 0.250
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.1 on 6 degrees of freedom
## Multiple R-squared: 0.833, Adjusted R-squared: 0.75
## F-statistic: 10 on 3 and 6 DF, p-value: 0.00948
Since summary() produces a list object as output, we can grab just the R2 value.
summary(my.lm)$r.squared ## [1] 0.8333 Normally, we would (somewhat loosely) interpret this as telling us that about 83% of the variation in the response y is explained by the model. Notice that there is also an "adjusted r-squared” value given by summary(). This tells us that only 75% of the variation is explained by the model. Which is right? ### The problem with R-squared Models that have many terms will always give higher R2 values, just because more terms will slightly improve the model fit to the given data. The unadjusted R2 is wrong. The calculation for adjusted R2 is intended to partially compensate for that “overfit,” so it's better. It's nice that R shows us both values, and a pity that Excel won't show the adjusted value. The only way to get an adjusted R2 in Excel is to run the Regression analysis; otherwise, we have to calculate adjusted R2 manually. Both R2 and adjusted R2 are measures of how well the model explains the given data. However, in industry we usually want to know something a little different. We don't build regression models to explain only the data we have; we build them to think about future results. We want R2 to tell us how well the model predicts the future. That is, we want a predictive R2. Minitab has added the ability to calculate predictive R2 in Minitab 17, and has a nice blog post explaining this statistic. ### Calcuting predictive R-squared Neither R nor Excel provide a means of calculating the predictive R2 within the default functions. While some free R add-on packages provide this ability (DAAG, at least), we can easily do it ourselves. We'll need a linear model, created with lm(), for the residuals so we can calculate the “PRESS” statistic, and then we need the sum of squares of the terms so we can calculate a predictive R2. Since the predictive R2 depends entirely on the PRESS statistic, we could skip the added work of calculating predictive R2 and just use PRESS, as some authors advocate. The lower the PRESS, the better the model is at fitting future data from the same process, so we can use PRESS to compare different models. Personally, I'm used to thinking in terms of R2, and I like having the ability to compare to the old R2 statistic that I'm familiar with. To calculate PRESS, first we calculate the predictive residuals, then take the sum of squares (thanks to (Walker’s helpful blog post) for this). This is pretty easy if we already have a linear model. It would take a little more work in Excel. pr <- residuals(my.lm)/(1 - lm.influence(my.lm)$hat)
PRESS <- sum(pr^2)
PRESS
## [1] 19.9
The predictive R2 is then (from a helpful comment by Ibanescu on LikedIn) the PRESS divided by the total sum of squares, subtracted from one. The total sum of squares can be calculated directly as the sum of the squared residuals, or obtained by summing over Sum Sq from an anova() on our linear model. I prefer using the anova function, as any statistical subtleties are more likely to be properly accounted for there than in my simple code.
# anova to calculate residual sum of squares
my.anova <- anova(my.lm)
tss <- sum(my.anova$"Sum Sq") # predictive R^2 pred.r.squared <- 1 - PRESS/(tss) pred.r.squared ## [1] 0.5401 You'll notice that this is smaller than the residual R2, which is itself smaller than the basic R2. This is the point of the exercise. We don't want to fool ourselves into thinking we have a better model than we actually do. One way to think of this is that 29% (83% – 54%) of the model is explained by too many factors and random correlations, which we would have attributed to our model if we were just using Excel's built-in function. When the model is good and has few terms, the differences are small. For example, working through the examples in Mitsa's two posts, we see that for her model 3, R2 = 0.96 and the predictive R2 = 0.94, so calculating the predictive R2 wasn't really worth the extra effort for that model. Unfortunately, we can't know, in advance, which models are “good.” For Mitsa's model 1 we have R2 = 0.95 and predictive R2 = 0.32. Even the adjusted R2 looks pretty good for model 1, at 0.94, but we see from the predictive R2 that our model is not very useful. This is the sort of thing we need to know to make correct decisions. ### Automating In R, we can easily wrap these in functions that we can source() and call directly, reducing the typing. Just create a linear model with lm() (or an equivalent) and pass that to either function. Note that pred_r_squared() calls PRESS(), so both functions have to be sourced. pred_r_squared <- function(linear.model) { lm.anova <- anova(linear.model) tss <- sum(lm.anova$"Sum Sq")
# predictive R^2
pred.r.squared <- 1 - PRESS(linear.model)/(tss)
return(pred.r.squared)
}
PRESS <- function(linear.model) {
pr <- residuals(linear.model)/(1 - lm.influence(linear.model)\$hat)
PRESS <- sum(pr^2)
return(PRESS)
}
Then we just call the function to get the result:
pred.r.squared <- pred_r_squared(my.lm)
pred.r.squared
## [1] 0.5401
I've posted these as Gists on GitHub, with extra comments, so you can copy and paste from here or go branch or copy them there. | 13,365 | 54,492 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 127, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.935017 |
https://www.cleancss.com/convert-units/units-of-volume/usqt-l/cup | 1,723,732,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00621.warc.gz | 554,848,101 | 18,200 | # Convert USQT-L to CUP
1
US Quart (liquid)s
=
4.0000000008
Cups
How To Calculate US Quart (liquid)s To Cups
To convert us quart (liquid)s to cups you simply multiply your us quart (liquid)s by 4.0000000008. The formula would look like this:
Ycup = Xusqt-l * 4.0000000008
#### 1 US Quart (liquid)s equals
Litres 0.946 Mililitres 946.353 Meters Cubeds 946.353 Pints 1.665 Gallons 0.215 US Fluid Ounces 32 UK Fluid Ounces 33.307 UK Quarts 0.833 US Quart (dry)s 0.859 Cups 4 Tablespoons 64 Teaspoons 192
#### US Quart (liquid)s To Cups Conversion Table
From To
1 usqt-l4.0000000008 cup
2 usqt-l8.0000000017 cup
3 usqt-l12.0000000025 cup
4 usqt-l16.0000000034 cup
5 usqt-l20.0000000042 cup
6 usqt-l24.0000000051 cup
7 usqt-l28.0000000059 cup
8 usqt-l32.0000000068 cup
9 usqt-l36.0000000076 cup
10 usqt-l40.0000000085 cup
11 usqt-l44.0000000093 cup
12 usqt-l48.0000000101 cup
13 usqt-l52.000000011 cup
14 usqt-l56.0000000118 cup
15 usqt-l60.0000000127 cup
16 usqt-l64.0000000135 cup
17 usqt-l68.0000000144 cup
18 usqt-l72.0000000152 cup
19 usqt-l76.0000000161 cup
20 usqt-l80.0000000169 cup
30 usqt-l120.0000000254 cup
40 usqt-l160.0000000338 cup
50 usqt-l200.0000000423 cup
60 usqt-l240.0000000507 cup
70 usqt-l280.0000000592 cup
80 usqt-l320.0000000676 cup
90 usqt-l360.0000000761 cup
100 usqt-l400.0000000845 cup | 527 | 1,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.399847 |
http://slideplayer.com/slide/1545258/ | 1,508,571,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00340.warc.gz | 317,641,187 | 22,894 | # Est+Opt CIRM 18/8/2009 A. ILIADIS 1 Estimation + optimization in PK modeling Introduction to modeling Estimation criteria Numerical optimization, examples.
## Presentation on theme: "Est+Opt CIRM 18/8/2009 A. ILIADIS 1 Estimation + optimization in PK modeling Introduction to modeling Estimation criteria Numerical optimization, examples."— Presentation transcript:
Est+Opt CIRM 18/8/2009 A. ILIADIS 1 Estimation + optimization in PK modeling Introduction to modeling Estimation criteria Numerical optimization, examples http://pharmapk.pharmacie.univ-mrs.fr/
Est+Opt CIRM 18/8/2009 A. ILIADIS 2 Real process and mathematical model Fitted model Fitted model Real process Math. model
Est+Opt CIRM 18/8/2009 A. ILIADIS 3 Functional scheme Measurement noise Measurement noise PK model Equivalence criterion Equivalence criterion Nonlinear programming Nonlinear programming Administration protocol Administration protocol A priori information A priori information + Observation Prediction PK process
Est+Opt CIRM 18/8/2009 A. ILIADIS 4 Mathematical modeling Models are defined by : - their structure ( number and connectivity of compartments, etc ) expressed by mathematical operations involving adjustable parameters : Ex :1-cpt, exponential structure, parameters : - the numerical value of parameters used : CHARACTERIZATION Structure CHARACTERIZATION Structure MODELING System Identification MODELING System Identification ESTIMATION Parameters ESTIMATION Parameters +
Est+Opt CIRM 18/8/2009 A. ILIADIS 5 Checking identifiability Structural identifiability : given a hypothetical structure with unknown parameters, and a set of proposed experiments (not measurements !), would the unknown parameters be uniquely determinable from these experiments ? Parametric identifiability : estimating the parameters from measurements with errors and optimize sampling designs. : structural non-identifiable Non-consistent estimate 2 2 1 1
Est+Opt CIRM 18/8/2009 A. ILIADIS 6 Structural identifiability It depends on the observation site ! Solutions : Grouping : But ONLY identifiable parameters : Setting : 2 2 1 1 2 2 1 1
Est+Opt CIRM 18/8/2009 A. ILIADIS 7 Functional scheme (dynamic) Linear / p model : no loop, one stage estimation. Nonlinear / p model : many loops until convergence. Measurement noise Measurement noise PK model Equivalence criterion Equivalence criterion Nonlinear programming Nonlinear programming Administration protocol Administration protocol A priori information A priori information + PK process Arbitrary initial values Arbitrary initial values
Est+Opt CIRM 18/8/2009 A. ILIADIS 8 Iterations, parameter convergence Ex : Fotemustine neutrophil toxicity : Nonlinear modeling : Optimized final values Arbitrary initial values
Est+Opt CIRM 18/8/2009 A. ILIADIS 9 Errors in the functional scheme The existing errors : experimental, structural, parametric. Residual error : experimental, structural (model misspecification). Residual error Initial parametric error (canceled at the convergence)
Est+Opt CIRM 18/8/2009 A. ILIADIS 10 Parametric and output spaces Observation Comparison Parametric space Output space Prediction PK process PK model Real process Artificial mechanism Random component Precision of estimates Measurement error
Est+Opt CIRM 18/8/2009 A. ILIADIS 11 Optimal estimation Estimation is the operation of assigning a numerical values to unknown parameters, based on noise-corrupted observations. Organization of the variables : The observed drug concentrations over time, ( dimensional vector). The random parameters to be estimated, ( dimensional vector). Consider the joint pdf and then : the marginal is called prior pdf [ the marginal is not of interest ]. the conditional is called posterior pdf : the conditional leads to the likelihood function : MAP MLE
Est+Opt CIRM 18/8/2009 A. ILIADIS 12 Maximum a posteriori (MAP) Design : A reasonable estimate of would be the mode of the posterior density for the given observation : Ex : if The role of the dispersion.
Est+Opt CIRM 18/8/2009 A. ILIADIS 13 Maximum likelihood (MLE) Design : After the observation has been obtained, a reasonable estimate of would be, the value which gives to the particular observation the highest probability of occurrence : Ex : if The role of the precision.
Est+Opt CIRM 18/8/2009 A. ILIADIS 14 The influence of x on the conditional
Est+Opt CIRM 18/8/2009 A. ILIADIS 15 The influence of y on the likelihood
Est+Opt CIRM 18/8/2009 A. ILIADIS 16 MLE criterion for single - output Initial form : Hypotheses : H0 : The model is an exact description of the real process. Error Output H1 : Additive error : H2 : Normal error : H3 : Independence :
Est+Opt CIRM 18/8/2009 A. ILIADIS 17 Variance heterogeneity The regression model : assumes that Need to relax this assumption (particularly when the model is highly nonlinear). Transformed models Find a transformation function under which the error assumptions hold, i.e. : where Box – Cox transformations : Other transformations : John – Drapper, Carroll, Huber, etc. Estimate !
Est+Opt CIRM 18/8/2009 A. ILIADIS 18 General form of the MLE criterion For available observed data and under the H3 hypothesis the estimator becomes : Where : if is the criterion function to be minimized. The 1st term is known as the extended SE term. The 2nd term is called the weighted SE term. It is the only one involving observed data and it is weighted by the uncertainty of experiment.
Est+Opt CIRM 18/8/2009 A. ILIADIS 19 Criterion and error variance model After introducing the error variance model : is minimum along the direction when : or with Then : Nonlinearly unconstrained optimization : Find : Assumptions :is computable for all and analytic solution does not exist.
Est+Opt CIRM 18/8/2009 A. ILIADIS 20 Iterative solutions Solution for the nonlinear optimization problem Sequentially approximate starting from an initial value and converging towards a stationary point. Design a routine algorithm generating the converging sequence : Terminology : is the initialization and obtaining from is an iteration. Assign :
Est+Opt CIRM 18/8/2009 A. ILIADIS 21 Taylor's expansion for smooth multivariate function: Construct simple approximations of a general function in a neighborhood of the reference point. With a vector of unit length supplying the direction of search, and a scalar specifying the step length: Associate successive approximations to iterations : Approximation of functions
Est+Opt CIRM 18/8/2009 A. ILIADIS 22 Direction of search Linear approx. of : The scalar gives the rate of change of at the point along the direction. To reduce, move along the direction opposite to : the descent direction. Quadratic approx. of : The scalar involves the second derivative of. characterizes an ellipse. To reduce, move along the direction targeting the center of ellipse : : the Newton direction
Est+Opt CIRM 18/8/2009 A. ILIADIS 23 Line search Minimization directions : moving along the Newton direction for quadratic surfaces, near. Elsewhere, move along the descent direction. Line search : to complete the iteration search for in the direction of search : or Minimization direction 3.8 3.4 4.5
Est+Opt CIRM 18/8/2009 A. ILIADIS 24 Families of algorithms Practical : Approximate derivatives by finite-differences instead analytical calculation. Classify : Twice-continuously differentiable : Second derivative : quadratic model of, compute and invert (not numerically stable, time consuming processing). First derivative : quadratic model of, approximate : without inverting but directly from by finite-differences of. quasi-Newton methods (appropriate in many circumstances) : BFGS, DFP,... Non-derivative : linear model of. Approximate by finite-differences of (for smooth functions) : Powell, Brent,... No assumptions on differentiability : heuristic algorithms : NMS, Hooke-Jeeves,...
Est+Opt CIRM 18/8/2009 A. ILIADIS 25 The information matrix The Fisher information matrix : For MLE estimation : Cramér-Rao inequality : is a lower bound of the covariance matrix, evaluating the precision of. In practice : With the vector of the sampling times, Obtain the sensitivity matrix with elements, Set and the order diagonal matrices having as elements and respectively.
Est+Opt CIRM 18/8/2009 A. ILIADIS 26 Covariance (precision) matrix The order precision matrix is : Dependence on the sampling protocol: Graphic interpretation of the precision matrix : is symmetric, and, if, it is also definite positive (by construction). If, then is an a dimensional ellipsoid. The volume of the ellipsoid is : Sums of weighted products of sensitivity functions over the available sampling times The lowest, the most precise
Est+Opt CIRM 18/8/2009 A. ILIADIS 27 Check the structural identifiability The sensitivity matrix depends : On the experiment (not measurements !) and On the model parametrization (structural and parametric). Ensure definite-positivity of the sensitivity matrix : It must be of full rank for any numeric value of the parameters, e.g., for the arbitrary initial values (several). Ex : Observation in the central cptfree # central cpt fixed #peripheral cpt ? ?
Est+Opt CIRM 18/8/2009 A. ILIADIS 28 Simulation in optimization 2-cpt model : Administration : IV bolus Observation : Horizon nbr Heteroscedastic
Est+Opt CIRM 18/8/2009 A. ILIADIS 29 Performances of algorithms
Download ppt "Est+Opt CIRM 18/8/2009 A. ILIADIS 1 Estimation + optimization in PK modeling Introduction to modeling Estimation criteria Numerical optimization, examples."
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Mathematics
(Chapter 7) (Integrals)
(Class XII)
Exercise 7.4
Integrate the functions in Exercises 1 to 23.
Question 1:
Let x3 = t
3x2 dx = dt
Question 2:
Let 2x = t
2dx = dt
1
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 3:
Let 2 x = t
dx = dt
Question 4:
Let 5x = t
5dx = dt
2
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 5:
Question 6:
Let x3 = t
3x2 dx = dt
3
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 7:
From (1), we obtain
Question 8:
Let x3 = t
3x2 dx = dt
4
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 9:
Let tan x = t
sec2x dx = dt
Question 10:
5
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 11:
Question 12:
6
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 13:
7
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 14:
8
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 15:
9
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 16:
Equating the coefficients of x and constant term on both sides, we obtain
4A = 4 A = 1
A+B=1B=0
Let 2x2 + x 3 = t
(4x + 1) dx = dt
Question 17:
Equating the coefficients of x and constant term on both sides, we obtain
10
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 18:
Equating the coefficient of x and constant term on both sides, we obtain
11
Mathematics
(Chapter 7) (Integrals)
(Class XII)
12
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Substituting equations (2) and (3) in equation (1), we obtain
13
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 19:
Equating the coefficients of x and constant term, we obtain
2A = 6 A = 3
9A + B = 7 B = 34
6x + 7 = 3 (2x 9) + 34
14
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 20:
Equating the coefficients of x and constant term on both sides, we obtain
15
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Using equations (2) and (3) in (1), we obtain
16
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 21:
Let x2 + 2x +3 = t
(2x + 2) dx =dt
17
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 22:
Equating the coefficients of x and constant term on both sides, we obtain
18
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 23:
Equating the coefficients of x and constant term, we obtain
19
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Using equations (2) and (3) in (1), we obtain
20
Mathematics
(Chapter 7) (Integrals)
(Class XII)
Question 24:
equals
(A) x tan1 (x + 1) + C
(B) tan1 (x + 1) + C
(C) (x + 1) tan1x + C
(D) tan1x + C
Question 25:
equals
(A)
(B)
(C)
(D)
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# Arithmetic : Ratio and Proportion
Preparation Just what you need to know !
Ratio
A ratio is a number (like a fraction) that compares two numbers by division.
The ratio of a to b may be written in different ways such as a to b, or a : b, or a ÷ b, or a / b.
The numbers a and b are called the terms of the ratio.
Since a ratio a : b is a fraction a / b, the term b cannot be zero.
Ratios can be treated just like fractions.
Since the fraction a / b is different from the fraction b / a, the order of the terms in a ratio is important.
For example, the ratio of the number of sides in a pentagon to the number of sides in an octagon is 5 : 8 and not 8 : 5.
To find the ratio of two quantities, they must be expressed in the same unit of measure.
Example
What is the ratio of 5 feet to 9 inches?
Solution. 5 feet = 5 × 12 inches = 60 inches since 1 foot = 12 inches.
So, the ratio is 60 inches / 9 inches = 20/3 or 20 : 3
Note that the units cancel out in the fraction above to give the ratio as a pure number without any units.
Note that a division of two distinct quantities with different units is usually a rate and not a ratio.
For example, 100 miles / 2 hours is a rate of speed of 50 miles per hour.
GMAT Math Review - Arithmetic : Index for Ratio & Proportion
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Sometimes a text value might repeat in a range more than once. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy, Download Extract Number from String Excel Template, New Year Offer - All in One Excel VBA Bundle (35 Courses with Projects) View More, You can download this Extract Number from String Excel Template here –, All in One Excel VBA Bundle (35 Courses with Projects), 35+ Courses | 120+ Hours | Full Lifetime Access | Certificate of Completion, #1 – Extract Number from the String at the End of the String, #2 – Extract Numbers from Right Side but Without Special Characters, #3 – Extract Numbers from any Position of the String, Extract Number from String Excel Template. If you want to check if a cell values is between two values or checking for the range of numbers or multiple values in cells, at this time, we need to use AND or OR logical function in combination with the logical operator and IF function. “How do I separate numbers coming before a text and in the end of text using excel Formula. DATE Formula in Excel. Excel. Microsoft Excel has a special function to convert a string to number - the VALUE function. This example used IF and COUNTIF functions to check for a value in a Cell and return required value. Date data type range from 1/1/100 to 31/12/9999. Even simple formulas such as =5+2. Join our mailing list (it's free!) Excel Boot Camp. Get smarter in just five minutes! The Excel ISNUMBER function is categorized under Information functions. I need to return ONLY formula with the number of months in full. Excel Formula Training. Thanks so much for. | Privacy Policy | Term Of Service | RSS, Return Value If Cell Contains Certain Value, How to Display Axis Label in Millions (M) or Thousand (K) in Excel, How to Split Multiple Lines in One Cell into Separate Rows or Columns in Excel, How to Sum Numbers by Formula if Cells Are Not Equal to Certain Value in Excel, How to Sum Numbers by Formula if Cells Are Equal to A Certain Value in Excel, How to Count Row That Contain Specific Value in Excel, How to Sum if Cell Contains Text in Another Column, How to Count Occurrences in Entire Workbook in Excel, How to Count Numbers Nth Digit Equals to Specific Number in Excel, How to Count Numbers that begin with Specific Value in Excel, How to Count Number by Range with COUNTIFS Function in Excel, How to Count Matches between Two Columns in Excel. It checks if Excel contains a number in a cell or not. Return smallest number with criteria This tutorial shows how to get the smallest number from a list with a specific criteria using the SMALL and IF functions . Shortcuts. Following is the Excel formula to find If Cell Contains Specific Text Then Return Value. So check only "XG" in all cells and return "Yes" if found and return "No" If not. Formula 1. IF Pros IF has been around forever and many Excel users are already comfortable using it, which will make it easy to adopt for the use of returning references. ISNUMBER in excel is a logical function in excel which we use to find out that if the target or the cell being referred to is blank or not, the method of using this formula is =ISNUMBER ( Reference Cell) the argument reference cell is the cell which we want to check or identify, if we have for example =ISNUMBER (T1XT) the output is false as the argument does not only has numbers in it. You can use a formula based on the IF function, the ISNUMBER function and the SEARCH function to achieve the result of return a value if Cell contains a specific value. Since we have supplied all the possible numbers to the array, the resulting arrays also should contain the same numbers. To avoid multiple helper columns, let’s combine the formula in a single cell. So now we got a numerical position, now let’s find the total number of characters in the cell. I'm trying to normalize data by reading a custom number format. Use COUNTBLANK to count empty or blank cells in a range in Excel. I have a cell (say Sheet2 A1) that references a cell in another sheet (same workbook), something like: =Sheet1!C4 It pulls back the text from that cell, the heading for a table, say "Size". = MATCH (C6,4:4,0) The formula uses the Excel MATCH function to return the column number for a specific value ("Cereal") that is captured in row 4. Cell B3 correctly returns the column letter F. Cell B4 contains the value I wish to find in column F, for example 'H' What formula is required in cell B5 to return the row number of the contents of cell B4 (value 'H') in column F? Extract text from a cell in Excel. This post will guide you how to return a value if a cell contains a certain number or text string in Excel. So, like this, we can get the numbers from the right-hand side when we have a common letter before the number starts in the string value. We can use the ROWS function to return the count of the number of rows in the source range or table. Tom September 24, 2018 at 8:17 amLog in to Reply. In this article, we will show you the 3 ways to extract numbers from a string in excel. Now we have total characters and positions of underscore before the numerical value. We then used some formulas to compare the information and match the number to the name. It is aligned on the right side of the cell to indicate it is now a value that can be used in calculations. Method 2:Extract number only from text strings with VBA code . Formula to Determine If Cell Contains Specific Text. This ROW formula: ROW(\$1:\$12) returns an array of values from 1 to 12 like this {1;2;3;4;5;6;7;8;9;10;11;12}. Most of the time you will pass references of a cell or a formula as values to check if it's a number or not. If you don’t know how Excel’s VLOOKUP formula works, then give it a revision (from the image below) 2) Compare two columns and return value from the third column (using INDEX and MATCH functions) We shall work again with the same data but this time we shall use this formula in cell E2: =IFERROR(INDEX(B2:B16, MATCH(D2,A2:A16,0)), "") Note the INT function rounds down, so negative numbers become more negative. The PRODUCT function is a built-in function in Excel that is categorized as a Math/Trig Function. Press Enter. Enter a number in cell A1. This post will guide you how to return a value if a cell contains a certain number or text string in Excel. Method 2: Use the “CELL” formula to return the number format code. We welcome your comments and questions about this lesson. EXCEL. Between formula in Excel for Numbers. Best and Clear Explanation! OPTION 1: Using a combination of MIN, MAX & AND function. Method 1: Extract number only from text strings with formula. You can see other characteristics in the formula bar, all alphanumeric have quotes in front of them, while no quotes in a number. Excel formula to calculate hours worked and overtime [with template] Excel Formula to Add Percentage Markup [with Calculator] How to find the 5 most frequent numbers in Excel (3 Formulas) How to find text in an Excel range and return cell reference (3 ways) How to reference cell in another Excel sheet based on cell value! As a financial analyst All formulas do the same thing (addition, multiplication, subtractiondivision & sum) I checked and made sure that all cells were formatted as currency or accounting and neither worked. The formula uses the Excel MATCH function to return the column number for a specific value ("Cereal") that is captured in row 4. The answer is simple: The CELL() formula returns Beginner. Cell B2 correctly returns the column number 6. is there anyway this formula can be revised to return like for example: found word "wow" and return the word into the cell instead of TRUE/FALSE? For example, the city with its pin code below is the sample of the same. Tweet... Our Comment Policy. The number 30 appears in cell B3. Formulas are equations that can perform calculations, return information, manipulate the contents of other cells, test conditions, and more. LEFT: Returns the specified number of characters from the start of a text string. Ben November 3, 2018 at 6:32 amLog in to Reply. The ISNUMBER syntax requires just one argument: =ISNUMBER (value) Where value is the value you want to test. thank you excel excel-formula match string-matching formulas example: Number Return result of the Month in full 1 January 2 February 12 December 7 July Thank you!! Return smallest number with criteria with SMALL and IF functions. Mastering the basic Excel formulas is critical for beginners to become highly proficient in financial analysis Financial Analyst Job Description The financial analyst job description below gives a typical example of all the skills, education, and experience required to be hired for an analyst job at a bank, institution, or corporation. The function accepts both a text string enclosed in quotation marks and a reference to a cell containing the text to be converted. 2. Method 3: Extract number only from text string with Kutools for Excel . Return Whole Number From Formula. We will use the LEFT & VALUE functions to make a formula that will convert text into numbers. For custom Excel number formats, the CELL function may return other values, and the following tips will help you to interpret them: ... How to use the CELL function in Excel - formula examples. We don't welcome spam. Excel functions, formula, charts, formatting creating excel dashboard & others As an example, let’s assume there is an array “B2:F7” then in the COLUMNS function (=COLUMNS (B2:F7)) will return … Usually, it is represented by a cell reference, but you can also supply a real value or nest another function inside ISNUMBER to check the result. You probably wonder, why it’s only the second of our four methods and why do we still need three other methods? You can apply this formula, please enter it into a cell where you need: =MATCH(E1,A1:A10,0), and then press Enter key to get the row number you need, see screenshot: Notes: 1. numbers only. Between formula in Excel for Numbers. The Excel INT function returns the integer part of a decimal number by rounding down to the integer. Then MIN function in excel returns the smallest number among two, so the formula reads below. We have seen from the right side extraction, but this is not the case with all the scenarios, so now we will see how to extract numbers from any position of the string in excel. Check if cells contain a number. eval(ez_write_tag([[300,250],'excelhow_net-medrectangle-3','ezslot_1',132,'0','0'])); You will see that the text string “learning excel” will be returned in the Cell C1. Active 1 year, 2 months ago. In this example the formula will return the value captured in cell C7 (1,400) given it contains the largest number in the selected range (C5:C9). At the moment, I'm using a simple COUNTIF formula to return the number of times each record is duplicated (based on a concatenated value). Formula will be: Within Text is in which text we need to find the mentioned text, so select cell reference. =IF (COUNTIF (A1,"*Specific Text*"),"Yes","No") We just need to know the number of texts to extract. This is one of Excel's most powerful features! Let us see an example to understand it. 2. Points 19,050 Trophies 1 Posts 6,316. So apply the FIND function in excel. Basic Excel Formulas Guide. See below the sample workbooks received for this request. COUNTIF The Microsoft Excel PRODUCT function multiplies the numbers and returns the product. In this case it would return 1. Enter a decimal number (0.2) in cell B1 and apply a Percentage format. In combination with other Excel functions, it is capable of much more. and get updates whenever new lessons are added! List of Shortcuts. Shortcut Training App. I have this array formula: ={SUM(IF(ISERROR(SEARCH(Source!K2:L13,F2)),0,1))} The formula simply looks for all possible matches of words from K2:L3 in a phrase in F2. Again, users of Excel 2019 and earlier would need to either wrap the formula in another formula that uses the reference returned by IF, like SUM (shown below), or define a name. Login details for this Free course will be emailed to you, This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. Return smallest number with criteria using SMALL and IF functions. ;) Below is the formula to extract the numbers from any position of the string. Read the entire article to know this technique. The day of month’s last day is always the total number of days in a month. =MIN (SEARCH ({0,1,2,3,4,5,6,7,8,9},A2&”0123456789″)) So now we got a numerical position, now let’s find the total number of characters in the cell. All formulas return 0 IN Excel 2003 every formula is returning the answer 0. EXCEL FORMULA 1. This will return the total number of characters in … The cell formula does that. OPTION 1: Using a combination of MIN, MAX & AND function. Apply the LEN function in excel to get the total length of the text value. How do I check if a Cell contains specific text and then return another specific text in another cell with a formula in Excel. When you select a cell, Excel shows the value or formula of the cell in the formula bar. For example, if the given value is a text, date, or time, it will return FALSE. As a worksheet function, the PRODUCT function can be entered as part of a formula in a cell of a worksheet. How do I VLOOKUP between two dates and return corresponding value with lookup formula in Excel. To edit a formula, click in the formula bar and change the formula. Very useful excel formula to return value in another cell if cell contains some text, value or number. The round down formula in Excel requires you to reference a number and then specify how many units (decimal places) you want to round it to. Now LEN – Position of the Numerical value will return the number of characters required from the right side, so apply the formula to get the number of characters. BottomRow = Data!A10:J10. Let me show you how I get these row numbers. Example: 40 hits = 1 point Thanks in advance for the formula info. Hard Coded Cell Reference. In the example below, you have the start of the range in Column A, end of the range in Column B and the value to be evaluated in Column C. You need to check whether the number entered in Column C is in between the numbers in Column A & Column B using a creatively formulated BETWEEN formula in Excel … Viewed 14k times 4. In this case, we know we have to extract Zip code from the right-hand side of the string. You probably wonder, why it’s only the second of our four methods and why do we still need three other methods? If the data is a number or is a formula that returns a number as output, a value of TRUE is returned by the function — the example in row 1 in the image above. You can search for a text and return the required … Just like this: Type this formula into the formula box of cell C1, and then press Enter key in your keyboard. 2. For example 125EvenueStreet and LoveYou3000 etc.” To extracting text, we use RIGHT, LEFT, MID and other text functions. We just need to know the number of texts to extract. CFA® And Chartered Financial Analyst® Are Registered Trademarks Owned By CFA Institute.Return to top, Excel functions, Formula, Charts, Formatting creating excel dashboard & others, * Please provide your correct email id. And Here we will do the same first. Super Moderator. 14 years ago I have a situation in which several (hundreds) of sheets in the same workbook require the same formula to be enterred. Excel has a built-in formula to return the number format code in Excel. And Here we will do the same first. Extract text from a cell in Excel. EXCEL. It returns the sum of how many words where found in F2. .free_excel_div{background:#d9d9d9;font-size:16px;border-radius:7px;position:relative;margin:30px;padding:25px 25px 25px 45px}.free_excel_div:before{content:"";background:url(https://www.wallstreetmojo.com/assets/excel_icon.png) center center no-repeat #207245;width:70px;height:70px;position:absolute;top:50%;margin-top:-35px;left:-35px;border:5px solid #fff;border-radius:50%}. This will eliminate all the supporting columns and reduce the time drastically. In our example, we want to check if texts from column B contain any number and return a result in column C. The formula looks like: =COUNT(FIND({0,1,2,3,4,5,6,7,8,9}, B3)) > 0 . This obscure technique is sometimes called "dereferencing", because it stops INDEX from handling results as cell references, and subsequently dropping all but the first item in the array. First, we need to identify the position of the underscore character. Formulas are the key to getting things done in Excel. Our readers get a lot of value out of the comments and answers on our lessons and spam hurts that experience. In the above example, we have the city name and Zip code together. Dynamic Formula. This can be done by using the FIND method. The VALUE function can even recognize a number surrounded by some "extra" characters - it's what none of the previous … Text:It is the text string from which you need to extract the characters. Points 11 Posts 3. In the above formula, E1 is the value that we want to look for its row number, and A1:A10 is the range of cells where we want to search. Select OK to complete the function and return to the worksheet. But one of the problems is we don’t know exactly how many digits we need from the right-hand side of the string. Excel Formula to return column number of a cell referenced in another cell Hi, I know this should be simple but just can't seem to get the formula quite right and google search isn't helping. When we get the data, it follows a certain pattern, and having all the numbers at the end of the string is one of the patterns. This is true whether I enter the formula or use the sum sign on the toolbar. While working or calculating dates in excel one should know the Date function. How do I check if a Cell contains specific text and then return another specific text in another cell with a formula in Excel. Now we need to identify totally how many characters we have in the entire text. The annualized rate of return formula is equal to Current value upon original value raise to the power one divided by number of years, the whole component is then subtracted by one. Excel formula for this Criteria is: = IF (A1="My Text To Check", "My Text To Return", "NOT My Text") This formula will check If Cell Contains Text Then Return Value. Excel allows a user to check if a cell contains a number by using ... while in column C we want to return TRUE if the cell contains number and FALSE if not. Convert string to number in Excel. This has been a guide to Extract Number From String in Excel. If the data is a number or is a formula that returns a number as output, a value of TRUE is returned by the function — the example in row 1 in the image above. Use COUNTBLANK to count empty or blank cells in a range in Excel. You see from the following image that text “Apple” is repeating itself 3 times in the range A1:A11. I have used this formula in the cell D6: {=… The ISNUMBER function's job is to determine if the data in a certain cell is a number or not. It can be used as a worksheet function (WS) in Excel. Instructions in this article apply to Excel for Microsoft 365, Excel 2019, 2016, 2013, and 2010. What I need to do is return the column … In the previous example, we have found a special character position, but here we don’t have that luxury. In this If formula, we have three parameters. Let’s you want to get total days in a month for the current month and you want this formula to return it automatically every time month changes. Jan 5th 2008 #2; Hi. Hi. 2. This topic provides an introduction to formulas and functions in Excel. This formula uses the Excel LARGE function, with number one (1) as the criteria, to return the highest number from a selected range. Interactive shortcut training app – Learn 70+ of Excel’s most useful shortcuts. Formulas are equations that can perform calculations, return information, manipulate the contents of other cells, test conditions, and more. Splitting single-cell values into multiple cells, collating multiple cell values into one, is the part of data manipulation. Here we discuss the top 3 ways of extracting the numbers from strings in excel along with practical examples and a downloadable excel template. Use concatenation to combine text and numbers in Excel formulas. To ensure that you are working with the correct cell addresses referenced in the formulas, you can press F2 on the keyboard. The Excel SEARCH function returns the number of the starting location of a substring in a text string.The syntax of the SEARCH function is as below:= SEARCH (find_text, within_text,[start_num])… Excel ISNUMBER function Very helpful. Email Address * First Name * I want to learn more about: Microsoft Excel . Rate of Return Formula – Example #3. How to lookup a value if the date fall between two dates in Excel. pogo202; Jan 5th 2008; pogo202. num_chars:It specifies the number of characters from the left which you want to extract. The SUMPRODUCT formula in cell C18 looks like this: In this accelerated training, you'll learn how to use formulas to manipulate text, work with dates and times, lookup values with VLOOKUP and INDEX & MATCH, count and sum with criteria, dynamically rank … METHOD 1. Alphanumeric is a number in appearance, but Excel acknowledges it as text. For example 125EvenueStreet and LoveYou3000 etc.” To extracting text, we use RIGHT, LEFT, MID and other text functions. With the inbuilt info_types, the CELL function can return total 12 different parameters about a cell. How to achieve it. You can simply use the MONTH Formula in Excel given below: = MONTH (B3) I have two separate spreadsheets. It will return TRUE if the value is a number and if not, a FALSE value. Instead of having so many helper columns, we can apply the formula in a single cell itself. Thanks. This ROW formula: ROW(A1) returns the row number of a reference. Very useful formula to check cell contains criteria. The formula will only return the column number of … So, we have got positions of underscore character for each cell. Below we have explained the different ways of extracting the numbers from strings in excel. Now apply the RIGHT function in excel to get only the numerical part from the string. Join our mailing list. The formula will only return the column number of the first occurrence of the specific value. To force the formula to return a number rather than a numeric string, nest it into the VALUE function: =VALUE(RIGHT(A2,LEN(A2) - MIN(SEARCH({0,1,2,3,4,5,6,7,8,9}, A2&"0123456789")) +1)) How the position of the 1st digit is calculated Excel spreadsheet formulas usually work with numeric data; you can take advantage of data validation to specify the type of data that should be accepted by a cell i.e. List of 200+ Excel shortcuts. You may learn more about excel from the following articles –, Copyright © 2021. Select cell A3 in the spreadsheet. The ISNUMBER function's job is to determine if the data in a certain cell is a number or not. Growing list of Excel Formula examples (and detailed descriptions) for common Excel tasks. For the SEARCH function in excel, we have supplied all the possible starting numbers of numbers, so the formula looks for the position of the numerical value. Instructions in this article apply to Excel for Microsoft 365, Excel 2019, 2016, 2013, and 2010. does excel have a formula to calculate the return the day number for a condition such as the 1st Mon hi I'm trying to get the day number (1-31,etc) on a spreadsheet corresponding to … In this example, we need to find the position of the underscore, so enter underscore in double-quotes. So, here is the formula for the row_num argument in INDEX: =SEQUENCE(ROWS(tblData)) COUNTA () evaluates all values, including text and even an empty string, "". So Below the formula below will find the numerical position. One of the common things before the numerical value starts is the underscore (_) character. Edit a Formula. Excel formula to return a cell's custom format. There are multiple numbers of data types in excel and Date is one of them. To increase the number in cell A1 by 20%, multiply the number by 1.2 (1+0.2). You should notice that the above formula returns a relative reference, not an absolute reference to … Its length is 8 bytes. Because the value that you want to return is a number, you can use a simple SUMPRODUCT () formula to look for the Name “James Atkinson” and the Product “Milk Pack” to return the Qty. Excel IF formula for range of numbers. Don’t turn off your computer by looking at the formula; I will decode this for you. The formula for same can be written as:-In this formula, any gain made is included in formula. Assuming that you want to check if a given Cell such as B1 contains a text string “excel”, if True, returns another text string “learning excel” in Cell C1. Hard coded formula. “How do I separate numbers coming before a text and in the end of text using excel Formula. The answer should be 3! The criteria are “Name” and “Product,” and you want them to return a “Qty” value in cell C18. Here I introduce some ways for you to extract only numbers quickly and easily in Excel. For this, we need to employ various functions of excel. With the help of the text function in excel “Left, MID, and Right,” we are able to extract part of the selected text value or string value. Excel formula to return the row number I have named data ranges as follows: AllData = Data!A1:J10. And if you want to check the range of cells B1:B4, you need to drag the AutoFill Handle down to other cells to apply this formula. Start Your Free Excel Course. This topic provides an introduction to formulas and functions in Excel. CFA Institute Does Not Endorse, Promote, Or Warrant The Accuracy Or Quality Of WallStreetMojo. Ask Question Asked 6 years, 10 months ago. james April 6, 2019 at 4:23 pmLog in to Reply. What is the text we need to find is the Find Text argument? To supply the number of characters needed for the RIGHT function, we need to minus the Total Characters with Underscore Position. The last argument is not required, so leave it as of now. Similarly, this formula: = SUM(INDEX( data,N(IF(1,{1,3,5})))) correctly returns 60, the sum of 10, 20, and 30. How the Excel round down formula works. In order to make the formula dynamic, we can use other supporting functions like “Find and LEN.” However, the extraction of only numbers with the combination of alpha-numeric values requires an advanced level of formula knowledge. Excel functions, formula, charts, formatting creating excel dashboard & others. Cell B1 contains the value I wish to find in the bottom row, for example 'E' Cell B2 correctly returns the column number 6. Use concatenation to combine text and numbers in Excel formulas. This post will guide you how to do a VLOOKUP between two dates and return corresponding value in Excel. Then MIN function in excel returns the smallest number among two, so the formula reads below. Operator Precedence If value is numeric, the function returns TRUE. Formula needed to return a set number if time is before a certain period Hello everyone I have a timestamp from a thumbprint reader that I would like to tell me if staff have entered the work place by our in-house taxi service or not, the taxi only travels at 8am so if staff arrive at work before 10am then we can deduct money from their salary slip. Assume we have the same data, but this time we don’t have any special character before the numerical value. I discovered the reason for this is that the value of 31 is not in fact a whole number, the number being returned is actually 31.3959606499993 Ho do I get Excel … Here is the simple formula to check if cell contains specific text or not. Use the formula: =ISNUMBER(FIND(D4,C4)) As you can see First employee doesn't belong to "XG" so the formula returns "No" using the FIND function. The characteristics, alphanumeric by default is left aligned while numbers are by default is right aligned. Although flexible, the formula has one flaw--it can't handle a blank cell. Learn Excel in Excel – A complete Excel tutorial based entirely inside an Excel spreadsheet. We can return the row number of that text in the range. Excel automatically recalculates the value of cell A3. Excel has a built-in formula to return the number format code in Excel. 1. Here the Excel formula to Return Value If Cell Contains Specific Text : =IFERROR(IF(SEARCH(B2,A2,1)>0,A2,0),"") ... Parameter 3: 1=Character number which you want start you search in within_text from the left. It works this way because this is an array formula entered with CTRL+SHIFT+ENTER. Jan 5th 2008 #1; I cannot get the cell to show the following; if I have 40 units it would need to equal one point. Video: Return Value If Cell Contains Certain Value, © Copyright 2017 - 2020 Excel How All Rights Reserved. This will return the total number of characters in the supplied cell value. From 31-Mar-2017, it returns 31 which is the totals number of days in the March month. Return column number of … The cell formula does that. One with a list of names and numbers and one with a list of some of the names on the first list.I want to tell it to match the names on the spreadsheets and if they match, then put the number on page one next to the name on page two.Is this possible?For this problem we received some sample workbooks. Report Content ; Wigi.
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Mise En Place is located at: 1639 11th Street, Suite 107, Santa Monica, CA 90404 | 6,899 | 30,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | longest | en | 0.786113 |
industrialenergyconservation.com | 1,369,184,571,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700984410/warc/CC-MAIN-20130516104304-00056-ip-10-60-113-184.ec2.internal.warc.gz | 139,262,075 | 4,164 | ## Conversions
Wt. of Water = 8.34 lb./gal. or 62.37 lb./cu. Ft. @ 60oF
1 lb./sq. in. = 2.31 ft. water (1.0 sp. Gr.)
1 ft. water = .433 psi
Atmospheric Press = 14.7 psi or 33.9 ft. fresh water @ sea level
1 inch of mercury (Hg) vacuum = 1.133 ft. of water
Dynamic Suction Lift (ft.) = vacuum in inches of Hg. divided (0.883 x sp. gr.)
## Cost of Pumping Water
Cost per 1000 gallons pumped:
.189x Power Cost per kilowatt-hour x head in feet
Pump Eff. X Motor Eff. x 60
## Formulas
Head (ft.) = psi x 2.31 = In. Hg. BHP (Centrifugal) = CPM x head (ft.) x sp. gr. sp. gr. sp. gr. X .88 3960 x pump eff.
PSI = Head (ft.) x sp. gr. = .49 x in. Hg. BHP (Positive Disp.) = GPM x PSI 2.31 1715 x pump eff.
Velocity (ft./sec.) = 0.4085 x gpm Specific Gravity = Wt. of Liquid [i.d. (in.) of pipe]2 Wt. of Water
## Net Positive Suction Head (NPSH)
NPSHR = head required at the eye of the pump’s impeller to prevent cavitation, a function of the pump design.
NPSHA = head available from the suction side of the system, must always equal or exceed the required (NPSHR ) to prevent cavitation.
NPSHA FACTORS:
POSITIVE FACTORS NEGATIVE FACTORS
(1) absolute pressure on liquid (1) vapor pressure of the liquid, absolute
(2) static suction head (2) friction & entrance losses
(3) static suction lift
AFFINITY LAWS: (Effect on centrifugal pumps with limited change of speed (RPM) or impeller diameter.
(1) Capacity (Q) varies directly as the speed (N) or impeller diameter (D).
(2) Head (H) varies as the square of the speed or impeller diameter.
(3) BHP varies as the cube of the speed or impeller diameter.
LAW 1: Q1 = N1 or D1 LAW 2: H1 = ( N1 ) 2 or ( D1 ) 2 LAW 2: BHP1 = ( N1 ) 3 or ( D1 ) 3 Q2 N2 D2 H2 N2 D2 BHP2 N2 D2
Cubic feet x 7.48025 = U.S. Gallons 1 Gallon water = 8.3453 lbs.
U.S. Gallon x .1336 = 1.0 Cubic Feet 1.0 Litre = .2642 Gallons | 669 | 2,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-20 | latest | en | 0.72832 |
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# MidtermPracticeTestAnswers - M idterm Practice test with...
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Midterm Practice test with answers 1. A triangle wave A. contains odd partials only B. contains even partials only C. contains an error D. contains both odd and even partials 2. A sawtooth wave has a period of 10 milliseconds. Its frequency is 10msec = .01 sec = 1/100 sec inverse of 1/100 = 100Hz 3. What is the decimal equivalent of this binary number: 0010 1100 32 + 8 + 4 = 44 4. How many bits in 7 bytes? 7 * 8 bits/byte = 56 5. The period of a 1000 Hz wave is 1/1000 cycles per second = .001 sec A. .1 millisec B. .1 Hz C. 10 msec. D. .001 sec.
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6. If there are 30 beats per minute and 8 notes to the beat, how long does each note last? 30 bpm /60 seconds per minute = 1/2 beat per second so period of a beat is 2 seconds per beat. 2 divided by 8 notes per beat gives 1/4 or .24 seconds per note
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation2.11 Speed and Distance Word Problems
Steps:
1. Read the problem carefully and note down what is given and what is required.
2. Select a letter letter or variable x or y or z to represent the unknown quantity asked for.
3. Represent the word statements of the problems in the symbolic language step by step.
4. Look for quantities which are equal as per conditions given and form an equation.
5. Solve the equation.
6. Verify the result for making sure that your answer satisfies the requirements of the problems.
Examples:
1. John walked to school at an average speed of 3 miles an hour and jogged back along the same route at 5 miles an hour. If his total traveling time was 1 hour, what was the total number of miles in the round trip?
Solution:
Total distance = Total rate * total time
Let 'x' be the distance from home to school.
Time Rate Distance To School x/3 3 miles/hr x Back x/5 5 miles/hr x Total 5 miles/hr 2x
x/3 + x/5 = 1
5x + 3x = 15
8x = 15
x = 15/8 miles
Total distance = 2x = 2 (15/8) = 3 3/4
2. In a stream the current flows at the rate 4 miles/hr. For a boat the time taken to cover a certain distance upstream is 5 times the time it takes to cover the same distance downstream. Find the speed of the boat in still water.
Solution:
Let the speed of the boat in still water = x miles/hr.
Speed of the boat downstream will be greater than its speed in still water, due to favorable current of water.
\ Speed of the boat downstream = Speed of the boat + Speed of water current.
= (x+4) miles/hr.
Speed of the boat upstream will be less than its speed in still water, due to opposing current of water.
\Speed of the boat upstream = Speed of the boat - Speed of the current.
= (x-40)miles/hr.
Let the time taken to cover the distance downstream = t hrs.
\Time taken to cover distance upstream = 5t hr.
Now, distance traveled downstream = Speed * Time
= (x+4)t miles.
Distance traveled upstream = (x-4)5t miles.
Since distance traveled both ways are same.
(x+4)t = (x-4)5t
Dividing both sides by 't'(¹0).
(x+4) = 5(x-4)
x+4 = 5x-20
Transposing 5x and 4
5x - x = 20 + 4
4x = 24
x = 24/4 = 6
Directions: Solve the following word problems. Write and solve at least four more word problems involving speed and distance.
Q 1: Jessica drove her car at a certain speed for the first 4 hours and increased its speed by 10 miles/hr, for the next two hours. If the total distance traveled by her was 500 miles, find the speeds at which Jessica drove her car at different times.50 and 60 miles/hr respectively80 and 90 miles/hr respectively60 and 80 miles/hr respectively Q 2: In a stream whose current flows at the speed of 3 miles/hr, a man rowed upstream and returned to his starting point in 6.25 hr. If the man rows at a speed of 5 km/hr, in still water, how long did he row upstream.3 hours6 hours5 hours Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only! | 842 | 3,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-04 | latest | en | 0.924104 |
https://www.physicsforums.com/threads/firing-cannons-projectile-motion.750859/ | 1,531,838,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589726.60/warc/CC-MAIN-20180717125344-20180717145344-00144.warc.gz | 947,671,155 | 18,067 | # Homework Help: Firing cannons projectile motion
1. Apr 26, 2014
### Psychros
1. The problem statement, all variables and given/known data
A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)
2. Relevant equations
Time= 0=v0T-½gT2 T<0
t= (vOy+√(vOy2-2gy))/g
distance/range x=x0+V0xt
x/y velocity- v0x=Vcos(θ); v0y=vsin(θ)
3. The attempt at a solution
This is a practice problem (with different values than original), and the provided answer is 32.9m/s
i set the origin at the cannon
Since the velocity magnitude = √(Vcos(θ)2+Vsin(θ)2)
I tried variations along the line of V0cos(30)2+V0sin(30)2= 1.3 V02
in order to split the velocity into xy components.
Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.
Any and all help very appreciated, thank you.
2. Apr 26, 2014
### SteamKing
Staff Emeritus
I'm not sure what you are trying to do here. The angle at which the projectile strikes the ground may not be equal to the initial firing angle, i.e. 30 degrees.
The best way to approach this problem is to find the maximum altitude which the projectile reaches after it is fired. At this point, the vertical component of the velocity is zero. After reaching the maximum altitude, the projectile falls under the force of gravity. Knowing the magnitude of the final velocity, you should be able to determine the duration of the fall and the resulting vertical velocity component. Remember, since there is no air resistance, the horizontal component of the velocity remains unchanged during the entire flight of the projectile.
3. Apr 26, 2014
### Psychros
That is brilliant, thank you. However, Without knowing distance (x), time, or hard value for velocity (since it's expressed as a relation to the initial velocity) i'm not sure how to go about this.
4. Apr 26, 2014
### SammyS
Staff Emeritus
Try using conservation of energy.
5. Apr 26, 2014
### Psychros
Hmmm I don't think we've learned that yet..
6. Apr 26, 2014
### SammyS
Staff Emeritus
In that case,
What can you conclude about the x component of the velocity?
What kinematic equations do you know regarding the vertical component of the velocity ?
7. Apr 26, 2014
### Psychros
X component will be the same through it's entirety.
y=y0+v0y(t)+½at2
or
Vy=V0y+gt
Vy=1.3V0y-9.81t
But Theoretically if we were starting at the maximum, V0y would = 0,
Vy/-9.81=t?
I'm getting stuck down either avenue, I do think i'm missing something intuitive here..
8. Apr 26, 2014
### SammyS
Staff Emeritus
vy ≠ 1.3 v0y . I'm assuming that you mean vy to be the y component of the final velocity.
How is speed obtained from components of velocity ?
Also, there's another kinematic equation. One that doesn't involve time, t.
Edited:
[STRIKE](vy)2 = (v0y)2 - g(Δy)[/STRIKE]
(vy)2 = (v0y)2 - 2g(Δy)
Last edited: Apr 27, 2014
9. Apr 27, 2014
### Psychros
Do you mean speed as in magnitude of velocity?
Also, should (vy)2 = (v0y)2 - g(Δy)
have a -2g(Δy)?
I apologize for being slow, but i'm not sure how to use this either since I don't know the change in height from the maximum
10. Apr 27, 2014
### SammyS
Staff Emeritus
Right. There should be a 2 in there.
Then, ...
How is speed obtained from components of velocity ?
11. Apr 27, 2014
### Psychros
Since speed is the magnitude of velocity,
by components of velocity I imagine you mean Vx and Vy,
or √(Vcos(θ)2+Vsin(θ)2)?
12. Apr 27, 2014
### SammyS
Staff Emeritus
Yes. Of course, sin2(θ) + cos2(θ) = 1
Anyway, $\ v_0=\sqrt{(v_{0x})^2+(v_{0y})^2}\ ,\$ so that $\ (v_0)^2=(v_{0x})^2+(v_{0y})^2\ .\$
And in general $\ v^2=(v_{x})^2+(v_{y})^2\ .\$ Right?
Now, what do you know regarding the x component of velocity for this projectile?
13. Apr 27, 2014
### Psychros
sorry, that's what i meant when i put √(V0cos(θ)2+V0sin(θ)2), that it's = V0.
I know that it doesn't change, but i'm not sure how to find its' magnitude in this situation.
14. Apr 27, 2014
### SammyS
Staff Emeritus
So then $v_{0x}=v_x\$ .
Regarding your kinematic equation with the $(v_{y})^2\,$, what is it missing for it to have v2 and v02 in it ? | 1,322 | 4,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-30 | latest | en | 0.90706 |
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