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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://metanumbers.com/16276 | 1,600,938,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00549.warc.gz | 444,541,802 | 7,449 | ## 16276
16,276 (sixteen thousand two hundred seventy-six) is an even five-digits composite number following 16275 and preceding 16277. In scientific notation, it is written as 1.6276 × 104. The sum of its digits is 22. It has a total of 4 prime factors and 12 positive divisors. There are 7,488 positive integers (up to 16276) that are relatively prime to 16276.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 16 thousand 276 sixteen thousand two hundred seventy-six
## Notation
Scientific notation 1.6276 × 104 16.276 × 103
## Prime Factorization of 16276
Prime Factorization 22 × 13 × 313
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 8138 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 16,276 is 22 × 13 × 313. Since it has a total of 4 prime factors, 16,276 is a composite number.
## Divisors of 16276
1, 2, 4, 13, 26, 52, 313, 626, 1252, 4069, 8138, 16276
12 divisors
Even divisors 8 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 30772 Sum of all the positive divisors of n s(n) 14496 Sum of the proper positive divisors of n A(n) 2564.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 127.577 Returns the nth root of the product of n divisors H(n) 6.34707 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 16,276 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 16,276) is 30,772, the average is 25,64.,333.
## Other Arithmetic Functions (n = 16276)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7488 Total number of positive integers not greater than n that are coprime to n λ(n) 312 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1888 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 7,488 positive integers (less than 16,276) that are coprime with 16,276. And there are approximately 1,888 prime numbers less than or equal to 16,276.
## Divisibility of 16276
m n mod m 2 3 4 5 6 7 8 9 0 1 0 1 4 1 4 4
The number 16,276 is divisible by 2 and 4.
• Deficient
• Polite
## Base conversion (16276)
Base System Value
2 Binary 11111110010100
3 Ternary 211022211
4 Quaternary 3332110
5 Quinary 1010101
6 Senary 203204
8 Octal 37624
10 Decimal 16276
12 Duodecimal 9504
20 Vigesimal 20dg
36 Base36 ck4
## Basic calculations (n = 16276)
### Multiplication
n×i
n×2 32552 48828 65104 81380
### Division
ni
n⁄2 8138 5425.33 4069 3255.2
### Exponentiation
ni
n2 264908176 4311645472576 70176341711646976 1142190137698766181376
### Nth Root
i√n
2√n 127.577 25.3425 11.295 6.9552
## 16276 as geometric shapes
### Circle
Diameter 32552 102265 8.32234e+08
### Sphere
Volume 1.80606e+13 3.32893e+09 102265
### Square
Length = n
Perimeter 65104 2.64908e+08 23017.7
### Cube
Length = n
Surface area 1.58945e+09 4.31165e+12 28190.9
### Equilateral Triangle
Length = n
Perimeter 48828 1.14709e+08 14095.4
### Triangular Pyramid
Length = n
Surface area 4.58834e+08 5.08132e+11 13289.3
## Cryptographic Hash Functions
md5 1ab4eabb60df171d0d442f0c7fb875a0 192bb6d28dae0d0f234b56943a85566465325449 195f4798a7ab88efbd42f12758a690db92f7b9ed7a8d59de9f72f7ab03af8029 88ae931c462c1ea24610dc5849b50a91cd4a1baac6ff6652f5197aa8c7e18110bb897e2908fa65d6c7df32a2e8d72fbd2c4593e7e111d9f2a00a80faa5e90ba4 22278c01937d6ac20575d59256adeebf68b3effa | 1,455 | 4,058 | {"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-2020-40 | longest | en | 0.800729 |
https://www.teacherspayteachers.com/Product/Cowboys-and-Cowgirls-Domino-Add-the-Room-2372044 | 1,547,767,492,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659417.14/warc/CC-MAIN-20190117224929-20190118010929-00218.warc.gz | 956,390,186 | 18,549 | Cowboys and Cowgirls Domino Add the Room
Subject
Resource Type
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Product Description
This resource provides fun and exciting western themed math centers for your little learners to practice addition facts to 10. This packet has 5 sets of Domino Add the Room cards. Each set has 10 cards. There is a recording page for each set of cards. The cards and recording sheets have cowboys and cowgirls graphics. The graphics include kids riding horses, kids wearing western boots, kids wearing western hats, kids with lariats, a kid sheriff, and horses.
Print, laminate, and cut out the equation cards. Make copies of the recording pages.
Hang the cards in places all around the room. Give each student a clipboard, pencil, and recording page. Learners search the room for cards with dominoes. When they find a card, they write the equation and sum represented by the number of dots on the domino in the correct space on the recording sheet.
For example, if a card has a domino with 5 dots on the left side and 2 dots on the right side, learners write 5 + 2 = 7 in the correct space on the recording sheet.
Cowboys and Cowgirls Domino Add the Room helps learners develop addition fluency and supports the Math Common Core Standards for Operations and Algebraic Thinking.
The graphics used in Cowboys and Cowgirls Domino Add the Room are from Scrappin Doodles, Whimsy Clips, My Cute Graphics, and First Grade Brain.
www.scrappindoodles.com
www.whimsyclips.com
www.mycutegraphics.com
Thank you for looking at Cowboys and Cowgirls Domino Add the Room. I hope you and your learners enjoy these activities.
Happy Trails!
Kamp Kindergarten
Please click on the links below to view other resources you may enjoy.
Cowboys and Cowgirls Dice Add the Room
Cowboys and Cowgirls Addition Math Centers
Cowboys and Cowgirls Add the Room
Cowboys and Cowgirls Subtract the Room
Cowboy Christmas Math Centers (Numerals to 20)
Total Pages
30 pages
N/A
Teaching Duration
1 Week
Report this Resource
\$5.00 | 471 | 2,061 | {"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-2019-04 | latest | en | 0.888117 |
https://reference.wolfram.com/language/ref/LyapunovSolve.html | 1,726,277,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00862.warc.gz | 436,992,012 | 14,829 | # LyapunovSolve
LyapunovSolve[a,c]
finds a solution of the matrix Lyapunov equation .
LyapunovSolve[a,b,c]
solves .
LyapunovSolve[{a,d},c]
solves .
LyapunovSolve[{a,d},{b,e},c]
solves .
# Details
• LyapunovSolve solves the continuous-time Lyapunov and Sylvester equations.
• LyapunovSolve works on both numerical and symbolic matrices.
# Examples
open allclose all
## Basic Examples(1)
Solve the Lyapunov equation :
## Scope(7)
Solve a Lyapunov equation:
Verify the solution:
Solve :
Solve for coefficient matrices with different dimensions:
Solve :
Solve :
Solve the Lyapunov equation with symbolic coefficients:
Obtain the symbolic solution of :
## Applications(7)
Test the stability of by checking if the solution of is positive definite for a negative definite :
As expected, the eigenvalues are in the left half-plane:
An unstable system:
Compute the controllability Gramian of a stable continuous-time system:
Compute the observability Gramian of a stable continuous-time system:
Compute the norm of an asymptotically stable continuous-time system:
Compute the feedback gains that place poles at desired locations:
Verify the solution:
For MIMO systems, the feedback gains are not unique:
Construct an observer for a StateSpaceModel:
First, choose an appropriate and such that the Lyapunov equation yields a nonsingular solution:
Then construct the observer as , , where is the observer state vector, is the output, is the input, and is the estimated state vector:
Compute the estimated state trajectories for a UnitStep input:
Compute the actual state trajectories for a UnitStep input:
Plot the actual and estimated states:
## Properties & Relations(5)
The equation , with a negative definite , yields a unique positive definite solution if and only if the eigenvalues of are in the closed left half-plane:
A stable system:
The definite integral is the solution to if is asymptotically stable:
Compute the infinite-horizon quadratic cost for the asymptotically stable system :
Compute using direct integration:
Solve the matrix equation :
LinearSolve gives the same solution:
Solve the Lyapunov equation using LinearSolve:
LyapunovSolve gives the same solution:
Wolfram Research (2010), LyapunovSolve, Wolfram Language function, https://reference.wolfram.com/language/ref/LyapunovSolve.html.
#### Text
Wolfram Research (2010), LyapunovSolve, Wolfram Language function, https://reference.wolfram.com/language/ref/LyapunovSolve.html.
#### CMS
Wolfram Language. 2010. "LyapunovSolve." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/LyapunovSolve.html.
#### APA
Wolfram Language. (2010). LyapunovSolve. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/LyapunovSolve.html
#### BibTeX
@misc{reference.wolfram_2024_lyapunovsolve, author="Wolfram Research", title="{LyapunovSolve}", year="2010", howpublished="\url{https://reference.wolfram.com/language/ref/LyapunovSolve.html}", note=[Accessed: 13-September-2024 ]}
#### BibLaTeX
@online{reference.wolfram_2024_lyapunovsolve, organization={Wolfram Research}, title={LyapunovSolve}, year={2010}, url={https://reference.wolfram.com/language/ref/LyapunovSolve.html}, note=[Accessed: 13-September-2024 ]} | 831 | 3,330 | {"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-2024-38 | latest | en | 0.790283 |
http://en.wikipedia.org/wiki/Lax_equivalence_theorem | 1,405,079,415,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776426734.39/warc/CC-MAIN-20140707234026-00064-ip-10-180-212-248.ec2.internal.warc.gz | 37,337,585 | 10,699 | # Lax equivalence theorem
In numerical analysis, the Lax equivalence theorem is the fundamental theorem in the analysis of finite difference methods for the numerical solution of partial differential equations. It states that for a consistent finite difference method for a well-posed linear initial value problem, the method is convergent if and only if it is stable.[1]
The importance of the theorem is that while convergence of the solution of the finite difference method to the solution of the partial differential equation is what is desired, it is ordinarily difficult to establish because the numerical method is defined by a recurrence relation while the differential equation involves a differentiable function. However, consistency—the requirement that the finite difference method approximate the correct partial differential equation—is straightforward to verify, and stability is typically much easier to show than convergence (and would be needed in any event to show that round-off error will not destroy the computation). Hence convergence is usually shown via the Lax equivalence theorem.
Stability in this context means that a matrix norm of the matrix used in the iteration is at most unity, called (practical) Lax-Richtmyer stability.[2] Often a von Neumann stability analysis is substituted for convenience, although von Neumann stability only implies Lax-Richtmyer stability in certain cases.
This theorem is due to Peter Lax. It is sometimes called the Lax–Richtmyer theorem, after Peter Lax and Robert D. Richtmyer.[3]
## References
1. ^ Strikwerda, John C. (1989), Finite Difference Schemes and Partial Differential Equations (1st ed.), Chapman & Hall, pp. 26, 222
2. ^ Smith, G. D. (1985), Numerical Solution of Partial Differential Equations: Finite Difference Methods, 3rd ed., Oxford University Press, pp. 67–68
3. ^ Lax, P. D.; Richtmyer, R. D. Survey of the stability of linear finite difference equations. Comm. Pure Appl. Math. 9 (1956), 267--293 MR 79204, doi:10.1002/cpa.3160090206 | 442 | 2,023 | {"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} | 3.453125 | 3 | CC-MAIN-2014-23 | longest | en | 0.880487 |
https://www.thestudentroom.co.uk/showthread.php?t=27323 | 1,537,633,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158450.41/warc/CC-MAIN-20180922142831-20180922163231-00417.warc.gz | 900,464,382 | 38,223 | You are Here: Home
# M2 Mock 2000 Edexcel Q... watch
1. This question is given on the M2 Mock 2000 Paper. Solution of whole paper would be nice but if not, could you answer this q?
1) A Smooth sphere is moving with speed 'U' in a straight line on a smooth horizontal plane. It strikes a fixed smooth vertical wall at right angles. The coefficient of restitution between the sphere and the wall in 0.5.
Find the fraction of the Kinetic Energy of the sphere that is lost as a result of the impact.
(5 Marks)
2. it will rebound off the wall with velocity: -eu = -u/2
initial K.E. = (1/2)mU^2
final K.E. = (1/2)m(-U/2)^2 = (1/8)mU^2
K.E. lost = (mU^2/2) - (mU^2/8) = (mU^2)/4
fraction of K.E. lost = (mU^2/4) / (mU^2/2) = (1/4) / (1/2) = 1/2
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 458 | 1,533 | {"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-2018-39 | latest | en | 0.882368 |
https://www.netshock.co.uk/category/six-sigma/ | 1,556,059,569,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578613888.70/warc/CC-MAIN-20190423214818-20190424000818-00269.warc.gz | 753,271,191 | 17,570 | ## Six Sigma Series: Wrapping our DMAIC up
Below is a model that I’ve put together to summarize the various pieces of each stage. I’ve intentionally not linked to any of the improvement and control elements. The improvement…
## 6 Sigma Series: calculating the correlation coefficient
Finding the relationship between input variables and the output from the process can be vital to implementing a successful Six Sigma strategy. In the below chart, we can say that…
## 6 Sigma Series: Making Decisions; Confidence Scores
Decision making from analysis in Six Sigma is the most important part of any project. It’s great that you’ve collected lots of tasty data but if you don’t take the…
## 6 Sigma Series: finding the important X’s and sampling data
Our next task is to reduce all the inputs that we’ve identified to just those few that impact variability the most. We can start to strip down the list of…
## 6 Sigma Series: Capability index & quantifying the capability of the process to meet specification
In Six Sigma, we use capability indices to directly compare the voice of the process to the voice of the customer. This helps us to quantify the capability of the…
## 6 Sigma Series: Calculating yield, defects and the Z score
Capability is the term used to define how well a process matches the voice of the customer. To find this out, we need to be able to calculate the defect…
## 6 Sigma Series: Analyzing to deliver maximum value with Six Sigma
To truly deliver value, we first need to understand what value is from our customers’ perspective. We call this definition of value the Voice Of the Customer (VOC) and we…
## 6 Sigma Series: Getting statistical with 6 Sigma: standard deviation and short term variance
Six Sigma deploys a number of statistical methods to understand current processes. As we have discussed previously, variation in output costs the business, both directly and indirectly so we need…
## 6 Sigma Series: Diagramming for definition in Six Sigma
To identify all the X’s (inputs) in our process that impact the Y’s (the critical output(s)) through brainstorming alone can be an impossible task. So, we try to add some…
## 6 Sigma Series: Using process mapping and value streams to identify problems
Identifying problems in Six Sigma starts with a very AS-IS detailed process map (as discussed here). For each step in our process maps, we should be adding: Operation time +… | 507 | 2,424 | {"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-2019-18 | latest | en | 0.876786 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=1418&CurriculumID=2&NQ=2&Num=1.12 | 1,720,777,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00830.warc.gz | 14,619,692 | 3,766 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
A numberline shows the numbers in order. Numbers to the right of zero are positive numbers. Numbers to the left are negative numbers. A numberline has arrows because the numbers keep going forever. You can use the numberline to practice the addition and subtraction. Method: Draw a number line with numbers. To find difference, we first go forward till we reach the bigger number and then backward till we reach the smaller number. To find sum, we first go till the first number and then move further from there to reach the next number. Directions: Answer the following questions. Also write at least ten examples with subtraction and addition using number line.
Q 1: Positive numbers are numbers on the right of zero on the number line.FalseTrue Question 2: This question is available to subscribers only! | 208 | 975 | {"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} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.852368 |
https://philoid.com/question/29159-if-angle-between-two-tangents-drawn-from-a-point-p-to-a-circle-of-radius-a-and-center-o-is-90-then | 1,723,783,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00216.warc.gz | 358,905,825 | 8,658 | ##### If angle between two tangents drawn from a point P to a circle of radius a and center O is 90°, then
True
Let us consider a circle with center O and tangents PT and PR and angle between them is 90° and radius of circle is a
To show :
Proof :
In OTP and ORP
TO = OR [ radii of same circle]
OP = OP [ common ]
TP = PR [ tangents through an external point to a circle are equal]
OTP ORP [ By Side Side Side Criterion ]
TPO = OPR [Corresponding parts of congruent triangles are equal ] [1]
Now, TPR = 90° [Given]
TPO + OPR = 90°
TPO + TPO = 90° [By 1]
TP0 = 45°
Now, OT TP [ As tangent at any point on the circle is perpendicular to the radius through point of contact]
OTP = 90°
So POT is a right-angled triangle
And we know that,
So,
[As OT is radius and equal to a]
Hence Proved !
5 | 236 | 809 | {"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} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.884862 |
http://openbooks.library.umass.edu/funee/part/appendices/ | 1,721,491,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00005.warc.gz | 26,542,099 | 15,846 | C = A + jB
C1 + C2 =
C2 – C2 =
C1 x C2 =
C1 / C2 = | 32 | 54 | {"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-2024-30 | latest | en | 0.827153 |
https://www.evan-moor.com/p/19140/take-it-to-your-seat-math-centers-fourth-grade | 1,544,498,813,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823550.42/warc/CC-MAIN-20181211015030-20181211040530-00332.warc.gz | 886,674,473 | 25,030 | # Take It to Your Seat: Math Centers, Grade 4 - Teacher Reproducibles, Print EMC 3074
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Description
EMC Number: 3074
Page Count: 160
EAN: 9781609637835
Help your fourth graders master key math skills, and provide differentiated math practice through fun, hands-on activities in these 12 centers. Each full-color center focuses on a skill from a math domain, such as numbers and counting, operations, measurement and data, geometry, and math vocabulary.
The easy-to-assemble centers are stored in pocket folders, making them easy to use at a table, desk, or quiet area in the classroom. After a teacher or classroom aide models how to use a center, students can complete the activity independently, in pairs, or with the teacher, who may use the task to informally assess a student’s understanding.
Each of the 12 centers includes:
• full-color center mats and task cards
• an overview with lesson objectives
• a student direction page that explains the center activity
• a response form or written practice activity
• a center checklist to record each student’s progress
Take It to Your Seat: Math Centers has been updated to address the new advanced standards, and provides:
• an increase in the variety and complexity of activities to practice each skill in alignment with current standards.
• leveled tasks within some centers so that tasks progress in difficulty from level 1 to level 2.
• an illustrated math concept or rule for every center to support visual learners and keep the focus on the targeted math concept or skill.
The 12 centers cover the following fourth grade math skills:
Numbers and Counting
• Describe patterns of numbers and shapes by identifying the rule that the pattern follows
• Read and write multi-digit whole numbers, using numerals, number names, and expanded forms
• Use place value to round multi-digit numbers
• Identify equivalent fractions based on visual models
• Compare fractions, using greater than, less than, and equal to symbols
• Use the decimal form of fractions that have denominators of 10 and 100
Operations
• Build fluency with multiplication facts to 12
• Use the four operations to solve multistep word problems
Measurement and Data
• Know the relative sizes of linear units within measurement systems (U.S. customary and metric)
• Organize, display, and interpret data on a line plot
Geometry
• Identify right, acute, and obtuse angles
Math Vocabulary
• Understand unique math vocabulary
This resource contains teacher support pages, reproducible student pages, and an answer key.
Correlations
View Activities View Standards
Sampler
### Daily Common Core Math Practice, Grades 1–6
Completely revised for Common Core, the daily practice activities are ideal for warm-up, review, or homework. View weekly units and home–school connection activities for each grade level.
WhitePaper
### The Use of Practice as an Effective Teaching Strategy: Math
The purpose of this paper is to provide evidence from research studies that support frequent, focused practice as an effective strategy for teaching elementary school children mathematics skills. This paper reviews Daily Math Practice and Daily Word Problems in this context.
### Customers who bought this item also bought
Review Snapshot® by PowerReviews | 685 | 3,326 | {"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} | 3.375 | 3 | CC-MAIN-2018-51 | longest | en | 0.89503 |
https://physics.stackexchange.com/questions/267681/what-is-the-difference-between-kilo-watt-and-kilo-volt-ampere | 1,571,751,532,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822098.86/warc/CC-MAIN-20191022132135-20191022155635-00423.warc.gz | 645,369,360 | 29,072 | # What is the difference between kilo-watt and kilo-volt-ampere? [closed]
What is Kilo-Volt_Ampere used for? What is kW used for?
## closed as unclear what you're asking by ACuriousMind♦, user36790, knzhou, Diracology, GertJul 13 '16 at 1:48
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.
• There is no difference. A Watt is equivalent to a volt-ampere - the unit of power. – lemon Jul 12 '16 at 14:59
• @lemon Yes there is. A coil with 14mH inductance and zero resistance will draw a current of 50A when connected to 220V 50Hz mains, enough to blow a domestic fuse. Yet the coil itself doesn't dissipate any power (0W), the (220*50) 11kVA is reactive power. – Previous Jul 12 '16 at 17:29
• I do not have enough rep to vote to re-open, but it's pretty clear what the OP is asking. The question is straightforward and is in the title, and the body is about as clear as it could be. I'm not sure how anybody couldn't be clear on what is being asked. That said it might be a better question for EE than physics. – Jason C Jul 16 '16 at 18:47
VA is used for apparent power and reactive power. Watt is used for active power. An inductor or a capacitor in an AC circuit does not dissipate energy, because the current and the voltage are 90° out of phase: energy flows into them during half the cycle, but it flows back during the next half cycle.
The power lines have to supply that current however, and they have transport losses since their resistance is not zero. The dissipated power in a supply line is equal to I²R: if the current is I+d during the first half of the cycle and I-d during the second half, the dissipated power during the whole cycle is: (I+d)²R/2 + (I-d)²R/2 = (I²+d²)R.
The rated VA of a load also determines the required current capacity of the line, the fuse needed to protect it etc...
The ratio of active power to apparent power in a circuit is called the power factor, or sometimes the "$cos \phi$". Large customers (factories,...) may be charged extra when their $cos \phi$ is too low. Usually there is an excess of inductive load (electric motors for example), and the $cos \phi$ will be corrected by adding capacitive load.
Power ratings for transformers, AC motors, AC motor controllers are usually expressed in (k)VA. Purely resistive loads like heating elements will have a rating in (k)W.
You can reference these links: https://en.wikipedia.org/wiki/Volt-ampere
https://en.wikipedia.org/wiki/Watt
The prefex Kilo simply means 1000. So a Kw is 1000 watts and a KVA is 1000 volt-amperes.
Hope this helps! | 728 | 2,825 | {"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} | 3.375 | 3 | CC-MAIN-2019-43 | latest | en | 0.967984 |
http://clay6.com/qa/8331/find-the-equation-of-the-tangent-and-normal-to-the-hyperbola-4x-y-64-which- | 1,516,538,781,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890582.77/warc/CC-MAIN-20180121120038-20180121140038-00185.warc.gz | 71,286,593 | 26,903 | # Find the equation of the tangent and normal to the hyperbola $4x^{2}-y^{2}=64$ ,which are parallel to $10x-3y+9=0$
Toolbox:
• Equation of any tangent to
• $\quad(i)$ The parabola $y^2=4ax$ is of the form $y=mx+\large\frac{a}{m}$
• $\quad(ii)$ The ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 is of the form y=mx\pm\sqrt{a^2m^2+b^2} • \quad(iii) The hyperbola \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2-b^2}$
Step 1:
$4x^2-y^2=64$
The above equation is divided by $64$
$\large\frac{x^2}{16}-\frac{y^2}{64}$$=1 a^2=16,b^2=64 Step 2: The tangent to \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2-b^2}$
Here $a^2=16$,$b^2=64$
Any tangent to the given hyperbola is of the form $y=mx\pm\sqrt{16m^2-64}$
Step 3:
The required tangents are parallel to $10x-3y+9=0$ whose slope is $\large\frac{10}{3}$$=m The equations of the tangents are y=\large\frac{10x}{3}$$\pm\sqrt{16\times \large\frac{100}{9}-\normalsize 64}$ | 442 | 983 | {"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} | 4.5 | 4 | CC-MAIN-2018-05 | longest | en | 0.595954 |
http://wmbriggs.com/blog/?p=12791 | 1,419,103,443,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770324.129/warc/CC-MAIN-20141217075250-00036-ip-10-231-17-201.ec2.internal.warc.gz | 301,265,932 | 17,914 | # William M. Briggs
### Statistician to the Stars!
Mistakes can happen
I was wrong about how belief in axioms are held. I was not wrong about the beliefs themselves, which is to say, the axioms which we all know and love are still true.
But I used to say, and say wrongly, that we knew axioms were true based on “no evidence” save our intuition, or on faith. This is wrong and bone-headed, as I should have recognized. Actually, we know that axioms are true the same way we know any universal is true: via the evidence of observations.
This dawned on me when I was reading Ed Feser’s new book Scholastic Metaphysics when I was reminded of the Aristotelian and Scholastic dictum “there is nothing in the intellect that was not first in the senses”. If that’s true (and it is) then the sort of a priori knowledge I had in mind, which was the fully formed idea being present to us (somehow), must be wrong.
I was originally convinced of that kind of a priori knowledge by a beauty of an argument by David Stove which appeared in his The Rationality of Induction. The argument is still right if only a few words in it are changed. It purports to prove, and does prove, that our knowledge of logic cannot be had empirically. And that’s so; at least, that is true if we mean all our knowledge. His argument went something like this.
In order to know via experience the validity of (say) the schema A = “For all x, all F, all G, either ‘x is F and all F are G’ is false, or ‘x is G’ is true”, we could make observations like O1 = “David is bald and David is a person now in this room and all persons in this room are bald.” But in order to get from O1 to A; that is, to know A is necessarily true, we have to already know that O2 = “O1 confirms A”, and that is to have non-empirical logical knowledge. Or you could insist at O2 was learnt by experience, but that would require knowing some other logical knowledge, call it O3, which somehow confirms O2. And then there would have to be some O4 which somehow confirms O3, and so on. There cannot be an infinite regress—the series must stop somewhere, at a point where we just know (my guess is O2)—so we must possess some innate logical knowledge.
Perhaps this sketch isn’t fully convincing, but it is in the context of the rest of the chapter, which contains several other proofs that our knowledge of logic cannot be (fully) empirical. (I have a more thorough summary coming in my book—which is still in the works; see more examples below.)
Anyway, the conclusion is still true. Consider the major premise in the old standby “All men are mortal”, which is also unobserved, but still true. Our senses confirm instances, and we extrapolate to the universal. Consider my favorite mathematical axiom that “For all natural numbers x, y, if x = y then y = x”. Again, our senses confirm instances, and we extrapolate to the universal, which we call an “axiom.”
The Scholastic approach solves problems the rationalist a priori raises. If there is rationalistic knowledge we “just know”, then does everybody know all of this knowledge or just a few? And if not everybody knows everything, who or what decides who gets what knowledge and who doesn’t? Too confusing. I should have thought of that.
Much more satisfying is the Scholastic approach. We observe, do we not, that not everybody knows Peano, to pick just one. But they can be brought to its knowledge, and brought pretty quickly, by giving them a few experiences where x = y and y = x. This explains the difficulties over the Axiom of Choice, too (look it up); agreement is harder to come by. Solves the “problem” of induction, too.
Of course I speak of the Aristotelian and not Platonic universals, about which there isn’t space to disagree here. Except to insist that because we are sometimes wrong in our judgement of universals/axioms, it does not imply always wrong. Intuition is a tool that can be misused like any other.
Last of course: The above is not a proof that we do not (all or those non-defective) come pre-wired according to some scheme so that we have the ability to come to universals. Let him that readth understand.
Very little thus has to change in what I wrote before about axioms, but there must still be changes. It is not “no evidence save our intuition or faith” but “the evidence of our senses and our intuition (or faith).” I’ll put this up in the Classic Posts page as a warning. Still unsure whether the language of faith and universals is best. I’m leaning on “yes.” More later.
Teaching
My plan was to write posts about the class discussions of the day before. There is no day before yet. But I talk about axioms on the first day, so this is a natural place.
More Stove
We can learn from observation the following argument is invalid: “‘All men are mortal and David is mortal’ therefore ‘David is a man” if perchance we see David is not a man (maybe he’s a puppy). And we can learn from observation the invalidity of “‘All men are mortal and Peter is mortal’ therefore `Peter is a man” only if we see Peter is not a man (maybe he’s a cow). But we cannot learn the invalidity of “‘All men are mortal and X is mortal’ therefore ‘X is a man” through observation because we would have to measure every imaginable X, and that’s not possible. If we believe “‘All men are mortal and X is mortal’ therefore `X is a man” is unsound, and it surely is, this belief can be informed by experience but it cannot be solely because of it that we have knowledge of it. Another universal is born, though of a more complicated form.
Stove himself: “If an argument from P to Q is invalid, then its invalidity can be learnt from experience if, but also only if, P is true and Q is false in fact, and the conjunction P-and-not-Q, as well as being true, is observational. This has the consequence, first, that only singular judgments of invalidity can be learnt from experience; and second, that very few even of them can be so learnt.” And here’s the kicker: “If the premise P should happen to be false; or the conclusion Q should be true; or if the conjunction P-and-not-Q is not observational but entails some metaphysical proposition, or some scientific-theoretical one, or even a mere universal contingent like ‘All men are mortal': then it will not be possible to learn, by experience, the invalidity of even this particular argument” (pp. 155–156). The key is that :scarcely any of the vast fund of knowledge of invalidity which every normal human being possesses can have been acquired from experience.” Can we allow the hilarious pun universal universals?
Examples? The invalidity of the argument “Given ‘The moon is made of cheese’ therefore ‘Cats do not understand French'” cannot be learned from experience. Neither can “Given ‘Men can breathe underwater unaided’ therefore ‘The atmosphere is largely transparent to sunlight'”. In neither can we can ever observe the conjunct P-and-not-Q.
1. I have noted several times on Dr. Curry’s site that ‘Climate Science’ is the only science that is based on an axiom rather than theory.
The Central Axiom of Climate Science is: ‘The Temperature of the Earth (TOE) is controlled by the amount of CO2 in the atmosphere. Human civilization, as a byproduct of producing energy, is injecting massive quantities of CO2 into the atmosphere, raising the concentration of atmospheric CO2. This injection of CO2 is causing the TOE to rise at a historically unprecedented rate. The results of this rise in the TOE will prove catastrophic unless coordinated government action is taken to drastically reduce (90+%) the production of anthropogenic CO2. The reduction can be achieved most efficiently by taxing and regulating every human activity that produces a ‘carbon signature’.’
You will note that Climate Science, writ large, does NOT treat the above as a theory, subject to confirmation or rejection by observations. It, all of it, including the mitigation prescription, is treated as an axiom, as defined by Wikipedia: “An axiom, or postulate, is a premise or starting point of reasoning. As classically conceived, an axiom is a premise so evident as to be accepted as true without controversy.”
Any nominal scientist, no matter his credentials, who questions any part of the axiom, however mildly, instantly loses his status as ‘Climate Scientist’. Ask Dr. Curry, Dr. Rossiter, and Dr. Bengtsson, to cite three recent examples.
All other sciences propose a theory, collect and analyze related data, and, based on the observations, either confirm, adjust, or reject the theory. Climate Science collects data and adjusts the data as necessary to confirm the axiom. Or simply declares that ALL observations (too hot, too cold, too wet, too dry, too much snow, too little snow, more tropical storms, fewer tropical storms, more Antarctic ice, less Arctic ice, etc.) confirm the axiom. Only axiomatic science is ‘settled’.
2. This is an excellent post. The pithy ‘there is nothing in the intellect that was not first in the senses’ alone is worth the price of admission. Every once in a while Kant comes up in class, and while I don’t quite have a Randian aversion to the man (‘the first hippie’), I have a strong aversion to his overly ‘a priori’ way of looking at knowledge. The above pithy aphorism, and I’m willing to admit I may be rushing to judgment , seems to put the dagger into Kant’s ever-beating heart. I hope I don’t have this completely wrong.
I get what you say about logic, though. I’m still not sure that, somehow, that knowledge still had to come to us first through the senses. ‘Somehow’ is my wiggle word. Thank you!
3. Don’t be too hard on yourself Briggs; the axiom par excellence, the Priciple of Non-Contradiction, can only be known by intuition, and any other axiom that we know to be true is indirectly (at least usually) derived from it by reductio ad absurdum.
To use the Peano axiom as an example, to deny it implies that, in at least in some instance, a natural number x may be equal to some distinct natural number y, but y not be equal to x. But, since we have already agreed that x=y, that means that y must have the same value, and thus, be the same value of x; to say otherwise goes against the definition of what is meant by x=y. Thus, to deny Peano’s axiom implies that both y=x always and that y=x not always. But this violates the PNC, therefore, Peano’s axiom is true. Note however, that without knowledge that the PNC is true, we could affirm that y=x always and not always, so the PNC is absolutely critical to establishing the axiom.
So, we start with our knowledge of the PNC, which we cannot prove but know to be undeniably true, and from there, gain knowledge of other axioms by gaining knowledge of the natures of things, definitions, by experience, and then can gain knowledge of other axioms by seeing the truths that are implied by the definitions.
To go back to the example, we see that the definitions of “natural number”, “x”, “y”, and “=”, combined together, imply that “y=x” always. And since the PNC is true, we can conclude that the axiom is true.
To wrap up my (rather long-winded) comment, in one sense, we only know all of the axioms because of our intuition; because our knowledge of them is based on our intuitive knowledge of the PNC. But in a different sense, we can say that they require no more intuition than the PNC, and thus, it’s fallacious to conclude that they need any MORE intuition than the PNC, or that ALL axioms are impossible to prove in at least in some sense.
4. Sander van der Wal
16 June 2014 at 11:40 am
An axiom is just an axiom. All man are mortal is one. And because all man upto now have died, it is reasonable to add the proposition: ‘the axiom “all men are mortal” is true in this world.’
But an axiom, any axiom, is not *automatically* a true proposition of this world. Take the Geometries, Euclidian, Hyperbolic and Spherical. All three geometries have axioms about parallel lines, and they are all different. But that doesn’t mean that they are all valid as a description of the geometry of this universe, or parts of it. On a sphere, like Earth, Parallel lines at the Equator cross at the Poles. The large scale universe is either Euclidian, or Hyperbolic.
5. Faith is something that cometh from reading too, so I’m not sure why you’d put that with intuition. If by faith you mean “trust,” as a child has for a parent, then that is both learned and innate.
When we observe “wild people” we see that we see they do not develop a recognizable epistemology and so receive experience in a way that we simply can’t understand. Understanding experience, beyond “fire hurts” to “Tom is the kind of guy you’d like to see in higher office,” requires a social assimilation, and the development of critical abstract intelligence so as form understandings of unexperienced phenomena.
JMJ
6. van der Wal,
“On a sphere, like Earth, Parallel lines at the Equator cross at the Poles.”
This is incorrect. There are no parallel straight lines (geodesics) on the surface of a sphere although you can have curved parallel lines, e.g. lines of latitude. You should be referring to Euclid’s fifth postulate and not parallel lines per se.
7. I think the last post that I made on this site, your response to me was that you know your right “a priori”. I pretty much washed my hands of the discussion at that point.
You might want to consider the full implications and correlations of discarding the mystical notion of a priori knowledge.
8. Brandon Gates
17 June 2014 at 12:34 am
Briggs,
Speaking of homework:
http://www.amazon.com/Introduction-Error-Analysis-Uncertainties-Measurements/dp/093570275X
When I was in college, my mother gave me a poster-sized print of the book cover. I framed it and hung it on my wall for years. Thanks for the fond memories.
9. Jim’s right, too. There’s no such thing as a priori knwoledge.
JMJ
10. Sander van der Wal
17 June 2014 at 1:05 am
@Scotian
On the equator, the lines are parallel, as the angles are 90 degrees.
But you’r missing the point. You can define three different geometries by changing one axiom. All of these geometries are internally consistent. What differs is the truthfullness of the propositions you make about the geometries and their relation to the real world, or parts of it.
11. Brandon Gates
17 June 2014 at 2:05 am
JMJ,
There’s no such thing as a priori knowledge.
I’ll go as far as to say that it gets abused. But doesn’t exist? Does knowledge itself exist? [big hint] Do explain.
12. van der Wal,
“On the equator, the lines are parallel, as the angles are 90 degrees.”
That would be true in Euclidian space. The definition of parallel, as lines that never meet, remains. It is the fifth postulate that changes.
13. @Brandon,
It might come down to a difference of definitions…. but a priori knowledge is fully formed, conceptual knowledge that “appears” some how in the mind – bypassing observation. This is not logical inference or deductive reasoning, nor is it a predisposition to learn grammar/language etc.
I’m not sure what your cryptic post is meant to communicate. Do explain.
14. I’ve come to this late, but there is one point that seems to be neglected by most philosophers and cognitive scientists…. The human mind develops. I would doubt that a two year old or probably a four year old would do syllogisms…the question then is when do we develop this “innate” logical facility?
See, for example, the work of Piaget and Rochaut
http://psychology.emory.edu/cognition/rochat/Five%20levels%20.pdf
15. Brandon Gates
18 June 2014 at 11:15 pm
Jim S, there’s was this little fragment running around from Sunday’s Aquinas post (I think) which I can’t find now. Anyway, my caution to JMJ is that discarding any and all a priori knowledge may get him into a bind if he doesn’t watch out because it leaves open the possibility that God could have defined a different value for pi. | 3,697 | 15,928 | {"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-2014-52 | longest | en | 0.984105 |
https://dba.stackexchange.com/questions/286476/advice-on-database-design | 1,722,700,583,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00113.warc.gz | 149,929,028 | 43,525 | • context :
• Teachers create problems to their Students
• A Problem have Numerals(Questions)
• A Numeral(Question) have Steps to guide the student to the answer
• Choice represent the type of mathematical abstraction that fits better
• Solve: (unknown variable, starting expression)
• Simplify: (reference, target)
• Plot: (slope,b)
• Base Models and Relations:
• Problems: one to Many Numerals(Questions)
• Numerals: zero to Many Steps
• Steps:
• Requirements:
• Numerals and Steps now have a choice which where described above, currently just 3 choices but it's expected to grow. only one and at least one choice
• Depending on the choice the numeral or step will add new attributes
• for now choices do not have any relations between them
• choice in Numeral could be different to the choice in Step and vice versa (see Example)
• Example: did my best to make a close abstraction to show the idea, I've made a JSON Structure, almost identical to the structure used in my front end. Maybe the example is a little bit dumb but it gives a taste.
• My Attempts: For simplicity , Models are shown with the minimum attributes
• 1st:
-My concerning with this approach is that the amount of FK in the Numeral and Step model. Since only one choice is possible, i would have choices-1 amount of useless columns per record
• 2nd:
-I'm programmatically checking for the choice field in the Numeral or the Step, and then using the proper table to fill the data, this second attempt started as a solution to fix the empty columns in the 1st approach
-My concerning with this approach is that the amount of tables will grown 2*n, since for each choice i need to create the corresponding table for step or numeral
Edit: Thanks to @J.D for suggesting this idea. i've modified a bit because i need to keep track of the choices and their fields somewhere somewhere. I think this solution is easier to mantain and removes the need to add tables in case of new choices
• 3rd:
• What are ANumeral, BNumeral, CNumeral, AStep, BStep, CStep? Can you use some real examples so we understand more of your problem? Also, there are many questions and answers on subtypes; there's probably an answer there somewhere. The problem with database-design is that while it's mostly science, it's also partly art, and with a dash of opinion thrown in. This makes it hard to definitely give an answer to any design problem. The answer is usually, "It depends." Commented Mar 5, 2021 at 7:15
• will add some examples, working on it Commented Mar 5, 2021 at 14:13
As Colin touched on, it's hard to follow what schema problem you're trying to solve with just generalities. Unfortunately the devil's in the details with this kind of thing, since one detail could drastically change the best approach.
Looking at your database diagrams at a high level the tables `A`, `B`, and `C` in the first diagram look like the same entity in regards to their purpose, but varies in value of their columns, most likely?...same when comparing tables `ANumeral`, `BNumeral`, `CNumeral` to each other, or when comparing `AStep`, `BStep`, and `CStep` to each other, in your second diagram. The question is why couldn't each of these sets of tables be just one table each (or possibly even one table altogether)?...e.g. you likely could have just a single `StepAttributes` table that `Step` joins to.
I did notice in your example the only difference between two tables of the same set are the attribute data types. If that's your reasoning for making multiple tables, instead just structure your single attribute table (e.g. `StepAttribute`) to account for that which can be done two ways.
The first way by having three columns, one being the `AttributeName`, the next being the `AttributeValue` (which will be a string type field, like VARCHAR, to accommodate any value), and the third being the `AttributeDataType` which stores the native data type of the attribute so you know what to appropriately cast it to when you consume it.
The second way is you can have a denormalized attribute table that has a nullable column of each data type with generic column names, such as `DateAttributeValue`, `IntAttributeValue`, `VarCharAtrributeValue` etc, and a column to specify which data type the value is / column it's stored in. Some systems even design such a table like `DateAttributeValue1`, `DateAttributeValue2`, `DateAttributeValue3`, `IntAttributeValue1`, `IntAttributeValue2`, etc so they can support storing multiple attributes within the same row.
Thanks for updating your question with more details. So at the least it looks like each sub-table to `Numeral` vs `Step` are a bit redundant in your second design, and I understand why you normalized them that way based on your first design. You could actually combine the best of both worlds though and just have a single `Plot`, `Simplify`, and `Solve` table with the foreign key reference field in those tables going back to your `Numeral` and `Step` tables by storing both IDs in that column and calling it something like `NuneralOrStepId`. Then you'd need an additional field on those tables called like `IsForNumeral` or `ParentType` (come up with a better name than mine lol) which would distinguish which base table it joins back to. Then you have no waste in fields and reduce your redundancy as well.
It depends on how many of these sub-tables you anticipate having to generate and manage (right now 3 is completely manageable) but if you think that's going to grow into over 10 tables or so, I'd still consider my first suggestion with one generic table of `AttributeName` / `AttributeValue` pairs. You can add any additional columns to this structure that make sense as well, such as the `ParentType` column to distinguish which parent table the `Attribute` belongs to.
• Thanks for your ideas, I've added context and an example . Indeed A,B,C which are the choices varies in columns. In regards to the Numerals and Steps, they have different columns by default. So i think having two separate models fits better. Commented Mar 6, 2021 at 4:33
• @joaortizro Thanks I've updated my answer with additional suggestions and information.
– J.D.
Commented Mar 6, 2021 at 14:37
• Did my best to understand your first suggestion , i decided to make a 3rd design based on your ideas, Commented Mar 7, 2021 at 7:34
• @joaortizro Cool the only suggestions I have is your `belongsTo` column should probably be a boolean or int based field as that's more manageable than a string (in case the table name changes, etc). So if there will only be the two parent tables `Numeral` and `Step` then I'd make it a boolean field called `belongsToNumeral`. Otherwise, if more parent tables are possible in the future, then I'd have 1 meta-data table `AttributeParent` that had the name and a unique `ID` for each parent table, and then that field in your `Attribute` table would be an int referencing that `ID`. Also because...
– J.D.
Commented Mar 7, 2021 at 14:08
• Yes, indeed there should be a one to many between Numeral and Attribute , as well as Step and Attribure. Thank you so much for ur suggestions, really like the design. In regards to the `numeralOrStepId` working together with the `belongsTo` ,i will check it, in theory seems really good, since im using Django in my backend and i 've never seen a FK as a reference for two models but i guess is doable. I really appreciate your suggestions and effort , thank you. Commented Mar 8, 2021 at 1:42 | 1,727 | 7,468 | {"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} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.904985 |
http://thefinancebase.com/net-income-equal-stockholders-equity-1822.html | 1,488,136,190,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172050.87/warc/CC-MAIN-20170219104612-00386-ip-10-171-10-108.ec2.internal.warc.gz | 246,078,910 | 13,327 | # Does Net Income Equal Stockholders' Equity?
by Nola Moore
As you begin to analyze potential investments, you'll likely spend a lot of time staring at financial statements, trying to determine the potential profitability of a company. The net income and stockholders' equity are two concepts that are crucial to this determination. While both numbers are very similar in concept, the values they describe are very different, and it's vital to your analysis to understand the difference between the two.
## Net Income Defined
Net income, sometimes referred to as net profit, is simply defined as revenue minus expenses for a particular period. Revenue is all the cash that the company took in, and expenses are what it put out -- cost of goods sold, salaries, equipment and other expenditures. Depending on the report you're reading, net income may or may not include taxes and interest expenses, but this should be fairly easy to figure out. It is important to understand that net income is specific to the reporting period -- one year in the case of an annual report, and the calculation only includes the money that was made or spent during that year.
## Stockholders' Equity
Stockholders' equity is also referred to as shareholders' equity, or simply SE, and is the corporation equivalent to owner's equity in a sole proprietorship. SE is a snapshot of a moment in time, and is a bigger picture than net income because it looks at the company as a whole: everything it owns and everything it owes. SE adds up buildings, equipment, inventory and other assets and subtracts the total of debts, taxes and any other liabilities -- it is the estimated value of the company if it were liquidated today.
## Relationship Between Net Income and SE
When a company has a net income, it can do two things: pay that money out to shareholders or keep it for reinvestment into the business. Many companies do some combination of the two. The portion of net income that is kept for the business is known as retained earnings. These retained earnings become an asset of the company, and are added to the asset portion of the SE equation.
## Using Net Income and SE in Analysis
You can review both the net income and the SE over time to determine whether they are increasing or decreasing and why. You can also use these two numbers in ratios to compare the health of one company to another. Net income per share (net income compared to total outstanding common stock) is one such ratio. Another common ratio is the debt-to-equity ratio, where total liabilities are compared to SE. The return on equity compares the net income to SE for another valuable measurement. | 530 | 2,662 | {"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-2017-09 | latest | en | 0.97724 |
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Discover how many days in 15 months, include the articles, news, trends, analysis and practical advice about how many days in 15 months on alibabacloud.com
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### Go How to get a date in SQL SERVER 2005 (the last day of one months, the first day of the year, and so on)
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 1,050 | 4,523 | {"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-2022-49 | latest | en | 0.909924 |
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###### Question:
Question 1) [6 marks] Given the strong form problem: "" (2) + sin(2rx)u' (2) + cos(rr)u(z) = 0. 10 ult = 0) = 1 "(=4) =0 en = [. 4 [1 mark] Please describe what type of boundary conditions are applied and where. [4 marks] Derive the continuous weak: form equation for this problem_ [1 mark] Please define the solution space for W4 ad the test function space for
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Quantity Demanded Price $25$15 Complete the following table by calculating the price elasticity of demand between specified points and indicating whether inelastic, or unit elastic. (Hint: Use the midpoints formula.) Interval From P - $20 to P -$25 From P - $20 to P -$15 Price Elasticity of Deman...
##### Point) A tank in the shape of an inverted right circular cone has height 13 meters and radius 14 meters. It is filled with meters of hot chocolate_ Find the work required to empty the tank by pumping the hot chocolate over the top of the tank: Note: the density of hot chocolate is 6 = 1S10kglm?
point) A tank in the shape of an inverted right circular cone has height 13 meters and radius 14 meters. It is filled with meters of hot chocolate_ Find the work required to empty the tank by pumping the hot chocolate over the top of the tank: Note: the density of hot chocolate is 6 = 1S10kglm?... | 4,251 | 14,274 | {"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} | 2.890625 | 3 | CC-MAIN-2023-40 | latest | en | 0.724641 |
https://www.geeksforgeeks.org/sum-of-multiplication-of-triplet-of-divisors-of-a-number/?ref=rp | 1,653,768,128,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00178.warc.gz | 895,548,701 | 29,774 | # Sum of multiplication of triplet of divisors of a number
• Last Updated : 14 May, 2021
Given an array arr[] of integers of size n. For each element, you have to print the summation of multiplication of every triplet formed using divisors of this element.
Examples:
Input: arr[] = {4}
Output:
4 has three divisors 1, 2 and 4.
1 * 2 * 4 = 8
Input: arr[] = {9, 5, 6}
Output: 27 0 72
9 has three divisors 1, 3 and 9. 1 * 3 * 9 = 27
5 has two divisors 1 and 5. So, no triplet is formed .
Similarly, 6 has four divisors 1, 2, 3 and 6. (1 * 2 * 3) + (1 * 3 * 6) + (2 * 3 * 6) + (1 * 6 * 2) = 72
Naive approach: Store all the divisors of a number and apply brute force method and for every triplet, add multiplication of these three elements to the answer.
Efficient approach: This method is similar to Sieve of Eratosthenes, which we use for finding all prime numbers of a range.
1. First, we make an array sum1 which stores the sum of all divisors of number x at sum1[x]. So first iterate through all the numbers less than maximum_Element and add this number to the sum of divisors of multiple of this number. Hence we will be able to store sum of all divisors of any number at that number’s position.
2. After filling the sum1 array, it’s time to fill the second array sum2 which will store the sum of multiplication of every pair divisor of number x at sum2[x].For filling this we go for every number similar to step1 and add multiplication of this number with to its higher values.
3. For filling sum3 array we will go similarly for all numbers less than max_Element and will add sum of multiplication of divisors of j such that i is a divisor of j.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach``#include ``using` `namespace` `std;``#define ll long long``const` `ll max_Element = 1e6 + 5;` `// Global array declaration``int` `sum1[max_Element], sum2[max_Element], sum3[max_Element];` `// Function to find the sum of multiplication of``// every triplet in the divisors of a number``void` `precomputation(``int` `arr[], ``int` `n)``{`` ``// sum1[x] represents the sum of all the divisors of x`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)` ` ``// Adding i to sum1[j] because i`` ``// is a divisor of j`` ``sum1[j] += i;` ` ``// sum2[x] represents the sum of all the divisors of x`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)` ` ``// Here i is divisor of j and sum1[j] - i`` ``// represents sum of all divisors of`` ``// j which do not include i so we add`` ``// i * (sum1[j] - i) to sum2[j]`` ``sum2[j] += (sum1[j] - i) * i;` ` ``// In the above implementation we have considered`` ``// every pair two times so we have to divide`` ``// every sum2 array element by 2`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``sum2[i] /= 2;` ` ``// Here i is the divisor of j and we are trying to`` ``// add the sum of multiplication of all triplets of`` ``// divisors of j such that one of the divisors is i`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)`` ``sum3[j] += i * (sum2[j] - i * (sum1[j] - i));` ` ``// In the above implementation we have considered`` ``// every triplet three times so we have to divide`` ``// every sum3 array element by 3`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``sum3[i] /= 3;` ` ``// Print the results`` ``for` `(``int` `i = 0; i < n; i++)`` ``cout << sum3[arr[i]] << ``" "``;``}` `// Driver code``int` `main()``{` ` ``int` `arr[] = { 9, 5, 6 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Precomputing`` ``precomputation(arr, n);` ` ``return` `0;``}`
## Java
`// Java implementation of the approach``class` `GFGq``{` `static` `int` `max_Element = (``int``) (1e6 + ``5``);` `// Global array declaration``static` `int` `sum1[] = ``new` `int``[max_Element], sum2[] =`` ``new` `int``[max_Element], sum3[] = ``new` `int``[max_Element];` `// Function to find the sum of multiplication of``// every triplet in the divisors of a number``static` `void` `precomputation(``int` `arr[], ``int` `n)``{`` ``// sum1[x] represents the sum of all the divisors of x`` ``for` `(``int` `i = ``1``; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)` ` ``// Adding i to sum1[j] because i`` ``// is a divisor of j`` ``sum1[j] += i;` ` ``// sum2[x] represents the sum of all the divisors of x`` ``for` `(``int` `i = ``1``; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)` ` ``// Here i is divisor of j and sum1[j] - i`` ``// represents sum of all divisors of`` ``// j which do not include i so we add`` ``// i * (sum1[j] - i) to sum2[j]`` ``sum2[j] += (sum1[j] - i) * i;` ` ``// In the above implementation we have considered`` ``// every pair two times so we have to divide`` ``// every sum2 array element by 2`` ``for` `(``int` `i = ``1``; i < max_Element; i++)`` ``sum2[i] /= ``2``;` ` ``// Here i is the divisor of j and we are trying to`` ``// add the sum of multiplication of all triplets of`` ``// divisors of j such that one of the divisors is i`` ``for` `(``int` `i = ``1``; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)`` ``sum3[j] += i * (sum2[j] - i * (sum1[j] - i));` ` ``// In the above implementation we have considered`` ``// every triplet three times so we have to divide`` ``// every sum3 array element by 3`` ``for` `(``int` `i = ``1``; i < max_Element; i++)`` ``sum3[i] /= ``3``;` ` ``// Print the results`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``System.out.print(sum3[arr[i]] + ``" "``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{`` ``int` `arr[] = { ``9``, ``5``, ``6` `};`` ``int` `n = arr.length;` ` ``// Precomputing`` ``precomputation(arr, n);``}``}` `// This code has been contributed by 29AjayKumar`
## Python3
`# Python 3 implementation of the approach` `# Global array declaration``#global max_Element``max_Element ``=` `100005` `sum1 ``=` `[``0` `for` `i ``in` `range``(max_Element)]``sum2 ``=` `[``0` `for` `i ``in` `range``(max_Element)]``sum3 ``=` `[``0` `for` `i ``in` `range``(max_Element)]` `# Function to find the sum of multiplication of``# every triplet in the divisors of a number``def` `precomputation(arr, n):`` ` ` ``# global max_Element`` ``# sum1[x] represents the sum of`` ``# all the divisors of x`` ``for` `i ``in` `range``(``1``, max_Element, ``1``):`` ``for` `j ``in` `range``(i, max_Element, i):`` ` ` ``# Adding i to sum1[j] because i`` ``# is a divisor of j`` ``sum1[j] ``+``=` `i` ` ``# sum2[x] represents the sum of`` ``# all the divisors of x`` ``for` `i ``in` `range``(``1``, max_Element, ``1``):`` ``for` `j ``in` `range``(i, max_Element, i):`` ` ` ``# Here i is divisor of j and sum1[j] - i`` ``# represents sum of all divisors of`` ``# j which do not include i so we add`` ``# i * (sum1[j] - i) to sum2[j]`` ``sum2[j] ``+``=` `(sum1[j] ``-` `i) ``*` `i` ` ``# In the above implementation we have considered`` ``# every pair two times so we have to divide`` ``# every sum2 array element by 2`` ``for` `i ``in` `range``(``1``, max_Element, ``1``):`` ``sum2[i] ``=` `int``(sum2[i] ``/` `2``)` ` ``# Here i is the divisor of j and we are trying to`` ``# add the sum of multiplication of all triplets of`` ``# divisors of j such that one of the divisors is i`` ``for` `i ``in` `range``(``1``, max_Element, ``1``):`` ``for` `j ``in` `range``(i, max_Element, i):`` ``sum3[j] ``+``=` `i ``*` `(sum2[j] ``-` `i ``*`` ``(sum1[j] ``-` `i))` ` ``# In the above implementation we have considered`` ``# every triplet three times so we have to divide`` ``# every sum3 array element by 3`` ``for` `i ``in` `range``(``1``, max_Element, ``1``):`` ``sum3[i] ``=` `int``(sum3[i] ``/` `3``)` ` ``# Print the results`` ``for` `i ``in` `range``(n):`` ``print``(sum3[arr[i]], end ``=` `" "``)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:`` ``arr ``=` `[``9``, ``5``, ``6``]`` ``n ``=` `len``(arr)` ` ``# Precomputing`` ``precomputation(arr, n)` `# This code is contributed by``# Surendra_Gangwar`
## C#
`// C# implementation of the approach``using` `System;` `class` `GFG``{` ` ``static` `int` `max_Element = (``int``) (1e6 + 5);`` ` ` ``// Global array declaration`` ``static` `int` `[]sum1 = ``new` `int``[max_Element];`` ``static` `int` `[]sum2 = ``new` `int``[max_Element];`` ``static` `int` `[]sum3 = ``new` `int``[max_Element];`` ` ` ``// Function to find the sum of multiplication of`` ``// every triplet in the divisors of a number`` ``static` `void` `precomputation(``int` `[]arr, ``int` `n)`` ``{`` ``// sum1[x] represents the sum of all the divisors of x`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)`` ` ` ``// Adding i to sum1[j] because i`` ``// is a divisor of j`` ``sum1[j] += i;`` ` ` ``// sum2[x] represents the sum of all the divisors of x`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)`` ` ` ``// Here i is divisor of j and sum1[j] - i`` ``// represents sum of all divisors of`` ``// j which do not include i so we add`` ``// i * (sum1[j] - i) to sum2[j]`` ``sum2[j] += (sum1[j] - i) * i;`` ` ` ``// In the above implementation we have considered`` ``// every pair two times so we have to divide`` ``// every sum2 array element by 2`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``sum2[i] /= 2;`` ` ` ``// Here i is the divisor of j and we are trying to`` ``// add the sum of multiplication of all triplets of`` ``// divisors of j such that one of the divisors is i`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``for` `(``int` `j = i; j < max_Element; j += i)`` ``sum3[j] += i * (sum2[j] - i * (sum1[j] - i));`` ` ` ``// In the above implementation we have considered`` ``// every triplet three times so we have to divide`` ``// every sum3 array element by 3`` ``for` `(``int` `i = 1; i < max_Element; i++)`` ``sum3[i] /= 3;`` ` ` ``// Print the results`` ``for` `(``int` `i = 0; i < n; i++)`` ``Console.Write(sum3[arr[i]] + ``" "``);`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int` `[]arr = { 9, 5, 6 };`` ``int` `n = arr.Length;`` ` ` ``// Precomputing`` ``precomputation(arr, n);`` ``}``}` `// This code has been contributed by Ryuga`
## PHP
``
## Javascript
``
Output:
`27 0 72`
My Personal Notes arrow_drop_up | 3,991 | 11,512 | {"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-2022-21 | latest | en | 0.817284 |
https://www.wyzant.com/resources/answers/15359/9_x_31_5_x_3_12 | 1,529,393,010,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00332.warc.gz | 972,818,682 | 16,925 | 0
# /9/x/+31+5/x/+3=12
how do I solve this? | 23 | 45 | {"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-2018-26 | latest | en | 0.872114 |
https://www.engineersedge.com/engineering-forum/showthread.php/3426-Measured-running-tire-circumference-VS-Manufacturer-s-numbers?s=6933a0b5fc73d32c4e89adfa5e77f0ca | 1,660,655,175,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00653.warc.gz | 665,634,501 | 10,775 | # Thread: Measured "running" tire circumference VS Manufacturer's numbers
1. ## Measured "running" tire circumference VS Manufacturer's numbers
I have a dilemma that I can't seem to get my head around. . .
In calculating gearing options for a modified vehicle, I am using the measured distance from the ground to the center of the axle of a pneumatic drive tire.
I figure that is the lever arm and is the ONLY number that should be used to determine my distance traveled.
(Radius (distance from ground to axle center) X 2) X pi)) = "running" circumference.
However, several folks keep telling me that I must use the tire manufacturer's calculation for that size tire (which is not the same as the one I have measured).
While the differences are slight, they do change the final ratio, and I'd like to know if I am misunderstanding things, or am correct in my calculations.
2. Due to variations in manufacturing, the as-built 'pneumatic drive tire" will measure different than OEM specifications. Moreover, any loading applied to the tire may cause significant radial distortions at rest that may go away at rated velocity due to centrifugal force.
I recommend you use the manufacturer specifications. You may consider doing your calculations with the specified largest and smallest radius of the tire.
The circumference is equal to:
Pi x d
or
2 x pi x r
where pi = 3.14.157 and d = diameter and r = radius
3. I agree with Kelly.
No matter how much deflection the tire encounters, one rotation of the tire will travel the distance of its circumference (not including tire slippage). Think of a tracked vehicle (like a bulldozer or tank). One rotation of the track travels exactly the distance of the length of the track.
With a tire, however, the length of the "arm" as you mentioned will change due to loading and wheel speed (centrifugal force). As the length of the arm changes, so will the amount of torque being transferred to the ground. However, the distance traveled per tire rotation will be the same as the circumference of the tire.
4. In support of the above, the maximum rolling circumference and diameter is based upon the molded in tire belt. These belts, being either kevlar or steel (generally kevlar for road tires) are strong enough to limit the maximum tread diameter under centifugal force to the manufacturer's specification. This is the reason that super wide racing tires don't deform into a complete torus shape at high racing speeds.
5. While you can use the manufactures specs, keep in mind that those numbers are for new tires. As your tires wear overtime, you will end up with a smaller diameter and thus shorter circumference.
To the point of measuring, if you measure from the ground to the axle you will have inaccuracy because of the flat spot the tire forms with the ground. A better reading would be from any point around the tire that is not touching the ground. The best way, although much more hassle, would be to take the tire off the vehicle and measure.
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• | 672 | 3,157 | {"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} | 3.296875 | 3 | CC-MAIN-2022-33 | latest | en | 0.941374 |
travisphysicsaction.tumblr.com | 1,371,625,645,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708143620/warc/CC-MAIN-20130516124223-00088-ip-10-60-113-184.ec2.internal.warc.gz | 258,778,144 | 11,262 | Sometimes names aren’t what they seem in science and mathematics. A case-in-point I will use here is an equation that you know very well: the Pythagorean Theorem. The theorem was well known and understood long before Pythagoras and his secret cult. Here’s a good page on it. Besides the equation being used in the west and mideast, it was also used in China around the turn of the century. This theorem is an example of how scientific and mathematical knowledge know no borders culturally, politically, or otherwise. Why was it that Pythagoras was named after this most useful formula? Did he discover it, popularize it, or were his uses of it clever/smart/useful enough to merit applying his name to it? Is it the same type of situation that the Persian mathematician and astronomer al-Kashi, in some countries, has the ‘law of cosines’ named after him? In French the law of cosines — a more generalized version of the Pythagorean theorem — is known as “le théorème d’Al-Kashi”. This ambiguity of naming conventions in science and mathematics is a source of confusion. For instance, Galileo dealt with the concept of inertia at great length before Newton wrote it down in his famous book on mechanics called the Principia. Newton included inertia as his first law of motion. Why isn’t it called one of Galileo’s laws? Who deserves naming credit?
Also, as an extra credit challenge, can you derive the Pythagorean theorem from the figure above? Hints: find the areas of all the figures that have to fit into the area of the largest square.
Incidentally, since we’re now speaking of Galileo and inertia, and since one of the questions posed below on the images of Rosetta from three different reference frames, one thing that Galileo’s name stuck to was this technical-sounding tidbit: “Galilean Transformation”. This falls out of the Newtonian Laws of motion not depending on velocities — only accelerations. So it doesn’t matter what the velocity is — can we ever detect how fast we’re moving if we’re not comparing to something else (remember the train with no windows question from the homework?). This idea of being free to choose one’s frame of reference is the basis of Einstein’s Special Relativity.
1. travisphysicsaction posted this | 517 | 2,262 | {"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} | 2.984375 | 3 | CC-MAIN-2013-20 | latest | en | 0.951167 |
http://www.mathworks.com/matlabcentral/fileexchange/40804-contemporary-communications-systems-matlab-files/content/MatlabFiles%20for%20Contemporary%20Communicatios%20Books/fm2.m | 1,438,683,602,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042990609.0/warc/CC-MAIN-20150728002310-00108-ip-10-236-191-2.ec2.internal.warc.gz | 565,669,064 | 10,574 | Code covered by the BSD License
Contemporary Communications Systems Matlab Files
Omar Ruiz (view profile)
Matlab Files in this book
fm2.m
```% fm2.m
% Matlab demonstration script for frequency modulation. The message signal
% is m(t)=sinc(100t).
echo on
t0=.2; % signal duration
ts=0.001; % sampling interval
fc=250; % carrier frequency
snr=20; % SNR in dB (logarithmic)
fs=1/ts; % sampling frequency
df=0.3; % required freq. resolution
t=[-t0/2:ts:t0/2]; % time vector
kf=100; % deviation constant
df=0.25; % required frequency resolution
m=sinc(100*t); % the message signal
int_m(1)=0;
for i=1:length(t)-1 % Integral of m
int_m(i+1)=int_m(i)+m(i)*ts;
echo off ;
end
echo on ;
[M,m,df1]=fftseq(m,ts,df); % Fourier transform
M=M/fs; % scaling
f=[0:df1:df1*(length(m)-1)]-fs/2; % frequency vector
u=cos(2*pi*fc*t+2*pi*kf*int_m); % modulated signal
[U,u,df1]=fftseq(u,ts,df); % Fourier transform
U=U/fs; % scaling
[v,phase]=env_phas(u,ts,250); % demodulation, find phase of u
phi=unwrap(phase); % restore original phase
dem=(1/(2*pi*kf))*(diff(phi)/ts); % demodulator output, differentiate and scale phase
pause % Press any key to see a plot of the message and the modulated signal
subplot(2,1,1)
plot(t,m(1:length(t)))
xlabel('Time')
title('The message signal')
subplot(2,1,2)
plot(t,u(1:length(t)))
xlabel('Time')
title('The modulated signal')
pause % Press any key to see a plots of the magnitude of the message and the
% modulated signal in the frequency domain.
subplot(2,1,1)
plot(f,abs(fftshift(M)))
xlabel('Frequency')
title('Magnitude-spectrum of the message signal')
subplot(2,1,2)
plot(f,abs(fftshift(U)))
title('Magnitude-spectrum of the modulated signal')
xlabel('Frequency')
pause % Pres any key to see plots of the message and the demodulator output with no
% noise
subplot(2,1,1)
plot(t,m(1:length(t)))
xlabel('Time')
title('The message signal')
subplot(2,1,2)
plot(t,dem(1:length(t)))
xlabel('Time')
title('The demodulated signal')``` | 659 | 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} | 2.859375 | 3 | CC-MAIN-2015-32 | longest | en | 0.503966 |
http://www.bbc.co.uk/bitesize/quiz/q50566881 | 1,440,880,390,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064538.25/warc/CC-MAIN-20150827025424-00034-ip-10-171-96-226.ec2.internal.warc.gz | 319,759,791 | 14,489 | Home > Maths > Measures > Volume
# Volume - Test
1.
What is the volume of this cuboid? (Each cube has a side length of 1cm.)
2.
What is the volume of this box?
3.
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# Handicap from Different Tees
## 30 posts in this topic
Does anyone play where players compete against each other from different sets of tees based on their index? How do they compute players handicaps?
I played my first Men's League round this year and they have the low third play the tips (gold), the middle third play the blue and the high cappers play the whites. When I asked in the shop what my course handicap would be, they said that you round .5+ up and .4- down, but this seems inconsistent with the idea of looking at the course slope to calculate handicap.
How should the league be computing our handicaps based on the following
Gold 66.8/117
Blue 64.8/113
While 62.6/103
A index= 10.4 playing Gold
B index= 10.5 playing Blue
C index= 30.5 playing White
League Handicap
(my calculation based on info from the USGA site- I think section 3-5 and a supplemental paper explaining the logic of an adjustment for different tees)
A= 15 from Gold
B= 13 from Blue
C= 28 from White
or (per guys in the shop)
A= 10 from Gold
B= 11 from Blue
C= 31 from White
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Should be 11,11,28 ... Just google course handicap calculator, usga has one.
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Originally Posted by MEfree
Does anyone play where players compete against each other from different sets of tees based on their index? How do they compute players handicaps?
I played my first Men's League round this year and they have the low third play the tips (gold), the middle third play the blue and the high cappers play the whites. When I asked in the shop what my course handicap would be, they said that you round .5+ up and .4- down, but this seems inconsistent with the idea of looking at the course slope to calculate handicap.
How should the league be computing our handicaps based on the following
Gold 66.8/117
Blue 64.8/113
While 62.6/103
A index= 10.4 playing Gold
B index= 10.5 playing Blue
C index= 30.5 playing White
League Handicap
(my calculation based on info from the USGA site- I think section 3-5 and a supplemental paper explaining the logic of an adjustment for different tees)
A= 15 from Gold
B= 13 from Blue
C= 28 from White
or (per guys in the shop)
A= 10 from Gold
B= 11 from Blue
C= 31 from White
You're close, but I think you miscalculated B's course handicap. The guys in the shop are way off.
Should be 11,11,28 ... Just google course handicap calculator, usga has one.
Not quite.
You calculate each individuals course handicap as always, based on the slope for the set of tees they're playing. However, if they're competing against each other, you also need to adjust their handicaps relative to each other based on the difference in the course rating for each set of tees.
First step is to calculate the individual course handicaps based on their handicap indexes and the set of tees from which they're playing. Then you adjust those further based on the difference in course rating for the tees each is playing. In this case the results would be as follows:
A from Gold: Course Handicap = 11 Competition Handicap = 15
B from Blue: Course Handicap = 10 Competition Handicap = 12
C from White: Course Handicap = 28 Competition Handicap = 28
You see that A and B have strokes added to their handicaps to match the difference in each tee's course rating (rounded to the nearest whole).
From the USGA Handicap Manual:
c. Players Competing From Different Tees or Men and Women From Same Tees
(i) Different Tees: Men vs. Men; Women vs. Women; Women vs. Men
Different tees usually have different ratings. Since a USGA Course Rating reflects the probable scores of scratch golfers , the higher-rated course is more difficult, and the player playing from the set of tees with the higher USGA Course Rating receives additional stroke(s) equal to the difference between each USGA Course Rating , with the resulting figure rounded off to the nearest whole number (.5 or more is rounded upward). (See Decision 3-5/1 .)
Example 1: If men playing from the middle tees, from which the men's USGA Course Rating is 70.3, compete against men playing from the back tees, from which the men's USGA Course Rating is 72.6, the men playing from the back tees will add two strokes
(72.6 - 70.3 = 2.3 rounded to 2) to Course Handicap .
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Originally Posted by David in FL
You're close, but I think you miscalculated B's course handicap. The guys in the shop are way off.
Not quite.
You calculate each individuals course handicap as always, based on the slope for the set of tees they're playing. However, if they're competing against each other, you also need to adjust their handicaps relative to each other based on the difference in the course rating for each set of tees.
First step is to calculate the individual course handicaps based on their handicap indexes and the set of tees from which they're playing. Then you adjust those further based on the difference in course rating for the tees each is playing. In this case the results would be as follows:
A from Gold: Course Handicap = 11 Competition Handicap = 15
B from Blue: Course Handicap = 10 Competition Handicap = 12
C from White: Course Handicap = 28 Competition Handicap = 28
You see that A and B have strokes added to their handicaps to match the difference in each tee's course rating (rounded to the nearest whole).
From the USGA Handicap Manual:
c. Players Competing From Different Tees or Men and Women From Same Tees
(i) Different Tees: Men vs. Men; Women vs. Women; Women vs. Men
Different tees usually have different ratings. Since a USGA Course Rating reflects the probable scores of scratch golfers, the higher-rated course is more difficult, and the player playing from the set of tees with the higher USGA Course Rating receives additional stroke(s) equal to the difference between each USGA Course Rating, with the resulting figure rounded off to the nearest whole number (.5 or more is rounded upward). (See Decision 3-5/1.)
Example 1: If men playing from the middle tees, from which the men's USGA Course Rating is 70.3, compete against men playing from the back tees, from which the men's USGA Course Rating is 72.6, the men playing from the back tees will add two strokes
(72.6 - 70.3 = 2.3 rounded to 2) to Course Handicap.
Sorry, I got B's course handicap wrong......wrong slope. You're right and the guys in the shop were way off.
B's course handicap is 11 and his competition handicap is 13.
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Sorry, I got B's course handicap wrong......wrong slope. You're right and the guys in the shop were way off. B's course handicap is 11 and his competition handicap is 13.
Thanks David! Very interesting, and makes sense too. A and b getting same strokes from different tees does look weird on paper.
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Originally Posted by David in FL
Sorry, I got B's course handicap wrong......wrong slope. You're right and the guys in the shop were way off.
B's course handicap is 11 and his competition handicap is 13.
Thanks for the confirmation...last night I replied to an email that the league had sent out and wrote what I thought the handicaps should be and asking what they thought but have not gotten a reply yet. I realize that many might not be aware of the course rating adjustment when different tees are used, but the guys in the shop didn't even seem to know that slope factors into determining your course handicap...I tried explaining it briefly and all he said was that it seemed I knew more than him and said that it doesn't matter because it is the same for everyone.
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Thanks for the confirmation...last night I replied to an email that the league had sent out and wrote what I thought the handicaps should be and asking what they thought but have not gotten a reply yet. I realize that many might not be aware of the course rating adjustment when different tees are used, but the guys in the shop didn't even seem to know that slope factors into determining your course handicap...I tried explaining it briefly and all he said was that it seemed I knew more than him and said that it doesn't matter because it is the same for everyone.
They're out of their minds. In this example player A is giving player C 8 strokes more than he should. That's an ENORMOUS difference!
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David is spot on. It's NOT the same for everyone if you don't make the adjustment. Also, as in David's example, you add strokes for the players playing form the higher rated tee boxes. Depending on what tees most of the players are playing you can also leave the player's handicap who is playing the highest rated tee alone and subtract the difference in ratings from the handicaps for the players playing the lower rated tees. Decision 3-5/1 in the Handicap Manual.
You can print out the "handigram section 3-5" and take it to the shop.
BTW if I went to 10 pro shops, I bet 8 would have no idea how to do this.
Edit: It's actually sad the lack of understanding most clubs have regarding the handicap system. We use the USGA system at our club and it was a requirement for a representative to go to a USGA handicap seminar when it came to town and pass and open book test. I think you have a couple of years to get this done. I've gone a couple of times, well worth the half a day......and lunch is included.
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This web site it GREAT!!
My dad has been screwing me for years. I'm headed to NC next week to play their course and boy am I going surprise him. Not only does he have home course advantage but he's been taking way too many strokes from me.
In the past we have just looked at our indexes. His is 16.0 and mine is 9.1(damn knee surgery and yes I know my profile is wrong). Well, I used to just give him 7 strokes. NO MORE.
I play the maroons at 70.4/140 and he plays the greens at 66/122(he's 71yr old).
Computing for course handicap that's 17 for him and 11 for me. 6 strokes.
But then accounting for tee handicap it's 70.4-66=4.4
He's only getting 2 strokes next time we play. This should be fun!
Correct me if I'm wrong please before I make an ass of myself..............
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Originally Posted by Dormie1360
All's good. It's all friendly banter between us.........
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Originally Posted by MEfree
Thanks for the confirmation...last night I replied to an email that the league had sent out and wrote what I thought the handicaps should be and asking what they thought but have not gotten a reply yet. I realize that many might not be aware of the course rating adjustment when different tees are used, but the guys in the shop didn't even seem to know that slope factors into determining your course handicap...I tried explaining it briefly and all he said was that it seemed I knew more than him and said that it doesn't matter because it is the same for everyone.
IMO, this is basic knowledge that all pro shop employees should know, especially if they are in charge of tournaments/events. It makes me wonder what other stuff your guys don't know or what other wrong information they have passed on to other golfers.
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Originally Posted by SCfanatic35
IMO, this is basic knowledge that all pro shop employees should know, especially if they are in charge of tournaments/events. It makes me wonder what other stuff your guys don't know or what other wrong information they have passed on to other golfers.
I agree, but to be fair, I was not talking to a Pro or the league coordinator, just the people present who ring the register and check you in.
With that said, I have a number of pending issues that I would like to get resolved without coming off as an a*****e-
1. I got an email from the Director of Golf saying that he would get with the league coordinator to go over my questions, but have not heard back yet.
2. In addition to the handicap issue in my OP, I asked in the shop if I should post or would they post and was given their i pad to post (after a bit of uncertainty on their part). I later noticed that the last "local rule" on the league sheet said not to post, that they would do it. So in my Wednesday night email, I also asked them not to double post. This resulted in the following:
a. They posted my actual score of 76 instead of my ESC score of 75 BUT with a completely different course rating/slope of 70.0/113 (which as far as I know does not exist from any tees for this course) instead of 66.8/117.
b. After informing them of the double post, they replied " Not sure how the score got double posted as we posted the score on Friday . The rating/slope has been fixed. ". They also changed the date to the 12th, instead of the 10th when I actually played but did not remove either my original posting or their double post.
The score cards have no place for you to write your adjusted score, so my guess is that some league players either write their adjusted score as their actual score and/or those that write their actual score like I did have a higher handicap than they should (Unless those players don't notice when their score gets entered with a rating 3.8 above the course rating like mine did- still can't figure out how that happened.
3. The lack of red and yellow stakes is make for a lot of ambiguity in my opinion. As an example, hole 5 has two different water hazards right of the fairway. Neither is marked with stakes despite the fact that the local rules on the card say lateral hazards are marked with red stakes and water hazards are marked with yellow stakes. As I played 9 today, I asked one of the maintenance guys who was cutting some tall grass leading into the first hazard about this. He replied that the first hazard is lateral and the 2nd is a regular water hazard. I asked him how I would know this without stakes and he replied it is because the first runs lateral to the hole and the second hazard (which runs mostly lateral to the 5th hole but is direct for the 6th) is a regular water hazard because its border is made up of railroad ties. The only mention of railroad ties on the card is that they are an integral part of the course with no relief.
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Hi MeFree,
What golf course are you guys playing at?
The Slope numbers seems very low. Are they for a Nine hole Rating?
These Slopes seem typical of a short yardage course, just over 5000 yards.
As an example of adjustments for the different tees, we use a base from the tees which the majority of players are playing.
Example, in our league the majority of players would be playing your Blue Course then as per the Course rating these players would play to their Course Handicap and 2 stokes would be added to players at the Gold Course and 2 stokes would be subtracted from Players on the White Course.
Adjusting Handicaps Based off the White Course when there are players at White, Blue & Gold, adding strokes to the Blue & Gold players
would be as per USGA guidelines.
It sometimes gets tricky calculating, especially when sometimes there may only be players at Gold and White or Blue and White, or just Gold and Blue.
To keep it simple and accurate, I created an Excel spreadsheet with the calculations based from the Slope/Rating.
I update the Players Index and it calculates each players Course Handicap.
Here is an example. Notice - Some Players have a 3 stroke reduction when they move up to the White or back to the Black and Gold.
Home Course Calculations Name Index Red/Wht White Blu/Wht Blue Black Gold Tony 15.4 12 14 16 17 20 22 Ken na - - - - - - Mike na - - - - - - Mike 14.7 11 13 15 17 19 21 Randy 4.9 1 3 4 6 8 10 Dustin 2.5 -1 1 2 3 5 7 Curtis 21.4 18 21 23 24 26 29 Micheal 17 13 16 18 19 21 24 Joe 14.5 11 13 15 16 19 21 Gary 18.2 14 17 19 20 23 26
The formula for the Blue Course I use is (Index multiplied by Slope, then divide by 113)
The other course formulas add or subtract 2. We have Combo Tee courses and the diff is only 1 stroke.
I post these sheets on the board and everyone can view their Index and Course Handicaps and I keep a copy
handy when I enter scores into our Game Program.
I get questions from guys all the time as to Why, Why, Why do they lose strokes when they play the whites.
I just show them the sheet and say here your are. Saves a lot of headaches explaining the process and equations.
How's your golf game this summer?
Club Rat
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I agree that the slope is low as there is the possibility of a lost ball, OB or hazard on every hole, many on multiple sides. However, it is under 6 K and at altitude so it is short on the card.
My game is ok, but I have been making some big numbers- yesterday I was hit 7 GIR on the front and the fringe on another and still managed to be +7 with 20 putts and a 9 on the only green I missed, hitting it OB off the tee and then flying the green on the approach into a hazard when the wind died momentarily.
A new guy took over the league just before last weeks comp- I talked with him some yesterday but was more concerned with the double posting of my Wednesday score and the lack of stakes on many of the hazards and forgot to ask him how he calculated the HCs.
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Thanks Club Rat...it looks like you are very organized with this.
Originally Posted by Club Rat
Hi MeFree,
As an example of adjustments for the different tees, we use a base from the tees which the majority of players are playing.
Example, in our league the majority of players would be playing your Blue Course then as per the Course rating these players would play to their Course Handicap and 2 stokes would be added to players at the Gold Course and 2 stokes would be subtracted from Players on the White Course.
Club Rat
The way I understand it, if you have 3 sets of tees involved, the USGA allows you to add strokes to the guys playing the two back sets or subtract strokes from the two front sets. Does the USGA allow you to adjust off the middle set? While adjusting from the middle has the same effect from my course, I could see some scenarios where that comes up with a different result based on rounding.
i.e. Front= 69.0
Middle= 70.4
Back= 72.8
In this case the guys from the back should get an extra 4 shots relative to the guys in the front, but if you adjust from the middle it would only be 3 shots.
Another example
Front= 69.0
Middle= 70.6
Back= 72.2
Here, the guys on the back should get an extra 3 relative to the guys on the front, but would be getting 4 if you adjust from the middle.
Overall, I like the idea of adjusting from the middle and/or the most used set of tees so more guys are playing closer to their actual course handicap, but wonder about the rounding issues in some cases.
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I don't have an answer for you. I'm not sure the USGA contemplates 3 different tees competing at once, all against each other. The problem is just changing the handicaps based on one end of the slope (highest or lowest) adjust all players referencing that slope (highest or lowest). When you have 3 tees competing against each other, the middle player may not have the same number of strokes as if he were just playing against one other player.
In your second example, the handicaps based off of 69.0 would add 2 strokes to the middle player, and 3 strokes to the back player.....a difference of 1 stroke between the middle and back player. If the middle player were just competing against the back player, their handicap difference would be 2 strokes. So, although the middle and back players are competing against each other in both scenarios, their handicap adjustments are not the same.
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Originally Posted by Dormie1360
I don't have an answer for you. I'm not sure the USGA contemplates 3 different tees competing at once, all against each other. The problem is just changing the handicaps based on one end of the slope (highest or lowest) adjust all players referencing that slope (highest or lowest). When you have 3 tees competing against each other, the middle player may not have the same number of strokes as if he were just playing against one other player.
In your second example, the handicaps based off of 69.0 would add 2 strokes to the middle player, and 3 strokes to the back player.....a difference of 1 stroke between the middle and back player. If the middle player were just competing against the back player, their handicap difference would be 2 strokes. So, although the middle and back players are competing against each other in both scenarios, their handicap adjustments are not the same.
It really is an odd setup. We always flighted by handicap, so we were never competing against someone playing from a different tee.
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Every time we host a party at our home, we engage in a brief and spirited… let’s go with the word “discussion”.
In every household around the world, I suppose, this discussion plays out every time there’s a party. One side of the debate will worry about how to fit in the leftovers in the refrigerator the next day, while the other will fret about – the horror! – there not being enough food on the table midway through a meal.
There is, I should mention, no “right” answer over here. Each side makes valid arguments, and each side has logic going for it. Now, me, personally, I quite like the idea of leftovers, because what can possibly be better than waking up at 3 in the morning for no good reason, waddling over to the fridge, and getting a big fat meaty slice of whatever one may find in there? But having been a part of running a household for a decade and change, I know the challenges that leftovers can pose in terms of storage.
You might by now be wondering about where I am going with this, but asking yourself which side of the debate you fall upon when it comes to this specific issue is also a good way to understand why formulating the null hypothesis can be so very challenging.
Let’s assume that there’s going to be four adults and two kids at a party.
How many chapatis should be made?
Should the null hypothesis be: We will eat exactly 16 chapatis tonight
With the alternate then being: 16 chapatis will either be too much or too little
Or should the null hypothesis be: We will eat 20 chapatis or more
With the alternate being: We will definitely eat less than 20 chapatis tonight.
The reason we end up having a “discussion” is because we can’t agree on which outcome we would rather avoid: that of potentially being embarrassed as hosts, or the one of standing, arms exasperatedly akimbo, in front of the refrigerator post-party.
It is the outcome we would rather avoid that guides us in our formation of the null hypothesis, in other words. We give it every chance to be true, and if we reject it, it is because we are almost entirely confident that we are right in rejecting it.
What is “almost entirely“?
That is the point of the “significant at 1%” or “5%” or “10%” sentence in academic papers.
Which, of course, is another way to think about it. This set of the null and the alternate…
H0: We will eat 20 chapatis or more
Ha: We will eat less than 20 chapatis
… I am not ok rejecting the null at even 1%. Or in the language of statistics, I am not ok with committing a Type I error, even at a probability (p-value) of 1%.
A Type I error is rejecting the null when it is true. So even a 1% chance that we and our guests would have wanted to eat more than 20 chapatis* to me means that we should get more than 20 chapatis made.
At this point in our discussions (we’re both economists, so these discussions really do take place at our home), my wife exasperatedly points out that not once has the food actually fallen short.
Ah, I say, triumphantly. Can you guarantee that it won’t this time around? 100% guarantee?
No? So you’re saying there’s a teeny-tiny 1% chance that we’ll have too few chapatis?
Well, then.
Boss.
*Don’t judge us, ok. Sometimes the curry just is that good.
## 4 thoughts on “Team “Kam Nahi Padna Chahiye””
1. Don’t worry about the quantity, always keep in excess, it’s the sacred duty of the child in every Indian household to finish the food that’s left๐๐
2. Anurag Asawa says:
Exactly, I use the similar example. At least for the estimation/ estimated and estimates. How to decide the number of chapati to be made_ Estimation of the number of people, age of them, eating habits, timing of arrival etc are the point of discussion whenever we try to ESTIMATE the amount and eating material.
No way of fixing a significant level, it has to be highly significant, which is very unlikely. Then what level of significance would be acceptable?
Fight starts at home
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Worksheet 333
A typical small rescue helicopter, like the one in the Figure below, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?
The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)
Strategy
Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.
Solution for (a)
We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find Er . The angular velocity ω for 1 r.p.m is
Angular velocity
and for 300 r.p.m
Multiplication
The moment of inertia of one blade will be that of a thin rod rotated about its end.
Moment of Inertia - Rod end
The total I is four times this moment of inertia, because there are four blades. Thus,
Multiplication
and so The rotational kinetic energy is
Rotational energy
Solution for (b)
Translational kinetic energy is defined as
Kinetic energy ( related to the object 's velocity )
To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is
Division
Solution for (c)
At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:
Potential energy
Discussion
The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
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In the case of longitudinal harmonic sound waves, the wavelength can be calculated by the distance the point has traveled from the wave’s source, the ... more
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https://www.free-online-converters.com/marks/marks-into-carats-metric.html | 1,653,813,588,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00764.warc.gz | 883,113,686 | 9,244 | Free Online Converters > Convert Marks Into Carats (metric)
Here you can Convert units of Marks to Carats (metric) units, find all information about Marks. So, enter your unit's value in Left Column like Marks(if you use standard resolution on most non-HD laptops. FULL HD resolution starts at 1920 x 1080). Otherwise, if you use a lower value, enter the value in the box above. The Result / another converted unit value shell appears in the Left or below Column.
# Convert Marks Into Carats (metric)
Marks
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##### conversion Table / conversion Chart
1 Marks = 1244.1391 Carats (metric)
2 Marks = 2488.2782 Carats (metric)
3 Marks = 3732.4173 Carats (metric)
4 Marks = 4976.5564 Carats (metric)
5 Marks = 6220.6955 Carats (metric)
6 Marks = 7464.8346 Carats (metric)
7 Marks = 8708.9737 Carats (metric)
8 Marks = 9953.1128 Carats (metric)
9 Marks = 11197.2519 Carats (metric)
10 Marks = 12441.391 Carats (metric)
11 Marks = 13685.5301 Carats (metric)
12 Marks = 14929.6692 Carats (metric)
13 Marks = 16173.8083 Carats (metric)
14 Marks = 17417.9474 Carats (metric)
15 Marks = 18662.0865 Carats (metric)
16 Marks = 19906.2256 Carats (metric)
17 Marks = 21150.3647 Carats (metric)
18 Marks = 22394.5038 Carats (metric)
19 Marks = 23638.6429 Carats (metric)
20 Marks = 24882.782 Carats (metric)
21 Marks = 26126.9211 Carats (metric)
22 Marks = 27371.0602 Carats (metric)
23 Marks = 28615.1993 Carats (metric)
24 Marks = 29859.3384 Carats (metric)
25 Marks = 31103.4775 Carats (metric)
26 Marks = 32347.6166 Carats (metric)
27 Marks = 33591.7557 Carats (metric)
28 Marks = 34835.8948 Carats (metric)
29 Marks = 36080.0339 Carats (metric)
30 Marks = 37324.173 Carats (metric)
31 Marks = 38568.3121 Carats (metric)
32 Marks = 39812.4512 Carats (metric)
33 Marks = 41056.5903 Carats (metric)
34 Marks = 42300.7294 Carats (metric)
35 Marks = 43544.8685 Carats (metric)
36 Marks = 44789.0076 Carats (metric)
37 Marks = 46033.1467 Carats (metric)
38 Marks = 47277.2858 Carats (metric)
39 Marks = 48521.4249 Carats (metric)
40 Marks = 49765.564 Carats (metric)
41 Marks = 51009.7031 Carats (metric)
42 Marks = 52253.8422 Carats (metric)
43 Marks = 53497.9813 Carats (metric)
44 Marks = 54742.1204 Carats (metric)
45 Marks = 55986.2595 Carats (metric)
46 Marks = 57230.3986 Carats (metric)
47 Marks = 58474.5377 Carats (metric)
48 Marks = 59718.6768 Carats (metric)
49 Marks = 60962.8159 Carats (metric)
50 Marks = 62206.955 Carats (metric)
50 Marks = 62206.955 Carats (metric)
51 Marks = 63451.0941 Carats (metric)
52 Marks = 64695.2332 Carats (metric)
53 Marks = 65939.3723 Carats (metric)
54 Marks = 67183.5114 Carats (metric)
55 Marks = 68427.6505 Carats (metric)
56 Marks = 69671.7896 Carats (metric)
57 Marks = 70915.9287 Carats (metric)
58 Marks = 72160.0678 Carats (metric)
59 Marks = 73404.2069 Carats (metric)
60 Marks = 74648.346 Carats (metric)
61 Marks = 75892.4851 Carats (metric)
62 Marks = 77136.6242 Carats (metric)
63 Marks = 78380.7633 Carats (metric)
64 Marks = 79624.9024 Carats (metric)
65 Marks = 80869.0415 Carats (metric)
66 Marks = 82113.1806 Carats (metric)
67 Marks = 83357.3197 Carats (metric)
68 Marks = 84601.4588 Carats (metric)
69 Marks = 85845.5979 Carats (metric)
70 Marks = 87089.737 Carats (metric)
71 Marks = 88333.8761 Carats (metric)
72 Marks = 89578.0152 Carats (metric)
73 Marks = 90822.1543 Carats (metric)
74 Marks = 92066.2934 Carats (metric)
75 Marks = 93310.4325 Carats (metric)
76 Marks = 94554.5716 Carats (metric)
77 Marks = 95798.7107 Carats (metric)
78 Marks = 97042.8498 Carats (metric)
79 Marks = 98286.9889 Carats (metric)
80 Marks = 99531.128 Carats (metric)
81 Marks = 100775.2671 Carats (metric)
82 Marks = 102019.4062 Carats (metric)
83 Marks = 103263.5453 Carats (metric)
84 Marks = 104507.6844 Carats (metric)
85 Marks = 105751.8235 Carats (metric)
86 Marks = 106995.9626 Carats (metric)
87 Marks = 108240.1017 Carats (metric)
88 Marks = 109484.2408 Carats (metric)
89 Marks = 110728.3799 Carats (metric)
90 Marks = 111972.519 Carats (metric)
91 Marks = 113216.6581 Carats (metric)
92 Marks = 114460.7972 Carats (metric)
93 Marks = 115704.9363 Carats (metric)
94 Marks = 116949.0754 Carats (metric)
95 Marks = 118193.2145 Carats (metric)
96 Marks = 119437.3536 Carats (metric)
97 Marks = 120681.4927 Carats (metric)
98 Marks = 121925.6318 Carats (metric)
99 Marks = 123169.7709 Carats (metric)
100 Marks = 124413.91 Carats (metric)
101 Marks = 125658.0491 Carats (metric)
102 Marks = 126902.1882 Carats (metric)
103 Marks = 128146.3273 Carats (metric)
104 Marks = 129390.4664 Carats (metric)
105 Marks = 130634.6055 Carats (metric)
106 Marks = 131878.7446 Carats (metric)
107 Marks = 133122.8837 Carats (metric)
108 Marks = 134367.0228 Carats (metric)
109 Marks = 135611.1619 Carats (metric)
110 Marks = 136855.301 Carats (metric)
111 Marks = 138099.4401 Carats (metric)
112 Marks = 139343.5792 Carats (metric)
113 Marks = 140587.7183 Carats (metric)
114 Marks = 141831.8574 Carats (metric)
115 Marks = 143075.9965 Carats (metric)
116 Marks = 144320.1356 Carats (metric)
117 Marks = 145564.2747 Carats (metric)
118 Marks = 146808.4138 Carats (metric)
119 Marks = 148052.5529 Carats (metric)
120 Marks = 149296.692 Carats (metric)
121 Marks = 150540.8311 Carats (metric)
122 Marks = 151784.9702 Carats (metric)
123 Marks = 153029.1093 Carats (metric)
124 Marks = 154273.2484 Carats (metric)
125 Marks = 155517.3875 Carats (metric)
126 Marks = 156761.5266 Carats (metric)
127 Marks = 158005.6657 Carats (metric)
128 Marks = 159249.8048 Carats (metric)
129 Marks = 160493.9439 Carats (metric)
130 Marks = 161738.083 Carats (metric)
131 Marks = 162982.2221 Carats (metric)
132 Marks = 164226.3612 Carats (metric)
133 Marks = 165470.5003 Carats (metric)
134 Marks = 166714.6394 Carats (metric)
135 Marks = 167958.7785 Carats (metric)
136 Marks = 169202.9176 Carats (metric)
137 Marks = 170447.0567 Carats (metric)
138 Marks = 171691.1958 Carats (metric)
139 Marks = 172935.3349 Carats (metric)
140 Marks = 174179.474 Carats (metric)
141 Marks = 175423.6131 Carats (metric)
142 Marks = 176667.7522 Carats (metric)
143 Marks = 177911.8913 Carats (metric)
144 Marks = 179156.0304 Carats (metric)
145 Marks = 180400.1695 Carats (metric)
146 Marks = 181644.3086 Carats (metric)
147 Marks = 182888.4477 Carats (metric)
148 Marks = 184132.5868 Carats (metric)
149 Marks = 185376.7259 Carats (metric)
150 Marks = 186620.865 Carats (metric)
## how many Marks Into Carats (metric)
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# 2 A car moving at 40 km h is to be stopped by applying
Last updated: 6/30/2023
2 A car moving at 40 km h is to be stopped by applying brakes in the next 4 0 m If the car weighs 2000 kg what average force must be applied on it | 72 | 240 | {"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-2024-33 | latest | en | 0.955854 |
https://forum.powerscore.com/lsat/viewtopic.php?t=7870&p=58014 | 1,558,623,442,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257259.71/warc/CC-MAIN-20190523143923-20190523165923-00414.warc.gz | 481,239,747 | 9,411 | ## #17 - It is the mark of a superior conductor that he or she
Administrator
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Complete Question Explanation
Assumption. The correct answer choice is (D)
In this stimulus the conclusion is the opening sentence, which is a characteristic of "superior" conductors. In order to be considered "superior" a conductor must be have the authority to insist on more intense practice. She can only get this authority, according to the last sentence, if the orchestra respects her artistic interpretation of the current piece. Apply the Negate and Destory technique to the answer choices and see which answer destroys this argument.
Answer Choice (A): This answer choice is incorrect because when it is negated it does not have an affect on the argument. Whether or not the conductors devise different interpretations does not relate to the argument at hand because the argument mentions specifically "the current piece."
Answer Choice (B): This answer choice is incorrect because it is also unrelated to the argument at hand. Negated, it does not affect the argument. Whether or not the conductors are satisfied is irrelevant; the argument is concerned with how the orchestra views her.
Answer Choice (C): This answer choice is incorrect because it does not affect the stimulus when negated. The argument says the conductor has the authority "to insist" on extra practice. Therefore, if top orchestras are not ready, then the conductor must not be superior. It does not destroy the argument.
Answer Choice (D): This is the correct answer choice. If we negate this answer and say that top orchestras cannot appreciate the merits of interpretation before full realization, then how can a conductor ever win their respect? The two idea are contradictory: the conductor can only insist on extra practice if the orchestra appreciates his finished work. Negated, this answer choice destroys the argument.
Answer Choice (E): This answer choice is incorrect because it does not destroy the argument when negated. If top orchestras are always led by superior conductors, it does not affect the logic about what gives a conductor the authority that makes him "superior."
Nfontes93
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Joined: Wed Oct 28, 2015 12:19 pm
Points: 4
This question really has me stuck! I pick D, but only through POE; on its own, I don't really understand why it is the correct answer. Can someone explain this to me?
"It is the mark of a superior conductor that he or she has the authority to insist, even with a top orchestra, that rehearsal work must be intensified..." [Remainder of question content removed in order to comply with LSAC copyright restrictions.]
Clay Cooper
PowerScore Staff
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Hi NFontes,
Thanks for your question, it is a good one.
Answer choice D is correct because the author's argument doesn't make any sense unless one assumes that the orchestra can know, while still rehearsing, whether what they are rehearsing will be any good.
The author asserts that the authority of a top conductor cannot simply be claimed, but instead that it develops when an orchestra respects the conductor for the quality of the artistic interpretations he or she is currently pursuing; however, if answer choice D were not true and and it were the case that this orchestra cannot predict the quality of those interpretations before the rehearsal becomes a finished performance, this claim (about where the conductor derives there authority) would be absurd.
Thus, the negated form of answer choice D becomes a powerful attack on the author's argument about how conductors earn authority.
Lawyered
LSAT Apprentice
Posts: 23
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Can someone explain the Negate and Destroy Technique?
Do we negate the conclusion and destroy anything to the contrary?
Seems like we negate the answer choices and the one that's most incorrect/faulty to our stem is the right one???
Administrator wrote:Complete Question Explanation
Assumption. The correct answer choice is (D)
In this stimulus the conclusion is the opening sentence, which is a characteristic of "superior" conductors. In order to be considered "superior" a conductor must be have the authority to insist on more intense practice. She can only get this authority, according to the last sentence, if the orchestra respects her artistic interpretation of the current piece. Apply the Negate and Destory technique to the answer choices and see which answer destroys this argument.
Answer Choice (A): This answer choice is incorrect because when it is negated it does not have an affect on the argument. Whether or not the conductors devise different interpretations does not relate to the argument at hand because the argument mentions specifically "the current piece."
Answer Choice (B): This answer choice is incorrect because it is also unrelated to the argument at hand. Negated, it does not affect the argument. Whether or not the conductors are satisfied is irrelevant; the argument is concerned with how the orchestra views her.
Answer Choice (C): This answer choice is incorrect because it does not affect the stimulus when negated. The argument says the conductor has the authority "to insist" on extra practice. Therefore, if top orchestras are not ready, then the conductor must not be superior. It does not destroy the argument.
Answer Choice (D): This is the correct answer choice. If we negate this answer and say that top orchestras cannot appreciate the merits of interpretation before full realization, then how can a conductor ever win their respect? The two idea are contradictory: the conductor can only insist on extra practice if the orchestra appreciates his finished work. Negated, this answer choice destroys the argument.
Answer Choice (E): This answer choice is incorrect because it does not destroy the argument when negated. If top orchestras are always led by superior conductors, it does not affect the logic about what gives a conductor the authority that makes him "superior."
Alexandra Ruby
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HI Lawyered,
Yes, you are correct. You negate the answer choice, not the conclusion in the stimulus, and then plug the negated answer choice into the stimulus. If it destroys the argument, then that is the correct answer choice.
But it's really better practice not to go through all five answer choices using this technique. Rather, narrow your answer choices to 2-3 contenders and then use the negation technique on those remaining answer choices.
Hope this helps!
Lawyered
LSAT Apprentice
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Joined: Tue Jun 13, 2017 12:27 pm
Points: 25
Yes! Absolutely. Just making sure I had my steps right.
Thank you!
niki
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Hi,
I made this into a conditional relationship where if authority -> orchestra respects artistic interpretations currently pursued.
As such, when I negated answer choice D, it resulted in the contrapositive of the above conditional relationship. Thus, I knew D was the correct answer.
Was this a legitimate way of finding the answer choice?
Adam Tyson
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I think that is legit, niki! The "must" in the stimulus could be treated conditionally, and that would yield the relationship you diagrammed. The negation of D is a little challenging to twist into a conditional framework, in my view, because "before" isn't really conditional, just temporal, but if that got you to where you needed to be, in this case I would say it's okay that you went there. Just be careful about trying to force things into conditional relationships - just because you can doesn't mean you should, and sometimes it's better to just focus on the "gap" in the argument and look for something that connects the disconnected elements.
Good work!
Adam M. Tyson
PowerScore LSAT, GRE, ACT and SAT Instructor
Follow me on Twitter at https://twitter.com/LSATadam
hwkim93
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Hello.
I had a quick question regarding diagramming the first sentence of the stimulus.
I diagrammed it as follows:
superior conductor authority to insist
But I was wondering if the "mark" had instead been "surest mark" (which I saw in PT 3 Section 4 #9), would it have been correct to have diagrammed it with the sufficient and necessary conditions flipped?
authority to insist superior condcutor
My reasoning was that while "mark" simply entails that the authority to insist was a characteristic of a superior conductor that other professions could possibly hold, "surest mark" would imply that the characteristic is most characteristic of a superior conductor and thus be rightfully deemed a sufficient condition.
I hope that made sense.
Brook Miscoski
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kim,
I think that your concern is whether a slight wording difference could complicate the identification of the correct answer choice. I would like to address that before moving on to the question about "surest."
Even if the relationship between the conductor and authority were flipped, the second sentence still establishes the claim that authority depends on obtaining respect for artistic interpretations. Answer Choice D is still required (an assumption) for that claim, and it is still true that none of the other answer choices are good candidates. So, you shouldn't be worried about this issue, if you are.
Getting to the issue of how "surest" might change the scenario...
Currently, the author's claim is absolute--a superior conductor has "THE mark" (emphasis added). So it is fair to propose the conditional statement that you have made.
If you added "surest," the author's claim would be less absolute. I would not say that "the surest mark" is either necessary or sufficient. On December 1991 LR Section IV, #9, the author stated that the "surest mark" of self confidence was self-deprecation, and that self-deprecation was even more revealing than letting others poke fun. That lets you know that self-deprecation and the willingness to hear stories about yourself have a relationship to self confidence, and answer choice A to that question is the only answer choice that has anything to do with the relationship established by the stimulus. It is not a stellar example of conditional reasoning, which is why the question stem asks you what is "MOST supported" (emphasis added). | 2,296 | 10,647 | {"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-2019-22 | latest | en | 0.911891 |
https://www.convertunits.com/from/micrometer/to/vershok | 1,619,178,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00286.warc.gz | 773,471,613 | 16,921 | ## ››Convert micrometre to vershok
micrometer vershok
Did you mean to convert micrometre micron to vershok
How many micrometer in 1 vershok? The answer is 44450.
We assume you are converting between micrometre and vershok.
You can view more details on each measurement unit:
micrometer or vershok
The SI base unit for length is the metre.
1 metre is equal to 1000000 micrometer, or 22.497187851519 vershok.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between micrometres and vershok.
Type in your own numbers in the form to convert the units!
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## ››Definition: Micrometer
A micrometre (American spelling: micrometer, symbol µm) is an SI unit of length equal to one millionth of a metre, or about a tenth of the size of a droplet of mist or fog. It is also commonly known as a micron, although that term is officially outdated. It can be written in the expanded mathmatical notation (1×10-6 m)
The symbol µ is the "micro sign", which should look identical to the Greek letter mu (?) (the two may or may not look the same, depending on the font). The symbol "um" is sometimes used, when the µ and ? are not available, for example when using a typewriter.
The micrometre is a common unit of measurement for wavelengths of infrared radiation. Some people (especially in astronomy and the semiconductor business) use the old name micron and/or the solitary symbol µ (both of which were official between 1879 and 1967) to denote a micrometre. This practice persists in the face of official discouragement, perhaps to help disambiguate between the unit of measurement and the micrometer, a measuring device.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 557 | 2,318 | {"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} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.89168 |
https://en.m.wikipedia.org/wiki/%E2%A7%A0 | 1,586,490,646,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00356.warc.gz | 458,974,515 | 14,458 | # d'Alembert operator
(Redirected from )
In special relativity, electromagnetism and wave theory, the d'Alembert operator (denoted by a box: ${\displaystyle \Box }$), also called the d'Alembertian, wave operator, or box operator is the Laplace operator of Minkowski space. The operator is named after French mathematician and physicist Jean le Rond d'Alembert.
In Minkowski space, in standard coordinates (t, x, y, z), it has the form
{\displaystyle {\begin{aligned}\Box &=\partial ^{\mu }\partial _{\mu }=g^{\mu \nu }\partial _{\nu }\partial _{\mu }={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-{\frac {\partial ^{2}}{\partial x^{2}}}-{\frac {\partial ^{2}}{\partial y^{2}}}-{\frac {\partial ^{2}}{\partial z^{2}}}\\&={\frac {1}{c^{2}}}{\partial ^{2} \over \partial t^{2}}-\nabla ^{2}={\frac {1}{c^{2}}}{\partial ^{2} \over \partial t^{2}}-\Delta ~~.\end{aligned}}}
Here ${\displaystyle \nabla ^{2}:=\Delta }$ is the 3-dimensional Laplacian and gμν is the inverse Minkowski metric with
${\displaystyle g_{00}=1}$, ${\displaystyle g_{11}=g_{22}=g_{33}=-1}$, ${\displaystyle g_{\mu \nu }=0}$ for ${\displaystyle \mu \neq \nu }$.
Note that the μ and ν summation indices range from 0 to 3: see Einstein notation. We have assumed units such that the speed of light c = 1.
(Some authors alternatively use the negative metric signature of (− + + +), with ${\displaystyle g_{00}=-1,\;g_{11}=g_{22}=g_{33}=1}$.)
Lorentz transformations leave the Minkowski metric invariant, so the d'Alembertian yields a Lorentz scalar. The above coordinate expressions remain valid for the standard coordinates in every inertial frame.
## The box symbol (${\displaystyle \Box }$) and alternate notations
There are a variety of notations for the d'Alembertian. The most common are the box symbol ${\displaystyle \Box }$ (Unicode: U+2610 BALLOT BOX) whose four sides represent the four dimensions of space-time and the box-squared symbol ${\displaystyle \Box ^{2}}$ which emphasizes the scalar property through the squared term (much like the Laplacian). This symbol is sometimes called the quabla (cf. nabla symbol). In keeping with the triangular notation for the Laplacian, sometimes ${\displaystyle \Delta _{M}}$ is used.
Another way to write the d'Alembertian in flat standard coordinates is ${\displaystyle \partial ^{2}}$ . This notation is used extensively in quantum field theory, where partial derivatives are usually indexed, so the lack of an index with the squared partial derivative signals the presence of the d'Alembertian.
Sometimes the box symbol is used to represent the four-dimensional Levi-Civita covariant derivative. The symbol ${\displaystyle \nabla }$ is then used to represent the space derivatives, but this is coordinate chart dependent.
## Applications
The wave equation for small vibrations is of the form
${\displaystyle \Box _{c}u\left(x,t\right)\equiv u_{tt}-c^{2}u_{xx}=0~,}$
where u(x, t) is the displacement.
The wave equation for the electromagnetic field in vacuum is
${\displaystyle \Box A^{\mu }=0}$
where Aμ is the electromagnetic four-potential in Lorentz gauge.
The Klein–Gordon equation has the form
${\displaystyle \left(\Box +m^{2}\right)\psi =0~.}$
## Green's function
The Green's function, ${\displaystyle G\left({\tilde {x}}-{\tilde {x}}'\right)}$ , for the d'Alembertian is defined by the equation
${\displaystyle \Box G\left({\tilde {x}}-{\tilde {x}}'\right)=\delta \left({\tilde {x}}-{\tilde {x}}'\right)}$
where ${\displaystyle \delta \left({\tilde {x}}-{\tilde {x}}'\right)}$ is the multidimensional Dirac delta function and ${\displaystyle {\tilde {x}}}$ and ${\displaystyle {\tilde {x}}'}$ are two points in Minkowski space.
A special solution is given by the retarded Green's function which corresponds to signal propagation only forward in time[1]
${\displaystyle G\left({\vec {r}},t\right)={\frac {1}{4\pi r}}\Theta (t)\delta \left(t-{\frac {r}{c}}\right)}$
where ${\displaystyle \Theta }$ is the Heaviside step function. | 1,167 | 4,005 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 25, "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-2020-16 | latest | en | 0.736989 |
https://fr.mathworks.com/matlabcentral/cody/problems/1433-mirror-image-matrix-across-anti-diagonal/solutions/1891423 | 1,597,424,809,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739347.81/warc/CC-MAIN-20200814160701-20200814190701-00346.warc.gz | 321,813,176 | 15,975 | Cody
Problem 1433. Mirror Image matrix across anti-diagonal
Solution 1891423
Submitted on 4 Aug 2019 by Ed P.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = 3; y_correct = [1 2 3;2 3 2; 3 2 1]; assert(isequal(mirror_anti_diag(x),y_correct))
Y = 1 Y = 1 2 Y = 1 2 3 Y = 1 2 3 2 0 0 Y = 1 2 3 2 3 0 Y = 1 2 3 2 3 4 Y = 1 2 3 2 3 2 3 0 0 Y = 1 2 3 2 3 2 3 4 0 Y = 1 2 3 2 3 2 3 2 5
2 Pass
x = 1; y_correct = [1]; assert(isequal(mirror_anti_diag(x),y_correct))
Y = 1
3 Pass
x = 4; y_correct = [1 2 3 4; 2 3 4 3; 3 4 3 2; 4 3 2 1 ]; assert(isequal(mirror_anti_diag(x),y_correct))
Y = 1 Y = 1 2 Y = 1 2 3 Y = 1 2 3 4 Y = 1 2 3 4 2 0 0 0 Y = 1 2 3 4 2 3 0 0 Y = 1 2 3 4 2 3 4 0 Y = 1 2 3 4 2 3 4 5 Y = 1 2 3 4 2 3 4 3 3 0 0 0 Y = 1 2 3 4 2 3 4 3 3 4 0 0 Y = 1 2 3 4 2 3 4 3 3 4 5 0 Y = 1 2 3 4 2 3 4 3 3 4 3 6 Y = 1 2 3 4 2 3 4 3 3 4 3 2 4 0 0 0 Y = 1 2 3 4 2 3 4 3 3 4 3 2 4 5 0 0 Y = 1 2 3 4 2 3 4 3 3 4 3 2 4 3 6 0 Y = 1 2 3 4 2 3 4 3 3 4 3 2 4 3 2 7 | 671 | 1,079 | {"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} | 2.984375 | 3 | CC-MAIN-2020-34 | latest | en | 0.493206 |
https://www.physicsforums.com/threads/expectation-values-in-momentum-position-space.25006/ | 1,539,918,761,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00225.warc.gz | 1,023,738,118 | 13,750 | # Expectation values in momentum/position space?
1. May 10, 2004
### clumsy9irl
I'm currently, yet again, left clueless by a problem.
See: http://idefix.physik.uni-freiburg.de/~aufgabe/QMI2004/qm3/node1.html [Broken]
Ok, so they give the Psi(x) in position space, and the first question is to give the corresponding, normalized wave equation in momentum space.
You do a fourier transform, no?
Psi(p)=Int ((1/sqrt2pi) Psi(x) (e^ipx/hbar) dx
no?
From that, I get Psi(p)=(2hbar/(p sqrt2pi)) sin (pa/hbar)
or....?
After that, I tried to get the expectation value, <p>, but with poor results. Integrate Psi* Psi p dp, yeah? What I get is 4hbar^2/(2pi p^2) [ln|p| sin^2(pa/hbar)]
Now, I know this has to be wrong, but I have no idea what I'm doing to screw it up.
I tried to do <p^2> and <x> and it just gets worse, and messier. WHAT AM I DOING WRONG??
Pleease help. Thank you :)
edit: I think I've got it! Well, at least part of it. I've gotten my expectation values for p and x to be zero, and I think my x^2 and p^2 are wrong, but... I can't think of anything else... so, it's staying right now with
<p^2>= (2hbarp*sinpa)/pi
<x^2>= A^2 (2a^3/3)
Working on the rest, still... but coming a bit further?
Last edited by a moderator: May 1, 2017 | 400 | 1,256 | {"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} | 3.203125 | 3 | CC-MAIN-2018-43 | latest | en | 0.908668 |
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# If a, b, and c are positive numbers, is a < b < c?
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If a, b, and c are positive numbers, is a < b < c? [#permalink]
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13 Apr 2012, 21:21
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If a, b, and c are positive numbers, is a < b < c?
(1) ab = bc
(2) ac = bc
[Reveal] Spoiler: OA
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Re: Arithmetic [#permalink]
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13 Apr 2012, 23:14
Smita04 wrote:
If a, b, and c are positive numbers, is a < b < c?
(1) ab = bc
(2) ac = bc
statement 1:
b(a-c)=0
i.e. either b=0 or a=c
given b>0
hence a=c
=> a IS NOT LESS THAN c.
sufficient
statement 2:
similarly by above reasoning a=b
so NO
hence SUfficient
hence D
HOPE THIS HELPS...!!
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Re: Arithmetic [#permalink]
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14 Apr 2012, 00:45
If a, b, and c are positive numbers, is a < b < c?
(1) ab = bc --> since b>0 we can reduce given equation by b and we'll get a = c, so a < b < c is not true. Sufficient.
(2) ac = bc --> since c>0 we can reduce given equation by c and we'll get a = b, so a < b < c is not true. Sufficient.
Answer: D.
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Re: Arithmetic [#permalink] 14 Apr 2012, 00:45
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# If a, b, and c are positive numbers, is a < b < c?
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,052 | 3,425 | {"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-2017-09 | longest | en | 0.817339 |
https://homework.cpm.org/category/CON_FOUND/textbook/mc2/chapter/3/lesson/3.1.3/problem/3-31 | 1,726,526,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00289.warc.gz | 268,159,627 | 15,104 | ### Home > MC2 > Chapter 3 > Lesson 3.1.3 > Problem3-31
3-31.
Find the mean, range, and median of the values: $12$, $4$, $-2$, $0$, $9$, $-2$, $1$, $7$, $8$, $2$. Recall that the range is calculated by finding the difference between the largest and smallest data values.
The mean is the arithmetic average of the data set. The median is the middle number in a set of data arranged numerically. | 117 | 397 | {"found_math": true, "script_math_tex": 10, "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} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.83656 |
https://newey.me/fixing-the-ux-of-ip-addresses/ | 1,539,714,204,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510866.52/warc/CC-MAIN-20181016180631-20181016202131-00232.warc.gz | 754,164,730 | 6,504 | Remembering IP addresses has long been a pain point for system administrators. They’re long, clunky and difficult to remember - essentially a 12-digit phone number for internet-connected machines. Trying to remember the IPs of the endless reams of servers that you manage (was it x.x.x.y? or was it y.y.y.z?) is difficult - after all, that’s why we have DNS.
Trying to remember or communicate all 12 digits (which can be up to 20 syllables long!) is a cognitively demanding task. When we examine the limits of short term memory, it appears that the average person is capable of recalling between 5 and 9 abstract pieces of information - a figure which can easily be reached when trying to remember one or two IPs.
What if we could make IPs easier to remember by reducing the cognitive overhead? What if, instead of some abstract and oft-meaningless list of numbers, we could turn IPs into real, pronounceable words? What if, instead of 20 syllables, we could reduce an IP address to 8 syllables (or fewer) - and make it pronounceable?
There are several approaches that we can take to solve this problem:
• Map the digits of an IP to pronounceable syllables
• Use a pre-defined list of unique words to map IP chunks to words
In this blog post, we’ll look at each approach and see which works best.
## IPs as syllable groups
OK, so let’s look at our first option; mapping individual digits to syllables. Let’s look at some numbers. In the case of IPv4 addresses (I’ll refer to IPv4 addresses as IP addresses from this point) there are four groups of integers, ranging between 0 and 255 respectively. That gives us 256 possibilities for each digit - giving us 256⁴ possible combinations for an IP address, or 4,294,967,296 combinations. Obviously, there aren’t that many pronounceable syllables (or in fact, words in the English language) - so mapping a single syllable to an IP address as a whole is out of the question. That seems obvious.
However - what if we reduce each group in an IP address to an arbitrary base (e.g. base-16), and map each resulting digit to a syllable? Conveniently, in base-16, any integer between 0 and 255 can be represented by only two digits. For example, in base 16, the IP address 127.0.0.1 looks like this: 7F.0.0.1. The IP address 45.63.43.64 looks like this: 2D.3F.2B.40. Mapping these individual digits to syllables will give us four two-syllable “words”, which will hopefully make for a easier-to-pronounce IP address. Something like this, perhaps;
Digit Syllable Digit Syllable
0 ba 8 le
1 bo 9 lu
2 che 10 (Hex: A) mo
3 do 11 (Hex: B) na
4 fu 12 (Hex: C) ra
5 ra 13 (Hex: D) te
6 ko 14 (Hex: E) zo
7 ka 15 (Hex: F) zu
Using this format, the IP 127.0.0.1 would become kazu.ba.ba.bo, and so on. This is better! I’ll show a few more here;
IP Base-16 Number Syllable-Encoded Result
127.0.0.1
7F.0.0.1
kazu.ba.ba.bo
8.8.8.8
8.8.8.8
le.le.le.le
45.63.43.64
2D.3F.2B.40
chete.dozu.chena.fuba
However, these semi-meaningless strings of syllables are a little abstract - definitely still quite difficult to remember, even if they are a little easier to pronounce. This brings us to the next possibility.
## IPs as word pairs
Sticking with the a similar principle to before, it’s possible to map words to IP addresses. If we can come up with a long list of distinct words, then it’s possible to map an IP address to a unique phrase (à la “correct horse battery staple”). Better yet, if we can come up with a list of 65,536 different words (65,536 is 256²), we can reduce the number of words needed to represent an IP address by a factor of two - that is, we can represent an IP address with only two words! Notably, this is somewhat similar to the naming scheme for URLs at GFYCat (e.g. “https://gfycat.com/AthleticTinyBeauceron”).
Let’s give this a try. I have acquired a list of ~91,000 nouns from here. I have then converted each word to lowercase, ensured there were no duplicate words, selected 65,536 words from the list entirely at random (using the shuf -n 65536 Unix command), and written this out to a new file (available here). Now we can now start thinking about converting IPs to phrases.
We can essentially treat an IP address (4 groups of integers between 0-255) as a base-256 number with 4 digits. What we need to do is convert that base-256 number to base-65536 - which, conveniently, can represent an entire IP address in two digits. Once that’s done, it’s easy to look up the corresponding word for each digit. For example, in base-65536, the IP address 127.0.0.1 becomes 32512.1, 8.8.8.8 becomes 2056.2056, 45.63.43.64 becomes 11583.11072, and so on. If we have that list of 65,536 words handy, then it’s quite simple to look up the word corresponding with each digit in the base-65536 number, and therefore come up with a memorable phrase to refer to each IP uniquely. For example;
IP Base-65536 Number Word-Encoded Result
127.0.0.1
32512.1
kuomintang.aardvark
8.8.8.8
2056.2056
andorran.andorran
45.63.43.64
11583.11072
coffers.claytonia
I haven’t specifically mentioned IPv6 addresses in this blog post - mostly because it’s late and I should be in bed, but partly because exactly the same principles apply to IPv6 - in fact, if my back-of-the-metaphorical-envelope calculations are correct, it’s possible to represent an IPv6 address by using 8 unique words (assuming we’re using the same 65,536 word dictionary from earlier) - which is definitely an improvement on an incomprehensible 128-bit hexadecimal string. I’ve got a much better chance of remembering 8 words than I do of 32 hexadecimal characters - and let’s face it, any improvement we can make to the usability of IPv6 has to be a Good Thing™.
In fact, let’s try it here.
2001:0DB8:AC10:FE01::::
calibrations:astringents:pinpoint:xantippe::::
2001:19f0:5001:117:5400:ff:fe33:71c3
calibrations:borsch:exclusives:accentor: ...
What have we done? We’ve improved the pronounceability and memorability of the humble IP address by using base-n trickery and large dictionaries of memorable nouns to re-encode them into something a bit… nicer. Not only do these approaches reduce the number of syllables to encode an IP address, but they also increase pronounceability, memorability, and overall friendliness. Isn’t coffers.claytonia much easier to say than 45.63.43.64? I certainly think so. | 1,689 | 6,316 | {"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} | 2.8125 | 3 | CC-MAIN-2018-43 | longest | en | 0.953281 |
https://www.jiskha.com/questions/705155/light-waves-with-two-different-wavelengths-632-nm-and-474-nm-pass-simultaneously-through | 1,611,744,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821381.83/warc/CC-MAIN-20210127090152-20210127120152-00062.warc.gz | 825,791,772 | 5,850 | Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.63 x 10-5 m and strike a screen 1.30 m from the slit. Two diffraction patterns are formed on the screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern?
1. 👍
2. 👎
3. 👁
1. Let’s use the condition of diffraction minimum for one split of the width b
b•sinα =k1•λ1
b•sinα =k2•λ2,
Since we have superposition of two maxima
b•sinα is the same for two wavelengths,
k1•λ1= k2•λ2,
λ1/λ2 = k1/k2, 632/474 = 4/3.
Therefore k1=4, k2=3.
Now
sinα = k1•λ1/b =4•632•10^-9/ 7.63•10^-5 =0.033.
As the angle is very small tanα = sinα = 0.033.
tan α= x/L,
x =L• tanα =1.3•0.033 = 0.043 m = 4/3 cm
1. 👍
2. 👎
2. My mistake.
It should be
k1•λ1= k2•λ2,
λ1/λ2 = k2/k1, 632/474 = 4/3.
k1 =3 k2 = 4
sinα = k1•λ1/b =3•632•10^-9/ 7.63•10^-5 =0.0248
x=L• tanα =L• sinα =1.3•0.0248 =0.0323 m = 3.23 cm
1. 👍
2. 👎
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https://proofwiki.org/wiki/Definition:Differentiability_Class | 1,721,831,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00662.warc.gz | 406,875,807 | 12,439 | # Definition:Differentiability Class
This page is about Differentiability Class in the context of Real Analysis. For other uses, see Class.
## Definition
Let $f: \R \to \R$ be a real function.
Then $\map f x$ is of differentiability class $C^k$ if and only if:
$\dfrac {\d^k} {\d x^k} \map f x \in C$
where $C$ denotes the class of continuous real functions.
That is, $f$ is in differentiability class $k$ if and only if there exists a $k$th derivative of $f$ which is continuous.
If $\dfrac {\d^k} {\d x^k} \map f x$ is continuous for all $k \in \N$, then $\map f x$ is of differentiability class $C^\infty$.
## Specific Instances
### Class Zero
Differentiability class $C^0$ consists of the class of continuous real functions $C$ whether they be differentiable or not.
### Continuously Differentiable
A differentiable function $f$ is continuously differentiable if and only if $f$ is of differentiability class $C^1$.
That is, if the first order derivative of $f$ (and possibly higher) is continuous.
### Smooth Function
A real function is smooth if and only if it is of differentiability class $C^\infty$.
That is, if and only if it admits of continuous derivatives of all orders.
## Domain Restriction
Let $f: \R \to \R$ be a real function.
Let $S \subseteq \R$ be a subset of $\R$ on which the $n$th derivative of $f$ is continuous on $S$.
Then $\map f x$ is of differentiability class $C^n$ on $S$.
## Also known as
A real function in differentiability class $C^n$ can be described as being $n$ times differentiable.
Some authors use $C^{\paren n}$ for $C^n$.
Some use $\mathrm C^{\paren n}$.
Some use $\mathrm C^\omega$ for differentiability class $C^\infty$.
## Examples
### Class $C^0$ Function
Let $f$ be the real function defined as:
$\map f x = \begin {cases} 0 & : x < 0 \\ x & : x \ge 0 \end {cases}$
Then $f \in C^0$ but $f \notin C^1$.
### Class $C^1$ Function
Let $f$ be the real function defined as:
$\map f x = \begin {cases} 0 & : x < 0 \\ x^2 & : x \ge 0 \end {cases}$
Then $f \in C^1$ but $f \notin C^2$.
### Class $C^n$ Function
Let a real function $f$ be required that has the following properties:
$(1): \quad f \in C^n$
$(2): \quad f \notin C^{n + 1}$
where $C^k$ denotes the differentiability class of order $k$.
Then $f$ may be defined as:
$\map f x = \begin {cases} 0 & : x < 0 \\ x^{n + 1} & : x \ge 0 \end {cases}$
### Class $C^0$ Function with Derivative Discontinuous at Point
Let $f$ be the real function defined as:
$\map f x = \begin {cases} x^2 \sin \dfrac 1 x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
Then $f \in C^0$ but $f \notin C^1$.
## Also see
• Results about differentiability classes can be found here. | 835 | 2,695 | {"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} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.771095 |
http://www.umsu.de/wo/2012/582 | 1,521,351,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645538.8/warc/CC-MAIN-20180318052202-20180318072202-00468.warc.gz | 484,896,129 | 3,831 | ## Bader against contingent and occasional identity
In a nice little paper, "The Non-Transitivity of the Contingent and Occasional Identity Relations", Ralf Bader argues that if identity is relative to times or worlds, then it becomes non-transitive and thus no longer qualifies as real identity.
Following Gallois, Bader assumes that a proponent of occasional identity must insist that identity statements are always relativised to a time. Now he considers a case where between times t1 and t2, two objects B and D simultaneously undergo fission in such a way that one fission product of B fuses with one fission product of D. Of the three resulting objects A, C and E, one (C) is a fission product of both B and D. Bader argues that at the initial time t1, it is then true that A=C and C=E, but not that A=E. So identity at t1 is not transitive.
According to Bader, the same problem arises in the modal domain. Here we would have two objects B and C at world w1 such that at w2, B has two counterparts A and C, and D has counterparts C and E. Then at w1, A=C and C=E but not A=E.
(I'm not quite sure why Bader uses the fission-plus-fusion setup rather than a simple case of fusion. Suppose between t1 and t2, B and D fuse into a single object C. According to Bader, it should then be true at t1 that B=C and C=D without B=D.)
In any case, I think the argument does not work. The crucial question is how to interpret the name 'C' at t1 (or w1). The name is introduced to denote an object at t2 which is a counterpart of both B and D at t1. Bader assumes that at t1, 'C' picks out both B and D. Since 'A' uniquely picks out B and 'E' picks out D, one can then truly say 'at t1, A=C and C=E'. But a friend of occasional (or contingent) identity need not agree that this is how to evaluate 'C' in the context of 'at t1'.
In fact, Gallois (who seems to be Bader's main target) disagrees. On pp.106--109 of Occasions of Identity, Gallois argues that a proper name should never pick out more than one thing at any given time. He also suggests that in the context of t1, names like 'C' should be interpreted as Russellian definite descriptions, picking out whatever object is uniquely identical, at t1, to C at t2. Since the uniqueness clause is violated, 'C' is then empty at t1. So it is not true that at t1, A=C and C=E. Similarly, Gibbard and Stalnaker would arguably say that there are several individual concepts (or individuating functions) one might associate with the individual C at w2. Relative to one of them, C satisfies A=x at w1; relative to another, C satisfies x=E at w1. But we never get both A=x and x=E.
I am actually sympathetic to the view, assumed by Bader, that 'C' is multiply referring at t1. As I explained in the previous entry, I also think we shouldn't relativise identity statements to times and worlds. So the transitivity of identity can be expressed in the standard, untensed manner:
if A=C and C=E, then A=C.
But what shall we say about this if 'C' is multiply referring, picking out both A and E? In ordinary language, such cases happen all the time, since names are rarely unique. For example, what shall we say about
if London = the capital of England and London = the 15th largest city in Canada, then the capital of England = the 15th largest city in Canada?
The consequent is clearly false. One might argue that the antecedent is also false, since it is false on every resolution of the ambiguity. This would mean that the conditional is true. But I think we should allow for "mixed resolutions" where different occurrences of an ambiguous word are resolved differently. This way, the antecedent can be true. The whole conditional comes out true on some resolutions and false on others. I think truth-on-some-resolutions is a better candidate for truth than truth-on-all-resolutions, so if I had to tick either "true" or "false", I would still say the conditional is true.
The upshot is that even if we follow Bader and assume that terms like 'C' are multiply referring at t1 or w1 -- which Gallois, Gibbard and Stalnaker would arguably reject -- we do not automatically get false instances of the transitivity principle. It depends on the general rules for interpreting statements with multiply referring terms. Even if someone were to suggest that certain instances of the principle are false -- say, because mixed resolutions should be allowed and because truth is truth-on-all-resolutions -- I don't think we can conclude that this person does not mean identity by 'identity'. | 1,074 | 4,527 | {"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} | 2.890625 | 3 | CC-MAIN-2018-13 | latest | en | 0.948306 |
https://uaapgames.com/lotteries/best-answer-which-is-more-likely-rolling-a-total-of-8-when-two-dice-are-rolled-or-rolling-a-total-of-8.html | 1,660,352,535,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00376.warc.gz | 545,310,821 | 16,518 | # Best answer: Which is more likely rolling a total of 8 when two dice are rolled or rolling a total of 8?
Contents
## What is the probability of rolling a total of 8 on two dice?
Probabilities for the two dice
Total Number of combinations Probability
5 4 11.11%
6 5 13.89%
7 6 16.67%
8 5 13.89%
## Which is more likely rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled?
Which is more likely, rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled? Answer: 9 = 3+6 = 4+5 = 5+4 = 6+3, — 4 ways to get total of 9 points rolling two dice, so the probability that the outcome is 9 points is: 4/(6*6) = 1/9 = 0.111.
## What is the probability of getting a total of 6 or 8 on a roll of a pair of dice?
Two (6-sided) dice roll probability table
Roll a… Probability
6 15/36 (41.667%)
7 21/36 (58.333%)
8 26/36 (72.222%)
9 30/36 (83.333%)
## What is the expected sum of the numbers that appear when three fair dice are rolled?
The expectation of the sum is the sum of the expectation values for the three dice. But since they are all fair dice, they have the same expectation values. Hence E(S) = 3 E(s1). The outcomes for a single fair die are 1,2,3,4,5,6 each with probability 1/6.
## What are the outcomes of rolling 2 dice?
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36.
## What is the probability of rolling a total that is neither 7 nor 11?
Therefore Probablity of sum neither 7 nor 11 is 7/9.
## What is the probability that sum of outcomes on pair of dice is equal to 8?
So the probability is the sum of the five individual probabilities which is 5*(1/36)= 5/36. Therefore the probability that we get the sum as 8 when two dice are thrown is 5/36.
## What is the probability of getting a sum of 9 when two dice are thrown simultaneously?
probability 4/36 or 1/9. In order to get a sum of 9 with two dice, you would have to roll the pairs 4 & 5, 5 & 4, 3 & 6, or 6 & 3.
## What is the probability of getting 1 and 5 If a dice is thrown once?
So they are mutually exclusive events, therefore their probabilities add to 1. By symmetry we expect that each face is equally likely to appear and so each has probability = 1/6. The outcome of a 5 is one of those events and so has probability = 1/6 of appearing.
IT IS INTERESTING: What year did the casino open on the Gold Coast? | 744 | 2,635 | {"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} | 4.5 | 4 | CC-MAIN-2022-33 | latest | en | 0.92211 |
https://www.jiskha.com/display.cgi?id=1368549558 | 1,503,006,076,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104160.96/warc/CC-MAIN-20170817210535-20170817230535-00521.warc.gz | 933,220,550 | 3,559 | # Perform the indicated operation and simplify
posted by .
X^2-2x-15/x^2+x-12*2x^2-6x/x^3+3x^2
• Perform the indicated operation and simplify -
Brackets are absolutely essential here.
I am sure you meant
(x^2-2x-15)/(x^2+x-12)*(2x^2-6x)/(x^3+3x^2)
= (x-5)(x+3)/( (x+4)(x-3) ) * 2x(x-3)/( x^2(x+3))
= 2(x-5)/(x(x+4)) or (2x -10)/(x^2 + 4x) , x ≠ ±3,0
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1. divide and simplify x^2+x-56/x^1-1 / x-7/2x+2 2. find the product in simplest from for x^2+x-6/2x^2-x-3 * x^7-2x^6+4x^5/x^4+8x 3. perform the indicated operations and simplify: 2/3x +4/3x^2 y^4 -6/2x^4 y^5 4. perform the indicated …
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Jason pushes a lawnmower with a force of 160N at an angle of 56° to the lawn. If he pushes the lawnmower 95 meters, how much work has he done?
#### Solution
Using the formula W = Fd cosθ, we can plug in the values we know to get that W = (160) (95) cos56 = 8500J. | 87 | 285 | {"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} | 3.578125 | 4 | CC-MAIN-2019-09 | latest | en | 0.899494 |
http://cozybeehive.blogspot.com/2007/12/field-test-for-calculating-aerodynamic.html | 1,527,113,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00365.warc.gz | 75,656,698 | 26,026 | ## Wednesday, December 26, 2007
### 10A Field Test for Calculating Aerodynamic Drag Area?
We all know and have seen professional cyclists with bent, flat backs over the aerobars or knees closer to the top tube while pedaling. The lore of aerodynamics is that drag increases as square of the velocity, so the faster you go, the double becomes the drag against you. The biggest area exposed to the wind is the cyclist himself, so the largest gains in speed are made if one minimizes or streamlines this exposed area. In mathematical terms, its called CdA.
Traditionally, aero drag area of a cyclist was measured in wind tunnels. But it has been found that the energy expenditures of operating the wind tunnels overwhelm the costs of the measurements. It is a big investment.
Ordinary folks (or extraordinary folks) sit on their saddles wondering how they can calculate/quantify this area. I have many times.
But if you want to sink your head deeper into numbers, there is a paper I discovered written by James C. Martin et. al. (from the University of Utah at Salt Lake) in 2006 describing in detail a field test that could be done by anyone with :
1. Microsoft EXCEL
2. A Power meter like PT or SRM
3. A nice stretch of road or a velodrome
4. Some knack for taking measurements and a pinch patience
Aerodynamic Drag Area of Cyclists Determined With Field Based Measures
can now be accessed on the peer reviewed website, SportSci.Org in the Tests and Technology Section under Biomechanics category. The results can have slight errors compared to a wind tunnel (duh) but according to the paper, they are negligible.
Along with the free paper comes the EXCEL spreadsheet that you can play with!
What I like about this method is the fact that you can immediately get an idea of all the factors that play a role in dictating CdA. And you can get the experience being a bike scientist as well..
* * *
1. Since I don't have a power meter I did something very primitive one day. I was pedling into the wind on the hoods and wanted to see how much difference the drops would make. I focued on maintaining my pedaling force as best I could. As near as I could tell, I could do 2 mph faster when on the hoods given the same exertion. Rough but I believe the number.
2. Chris,
That has been the observation for 100 or so years :)
3. I meant to say 2mph faster when on the drops, but I am sure you figured that out. Not insignificant at all. There are many many recreational cyclist who never ride the drops.
4. Chris,
No sarcasm intended above.
I think some folks don't think about the drops during recreation. They just want to have fun you know. I have a feeling this is more an issue of discomfort, but I'm also believing that discomfort issues aside, the low speeds (~<18mph) involved in casual riding don't quite justify the use of the drops as much.
I tend to think that above a certain threshold forward velocity and perceived effort, its better to go on the drops in order to be more efficient, i.e, getting more out of doing the same.
5. Ron
Interesting post
I guess cycling uphill the wind drag is relatively minimal and its all about gravity.
6. I believe that being in the drops also gives you better handling and control of the bike.
Over a year ago I broadsided a big dog (Rottie mix) that charged our group. I was doing about 26 mph at the time. Rode over the pooch, spun the bike around, and although I thought I was going to be able to save it, high-sided the bike and landed on the grass (luckily). I thought the dog was going to come over and chomp on my leg but he just ran home.
I tacoed the rear wheel but, again, was able to maintain better control of the bike in spite of the (large and moving) speed-bump.
Later I heard the dog was fine, but I am sure he was sore for a while. I've never seen him again when we ride that road.
Just my 2 cents.
7. Luis,
Interesting story. I think you ride for a club. Do you have any provisions for taking action against the owner, should the dog bite off, I don't know...your shoe adjusting lever or something?
There was only once I was chased by an angry black dog, that was all uphill. No point of the hoods then. I wasn't in the best mood to dance the pedals like Pantani either. The best part is that I was doing hill repeats at the time, and I believe I broke my best time on the 2 mile climb. I'm sure the climb was tough on the dog's cardio system so he probably gave up.
Having said that, its nice to carry a small hand pump with you. Should the dog make any nasty moves, you have something to scare away or wield against an attack...
8. Hi Ron,
we don't have any legal provisions per se, so it's every man for himself. Of course I could've sued the owners of the dog, but, as someone said, "You can become the proud owner of a trailer." It seems that these situations usually occur in those type of areas.
Your encounter with a dog reminded me of American Flyers when Kevin Costner takes his "brother" out for a training ride.
Oops, time to get ready for our New Year's Ride. Gotta run... err, pedal :)
Happy New Year!
Luis
9. hi Ron,
me and my team are designing a 'Quad Rider' as a project.The seating arrangement of this rider is same as that of a normal car.The four people together provide the total power for cycling.This rider has no outer body like a car and the riders ride in open.
I wanted to ask you if there is any way in your knowledge to calculate the optimum distance between the drivers seat and the seat beside it for minimum aerodynamic drag.
10. Pranay,
I would think quad bicycles have been made before, if not commonplace. Google tells me so, atleast. You can check it out yourself.
I think you can focus on keeping the riders close to the ground, like in Recumbent designs, or somewhat reclined positions. But with four riders, you have to design with contraints.
Spacing arrangments must be centered around seating and pedalling comfort, you don't want people too close to each other so they can smell each other's sweat and tears, you know that kind of thing. :)
Brainstorm a few ideas out. You can send your concept to me or the bicycle design blog, if this is a student design.
Thank you. I read every single comment. | 1,425 | 6,244 | {"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} | 2.84375 | 3 | CC-MAIN-2018-22 | latest | en | 0.963313 |
http://ask.sagemath.org/question/55611/vector_fieldapply_map-before-and-after-vectordisplay-weird-behaviour/ | 1,618,144,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00082.warc.gz | 9,093,379 | 14,354 | vector_field.apply_map before and after vector.display (weird behaviour)
Hello,
Recently, I have come a cross a very weird behaviour with vector_field.apply_map function. Here is the case:
Consider the following code:
E.<r,th,ph>=EuclideanSpace(coordinates="spherical",start_index=0)
cart.<x,y,z>=E.cartesian_coordinates()
cartf=E.cartesian_frame()
spherf=E.spherical_frame()
E.set_default_frame(cartf)
Now, if I create a vector field, substitute ph with ph_1 and th with th_1, and then display the resultant vector as
var("r_1 th_1 ph_1")
v=E.vector_field([r_1,0,0], frame=spherf, chart=cart);
v.apply_map(lambda c:c.subs(ph==ph_1, th==th_1));show(v.display())
I get 0 printed
However,
If I do the same, but this time if I add show(v.display()) before calling apply_map function as
var("r_1 th_1 ph_1")
v=E.vector_field([r_1,0,0], frame=spherf, chart=cart); show(v.display())
v.apply_map(lambda c:c.subs(ph==ph_1, th==th_1));show(v.display())
I get
r_1 cos(ph)sin(th) e_x + r_1 sin(ph)sin(th) e_y + r_1 cos(th) e_z
r_1 cos(ph_1)sin(th_1) e_x + r_1 sin(ph_1)sin(th_1) e_y + r_1 cos(th_1) e_z
printed (as expected).
Why does the display function affect the substitution?
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Well, I would say that this is caused by a somewhat strange operation that you are asking for. When you write
v = E.vector_field([r_1,0,0], frame=spherf, chart=cart)
v.apply_map(lambda c:c.subs(ph==ph_1, th==th_1))
you are initializing the vector field with components in the frame spherf (orthonormal frame associated with spherical coordinates), while the second line asks for an apply_map in the default frame, which is cartf. But at this stage, such components are not known. Such a substitution is thus not very meaningful. If you add v.display() before v.apply_map, as in your second example, this triggers the computation of the components with respect to cartf, so that the substitution becomes meaningful. If you want to stick to the first example, then you should enforce apply_map to act on the components w.r.t spherf, by adding the argument frame=spherf:
v = E.vector_field([r_1,0,0], frame=spherf, chart=cart)
v.apply_map(lambda c:c.subs(ph==ph_1, th==th_1), frame=spherf)
Then everything is OK.
more
Your explanation makes sense, though I thought when the vector_field is called, as in the question, the corresponding representation of the field in spherical chart is updated.
( 2021-02-08 15:05:08 +0200 )edit | 699 | 2,495 | {"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} | 2.515625 | 3 | CC-MAIN-2021-17 | latest | en | 0.712422 |
https://www.andrew.cmu.edu/user/achoulde/94842/lectures/lecture07/lecture07-94842.Rmd | 1,713,542,517,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00434.warc.gz | 564,659,639 | 6,637 | --- title: Lecture 8 - Hypothesis testing in R author: Prof. Alexandra Chouldechova date: output: html_document: toc: true toc_depth: 5 fig_width: 5 fig_height: 5 --- ##Agenda - Testing differences in mean between two groups - QQ plots - Tests for 2 x 2 tables, j x k tables - Plotting confidence intervals Let's begin by loading the packages we'll need to get started {r} library(tidyverse) ## Exploring the birthwt data We'll begin by doing all the same data processing as in previous lectures {r} # Load data from MASS into a tibble birthwt <- as_tibble(MASS::birthwt) # Rename variables birthwt <- birthwt %>% rename(birthwt.below.2500 = low, mother.age = age, mother.weight = lwt, mother.smokes = smoke, previous.prem.labor = ptl, hypertension = ht, uterine.irr = ui, physician.visits = ftv, birthwt.grams = bwt) # Change factor level names birthwt <- birthwt %>% mutate(race = recode_factor(race, 1 = "white", 2 = "black", 3 = "other")) %>% mutate_at(c("mother.smokes", "hypertension", "uterine.irr", "birthwt.below.2500"), ~ recode_factor(.x, 0 = "no", 1 = "yes")) Over the past two lectures we created various tables and graphics to help us better understand the data. Our focus for today is to run hypothesis tests to assess whether the trends we observed last time are statistically significant. One of the main reasons we want to understand hypothesis testing is that it is important for our tables and figures to convey statistical uncertainty in any cases where it is non-negligible, and where failing to account for it may produce misleading conclusions. ### Testing differences in means One of the most common statistical tasks is to compare an outcome between two groups. The example here looks at comparing birth weight between smoking and non-smoking mothers. To start, it always helps to plot things {r, fig.align='center', fig.width = 5, fig.height = 4} # Create boxplot showing how birthwt.grams varies between # smoking status qplot(x = mother.smokes, y = birthwt.grams, geom = "boxplot", data = birthwt, xlab = "Mother smokes", ylab = "Birthweight (grams)", fill = I("lightblue")) This plot suggests that smoking is associated with lower birth weight. **How can we assess whether this difference is statistically significant?** Let's compute a summary table {r} # Notice the consistent use of round() to ensure that our summaries # do not have too many decimal values birthwt %>% group_by(mother.smokes) %>% summarize(mean.birthwt = round(mean(birthwt.grams), 0), sd.birthwt = round(sd(birthwt.grams), 0)) The standard deviation is good to have, but to assess statistical significance we really want to have the standard error (which the standard deviation adjusted by the group size). {r} birthwt %>% group_by(mother.smokes) %>% summarize(num.obs = n(), mean.birthwt = round(mean(birthwt.grams), 0), sd.birthwt = round(sd(birthwt.grams), 0), se.birthwt = round(sd(birthwt.grams) / sqrt(num.obs), 0)) #### t-test via t.test() This difference is looking quite significant. To run a two-sample t-test, we can simple use the t.test() function. {r} birthwt.t.test <- t.test(birthwt.grams ~ mother.smokes, data = birthwt) birthwt.t.test We see from this output that the difference is highly significant. The t.test() function also outputs a confidence interval for us. Notice that the function returns a lot of information, and we can access this information element by element {r} names(birthwt.t.test) birthwt.t.test$p.value # p-value birthwt.t.test$estimate # group means birthwt.t.test$conf.int # confidence interval for difference attr(birthwt.t.test$conf.int, "conf.level") # confidence level The ability to pull specific information from the output of the hypothesis test allows you to report your results using inline code chunks. That is, you don't have to hardcode estimates, p-values, confidence intervals, etc. {r} # Calculate difference in means between smoking and nonsmoking groups birthwt.t.test$estimate birthwt.smoke.diff <- round(birthwt.t.test$estimate[1] - birthwt.t.test$estimate[2], 1) # Confidence level as a % conf.level <- attr(birthwt.t.test$conf.int, "conf.level") * 100 **Example**: Here's what happens when we knit the following paragraph. {r, eval = FALSE} Our study finds that birth weights are on average r birthwt.smoke.diffg higher in the non-smoking group compared to the smoking group (t-statistic r round(birthwt.t.test$statistic,2), p=r round(birthwt.t.test$p.value, 3), r conf.level% CI [r round(birthwt.t.test$conf.int,1)]g) **Output**: Our study finds that birth weights are on average r birthwt.smoke.diffg higher in the non-smoking group compared to the smoking group (t-statistic r round(birthwt.t.test$statistic,2), p=r round(birthwt.t.test$p.value, 3), r conf.level% CI [r round(birthwt.t.test$conf.int,1)]g) One other thing to know is that t.test() accepts input in multiple forms. I like using the formula form whenever it's available, as I find it to be more easily interpretable. Here's another way of specifying the same information. {r} with(birthwt, t.test(x=birthwt.grams[mother.smokes=="no"], y=birthwt.grams[mother.smokes=="yes"])) Specifying x and y arguments to the t.test function runs a t-test to check whether x and y have the same mean. #### What is statistical significance testing doing? Here's a little simulation where we have two groups, a treatment groups and a control group. We're going to simulate observations from both groups. We'll run the simulation two ways. - First simulation (Null case): the treatment has no effect - Second simulation (Non-null case): the treatment on average increases outcome {r} set.seed(12345) # Function to generate data generateSimulationData <- function(n1, n2, mean.shift = 0) { y <- rnorm(n1 + n2) + c(rep(0, n1), rep(mean.shift, n2)) groups <- c(rep("control", n1), rep("treatment", n2)) data.frame(y = y, groups = groups) } Let's look at a single realization in the null setting. {r, fig.height = 5, fig.width = 7} n1 = 30 n2 = 40 # Observation, null case obs.data <- generateSimulationData(n1 = n1, n2 = n2) obs.data # Box plots qplot(x = groups, y = y, data = obs.data, geom = "boxplot") # Density plots qplot(fill = groups, x = y, data = obs.data, geom = "density", alpha = I(0.5), adjust = 1.5, xlim = c(-4, 6)) # t-test t.test(y ~ groups, data = obs.data) And here's what happens in a random realization in the non-null setting. {r, fig.height = 5, fig.width = 7} # Non-null case, very strong treatment effect # Observation, null case obs.data <- generateSimulationData(n1 = n1, n2 = n2, mean.shift = 1.5) # Box plots qplot(x = groups, y = y, data = obs.data, geom = "boxplot") # Density plots qplot(fill = groups, x = y, data = obs.data, geom = "density", alpha = I(0.5), adjust = 1.5, xlim = c(-4, 6)) # t-test t.test(y ~ groups, data = obs.data) More interestingly, let's see what happens if we repeat our simulation 10000 times and look at the p-values. We'll use a moderate effect of 0.5 instead of the really strong effect of 1.5 in this simulation. {r, cache = TRUE} NUM_ITER <- 10000 pvals <- matrix(0, nrow = NUM_ITER, ncol = 2) for(i in 1:NUM_ITER) { # Generate data obs.null <- generateSimulationData(n1 = n1, n2 = n2) obs.alt <- generateSimulationData(n1 = n1, n2 = n2, mean.shift = 0.5) # Record p-values pvals[i, 1] <- t.test(y ~ groups, data = obs.null)$p.value pvals[i, 2] <- t.test(y ~ groups, data = obs.alt)$p.value } pvals <- as.data.frame(pvals) colnames(pvals) <- c("null", "nonnull") # Plotting routine qplot(x = null, data = pvals, xlab = "p-value", xlim = c(0, 1), main = "P-value when treatment has 0 effect") qplot(x = nonnull, data = pvals, xlab = "p-value", xlim = c(0, 1), main = "P-value when treatment has MODERATE effect") Let's show both histograms on the same plot. {r} # Let's start by reshaping the data # This approach isn't the best one, but it works well for this simple case pvals.df <- data.frame(pvals = c(pvals$null, pvals$nonnull), case = c(rep("null", NUM_ITER), rep("nonnull", NUM_ITER))) # Plot ggplot(data = pvals.df, aes(x = pvals, fill = case)) + geom_histogram(alpha=0.75, position="identity") + xlim(0,1) ##### Why do the distributions look this way? Let's think back to what you learned about hypothesis testing in your statistics class. You probably learned that if we want to control Type I error (i.e., the likelihood of rejecting then the null hypothesis is true) at some level $\alpha$, we should reject the null when the observed p-value is less than $\alpha$. As a probability statement, this says that: $$\mathbb{P}_{H_0}(\text{p-value} \le \alpha) = \alpha$$ Now, the p-value is a random variable that depends on (i.e., gets its randomness from) the observed data used to compute it. The expression above may look familiar to you. Recall that for a random variable $X$, the CDF (cumulative distribution function) of $X$ is defined as $$F_X(x) = \mathbb{P}(X \le x)$$ So what we know is that, when the null is true, the p-value has the CDF $F(x) = x$ for $0 \le x \le 1$. This is the CDF of the **uniform distribution on the interval $[0,1]$**. Now, looking back at the blue histogram in our simulation, we see that this is exactly what the observed p-value distribution looks like. It looks exactly like a sample of uniform random variables from the interval $[0,1]$. Now how about the distribution of p-values when the null is FALSE? In your statistics class, you would have learned about the **power** of a test, often denoted as $\beta$, which is the likelihood of rejecting when the null is FALSE. i.e., the power is $$\beta = \mathbb{P}_{H_a}(\text{p-value} \le \alpha) ,$$ and we generally want $\beta$ to be much larger than $\alpha$ for all values of $\alpha$. If you look at the p-value distribution for the non-null setting, the power $\beta$ at some threshold $\alpha$ is given by the (normalized) area under the pink curve to the *left* of $\alpha$. This area under the blue curve is equal to $\alpha$. And because p-values tend to be smaller under the non-null, it is greater than $\alpha$ when the null is FALSE. #### What if sample is small and data are non-Gaussian? In your statistics classes you've been taught to approach the t-test with caution. If your data is highly skewed, you would need a very large sample size for the t-statistic to actually be t-distributed. When it doubt, you can run a non-parametric test. Here's how we run a Mann-Whitney U test (aka Wilcoxon rank-sum test) using the wilcox.test() function. {r} # Formula specification birthwt.wilcox.test <- wilcox.test(birthwt.grams ~ mother.smokes, data=birthwt, conf.int=TRUE) birthwt.wilcox.test # x,y specification with(birthwt, wilcox.test(x=birthwt.grams[mother.smokes=="no"], y=birthwt.grams[mother.smokes=="yes"])) In general, hypothesis tests in R return an object of class htest which has similar attributes to what we saw in the t-test. {r} class(birthwt.wilcox.test) Here's a summary of the attributes: name | description ------------------------------|--------------------------------- statistic | the value of the test statistic with a name describing it. parameter | the parameter(s) for the exact distribution of the test statistic. p.value | the p-value for the test. null.value | the location parameter mu. alternative | a character string describing the alternative hypothesis method | the type of test applied. data.name | a character string giving the names of the data. conf.int | a confidence interval for the location parameter. (Only present if argument conf.int = TRUE.) estimate | an estimate of the location parameter. (Only present if argument conf.int = TRUE.) ### Is the data normal? I would recommend using a non-parametric test when the data appears highly non-normal and the sample size is small. If you really want to stick to t-testing, it's good to know how to diagnose non-normality. #### qq-plot The simplest thing to look at is a normal qq plot of the data. This is obtained using the stat_qq() function. {r, fig.align='center', fig.width = 5, fig.height = 4} # qq plot p.birthwt <- ggplot(data = birthwt, aes(sample = birthwt.grams)) p.birthwt + stat_qq() + stat_qq_line() # Separate plots for different values of smoking status p.birthwt + stat_qq() + stat_qq_line() + facet_grid(. ~ mother.smokes) # qq plot for 115 observations of truly normal data df <- data.frame(x = rnorm(115)) ggplot(data = df, aes(sample = x)) + stat_qq() + stat_qq_line() If the data are exactly normal, you expect the points to lie on a straight line. The data we have here are pretty close to lying on a line. Here's what we would see if the data were right-skewed {r, fig.align='center', fig.width = 5, fig.height = 4} set.seed(12345) fake.data <- data.frame(x = rexp(200)) p.fake <- ggplot(fake.data, aes(sample = x)) qplot(x, data = fake.data) p.fake + stat_qq() + stat_qq_line() If you construct a qqplot and it looks like this, you should be carefully, particularly if your sample size is small. ### Tests for 2x2 tables Here's an example of a 2 x 2 table that we might want to run a test on. This one looks at low birthweight broken down by mother's smoking status. You can think of it as another approach to the t-test problem, this time looking at indicators of low birth weight instead of the actual weights. First, let's build our table using the table() function (we did this back in Lecture 5) {r} weight.smoke.tbl <- with(birthwt, table(birthwt.below.2500, mother.smokes)) weight.smoke.tbl It looks like there's a positive association between low birthweight and smoking status. To test for significance, we just need to pass our 2 x 2 table into the appropriate function. Here's the result of using fisher's exact test by calling fisher.test {r} birthwt.fisher.test <- fisher.test(weight.smoke.tbl) birthwt.fisher.test attributes(birthwt.fisher.test) As when using the t-test, we find that there is a significant association between smoking an low birth weight. You can also use the chi-squared test via the chisq.test function. This is the test that you may be more familiar with from your statistics class. {r} chisq.test(weight.smoke.tbl) You get essentially the same answer by running the chi-squared test, but the output isn't as useful. In particular, you're not getting an estimate or confidence interval for the odds ratio. This is why I prefer fisher.test() for testing 2 x 2 tables. #### Tests for j x k tables Here's a small data set on party affiliation broken down by gender. {r} # Manually enter the data politics <- as.table(rbind(c(762, 327, 468), c(484, 239, 477))) dimnames(politics) <- list(gender = c("F", "M"), party = c("Democrat","Independent", "Republican")) politics # display the data We may be interested in asking whether men and women have different party affiliations. The answer will be easier to guess at if we convert the rows to show proportions instead of counts. Here's one way of doing this. {r} politics.prop <- prop.table(politics, 1) politics.prop colSums(politics.prop) # Check that columns sum to 1 # Fix dimnames dimnames(politics.prop) <- list(gender = c("F", "M"), party = c("Democrat","Independent", "Republican")) # Output politics.prop By looking at the table we see that Female are more likely to be Democrats and less likely to be Republicans. We still want to know if this difference is significant. To assess this we can use the chi-squared test (on the counts table, not the proportions table!). {r} chisq.test(politics) There isn't really a good one-number summary for general $j$ x $k$ tables the way there is for 2 x 2 tables. One thing that we may want to do at this stage is to ignore the Independent category and just look at the 2 x 2 table showing the counts for the Democrat and Republican categories. {r} politics.dem.rep <- politics[,c(1,3)] politics.dem.rep # Run Fisher's exact test fisher.test(politics.dem.rep) We see that women have significantly higher odds of being Democrat compared to men. | 4,122 | 15,901 | {"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": 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} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.873733 |
http://bookshadow.com/weblog/2015/week/27/ | 1,611,711,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00223.warc.gz | 16,949,639 | 19,977 | # 档案日期2015的27
## 题目描述：
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w ...
## 题目描述：
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Can you solve it without using extra space?
## 题目描述：
Given a singly linked list, determine if it is a palindrome.
Could you do it in O(n) time and O(1) space?
## 解题思路：
1). 使用快慢指针寻找链表中点
2). 将链表的后半部分就地逆置 ...
## 题目描述：
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10 ...
## 题目描述：
Implement the following operations of a queue using stacks.
• push(x) -- Push element x to the back of queue.
• pop() -- Removes the element from in front of queue.
• peek() -- Get the front element.
• empty() -- Return whether the queue is ... | 295 | 1,053 | {"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-2021-04 | latest | en | 0.674286 |
https://dyinglovegrape.wordpress.com/2011/08/ | 1,540,001,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512501.27/warc/CC-MAIN-20181020013721-20181020035221-00091.warc.gz | 657,452,532 | 17,733 | ## Sequence which has countably many convergent subsequences.
### August 22, 2011
This post is about a question I just thought about that I thought had a pretty neat solution. The values we consider below are all in the Reals.
So, we’re all familiar with sequences which have subsequences which converge to two different values; for example, take $\{1,-1,1,-1,1,-1,\dots\}$ which has a subsequence converging to 1 and a subsequence converging to -1. Similarly, we can construct a sequence containing subsequences converging to three different values in a similar way; for example: $\{-1,0,1,-1,0,1, \dots\}$. Indeed, for any finite number, we can make a sequence containing subsequences converging to that number of different values.
The obvious next question is: can a sequence have subsequences which converge to a countable number of different values? What about an uncountable number of different values?
For the former question, I thought of this solution. First, enumerate your countable list of distinct values (think of the integers or the rationals, for example) as $\{q_{1}, q_{2}, \dots\}$. To construct the sequence, we do the following:
• Let $a_{1} = 0$.
• Let $a_{b} = 0$ if $b$ is not some positive power of a prime (for example, if $b$ is 15, 21, 35, 100, etc.)
• For $a_{k}$ where $k$ is a positive power of a prime, we do the following: If $k = p_{i}^{r}$ for $r\in {\mathbb N}$ and $p_{i}$ is the $i$-th prime (so $p_{1} = 2$, $p_{2} = 3$, $p_{3} = 5$, and so on) then set $a_{k} = q_{i}$ where $q_{i}$ is the $i$-th element from your list above.
How does this actually work? Let’s take a simple example. Let’s let our countable set be the set of natural numbers. This is easily ordered as $\{1,2,3,4,5,\dots\}$ and so, for this example, $q_{1} = 1$, $q_{2} = 2$, $q_{3} = 3$, and so forth. Our sequence is now:
$\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10},\dots\}$
$= \{a_{1}, a_{2}, a_{3}, a_{2^{2}}, a_{5}, a_{2\cdot 3}, a_{7}, a_{2^{3}}, a_{3^{2}}, a_{2\cdot 5}, \dots\}$
$= \{0, q_{1}, q_{2}, q_{1}, q_{3}, 0, q_{4}, q_{1}, q_{2}, 0, \dots\}$
$= \{0, 1, 2, 1, 3, 0, 4, 1, 2, 0, \dots\}$
and if we kept going like this,
$= \{0, 1, 2, 1, 3, 0, 4, 1, 2, 0, 5, 0, 6, 0, 0, 1, 7, 0, 8, 0, 0, 0, 0, 0, 3, 0,\dots\}$
You kinda see the pattern going on here. The 1 will appear infinitely many times, but the spaces between it grow exponentially. Same for 2, 3, 4, and so on. Thus, this sequence has the subsequence $\{n, n, n, n, \dots\}$ for any $n\in {\mathbb N}$ which obviously will converge to $n$.
Do you have another neat way to do this? Encoding this in the primes was my first thought, but you never know!
As for uncountable, my guess is no. Of course, thinking of the sequence as a SET, we would get the reals from the rationals which is uncountable, but as a SEQUENCE it is not so trivial I feel. I don’t want to spend too much time thinking about this (as analysis is calling me!) but I’m sure the proof isn’t too crazy.
Edit: Of course, a nice example exists where a sequence has subsequences converging to an uncountable number of distinct point. In fact, two nice examples exist, and both were provided to me by Brooke (as usual!).
Uncountable Example 1:
It was a bit difficult for me to see the first one, but after doing a nice little thought experiment everything became much clearer. Here it is:
Let $\{q_{1}, q_{2}, \dots\}$ be an enumeration of the rationals. The claim is that there is actually a subsequence which converges to any real number. Think about it for a minute.
Here’s a sketch for the proof. Take your favorite real number. Let’s start easy and just say we want a subsequence which converges to $\pi$. Let $a_{1}$ be the first element in our enumeration of the rationals above which is a distance less than 1 away from $\pi$; let’s say it’s $q_{k}$. Now let $a_{2}$ be the first element in our enumeration which is after $q_{k}$ and which is a distance less than $\frac{1}{2}$ away from $\pi$
After that, you just keep picking the "next" element from the list that’s $\frac{1}{2^{n}}$ away from $\pi$. There must be one due to the density of the rationals in the reals (or, assume not and see what happens!). We end up with $\{a_{1}, a_{2}, \dots\}$ as a subsequence of our enumeration of the rationals which converges to $\pi$. Obviously, replacing $\pi$ with any other real number works exactly the same.
Uncountable example 2:
This was one that Brooke presented to me that I liked quite a bit since it’s really easy to state. Here’s the sequence (broken up by lines for clarity):
$0.0\,,\, 0.1\,,\, 0.2\,,\, \dots\,,\, 0.9,$
$0.00\,,\, 0.01\,,\, 0.02\,,\, \dots\,, \,0.09\,,\, 0.10\,,\, 0.11\,,\, \dots\,,\, 0.99\,,$
$0.000\,,\, 0.001\,,\, 0.002\,,\, \dots\,,\, 0.999\,, \dots$
This converges to any real number in $[0,1]$. To see this is easy: decimal expand your chosen real number and pick the element "from each line" (as I’ve written it above) which is closest to it. They’ll be at most $10^{-j}$ away (for some suitable $j$ depending on the line you’re on) and thus a subsequence exists which converges to whatever element you picked.
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## Lim Sup: Life Lessons.
### August 22, 2011
I’ve always felt a little uneasy with Lim Sup and Lim Inf’s. Today while reading Rudin’s Complex book, I finally "got" it.
To find the lim sup of your sequence: take every convergent subsequence’s limit and find the sup of those limits.
This is, of course, exactly what the definition is, but it wasn’t "visual" to me until I thought about it this way. Is there something you learned early on that you weren’t quite clear with until much, much later?
## A Quick Note on Fatou’s Lemma.
### August 16, 2011
Above is a sweet picture of this French mathematician Pierre Fatou. Of course he had a nice mustache, but a few math things that are kind of neat about him: he was the first person to study the Mandelbrot set (though there is some controversy regarding the word "first" hereā¦) and he also enjoyed iterative and recursive processes before they became really cool, making him a computer science hipster. Also, he has a lemma in measure theory named after him. That’s what this post is about.
Lemma (Fatou’s): Let $\{f_{i}\}_{i}$ be a sequence of non-negative measurable functions on ${\mathbb R}$ with the Lebesgue measure $\mu$, and let $\displaystyle f(x) = \lim_{n\rightarrow\infty}\inf_{m \geq n} f_{n}(x)$ for all $x\in {\mathbb R}$. Then, $f$ is measurable and we have the inequality
$\displaystyle \int f\, d\mu \leq \lim_{n\rightarrow\infty}\inf_{m \geq n} \int f_{n}\, d\mu$.
## Nested Sequences of Measurable Sets.
### August 10, 2011
On nearly every practice qualifying exam that I’ve been studying from, the following question (in some guise) comes up:
Question: Let $\{E_{i}\}_{i=1}^{\infty}$ be a sequence of Lebesgue measurable sets in ${\mathbb R}$. Denote the Lebesgue measure by $\mu$
1. If $E_{n}\subseteq E_{n+1}$ for every $n\in {\mathbb N}$, then is it true that $\displaystyle \mu(\bigcup_{i=1}^{\infty} E_{i}) = \lim_{n\rightarrow\infty}\mu (E_{n})$? If not, add a criteria to make it true.
2. If $E_{n}\supseteq E_{n+1}$ for every $n\in {\mathbb N}$, then is it true that $\displaystyle \mu(\bigcap_{i=1}^{\infty} E_{i}) = \lim_{n\rightarrow\infty}\mu (E_{n})$? If not, add a criteria to make it true.
I’ll write up the solutions, since they are available elsewhere, but take a little bit of time to think about the second one if you haven’t. I’ll put the remainder of the post under a cut so that I don’t spoil it right away.
## Jordan’s Lemma.
### August 9, 2011
[This post is for those of you who are already comfy with doing some basic contour integrals in complex analysis.]
So you’re sitting around, evaluating contour integrals, and everything is fine. Then something weird comes up. You’re asked to evaluate an integral that looks like
$\displaystyle \int_{-\infty}^{\infty} e^{aix}g(x)\, dx$
for $g(x)$ is continuous. Eek. Don’t panic though, because Camille Jordan’s gonna help you out.
## Brief "Where are we going?" Motivation.
So far we’ve been talking about measure theory, which is nice to study just for the sake of measuring things, but when we talk about functions and measurements, we generally want to know the integral. The Riemann integral was wonderful because of its simplicity in defining it: you have a big sheet of wavy paper (the total area under a function) and you keep slicing the pieces until they look like rectangles — once they look sufficiently close to a rectangle, just use length-times-width, and add up all the rectangles. This is one interpretation of the Riemann integral, and, for the most part, it works wonderfully. But there are many functions where it fails miserably. This is where Lebesgue takes over.
So what’s different about Lebesgue? While Riemann cuts apart the domain, Lebesgue says, "Okay, how large is the set of x-values that have the range value $c\in {\mathbb R}$?" Specifically, we cut up the function into sets which have the same range value, then multiply that range value with the measure of the set.
As an example, take the step function $f$ defined on, say, $[-10,10]$ such that $f(x) = 1$ for $0\leq x < 1$ and $f(x) = 2$ for $1\leq x < 2$ and $f(x) = 0$ elsewhere. The Lebesgue integral will say, "Where does $f$ take on the value 0? Well, $[-10,0)\cup(2,10]$ which has measure of 18. Since the value is 0, we have $18\cdot 0 = 0$, so this contributes nothing to the integral. Where does $f$ take on the value 1? On $[0,1)$, which has a measure 1. So this contributes $1\cdot 1$ to the integral. Similarly, the function takes on a value 2 onn $[1,2)$, this has measure 1, so this contributes $1\cdot 2$ to the integral. No other values are taken on, so this integral is 0 + 1 + 2 = 3."
This extremely simple example shows the difference between Riemann and Lebesgue at the most fundamental level. To do Riemann would be much easier in this case: it’s just adding up the rectangles.
There is a slight hitch at this point. Suppose we naively tried to apply what we’ve just said to a more complex function like $f(x) = x$ defined, say, on $[0,2]$. We know (because this forms a triangle) that the area should be 2. At every number between $[0,2]$ in the range, though, the function only takes on the value at a single point which has measure 0. Thus, every one of these contributes 0 to the integral; and the sum of 0’s will be 0. So the integral is 0? That’s not right!
The problem here is that we must define the Lebesgue integral for functions which take on simple functions (which take on a countable number of values) and somehow approximate more general functions. As we know, a function is measurable if and only if there is a sequence of simple functions which converge uniformly to it; and this will give us the means to integrate general (measurable) functions: we will take the integrals of each element of the sequence, and then define that limit to be the integral of the measurable function. A number of questions will arise instantly, like, "Does this work? Does this make sense? Is this well-defined?", all of which can easily be put down with the elegant slashing of our proof-swords.
So, let’s get down to it.
## The Lebesgue Integral and Simple Functions.
We’ve already gone over simple functions (ones with a countable number of values, the pre-image of each of these being measurable making simple functions, themselves, measurable) so we might as well dive right in. We will begin by defining, after long last, the Lebesgue Integral for these simple functions. It will be defined exactly as we have motivated it above.
Definition. Let $f$ be a measurable simple function on a (measurable) set $A$ taking the values $a_{1}, a_{2}, \dots$. Let $A_{i}$ be defined as the set of all $x$ such that $f(x) = a_{i}$; that is, $A_{i}$ is the pre-image of $a_{i}$. Then we define
$\displaystyle \int_{A}f(x)\, d\mu$
to be exactly the sum
$\displaystyle \sum_{i=1}^{\infty} a_{i}\mu(A_{i})$
provided that the sum converges absolutely (that is, the sum converges with $|a_{i}|$ replacing $a_{i}$).
We call this the Lebesgue integral of $f$ over the set $A$.
Note that we have explicitly noted that $A$ must be measurable, though this is not specifically stated in the texts I am using. I don’t feel I am losing much in assuming this (expect, perhaps, for more pathological examples such as being able to integrate some function which is non-zero only on a measurable subset of the non-measurable set). Notice that we have also assumed that our sum must converge absolutely, and not just conditionally.
[Note: At this point, it is not entirely clear to me why this should be necessary; in finite cases, we will generally not worry about it, but perhaps there is some reasoning as to why in infinite measured sets, we need this absolute qualification. Feel free to weigh in here.]
It is reasonable to want to check the well-defined-ness of this, but I will not do this here since it is available where ever Lebesgue integrals are gone over. I will also not define the integral of $f + g$ or $\alpha f$ for $\alpha\in {\mathbb R}$, because these should be relatively obvious.
TO BE CONTINUED. | 3,873 | 13,329 | {"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": 106, "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-2018-43 | latest | en | 0.86175 |
https://github.com/sympy/sympy/wiki/GSoC-2015-Report-Shivam-Vats:-Fast-Series-Expansion | 1,531,975,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00193.warc.gz | 655,811,817 | 28,000 | Brief Information
Project : Fast series expansion using sparse representation.
Proposal : wiki page
Blog : blog
Email : shivamvats.iitkgp@gmail.com
I am Shivam Vats, currently a 3rd year undergraduate student of Mathematics and Computing at IIT Kharagpur.
Introduction
I was introduced to Sympy by Harsh Gupta who is himself a former GSoC student and senior to me in my department. I wanted to learn Python and he suggested Sympy as a great place to learn it by doing useful work. And indeed, it is one of the most beginner friendly projects out there. I started from scratch and was guided in by initial days by Sergey. Later on, I accidentally stumbled upon SymEngine and started contributing to it too.
Application Phase
I had experimented with quite a few modules of Sympy and most of the topics I was interested in (Plotting, polynomials, etc) were either too advanced for me or were not being actively developed. On the other hand, SymEngine was quite new and a lot of things needed to be done. Moreover, I like coding in C++ and Series expansion seemed to be an important functionality.
I had extensive discussions with Ondrej, during which he told me about the things he had tried and what seemed to work best. The very first decision we had to take was how to represent a series. Sympy does so as expressions, which is quite slow. Moreover, he also pointed out that a truncated series is effectively a polynomial. So, we finally decided to use polynomial operations, which meant we could reuse the polynomial module. Further, sparse representation was chosen as it gave decent execution timings over a large range of series sizes.
SymEngine lacked a Polynomial module that the time, which meant I had to write the parts required for my project. I was thus, expecting most of my work to be in SymEngine. Though I planned to implement fast expansion in Sympy too, as a proof of concept, I expected to finish it in the first month.
Coding
Fortunately for me, Sumith's proposal for Polynomials got accepted as well. This meant I could focus on series expansion. Sympy at the time, already had a limited expansion functionality implemented using Rings. It had been done by Mario, who had written optimised algorithms for the same. However, a larger porting of his work on series expansion had not been merged and lay in a long forgotten pull request. I started by porting all those functions in ring_series.py. This proved to be quite a daunting task as I was not aware of all the algorithms used and in many cases new functions had to be implemented to get equivalent behaviour.
One important point about Mario's code was that it was written for power series, meaning all the exponents in the series were positive integers. We wanted out algos to support the most general series, meaning a puiseux series( fractional and negative powers). This called for rewriting all the functions. The possibility of fractional powers meant we had to change the way we calculated the order of the series. Some of the algorithms are semi-numerical in nature, which make use of the fact that the series is a power series. This was done in #9495 with major help from Mario who also pointed our the flaws in my method.
The way we create a new sparse polynomial is with something like ring('x', QQ), which basically creates a polynomial object in the variable x with rational coefficients. We can of, course choose any number of variables and different types of coefficients. The next step was to have series involving symbolic terms- things like sin(x + a), when you expand wrt say, x. This was partly done in #9575. Polynomials, do in fact have an EX domain, which allows symbolic coefficients, but at the cost of significant slow down. Since, a major goal of my project was to improve execution speed, this was not acceptable.
What I then did was simple. I used the definition of a variable. For a ring, a variable is something that was specified at the time of creation and a coefficient is anything multiplied with or added to any variable. So, instead of creating a new coefficient domain, I added all the required symbolic coefficients as variables to the ring. For example, for sin(x + a), this means adding sin(a) and cos(a) as variable to the ring. Problem solved?
Not yet! It turned out that the restriction of a polynomial to positive powers was hard coded in polynomials which made operations involving different series very difficult. Eventually, I had to remove the restrictions, compromising mathematical accuracy in exchange for practical usability. So, now I had the elementary operations ready. But they weren't too useful for most users. We needed the equivalent of Sympy's existing series function that could expand an arbitrary expression. This was done in #9775 and was definitely the most challenging part of my project.
I had proposed in my proposal to write classes for different types of series and I did start writing them in #9614. But I eventually realised that they weren't really needed. If the rs_series could be smart enough to know which ring_series function to call and with the right arguments, I was done. The way the functions are evaluated, different series have different rings associated with them. Thus, while operating on two or more different series, they need to be converted appropriately. A lot of similar conversions and checks are done in rs_series to make sure that the answer is correct and uses the minimum possible operations.
I kept running into unforeseen difficulties during the course of the project and the timeline in my proposal became irrelevant after a point. That is partly also because I started developing ring_series as a potential replacement of the present series method, instead of implementing it as a proof of concept only. As for SymEngine I could not do any work related to series expansion as the polynomial module couldn't be completed. So, I mostly helped out on a few issues not related to my project.
Future Plans
Ring series seems very promising with excellent speed ups. I intend to continue developing it. The first target is the make rs_series work with all the ring_series functions. There are a number of bugs and I am sure many more are to be caught. I am also working on detailed documentation so that interested people can help. The important thing is to get people to start using it.
There is a lot of work in porting the code to SymEngine.
Conclusion
I had a great time working with Sympy during the summer. The whole GSoC experience is full of life lessons, of which, coding is a minor part. Community building is definitely the most important aspect of open source development; because at the end of the day all of us want to make new friends :) | 1,387 | 6,744 | {"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} | 2.828125 | 3 | CC-MAIN-2018-30 | latest | en | 0.985597 |
http://semantic-domain.blogspot.com/2012/08/pattern-compilation-made-easy.html?showComment=1344329109419 | 1,539,699,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510754.1/warc/CC-MAIN-20181016134654-20181016160154-00113.warc.gz | 337,186,063 | 23,122 | ## Monday, August 6, 2012
One of the most practically useful features of functional programming languages is pattern matching, and one of the most important quality-of-implementation features for an implementation is the exhaustiveness checking for pattern matches. In this post, I will give the simplest algorithm for coverage checking that I know.
I previously gave one in my paper Focusing on Pattern Matching. I was never completely satisfied with that algorithm, since it was harder to understand than I liked. It worked by giving a judgment to prove that no match would fail, rather than directly proving that all matches covered. The negation made thinking about extensions (such as pattern matching for dependent types) harder than it should be, and it also made it harder to get a match compilation algorithm.
So in this blog post, I'll give a better algorithm. It's super-simple, and is about as naive as it should be -- there are only six rules in total. It's not totally industrial strength, but (a) it's certainly good enough to use in a prototype compiler, and (b) it's easy enough to beef up.
There's also actually some cool proof theory yet to be worked out lurking here. I'll point it out as we go along. $$\newcommand{\cover}[2]{{#1} \;\mathsf{covers}\;{#2}} \newcommand{\elaborate}[3]{{#1} \;\mathsf{covers}\;{#2} \leadsto {#3}} \newcommand{\To}{\Rightarrow}$$
First, I'll give the syntax. $$\begin{array}{llcl} \mbox{Types} & A & ::= & 1 \bnfalt A \times B \bnfalt A + B \bnfalt A \to B \bnfalt o \\ \mbox{Patterns} & p & ::= & () \bnfalt (p, p) \bnfalt \inl(p) \bnfalt \inr(p) \bnfalt \_ \bnfalt x \\ \mbox{Pattern Lists} & ps & ::= & \cdot \bnfalt p, ps \\ \mbox{Branch Lists} & \Pi & ::= & \cdot \bnfalt ps \To e; \Pi \\ \mbox{Contexts} & \Gamma & ::= & \cdot \bnfalt u:A, \Gamma \end{array}$$
The types are sums and products, plus a type $o$ to indicate functions and other non-matchable types. The patterns are just what you'd expect -- $()$ for the unit pattenrn, $(p, p')$ for pairs, and $\inl(p)$ and $\inr(p)$ for the left and right branches of sums. (Of course, there is no pattern for the empty type!) We also have wildcard $\_$ and variable $x$ patterns.
Since our algorithm works inductively, we will destructure patterns into their components, and we will need to keep track of lists of patterns. A pattern list (with metavariable $ps$) is just the obvious thing --- a comma-separated list of patterns.
We also need branch lists -- a pattern match clause consists of a list of of patterns and expressions. So we use branch lists to say what body is associated with each pattern list (a pattern list, since we are going to destructure the pattern given for each arm).
We will also write $\overrightarrow{ps \To e}$ to indicate branch lists. If I write a list like $\overrightarrow{(p_1, p_2), ps \To e}$, then I mean that every element of the branch list starts with a pair pattern, and similarly for the other constructors.
This algorithm keeps track of the type of the value being destructured, so we introduce a typed context $\Gamma$ to keep track of the types of the expressions being destructured.
Without further delay, here's the coverage judgment $\cover{\Pi}{\Gamma}$. You can read this as algorithm bottom-up, in the usual logic-programming style. $$\array{ \rule{ \cover{\overrightarrow{ps' \To e'}}{\Gamma} } { \cover{\overrightarrow{\_, ps' \To e'}}{u:A, \Gamma} } \\[1em] \rule{ \cover{ \overrightarrow{ps' \To e'},\;\; (\_, ps \To e), \overrightarrow{ps'' \To e''} }{u:A, \Gamma} } { \cover{ \overrightarrow{ps' \To e'},\;\; (x, ps \To e), \overrightarrow{ps'' \To e''} }{u:A, \Gamma} } \\[1em] \rule{ \cover{ \overrightarrow{ps \To e},\;\; \overrightarrow{ps' \To e'} }{\Gamma}} { \cover{ \overrightarrow{(), ps \To e},\;\; \overrightarrow{\_, ps' \To e'} }{u:1, \Gamma}} \\[1em] \rule{ \cover{ \overrightarrow{p_1, p_2, ps \To e},\;\; \overrightarrow{\_, \_, ps' \To e'} }{a:A, b:B, \Gamma}} { \cover{ \overrightarrow{(p_1,p_2), ps \To e},\;\; \overrightarrow{\_, ps' \To e'} }{u:A \times B, \Gamma}} \\[1em] \rule{ } { \cover{ \overrightarrow{\_, ps'' \To e''} }{u:0, \Gamma}} \\[1em] \rule{ \cover{ \overrightarrow{p, ps \To e},\;\; \overrightarrow{\_, ps'' \To e''}}{a : A, \Gamma} \\ \cover{ \overrightarrow{p', ps' \To e'},\;\; \overrightarrow{\_, ps'' \To e''}}{b : B, \Gamma}} { \cover{ \overrightarrow{\inl(p), ps \To e},\;\; \overrightarrow{\inr(p'), ps' \To e'},\;\; \overrightarrow{\_, ps'' \To e''} } {u:A + B, \Gamma}} }$$
The first two rules handle wildcards and variables. The first rule says that if the lead pattern of every pattern list in the branch set is a wildcard, we can just drop it and keep going. The second rule says that you can turn any lead variable pattern into a wildcard pattern.
The third rule handles the unit pattern. It is nice and easy --- it says that if every lead pattern is either a unit or wildcard pattern, we can drop it and keep going.
The fourth rule handles tuples, and is almost as easy. It says that if we have a lead pattern $(p_1, p_2), ps \To e$, we can rewrite it to $p_1, p_2, ps \To e$. We also have to remember to double any lead wildcard pattern $\_, ps' \To e'$ into $\_, \_, ps' \To e'$, since we are destructuring a product into two subpatterns.
The fifth rule handles the void type. It just succeeds, since there are no patterns of empty type!
The sixth rule handles sums. It says that if we are scrutinizing a value of type $A + B$, then we can separate the branch list into those branches handling the left case $\inl(p), ps \To e$ and those branches handling the right case $\inr(p'), ps' \To e'$, and then check then at $A$ and $B$ respectively. Here, we have to remember to send wildcards $\_, ps'' \To e''$ to both sides, since a wildcard can match both the left- and right-cases.
And that's it!
The open problem in proof theory is a bit of slight of hand in my notation. You understand exactly what I mean when I take a meta-pattern like $\overrightarrow{(p_1, p_2), ps \To e}$, and turn it into $\overrightarrow{p_1, p_2, ps \To e}$. However, formalizing this is trickier than it looks, and it's not known how to give a proof-theoretic characterization of this kind of pattern (according to Rob Simmons, who I asked early this year about it).
As a bonus, I'll also show how coverage checking can also give you a compilation algorithm for pattern matching. The judgment $\elaborate{\Pi}{\Gamma}{t}$ tells you how to turn a list of branches $\Pi$ into a nested sequence of tuple destructurings and case-statements. $$\array{ \rule{ \elaborate{\overrightarrow{ps' \To e'}}{\Gamma}{t} } { \elaborate{\overrightarrow{\_, ps' \To e'}}{u:A, \Gamma}{t} } \\[1em] \rule{ \elaborate{\overrightarrow{ps' \To e'}}{\Gamma}{t} } { \elaborate{\overrightarrow{\_, ps' \To e'}}{u:A, \Gamma}{t} } \\[1em] \rule{ \elaborate{ \overrightarrow{ps' \To e'},\;\; (\_, ps \To \letv{x}{u}{e}), \overrightarrow{ps'' \To e''} }{\Gamma}{t} } { \elaborate{ \overrightarrow{ps' \To e'},\;\; (x, ps \To e), \overrightarrow{ps'' \To e''} }{\Gamma}{t} } \\[1em] \rule{ \elaborate{ \overrightarrow{ps \To e},\;\; \overrightarrow{ps' \To e'} }{\Gamma}{t} } { \elaborate{ \overrightarrow{(), ps \To e},\;\; \overrightarrow{\_, ps' \To e'} }{u:1, \Gamma}{\letv{()}{u}{t}} } \\[1em] \rule{ \elaborate{ \overrightarrow{p_1, p_2, ps \To e},\;\; \overrightarrow{\_, \_, ps' \To e'} }{a:A, b:B, \Gamma}{t} } { \elaborate{ \overrightarrow{(p_1,p_2), ps \To e},\;\; \overrightarrow{\_, ps' \To e'} }{u:A \times B, \Gamma}{\letv{(a,b)}{u}{t}} } \\[1em] \rule{ } { \elaborate{ \overrightarrow{\_, ps'' \To e''} }{u:0, \Gamma}{\abort(u)} } \\[1em] \rule{ \elaborate{ \overrightarrow{p, ps \To e},\;\; \overrightarrow{\_, ps'' \To e''}}{a : A, \Gamma}{t_1} \\ \elaborate{ \overrightarrow{p', ps' \To e'},\;\; \overrightarrow{\_, ps'' \To e''}}{b : B, \Gamma}{t_2} } { \elaborate{ \overrightarrow{\inl(p), ps \To e},\;\; \overrightarrow{\inr(p'), ps' \To e'},\;\; \overrightarrow{\_, ps'' \To e''} } {u:A + B, \Gamma} {\case{u}{a}{t_1}{b}{t_2}} } }$$
Note that it is exactly the same judgment as before --- coverage checking computes exactly the information we need, and so code generation comes along for the ride! | 2,533 | 8,207 | {"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} | 3.234375 | 3 | CC-MAIN-2018-43 | longest | en | 0.90754 |
https://www.studyadda.com/index.php?/notes/2nd-class/mathematics/spatial-understandings/shapes-and-spatial-understanding/6081 | 1,524,268,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944848.33/warc/CC-MAIN-20180420233255-20180421013255-00086.warc.gz | 882,654,101 | 16,528 | # 2nd Class Mathematics Spatial understandings Shapes and Spatial Understanding
## Shapes and Spatial Understanding
Category : 2nd Class
LEARNING OBJECTIVES
• identify 2-Dimensional (2-D) shapes.
• study the properties of 2-D shapes.
• identify some of the 3-D shapes.
• distinguish between straight and curved lines.
QUICK CONCEPT REVIEW
SHAPES
When we look around, we find objects of different shapes. Some objects you also carry in your school bags. For example: pencil, pen lunch box, ruler etc.
Do you know their shapes?
Let us discuss some of these shapes in detail.
1. You all must have played LUDO with your friends. Do you know shape of its dice?
Dice is having shape of a cube.
2. Match box, that we use for lighting fire has a shape of a cuboid.
3. Cold drink cans have shape of a cylinder.
4. Ice-cream cones are having shape of a cone.
5. Football has a shape of a sphere.
The shapes we have discussed above are 3-Dimensional shapes. These are also known as solid shapes. Flat shapes like circle, triangle, square and rectangle are also called 2 - Dimensional shapes. Now, observe how we can make 2-D (2-Dimensional) shapes from 3-D (3-Dimensional) shapes by simply using pencil. Whichever shapes you want to draw, put it on the paper and draw its lower bottom. The shapes thus formed are 2-D (2-Dimensional) shapes.
Let us understand with the help of examples given below.
1.
2.
3.
4.
Rectangle
A Rectangle has four sides. Two opposite?s sides are equal and all the four angles are also equal.
Square
A Square has four sides. All the sides are equal Four angles are also equal.
Triangle
It has sides and three and angles. Sides and angles can be different or equal.
Circle
It has no sides and it is round in shape.
LINES
When we move a pencil along one of the edges of a scale we get aLINE.
1. Straight Line
If you stretch a thread tightly, you get a straight line squares, rectangles and triangles are made of straight lines.
Horizontal, Vertical and Slanting lines
Look at door of your classroom. Along the floor, we see a line. It is called horizontal or sleeping line. The lines along the wall are vertical lines.
The lines that are neither vertical nor horizontal are called slanting lines.
We can draw a slanted line by placing a ruler on the paper.
2. Curved line
The curved line is drawn without the use of a ruler. It is not a straight line. If you hold the thread loosely, you get a curved line.
Examples of curved lines
Circles and ovals are made curved lines.
ANGLE
An angle is formed by joining two lines.
Amazing Fact
• Squircle is a type of shape which is essentially a combination of a circle and a square having properties of a both, Lately, Squircles have found widespread use in modern car designs.
Real Life Examples
We can find a lot of example of solid shapes in our surroundings.
For example
• Coins are circular in shape and books that we read are cuboid in shape.
• The bricks that are used for making houses are cuboid in shape, the ball with which we play is a sphere.
Historical preview
The invention of a wheel (a circle shape) was one of the most important inventions in human history.
#### Other Topics
LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!
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You will be redirected in 3 sec | 771 | 3,348 | {"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} | 5 | 5 | CC-MAIN-2018-17 | latest | en | 0.94919 |
https://brilliant.org/problems/who-is-going-to-expand/ | 1,508,259,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822145.14/warc/CC-MAIN-20171017163022-20171017183022-00765.warc.gz | 660,400,104 | 18,030 | # Who is going to expand?
Algebra Level 4
$(1-x)(1-2x)(1-4x)\cdots \left(1-2^{101}x\right)$
What is the coefficient of $$x^{101}$$ in the expansion of the above?
× | 63 | 167 | {"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} | 2.65625 | 3 | CC-MAIN-2017-43 | longest | en | 0.80483 |
https://www.tomshardware.com/reviews/a-cool-bunch,516-2.html | 1,568,813,442,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573289.83/warc/CC-MAIN-20190918131429-20190918153429-00158.warc.gz | 1,020,829,675 | 39,343 | # A Cool Bunch: How To Put A Lid On The Die Temperature Of Your Athlon
## The Theory Behind The Dream Cooler
It's not hard to sum up what users expect of their dream cooler: it should be easy to mount and should run quietly. And it should offer enough extra cooling capacity so that you can upgrade your processor once or twice.
### Specifications Of The Processor Maker
The most important design specifications given by AMD for an Athlon cooler and for the computer case are as follows:
• Maximum total weight: 300 grams (for coolers that are affixed to the socket with a clip);
• Maximum external dimensions: 60 x 80 x 60 millimeters;
• The internal temperature, TA , immediately surrounding the cooler must not exceed 40° Celsius, no matter what the operating state of the CPU is.
### General Physical Requirements
The cooling capacity/heat conduction provided by a cooler can be described by the following (extremely simplified) equation:
Iw = G*TD -TA with G=κ (A/l)
whereas
Iw : Heat flux
G: Heat conductivity coefficient
κ: Thermal conductivity of the cooler material
A: Surface through which heat flows (contact surface between die and cooler)
l: Path taken by heat flux
TD : Maximum acceptable die temperature according to specifications
TA : Temperature of the air surrounding the fan
According to this equation, the driving force for the heat flux Iw (Iw stands for the heat dissipated by the processor) is the difference in temperature between the die and the internal case temperature. For a given difference in temperature, the heat flux will be greater (in other words, more heat will be dissipated) the higher the thermal conductivity coefficient (or rather, the lower the thermal resistance) of the cooler is. And for the thermal conductivity coefficient to be high, the thermal conductivity of the cooler material needs to be as high, the contact surface as large and the path taken by the heat as short as possible.
Summary | 412 | 1,954 | {"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} | 2.671875 | 3 | CC-MAIN-2019-39 | longest | en | 0.893781 |
https://socratic.org/questions/how-do-you-factor-7x-2-9x-2 | 1,582,513,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00061.warc.gz | 536,627,926 | 6,159 | # How do you factor 7x^2 - 9x + 2?
Jun 6, 2015
You find its roots and turn them into factors by equaling them to zero.
Let's find the roots using Bhaskara:
$\frac{9 \pm \sqrt{81 - 4 \left(7\right) \left(2\right)}}{14}$
$\frac{9 \pm \sqrt{81 - 56}}{14}$
$\frac{9 \pm \sqrt{25}}{14}$
$\frac{9 \pm 5}{14}$
${x}_{1} = 1$, which, equaled to zero is $x - 1 = 0$
${x}_{2} = \frac{2}{7}$, which, equaled to zero is $7 x - 2 = 0$
Thus, we can factor $7 {x}^{2} - 9 x + 2$ this way:
$\left(x - 1\right) \left(7 x - 2\right)$
Jun 6, 2015
$f \left(x\right) = 7 {x}^{2} - 9 x + 2.$
Since (a + b + c) = 0, one factor is (x - 1) and the other is $\left(- \frac{c}{a} = - \frac{2}{7}\right)$
f(x) = (x - 1)(7x - 2) | 326 | 712 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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} | 4.5625 | 5 | CC-MAIN-2020-10 | latest | en | 0.722099 |
https://www.bankersadda.com/quantitative-aptitude-quiz-for-fci-phase-i-2022-08th-november/ | 1,669,467,075,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00082.warc.gz | 709,588,811 | 115,553 | Latest Banking jobs »
# Quantitative Aptitude Quiz For FCI Phase I 2022- 08th November
Directions (1-15):- In each of the following questions, two equations (I) and (II) are given. Solve the equations and mark the correct option:
(a) if x>y
(b) if x≥y
(c) if x<y
(d) if x ≤y
(e) if x = y or no relation can be established between x and y.
Q1.
I. x²-25x+156=0
II. y²-29y+210=0
Q2.
I. x²=196
II. y=√196
Q3.
I. x²+12x+35=0
II. y²+14y+48=0
Q4.
I. 3x² + 23x + 30 = 0
II. y²+15y+56 =0
Q5.
I. x²+17x+72=0
II. y²+13y+42=0
Q6.
I. x²+17x+72=0
II. y²+11y+30=0
Q7.
I. 3x²-23x+40=0
II. 5y²-17y+14=0
Q8.
I. x²-26x+168=0
II. y²-29y+208=0
Q9.
I. x³+340=2537
II.y²+23=192
Q10.
I. x²+48x+575=0
II. y²+44y+483=0
Q11. I. x²+23x+132=0
II. y²+21y+110=0
Q12. I. 3x²+20x+32=0
II. 5y²+23y+24=0
Q13. I. x²-29x+208=0
II. y²-21y+108=0
Q14. I. x²+30x+224=0
II.y²+35y+306=0
Q15. I. x= ∛4096
II. y² =256
Solutions | 499 | 902 | {"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-2022-49 | latest | en | 0.221218 |
www.superdigitalhealth.com | 1,702,169,920,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00730.warc.gz | 968,990,118 | 27,023 | # How much is 165 F to C?
others
## 165 F to C Formula
165 F = 73.89 C .
The 165 Fahrenheit to Celsius formula is: [°C] = ([165] − 32) x 5 ⁄ 9. Therefore, we get:
165 F to C = 73.889 °C
165 F in C = 73.889 Celsius
As a side memo: the unit of temperature Fahrenheit calls after the German physicist Daniel Gabriel Fahrenheit.
In turn, the temperature Celsius unit terms after the Swedish astronomer Anders Celsius.
## 165 °Fahrenheit to Celsius Conversion
To convert Fahrenheit, start by deducting 32 from 165.
Then multiply 133 by five over 9 to obtain 73.889 degrees Celsius.
More accessible, however, is using our converter above.
Similar temperature conversions on our website include:
167 Celsius to Fahrenheit
168 Fahrenheit to Celsius
## What is 0 Degrees in Fahrenheit to Celsius?
So far, we have cast off the formula to change 165 °F to Celsius.
However, in daily life, the estimating formula explained on our home page sometimes suffices.
With that the approximate Celsius temperature is (165 – 30) / 2 = 67.5 °C.
In any case, a precise thermometer which displays both temperature units recommend.
Temperatures in degrees Fahrenheit and Temperatures in degrees Celsius mean the same if the word degree omits.
Thus: 165 Fahrenheit = 165 degrees Fahrenheit, 165 degrees Celsius = 165 Celsius.
165 Fahrenheit in other temperature units is:
• Newton: 24,383°N
• Kelvin: 347.039 K
• Reaumur: 59.111 °D
• Romer: 46.292 °Ro
• Delisle: 39.167°From
• Rankine: 624.67 °R
Ambient temperatures, such as refrigerators or human body temperatures, are usually expressed in °F or °C.
In contrast, the absolute temperature stipulated by Lord Kelvin is mainly used for scientific purposes.
## What is a Fahrenheit?
When using a thermometer, we must mark a scale on the tube wall with numbers. But, first, we have to define a temperature scale. A temperature scale measures temperature relative to a starting point (0 or zero) and a unit of measure.
These numbers are arbitrary, and many different schemes have been used historically. For example, this was complete by critical some physical incidences at given temperatures, such as water freezing and boiling points, and defining them as 0 and 100, respectively.
## What is Celsius?
Degrees Celsius is a unit of temperature measurement erroneously known as degrees Celsius and represented by the symbol °C. This unit pays homage to its creator, the Swedish physicist and astronomer Anders Celsius. It is equivalent in heat intensity to the kelvin scale so it can define with the following formula:
Temperature (°C) = Temperature (K) – 273.15
Paradoxically, William Thompson, creator of the Kelvin scale, created it based on the Celsius scale since it was later.
On the other hand, there is another temperature scale called the Fahrenheit scale. The conversion from degrees Celsius to degrees Fahrenheit is complete using the following formula:
Temperature (°F) = 1.8 x Temperature (°C)+ 32
The Celsius degree scale places its zero point (0) at about 0.01 degrees below the triple point of water: that in which the three states of matter, solid, liquid and gas, coexist in equilibrium.
## FAQs
• Which is colder, 165 °C or 165 °F?
• What do 165 degrees Fahrenheit mean?
• How much is 165 degrees Fahrenheit in Celsius?
• Which is warmer, 165 °C or 165 °F?
• What is 165 ° in Celsius?
• What are 165 degrees in Celsius?
• 165 °F is what °C?
## Conclusion
Sometimes you will come across Fahrenheit temperatures, but you must have them in Celsius. In this case, 165 Fahrenheit = 73.89ºC. This result obtains by performing a calculation that we previously taught you. However, a simple solution is to use our converter. | 883 | 3,702 | {"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} | 3.375 | 3 | CC-MAIN-2023-50 | latest | en | 0.766881 |
http://mathhelpforum.com/advanced-statistics/279237-expectancy-number-trials.html | 1,516,291,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887423.43/warc/CC-MAIN-20180118151122-20180118171122-00436.warc.gz | 225,980,501 | 14,721 | # Thread: Expectancy of number of trials
1. ## Expectancy of number of trials
Five cards are drawn out of a stock of cards. If the cards that have been drawed contain four cards of the same type (Ace, King,same number,queen,etc.),we say we have a a quartet. Repeating the process until all 13 possible quartets are received, what is the expected number of repetitions required?
2. ## Re: Expectancy of number of trials
I assume there are 52 cards, because if there are more or less than 52, we would need to know the number of cards of each type in the "stock". I assume that each time the process is repeated, the five cards pulled the previous time are replaced (otherwise, we would never be able to get all 13 quartets).
It is impossible to achieve all 13 quartets in less than 13 repetitions. So, if we are to have a chance at succeeding on the 13th pull, we must have 12 of the quartets already pulled and the 13th quartet will be the 13th pull. Similarly, if we pull the 13th quartet on the 14th pull, then in the first 13 pulls, we got only 12 of the quartets, but not the 13th. Let's consider the probability that in n pulls, we pull 12 of the quartets, but not the 13th:
$\dbinom{13}{12}\dbinom{n}{12} \left( \dfrac{ 48\dbinom{5}{4} }{ \dbinom{52}{5} } \right)^{12} \left(1 - \dfrac{ 48\dbinom{5}{4} }{ \dbinom{52}{5} } \right)^{n-12}$
Then the probability that the next pull is the 13th quartet is: $\dfrac{ \dbinom{5}{4} }{ \dbinom{52}{5} }$
So, the expected value is given by:
\displaystyle \begin{align*} & \sum_{n \ge 12}\dbinom{13}{12}\dbinom{n}{12} \left( \dfrac{ 48\dbinom{5}{4} }{ \dbinom{52}{5} } \right)^{13} \left(1 - \dfrac{ 48\dbinom{5}{4} }{ \dbinom{52}{5} } \right)^{n-12}(n+1) \\ = & \dbinom{13}{12} \left( \dfrac{ 48\dbinom{5}{4} }{ \dbinom{52}{5} } \right)^{13} \sum_{n \ge 12} \dbinom{n}{12} \left(1 - \dfrac{ 48\dbinom{5}{4} }{ \dbinom{52}{5} } \right)^{n-12}(n+1) = 1830101\end{align*}
http://www.wolframalpha.com/input/?i...infinity%7D%5D
Edit: I missed a factor of 48.
3. ## Re: Expectancy of number of trials
I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is
13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!).
Do you see a way of computing the expectancy by using indicators?
4. ## Re: Expectancy of number of trials
Originally Posted by hedi
I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is
13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!).
Do you see a way of computing the expectancy by using indicators?
I apologize. I missed a factor of 48 for the 5th card that was not part of the quartet. I do not understand what you mean by "indicators". I updated my calculations.
5. ## Re: Expectancy of number of trials
Also, you can try the experiment with a smaller deck to see if the formula is correct. For example, suppose you have a deck with 24 cards: 4x2, 4x3, 4x4, 4x5, 4x6, 4x7. Find the expected number of pulls to get all six quartets. According to my formula, that would be:
\displaystyle \begin{align*} & \sum_{n \ge 5}\dbinom{6}{5}\dbinom{n}{5} \left( \dfrac{ 20\dbinom{5}{4} }{ \dbinom{24}{5} } \right)^{6} \left(1 - \dfrac{ 20\dbinom{5}{4} }{ \dbinom{24}{5} } \right)^{n-5}(n+1) \\ = & \dbinom{6}{5} \left( \dfrac{ 20\dbinom{5}{4} }{ \dbinom{24}{5} } \right)^{6} \sum_{n \ge 5} \dbinom{n}{5} \left(1 - \dfrac{ 20\dbinom{5}{4} }{ \dbinom{24}{5} } \right)^{n-5}(n+1) = 15301.44\end{align*}
Wolfram|Alpha: Computational Knowledge Engine
6. ## Re: Expectancy of number of trials
Yes, but the 12 quarters in the first n trials can be permutated so there must be a factor of 12!.Similar consideration in computing the probability of a quarter in a single trial.
7. ## Re: Expectancy of number of trials
Originally Posted by hedi
Yes, but the 12 quarters in the first n trials can be permutated so there must be a factor of 12!.Similar consideration in computing the probability of a quarter in a single trial.
It is still not correct. I just tried it for the 24 card deck that I showed above, and it turns out that the expected number should be MUCH lower. I ran it 50 times and got an average of 412.611111 repetitions. So, I will need to rethink my solution. It is obviously not correct. Multiplying by 5! certainly would not help, as that would make it an even larger number.
8. ## Re: Expectancy of number of trials
It must be the stock with 52 cards not less.This experiment shouid be repeatedly run and see if the first time all quartets appear is close to our result.If you take less cards the expectancy will obvioulisly be smaller.
9. ## Re: Expectancy of number of trials
Originally Posted by hedi
It must be the stock with 52 cards not less.This experiment shouid be repeatedly run and see if the first time all quartets appear is close to our result.If you take less cards the expectancy will obvioulisly be smaller.
You misunderstand my point. I calculated a NEW expected value using a 24 card deck (4 of each card in the deck). I used the same method, but smaller numbers. The expected value was WAYYY too high. A computer program running a monte carlo simulation of this problem would take a long time. I used a smaller subset of the problem to be able to actually run the simulation. The fact that the method is wrong for a 24-card deck shows that it is wrong for a 52-card deck. There is no reason to run the simulation for the 52-card deck to prove the same results. The method I used was flawed.
10. ## Re: Expectancy of number of trials
Let's try to solve this using recursion:
P(n,r) = Probability of getting r distinct quartets in n repetitions (the multiplicity of each of the r quartets is at least 1)
p = Probability of getting a specific quartet (for example, 4 aces and another card) in a single repetition. We know that $p = \dfrac{48\cdot 5}{ \dbinom{52}{5} }$.
P(r) = Probability of getting any of r distinct quartets in a single repetition. We know that $P(r) = rp$
P(!r) = Probability of not getting any of r distinct quartets in a single repetition. We know that $P(!r) = 1-rp$
Then, we have the following recursion:
$P(n,r) = P(n-1,r)P(!(13-r))+P(n-1,r-1)P(14-r) = (1-(13-r)p)P(n-1,r)+(14-r)pP(n-1,r-1)$
where
$\forall n<r: P(n,r) = 0$
$\forall r>13: P(n,r) = 0$
$\forall 0 \le n \le 13: P(n,n) = \dfrac{13!}{n!}p^n$
We are trying to calculate the expected value:
$\displaystyle \sum_{n\ge 12}(n+1)pP(n,12)$
11. ## Re: Expectancy of number of trials
I tried working through this a little. It is messy as could be. I doubt you will find a closed solution. You may be able to find the answer using a statistics engine like R.
12. ## Re: Expectancy of number of trials
The expectancy can be computed as the sum of geometric expectancies of the geometric variables counting the number of experiments needed to obtain the next different quartet.This gives:
(1/48)*bin(52,5)Sum(1/n) from 1 to 13
13. ## Re: Expectancy of number of trials
Originally Posted by hedi
The expectancy can be computed as the sum of geometric expectancies of the geometric variables counting the number of experiments needed to obtain the next different quartet.This gives:
(1/48)*bin(52,5)Sum(1/n) from 1 to 13
That is still not correct unfortunately.
14. ## Re: Expectancy of number of trials
Originally Posted by SlipEternal
That is still not correct unfortunately.
Never mind. It turns out that my random function suffered from a predictability issue. It was generating far more "quartets" that in should have. I give up. Maybe Plato can help with this one. Now, I am getting ridiculously large expected values (several orders of magnitude larger). I tried using R to calculate the correct expected value. It did not work out.
15. ## Re: Expectancy of number of trials
Originally Posted by SlipEternal
Never mind. It turns out that my random function suffered from a predictability issue. It was generating far more "quartets" that in should have. I give up. Maybe Plato can help with this one. Now, I am getting ridiculously large expected values (several orders of magnitude larger). I tried using R to calculate the correct expected value. It did not work out.
No help from here. I simply have no idea what you all have been doing.
Because the question seems to me to be asking the same as the coupon collector problem.
Therefore, I must not understand the setup.
Page 1 of 2 12 Last | 2,501 | 8,494 | {"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} | 4.4375 | 4 | CC-MAIN-2018-05 | longest | en | 0.842643 |
https://findingthewayout.wordpress.com/category/prediction/ | 1,531,791,416,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589537.21/warc/CC-MAIN-20180717012034-20180717032034-00457.warc.gz | 641,623,922 | 15,962 | # Permutation symmetry and finding a noise model’
If we have a time series of data points $\{ x_i\},$ how can we tell if these come from some stationary distribution or if there is some secular process coupled to an uncertainty inducing step? Many statistical theorems start out by assuming stationarity. Suppose we ask for the probability of a model $P(f,\eta|x_i)$ where $x_i = f(t_i) + \eta_i,$ $t_i$ is the time coordinate’ of the $i^{\rm th}$ data point, and $\eta$ comes from a stationary distribution. How do we disentangle $f$ and $\eta?$ By definition, the $f$ likelihood should be invariant under replacing $\eta_i \rightarrow \eta_{\sigma(i)}, \sigma \in P_N,$ the permutation group on $N$ data points. In other words, $P(f,\eta|D) = P(f,\eta_\sigma|D),$ but then this is going to require $P(D|f,\eta) = P(D|f,\eta_\sigma)$ which suggests the trivial solution of just defining $P(D|f,\langle\eta\rangle)$ as the product over $\sigma,$ which seems a bit heavy-handed. It is easily achieved, of course, but this will be a computational nightmare. If this could be done, I think it would be a pretty strong constraint on what $\eta$ and $f$ could be.
One way to try to do this would be to do Monte-Carlo sampling over $P_N$ in calculating the cost function at each step of the MCMC for finding $f, \eta.$ We really want to evaluate something like
$- \log P(x|f,\langle\eta\rangle) \equiv \sum_{\sigma\in P_N} \sum_i (x_i - f(t_i) - \eta_{\sigma(i)})^2,$
and I think that for $N$ large enough the sum over $\sigma$ could be approximated by a probabilistic sampling over some permutations. The worry here is that $\eta$ will come out all constant, so the noise model preferred will be the trivial distribution, so all the variation will try to be fit by $f.$ The other extreme is that $f$ will be a constant and all the data will be fit by $\eta.$
The quick way might be to use something like singular spectrum analysis on each permuted summand and demand that the SSA interpolations all agree: Pick $\eta,$ compute SSA for some sample of permutations, vary $\eta$ in a standard MCMC to sample over possible $\eta.$ The end result should be a de-noised prediction for the actual functional dependence $f.$
# Ray Solomonoff’s universal distribution
Everyone knows about Kolmogorov complexity. Kolmogorov addressed randomness and came up with the Kolmogorov complexity $C$of a string (defined as the length of the shortest program that can reproduce the string) as a measure of its randomness. Solomonoff, two years before Kolmogorov, looked at the problem from a much deeper perspective in my opinion. He asked, in drastic paraphrase: How should you weight different programs that could generate a given string? And the two are related because it turns out that the function that Solomonoff picked is $C.$ [So why isn’t it called Solomonoff complexity, though Kolmogorov referenced Solomonoff in his paper? Because Kolmogorov was already famous.]
What I would like to understand is the following: How do you go about generating programs that could reproduce the string? Induction is a hoary field, but most of the work is done assuming that you have to decide between different programs, and very little, in proportion, addresses the question of how to go about generating such programs. That is actually a problem because the search space is, to put it mildly, large. Is there a fundamental process that is the optimal manner to induce programs given data? This must itself be a probabilistic process. In other words, the program output match to the desired string must be allowed to be imperfect, so the search space summation is not just over programs that reproduce a given string $S$
$\sum_{{\rm programs\ reproducing\ } S} \exp(-{\rm length}({\rm program}))$
but rather weighted in some way to balance fitting $S$ and reducing program length. Why does everything involve a balance between energy and entropy? Because there is nothing more fundamental in all of science.
So I think there should be a way to write something like
$\sum_{S',{\rm programs\ producing\ } S'} \exp(-{\rm length}({\rm program})) \exp(- {\rm mismatch}(S,S'))$
and then the question is: What is the universal form of the mismatch function? An interesting point here is that this process might work even with noisy $S$ since if the correct string is $0101010101$ but you read $S= 0111010101,$ then you’ll find a very short program that can reproduce the string $S' = 0101010101,$ and since the mismatch is only one bit this process would autocorrect your initial misread string.
We want symbols and rules for manipulation such that the resulting stream contains the known sequence of symbols. In other words, an inverse Godel problem: Given a stream of symbols, find a set of rules so that this stream is a proof.
# Normalizing gene expression values for classification
What’s the best way to normalize an array of gene expression values? First question: Best for what?
This depends on what you want to do with the array. The requirements for sample classification seem to me to be a little different than for, say, class discovery. What we want for sample classification is to normalize each array in a self-referential manner without taking into account the rest of the samples, because we want to be able to do the exact same normalization on any new sample that we might want to classify, given what we’ve learned from the training set. So suppose we normalize by total mRNA. Now all samples will be points on some high-dimensional hypersurface, but if some gene had high expression values in disease samples, the other genes in disease samples will now have depressed values because of the overall normalization. That doesn’t mean that those genes are biomarkers for that disease. On the other hand, we could arbitrarily declare that some genes are `household’ genes and normalize one (or several) to a fixed expression level, with rescalings of all the rest. This avoids the problem of the total mRNA normalization but exactly how do you decide that a gene is a household gene without taking into account the entire set of samples? And if you do pick the entire set of samples to decide what is a household gene (i.e. one that shows no correlation with the known classification, for example), then you are not normalizing in a sample-specific manner.
So, in some sense we want to live in the projective space associated with gene expression, and the relevant data is the gene expression values up to a constant. We’re picking charts on this projective space with all these different normalizations but what we should do is sample classification in a coordinate invariant manner. So how do we do this? Take any fixed gene, something with a median value for the normal samples for example (this is just bowing to experimental reality, and has no theoretical justification), and normalize all samples $S_i$ so that this gene is 1. Now $d(S_1,S_2) \equiv \sum ((S_{1i}-S_{2i})/\sigma_i)^2$ with $\sigma_i$ the uncertainty in $g_i$ expression is a measure of the distance in projective space. We could also compute the distance on a chart adapted to a sphere by normalizing each $S$ and then computing $\arccos S_1\cdot S_2,$ with appropriate $\sigma_i$ in various places of course.
With such a measure of distance, the first thing one might do is compute the distribution of distances between all the normal samples. It’s a common view that healthy people are all alike, every unhealthy person is unhealthy in his/her own way (with apologies to Tolstoy). So I would imagine that one would get more insight in seeing how the distribution of disease sample-to-normal samples distances looks different from the normal-to-normal distribution.
Up next: PCA on projective space??
# Stochastic grammars and grammatical evolution
I’ve been wondering how to use grammatical evolution to generate signaling networks. So first we have to think up some sort of grammar for signaling networks. What would be appropriate start symbols? Productions? Terminals?
Start: Gene
Transcription: Gene > Gene + RNA (constitutive expression) | Gene*TF | Gene*Inhibitor
Transcription: Gene*TF > Gene + RNA | Gene*TF[*Cofactor]^n | Gene*TF*Inhibitor
Transcription: Gene*TF*Cofactor > Gene + RNA
Signaling: Receptor > Receptor*SIgnal | Receptor*Blocker
Degradation: Any > Nothing
and so on
People have done this sort of thing before, obviously, but I’m wondering about how applying genetic mutation operators to a string of such productions will lead to the same sort of changes to gene networks that are actually observed. Not obvious to me …
What happens if you use a stochastic grammar? What’s the difference between a stochastic grammar applied many times to a fixed genome vs a deterministic grammar applied to a population of genomes? In biology, the binding of TFs may actually be stochastic, so perhaps we should encode the probability of a symbol in the genome going to a particular production in the genome itself.
# Robustness vs adaptability
So much of science depends on how well you can communicate. I’m now on the least favorite part of any research project from my perspective: The communication part.
Here’s the question that motivated this particular bit: Natural selection acts on the phenotype, i.e. on the adaptability and competence of a particular individual. The search for improvement works by mutating the genotype, and unfortunately(?), even if there is a beneficial mutation in the germline, the individual whose progeny will carry that mutation does not benefit at all in its lifetime. Now: If mutations alter the phenotype to be inherited, then the genotype that leads to the selected parent phenotype will not really play much of a role since mutations will provide the progeny with a phenotype that may not be as good. So there has to be some sort of interplay between mutation resistance and adaptability, but in principle a dynamical system that is adaptable may not be stable to parameter perturbations (due to mutations) and vice versa.
So I suggested a little thought experiment to my postdoc (Zeina Shreif): Imagine two parts of a network, one which receives some environmental input and another that gets some signal from this part and produces some output. If the system is adaptable, we expect that it should be able to maintain the same output for a change in input. However, how exactly can the output part of the network distinguish between a change in the input or a change in some parameter in the input part of the network? So I conjectured that in fact these two properties must be correlated, that statistically homeostatic networks will be robust with respect to parameter fluctuations.
It turns out that this is true. Heroic work by Zeina has demonstrated this all the way up to 50 node random networks! So this is an exciting (to me) result, and so it is worth it to try to write this in a way that people will understand it. The problem is that there are so many exciting ramifications that I am very reluctant to do the careful writing. Must be done 😦
# Inference and prediction: Dimon 0 Volcker 1
Jamie Dimon finally gets his. Chase posted a 2 billion \$ loss because of a trade that they didn’t do enough risk management on. Too funny.
From DealBook:
>> On Wall Street, few have been more outspoken about the pitfalls of the Volcker Rule than JPMorgan’s chief executive, Jamie Dimon. Mr. Dimon not only attacked the rule, he personally criticized Paul Volcker, the former Federal Reserve chairman and the regulation’s namesake.
“Paul Volcker by his own admission has said he doesn’t understand capital markets,” Mr. Dimon told Fox Business earlier this year. “He has proven that to me.” ….
Even Mr. Dimon had to admit Thursday’s disclosure was a setback for JPMorgan and other banks that want more flexibility when the final version of the Volcker Rule is issued. “It plays into the hands of a bunch of pundits but you have to deal with that and that’s life,” Mr. Dimon said Thursday on a conference call with analysts. …
“Just because we’re stupid doesn’t mean everybody else was,” he said. “There were huge moves in the marketplace but we made these positions more complex and they were badly monitored.”
“This may not have violated the Volcker Rule, but it violates the Dimon Principle.” <<
Apparently, Mr. Dimon doesn’t understand capital markets either.
On a constructive note, I think some of these problems arise due to an inadequate understanding of how to model probability distributions from a finite amount of data and how to automatically learn changing distributions. These issues are exactly the same as the conflict in evolution: You want stability to propagate your genome, but you need variation to allow for the emergence of new traits that can handle changes in the environment. It isn’t easy to come up with dynamical systems that meet both of these desiderata. | 2,865 | 12,944 | {"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": 47, "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-2018-30 | latest | en | 0.876813 |
https://www.math-forums.com/threads/convert-to-rectangular-form-7.442238/ | 1,722,655,940,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00547.warc.gz | 677,225,931 | 11,321 | # Convert to Rectangular Form...7
Discussion in 'Geometry and Trigonometry' started by nycmathguy, Mar 19, 2022.
1. ### nycmathguy
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Cam you get me started with 11 and 13 by providing the first 2 steps? I will then try solving the problems on my own.
nycmathguy, Mar 19, 2022
2. ### MathLover1
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11.
r=3cos(2θ)
first apply the formula cos(2θ)=cos^2(θ)-sin^2(θ)
r= 3cos^2(θ)-3sin^2(θ)
then use
x=rcos(θ) and y=rsin(θ)
you continue
13.
r^2=8/(2-sin^2(θ))
From x=rcos(θ) and y=rsin(θ), we have that
cos(θ)= x/r
sin(θ)= y/r
tan(θ)= y/x
cot(θ)= x/y
The input becomes
r^2 =8/(2- y^2/r ^2)
you continue
MathLover1, Mar 19, 2022
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3. ### nycmathguy
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nycmathguy, Mar 19, 2022
4. ### MathLover1
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11. need to finish, answer cannot have r , only x and y
13. correct
MathLover1, Mar 19, 2022
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5. ### nycmathguy
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Can you finish 11 for me? I can't further simplify.
nycmathguy, Mar 19, 2022
6. ### MathLover1
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11.
substitute
........ both sides raise to power of 3
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Ceck the user's input and display a message indicating wheter it is the correct answer.
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Starting Out With Visual Basic 2012 (6th Edition)
Q:
Addition TutorCreate an application that generates two random integers, each in the range of 100 through 500. The numbers should be displayed as an addition problem on the application’s form, such as247 + 129 = ?The form should have a text box for the user to enter the problem’s answer. When a button is clicked, the application should do the following:• Check the user’s input and display a message indicating whether it is the correct answer.• Generate two new random numbers and display them in a new problem on the form. | 253 | 1,150 | {"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-2016-30 | latest | en | 0.845202 |
https://endless-sphere.com/forums/viewtopic.php?f=2&t=108432 | 1,604,129,178,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00112.warc.gz | 287,911,036 | 13,167 | ## FEM Dropout analysis - max force ?
qwerkus 10 kW
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### FEM Dropout analysis - max force ?
Hello,
I'm trying to setup a simple FEM static analysis of a bike dropout to figure out metal breaking point. Model is done and software seems to work. Though I could use some help with the physics. For an instance, what's the max. force a 12mm hub motor axle edge would put on a dropout face if the motor has a torque rating of 100 n.m-2 ?
force.png (13.45 KiB) Viewed 397 times
I get that 100n.m-2 means a force of 100N perpendicular to the axle, applied at a distance of 1m. Now if I take this same force 5mm from the axle center, I get 1000/5*100 = 20 000N which seems just silly. What am I doing wrong ?
qwerkus
serious_sam 1 kW
Posts: 436
Joined: Mar 05 2017 8:07am
Location: Australia
### Re: FEM Dropout analysis - max force ?
Just so we're all speaking the same language, the unit of torque is Nm (Newton metres). Not n.m-2.
So 100Nm @ 5mm is 100*1000/5=20 000N. You are correct. Since it is reacting in 2 places, each load will be 10kN.
The problem you will have is that the force is acting on a line. Since a line has no area, and stress=force/area, in a static linear analysis you end up with a thing called a singularity (effectively a point of infinite stress). So the result at the point of contact is not realistic. Sharp internal corners have the same effect. This is one limitation of static linear analyses.
In reality, materials and contacts are non-linear, and localised yielding allows for stress relief, and things don't turn into black holes...
There's a thing called St Vennant's principal, which basically says that if you ignore that localised area of unrealistic stress, the rest of the results away from that area are probably fine. Experience counts when setting up and interpreting FEA.
E-HP 1 MW
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Joined: Nov 01 2018 9:20pm
### Re: FEM Dropout analysis - max force ?
Maybe I'm not reading the inputs correctly, but if it's a 12mm axle, why are you using 5mm in your calculation instead of 6mm?
serious_sam 1 kW
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Location: Australia
### Re: FEM Dropout analysis - max force ?
Since the area of concern is AF10mm, the distance is slightly above 5mm. But in reality, even the geometry is not so simple, because the thread is truncated which creates a serated contact line. For intents and purposes, simplifying and approximating gives ROM results.
E-HP 1 MW
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Joined: Nov 01 2018 9:20pm
### Re: FEM Dropout analysis - max force ?
serious_sam wrote:
Sep 25 2020 10:45pm
Since the area of concern is AF10mm, the distance is slightly above 5mm. But in reality, even the geometry is not so simple, because the thread is truncated which creates a serated contact line. For intents and purposes, simplifying and approximating gives ROM results.
I get 5.45, but doesn't make much of a difference in the calc.
qwerkus 10 kW
Posts: 590
Joined: Jul 22 2017 4:00am
### Re: FEM Dropout analysis - max force ?
serious_sam wrote:
Sep 25 2020 10:07pm
Just so we're all speaking the same language, the unit of torque is Nm (Newton metres). Not n.m-2.
So 100Nm @ 5mm is 100*1000/5=20 000N. You are correct. Since it is reacting in 2 places, each load will be 10kN.
The problem you will have is that the force is acting on a line. Since a line has no area, and stress=force/area, in a static linear analysis you end up with a thing called a singularity (effectively a point of infinite stress). So the result at the point of contact is not realistic. Sharp internal corners have the same effect. This is one limitation of static linear analyses.
In reality, materials and contacts are non-linear, and localised yielding allows for stress relief, and things don't turn into black holes...
There's a thing called St Vennant's principal, which basically says that if you ignore that localised area of unrealistic stress, the rest of the results away from that area are probably fine. Experience counts when setting up and interpreting FEA.
Thanks a lot for your explanations. This proves to be less simple than expected but also a lot more fun! I ran a sim for backwards rotation (regen) on the dropouts with 8mm 304 type stainless steel, using 10kn top force (axle to dropout) on a 16mm2 surface and 5Kn bottom force (anti-turn washer) on a 32mm2 surface. Not sure how realistic this is, but it doesn't look great for now, as I get over 1035Mpa von mises stress and 0.2mm total displacement. That's 5 times more than the yield limit for that steel...
fem1.jpg (173.19 KiB) Viewed 366 times
fem2.jpg (436.4 KiB) Viewed 366 times
Still some optimizations to do...
serious_sam 1 kW
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Location: Australia
### Re: FEM Dropout analysis - max force ?
The displacement is probably realistic. The peak stress is like I described, an area of local yielding, which can't be simulated in a linear analysis. And a sharp internal edge causes the software to miscalculate. I'm fairly certain that if you halved the mesh size, you would approx double that peak stress seen. Each refinement of the mesh in that area would produce a further increase in stress, to infinity.
To fix the stress, you can add a small fillet in that area. To fix the deflection, you might want to consider a clamping bolt.
serious_sam 1 kW
Posts: 436
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Location: Australia
### Re: FEM Dropout analysis - max force ?
Just another hint. Not sure what software you're using, so I'm not sure what level of control you have, but having such a fine mesh over the entire part is costing significant computational time. If you can, try running a much coarser mesh, and only refine it around the areas of interest.
If you zip your CAD geometry (step or parasolid) then rename the file extension to pdf, you can upload it here as an attachment, and I can take a quick look in Ansys if you want. Give you a point of reference to compare to anyway.
qwerkus 10 kW
Posts: 590
Joined: Jul 22 2017 4:00am
### Re: FEM Dropout analysis - max force ?
serious_sam wrote:
Sep 26 2020 4:23am
Just another hint. Not sure what software you're using, so I'm not sure what level of control you have, but having such a fine mesh over the entire part is costing significant computational time. If you can, try running a much coarser mesh, and only refine it around the areas of interest.
If you zip your CAD geometry (step or parasolid) then rename the file extension to pdf, you can upload it here as an attachment, and I can take a quick look in Ansys if you want. Give you a point of reference to compare to anyway.
Thanks for the useful hints. I'm doing everything on freecad. Followed your advice and added a 0.5mm filled. Also used mesh region to speed things up. Results as you predicted: the filled produces a finer meshing which results in increase von Mises stress. See picture.
fem3.jpg (210.21 KiB) Viewed 347 times
File too big for this forum. Send me a pm to this address and I send you the step file: qwerkus at gmail com.
Yes, a clamping bolt would be the optimal solution, but I try to keep this as simple as possible. The whole point of this modular dropout would be to figure out a design/material strong enough to take 1000W hubs without annoying torque arms.
serious_sam 1 kW
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Location: Australia
### Re: FEM Dropout analysis - max force ?
Email sent.
serious_sam 1 kW
Posts: 436
Joined: Mar 05 2017 8:07am
Location: Australia
### Re: FEM Dropout analysis - max force ?
Here's some preliminary results.
Setup info:
- I added the axle geometry, and applied 50Nm torque over 20 substeps. Since there are 2 dropouts, then the total load would be 100Nm combined.
- The dropout is restrained at the 3x mounting holes.
- Applied µ=0.2 friction to the contact between axle and dropout. In reality this would be slightly higher, but 0.2 is a conservative estimate.
- Mesh size is 4mm, refined to 0.5mm in the contact areas.
- The dropout material is 304SS bi-linear (YS=240MPa, UTS=580MPa, e=55%). This is a simplified way of modelling the material in a non-linear manner, which approximates the plastic deformation process in a conservative and simple manner.
- The axle material is linear steel. Since we are analysing the dropout, we simplify the analysis in a conservative manner by using a linear material for the axle. It provides the stiffness of steel, but ignores any yielding, since we don;t actually know the strength of the axle anyway.
- The analysis is run as non-linear, with the 50Nm load applied over 20 substeps.
- The dropout material begins to yield (>240MPa = red in the image below) at around substep 14, which equates to 35Nm each dropout, or 70Nm for a pair.
- This setup has no axle nut applying friction. An axle nut would slightly increase the load capacity.
- This model doesn't have the axle threads modelled. Threads increase the stress at the loaded contact areas.
- I would seriously recommend adding a clamp. Without the clamp, there is small movement. If you run regen, then you will definitely have problems with axle nuts coming loose, and eventually wear, which may lead to failure.
If I have some time over the next couple of days, I will rerun the same analysis with a small clamping screw just to compare.
01stress.jpg (339.04 KiB) Viewed 290 times
01displacement.jpg (342.69 KiB) Viewed 290 times
serious_sam 1 kW
Posts: 436
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Location: Australia
### Re: FEM Dropout analysis - max force ?
Here's the same analysis, with the following modifications:
- I added a small clamping bolt, screwed into the body of the dropout (M5, A2-70 stainless, 5.1Nm torque, 3.93kN preload)
- Analysis broken into 2 steps: 1st step=bolt preload; 2nd step=torque application.
- Increased torque to 100Nm (again over 20 substeps)
Results:
- The dropout material begins to yield (>240MPa = red in the image below) at around substep 12, which equates to 60Nm each dropout, or 120Nm for a pair.
- So just adding a small clamping bolt increases torque handling from 35Nm to 60Nm - nearly double. AND, there is minimal relative movement, so regen will not cause the same issues. Overall a much safer and stronger arrangement.
Dropout 02 stress.jpg (454.74 KiB) Viewed 257 times
Dropout 02 displacement.jpg (418.83 KiB) Viewed 257 times
John in CR 100 GW
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### Re: FEM Dropout analysis - max force ?
Newton meters in the dropout is crazy newtons of force at the small radius, so yes it's thousands of newtons of force in those dropouts.
DogDipstick 10 kW
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Location: Fleetwood Pa
### Re: FEM Dropout analysis - max force ?
wHY STAINLESS?
83.1v of Ironhorse XC.. by Chevy Broke 10 horsies (..about 80% healed..).. Waddyamean? You cant tell me how many amperes/Ft.^2 of the plate ?!?!? WTF. (gottenymoem4115thangs?Yall?) Fabricator @ BSECo. Inc.
qwerkus 10 kW
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### Re: FEM Dropout analysis - max force ?
serious_sam wrote:
Sep 29 2020 5:24am
Results:
- The dropout material begins to yield (>240MPa = red in the image below) at around substep 12, which equates to 60Nm each dropout, or 120Nm for a pair.
- So just adding a small clamping bolt increases torque handling from 35Nm to 60Nm - nearly double. AND, there is minimal relative movement, so regen will not cause the same issues. Overall a much safer and stronger arrangement.
Dropout 02 stress.jpg
Dropout 02 displacement.jpg
Wow - awesome feedback. Thanks a lot. This confirms diy tests from this forum: clamping bolt beats material strength in dropout design. Problem is manufacturing: drilling an M5 hole sideways in 8mm stainless is a pain. Only 1.5mm left on each side - a slight deviation from the drill, and you're through. Maybe M4 would be enough ?
Stainless is not a must, but it means no need for paint, which makes it cheaper for small series. CroMoly could also work.
I wouldn't worry too much about threads causing cracks: the hubs I have in mind are not threaded at the dropout holding zone.
Another option left open by the modularity of the dropout is a closed axle hole. That would mean that you'd have to remove the entire dropout when repairing a flat and the absence of clamping force will always allows for some wiggle in regen mode. But it could be manageable if we reduce the amount of bolts holding the dropout to 2 per side. What does your sim software says about this ?
Finally: send me a paypal address or equivalent - there is no way this kind of help remains unpaid.
Last edited by qwerkus on Sep 29 2020 2:08pm, edited 1 time in total.
John in CR 100 GW
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Joined: May 20 2008 12:58am
### Re: FEM Dropout analysis - max force ?
serious_sam wrote:
Sep 29 2020 5:24am
- So just adding a small clamping bolt increases torque handling from 35Nm to 60Nm - nearly double.
"Nearly double" is not anywhere close to the real world benefit. There is never precision machining of the dropouts to the axle size, and a press fit would be impractical anyway, so there is always play. The forces are then more focused at 2 edges of the flats instead of across the entire face of the axle flat, and with regen the only thing to prevent rocking within the play are the axle nuts, which quickly loosen due to the alternating force.
Through hole torque arms suffer from the same looseness issue.
serious_sam 1 kW
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### Re: FEM Dropout analysis - max force ?
John in CR wrote:
Sep 29 2020 12:23pm
serious_sam wrote:
Sep 29 2020 5:24am
- So just adding a small clamping bolt increases torque handling from 35Nm to 60Nm - nearly double.
"Nearly double" is not anywhere close to the real world benefit.
Agreed on the additional benefits of the clamp. The nearly double increase is purely double of the torque to reach the yield limit of the dropout.
serious_sam 1 kW
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Location: Australia
### Re: FEM Dropout analysis - max force ?
qwerkus wrote:
Sep 29 2020 7:25am
Wow - awesome feedback. Thanks a lot...send me a paypal address or equivalent - there is no way this kind of help remains unpaid.
No worries. It's right in my wheelhouse, so it's quick and easy to do. You don't need to pay for free advice.
qwerkus wrote:
Sep 29 2020 7:25am
Maybe M4 would be enough ?
You could go M4. A2-70 stainless is too weak though. Class 8.8 is right on the limit. It would be ok if you go to class 12.9 socket head and torque to 4Nm. Need to have minimum 8mm engagement in the dropout.
serious_sam 1 kW
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Location: Australia
### Re: FEM Dropout analysis - max force ?
Something related:
viewtopic.php?f=2&t=68843
Torque-arm-clamping-iso.jpg (19.16 KiB) Viewed 146 times | 3,936 | 14,880 | {"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} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.907313 |
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Jayesh Vashishtha
4 months ago
Dear student,
As we know that,
Rate=k[conc.]^order(here order is 2)
so, Rate=k[conc.]^2 and the graph between rate and conc^2 will be same as y =mx. i.e straight line.
similarly, for rate vs[conc.]^3 will be straight line.
Regards | 150 | 505 | {"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} | 2.828125 | 3 | CC-MAIN-2022-21 | latest | en | 0.86935 |
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You are working on a laptop connected to a $100 \text{Mbps}$ Ethernet LAN. You need a $2 \text{GB}$ file that is on the server in the same LAN. The entire file is also on your pen drive but you have left the pen drive in another room. You have a dog sitting beside you, that is trained to bring the pen drive to you. The average speed of the dog is $20 \text{km/hour}$. Up to what distance (in meters) does the dog have the higher data rate than $100 \text{Mbps}$ Ethernet? (Assume $1K=1000$, up to 2 decimal places)
(Note: The dog should be able to go and bring the pen drive, before the transfer on the LAN completes. Assume continuous data transmission on the LAN(no packetization required)).
I got the answer as 888.88. Looking at these comments, I got that I needed to divide it by 2... I understood the calculations, but I just can't wrap my head around this line
Up to what distance (in meters) does the dog have the higher data rate than 100Mbps Ethernet?
444.44m would be the maximum distance the dog could go bring the pen drive. But what does that have to do with the dog having "higher data rate"? Lol. This might be a silly doubt, I realise, but can someone please help me get this line?
@JashanArora Could you eplain me how 160 secs will be the RTT for the dog? i am not able to get it.
Because the dog has to not just reach the pen-drive, but also be able to bring it back. That's a round-trip, not one-way.
Time taken by dog to go and comeback,
T(dog) = t(go) + t(come) = 2 * t(go) = 2 * d/ 20kmph = 18d/50 sec
Within this time 2GB data has to be transfered,
T (transfer) = data size/link bandwidwth = 2GB/100Mbps = 160 sec
Now,
T(dog) < T(transfer)
=> 18d/50 < 160
=> d = 160 * 50 /18
=> d = 444.444
or d = 444 m
1
823 views | 504 | 1,770 | {"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-2022-49 | latest | en | 0.944836 |
https://www.java-forums.org/java-2d/26726-clicking-oval-jpanel.html | 1,519,312,373,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814124.25/warc/CC-MAIN-20180222140814-20180222160814-00268.warc.gz | 906,878,270 | 15,973 | # Thread: Clicking on an oval in JPanel
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## Clicking on an oval in JPanel
I am making a paint applet, which draws shapes on a 'canvas' by storing shapes into an arraylist (using the Graphics class).
I need to be able te erase complete shapes by clicking them with the eraser tool.
I managed to make this work for rectangles, filled rectangles and lines. I am not managing to do this with ovals. This because the radius is not equal along the entire circumference. How can I calculate whether a user clicked withing the oval, or on the (variable) stroke of an open oval?
For an example this is what the code for the filled rectangle:
Java Code:
// p1 and p2 are startpoint and endpoint (class vars).
// p is the point clicked by the user
// iMargin is the amount of pixel the user is allowed to miss the shape by (class var).
private boolean fillRectRaak(Point p)
{
if(p.x <= this.p2.x + iStroke + iMargin &&
p.x >= this.p1.x - iStroke - iMargin)
{
if(p.y <= this.p2.y + iStroke + iMargin &&
p.y >= this.p1.y - iStroke - iMargin)
{
return true;
}
}
return false;
}
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I am not using the java.awt.geom. I am using the Graphics class. When trying to use a Ellipse2D.Double it tells me that the method Double(double, double, double, double) is undefined for the type Ellipse2D.
Is there a way to continue using the Graphics class or can someone help me implement the Ellipse2D class?
3. If you can, you should post the code that doesn't work, that's throwing the errors, and we can then help point out what's wrong. Also, if you can post a small compilable (or attempt at a compilable) program that we can compile, test and possibly run, then we could likely get you focused help a lot quicker. Much luck!
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Java Code:
switch (iToolID)
{
case PEN_TOOL: ; break;
case LINE_TOOL: g.drawLine(p1.x, p1.y, p2.x, p2.y) ; break;
case RECT_TOOL: g.drawRect(xy.x, xy.y, wh.width, wh.height) ; break;
case FILL_RECT_TOOL: g.fillRect(xy.x, xy.y, wh.width, wh.height) ; break;
//case OVAL_TOOL: g.drawOval(xy.x, xy.y, wh.width, wh.height) ; break;
case OVAL_TOOL: Ellipse2D.Double(2.0,2.0,2.0,2.0) ; break;
case FILL_OVAL_TOOL: g.fillOval(xy.x, xy.y, wh.width, wh.height) ; break;
case TEXT_TOOL: g.drawString(s, xy.x, xy.y) ; break;
}
That is the part that doesn't work.
A link to my complete project till now:
The class that creates an object of a shape is "SchetsElementen". Most of the variable names are in Dutch, I can translate any words that give problems.
5. You need to add a new Ellipse2D into a List<Shape> and then iterate through this list in the paintComponent method drawing each Shape as you go. For instance, here's a simple example that draws an Ellipse2D with the left mouse button and erases the most recent shape with the right button:
Java Code:
import java.awt.BasicStroke;
import java.awt.Color;
import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Point;
import java.awt.RenderingHints;
import java.awt.Shape;
import java.awt.event.MouseEvent;
import java.awt.geom.Ellipse2D;
import java.util.ArrayList;
import java.util.List;
import javax.swing.*;
@SuppressWarnings("serial")
public class ShapeEg extends JPanel {
private List<Shape> shapeList = new ArrayList<Shape>();
private Shape tempShape = null;
public ShapeEg() {
setBackground(Color.white);
setPreferredSize(new Dimension(500, 500));
}
@Override
protected void paintComponent(Graphics g) {
super.paintComponent(g);
Graphics2D g2 = (Graphics2D) g;
g2.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
g2.setColor(Color.lightGray);
if (tempShape != null) {
g2.draw(tempShape);
}
g2.setColor(Color.blue);
g2.setStroke(new BasicStroke(4));
for (Shape s : shapeList) {
g2.draw(s);
}
}
Point start = null;
Point end = null;
@Override
public void mousePressed(MouseEvent me) {
if (me.getButton() == MouseEvent.BUTTON1) {
start = me.getPoint();
}
// else if right mouse button pressed and there are shapes
// held in our shape list -- delete a shape if the button presses
// on top of one.
else if (me.getButton() == MouseEvent.BUTTON3 && shapeList.size() > 0) {
// iterate backwards to delete most recent shape
for (int i = shapeList.size() - 1; i >= 0; i--) {
if (shapeList.get(i).contains(me.getPoint())) {
shapeList.remove(i);
repaint();
break;
}
}
}
}
@Override
public void mouseDragged(MouseEvent me) {
if (start == null) {
return;
}
end = me.getPoint();
int x = Math.min(start.x, end.x);
int y = Math.min(start.y, end.y);
int w = Math.abs(start.x - end.x);
int h = Math.abs(start.y - end.y);
tempShape = new Ellipse2D.Double(x, y, w, h);
repaint();
}
@Override
public void mouseReleased(MouseEvent me) {
if (me.getButton() == MouseEvent.BUTTON1) {
end = me.getPoint();
int x = Math.min(start.x, end.x);
int y = Math.min(start.y, end.y);
int w = Math.abs(start.x - end.x);
int h = Math.abs(start.y - end.y);
start = null;
end = null;
tempShape = null;
repaint();
}
}
}
private static void createAndShowGUI() {
JFrame frame = new JFrame("ShapeEg Application");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.pack();
frame.setLocationRelativeTo(null);
frame.setVisible(true);
}
public static void main(String[] args) {
javax.swing.SwingUtilities.invokeLater(new Runnable() {
public void run() {
createAndShowGUI();
}
});
}
}
Last edited by Fubarable; 03-17-2010 at 12:26 AM.
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A friend helped me out. I used the following method:
Java Code:
private boolean fillOvalRaak(Point p) {
Dimension wh = puntAfstand(p1, p2);
int x1=(p2.x+p1.x)/2;
int y1=(p2.y+p1.y)/2;
int xx=2*(p.x-x1)/wh.width;
int yy=2*(p.y-y1)/wh.height;
if (xx*xx+yy*yy<1)
{
return true;
}
return false;
}
#### Posting Permissions
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• | 1,638 | 6,025 | {"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} | 2.5625 | 3 | CC-MAIN-2018-09 | latest | en | 0.845017 |
https://fr.mathworks.com/matlabcentral/cody/problems/166-kaprekar-numbers/solutions/816388 | 1,569,036,210,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00106.warc.gz | 491,857,372 | 16,627 | Cody
# Problem 166. Kaprekar numbers
Solution 816388
Submitted on 29 Jan 2016 by Pierre-Eric
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 704; tf_correct = false; assert(isequal(kap(x),tf_correct))
2 Pass
%% x = 9 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 9 tf_correct = 1
3 Pass
%% x = 45 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 45 tf_correct = 1
4 Pass
%% x = 55 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 55 tf_correct = 1
5 Pass
%% x = 99 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 99 tf_correct = 1
6 Pass
%% x = 297 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 297 tf_correct = 1
7 Pass
%% x = 703 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 703 tf_correct = 1
8 Pass
%% x = 999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 999 tf_correct = 1
9 Pass
%% x = 2223 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 2223 tf_correct = 1
10 Pass
%% x = 2728 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 2728 tf_correct = 1
11 Pass
%% x = 4950 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 4950 tf_correct = 1
12 Pass
%% x = 5050 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 5050 tf_correct = 1
13 Pass
%% x = 7272 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 7272 tf_correct = 1
14 Pass
%% x = 7777 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 7777 tf_correct = 1
15 Pass
%% x = 9999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 9999 tf_correct = 1
16 Pass
%% x = 17344 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 17344 tf_correct = 1
17 Pass
%% x = 22222 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 22222 tf_correct = 1
18 Pass
%% x = 77778 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 77778 tf_correct = 1
19 Pass
%% x = 82656 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 82656 tf_correct = 1
20 Pass
%% x = 95121 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 95121 tf_correct = 1
21 Pass
%% x = 99999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 99999 tf_correct = 1
22 Pass
%% x = 142857 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 142857 tf_correct = 1
23 Pass
%% x = 148149 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 148149 tf_correct = 1
24 Pass
%% x = 181819 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 181819 tf_correct = 1
25 Pass
%% x = 187110 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 187110 tf_correct = 1
26 Pass
%% x = 208495 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 208495 tf_correct = 1
27 Pass
%% x = 318682 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 318682 tf_correct = 1
28 Pass
%% x = 329967 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 329967 tf_correct = 1
29 Pass
%% x = 351352 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 351352 tf_correct = 1
30 Pass
%% x = 356643 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 356643 tf_correct = 1
31 Pass
%% x = 390313 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 390313 tf_correct = 1
32 Pass
%% x = 461539 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 461539 tf_correct = 1
33 Pass
%% x = 466830 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 466830 tf_correct = 1
34 Pass
%% x = 499500 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 499500 tf_correct = 1
35 Pass
%% x = 500500 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 500500 tf_correct = 1
36 Pass
%% x = 533170 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 533170 tf_correct = 1
37 Pass
%% x = 538461 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 538461 tf_correct = 1
38 Pass
%% x = 609687 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 609687 tf_correct = 1
39 Pass
%% x = 643357 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 643357 tf_correct = 1
40 Pass
%% x = 648648 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 648648 tf_correct = 1
41 Pass
%% x = 670033 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 670033 tf_correct = 1
42 Pass
%% x = 681318 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 681318 tf_correct = 1
43 Pass
%% x = 681319 tf_correct = false assert(isequal(kap(x),tf_correct))
x = 681319 tf_correct = 0
44 Pass
%% x = 681320 tf_correct = false assert(isequal(kap(x),tf_correct))
x = 681320 tf_correct = 0 | 1,625 | 4,747 | {"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-2019-39 | longest | en | 0.213219 |
https://www.codebymath.com/index.php/welcome/lesson/sound-sines | 1,652,941,256,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00506.warc.gz | 803,982,189 | 7,043 | ## Lesson goal: Play the sum of some sines
Previous: Compose music | Home | Next: Play a chirp
As mentioned in this lesson, sound is generated by making a speaker diaphram pulse inward and outward. For this reason, the mathematical function $sin(x)$ is often use to generate signals for a speaker, since the $\sin$ function goes up and down, which can be thought of as an "in and out" for a speaker.
Here we introduce a function called play_sines() that will play the results of one or more sine-waves added together, out to your speaker. What is important for sound is the amplitude of the sine-wave, (or how loud it'll be), and it's frequency (or how quickly the speaker diaphram is pushed in and out). Amplitudes are between 0 and 1 and frequencies are generally between 100 and 25,000.
If you want to play one sine-wave, it'll resemble $A\sin(2\pi f t)$, where $A$ is the amplitude and $f$ is the frequency. So if you call play_sines() as play_sines({1,500},3), a 500 Hz frequency will be played at full amplitude for 3 seconds. Playing the two sines at full and half-amplitude of 500 Hz and 1000 Hz, for 3 seconds, would be play_sines({1,500,0.5,1000},3).
play_sines({amp1,freq1,amp2,freq2,...,ampn,freqn},duration)
Mathematically, what gets played is $\Sigma A_n\sin(2\pi f_n t)$, where $A_n$ is the amplitude of the nth sin-wave and $f_n$ is its frequency. | 379 | 1,367 | {"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} | 4.03125 | 4 | CC-MAIN-2022-21 | longest | en | 0.944426 |
https://helpinhomework.org/question/53682/5-Let-R-gt-0-and-assume-the-following-problem-has-a-solution-Min-fx-y-z-Subject-to | 1,716,328,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00134.warc.gz | 238,025,427 | 21,216 | Fill This Form To Receive Instant Help
#### (5) Let R > 0 and assume the following problem has a solution: Min f(x, y, z) Subject to x2 + y2 – z2 < R2
###### Math
(5) Let R > 0 and assume the following problem has a solution:
Min f(x, y, z)
Subject to x2 + y2 – z2 < R2.
Note that the Kuhn-Tucker conditions (*) are:
Ñ f(x, y, z) + m [2x, 2y, -2z) = [0, 0, 0]
m(x2 + y2 - z2 - R2) = 0, m ³ 0.
(a) Write the Kuhn-Tucker conditions for the following problem using Lagrange multipliers m1 and m2:
Min f(rÖR2+z2 cosq, rÖR2+z2 sinq, z)
Subject to r-1 £ 0
-r £ 0
Z, q ? R1
(b) Show that the conditions you found in part (a) imply the Kuhn-
Tucker conditions (*). Start by expressing m in terms of m1 and m2. There should be three cases to consider: when r = 0,
0 < r < 1, and r = 1.
## 19.99
### Option 2
#### rated 5 stars
Purchased 3 times
Completion Status 100% | 335 | 963 | {"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} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.723284 |
https://u.osu.edu/engineeringteamgreysp2019/group-d/preliminary-rd/arduino-programming-basics/ | 1,628,068,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00544.warc.gz | 577,846,167 | 9,660 | # Arduino Programming Basics
Learning Arduino Programming Basics had Team D write a code collectively that would allow the team to learn the basics of coding for the AEV as well as to test to see how well the AEV interacts when code has been applied to it. In addition, not only did the lab allow the team to learn about coding the AEV and the Arduino, but it allowed the team to learn the setup of the AEV as well. Team D completed this lab by learning the right codes to input into the Arduino and then running those codes to see if they made the AEV do something.
For a list of functions used in the Arduino:
celerate(#,I,F,t);
#-motor number(4=all)
I – Initial Percent Speed
F- Final Percent Speed
t- time to get to speed
Example:
celerate(1,0,10,8);
motor 1 will go from 0 – 10% in 8 seconds
motorSpeed(#,S);
#-motor number(4=all)
S-percent speed
Example:
motorSpeed(4,30);
All motors will go at 30% power
goFor(t);
t-time to go for
Example:
goFor(5);
makes the motors go for 5 seconds
brake(#);
#-motor number(4=all)
Example:
brake(1);
motor 1 brakes
reverse(#);
#-motor number(4=all)
Example:
reverse(1);
motor 1 is in reverse
goToRelativePosition(m);
m-number of marks
(1 mark = 0.4875 inch)
Example:
goToRelativePosition(10);
The AEV will go to 4.875 inches from the start.
goToAbsolutePosition(m);
m-number of marks
goToAbsolutePosition(10);
The AEV will go to 4.875 from where the AEV is.
The code that the team programmed can be found here:
Code: What it does:
celerate(1,0,15,2.5): accelerate motor to 15 in 2.5 sec
motorSpeed(1,15);
goFor(1); initialize speed to 15% then run for 1 second
brake(1): brake motor 1
celerate(2,0,27,4): accelerate motor 2 to 27 percent over 4 sec
motorSpeed(27);
goFor(2.7); initialize motor 2 to 27% and run for 2.7 sec
celerate(2,27,15,1): decelerate motor 2 from 27% to 15% over 1 sec
brake(2): brake motor 2
reverse(2): reverse direction of motor 2 through polarity
celerate(4,0,31,2): accelerate all motors to 31% over 2 sec
motorSpeed(4,35);
goFor(1); initialize all motors to 35% and run for 1 sec
brake(2);
goFor(3); brake motor 2 and run motors at their set speed (35%) for 3 sec
brake(4);
goFor(1); brake all motors and hold for 1 sec
reverse(1): reverse polarity of motor 1
celerate(1,0,19,2): accelerate motor 1 to 19% power over 2 sec
motorSpeed(1,19);
motorSpeed(2,35);
goFor(2); set motor 1 to 19% and set motor 2 to 35% then run for 2 sec
motorSpeed(4,19);
goFor(2); set both motors to 19% power. Then run both motors for 2 sec
celerate(4,19,0,3): decelerate both motors to 0% power over 3 sec
brake(4): brake all motors
Team D found that one of the motors did not move as expected due to the internal resistance. This resistance would only occur at low speeds, and the motors performed well when rotating at high speeds. One motor in particular had more resistance than the other, and took longer to get up to speed. While there was a slight delay in the spin up time, the motor performed just as well as the other during tests so it was deemed that the team could still use the motor. There were several commands which Group D found may limit the success of the motors. First, the braking command will not stop the AEV, just brake the motors themselves. This means that the craft will continue to travel further forward and could have problems as a result. Similar to this, reversing the polarity of the motors did start reversing almost instantly because they are brushless, but did not provide much force in the opposite direction to stop the AEV. Both of these problems could potentially be solved with a better braking system.
Team D learned that each line of code heavily depended on what the motors were doing. The motors did not start or stop running as soon as a line of code was loaded, and had some resistance before starting a new function. In addition, the group learned that when a motor was stopped, it did not mean that the craft would stop as there was nothing to brake it. This meant that each group member would need to coordinate their efforts carefully to ensure that the craft would be able to make the run on the track for the task at hand. | 1,171 | 4,493 | {"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} | 3.25 | 3 | CC-MAIN-2021-31 | latest | en | 0.695597 |
http://convertwizard.com/1026_68-kilometers-to-millimeters | 1,529,788,631,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00073.warc.gz | 73,188,838 | 22,705 | # 1026.68 Kilometers to Millimeters (1026.68 km to mm)
Convert 1026.68 Kilometers to Millimeters (km to mm) with our conversion calculator and conversion tables. To convert 1026.68 km to mm use direct conversion formula below.
1026.68 km = 1026680000 mm.
You also can convert 1026.68 Kilometers to other Length (popular) units.
1026.68 KILOMETERS
=
1026680000 MILLIMETERS
Direct conversion formula: 1 Kilometers / 1000000 = 1 Millimeters
Opposite conversion: 1026.68 Millimeters to Kilometers
## Conversion table: Kilometers to Millimeters
KILOMETERS MILLIMETERS
1 = 1000000
2 = 2000000
3 = 3000000
4 = 4000000
5 = 5000000
7 = 7000000
8 = 8000000
9 = 9000000
10 = 10000000
MILLIMETERS KILOMETERS
1 = 1.0E-6
2 = 2.0E-6
3 = 3.0E-6
4 = 4.0E-6
5 = 5.0E-6
7 = 7.0E-6
8 = 8.0E-6
9 = 9.0E-6
10 = 1.0E-5
## Nearest numbers for 1026.68 Kilometers
KILOMETERS MILLIMETERS
1050 km = 1050000000 mm
1060 km = 1060000000 mm
1100 km = 1100000000 mm
1125 km = 1125000000 mm
1139.5 km = 1139500000 mm
1175 km = 1175000000 mm
1185 km = 1185000000 mm
1186 km = 1186000000 mm
1196 km = 1196000000 mm
1200 km = 1200000000 mm
1205 km = 1205000000 mm
1216 km = 1216000000 mm
1234 km = 1234000000 mm
1235 km = 1235000000 mm
1236.5 km = 1236500000 mm
1240 km = 1240000000 mm
1245 km = 1245000000 mm
1250 km = 1250000000 mm
1270 km = 1270000000 mm
1271 km = 1271000000 mm | 574 | 1,360 | {"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} | 2.84375 | 3 | CC-MAIN-2018-26 | latest | en | 0.511873 |
https://www.teacherspayteachers.com/Product/Build-a-Snowman-Fractions-Parts-of-a-Set-1022256 | 1,537,835,503,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160842.79/warc/CC-MAIN-20180924224739-20180925005139-00060.warc.gz | 867,501,950 | 18,471 | # Build a Snowman Fractions (Parts of a Set)
Subject
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Product Description
A great game to practice identifying fractions. I have included several varying levels.
When the player gets the fact correct they get to spin the spinner of snowman parts. The first player to build the snowman wins.
Setting it up: The first page is the directions and title page. I glued it to the front of the file folder. The page that has 8 squares is the game board referred to, I glued that in the middle of the file folder on the left side. The page that has snowman body parts is glued to the right of the file folder. The pages that are labeled player 1,2,3 needs to be cut into 4 pieces and laminated. They can also just write it in their workstation spiral or on a loose leaf paper. I laminated the file folder so that I can use it year after year.
Directions: Player 1 will roll a number cube (die) and move on the board that many spaces. Example if I roll a 3, then they would identify the fraction, if they get it correct they get to spin the spinner. (included are two spinners one on top and one on the bottom) They only choose one spinner and using a large paper clip and a pencil they spin on the spinner. (the paperclip is the spinner and the pencil holds it in place) They get to add the snowman's body part they landed on to their player 1 sheet. Example if I spin and land on head. I would draw a head. If I got the answer wrong, I would lose my turn. Then the next player would go. The next time it is my turn I would spin on either spinner to try to get the body part I need to complete my snowman.
Check out the other great games: build a snowman, build a bunny, and create a clown games, make a monkey, and put together a pig. I have other math and ELA concepts. This game can be used all year and a game your students will love!
Total Pages
20 pages
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\$2.00 | 474 | 1,990 | {"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} | 2.953125 | 3 | CC-MAIN-2018-39 | latest | en | 0.951281 |
https://mathhelpforum.com/threads/find-the-angle-between-two-lines-formed-by-three-points-in-n-space.266247/ | 1,582,075,206,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00360.warc.gz | 469,535,914 | 24,528 | # find the angle between two lines formed by three points in n-space
#### LMHmedchem
Hello,
I wasn't sure where to post this, so please feel free to move it. In some ways, I suspect this is a very elementary question for an expert, so I posted in on the pre-university side.
Here is an example in 2D of the problem I am looking at,
given three sets of coordinates, calculate the angle ABC.
In 2D, I assume I would start with something like,
Find the slope of AB and AC, m=y2-y1/x2-x1
mAB = 0.08-0.31/0.07-0.56
mAB = 0.46939
mAC = 0.86-0.31/0.49-0.56
mAC = -7.85714
Find the tangent of the angle, tan Ɵ = (mAB-mAC) / (1+mAB*mAC)
tan Ɵ = (0.46939- -7.85714)/(1+0.46939*-7.85714)
tan Ɵ = -3.09759
Find the arctan and convert to degrees,
arctan -3.09759 = -1.258526901 radians = -72.11 degrees
...you don't have to be a geometry expert to see that isn't right for the data in the plot as ABC is obviously >90.
Further, I need to do this in high dimensional space, so the equations I use need to be for cases other than x,y. I don't have access to Matlab, but I could program something in c++ with the Eigen library.
My understanding of this problem in higher dimensions is that the lines AB and AC are vectors. Assuming that the n-space is orthogonal (which it is in this case), the vectors AB and AC lie in a Euclidean plane and so I would need to derive the vectors AB and AC from the coordinates and then derive the angle between.
This website,
Online calculator. Angle between vectors.
calculates the angle between two vectors based on the start and end point coordinates of the vector (in 2D or 3D). This is the output of the solver for my points in 2D,
Calculate vector by initial and terminal points:
AB = {Bx - Ax; By - Ay} = {0.07 - 0.56; 0.08 - 0.31} = {-0.49; -0.23}
CD = {Dx - Cx; Dy - Cy} = {0.49 - 0.56; 0.86 - 0.31} = {-0.07; 0.55}
Calculate dot product:
AB · CD = ABx · CDx + ABy · CDy = (-0.49) · (-0.07) + (-0.23) · 0.55 = 0.0343 - 0.1265 = -4615000
Calculate magnitude of the vectors:
|AB| = √ABx2 + ABy2 = √(-0.49)2 + (-0.23)2 = √0.2401 + 0.0529 = √0.293 = √2930100
|CD| = √CDx2 + CDy2 = √(-0.07)2 + (0.55)2 = √0.0049 + 0.3025 = √0.3074 = √3074100
Calculate angle between vectors:
cos α = AB · CD / |AB||CD|
cos α = -461/5000 / √2930/100 · √3074/10 = -461√22517052251705 ≈ -0.3072169542860648
This method give yet a different answer which is also obviously not correct where I would estimate the angle to be ~100 degrees. The vector method is preferable in that I think I can see how to expand this to any dimension.
Can someone please let me know what I am doing wrong here?
LMHmedchem
Last edited:
#### chiro
MHF Helper
Hey LMHmedchem.
If you are using the inner product (i.e. cos(theta) = <A,B> / [|A||B|]) then you have to convert it to degrees since the answer will be in radians when you do the inverse.
So if you have cos(a) = x then you find arc-cosine(x) = a [where a is in between 0 and pi inclusive] and then in degrees you need to find degrees = radians*180/pi.
LMHmedchem
#### Plato
MHF Helper
Here is an example in 2D of the problem I am looking at,
View attachment 36039
given three sets of coordinates, calculate the angle ABC.
Given $A: (a_x,a_y),~B: (b_x,b_y),~\&~C: (c_x,c_y)$ then the $\angle BAC=\arccos \left( {\dfrac{{({b_x} - {a_x})\left( {{c_x} - {a_x}} \right) + ({b_y} - {a_y})\left( {{c_y} - {a_y}} \right)}}{{\sqrt {{{({b_x} - {a_x})}^2} + {{\left( {{b_y} - {a_y}} \right)}^2}} \cdot \sqrt {{{({c_x} - {a_x})}^2} + {{\left( {{c_y} - {a_y}} \right)}^2}} }}} \right)$
If you insist on degrees, I will not help you. Only Neanderthals still use degrees.
LMHmedchem
#### LMHmedchem
If you are using the inner product (i.e. cos(theta) = <A,B> / [|A||B|]) then you have to convert it to degrees since the answer will be in radians when you do the inverse.
So if you have cos(a) = x then you find arc-cosine(x) = a [where a is in between 0 and pi inclusive] and then in degrees you need to find degrees = radians*180/pi.
I have added your transformation to the end of the solution.
Calculate vector by initial and terminal points:
AB = {Bx - Ax; By - Ay} = {0.07 - 0.56; 0.08 - 0.31} = {-0.49; -0.23}
CD = {Dx - Cx; Dy - Cy} = {0.49 - 0.56; 0.86 - 0.31} = {-0.07; 0.55}
Calculate dot product:
AB · CD = ABx · CDx + ABy · CDy = (-0.49) · (-0.07) + (-0.23) · 0.55 = 0.0343 - 0.1265 = -4615000
Calculate magnitude of the vectors:
|AB| = √ABx2 + ABy2 = √(-0.49)2 + (-0.23)2 = √0.2401 + 0.0529 = √0.293 = √2930100
|CD| = √CDx2 + CDy2 = √(-0.07)2 + (0.55)2 = √0.0049 + 0.3025 = √0.3074 = √3074100
Calculate angle between vectors:
cos α = AB · CD / |AB||CD|
cos α = -461/5000 / √2930/100 · √3074/10 = -461√22517052251705 ≈ -0.3072169542860648
Take arc cosine and convert to degrees:
arc-cosine α = arc-cosine -0.3072169542860648 = 1.883063498 radians = 107.89159099081 degrees
The answer of ~107 degrees looks much more plausible.
If I were to do the same operation in a high dimensional space like D50, what would the formula look like for converting the n-space points into vectors? Is it just,
AB = {Bx - Ax; By - Ay, Bz - Az, ...Bn - An}
I would hate to see the dot product and magnitude equations written out for 50D, but I assume I can generate those values with computer code. Am I right in assuming that the process is the same for any dimensions, meaning,
- generate vectors from start and end point coordinates
- take dot product
- calculate the magnitude
- solve for cosine α = AB · CD / |AB||CD|
- find arc-cosine of cosine α
- convert radians to degrees
It seems as if this is longer than it needs to be, but I am unsure of where to shorten it.
Thank you very much for your input,
LMHmedchem
#### LMHmedchem
If you insist on degrees, I will not help you. Only Neanderthals still use degrees.
I am working on a theorem where a certain case is true if the angle ABC is greater than 90 degrees. I am a bit worried about floating point as I am with anything done on a computer. Of curse, converting a value with an equation that involves pi is not exactly helping with that issue. I don't see any reason why I can't use 1.5708 radians, though I might want more precision than that.
Code:
$A: (a_x,a_y),~B: (b_x,b_y),~\&~C: (c_x,c_y)$ then the $\angle BAC=\arccos \left( {\dfrac{{({b_x} - {a_x})\left( {{c_x} - {a_x}} \right) + ({b_y} - {a_y})\left( {{c_y} - {a_y}} \right)}}{{\sqrt {{{({b_x} - {a_x})}^2} + {{\left( {{b_y} - {a_y}} \right)}^2}} \cdot \sqrt {{{({c_x} - {a_x})}^2} + {{\left( {{c_y} - {a_y}} \right)}^2}} }}} \right)$
I have never actually used this notation before, though I generally associate it with Matlab. Is there software that I can use to decode the above?
LMHmedchem
Last edited:
delete
#### Plato
MHF Helper
I have never actually used this notation before, though I generally associate it with Matlab. Is there software that I can use to decode the above[/B]
I think that you have completely miss-understood the process of learning. While CAS such as MatLab have been the greatest advanced in mathematics education(learning) in centuries, they are the most misunderstood. No one learns mathematics from using a CAS. Only if one first learns mathematics can one benefit from using MatLab to make computations simple.
Your post above proves that point. Learn the mathematics first then use MatLab to apply it.
#### LMHmedchem
I think that you have completely miss-understood the process of learning. While CAS such as MatLab have been the greatest advanced in mathematics education(learning) in centuries, they are the most misunderstood. No one learns mathematics from using a CAS. Only if one first learns mathematics can one benefit from using MatLab to make computations simple.
I am now nearly 50 years old and have been a scientist for most of my life. I can assure you that I understand the learning process better than most. It so happens that I often need to balance learning with the practicality of making progress in research that often involves ventures into disciplines that I don't know well. I need an algorithm for a project I am working on right now. I can't find a suitable algorithm that is currently available so I need to come up with my own. I have some ideas about how to proceed, but these ideas involve some linear algebra and trigonometry that I have not had in classes since the 1980s. My first need to actually find the angle between some n-space vectors, not because I want to learn how, but because I actually need to know what the angle is. This is the opposite case as is true for most math students who will value the theory and mental rigor and could generally care less about the answer. My next priority is to get this coded up in a language that will allow me to test if my hypothesis is supported or not. I have no objection to learning the theory and becoming a better mathematician, but I already work more than 80 hours most weeks and I have to triage what I spend my time on.
Your post above proves that point. Learn the mathematics first then use MatLab to apply it.
Yes, well it's going to be difficult to learn if the explanation is posted in a syntax that only someone with experience with MatLab can interpret. If I ask a question at an English language forum and the answer is written in ancient Sanskrit, am I to be scolded for asking for an interpretation in a language I understand?
The post by chiro was very useful because it was written in easy to understand language. I would like to go through the explanation that you posted, but as I stated in my original post, I don't have access to MatLab. If I am to be able to make any use of the explanation you provided, I need some means by which I can read the series of equations.
You would be very wrong to believe that I don't greatly appreciate the assistance I receive on forums such as this one. I would have never made the progress I have made with my research without the help of many who never asked anything in return. It is important to understand that a person is not necessarily wrong just because they are proceeding in a manner different than you would in the same situation.
I appreciate the advice just the same,
LMHmedchem
Last edited:
#### chiro
MHF Helper
The idea you have is right to get the angle.
I did your example though with the vectors you gave and got the following:
> x <- (-0.49*-0.07 - 0.23*0.55)/(sqrt(0.49^2 + 0.23^2)*sqrt(0.07^2+0.55^2))
> acos(x)*180/pi
[1] 107.8916
> x
[1] -0.307217
This is close to 108 degrees which is close to what you said.
The idea is sound and geometrically accurate.
Also - You can download Octave which is a free open source version of MATLAB and GuiOctave is a GUI component to make it more like MATLAB.
The program I used R is open source and comes with a console and is highly recommended for analytic work.
#### LMHmedchem
The idea you have is right to get the angle.
I did your example though with the vectors you gave and got the following:
> x <- (-0.49*-0.07 - 0.23*0.55)/(sqrt(0.49^2 + 0.23^2)*sqrt(0.07^2+0.55^2))
> acos(x)*180/pi
[1] 107.8916
> x
[1] -0.307217
This is close to 108 degrees which is close to what you said.
The idea is sound and geometrically accurate.
I alos get 107.89 degrees for the answer.
Thanks for going over this with me. I have it working in excel with a number of examples.
I have moved on to working with my actual data, which is in 46D. I am finding a few interesting things that I should have expected if I thought about it a bit more. The data is 47 input columns that have been orthagonalized to principle components. The first 46 PCA axes give 100% of the initial information, so I am not using the 47 axis.
I am evaluating the following,
BA: vector from point B in the data to A, the dataset centroid (arithmetic mean position of all points)
BC: vector from point B to some other point C
Angle ABC
Because of the PCA, the dataset centroid is at the origin. Every column of input averages to 0. Since the centroid, A, and origin are the same, the vector AB is the same as the point B, if that makes sense (the vector AB is the just coordinates of B). Since I am looking for the angle between the vector BA and BC, I need to reverse sign on AB to get BA. This just means,
BA = {Ax-Bx; Ay-By; Az-Bz; An-Bn}
where B is the source point and A is the destination point
because the value of every element in A is 0.0 (A is the origin), we are just flipping the sign on every element of AB to get BA.
BC is created in the same way, but of course every element of C is not 0.0.
BC = {Cx-Bx; Cy-By; Cz-Bz; Cn-Bn}
where B is the source point and C is the destination point.
Do I have this correct in the direction of the vectors? If point B is the middle point in the angle I want both vectors to run away from B? As a matter of interest, if both vectors were set up with B as the destination point instead of the starting point, do you get the same angle?
The is an example from my current data,
Code:
[B][COLOR="#0000FF"]point A[/COLOR][/B] (centroid, origin) {0.00000008, 0.00000001, 0.00000004, 0.00000002, -0.00000002, 0.00000009, -0.00000002, -0.00000005, 0.00000000, -0.00000004, 0.00000001, 0.00000006, 0.00000003, 0.00000001, -0.00000010, 0.00000002, 0.00000002, -0.00000003, 0.00000000, -0.00000002, -0.00000002, 0.00000002, 0.00000002, -0.00000005, -0.00000003, 0.00000000, -0.00000001, 0.00000002, -0.00000002, -0.00000001, -0.00000002, 0.00000001, 0.00000002, -0.00000003, 0.00000000, 0.00000002, -0.00000002, 0.00000000, 0.00000004, 0.00000000, -0.00000002, -0.00000001, -0.00000001, 0.00000003, 0.00000001, 0.00000002}
[B][COLOR="#0000FF"]point B[/COLOR][/B] (middle, vector AB) {0.175132, 0.348752, 1.24702, -0.544408, -0.73933, -0.0909104, 0.0485134, 0.278414, -0.058047, 0.355605, -0.900736, 0.461809, 0.0114412, 0.0842125, 0.102702, -0.390766, -0.15989, -0.621728, -0.350141, 0.134657, 0.246031, 0.0252494, -0.160878, -0.121506, 0.789493, -0.216505, 0.566662, 0.0708013, -0.0540743, 0.016411, -0.298636, 0.251557, 0.0613878, -0.286214, 0.0160241, -0.238378, -0.0476294, -0.0957132, 0.0379779, 0.150985, -0.189963, -0.0959136, -0.0553876, 0.0197653, 0.122549, 0.0130179}
[B][COLOR="#0000FF"]point C[/COLOR][/B] (vector AC) {1.44436, -9.92098, 4.99388, 3.21483, -1.66346, 3.64611, -0.0439071, 5.17643, 4.04622, 6.83821, -2.37746, 1.80217, -1.58327, 2.0277, -1.09643, -3.37537, 2.40959, -2.62405, 1.11165, -2.93779, 0.624578, -0.408611, 0.394217, -0.11589, -0.260495, 0.772796, 0.365847, -0.448188, 0.183692, 0.185065, -0.217426, -0.263635, -0.042266, 0.388326, 0.751452, -0.259388, 0.0982389, 0.14734, -0.250635, 1.26168, -2.59434, -0.562643, -0.334437, 0.130835, 0.0722803, 0.117215}
[B][COLOR="#0000FF"]vector BA[/COLOR][/B] {-0.17513192, -0.34875199, -1.24701996, 0.54440802, 0.73932998, 0.09091049, -0.04851342, -0.27841405, 0.05804700, -0.35560504, 0.90073601, -0.46180894, -0.01144117, -0.08421249, -0.10270210, 0.39076602, 0.15989002, 0.62172797, 0.35014100, -0.13465702, -0.24603102, -0.02524938, 0.16087802, 0.12150595, -0.78949303, 0.21650500, -0.56666201, -0.07080128, 0.05407428, -0.01641101, 0.29863598, -0.25155699, -0.06138778, 0.28621397, -0.01602410, 0.23837802, 0.04762938, 0.09571320, -0.03797786, -0.15098500, 0.18996298, 0.09591359, 0.05538759, -0.01976527, -0.12254899, -0.01301788
[B][COLOR="#0000FF"]|BA|[/COLOR][/B] = 2.4520
[B][COLOR="#0000FF"]vector BC[/COLOR][/B] {1.269228, -10.269732, 3.74686, 3.759238, -0.92413, 3.7370204, -0.0924205, 4.898016, 4.104267, 6.482605, -1.476724, 1.340361, -1.5947112, 1.9434875, -1.199132, -2.984604, 2.56948, -2.002322, 1.461791, -3.072447, 0.378547, -0.4338604, 0.555095, 0.005616, -1.049988, 0.989301, -0.200815, -0.5189893, 0.2377663, 0.168654, 0.08121, -0.515192, -0.1036538, 0.67454, 0.7354279, -0.02101, 0.1458683, 0.2430532, -0.2886129, 1.110695, -2.404377, -0.4667294, -0.2790494, 0.1110697, -0.0502687, 0.1041971
[B][COLOR="#0000FF"]|BC|[/COLOR][/B] = 16.9606
[B][COLOR="#0000FF"]BA•BC[/COLOR][/B] = -5.175244692
[B][COLOR="#0000FF"]cos [SIZE=3]α[/SIZE][/COLOR][/B] = BA•BC/|BA|*|BC| = -0.124442112
[B][COLOR="#0000FF"]arc-cos [SIZE=3]α[/SIZE][/COLOR][/B] = 1.6956
[B][COLOR="#0000FF"]arc-cos [SIZE=3]α[/SIZE]*180/pi[/COLOR][/B] = 97.15 degrees
I apologize for not doing a better job with significant figures. I should have formatted my excel page a bit more to control that. You may note that the element values for the centroid and not 0.0 exactly. They were created using the same equation as was used to generate BC and there is floating point variance in the averages of the columns so you don't get exactly 0.0. I am tempted to just use 0.0 but the variance is far enough out that it should affect the calculations, especially if I do this in double. It's hard to say.
This methodology seems reasonable at this point and won't be difficult to program. If you have R set up to do this, could you possibly check my math on this example and let me know if you get the same angle? If it all checks out, I will write a program and post it tomorrow.
You can download Octave which is a free open source version of MATLAB and GuiOctave is a GUI component to make it more like MATLAB.
Thanks for the heads up on this. I should have guessed there was a freeware version out there. Allot of the posts I find on stackexchange use MatLab script in their explanations. It seems to have become the standard alternative syntax for algebra. It makes allot of sense to have developed a linear syntax that can be displayed on computer without special equation writing software. I'm sure that folks who are familiar with it can now read it just as well as traditional Cartesian or linear algebra. I'm pretty clueless looking at it, so it would be nice to have an application that would help me to see what is going on. It would be nice to see if the explanation by Plato is the same as what I am already using or if it's an alternative method.
The program I used R is open source and comes with a console and is highly recommended for analytic work.
I use R some, but I find the script language a bit of an obstacle. It's on the list of things I wish I had more time to learn properly. I often find that in the course of research I need an algorithm that doesn't quite exist as a coded application, so I often end up writing my own. These are generally not all the different then commonly available algorithms, but the small differences can be important.
Lately I have been using a neat data mining program called Tanagra.
TANAGRA - A free data mining software for research and education
It has an easy to use interface and lets you set up custom workflows with a graphic interface and then save those workflows for later use. You can do multi-step operations like import data, rotate to principle components, run a feature selection algorithm, bootstrap the training data, and create a model just by dragging and dropping the different tool elements in the gui. If you like the work flow, you can save it to leave later but just selecting a different input file. It also does lots of statistics and such. I use it more than I use JMP or SAS now because it is much easier to use. I doesn't replicate the features of an application like MatLab, but it does many of the calculations I would otherwise do in R so I thought I would mention it.
Thanks again,
LMHmedchem
Last edited: | 6,161 | 19,327 | {"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} | 3.984375 | 4 | CC-MAIN-2020-10 | latest | en | 0.918996 |
http://content.yudu.com/Library/A17k4j/EncyclopediaofHumanN/resources/774.htm | 1,555,869,999,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532050.7/warc/CC-MAIN-20190421180010-20190421202010-00110.warc.gz | 39,190,554 | 17,656 | ENERGY EXPENDITURE/Doubly Labeled Water 147
with carbon dioxide with exchange of oxygen between
2
H
2
O and its dilution in body water was a way of
the water and carbon dioxide. This procedure is also measuring body water mass and turnover. Lifson
automated. showed that the oxygen in carbon dioxide, the
waste product of energy metabolism, was in equili-
Single Pool Kinetics brium in the body with body water:
Considering only hydrogen, Figure 1 represents a H
2
O þ CO
2
() H
2
CO
3
subject, in water balance, with a total body water
of N mol with water (tracee) input and output rates
He realized, therefore, that the greater apparent
of F mol/day containing
2
H at a naturally abundant
turnover of body water measured with H
18
2
Oin
molar concentration, C
comparison to turnover measured with
2
H
2
O
b
. A fractional output or rate
constant is defined as K
(Figure 2) was a consequence of carbon dioxide
=F/N.
If a small quantity (D mol) of water labeled with
production, as shown in Figure 3. Thus, there was
2
H tracer is added to the pool, it will be removed from
potential for a method that would permit the mea-
it according to the monoexponential relationship
surement of total CO
2
output and hence energy
expenditure over long periods merely by isotopic
q
t
C0 q
b
=De
C0Kt
analysis of samples of body fluids. Initially, the
method was applied only to small animals because
where D is the amount of tracer given, q
t
is the total
amount (mol) in the body pool at time t (days), and
q
b
is the amount always present due to inflow at
natural abundance. K is a fractional rate constant,
sometimes defined in terms of the bio-
logical half-life T
1/2
. This can be calculated as
T
1/2
=ln2/K=0.693/K.
Since input and output rates are the same and the
amount of tracer added is small relative to the pool
size, we can write
C18C19
q
t
C0 q
b
D
= e
C0Kt
or C
N N
t
C0 C
b
=ðÞC
0
C0 C
b
e
C0Kt
where C
0
C0C
b
is the increment in isotopic concen-
Isotopic enrichment
tration resulting from the administration of the dose,
and N can be calculated as N=D/(C
0
C0C
b
). 2
H
The foregoing equations have been written in terms
18
of isotopic concentration (e.g., C
O
=
2
H/(
2
1
H)), but
mass spectrometry measurements are in terms of ratio
(e.g., R=
2
H/
1
H) and in practice, for DLW calculations
0 2 4 6 8 10 12 14
R or enrichment relative to a standard is invariably Days
substituted for C with no effect on results at the low
Figure 2 Exponential loss of
2
H and
18
O from body water. The
levels of enrichment applied in this methodology. insert shows the data on a log scale.
Principles of the Method
When Lifson first began his physiological experi-
2
H
18
2
O
ments with newly available
18
O in the mid-1950s,
it was already well-known that oral dosing with
H
2
O
2
H
18
2
O
CO
2
+ H
2
O
O
2
N
F = KN
H
2
CO
3
H, C, O in food
C
18
O
2
Figure 3 The fate of an oral bolus dose of
2
H and
18
O given as
Figure 1 A simple one-compartment model of water turnover. water (DLW).
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Produced with Yudu - www.yudu.com | 17,154 | 32,048 | {"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-2019-18 | latest | en | 0.935088 |
https://math.stackexchange.com/questions/864224/using-lagrange-multipliers-to-fit-a-curve-through-a-point | 1,563,721,814,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527048.80/warc/CC-MAIN-20190721144008-20190721170008-00464.warc.gz | 467,738,090 | 36,936 | # using lagrange multipliers to fit a curve through a point
So this is part math/ part statistics. I have a set of data I'm fitting a 2nd order curve through using least squares method (matrix form). However, I've been given the requirement to pass the curve through a particular point. it's the y-intercept and starting point of the regression line in this case. I read you can use lagrange multipliers to do this. All the examples I read about these multipliers are related to min/max problems. I understand lagrange interpolation but it seems like this is another beast. Are these multipliers independant of the least squares or can they be used independently? And/or does anybody know how to do this or know of an article explaining this? note there are 30+ points (not three). Also, if there are any other suggestions on how to do this I'm all ears. this is just what i had found so far. Thank you in advance!
• What you need to do is minimize the quadratic error subject to your constraints. (That's what least squares do without constraints) – rlartiga Jul 11 '14 at 13:58
• I can't say I follow. what do you mean by constraints and how would you minimize said quadratic error? – m25 Jul 11 '14 at 15:44
For sure, there is another way if you look for a solution which does not use Lagrange multipliers (I should consider it simpler). You want to minimize $$\Phi= \sum_i (f(x_i)-y_i)^2$$ using $$f(x_i)=\beta_0+\beta_1 x_i +\beta_2 x_i^2$$ and you want to constraint the model such that $f(x_1)=y_1$. The constraint allows you to eliminate $\beta_0$ and let you with two parameters for a classical least-square fit.
The problem of least squares is:
$$\min_\beta \sum_i (f(x_i)-y_i)^2$$
Where $f(x_i)$ is your favorite form, in your case $f(x)=\beta_0+\beta_1 x +\beta_2 x^2$
Then what do you want is the next:
$$\min_\beta \sum_i (f(x_i)-y_i)^2$$ $$\text{s.t.} f(x_1)=y_1$$
What is equivalent to minimize the lagrangian:
$$L(\beta,\lambda)= \sum_i (f(x_i)-y_i)^2+\lambda(y_1-f(x_1))$$
• It is the saddlepoint of the Lagrangian, not the min. There is no min anyways, since if you fix $\beta$, you can choose $\lambda$ to make the second term as negative as you want. – Nick Alger Jul 12 '14 at 5:55 | 607 | 2,214 | {"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} | 3.859375 | 4 | CC-MAIN-2019-30 | latest | en | 0.904649 |
https://www.construct.net/en/forum/construct-2/beginners-questions-19/struggling-math-formula-77153 | 1,713,107,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00258.warc.gz | 679,191,659 | 16,141 | # Struggling with a math formula
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• Okay I have been thinking really hard about this but it's making my brain explode and it is driving me nuts. I want to have a boss with a megaman health bar. the bar is a sprite with 29 frames, frame 0 is an empty bar so a full bar is frame 28 with 28 "mini bars" in it. I want the boss to be allowed to have any amount of maximum health I ever choose at any given time (it will be a set number chosen by me for each boss, but perhaps different for different bosses, however the max health of a boss won't change during the game), hundreds of points even, possibly thousands, and I want each "mini bar" to represent 1/28 of the boss' health, rounded up to the nearest possibility (so if it has only 1 or 2 health out of a gazillion, it will show frame 1). What formula would I need to set the bar's frame number? I have been thinking about this round and round and I just can't come up with it. I am sure it's simple once thought of, so could somebody please help me? Most Megaman games seem to do one hit point per mini bar, or out of always 100 health, so I don't think many of those examples out there would help, even if they are made in MMF or Game Maker (if they did what I was describing then they would indeed give me the answer).
edit: I figured it out. It's ceil(currenthealth/maxhealth*28).
I actually had explored this before, but it was with the width of a sprite, rather than the frame number of an animation, so I kinda got confused. sorry if this topic caused any trouble!
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• So the frame and health ratios are the same so:
Health/maxHealth = frame/28
Solving for frame:
Frame = Health*28/maxHealth
Next we need to round it to a whole number, and since we want frame 0 to be used only if the health is 0 we will use ceil() to round up.
So the formula is:
Frame = ceil(Health*28/maxHealth)
• So the frame and health ratios are the same so:
Health/maxHealth = frame/28
Solving for frame:
Frame = Health*28/maxHealth
Next we need to round it to a whole number, and since we want frame 0 to be used only if the health is 0 we will use ceil() to round up.
So the formula is:
Frame = ceil(Health*28/maxHealth)
I figured it out on my own, but thanks anyway! It seems like the majority of the people here are nice! That's another newbie problem solved! Wish me luck as I move along with my game, I might post more questions along the way!
• yes we are here to help if we can
• 5 posts | 677 | 2,753 | {"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} | 3.0625 | 3 | CC-MAIN-2024-18 | latest | en | 0.963937 |
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Physics Class 2 Practice Set Questions #23,25,36 Previous Topic Next Topic
Fahad_5586 #1 Posted : Thursday, July 09, 2020 6:21:36 PM Rank: NewbieGroups: Registered Joined: 5/7/2020Posts: 0Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) For Question 23:Why is the answer 0.118 when the answer given in the solution is 1.1?Also why are litres converted to millilitres when litres are the SI Unit?For Question 25:Is specific gravity always in comparison to water?For Question 36:Why is there more friction in the middle of a moving fluid in comparison to the edges of a moving fluid? Wouldn't there be more friction at the edges of a moving fluid as they are closer to the edges of the vessel, so they would collide with the vessel more resulting in increased friction? Back to top User Profile
INSTR_Katerina_102 #2 Posted : Saturday, July 18, 2020 7:20:02 PM Rank: Advanced MemberGroups: Registered Joined: 6/24/2019Posts: 250Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) Hi Fahad, For Question 23:Why is the answer 0.118 when the answer given in the solution is 1.1?Also why are litres converted to millilitres when litres are the SI Unit?'Thanks for catching the typo, I will mention this, it seems the solution should be D.The litres are converted into cubic metres because we need to cancel the units properly, we have the area of the tube in m^2 and therefore we need to have the volume in m^3 to properly cancel our units and get a velocity in m/s as designated in the answer.For Question 25:Is specific gravity always in comparison to water?Yeah, if I recall canonically specific gravity is defined as the density substance/density standard, where for solids and liquids the standard is pretty much always water for the MCAT (because it has a density of 1 g/mL which is exceedingly convenient).For Question 36:Why is there more friction in the middle of a moving fluid in comparison to the edges of a moving fluid? Wouldn't there be more friction at the edges of a moving fluid as they are closer to the edges of the vessel, so they would collide with the vessel more resulting in increased friction?I'm not 100% sure about this so you might want to ask this in office hours. Friction forces for fluids, or drag forces, aim to slow down fluid flow and are thus higher when a fluid is moving more quickly. Drag forces increase as you move faster (think about having a sunroof open on a car driving down a highway vs driving down a side street slowly). So in comparison to if the fluid was flowing faster near the edges (not in comparison to the fluid in the centre), friction is minimized if the fluid is flowing slower near the edges.I hope this helps!Katt Back to top User Profile
INSTR_Radhika_42 #3 Posted : Monday, July 20, 2020 11:46:21 AM Rank: MemberGroups: Registered Joined: 6/15/2019Posts: 21Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) Great questions! I just wanted to take a moment to add on to some of the information mentioned:Question 25:Yes, specific gravity (spgr) is always in relation to water. When this concept was founded, (reasonably) pure water was one of the most commonly available pure fluids. Scientists were looking for a way to relate the density of fluids to a standard and water seemed like a good choice. In medicine, you may be spgr in relation to urinary tests which compare solute content that thereby changes urine density.Question 36:This is a great question and should be thought of microscopically. So picture a circular cross-section of a tube with concentric layers of fluid. Now, think about the outermost layer - the layer in contact with the surface of the tube. The fluid molecules touch and slide past the inner surface of the piping; the fluid molecules are attracted to the surface of the material they are flowing within just as they are attracted to one another (- this is why we have viscosity). Generally, due to the attraction of the molecules to the surface of the pipe, we can expect friction which ultimately results in a slower speed. Now imagine the innermost layer of fluid, these molecules only interact with other fluid molecules - they don't touch the surface at all. As a result, there is reduced friction and a faster speed results.Hope this helps! Back to top User Profile
INSTR_Faisal_57 #4 Posted : Monday, July 20, 2020 10:04:35 PM Rank: NewbieGroups: Joined: 6/7/2020Posts: 0Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) Hi Fahad,Regarding specific gravity. While the above posts are correct that it is typical to use water as the standard for specific gravity, it is not necessarily the case. Specific gravity is simply a relative density concept: take the density of something relative to something else. That "something else" can anything.Every now and then you will encounter a question where you're given a specific gravity that is not relative to water, just as I did when I wrote the MCAT. Back to top | Edit by user User Profile
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Clean Slate theme by Jaben Cargman (Tiny Gecko) | 1,251 | 5,312 | {"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} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.917943 |
https://housecleaningwestpalm.com/a-good-will-also-cause-the-equilibrium-consumption/ | 1,624,557,538,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00298.warc.gz | 282,176,549 | 9,158 | a. A price reduction decreases the price of eachunit purchased. “Buy one, get one free” deal only reduces the price of secondunit purchased, reducing the price of the second book to zero. This offer doesnot alter the price of units below one book and above two books. With reference to figure15, in the beginning, the consumer faces a budget line connecting points A andis in equilibrium at point C. Book with a 50% discount (say book A which willcost 25£) is represented by point C. The consumer decides the best outcome isto buy Book A instead of a full priced book of 50£(say book B). Point Drepresents the point at which the consumer buys Book B, however, it isillustrated that the consumer prefers bundle C (Book A) to bundle D (Book B),since it lies on a higher indifference curve.
a. The clerk may have sensed dis-satisfaction fromthe consumer after having the “buy one get one free” deal rejected. A change inthe price of a good will definitely affect the satisfactory level of theconsumer. Change in price of good will also cause the equilibrium consumptionbundle to change. A reduction in price of good X will cause the budget line toturn counter clockwise. Illustratedfrom the figure above, if the beginning equilibrium is at point A, when thenprice of a book (X) falls to P^1x, the consumersopportunity set expands. With the new opportunity set, the consumer can attaina greater satisfactory level at point B.
This can be seen from the movement ofpoint A to point B into a new equilibrium. GoodX (Books) and Y (Other goods) are known as substitute goods. This implies thatan increase in the price of Books leads to an increase in consumption of othergoods vice versa. Since Books and Other goods are substitutes, a 40% discounton the price of Books (X) will cause the consumer to move from point A to pointB, whereby there are less Y consumed. (Y0 reduced to Y1). Toconclude, this change in price of good, in this case, the 40% discount on anybook, has expanded the consumers opportunity set and increase his satisfactorylevel.
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Check it out | 490 | 2,157 | {"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} | 2.796875 | 3 | CC-MAIN-2021-25 | latest | en | 0.914546 |
http://www.tumblr.com/search/hijira | 1,412,244,997,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663739.33/warc/CC-MAIN-20140930004103-00207-ip-10-234-18-248.ec2.internal.warc.gz | 1,011,732,752 | 23,555 | Earth, spacious.
If it has become difficult to follow the Right Way where you live, then migrate, you will find that Allah’s earth is spacious. Those who endure with patience will be rewarded without measure. Quran 39:10
In two weeks I will be in D.C. Starting over, starting fresh. Here’s to a crisp page in the first draft of a new chapter.
Character Charts
So starting working on character charts for my Camp NaNoWriMo novel yesterday. Got Emily’s almost done, started on basic stats for the others cause I felt like it. Just age, birthday, height, weight, hair and eye colors. Really put Siri to work doing this, mostly in converting the height and weight measurements into metric (the story is set in England, where they use metric, so I figured knowing the metric equivalents was important). There are equations to convert those, but as I stink at math (or more likely cause I’m lazy) I let Siri do it for me. Then I took whatever number she gave me and rounded it - the height to one decimal place and the weight to a whole number. (Interestingly, after all these years, and all the math I’ve forgotten, I still remember how to round). I also used Siri to find out the zodiac signs for each character once I determined their birthdays. Knowing their signs isn’t relevant to the story really, and I don’t believe in astrology, but I figured, hey, if somebody ever asks at least I know (of course, if I give them the birthdate they can always look it up too). Your sign does usually say something about your personality as well. (The descriptions I’ve read of my sign, Scorpio, actually seem surprisingly dead-on).
I also did other research…I looked at an ideal weight chart by height to determine the weights of my characters, so they’d be reasonable. Also, for Chelsea, my character who was raised Muslim, I found a converter which converts Gregorian dates to their equivalent in the Hijira, or Islamic calendar, and noted that date as well. (Surprisingly, the converter actually took my far-future year of 2332). Said date is the 17th day of Rabi II (the fourth month), 1763 AH (after Hijira, hijira being Mohammed’s flight from Mecca to Medina in 622 AD, the event which marks the beginning of the Islamic calendar).
Anyway, lots more work to do, but just wanted to share my progress. | 512 | 2,299 | {"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-2014-41 | latest | en | 0.959702 |
http://bootmath.com/show-that-the-equation-a_1ealpha_1x-a_2ealpha_2x-cdots-a_nealpha_nx-0-has-at-most-n-1-real-roots.html | 1,534,558,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213264.47/warc/CC-MAIN-20180818021014-20180818041014-00143.warc.gz | 57,309,827 | 5,755 | # Show that the equation $a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$ has at most $n – 1$ real roots.
For non-zero $a_1, a_2, \ldots , a_n$ and for $\alpha_1, \alpha_2, \ldots , \alpha_n$ such that $\alpha_i \neq \alpha_j$ for $i \neq j$, show that the equation
$$a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$$
has at most $n – 1$ real roots.
I thought of applying Rolle’s theorem to the function $f(x) = a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx}$ and reach a contradiction, but I can’t find a way to use it.
#### Solutions Collecting From Web of "Show that the equation $a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$ has at most $n – 1$ real roots."
Hints: Between any two roots of $f(x)$, there lies a root of $f'(x)$ by Rolle’s theorem.
Now, we shall use induction and the above fact to do your question.
For $n=1$- no real root.
Let’s consider $n=2$ in detail for the idea of the general case.
$f(x)=a_1e^{\alpha_1x}+a_2e^{\alpha_2x}$.
Let $g(x)=e^{-\alpha_1x}f(x)=a_1+a_2e^{(\alpha_1-\alpha_2)x}$. What’s $g'(x)?$ How many real roots does $g'(x)$ have? (answer: $0$ real roots). So, $g(x)$ can have at most one real root- which means $f$ can have at most one real root.
Now, use induction for the general case after considering $g(x)=a_1e^{-\alpha_1x}f(x)$. (Note that the real roots of $f$ and $g$ coincide.) | 544 | 1,417 | {"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} | 4.15625 | 4 | CC-MAIN-2018-34 | latest | en | 0.736579 |
https://www.javaguides.net/2023/11/python-program-to-print-identity-matrix.html | 1,716,373,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00484.warc.gz | 745,109,621 | 47,467 | # 1. Introduction
An identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. It is a key concept in linear algebra and has the property that any matrix multiplied by the identity matrix is unchanged. This post presents a Python program to print an identity matrix of a given size.
An identity matrix is defined as a square matrix with ones on the main diagonal and zeros in all other positions. The main diagonal is the diagonal that runs from the top left corner to the bottom right corner.
# 2. Program Steps
1. Define the size of the matrix.
2. Iterate over rows and columns to print the appropriate values.
3. Use nested loops: the outer loop for rows and the inner loop for columns.
4. Print '1' when the row and column indices are equal, and '0' otherwise.
# 3. Code Program
``````# Function to print an identity matrix of size n
def print_identity_matrix(n):
# Iterate over each row
for i in range(n):
# Iterate over each column
for j in range(n):
# Print 1 if the current row and column are the same (diagonal), else 0
print(1 if i == j else 0, end=" ")
# Move to the next line after each row
print()
# Size of the identity matrix
matrix_size = 4
# Print the identity matrix
print_identity_matrix(matrix_size)
``````
### Output:
```1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
```
### Explanation:
1. print_identity_matrix is a function that prints an identity matrix of size n.
2. The function uses two nested for loops, where i represents each row and j represents each column.
3. In the inner loop, print(1 if i == j else 0, end=" ") is used to print 1 if the row index i is equal to the column index j (indicating a diagonal element), otherwise 0.
4. end=" " specifies that each print statement ends with a space instead of a newline character, allowing the elements of a row to be printed on the same line.
5. After the inner loop finishes printing a row, print() is called to move to the next line before printing the next row.
6. Calling print_identity_matrix(matrix_size) with matrix_size set to 4 prints a 4x4 identity matrix.
7. The output is a 4x4 identity matrix with 1s on the diagonal and 0s elsewhere, formatted with spaces between elements and newlines after each row. | 550 | 2,225 | {"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-2024-22 | longest | en | 0.813545 |
https://www.wbez.org/stories/is-tilden-high-spending-too-much/74918778-e2c0-4635-bbf5-a0684db7af66 | 1,590,818,712,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00268.warc.gz | 965,954,943 | 63,025 | # Is Tilden High spending too much?
School budgets are reportedly shrinking across the city this year and it’s causing a lot of uproar. But as the debate over Chicago Public Schools’ current budget continues, Curious City is digging into a question asked by a mysterious commenter.
The original question came from a questioner who goes by “Tim.”
I recently learned the annual budget for Tilden High School. I then divided that number by the total number of students at Tilden and learned that Chicago is spending \$175,000 for each student per year. That is 5 to 6 times the cost of sending a kid to the best private high schools in the state. How do we justify the absurd amount of money Chicago spends to educate a child?
I made several attempts to reach Tim, but heard nothing back. Then I saw the following comment pop up on a recent story from a commenter named “greg”:
So I tried e-mailing “greg” to see if he was our questioner. Again, no luck.
Our mysterious commenter’s question about the “absurd amount of money” spent at Tilden is a loaded one — and one we can’t (and maybe shouldn’t) answer for you. We can, though, get across some basic information, such as how much CPS spent to educate children at Tilden and how that compares to other schools. At that point, you (as well as “Tim” and “greg”) can start a discussion about what spending — if any — is absurd.
I called Allan Odden, a researcher at the University of Wisconsin-Madison and an expert in school financing. Here’s what he told me about that \$175,000 figure:
“There’s no high school in America that spends that amount,” Odden said. “It’s not even in the same ballpark.”
Then what does Tilden High School spend per student every year?
I tried to follow Tim, or… greg’s, math. He says he took the annual budget for Tilden High School and divided it by the number of students.
CPS breaks out each individual school’s budget in the budget book and using those numbers, here’s my back-of-the-envelope calculations for the last three year’s budgets:
Odden and others I spoke to say these numbers are closer to what they expected.
On average, public high schools in America spend around \$12,000 per student according to the most recent data collected by the U.S. Department of Education. But recently, Tilden — a struggling school in Chicago’s New City area — has gotten extra support from the federal government in the form of a \$6 million School Improvement Grant.
These grants, launched by in 2009 by the Obama administration, are targeted at the lowest-performing high schools and meant to help reboot the school to improve academic outcomes. The interventions typically involve firing all the staff and hiring new.
Tilden was one of 13 schools awarded grants in the 2011-2012 school year. Initially, it was considered a “transformation” school and no staff was fired. But in the second year of the grant, the 2012-2013 school year, CPS decided to implement a “turnaround” at Tilden instead, firing the entire staff.
CPS officials say the school has gotten an additional \$1.9 million each year of the grant and is expected to get the last installment this coming year. This is likely why the amount spent per student is higher than the district and state average.
But the amount Tilden is currently getting is certainly not three times what private school tuition costs. Though tuition rates vary, private high schools in Chicago charge anywhere from about \$9,000 per year to \$30,000 a year.
At the prestigious University of Chicago Laboratory Schools, where Mayor Rahm Emanuel sends his children, tuition for the upcoming school year is \$28,290 for high school, \$27,096 for middle school, and \$25,296 for elementary school. It charges \$17,832 for half-day preschool. These figures don’t include additional fees for books, gym clothes, after-school programs and other extracurricular programs.
School spending across Illinois varies widely, with some districts spending more than \$20,000 per student per year, while others spend \$6,000. That’s because revenue for schools is mostly raised through local property taxes. Areas with tony homes and other high-value properties can generate lots of revenue. Areas with less valuable property are more limited.
To make up for disparities, the state sets a foundation. In 1997, it created a committee to help figure out what the foundation level should be — in other words, what it takes to adequately educate a student. Sylvia Puente, executive director of the Latino Policy Forum, is the chair of that committee, the Education Funding Advisory Board.
For the coming school year (2013-2014), EFAB recommended the state provide every district a minimum of \$8,672 per student. Puente tells me the number is generated using a complicated formula that looks at low-spending, high-performing districts.
But in every year except one, the Illinois General Assembly set the foundation amount lower than what EFAB recommended. In the most recent state budget, foundation funding is \$6,119.
However, most districts do not get the full \$6,119. For example, Chicago is getting \$5,720 per student from the state. The district then use local property taxes to generate more.
Private school funding is a different story. Parents pay tuition and schools are better able to make up for income disparities through charitable donations and alumni networks. But John Pantle, Finance and Advancement Consultant at the Archdiocese of Chicago, said Catholic schools do look at median family income in the surrounding areas when setting tuition rates.
“You need to meet the market,” said Pantle, the financial consultant at the Archdiocese of Chicago. “Any more than you would open a Cadillac dealership in an area that families couldn’t afford to buy a luxury car. Same thing with schools. St. Clements can charge a little more because it’s in the heart of Lincoln Park.”
But Pantle said in order to keep Catholic schools accessible to lower income families, the Archdiocese often steps in to cover the cost of education in areas where tuition is intentionally kept low. He said family discounts and individual scholarships are also provided.
“We try to balance it and it’s a constant struggle,” Pantle said.
How much money makes it to a school?
Once public school districts figure out how much they will get from state, federal and local sources, it all goes into one big pot and then district leaders decide how to distribute money to individual schools.
This year, the pot of money for Chicago Public Schools is about \$5.6 billion to serve 405,519 students. Applying our commenter’s back-of-the-envelope calculation means CPS is spending roughly \$13,789 per student.
But that crude calculation doesn’t tell the whole story.
For starters, a decent chunk of the money never makes it to the school level. It’s spent on salaries at the central office, consulting contracts, and mid-level management. For example, the legal department at the central office is budgeted \$12.9 million.
According to budget documents, \$3.4 billion of the \$5.6 billion operating budget is going directly to schools and most of that goes to pay teacher salaries. There are some services that do directly affect schools, but are part of a separate departments that operate citywide, such as a traveling nurse or curriculum specialist.
If you divide that \$3.4 billion by the 405,519 students enrolled in CPS schools, the per student amount drops to roughly \$8,536 per student.
But again, that doesn’t mean every school in Chicago gets \$13,000 or \$8,500 per student. Some get more, some get less.
Typically, elementary schools get less because they cost less to operate, mostly because they need less staff per student. There are also federal grants earmarked for specific schools and high-needs populations that can increase or decrease how much a school gets. This is the case for Tilden and the School Improvement Grant.
CPS also divides its revenues unevenly based on it’s district policies. For example, more money and positions go to selective enrollment and magnet schools.
After money makes it to the school
The amount a school gets from either parents or taxpayers varies widely from school to school, but what does the money get spent on?
“What you buy for education, like anything else, you buy quality,” Puente said.
On average, schools and districts spend about 80 percent of their budget on people. That includes salaries, benefits and retirement costs for teachers, clerks and administrators.
“No matter what, whether it’s \$10,000 a student or \$15,000 a student or less, the best use of those dollars is getting the very best people in the building,” said Mike Milkie, CEO and Superintendent of the high-performing Noble Street Charter School network that operates schools across Chicago.
Milkie said Noble spends about \$11,500 per student at the network’s 14 high schools. Charter schools are publicly-funded, but privately operated. Some of the money Milkie refers to comes from private sources and includes facility costs and other expenses not directly related to student instruction.
Charter schools have argued that getting less money per student has limited their ability to hire and keep great teachers. Charter school teachers are not part of the Chicago Teachers Union and typically make less money, according to the Illinois Network of Charter Schools, a group advocating for charters.
As with their public counterparts, private schools also spend the majority of their resources on personnel, says Pantle, the financial consultant at the Archdiocese of Chicago. However, average salaries at area Catholic schools run much lower than local public schools. Pantle pegged the average teacher salary at about \$35,000 per year, not including benefits. The average teacher salary at CPS, according to the district, is \$74,839.
To go back to the original question, \$175,000 per child could essentially buy the equivalent of two separate teachers dedicated solely to each student and still have money left over for supplies or other operating costs.
Since we’re at it, there’s also “hidden money”
Dividing a school’s individual budget by the number of students enrolled is an accurate and fair way to estimate how much is spent per student, Odden, the UW-Madison researcher, said. But if we’re talking back-of-the-envelope math, we could use a slightly larger envelope and include funding sources that our questioner (and possibly most of us) aren’t aware of.
These are essentially the “offshore bank accounts” of public education. And as states and districts slash spending, they’ve become more and more prevalent.
Most of the “hidden money” can be found in more affluent areas of the city, where highly-educated, middle- and upper-income families fundraise to provide extra services like art, music and additional technology, such as iPads. The groups usually operate as 501(c)3 non-profit organizations and can fundraise large sums.
These funding streams are entirely separate from a school’s public budget, but, in light of recent cuts, they have filled in what some may consider basic services, such as a full-time physical education or world language teacher.
But parent fundraising groups are not the only “offshore accounts” in school budgets. There are all kinds of grants, donations and philanthropic efforts that provide valuable services, but don’t show up in a school’s budget.
At Lindblom Math and Science Academy in the West Englewood neighborhood, Principal Alan Mather tells me about the school’s partnership with Baxter International. The corporation provides biotechnology courses and professional development, as well as funding for a few positions.
“There is no funding stream that you see from that, but there’s a huge benefit to the student body because of that relationship,” Mather said.
Lindblom is one of the city’s ten selective enrollment high schools, which gets extra resources from the CPS Board of Education. Still, Mather says he is constantly seeking out local partnerships to fill gaps and provide disadvantaged students new opportunities.
“That kind of stuff doesn’t have the same price tag.” Mather said. “It’s invaluable.”
Tilden is benefiting from a new partnership that also doesn’t show up in their budget, but is indirectly paid for by the federal government. In 2012, Columbia College received a \$3 million dollar innovation grant from the U.S. Department of Education to implement a “Convergence Academy” inside Tilden and one other school. The money will come through Columbia — not CPS, not Tilden.
Back to the question
“Tim” (and later, “greg”) got us onto this tangent on school spending. The skinny is that within Chicago, schools get anywhere from \$6,000 per student to \$17,000 per student. State data indicate average spending is about \$12,000 per student.
As for the area’s private high schools? Total costs there are not entirely clear, but tuition tops out at \$30,000.
Again, it’s not up to Curious City to say any of these amounts are absurd, but “Tim” (whoever you are): No area school — public or private — spends \$175,000 per student. | 2,733 | 13,177 | {"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} | 2.5625 | 3 | CC-MAIN-2020-24 | longest | en | 0.965161 |
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# VECTOR CALCULUS - PowerPoint PPT Presentation
13. VECTOR CALCULUS. VECTOR CALCULUS. 13.8 Stokes’ Theorem. In this section, we will learn about: The Stokes’ Theorem and using it to evaluate integrals. STOKES’ VS. GREEN’S THEOREM. Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem.
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## VECTOR CALCULUS
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13
VECTOR CALCULUS
VECTOR CALCULUS
13.8
Stokes’ Theorem
In this section, we will learn about:
The Stokes’ Theorem and
using it to evaluate integrals.
STOKES’ VS. GREEN’S THEOREM
• Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem.
• Green’s Theorem relates a double integral over a plane region D to a line integral around its plane boundary curve.
• Stokes’ Theorem relates a surface integral over a surface S to a line integral around the boundary curve of S (a space curve).
INTRODUCTION
• The figure shows an oriented surface with unit normal vector n.
• The orientation of Sinduces the positive orientation of the boundary curve C.
INTRODUCTION
• This means that:
• If you walk in the positive direction around Cwith your head pointing in the direction of n, the surface will always be on your left.
STOKES’ THEOREM
• Let:
• S be an oriented piecewise-smooth surface bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation.
• F be a vector field whose components have continuous partial derivatives on an open region in that contains S.
• Then,
STOKES’ THEOREM
• The theorem is named after the Irish mathematical physicist Sir George Stokes (1819–1903).
• What we call Stokes’ Theorem was actually discovered by the Scottish physicist Sir William Thomson (1824–1907, known as Lord Kelvin).
• Stokes learned of it in a letter from Thomson in 1850.
STOKES’ THEOREM
• Thus, Stokes’ Theorem says:
• The line integral around the boundary curve of Sof the tangential component of F is equal to the surface integral of the normal component of the curl of F.
STOKES’ THEOREM
Equation 1
• The positively oriented boundary curve of the oriented surface S is often written as ∂S.
• So,the theorem can be expressed as:
STOKES’ THEOREM, GREEN’S THEOREM, & FTC
• There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental Theorem of Calculus (FTC).
• As before, there is an integral involving derivatives on the left side of Equation 1 (recall that curl F is a sort of derivative of F).
• The right side involves the values of F only on the boundaryof S.
STOKES’ THEOREM, GREEN’S THEOREM, & FTC
• In fact, consider the special case where the surface S:
• Is flat.
• Lies in the xy-plane with upward orientation.
STOKES’ THEOREM, GREEN’S THEOREM, & FTC
• Then,
• The unit normal is k.
• The surface integral becomes a double integral.
• Stokes’ Theorem becomes:
STOKES’ THEOREM, GREEN’S THEOREM, & FTC
• This is precisely the vector form of Green’s Theorem given in Equation 12 in Section 12.5
• Thus, we see that Green’s Theorem is really a special case of Stokes’ Theorem.
STOKES’ THEOREM
• Stokes’ Theorem is too difficult for us to prove in its full generality.
• Still, we can give a proof when:
• S is a graph.
• F, S, and C are well behaved.
STOKES’ TH.—SPECIAL CASE
Proof
• We assume that the equation of Sis: z = g(x, y), (x, y) Dwhere:
• g has continuous second-order partial derivatives.
• D is a simple plane region whose boundary curve C1 corresponds to C.
STOKES’ TH.—SPECIAL CASE
Proof
• If the orientation of S is upward, the positive orientation of C corresponds to the positive orientation of C1.
STOKES’ TH.—SPECIAL CASE
Proof
• We are also given that:
• F = P i + Q j + R kwhere the partial derivatives of P, Q, and R are continuous.
STOKES’ TH.—SPECIAL CASE
Proof
• S is a graph of a function.
• Thus, we can apply Formula 10 in Section 12.7 with F replaced by curl F.
STOKES’ TH.—SPECIAL CASE
Proof—Equation 2
• The result is:
• where the partial derivatives of P, Q, and Rare evaluated at (x, y, g(x, y)).
STOKES’ TH.—SPECIAL CASE
Proof
• Suppose
• x =x(t) y =y(t) a ≤t ≤b
• is a parametric representation of C1.
• Then, a parametric representation of Cis: x =x(t) y =y(t) z =g(x(t), y(t)) a ≤t ≤b
STOKES’ TH.—SPECIAL CASE
Proof
• This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows:
STOKES’ TH.—SPECIAL CASE
Proof
• We have used Green’s Theorem in the last step.
STOKES’ TH.—SPECIAL CASE
Proof
• Next, we use the Chain Rule again, remembering that:
• P, Q, and R are functions of x, y, and z.
• z is itself a function of x and y.
STOKES’ TH.—SPECIAL CASE
Proof
• Thus, we get:
STOKES’ TH.—SPECIAL CASE
Proof
• Four terms in that double integral cancel.
• The remaining six can be arranged to coincide with the right side of Equation 2.
• Hence,
STOKES’ THEOREM
Example 1
• Evaluate where:
• F(x, y, z) = –y2i + x j + z2k
• C is the curve of intersection of the plane y + z = 2 and the cylinder x2 + y2 = 1. (Orient C to be counterclockwise when viewed from above.)
STOKES’ THEOREM
Example 1
• The curve C (an ellipse) is shown here.
• could be evaluated directly.
• However, it’s easier to use Stokes’ Theorem.
STOKES’ THEOREM
Example 1
• We first compute:
STOKES’ THEOREM
Example 1
• There are many surfaces with boundary C.
• The most convenient choice, though, is the elliptical region S in the plane y + z = 2 that is bounded by C.
• If we orient S upward, C has the induced positive orientation.
STOKES’ THEOREM
Example 1
• The projection D of S on the xy-plane is the disk x2 + y2≤ 1.
• So, using Equation 10 in Section 12.7 with z =g(x, y) = 2 – y, we have the following result.
STOKES’ THEOREM
Example 2
• Use Stokes’ Theorem to compute where:
• F(x, y, z) = xz i + yz j + xy k
• S is the part of the sphere x2 + y2 + z2 = 4 that lies inside the cylinder x2 + y2 =1 and above the xy-plane.
STOKES’ THEOREM
Example 2
• To find the boundary curve C, we solve: x2 + y2 + z2 = 4 and x2 + y2 = 1
• Subtracting, we get z2 = 3.
• So, (since z > 0).
STOKES’ THEOREM
Example 2
• So, C is the circle given by: x2 + y2 = 1,
STOKES’ THEOREM
Example 2
• A vector equation of C is:r(t) = cos t i + sin t j + k 0 ≤t ≤ 2π
• Therefore, r’(t) =–sin t i + cos t j
• Also, we have:
STOKES’ THEOREM
Example 2
• Thus, by Stokes’ Theorem,
STOKES’ THEOREM
• Note that, in Example 2, we computed a surface integral simply by knowing the values of F on the boundary curve C.
• This means that:
• If we have another oriented surface with the same boundary curve C, we get exactly the same value for the surface integral!
STOKES’ THEOREM
Equation 3
• In general, if S1 and S2 are oriented surfaces with the same oriented boundary curve Cand both satisfy the hypotheses of Stokes’ Theorem, then
• This fact is useful when it is difficult to integrate over one surface but easy to integrate over the other.
CURL VECTOR
• We now use Stokes’ Theorem to throw some light on the meaning of the curl vector.
• Suppose that C is an oriented closed curve and v represents the velocity field in fluid flow.
CURL VECTOR
• Consider the line integral and recall that v ∙T is the component of vin the direction of the unit tangent vector T.
• This means that the closer the direction of v is to the direction of T, the larger the value of v ∙T.
CIRCULATION
• Thus, is a measure of the tendency of the fluid to move around C.
• It iscalled the circulation of v around C.
CURL VECTOR
• Now, let: P0(x0, y0, z0) be a point in the fluid.
• Sa be a small disk with radius a and center P0.
• Then, (curl F)(P) ≈ (curl F)(P0) for all points P on Sa because curl F is continuous.
CURL VECTOR
• Thus, by Stokes’ Theorem, we get the following approximation to the circulation around the boundary circle Ca:
CURL VECTOR
Equation 4
• The approximation becomes better as a→ 0.
• Thus, we have:
CURL & CIRCULATION
• Equation 4 gives the relationship between the curl and the circulation.
• It shows that curl v ∙n is a measure of the rotating effect of the fluid about the axis n.
• The curling effect is greatest about the axis parallel to curl v.
CURL & CIRCULATION
• Imagine a tiny paddle wheel placed in the fluid at a point P.
• The paddle wheel rotates fastest when its axis is parallel to curl v.
CLOSED CURVES
• Finally, we mention that Stokes’ Theorem can be used to prove Theorem 4 in Section 12.5:
• If curl F = 0 on all of , then F is conservative.
CLOSED CURVES
• From Theorems 3 and 4 in Section 12.3, we know that F is conservative if for every closed path C.
• Given C, suppose we can find an orientable surface S whose boundary is C.
• This can be done, but the proof requires advanced techniques.
CLOSED CURVES
• Then, Stokes’ Theorem gives:
• A curve that is not simple can be broken into a number of simple curves.
• The integrals around these curves are all 0.
CLOSED CURVES
• Adding these integrals, we obtain: for any closed curve C. | 2,634 | 9,449 | {"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} | 4.4375 | 4 | CC-MAIN-2018-51 | latest | en | 0.85895 |
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42
PAGE: 402 SET: Exercises PROBLEM: 42
The equation is .
(Write the equation in the form of y = mx + b)
Compare the equation with slope-intercept form y = mx + b.where m is slope and b is y-intercept.
slope(m) = 6 and y-intercept is(b) = 7.
y-intercept is 7, so the line crosses the y-axis at (0, 7)
Using slope find the next point.
Slope(m) = = .
• Plot the point (0, 7)
• Start at point (0, 7),since the rise is 6. Move 6 units down.
• The run is 1. Move 1 unit left then plot the point (, 1)
• Draw a line through these points.
graph.
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# Tutor profile: Alexus L.
Alexus L.
## Questions
### Subject:Linear Algebra
TutorMe
Question:
Prove that if A is an nxn matrix, then det(A^-1) =(detA)^-1
Alexus L.
AA^-1 = I and det I =1 using the Theorem det(AB)=det(A)det(B) we get det(A)det(A^-1) = det(AA^-1) = det(I) = 1 Therefore det(A^-1) = 1/(detA) = (detA)^-1
### Subject:Chemistry
TutorMe
Question:
The number 0.0043015 has how many significant figures? Also, what is its scientific notation?
Alexus L.
The number has 5 significant figures seeing as though leading zeros do not matter. The zero in the middle is included. The scientific notation for this number would be 4.3015 x 10^3. In order to keep the same number of significant figures the numbers behind the 3 must be included. If it is rounded to 4.3 x 10^3 the number of significant figures will go down to 2.
### Subject:Calculus
TutorMe
Question:
A flat piece of ice is shaped like a square. When the ice melts the area of the sheet is decreasing at a rate of 0.25 m^2/sec. What is the decreasing rate of the length when the area of the ice is 25 m^2 ?
Alexus L.
The area of the square with respect to time can be defined as : A(t)= lw where l=length of the square and w= width of the square. Since we are dealing with a square we will simplify our equation to A(t)= l^2 since the our sides are all the same size. From the problem we can see that the rate of decrease for our are (dA/dt) = -0.25 and we are asked to find (dl/dt) when A(t)=25. Next we must differentiate our equation with respect to t. Our new equation then becomes dA/dt = (2)(l)(dl/dt) Finally, we must go back to our original equation and plug in to find our length since we were given the area at that specific time. 25= l^2 so, l=5 The rate of decrease for our length is then (-0.25) = (2)(5)(dl/dt) so, dl/dt = -0.025.
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https://byjusexamprep.com/gate-me/rolling-contact-bearing | 1,718,254,335,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00569.warc.gz | 128,380,900 | 33,545 | # Rolling Contact Bearing
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Rolling contact bearings are also known as antifriction bearings because of their reduced friction. Before discussing rolling contact bearing, let’s explain what a bearing is. A bearing is a mechanical component that decreases friction between moving parts by restricting relative motion to only the desired motion. The bearing’s design may allow for free linear movement of the moving part or free rotation around a fixed axis. It may also inhibit motion by controlling the vectors of normal forces acting on the moving parts. By reducing friction, most bearings make it easier to achieve the desired motion.
The contact between the bearing surfaces in rolling contact bearing is rolling rather than sliding, as in sliding contact bearings. Standard sliding bearings have a high coefficient of friction and start at rest with almost metal-to-metal contact. The decreased starting friction of a rolling contact bearing over a sliding bearing is a significant advantage. Let’s take a closer look at rolling contact bearing and its various types.
Table of content
## What is Rolling Contact Bearing?
A rolling contact bearing carries a load by sandwiching rolling elements (such as balls or rollers) between two concentric, grooved rings known as races. The rolling elements roll with extremely little rolling resistance and minimal sliding due to the relative motion of the races. Sets of logs arranged on the ground with a large stone block on top are one of the earliest and best-known rolling-element bearings.
As the stone is pulled, the logs roll along the ground with little sliding friction. As each log emerges from the back, it is transported to the front, where the block meets it. A typical example of a Rolling contact bearing is laying numerous pens or pencils on a table and placing an item on top that might be used to approximate such a bearing. A shaft is inserted into a much bigger hole in a rolling element rotary bearing, and cylinders known as
ollers snugly fill the space between the shaft and the hole.
## Types of Rolling Contact Bearing
Cages are found in the majority of rolling contact bearings or rolling-element bearings. The cages reduce friction, wear, and bind by preventing the elements from rubbing against each other. Following are the five different types of rolling bearing elements:
• Balls
• Cylindrical rollers
• Spherical rollers
• Tapered rollers
• Needle rollers.
### Ball Bearing
The ball bearing is a prevalent type of rolling contact bearing or rolling-element bearing. The inner and outer races of the bearing are where the balls roll. Each race has a groove that is normally designed such that the ball fits loosely. In theory, the ball makes touch with each race across a relatively small region. On the other hand, a load on an endlessly small spot would result in infinitely high contact pressure.
In practice, the ball deforms (flattens) little where it comes into contact with each race, similar to how a tyre flattens where it comes into contact with the road. Where each ball presses against it, the race likewise yields slightly. As a result, in this type of rolling contact bearing, the contact between the ball and the race is finite in size and pressure.
#### Cylindrical Roller Bearing
In this type of rolling contact bearing, the cylinders are used as the shells. The cylinders of standard roller bearings are somewhat longer than the diameter. Roller bearings have a higher radial load capacity than ball bearings but a lower axial load capacity and higher friction. A considerable amount of the load is often carried by fewer than half of the total number of rollers.
Compared to a ball bearing or a spherical roller bearing, the bearing capacity often lowers quickly if the inner and outer races are misaligned. The outer load is constantly re-distributed among the rollers in radial bearings, as it is in all radial bearings.
### Spherical Roller Bearing
The outer ring of a spherical roller bearing has a spherical internal form. The rollers have a thicker centre section and a thinner end section. As a result, both static and dynamic misalignment can be accommodated by spherical roller bearings.
However, because there will be some sliding between rolling parts and rings, spherical rollers are difficult to produce and thus expensive. The bearings have higher friction than an ideal cylindrical or tapered roller bearing.
### Tapered Roller Bearing
Conical rollers roll on conical races in tapered roller bearings. Most rolling contact bearings can only take radial or axial loads. However, tapered roller bearings can take both and generally withstand higher loads than ball bearings due to their larger contact area. For example, most wheeled land vehicles’ wheel bearings are tapered roller bearings.
The disadvantages of tapered roller bearings are that, due to manufacturing complexities, tapered roller bearings are usually more expensive than ball bearings; additionally, the tapered roller acts as a wedge under heavy loads, and bearing loads tend to try to eject the roller; and, in comparison to ball bearings, the force from the collar that keeps the roller in the bearing adds to bearing friction.
### Needle Roller Bearing
Needle roller bearings have cylinders that are incredibly long and thin. The ends of the rollers frequently taper to points, which are employed to keep the rollers captive; alternatively, the rollers may be hemispherical and not captive but held by the shaft or a similar arrangement.
In this type of rolling contact bearing, the outside diameter of the bearing is only slightly larger than the hole in the middle due to the thin rollers. On the other hand, the small-diameter rollers must bend dramatically where they contact the races, causing the bearing to fatigue fast.
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# Inverse Bine Pose
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
6 replies to this topic
Posted 22 June 2014 - 07:48 PM
Hi guys,
I'm having some problems implementing basic skeletal animation. I'm not concerned with skinning at the moment, just bone movements. I understand the concept, but am missing something.
I have a simple setup like this:
root->child
float4x4 transforms[2];
// root
matrixTranslation(&translation,&pose.joints[0].pos); // the pose's root joint position
matrixRotationQuaternion(&rotation,&pose.joints[0].q); // the pose's root joint orientation
matrixMultiply(&transforms[0],&rotation,&translation); // combine
// child
matrixTranslation(&translation,&pose.joints[1].pos); // the pose's child joint position
matrixRotationQuaternion(&rotation,&pose.joints[1].q); // the pose's child joint orientation
matrixMultiply(&temp,&rotation,&translation); // combine
matrixMultiply(&transforms[1],&temp,&transforms[0]); // incorporate parent's transformation
// then I would use the transforms I made to transform the actual vertices
I don't know where to take into account the inverse_bind_pose, or if I'm doing it right at all.
I also don't understand why there is a position/translation in the pose joint; when the parent joint is rotated doesn't that automatically change the child's position/translation since the imaginary bone between them is rigid ?
I can give more code and structs if needed.
Thanks.
edit: title should be Bind pose, not Bine
Edited by Endemoniada, 22 June 2014 - 07:50 PM.
### #2NumberXaero Prime Members
Posted 22 June 2014 - 08:32 PM
I also don't understand why there is a position/translation in the pose joint; when the parent joint is rotated doesn't that automatically change the child's position/translation since the imaginary bone between them is rigid ?
Not really sure what you mean here exactly, parent transforms affect child transforms, as youve done, where is this "pose joint" thats confusing you coming from?
Bones dont really exist, theyre just the space between a parent transform and a child transform, where each transform is a joint that affects the child joints below it.
Ex, an arm: shoulder joint <-> elbow joint <-> wrist joint, each "<->" would be a bone i guess
Edited by NumberXaero, 22 June 2014 - 08:34 PM.
### #3Samith Members
Posted 22 June 2014 - 09:13 PM
The way it works is like this: the animation stream will provide the location of each bone in parent-space, which is the space relative to the bone's parents. Then, you have to create a buffer of model-space transforms for each bone by transforming each bone by it's parent's model-space transform. Then, once you have model space transforms, you need to determine how the current model space transform differs from the original pose's model space transform for each bone. This difference is what you use to transform your verts.
In some (very) pseudo code:
bone[0] = root;
bone[1] = anim_stream[1];
bone[2] = anim_stream[2];
...
bone[n] = anim_stream[n];
model_space_bone[n] = model_space_bone[parent_of(n)] * bone[n];
// ok, so what can we tell from model_space_bone[n]?
// we know the original pose's model space transformation for bone[n]
// therefore model_space_bone[n] = D * bone_pose[n] where bone_pose[n] is the
// transform of the bone in it's original pose.
// this means the current model_space_bone[n] is actually just some
// transformation (D) from the original model space position. We need to
// find D, because that's what we'll use to transform our verts
// doing some matrix math:
model_space_bone[n] * inv_bone_pose[n] = D * bone_pose[n] * inv_bone_pose[n];
D = model_space_bone[n] * inv_bone_pose[n]
bone_pose_space_bone[n] = D = model_space_bone[n] * inv_bone_pose[n]
// vertex transformation uses bone_pose_space_bone[n], as this transformation
// represents the transformation of that bone from it's original pose
// to the pose it's in currently
projected_vert = proj_mtx * view_mtx * model_mtx * bone_pose_space_bone[bone_idx] * vert_pos;
### #4Buckeye GDNet+
Posted 22 June 2014 - 09:18 PM
First, you may want to review the subject in this article about animation using matrices.
Second, it's not clear where you're getting the translation/rotation data from which you're creating matrices.
That being said, the inverse bind pose is the matrix inverse of a bone's world transform, which you appear to be calculating for the child in your OP - IF the child's position and rotation are relative to it's parent.
[ninja'd by Samith - ]
Edited by Buckeye, 22 June 2014 - 09:20 PM.
Please don't PM me with questions. Post them in the forums for everyone's benefit, and I can embarrass myself publicly.
You don't forget how to play when you grow old; you grow old when you forget how to play.
Posted 22 June 2014 - 09:25 PM
I guess I'll try to understand it after I actually get it to work.
This is what I'm working with (from Game Engine Architecture):
struct Joint {
float4x4 inv_bind_pose; // matrix
int parent;
};
struct Skeleton {
int joint_count;
Joint* joints;
};
struct JointPose {
float4 rotation; // quaternion
float3 translation;
};
struct SkeletonPose {
int joint_count;
JointPose* joints;
};
// it's already confusing because of the different joint types
1) Are the inv_bind_pose matrices relative to the root joint (which would have it's inv_bind_pose as identity) ?
2) Are the rotations and translations in the JointPose structs relative to their parents ?
3) How do I put it all together ?
Thanks
### #6NumberXaero Prime Members
Posted 22 June 2014 - 10:11 PM
1) Its the transform of the joint at the time it was bound to the mesh, inverted. World coordinates.
2) Yes
3) ....
I use world/local terminology (local = parent relative, world = global = model)
A) Using the joint local transforms, compute each joints world transform, childJointWorld = parentJointWorld * childJointLocal
-> Convert to matrices first, or use quaternions/vector3 then convert, your choice, in the end youll need a world matrix to combine with the jointInvBindPose matrix
QWorld = QParentWorld * QLocal // rotate part
TWorld = TParentWorld + (QParentWorld * (SParentWorld * TLocal) // translate part
SWorld = SParentWorld * SLocal // scale part
-> If a joint has no parent, world = local
B) Now you have all joint world transforms, lets call this next one jointSkinMatrix = jointWorld * jointInvBindPose
C) The model probably has skin matrix / bind matrix / bind shape matrix, whatever its called this gets applied to all the original vertices, so you can do this once and store the modified model
Long version....
vec3 skinnedVertex = vec3(0, 0, 0)
For Each Vertex v
For Each Joint Affecting Vertex v
skinnedVertex += ((jointWorld * jointInvBindPose) * (bindShapeMatrix * v)) * jointWeight
Again, some of this depends on matrices or quaternion use, compute on gpu or cpu, but thats the general idea. Normals, same thing, but wouldnt be translated, and would have to be normalized for correct lighting.
Edited by NumberXaero, 22 June 2014 - 10:13 PM.
Posted 23 June 2014 - 08:38 PM
Hey guys,
After reading your posts, going over Buckeye's article (which I'll have to get back to since it's more advanced than what I want right now), and reading a paper by Frank Luna, I was able to get it working.
I wanted to get a simple system up and running before using an importer/exporter. I load a model (.obj) that looks like three bones and create the joints by hand.
Everything seems to be fine, I can setup my bind pose structures, then put it in any other pose I like (by rotating the individual joints.)
I'm doing it like this:
for(i=0;i<n;i++)
Ci = Li * Pi
where C is the combined matrix, L is the local matrix, and P is the parent's local matrix
then I finalize them:
for(i=0;i<n;i++)
Fi = Oi * Ci
where F is the final matrix, O is the offset matrix, and C is computed above.
For the offset matrix I use the inverse_bind_matrix. Are those normally the same thing ?
There is something that is nagging me. I don't understand why I would need translations in each pose, if I'm not using scaling it seems I can set them up along with the inverse_bind_pose in the skeleton since they will never change. Maybe I'm missing something ?
Thanks a lot.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 2,138 | 8,813 | {"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} | 3.0625 | 3 | CC-MAIN-2017-17 | latest | en | 0.805282 |
http://math.stackexchange.com/questions/356219/how-does-c-small-imply-setcop-locally-small | 1,467,206,121,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397744.64/warc/CC-MAIN-20160624154957-00127-ip-10-164-35-72.ec2.internal.warc.gz | 189,440,606 | 17,058 | # How does $C$ small imply $Set^{C^{op}}$ locally small?
I read in some notebook that $C$ small implies $Set^{C^{op}}$ locally small, but I don't see what is the reasoning used, because the Yoneda lemma is not mentioned so that it is probably not needed...
Could somebody propose me a track?
-
When $C$ is a small category and $D$ is a locally small category, then $D^C$ is locally small. The reason is that for functors $F,G : C \to D$ the set of natural transformations $F \to G$ embeds into $\prod_{x \in \mathrm{ob}(C)} \hom(F(x),G(x))$, which is a product of sets, therefore also a set.
Now $\hom(F(x),G(x))$ is in $D$. So, this argument also uses that $D$ is locally small. E.g. let $D$ be a big monoid, and $C:=1$ the terminal category... – Berci Apr 9 '13 at 19:11
Thank you very much Martin and Berci! Yes, I agree with Berci, this last part uses the fact that $D = Set$ so that its $hom$ are sets. – almaus Apr 9 '13 at 19:19 | 292 | 939 | {"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} | 2.6875 | 3 | CC-MAIN-2016-26 | latest | en | 0.908634 |
https://www.tutorialspoint.com/custom-length-matrix-in-python | 1,638,551,572,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00381.warc.gz | 1,163,819,364 | 10,967 | # Custom length Matrix in Python
PythonServer Side ProgrammingProgramming
Sometimes when creating a matrix using python we may need to control how many times a given element is repeated in the resulting matrix. In this articled we will see how to create a matrix with required number of elements when the elements are given as a list.
## Using zip
We declare a list with elements to be used in the matrix. Then we declare another list which will hold the number of occurrences of the element in the matrix. Using the zip function we can create the resulting matrix which will involve a for loop to organize the elements.
## Example
Live Demo
listA = ['m', 'n', 'p','q']
# Count of elements
elem_count = [1,0,3,2]
# Given Lists
print("Given List of elements: " ,listA)
print("Count of elements : ",elem_count)
# Creating Matrix
res = [[x] * y for x, y in zip(listA, elem_count)]
# Result
print("The new matrix is : " ,res)
## Output
Running the above code gives us the following result −
Given List of elements: ['m', 'n', 'p', 'q']
Count of elements : [1, 0, 3, 2]
The new matrix is : [['m'], [], ['p', 'p', 'p'], ['q', 'q']]
## with map and mul
In this approach we use the mul method from operator module in place of zip method above. Also the map function applies the mul method to every element in the list hence the for loop is not required.
## Example
Live Demo
from operator import mul
listA = ['m', 'n', 'p','q']
# Count of elements
elem_count = [1,0,3,2]
# Given Lists
print("Given List of elements: " ,listA)
print("Count of elements : ",elem_count)
# Creating Matrix
res = list(map(mul, listA, elem_count))
# Result
print("The new matrix is : " ,res)
## Output
Running the above code gives us the following result −
Given List of elements: ['m', 'n', 'p', 'q']
Count of elements : [1, 0, 3, 2]
The new matrix is : ['m', '', 'ppp', 'qq']
Published on 26-Aug-2020 06:56:11 | 516 | 1,907 | {"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} | 3.109375 | 3 | CC-MAIN-2021-49 | longest | en | 0.756203 |
https://www.physicsforums.com/threads/falling-object.14372/ | 1,490,428,175,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188891.62/warc/CC-MAIN-20170322212948-00035-ip-10-233-31-227.ec2.internal.warc.gz | 930,901,966 | 15,090 | # Falling Object
1. Feb 12, 2004
### Alice
Hey
This seems to be way easy, but i'm just not gettng where to start it. A stone is dropped from the top of a cliff, it is seen to hit the ground below after 3.66 s, how high is the cliff? Anyone know? Thanks.
2. Feb 12, 2004
### Tom Mattson
Staff Emeritus
That means that vi=0, and the yi the height of the cliff, which is unknown. Implicit in this is that the acceleration of the stone is that of gravity.
That means that Δt=3.66s. Implicit in that is that yf=0, which is ground level.
Can you find an equation that relates those quantities?
3. Feb 12, 2004
### svtec
Mentor Edit: Please don't post complete solutions. Thank you.
Last edited by a moderator: Feb 12, 2004
4. Feb 12, 2004
### Alice
Tom-
I pretty much got that far on my own. I just don't know which formula ot use. The one that looks the closes would be
/\=Delta
/\X=X-Xo=VoxT+1/2AxT^2
and
Vx^2-Vox^2=2Ax/\x
but i'm not sure. I'm having a really hard time figuring out which formula goes where. Thanks.
-Alice
5. Feb 12, 2004
### turin
This one always works for a constant acceleration. And, of course, that "X" doesn't have to be horizontal.
6. Feb 13, 2004
### Tom Mattson
Staff Emeritus
Right, the equation also holds for "y".
Alice, look at the equation that turin referred to, rewrite it for y (just replace x with y), and verify that you do in fact have all the information to solve the problem. Then, plug in the numbers. | 432 | 1,474 | {"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-2017-13 | latest | en | 0.948248 |
http://math.stackexchange.com/questions/247150/how-to-simplify-this-equation-1-cdot-n-2n-1-3n-2-cdots-in-i/247171 | 1,454,952,041,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153736.68/warc/CC-MAIN-20160205193913-00217-ip-10-236-182-209.ec2.internal.warc.gz | 141,897,786 | 17,146 | # How to simplify this equation: $1\cdot N + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \dots + N \cdot1$? [duplicate]
$$1 (N) + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \cdots + N (1)$$
I need to write it in simplest form?
here 1(N) means 1 multiply by N
-
## marked as duplicate by Martin Sleziak, Najib Idrissi, RecklessReckoner, 6005, jkabrgDec 16 '15 at 15:31
Sorry, but does $N$ stand for a complex number? – S. Snape Nov 29 '12 at 10:05
Babak, $N$ must be a natural number; $i$ is intended as an index, not the square root of $-1$. So the sum is $\sum_{k=0}^{N-1} (1+k)(N-k)$. (Ah, now I see why you asked that. Grijesh, the tags you've chosen are completely inappropriate.) – Rhys Nov 29 '12 at 10:12
I am tiring to answer some other problem related to computer scince – Grijesh Chauhan Nov 29 '12 at 10:17
Apart from the post already mentioned by Normal Human, some other posts which are linked there might be interesting for you, too. – Martin Sleziak Dec 16 '15 at 14:12
$$\sum_{k=1}^Nk(N-k+1)=\sum_{k=1}^N kN-\sum_{k=1}^Nk^2+\sum_{k=1}^Nk=N\,\frac{N(N+1)}2-\frac{N(N+1)(2N+1)}6+\frac{N(N+1)}2=\frac{N(N+1)}2\left(N-\frac{2N+1}3+1\right)=\frac{N(N+1)(N+2)}6.$$
-
There is a typo. It should be N+2 in place of N+4. – Gautam Shenoy Nov 29 '12 at 10:49
@Gautam Shenoy - thanks , Martin Argerami - thanks! – Grijesh Chauhan Nov 29 '12 at 10:52
@draks 10/6 = 1 floor function – Grijesh Chauhan Nov 29 '12 at 12:20
how are you expanding $\sum_{k=1}^N kN$ to $N\,\frac{N(N+1)}2$ ? – david_adler Jun 13 '15 at 16:29
It's the basic identity $\sum_{k=1}^Nk=\frac{N(N+1)}2$, multiplied by $N$. – Martin Argerami Jun 13 '15 at 17:37
Your sum can be written as
$$\sum_{k=1}^Nk(N-k+1) = N\sum_{k=1}^{N}k - \sum_{k=1}^N{k^2} + \sum_{k=1}^N{k}$$
$$= \frac{N(N+1)^2}{2} - \frac{N(N+1)(2N+1)}{6}$$
-
thank Gautam Shenoy ! – Grijesh Chauhan Nov 29 '12 at 10:53
A more complex but fun alternative solution using Finite Calculus:
\begin{align} \sum_{x=1}^N x \left[(N+1)-x\right] & = \sum_{1}^{N+1} \left[ Nx - x(x-1)\right] \delta x = \sum_{1}^{N+1} \left( Nx^{(1)} - x^{(2)} \right) \delta x \\ & = \left[ \frac{N}{2}x^{(2)} - \frac{1}{3} x^{(3)} \right]_{1}^{N+1} = \left[ \frac{N}{2}x(x-1) - \frac{1}{3} x(x-1)(x-2) \right]_{1}^{N+1}\\ & = \left[ \frac{N(N+1)^2}{2} - 0 \right] - \left[ \frac{1}{3}(N+1)N(N-1) - 0 \right] \\ & = \frac{N(N+1)(N+2)}{6}. \end{align}
-
@ffrenz : Thanks frenz!! – Grijesh Chauhan Nov 29 '12 at 12:41 | 1,078 | 2,439 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-07 | longest | en | 0.766984 |
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# Set No.
## III B.Tech. I-Semester Regular Examinations, November-2003.
1
AERODYNAMICS-I
(Aeronautical Engineering)
Time: 3 hours Max. Marks: 80
All questions carry equal marks
---
1.a) What are the different types of the flow? Describe by giving examples.
b) Starting from the basic, derive the energy equation for the adiabatic horizontal flow
system.
2. A spinning cylinder is placed in uniform flow. Draw the flow patterns at different
velocities and derive the expression for the lift produced on the cylinder.
## 3. A long right circular cylinder of diameter a is set horizontally in a stream of velocity u
and caused to rotate at angular velocity ω. Obtain an expression it terms of ω and u
for the ratio of pressure difference between the top and the bottom of the cylinder to
the dynamic pressure of the stream. Describe briefly the behavior of the stagnation
lines of such a system as ω is increased from zero, keeping u constant.
4. A small symmetrical Zhukovsky Aerofoil has a thickness ratio of 0.1 and is fitted with
pressure holes on both the surfaces at 12% of the chord behind the leading edge so
that it can be used as yaw meter. Estimate the pressure difference which would be
found when it is set 1.5o incidence to the flow Assume 2-D flow and express the result
in terms of pressure rise at the stagnation point.
y X X
5. The camber line of a thin circular arc aerofoil is given by = 4c 1 − Find
c c c
the load distribution a incidence α show that the zero lift angle α0 is equal to-2h and
sketch the load distribution at the incidence showing clearly the separate effects of
camber and incidence. Compare the lift curves of this with that of a flat plate.
6. Determine the expression for the vortex drag for modified elliptical loading and find
the condition for the minimum vortex drag condition.
7. A monoplane weighing 7.36 X 104 N has elliptical wings 15.23 m in span. For a
speed of 90m/s in straight and level flight at low altitude find
(a) Induced (vortex) drag (b) Circulation round sections half way along the wings.
## 8. Write short note on the following:
a) Charge in velocity in a vortex flow
b) Limitation of the lifting line
c) Carafoli profiles.
@@@@@
Set No.
Code No: 312103
## III B.Tech. I-Semester Regular Examinations, November-2003.
2
AERODYNAMICS-I
(Aeronautical Engineering)
Time: 3 hours Max. Marks: 80
All questions carry equal marks
---
1. Derive the general continuity equation in Cartesian coordinates. Write this equation
for different types of the flows stating clearly the assumptions made.
2. Derive the expressions for the stream function, velocity potential. Local velocity and
the stagnation points of the resultant a source and a sink are placed near to each other.
3. Show that a straight line of length 4a can be obtained from the conformal
transformation of a circle of radius a. A flat plate is inclined at 3o to a uniform stream.
Working from the first principal and using the Zhukovsky hypothesis calculate the lift
coefficient of the plate.
## 4. What is Kutta-Zhukovsky transformation? How a circle can be transformed into a
cambred aerofoil.
5. Derive the expressions for the lift, moment and centre of pressure for a general thin
aerofoil. Prove that the aerodynamic center lies at the 25% chord from the leading
edge.
6. What is the effect of the starting and trailing vortex on the lift distribution on a wing,
what is the reason for the formation of the vortex?
7. What is the effect of the aspect ratio on the drag on the wing? A glider has wings of
elliptical planform of aspect ratio 6. The total drag is giving by C D = 0.02 + 0.06 CL2.
Find the change in the minimum angle of glide if the aspect ratio is increased to 10.
## 8. Write short note on the following:
a) Change in velocity in a vortex flow
b) Limitations of the lifting line
c) Carafoli profiles.
@@@@@
Set No.
Code No: 312103
## III B.Tech. I-Semester Regular Examinations, November-2003.
3
AERODYNAMICS-I
(Aeronautical Engineering)
Time: 3 hours Max. Marks: 80
All questions carry equal marks
---
1. Derive the equation for streamline flow showing the conservation of momentum.
What is the form of this equation if the flow is incompressible?
2. Describe the behaviour of the resultant flow when a source flow is placed in the
uniform flow. Derive the expressions for the stream function, velocity potential, local
velocity and the stagnation points of the flow. Give practical example also.
## 3. Calculate the theoretical lift coefficient of a Zhukovsky aerofoil having a thickness
ratio of 0.2 and 2% camber, set at 4o incidence in a 2-D irrotational flow. Derive your
formula for the lift coefficient.
4. Show that the moment coefficient of thin aerofoil can be estimated from the shape of
its camber line. Derive expressions giving moment coefficient in terms of a Fourier
series defining the slope of the camber line.
5.a) What is circulation? Find the expression for the lift on an aerofoil in terms of
circulation.
b) State Kutta-Zhukovsky theorem.
6. Explain the formation of starting vortex, trailing vortex, bound vortex and horse shoe
vortex.
7. A monoplane weighing 20 X 104 N has a span of 40 m and is flying at 50m/s with its
tail plane level with its wings and 6.5 m above the ground. Calculate the change in
downwash angle at the tail plane which is 14m behind the CP of the wings.
## 8. Write short note on the following:
a) Biot Savart Law
b) Blaaius Theorem
c) Limitations of the lifting line.
@@@@@
Set No.
Code No: 312103
## III B.Tech. I-Semester Regular Examinations, November-2003.
4
AERODYNAMICS-I
(Aeronautical Engineering)
Time: 3 hours Max. Marks: 80
All questions carry equal marks
---
∂v ∂u
1. What is circulation and vorticity? Prove that vorticity ζ = − .
∂x ∂y
2. What is a doublet and, how it is formed and discuss its properties and draw the stream
line and equipotential lines for it.
3. A 20% Zhukovsky aerofoil is set with its chord at zero incidence to a 2-D irrotational
airstream of 10 m/s has a lift coefficient of 0.3. Estimate the lift coefficient at 5o
incidence and velocity of the airflow just outside the boundary layer at the nose at this
altitude.
## 4. What is Kutta-Zhukovsky transformation? How a circle can be transformed into a
cambred aerofoil.
y X X
5. The camber line of a thin circular arc aerofoil is given by = 4c 1 −
c c c
Find the load distribution at incidence α show that the zero lift angle α0 is equal to-2h
and sketch the load distribution at the incidence showing clearly the separate effects
of camber and incidence. Compare the lift curves of this with that of a flat plate.
6. What is downwash, how does it affect the lift on a wing find the expression for the
downwash for elliptical distribution.
7. What is the effect of the aspect ratio on the drag on the wing? A glider has wings of
elliptical planform of aspect ratio 6. The total drag is giving by C D = 0.02 + 0.06 CL2.
Find the change in the minimum angle of glide if the aspect ratio is increased to 10.
## 8. Write short note on the following:
a) Biot Savart Law
b) Blaaius Theorem
c) Limitations of the lifting line.
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https://eduhawks.com/tag/grid/ | 1,632,040,851,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00440.warc.gz | 275,557,858 | 32,874 | Categories
## The graph of y = x2 is shown below. If graphed on the same grid, which of the following could be the graph of y = 0.4×2?
1) The answer is: [B]: r = 5 .
__________________________
Explanation:
__________________________
Given: 7r − 7 = 2r + 18 ; Round your answer to the nearest tenth, if necessary.
____________________________
Since “r” is the only variable given, let us assume we want to solve for “r” (instead of “x”).
___________________________
→ Subtract “2r” from EACH SIDE of the equation; and & add “7” to EACH SIDE of the equation:
_____________
→ 7r − 7 − 2r + 7 = 2r + 18 − 2r + 7 ; to get: → 5r = 25 ;
_____________
→ Now, divide EACH SIDE of the equation by “5”; to isolate “r” on one side of the equation; and to solve for “r” :
______________
→ 5r / 5 = 25 / 5 → r = 5 → which is: “Answer choice: [B]”.
_________________
Let us check our answer, by plugging in “5” for “r” in the original equation:
_________________
→ 7r − 7 = 2r + 18 ; → 7(5) − 7 =? 2(5) + 18? ;
______________________
→ 35 − 7 =? 10 + 18 ?; → 28 =? 28? Yes!
______________________
2) The answer is: [D]: x = 2 .
_____________
Explanation:
_____________
Given: 2x + 12 = 18 − x ; Solve for “x” (round to nearest tenth, if necessary).
_______________
→ Add “x” to EACH SIDE of the equation, & subtract “12” from EACH SIDE of the equation: → 2x + 12 + x − 12 = 18 − x + x − 12 ;
______________
→ To get: 3x = 6 ; → Divide EACH SIDE of the equation by “3”;
to isolate “x” on one side of the equation; and to solve for “x”:
_____________
→ 3x / 3 = 6 / 3 ; → x = 2 ; which is: “Answer choice: [D]”.
______________
Let us check our answer, by plugging in “2” for “x” in the original equation:
________________
→ 2x + 12 = 18 − x ; → 2(2) + 12 =? 18 − 2 ?
________________
→ 4 + 12 =? 18 − 2 ? ; → 16 =? 16? Yes!
________________________________
3) The answer is: [A]: x = -3 .
_____________
Explanation:
________________
Given: 8x − 3 = 15x + 18 ; Solve for “x”. Round your answer to the nearest tenth, if necessary.
_________________
→ Subtract “8x” from EACH SIDE of the equation, & add “3” to EACH SIDE of the equation:
_______________
→ 8x − 3 − 8x + 3 = 15x + 18 − 8x + 3 ; to get:
_______________
→ 0 = 7x + 21 ; → Subtract “21” from EACH SIDE of the equation;
_______________
→ 0 − 21 = 7x + 21 − 21 ; to get:
_______________
→ -21 = 7x ; Now divide EACH SIDE of the equation by “7”;
to isolate “x” on one side of the equation; & to solve for “x”:
_______________
→ = -21 / 7 = 7x / 7 ; → -3 = x ; which is “Answer choice: [A].”
_________________
Let us check our answer, by plugging in “-3” for “x” in the original equation:
________________
→ 8x − 3 = 15x + 18 ; → 8(-3) − 3 =? 15(-3) + 18 ?;
________________________
→ -24 − 3 =? -45 + 18 ? ; → -27 =? -27? Yes!
___________________________
4) The answer is: [C]: y = 11 .
_____________
Explanation:
____________
Given: 6y − 6 = 4y + 16 ; Solve for “y”; Round to the nearest tenth, if necessary.
____________
(Note: Since “y” is the only variable given; assume we are to solve for “y” instead of “x”).
____________
→ Subtract “4y” from EACH SIDE of the equation, & add “6” to EACH SIDE of the equation; → 6y − 6 − 4y + 6 = 4y + 16 − 4y + 6 ; to get:
_______________
→ 2y = 22 ; Now, divide EACH SIDE of the equation by “2”; to isolate “y” one side of the equation; and to solve for “y” ;
_______________
→ 2y / 2 = 22 / 2 ; → y = 11 → which is “Answer choice: [C]”.
_______________________________
Let us check our answer, by plugging in “11” for “y” in the original equation:
___________________
→ 6y − 6 = 4y + 16 ; → 6(11) − 6 =? 4(11) + 16 ?
_______________________
→ 66 − 6 =? 44 + 16 ? → 60 =? 60 ? Yes!
__________________
5) The answer is: [B]: x = -11 .
_____________________
Explanation:
_________________
Given: 3(x − 4) = 5(x + 2) ; Solve for “x”. Round to the nearest tenth, if necessary.
___________
→Note the “distributive property of multiplication”:
_____________
a*(b + c) = ab + ac ; and: a*(b − c) = ab − ac ;
_______________
→ Let us expand EACH SIDE of our given equation.
____________
3(x − 4) = (3*x) − (3*4) = 3x − 12;
______________________________
→Now let us expand the “right-hand side” of the given equation:
____________
→ 5(x + 2) = (5*x) + (5*2) = 5x + 10 ;
______________
→Now, we can rewrite the original equation:
_______________
→ 3(x − 4) = 5(x + 2) ; by substituting the expanded values for EACH SIDE of the question: → 3x − 12 = 5x + 10 ;
__________________
→ Subtract “3x” from EACH SIDE of the equation; and add “12” to EACH SIDE of the equation: → 3x − 12 − 3x + 12 = 5x + 10 − 3x + 12 ; to get:
________________
→ 0 = 2x + 22; → Now subtract “22” from EACH SIDE of the equation:
______________
→ 0 − 22 = 2x + 22 − 22 ; to get: → -22 = 2x ;
__________
→ Divide EACH SIDE of the equation by “2”; to isolate “x” on one side of the equation; & to solve for “x” ;
_____________
→ -22 / 2 = 2x /2 ; → -11 = x ; which is “Answer choice: [B]”.
______________
Let us check our answer, by plugging in “-11” for “x” in the original equation:
___________
→ 3(x − 4) = 5(x + 2) ; → 3(-11 − 4) =? 5(-11 + 2) ? ;
_______________________
→3(-15) =? 5(-9) ? ; → -45 =? -45 ? Yes!
_____________________________________________
Categories
## Triangle PQR is transformed to similar triangle P’Q’R’. A coordinate grid is shown from negative 10 to 0 to positive 10 on both x- and y-axes at increments of 1. A triangle PQR has P at ordered pair 4, 4, Q at ordered pair 8, 4, R at ordered pair 4 and 8. A polygon P prime Q prime R prime has P prime at ordered pair 1,1, Q prime at ordered pair 2,1, R prime at ordered pair 1, 2. What is the scale factor of dilation?
The length of the legs, PQ and PR are 4 units.
The lengths of the legs, P’Q’ and P’R’ are 1 unit.
Since the legs (and triangle) got smaller we know that the scale factor is between 0 and 1.
So.. 4 * x = 1
divide both sides by 4
x = 1/4
The scale factor is 1/4
Categories
## 23. A grid shows the positions of a subway stop and your house. The subway stop is located at (-5, 2) and your house is located at (-9, 9). What is the distance, to the nearest unit, between your house and the subway stop?
23. A grid shows the positions of a subway stop and your house. The subway stop is located at (-5, 2) and your house is located at (-9, 9). What is the distance, to the nearest unit, between your house and the subway stop?
Categories
## (20 pts) What is the range of the relation? Coordinate grid with labeled points at (negative one, negative two), (negative four, negative three), (negative two, zero), (two, two) A. {–4, 4} B. {–4, –2, –1, 0, –2} C. {–3, –2, 0, 2} D. {–4, –2,–1, 2}
(20 pts) What is the range of the relation? Coordinate grid with labeled points at (negative one, negative two), (negative four, negative three), (negative two, zero), (two, two) A. {–4, 4} B. {–4, –2, –1, 0, –2} C. {–3, –2, 0, 2} D. {–4, –2,–1, 2}
Categories
## ?__________ is the term that describes devices that automate many industrial tasks by receiving digital commands and converting them into real-world actions such as shutting down a power grid or opening a valve in a fuel pipeline.
Remote terminal units (RTUs) is the term that describes devices that automate many industrial tasks by receiving digital commands and converting them into real-world actions such as shutting down a power grid or opening a valve in a fuel pipeline. RTUs are the essential part of an intelligent power grid, also called smart grid, but also intelligent production system.
Categories
## What is the distance between (–6, 2) and (8, 10) on a coordinate grid?
What is the distance between (–6, 2) and (8, 10) on a coordinate grid?
Categories
## A gardener is planning to fill her garden with mulch. She plots it on a grid to plan how much she will need. The garden is in the shape of a rectangle with vertices at (3, 9) (5, 9) (3, 3) (5, 3). One bag of mulch covers 5ft2 and costs \$5.00. How much will it cost her to cover her garden?
A gardener is planning to fill her garden with mulch. She plots it on a grid to plan how much she will need. The garden is in the shape of a rectangle with vertices at (3, 9) (5, 9) (3, 3) (5, 3). One bag of mulch covers 5ft2 and costs \$5.00. How much will it cost her to cover her garden?
Categories
## Marissa researched the cost to have custom T-shirts printed by several local and online vendors. She found that each store’s charge for the job could be modeled by a linear function that combined a flat charge for artwork with a per T-shirt rate. Marissa plotted the functions on a coordinate grid and found that two of the functions produced lines that had the same y-intercept. How should Marissa interpret this result? A:Those two vendors charge the same rate per shirt. B:Those two vendors charge the same flat rate for artwork. C:Those two vendors will have the same total cost to produce one shirt. D:Those two vendors will charge the same total cost for any size job.
C. Yes, because the population values appear to be normally distributed.
Step-by-step explanation:
Given is a graph which shows the distribution of values of a population
The graph has the following characteristics
i) Bell shaped
ii) symmerical about mid vertical line
iii) Unimodal with mode = median =mean
iv) As x deviates more from the mean probability is decreasing and also curve approaches asymptotically the x axis
Hence we find that the curve is a distribution of normal
Option C is right
C. Yes, because the population values appear to be normally distributed.
Categories
## Which equation does the graph below represent? A coordinate grid is shown. The x-axis values are from negative 5 to positive 5 in increments of 1 for each grid line, and the y-axis values are from negative 15 to positive 15 in increments of 3 for each grid line. A line is shown passing through the ordered pairs negative 4, 12 and 0, 0 and 4, negative 12. y = fraction negative 1 over 3x y = −3x y = 3x y = fraction 1 over 3x
127=7 (mod n) means when 127 is divided by n, the division leaves a remainder of 7.
The statement is equivalent to
120=0 (mod n), meaning that n divides 120.
All divisors of 120 will satisfy the statement because 120 divided by a divisor (factor) will leave a remainder of 0.
Factors of 120 are:
n={1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120}, |n|=16.
You can count how many such values of n there are, and try to check that each one satisfies 127=7 mod n.
Categories
## (02.02 MC) Figure 1 and figure 2 are two congruent parallelograms drawn on a coordinate grid as shown below: Figure 1 is at (5, 8) (4, 4) (6, 2) (7, 6) . Figure 2 is at (-5, -2) (-7, -4) (-4, -6) (-6, -8) . Which two transformations can map figure 1 onto figure 2? (6 points) A) Reflection across the y-axis, followed by translation 10 units down B) Reflection across the y-axis, followed by reflection across x-axis C) Reflection across the x-axis, followed by reflection across y-axis D) Translation 11 units left, followed by translation 10 units down
(02.02 MC) Figure 1 and figure 2 are two congruent parallelograms drawn on a coordinate grid as shown below: Figure 1 is at (5, 8) (4, 4) (6, 2) (7, 6) . Figure 2 is at (-5, -2) (-7, -4) (-4, -6) (-6, -8) . Which two transformations can map figure 1 onto figure 2? (6 points) A) Reflection across the y-axis, followed by translation 10 units down B) Reflection across the y-axis, followed by reflection across x-axis C) Reflection across the x-axis, followed by reflection across y-axis D) Translation 11 units left, followed by translation 10 units down
Categories
## Rectangle J’K’L’M’ shown on the grid is the image of rectangle JKLM after transformation. The same transformation will be applied on trapezoid STUV, as shown below: What will be the location of T’ in the image trapezoid S’T’U’V’? (12, 6) (12, 7) (16, 7) (16, 6)
Answer: The measure of an acute base angle of the trapezoid BPQC is 77°.
Step-by-step explanation: As given in the question, the isosceles trapezoid BPQC is a part of the isosceles triangle ABC with same base BC. And, the vertex angle, m∠A = 26°.
We are to find the measure of an acute base angle of the trapezoid. That is, measure of angle B or angle C.
Since ΔABC is an isosceles triangle, so
Now, using the angles sum property of a triangle, we have
So, m∠B = m∠C = 77°.
Thus, the measure of an acute base angle of the trapezoid BPQC is 77°.
Categories
## Triangle RST is dilated to create triangle UVW on a coordinate grid. You are given that angle T is congruent to angle W. What other information is required to prove that the two triangles are similar? please help!!!! A. Segments RS and UV are congruent. B. Angle S is congruent to angle V C. Segment RT is congruent to segment UW, and segment ST is congruent to segment VW D. Angle S is congruent to angle U
From (0, 0), the points A and D are at (-1, -1) and (2, -1) respectively.
Since the scale factor is 2, every point will be moved twice a distance in both directions. (x and y).
So, multiplying both coordinates by 2, we will get the dilated points.
Hence, A’ will be at (2 × (-1), 2 × (-1)) = (-2, -2)
D’ will be at (2 × 2, 2 × (-1)) = (4, -2).
Categories
## What are the dimensions of the rectangle shown below? Remember to use the axes on the coordinate grid to help you. A. 6 units × 3 units B. 7 units × 4 units C. 10 units × 4 units D. 14 units × 8 units
Gven that a
district manager rewards sales teams based on overall sales generated
in a month and that the data for earnings are shown in the table, where Low
represents the lowest sales and High represents the highest sales
generated by a single sales team member.
Team Low High Range Mean Median IQR σ
Team X 1970 2970 1200 2571.9 2684 426.3 313.8
Team Y 250 375 125 315.8 311 59 37.8
Team Z 950 1900 950 1529.9 1473 276 180.7
Part A: A data set, say X, is said to be more consistent than another data set, say Y, if data set X is less spread out than data set Y.
The range is used to determine the spread of the data. The range is calculated by subtracting the lowest data value from the highest data value.
From the data presented in the table, above, Team Y has the smallest spread (range) compaired to the other teams and hence is the most consistent team.
Therefore, if the manager wants to award the sales team that has the most consistent earnings among its team members, he should choose Team Y.
Part B: Average also known as mean is a measure of central tendency which measure the center of a given dataset be taking the sum of the dataset and dividing the result by the number of the dataset.
From the data presented in the table above, the team with the highest mean is Team X with a mean of 2571.9 compared to a mean of 1529.9 of Team Z and 315.8 of Team Y.
Therefore, if the manager wants to award the sales team with the highest average earnings, it should choose Team X.
Categories
## A shoe store owner is designing the layout of the displays in his store. The floor area for the store is 1,600 square feet. The owner wants to have circular displays for the shoes, each with a radius of 6 feet, and a 48-square-foot rectangular cashier’s station in the corner. Customers should be able to easily walk around the shoe displays, so there must be at least a 4-foot walkway around each. The owner started drawing the layout shown. The distance between grid lines is 4 feet. Which statement about the possible designs is true, given the owner’s requirements? Three circular displays can be placed in one row, exactly as shown on the given grid. Two circular displays can be placed so they are each the same distance from the cashier’s station. The maximum number of circular displays that can be used is 4. The maximum number of circular displays that can be used is 5.
\$2.43 per gallon of gas.
Step-by-step explanation:
We have been given that the amount of money it takes to fill up a gas tank varies directly with the number of gallons bought.
We know that the equation for two directly proportional quantities is in form: , where,
k = The constant of variation and y is directly proportional to x.
To find the constant of variation for our given equation, we will substitute the given values in above equation as:
Let us divide both sides of our equation by 9 gallons.
Therefore, the constant of variation is \$2.43 per gallon of gas.
Categories
## A garage floor measures 150 feet by 120 feet. A scale drawing of the floor on grid paper uses a scale of one unit 15 feet what are the dimensions of the drawing
A garage floor measures 150 feet by 120 feet. A scale drawing of the floor on grid paper uses a scale of one unit 15 feet what are the dimensions of the drawing
Categories
## When a figure is translated on a coordinate grid, what conclusion can you draw from the pre-image and image? answers: The angles remain the same, and the shape of the figure changes. The side lengths remain the same, and the orientation of the figure changes. The angles remain the same, and the size of the figure changes. The side lengths remain the same, and the position of the figure changes.
Step-by-step explanation:
Given: ABCD ≅ GHIJ and BCDA ≅ RSTU such that GH = 6 cm, HI = 8 cm, IJ = 10 cm, and GJ = 9 cm
As ABCD ≅ GHIJ
then its corresponding sides should be equal.[∵corresponding parts of congruent figures are equal]
i.e. AB=GH=6 cm, BC=HI=8 cm, CD=IJ=10cm , AD=GJ=9cm…………..(1)
Now ,also BCDA ≅ RSTU then
BC=RS=8 cm, CD=ST=10 cm, DA=TU=9 cm, AB=UR=6 cm………………..[from(1)]
Hence, we got ST or TS= 10 cm.
Categories
## When a figure is translated on a coordinate grid, what conclusion can you draw from the pre-image and image? answers: The angles remain the same, and the shape of the figure changes. The side lengths remain the same, and the orientation of the figure changes. The angles remain the same, and the size of the figure changes. The side lengths remain the same, and the position of the figure changes.
Step-by-step explanation:
Given: ABCD ≅ GHIJ and BCDA ≅ RSTU such that GH = 6 cm, HI = 8 cm, IJ = 10 cm, and GJ = 9 cm
As ABCD ≅ GHIJ
then its corresponding sides should be equal.[∵corresponding parts of congruent figures are equal]
i.e. AB=GH=6 cm, BC=HI=8 cm, CD=IJ=10cm , AD=GJ=9cm…………..(1)
Now ,also BCDA ≅ RSTU then
BC=RS=8 cm, CD=ST=10 cm, DA=TU=9 cm, AB=UR=6 cm………………..[from(1)]
Hence, we got ST or TS= 10 cm.
Categories
## Smart Grid Technology and Applications
### Smart Grid Technologies
#### Introduction
Technologies are always helpful for the people on the earth. It helps in all type of fields that are connected with the human to live healthy living standards. Human in the past as everyone knows it was bound to use slow methods of working. Those methods were only handmade as well at that time. That takes a lot of time to do any work or any difficult type of work related to living styles.
In the past when there is the very low amount of energy people just have the use of sunlight. Sun was the only source of living for the crops to grow for heat and energy as well. Benjamin Franklin in 1800s invented a connection between electricity and energy from the sun. He created plates and wires to connect these energies altogether so that they can be used in industries and houses for regular use. Electricity made a life of people so easy and comfortable.
It made life’s working efficient and more progressed. Electricity not only generated by sunlight now a day it can be created easily and efficiently by water reserves and coal as well. The electricity generated by these sources also helpful for the people. It has cheaper rates as compared to electricity produced by coal and furnace oil. Sunlight electricity also called as solar energy is cheap and more resistible in use. Solar energy minimizes the cost of bills and makes people more comfortable and loyal to the use of solaria energy now a day instead of electricity created from furnace oil or coal resources.
#### Smart Grid Technologies
There is a new technology invented about seven to eight years ago which is called Solar energy. This energy works on the principles of heat and power extracted from the sun. With the passage of time, new technologies have been invented to minimize the cost and to avoid the distortion problems that occur due to the use of expensive electricity by the use of consumer. Now a day with the increasing economic and financial budget problems. A consumer should provide reasonable and affordable means of electricity, which can be easily and efficiently utilized by the consumers without any cost burden.
A new technology Smart Grid technology is introduced a few years ago. Grids are the set of wires that transmit electricity from one resource to another resource. It is basically a transfer of energy between receiver and sender. It is invented in the 1960s and now it is in an improved shape and giving a lot of profit and ease to the consumers and scientists as well in providing the electricity transmission to their consumers properly and efficiently.
These electric grids can connect themselves to the millions of miles away from the main connection. Now this time this technology is so much improved that this can connect 3000,000 miles of transmission lines. It has units, which consumes and produces electricity at one place, which is about more than about 9200 in number. It also has more than 1 million of generating capacity. Now this time scientists are working on the method, which can enhance the working efficiency of a grid by increasing its connecting capacity to handle the groundswell technology dependent on it. (smartgrid.gov)
#### Context
The electrical grid named, as Smart Grids are responsible for transmission of electrical signals or molecules from building the unit to consumers place. This transmission is happened by the use of wires formed of copper usually. These wiring must be efficient and proper so that no interruption will occur in the transmission of electricity from grids. There are three main resources or components that make up a grid to possibly work these are Electricity Generation Sources, Transmission System and Distribution System.
Electricity generation sources are the sources from which electricity is generated and transmitted to the end user. These sources include copper wires, Generators, Transformers and heavy power units in which electricity is consumed and supplied to the consumers for regular use. These units had a large storage capacity that it can store a large number of electricity units in it. Electricity transmission system makes the transmission of energy smooth and efficient. Energy is transmitted in a bulk of amount at one time.
Nearly 100 K and above amount of electrical signals are transmitted at one time from powerhouses to consumers. This transmission has happened under a system, which is called SCADA systems. This system is responsible for controlling and monitoring the methods and working process of electricity making and transmission in Grids. (Mavridou, 2012)
#### Need of Smart Grid
The need for using of smart grid became necessary to fulfill the need of increasing demand of electricity with the passage of time. It increases the efficiency of using energy and controlling the usage of energy requirements. A smart grid makes produces technology resources that can generate more and more of electricity by the use of energy. It is an essential part of marinating the developmental and proper usage of Plug-In Hybrid Electric Vehicles (PHEVs) which is used for storage of electricity in the grid.
Smart meters are used in Smart grids, which save the energy for the rest of working, which reduces the expensive use of electricity by the consumers and makes them feel relax and lessen their burdens of the high amount of bills to be paid by them. The electric grid or smart grid makes the user know its usage of electricity. It makes people aware of their budget utilizing bills. This technology is not very expensive and should be improved by working on it. The labours or persons connected with the making of these grids must be encouraged and supported by incentives so that they can improve the efficiency of their working. This technology is very sensitive and caring to be performed in the field. (eesi.org, 2009p)
#### Giving Consumers Control
This technology is not just about modernism and development in electricity production methods. It is actually a process or an authority provided to you by the technology that you can keep a choice of using your energy. If a person manages his, other activities like working habits and follower a timetable to control wastage of time. Same as Gris system allowed a user to keep an eye on the usage of electricity by him/her.
The consumer can check frequently that how much energy is utilizing them. Which product is using more units of electricity so that it can be maintained? Which unit is utilizing a less quantity of units so its use can be made continues and long lasting? Now a day Smart meters are introduced by the government through which you can frequently check your usage of voltages of electricity. It also estimates the cost of electricity consumed by you so that you can control your excessive use of electricity as well as can control your bills.
Sometimes electricity becomes expensive because of any mean like increasing the prices of furnace oil or increasing the prices of coal or copper wires at the international markets. Country purchases it inexpensive rates and charges it from consumers in the form of taxes. You can when having knowledge of your users can save money or save units of electricity to avoid facing of heavy amounted bills.
#### Reactive Power, Voltage And Frequency
The Smart Grid technology is very efficient and fast. It works in the principles of affectionate working with precision and time saving methods. Smart grids use smart meters, which enhances the use of that electricity which utilizes more energy with less cost. Smart grid consumes a very large amount of energy to make low cost and effective amount of electricity. It has a high frequency of usage of voltage for the consumption of energy resources.The voltage depends on its working became very efficient and consumes less cost.
With the introduction of new technology, generation will use low-cost units with high working efficiency. People will use many small units of electricity, which costs less as compared to use one single big unit of voltage, which costs high to their meter. The consumption of electricity will be increased by the consumer at low cost. There will be provided many services in near future, which offers cost-saving tools to consumers.
There will be offered a large capacity consuming devices, which can store a large amount of voltage data at one time and at one place for future use. The voltage may fluctuate when we use a heavy machinery and runs something very heavy consuming appliance on electricity. It sometimes causes electricity failure and disturbance in the transmission of flow of electricity. (D’hulst, 2018)
#### Post-Fault Restoration, Loss Of Generation
Restoration means storing some amount of energy or voltages for future use in time of emergency. The restoration of the network can be implemented by the use of changing the status of loops and switches of the loops so that it can meet up the demand of required voltages in the system. The current technology method is not spread after the entrance of new channel distribution in it. Meanwhile, the operation power consumption, the quantity of current passed through the copper wires and the changes in the contribution of working capacities cause changes in restoration demands for the system. By changing the places and styling of fitting of loops and switches connected to the loops, it also changes the art of working and methods of working also are changed.
It also changes the topological positions of a system in the distribution network, which results in the change in power of supply consumption and supply restoration in the system. There should be happened a change in the restoration services and improvements should be implemented in the operations reliability and working efficiency of systems working. The new generation is losing its latest methods and stamina to use those methods because of overpowered supply of some electric appliances at home and at industries as well. (Xiaoyu, 2014)
The amount of load used by any area or any other place depends on the usage capacity of that place. If the area is commercial then there will be a continues use of heavy pieces of machinery which works on obvious electricity and utilizes many units of electricity. These heavy types of machinery are bound; you can say depends only on the consumption of high units of electricity to run. They can be made low consuming machines by changing some parts in it, which consumes high voltage of energy.
There is an option, which can utilize our energy consumption, which is a use of solar energy. This energy, which is introduced many years ago cost more just at the time of fixing. However, later it consumes the very low amount of electricity, which makes its use more efficient and relaxing for the consumer. If we talk about the electricity use in residential are then we can easily click this point that there will be a less use of electricity consumption as compared to the commercial area. With the increase in demand a load of devices on the electricity usage, this also increases the storage capacity and making of high capacity energy is also become essential to meet the needs. The less or more demands of energy voltage can be controlled by proper use of appliances and energy consuming devices so that wastage of electricity can be taken into account and it fulfills the need of technology where it is needed.
Load flow is an important tool, which is being considered by the engineers while making a plan to make the best system to run the best system in working. The exchange of power between the companies or transmitters is the essential and most prominent portion of the efficient working off-grid system. If we started doing the analysis of power flow, it will take a lot of time and energy of persons as well power flow can be controlled by using any method. The conversion of renewable resource energy into a distribution network of energy changes the entire structure of electric networking.
The issues are many such as improper wiring may be implemented in the system, which can cause interruption and distortion in the flow of electricity among the wires. It causes a disturbance in energy transmission as well. Second, one may be the disturbance in weather sometimes weather disturbs so much the working efficiency of energy flow which causes problems in working. Sometimes due to water, entrance in the system there may be happened a spark in the system, which damages a lot of part of a system and in result failure, happened in the system and load became doubled at one place or at different places. This issue can be resolved by taking precautionary measures in this field of work. Ravadanegha), 2015
#### Smart Meter Connectivity
Now with the passage of time and with the increase in the technology there comes the smart meters, which are cost-effective and consumes less energy also saves the time and measurements of reading of consumers. Smart meters can measure energy more efficiently and precisely and also consumes less energy. The smart meters connected with solar plates are very efficient and applicable in many ways easily and accurately.
Now a day people are moving to insert smart panels with solar plates especially in their houses. It saves many costs on bills in-house. Thus makes money save for further uses. This system is also implemented now a day by industries also adapting this solar systems energy use because it is cost minimizing and timesaving as well. An estimate says that it saves on a yearly basis up to 40 billion-energy cost all over the world. It is a long time investment for the households and as well as for industry users.
Smart meters along with solar energy are easy to apply and easy to emit out if you did not satisfied with the use of it. Millions of people now a day using this system and making them satisfied with this. (Ingrams, 2017)
#### Synopsis
The overall scenario of this report states that the use of Smart Grid systems along with solar energy a renewable energy is very useful and efficient. It saves a lot of time and energy consumption as well. People have gained trust in this system. This system makes the use of them easy and trustful. The proper wiring helps in proper transmission of energy, keeps the flow proper, and smooth as well. The improper connection between wires, which are mostly used, made of copper causes electricity failure and damage as well. The damage caused by electricity can become very dangerous and time taking as well which causes trouble in people’s life.
##### References
• D’hulst, R. (2018, February 09). Voltage and Frequency Control for Future Power Systems: the ELECTRA IRP Proposal. Retrieved from http://orbit.dtu.dk/files/113381466/ElectraWP3_EDST_submitted2.pdf
• org. (2009, January 8). Smart Grid: How Does It Work and Why Do We Need It? Retrieved from http://www.eesi.org/briefings/view/smart-grid-how-does-it-work-and-why-do-we-need-it
• Ingrams, S. (2017, August 12). Smart meters and solar panels: what’s the sticking point? Retrieved from https://www.which.co.uk/news/2017/08/smart-meters-and-solar-panels-whats-the-sticking-point/
• Mavridou, A. (2012, January). Block diagram of typical smart grid components and connections. Retrieved from https://www.researchgate.net/figure/Block-diagram-of-typical-smart-grid-components-and-connections_fig1_278029337
• Ravadanegha), S. N. (2015). Journal of Renewable and Sustainable Energy. | 8,550 | 34,169 | {"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} | 4.59375 | 5 | CC-MAIN-2021-39 | latest | en | 0.698232 |
https://math.stackexchange.com/questions/2631431/find-all-possible-values-of-x-y-a-b-if-x-y-in-mathbb-z-such-that-a | 1,560,912,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998882.88/warc/CC-MAIN-20190619023613-20190619045613-00311.warc.gz | 509,624,011 | 34,098 | # Find all possible values of $x, y, A, B$ if $x, y \in \mathbb Z_+$ such that $A = \frac{2y}{x(x-y)}$ and $B = \frac{(y -x)(y + 1)}{2y^2}$
Find all possible values of $x, y, A, B$ if $x, y \in \mathbb Z_+$ such that $A = \frac{2y}{x(x-y)}$ and $B = \frac{(y -x)(y + 1)}{2y^2}$ are integers.
My solution: Obviously $x \neq y$ and $x, y \ge 1$. Moreover $A \neq 0$ and $B \neq 0$. Since $A$ and $B$ have to be integers their product $AB = -\frac{y+1}{xy}$ also has to be an integer. Now $GCD(y, y + 1) = 1$, so $y = 1$ since otherwise $\frac{y + 1}{y}$ wouldn't be an integer. So we can substitute $y = 1$ into $A$ and get $A = \frac{2}{x(x-1)}$. Only two values of $x$ are possible so we can check that $x = 2$ is fine both for $A$ and $B$.
Can someone please verify my solution and tell whether it is correct? | 324 | 813 | {"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} | 3.765625 | 4 | CC-MAIN-2019-26 | latest | en | 0.745363 |
https://forum.allaboutcircuits.com/threads/help-with-using-functions-to-calculate-projectile-height-on-impact.173361/ | 1,632,807,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00414.warc.gz | 309,131,187 | 21,892 | # help with using functions to calculate projectile height on impact
#### Killerbee65
Joined May 15, 2017
250
Hello all!
I recently got an assignment from my pressor that ask me to Write a program that computes the duration of a projectile’s flight and its height above the ground when it reaches the target.
He gave us the following equations to work with:
a. time = distance/ velocity x cos(0)
b. height = velocity x sin(0) x time - (g x time^2)/2
distance, velocity, and angle are all inputs from the user and g is a gravitational consistency of 32.17
/*
Name: Erik Diaz
Course : Computer science and programming
Section : 07
Date : 09 / 23 / 2020
ID:lab 5b
Purpose:
Write a program that computes the duration of a projectile’s flight and its height above the ground when it reaches the target.
Analysis/ Data:
1. inputs: angle, distance, velocity (float)
2. output: time in seconds, height (float)
3. equations
a. time = distance/ velocity x cos(0)
b. height = velocity x sin(0) x time - (g x time^2)/2
Typical-
input:
expected output:
Results:
Boundry-
input:
expected output:
results:
Extreme-
input:
expected output:
results:
*/
#include <iostream>
#include <cmath>
using namespace std;
float velocity, target_distance, angle;
float grav_const = 32.17;
float inPuts( ) {
cout << "please enter the velocity of the projectile: ";
cin >> velocity;
cout << "it's distance: ";
cin >> target_distance;
cout << "The angle: ";
cin >> angle;
return velocity, target_distance, angle;
}
float time ( ) {
inPuts( );
return target_distance/ velocity * cos(angle);//equation for time
}
float height() {
inPuts();
time();
float height1 = velocity * sin(angle) * time() - (grav_const * (time() * time())) / 2;//equation for height
return height1;
}
int main() {
cout << time() << endl << height();
}
I'm using a different ide so I'm not sure if I copied and pasted correctly.
I keep getting consistent numbers when I input numbers yet they are wrong (ex: 1, 1, 1 for the inputs ->.54 for output or time). Also, the code consistently repeats 5 times yet I have no loops. I was tight for time so I only tested to see if the computer ran through each function.
#### RBR1317
Joined Nov 13, 2010
664
I keep getting consistent numbers when I input numbers yet they are wrong
If the numbers are wrong it may be because the equation for time needs some additional parentheses. Also, I am unfamiliar with this style of programming (not being a programmer); what language is it?
#### SamR
Joined Mar 19, 2019
3,498
32.17? Flying pigs? What are the units? I prefer 9.8 m/s^2 for gravity... And even when going up, it is falling down. Just a couple of hints...
#### RBR1317
Joined Nov 13, 2010
664
What are the units?
The units are baked into the equations provided by the professor. Not an issue here.
#### Killerbee65
Joined May 15, 2017
250
32.17? Flying pigs? What are the units? I prefer 9.8 m/s^2 for gravity... And even when going up, it is falling down. Just a couple of hints...
I know what you mean with the 32.17, I prefer 9.8m/s^2 but it is what my professor wants. But could you elaborate on your hint? I have an idea but I'm not sure my professor wants it to be too complicated.
#### SamR
Joined Mar 19, 2019
3,498
Dealing with projectiles, as soon as they leave the fixed mounted position gravity acts upon them determining their flight path. Hence even though they may be increasing in altitude they are also falling @ gravitational acceleration. Or else what goes up would not come down. Therefore they do not travel in a straight line. The US Navy spent billions of dollars calculating trajectories. In classical physics such things as air density changes with elevation, barometric pressures resultant frictional forces on projectile trajectories, diameter of projectile, shape of projectile, subsonic and supersonic frontal waves, etc. are ignored but were computed in the Naval calculations. Hence their need for computers such as ENIAC et al. It is much more complicated than gravity, angle, and velocity.
#### Wolframore
Joined Jan 21, 2019
2,342
32.17 ft/sec^2 is a conversion factor for gravity in English units. Useful if your bullets flight is measured in yards and feet.
#### drc_567
Joined Dec 29, 2008
1,154
32.17 ft/sec^2 is a conversion factor for gravity in English units. Useful if your bullets flight is measured in yards and feet.
g is plain gravitational acceleration ... $$32 ft/sec^2$$.
$$g_c$$ is the conversion factor between pound mass ... $$lb_m$$, and pound force .. $$lb_f$$ and is equal to $$32 [lb_m *ft]/[lb_f *sec^2]$$
The conversion factor is necessary when checking units in various formulae involving $$lb_m$$.
Last edited:
#### Wolframore
Joined Jan 21, 2019
2,342
Yeah that’s what I meant | 1,203 | 4,763 | {"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} | 3.46875 | 3 | CC-MAIN-2021-39 | latest | en | 0.748928 |
https://community.cesium.com/t/ellipsoid-over-time/2643 | 1,600,890,580,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212039.16/warc/CC-MAIN-20200923175652-20200923205652-00136.warc.gz | 309,665,809 | 6,464 | # ellipsoid over time
Hello,
i'm trying to draw an ellipsoid between 2 points at a certain clock time.
i tried several way, but none of them worked
i'm kinda lost, if someone can help me it would be very nice !
i tried like this :
var start = Cesium.JulianDate.fromDate(new Date(currentAttack._source.LogDate));
var stop = Cesium.JulianDate.addSeconds(start, duration, new Cesium.JulianDate());
var compostart = new Cesium.SampledPositionProperty();
Cesium.Cartesian3.fromDegrees(fromPoint.lon, fromPoint.lat));
var compostop = new Cesium.SampledPositionProperty();
Cesium.Cartesian3.fromDegrees(toPoint.lon, toPoint.lat));
var target = viewer.entities.add({
ellipsoid: new Cesium.EllipsoidGeodesic({
start: compostart,
end: compostop
})
});
can you help me ? or point me to the good direction ?
thanks very much, i'm very lost and cant find any clue on it
i found a solution on my own after all. can you give me some feedback on it, or at least if someone else is looking for something similar:
To stay in the pure Cesium world and since you’re using linear interpolation you can use the PolylinePipeline.generateArc to calculate the points. That is going to generate an array of points based on your specification in radians.
If you’re going to be doing non-linear interpolation I recommend using http://www.movable-type.co.uk/scripts/latlong.html’s library to calculate the heading/bearing between the 2 points based on varying distances traveled during the specific time interval.
The primary reason you want to use these methods instead of the method you used is that they both use spherical geometry into account when determining the in-between lat/lons.
thanks, i will look into it!
Thanks a lot Mike. it's working much better !
why this function is not into the documentation ? it's working awesomely and i wouldnt have find it if you didnt point me into the direction !
here is a sample of my code (i can't post it all because it's link to another application):
//generate a collection of position samples
var positions = calculatePositionSamples(att.fromPoint, att.toPoint, start, duration);
var target = viewer.entities.add({
description: att.name,
position: positions,
path: {
material: new Cesium.PolylineArrowMaterialProperty(getColor(att.name)),
width: 8,
trailTime: 15,
}
});
});
//major update thanks to Mike
function calculatePositionSamples(point, endPoint, startTime, duration) {
var property = new Cesium.SampledPositionProperty();
var positions = Cesium.PolylinePipeline.generateCartesianArc({
positions: Cesium.Cartesian3.fromDegreesArray([
point.lon, point.lat,
endPoint.lon, endPoint.lat
])
})
var deltaStep = duration / positions.length
var currentPosition;
for (var i = 0, since = 0; i < positions.length; i +=1, since += deltaStep)
{
Cesium.JulianDate.addSeconds(startTime, since, new Cesium.JulianDate()),
positions[i])
}
return property;
}
it's working very well when i got 2 very distant points but when i got 2 very close points i got some visbility issue.
here is a picture showing you what i mean : http://img42.com/zcX0d
you can see a huge arrow from france to the US and a very very small one from France to France.
is there another [magic] function like the last one that allow me to have a bigger curve (when i want to) to increase it's visibility ?
i think i can do it by changing the ellipsoid or the height, but i failed painfully when i tried some options blindly (i don't really understand the math behind ).
if you (or someone else) have a clue it would be awesome
can you share a picture of the close points and one of what you would like it to look like?
Hello Mike,
here is the issue i have (i made a red rectangle for you to see the very small arrow) : http://img42.com/wzGhq
as you can see you can barely see it. i know how to manage to isolate those i want to increase the visbility from the normal one (like the green on the example), but i have no clue how to increase its visbility.
here is an example made with gimp of what i would want to do: http://img42.com/SV3q2
Again thanks a lot for your Time!
To make what you want programmatically is going to be challenging. Probably involving some parametric equations.
You might be better off using a billboard when you’re so far zoomed out from a location. You can toggle the show/hide of the billboard, by attaching a method to the scene.preRender event. Check the distance of the points in screen space, if they are only a few pixels apart, then you hide the polyline and show the billboard.
Maybe someone else has some other ideas, but I think this will work well for you.
a long time ago i remember working with parametric equation, but my brain decide to forget this kind of thing pretty fast !
i will look a bit into it, but i don't think i will find a working thing
the idea of the billboard is nice but i cant predict the orientation of the arrow, but start and end point are geoposition generated by my backend (elasticsearch). so i dont think it will work. and i need somekind of animation too.
The animations can be handled by using SampledPointProperty. You can probably directly use the points you generated from the GenerateArc and space them out over your animation duration.
For the path at close proximity I I recommend going onto http://math.stackexchange.com/ and writing up the situation like a word problem. I’ve gotten some pretty awesome answers and equations from those users. Make sure to include the picture. | 1,245 | 5,476 | {"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} | 2.9375 | 3 | CC-MAIN-2020-40 | latest | en | 0.870527 |
http://math.mrozarka.com/calculus-1/lessons-2015-s2/unit-3/day-36 | 1,680,301,625,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949689.58/warc/CC-MAIN-20230331210803-20230401000803-00401.warc.gz | 31,667,249 | 11,637 | Day 36 - Extrema/Critical Numbers - 03.04.15
Update
Questions
Bell Ringer
1. Find the x-coordinate that has a horizontal tangent line for the following:
1. -2
2. 0
3. 2
4. 3
5. none of the above
2. Does the point in #1 occur at a maximum or a minimum on the graph?
1. maximum
2. minimum
3. neither
3. Find the x-coordinate in which the following function does not have a derivative:
1. 0
2. 2
3. -2
4. 1
5. none of the above
4. Does the point in #3 occur at a maximum or a minimum on the graph?
1. maximum
2. minimum
3. neither
Review
• N/A
Lesson
Exit Ticket
• N/A
Lesson Objective(s)
• How is the derivative used to locate the minimum and maximum values of a function on a closed interval?
Skills
1. Find critical numbers using differentiation.
2. Find extrema on a closed interval using differentiation.
In-Class Help Requests
Standard(s)
• APC.5
• Investigate derivatives presented in graphic, numerical, and analytic contexts and the relationship between continuity and differentiability.
• The derivative will be defined as the limit of the difference quotient and interpreted as an instantaneous rate of change.
• APC.6
• The student will investigate the derivative at a point on a curve.
• Includes:
• finding the slope of a curve at a point, including points at which the tangent is vertical and points at which there are no tangents
• using local linear approximation to find the slope of a tangent line to a curve at the point
• defining instantaneous rate of change as the limit of average rate of change
• approximating rate of change from graphs and tables of values.
• APC.7
• Analyze the derivative of a function as a function in itself.
• Includes:
• comparing corresponding characteristics of the graphs of f, f', and f''
• defining the relationship between the increasing and decreasing behavior of f and the sign of f'
• translating verbal descriptions into equations involving derivatives and vice versa
• defining the relationship between the concavity of f and the sign of f "
• APC.9
• Apply formulas to find derivatives.
• Includes:
• derivatives of algebraic and trigonometric functions
• derivations of sums, products, quotients, inverses, and composites (chain rule) of elementary functions
• derivatives of implicitly defined functions
• higher order derivatives of algebraic and trigonometric functions
Past Checkpoints
• N/A | 562 | 2,387 | {"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} | 4.15625 | 4 | CC-MAIN-2023-14 | latest | en | 0.884722 |
https://the-algorithms.com/fr/algorithm/kosaraju | 1,726,250,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00727.warc.gz | 531,222,732 | 30,226 | #### Kosaraju
R
A
P
```/**
* Kosaraju's Algorithm implementation in Javascript
* Kosaraju's Algorithm finds all the connected components in a Directed Acyclic Graph (DAG)
* It uses Stack data structure to store the Topological Sorted Order of vertices and also Graph data structure
*
* Wikipedia: https://en.wikipedia.org/wiki/Kosaraju%27s_algorithm
*
*/
class Kosaraju {
constructor(graph) {
this.connections = {}
this.reverseConnections = {}
this.stronglyConnectedComponents = []
for (const [i, j] of graph) {
}
this.topoSort()
return this.kosaraju()
}
// Function to add a node to the graph (connection represented by set)
this.connections[node] = new Set()
this.reverseConnections[node] = new Set()
this.topoSorted = []
}
// Function to add an edge (adds the node too if they are not present in the graph)
if (!(node1 in this.connections) || !(node1 in this.reverseConnections)) {
}
if (!(node2 in this.connections) || !(node2 in this.reverseConnections)) {
}
}
dfsTopoSort(node, visited) {
for (const child of this.connections[node]) {
if (!visited.has(child)) this.dfsTopoSort(child, visited)
}
this.topoSorted.push(node)
}
topoSort() {
// Function to perform topological sorting
const visited = new Set()
const nodes = Object.keys(this.connections).map((key) => Number(key))
for (const node of nodes) {
if (!visited.has(node)) this.dfsTopoSort(node, visited)
}
}
dfsKosaraju(node, visited) {
this.stronglyConnectedComponents[
this.stronglyConnectedComponents.length - 1
].push(node)
for (const child of this.reverseConnections[node]) {
if (!visited.has(child)) this.dfsKosaraju(child, visited)
}
}
kosaraju() {
// Function to perform Kosaraju Algorithm
const visited = new Set()
while (this.topoSorted.length > 0) {
const node = this.topoSorted.pop()
if (!visited.has(node)) {
this.stronglyConnectedComponents.push([])
this.dfsKosaraju(node, visited)
}
}
return this.stronglyConnectedComponents
}
}
function kosaraju(graph) {
const stronglyConnectedComponents = new Kosaraju(graph)
return stronglyConnectedComponents
}
export { kosaraju }
// kosaraju([
// [1, 2],
// [2, 3],
// [3, 1],
// [2, 4],
// [4, 5],
// [5, 6],
// [6, 4],
// ])
// [ [ 1, 3, 2 ], [ 4, 6, 5 ] ]
``` | 600 | 2,204 | {"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} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.609913 |
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