url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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jaidenvsldv.blog2learn.com | 1,537,662,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.65/warc/CC-MAIN-20180923000827-20180923021227-00422.warc.gz | 128,512,041 | 5,859 | # Weighing of Material on Balances - An Overview
For example, the center of a ground bends in excess of locations closer to outside the house walls, and higher floors sway; even slight misalignment alongside the vertical to the center of your earth can introduce mistake.
Once the correct harmony and vessel for weighing a substance is selected the right weighing system should be chosen. There's two main approaches to weighing; i) utilizing the tare facility and ii) weighing by change.
I n p u t = O u t p u t + A c c u m u l a t i o n displaystyle mathrm Input =mathrm Output +mathrm Accumulation ,
W. Winfield made the candlestick scale for weighing letters and packages, needed after the introduction from the Uniform Penny Publish.[2] Postal employees could perform far more swiftly with spring scales than balance scales given that they can be read instantaneously and didn't should be carefully well balanced with Every single measurement.
Moisture -- Goods for being weighed are ideal kept at ambient temperature and humidity levels to decrease the chance of condensation/evaporation influencing readouts.
Density measurement of strong, liquid and viscous samples on an analytical or precision stability. Quickly, straightforward process with Density Package and balance application.
The ultramicrobalance is any weighing system that serves to ascertain the burden of scaled-down samples than is often weighed Using the microbalance—i.e., total amounts as compact as a person or a handful of micrograms. The ideas on which ultramicrobalances have been successfully manufactured contain elasticity in structural factors, displacement in fluids, balancing by way of electrical and magnetic fields, and mixtures of those.
In spite of which technique illustrated below is used correctly to weigh a sample, the sample, put in the weighing bottle set in the upturned cap within a beaker with a view glass placed on major, must be first dried inside the oven. Chances are you'll determine your sample by marking the beaker but Tend not to mark the weighing bottle.
Counting -- The scale is capable of recalling a reference weight as a way to estimate a bulk count of similarly-weighted goods.
Schedule testing, weighing tolerance, SOP’s and other stability operation relevant difficulties are going to be covered as A part of the Operational Qualification. The equilibrium should not be in usage prior to Validation stage was completed successfully, documented and summarized.
In precision balances, a more exact perseverance in the mass is given with the position of a sliding mass moved alongside a graduated scale. Technically, a equilibrium compares excess weight instead of mass, but, inside a supplied gravitational field (including Earth's gravity), the load of the object is proportional to its mass, Therefore the common masses utilised with balances are often labeled in units of mass (e.g. g or kg).
The sensor information vertical posture modifications when the pan is loaded which happens to be employed to vary The existing in the coil to return to its Preliminary situation. The greater weight is additional to the pan, the more latest is required to compensate it, this is digitized within the Exhibit.
Thank you for visiting . Now we have attempted to optimize your experience though on the location, but we seen that you will be applying an older version of an online browser. We wish to let you are aware that some characteristics on the website is probably not offered or may not get the job done as nicely as they'd on a check here more moderen browser version.
Sartorius Entris series of analytical balances from Wolflabs. The Entris has become specifically designed to provide just effective and trustworthy weighing ends in your day by day function. | 737 | 3,790 | {"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-2018-39 | longest | en | 0.925295 |
https://amigotutor.com/newfoundland-and-labrador/npv-calculation-example-with-inflation.php | 1,601,028,092,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400223922.43/warc/CC-MAIN-20200925084428-20200925114428-00278.warc.gz | 260,490,176 | 6,869 | # Example with inflation npv calculation
## Real NPV Calculator With Interest Inflation Rates Built In
NPV With Inflation Calculation Net Present Value. net present value is the present value of net cash example 1: even cash inflows: calculate the net present value of a project which npv and inflation; npv, 25/04/2014 · npv with taxation and inflation in the answer provided , calculate the net present value. (examples 2 and 3)).
Home > Business > Finance > Capital Budgeting > Excel NPV Function Excel NPV Function. correct financial definition of net present value. Example Inflation Learn about NPV With Inflation Calculation, Net Present Value & Inflation Treatment.
This worksheet demonstrates examples of using an Excel function to find the net present value of of inflation) and a series of the NPV calculation in Excel by NPV is used to analyze an investment decision and give company management a clear way to tell if the (as shown in the example above) A Primer on Inflation
Net present value - NPV. The idea of using the present value to estimate the worth of a sum of money received in the future can be extended to an arbitrary cashflow. Would the Net Present Value of a project with cash flows be larger with inflation factored in, or smaller?
(For example, if we have to To Calculate Net Present Value NPV = Present Value of Cash inflow Calculation of NPV in Case Inflation Future money is also less valuable because inflation erodes its What is net present value? that is the discount rate the company will use to calculate NPV.
Net present value is the present value of net cash Example 1: Even Cash Inflows: Calculate the net present value of a project which NPV and Inflation; NPV 25/02/2015 · NPV WACC and inflation. This would be faster for this example, However that is part of the calculation of the WACC and does not ever affect the cash flows.
Would the Net Present Value of a project with cash flows be larger with inflation factored in, or smaller? Net Present Value (NPV) is the For example, assume that an While net present value (NPV) calculations are useful when you are valuing investment opportunities
Real NPV Calculator With Interest Inflation Rates Built In
Net present value NPV - I Programmer. 25/02/2015 · npv wacc and inflation. this would be faster for this example, however that is part of the calculation of the wacc and does not ever affect the cash flows., net present value (npv) is the for example, assume that an while net present value (npv) calculations are useful when you are valuing investment opportunities).
kfknowledgebank.kaplan.co.uk/KFKB/Wiki Pages/International
NPV WACC and inflation OpenTuition. npv is used to analyze an investment decision and give company management a clear way to tell if the (as shown in the example above) a primer on inflation, this information needs adjusting to take account of selling price inflation of 4% per calculate the expected net present value of the investment project).
Real NPV Calculator With Interest Inflation Rates Built In
Excel functions for investment appraisal Mead In Kent. 8.4 other factors affecting npv and irr analysis. which basis of accounting is used to calculate the npv and irr for long-term investments, for example, the, net present value method ( examples) of how to use the npv. how to calculate net present value index if cost of capital is given instead of discount rate?).
Npv with taxation and inflation OpenTuition
ACCA FM (F9) Past Papers NPV aCOWtancy Textbook. what you should know about the discount rate. consider the following chart showing the sensitivity of net present value to changes in the for example, if the, making investment decisions using excel if the discount rate is 10% and inflation 15% the npv calculation must use: (1+0.10) example. a firm is).
Net present value NPV - I Programmer
NPV & Inflation NPV and Risk Modelling for Projects. this worksheet demonstrates examples of using an excel function to find the net present value of of inflation) and a series of the npv calculation in excel by, how not to use npv in how not to use npv in excel. let’s take an example. value of future cash flows calculation. because the net present value is simply).
Net present value is the present value of net cash Example 1: Even Cash Inflows: Calculate the net present value of a project which NPV and Inflation; NPV Example Net Present Value NPV calculations illustrate the concept. Discounted Cash Flow, Net Present Value, and inflation—values that can change as time
This worksheet demonstrates examples of using an Excel function to find the net present value of of inflation) and a series of the NPV calculation in Excel by Profiting From Cleaner Production Performing Net Present Value (NPV) Calculations For this example, the NPV is calculated to be -\$14,052 which means that the
Net present value calculations can be used for either acquisitions (as shown in the example above) A Primer on Inflation-Linked Bonds. 9. Net present value (NPV) use this discount rate in the NPV calculation to allow a direct A more simple example of the net present value of incoming
Would the Net Present Value of a project with cash flows be larger with inflation factored in, or smaller? EXAMPLE 2. Storm Co is Calculate the working capital flows for incorporation into the NPV calculation. Question 6 – NPV with inflation and Discounted Payback.
NPV Models - Treatment of Inflation. discount rate is derived from an STPR calculation, of the effects of the discount rate and inflation. For example, 25/04/2014 · Npv with taxation and inflation In the answer provided , Calculate the net present value. (examples 2 and 3)
Managers use many different terms to describe the interest rate in a net present value calculation, including the following: To evaluate the NPV of a capital Example 5-4: Calculate ROR for the investment that has the following Escalated and constant dollar ROR and NPV Escalated and Constant Dollar Cash
NPV & Inflation NPV and Risk Modelling for Projects | 1,329 | 6,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-40 | latest | en | 0.886099 |
https://acm.sustech.edu.cn/onlinejudge/problem.php?id=1255 | 1,718,745,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00156.warc.gz | 69,311,568 | 3,496 | 12551255 SUSTech Online Judge
Problem 1255 --[Bonus] Suffix Zero
## 1255: [Bonus] Suffix Zero
Time Limit: 3 Sec Memory Limit: 128 MB
Submit: 1028 Solved: 245
[Submit][Status][Web Board]
## Description
Neko thinks math is interesting but he has got stuck at a math problem, so he asks you for help.
Find the number of consecutive zeros at the end of $$n!$$ (in decimal representation).
## Input
The first line contains a single integer $$T(1{\leq}T{\leq}10^6)$$ —— the number of test case.
The $$2^{nd}$$ line to the $$(n+1)^{th}$$ line, each line contains a single integer $$n(0{\leq}n{\leq}10^{18})$$.
## Output
Print the number of consecutive zeros at the end of $$n!$$.
## Sample Input
3
5
3
10
## Sample Output
1
0
2
## HINT
For the third case, 10! = 3628800, which have 2 consecutive zeros at the end.
[Submit][Status] | 272 | 841 | {"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": 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} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.703801 |
https://www.physicsforums.com/threads/triple-integral-and-finding-the-average.562997/ | 1,511,604,670,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809746.91/warc/CC-MAIN-20171125090503-20171125110503-00416.warc.gz | 848,584,829 | 15,194 | # Triple Integral and finding the average
1. Dec 27, 2011
### gipc
Solve the Integral in cylindrical coordinates
∫∫∫ dxdydz/(sqrt( x^2 + y^2 + (h-z)^2)
B
Where B is the Ball with a Radius R around (0,0,0), and the parameter h is greater than R.
And then infer the average on that ball B with radius R of the distance opposite to the point (0,0,h).
I have these so far:
Using Cylindrical Coordinates,
∫∫∫B dV/√(x² + y² + (h-z)²)
= ∫(θ = 0 to 2π) ∫(z = 0 to R) ∫(r = -√(R² - z²) to √(R² - z²)) (r dr dz dθ) / √(r² + (h-z)²)
= 2π ∫(z = 0 to R) ∫(r = -√(R² - z²) to √(R² - z²)) r dr dz / √(r² + (h-z)²)
= 2 * 2π ∫(z = 0 to R) ∫(r = 0 to √(R² - z²)) r dr dz / √(r² + (h-z)²), by evenness
= 4π ∫(z = 0 to R) √(r² + (h - z)²) {for r = 0 to √(R² - z²)} dz
= 4π ∫(z = 0 to R) [√((R² - z²) + (h - z)²) - √(h - z)²] dz
= 4π ∫(z = 0 to R) [√((R² - z²) + (h - z)²) - (h - z)] dz, since h > R > r
= 4π ∫(z = 0 to R) [√(R² + h² - 2hz) - h + z] dz
= 4π [(-1/(3h)) (R² + h² - 2hz)^(3/2) - hz + z²/2] {for z = 0 to R}
= 4π {[(-1/(3h)) (R² + h² - 2hR)^(3/2) - hR + R²/2] - (-1/(3h)) (R² + h²)^(3/2)}
= 4π [(-1/(3h)) (h - R)³ - hR + R²/2 + (1/(3h)) (R² + h²)^(3/2)].
Now, is that solution correct? Is there a way so simplify things? And from there, how do I find the average?
2. Dec 27, 2011
### Simon Bridge
Isn't $\sqrt{x^2+y^2+z^2}$ the distance from the origin to the point (x,y,z)?
So $\sqrt{x^2+y^2+(h-z)^2}$ would be the distance from (x,y,z) to (0,0,h)?
So your integral is just summing all these distances ... how do you find an average value by integration?
Same kind of problem, slightly different way:
http://www.mathpages.com/home/kmath324/kmath324.htm
3. Dec 27, 2011
### SammyS
Staff Emeritus
It looks like you have only integrated over the top half of the Ball.
The limits on z should go from -R to R .
4. Dec 27, 2011
### gipc
I think the final answer is still correct, no? | 841 | 1,888 | {"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} | 4.125 | 4 | CC-MAIN-2017-47 | longest | en | 0.759412 |
https://www.reference.com/math/equation-a2-b2-c2-refer-75bd5ca08641d7fc | 1,472,694,861,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982956861.76/warc/CC-MAIN-20160823200916-00192-ip-10-153-172-175.ec2.internal.warc.gz | 922,496,088 | 21,033 | Q:
# What does the equation "a2 + b2 = c2" refer to?
A:
The equation "a2 + b2 = c2" refers to the Pythagorean theorem. With this theorem, it is possible to find the length of any side of a right triangle when given the length of the other two sides.
## Keep Learning
Credit: Cultura RM/OPIFICIO 42 Collection Mix: Subjects Getty Images
The Pythagorean theorem states the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. In this equation, the twos after each term are not indicators of multiplication by two but of squaring each term. Normally, mathematicians write them as superscripts above the regular line of the text. This formula provides the basis for the distance formula, which is useful in determining the distance between two points on a graph when given their coordinates.
Sources:
## Related Questions
• A:
Pythagoras often receives credit for the discovery of a method for calculating the measurements of triangles, which is known as the Pythagorean theorem. However, there is some debate as to his actual contribution the theorem.
Filed Under:
• A:
The analytical method of vector addition can be described as a combination of steps that utilize the Pythagorean theorem as well as trigonometric identities. These two concepts determine magnitude and direction of the resultant vector.
Filed Under:
• A:
The tangent-secant theorem states that if two secant segments share an endpoint outside of a circle, the product of one segment length and the length of its external segment equals the product of the other segment's length and the length of its external segment. It is a special subtype of another mathematical theorem: the power of a point theorem. | 366 | 1,728 | {"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} | 3.28125 | 3 | CC-MAIN-2016-36 | longest | en | 0.933548 |
https://www.chegg.com/homework-help/steel-pipe-inner-diameter-275-outer-diameter-3-fixed-c-sub-chapter-9-problem-46p-solution-9780132209915-exc | 1,529,599,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00231.warc.gz | 781,504,539 | 15,713 | # Mechanics of Materials (7th Edition) View more editions Solutions for Chapter 9 Problem 46PProblem 46P: The steel pipe has an inner diameter of 2.75 in. and an outer diameter of 3 in. If it is fixed at C and subjected to the horizontal 20-lb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the surface of the pipe.
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Chapter: Problem:
100% (7 ratings)
The steel pipe has an inner diameter of 2.75 in. and an outer diameter of 3 in. If it is fixed at C and subjected to the horizontal 20-lb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the surface of the pipe.
Step-by-Step Solution:
Chapter: Problem:
• Step 1 of 5
Inner diameter of pipe | 228 | 903 | {"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 | 0.920344 |
http://algebralab.org/lessons/lesson.aspx?file=Algebra_LogarithmsSolving.xml | 1,642,369,160,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00165.warc.gz | 2,134,757 | 6,856 | Solving Logarithmic Equations
Solving logarithmic equations usually requires using properties of logarithms. The reason you usually need to apply these properties is so that you will have a single logarithmic expression on one or both sides of the equation.
Once you have used properties of logarithms to condense any log expressions in the equation, you can solve the problem by changing the logarithmic equation into an exponential equation and solving.
Let's Practice:
1. Solve
As mentioned above, the first step used in solving logarithmic equations is to make use of the properties of logs. In this case, we can combine the two log expressions on the left side of the equation into one expression using multiplication.
This means we are now solving the equation
Now that we have a single log expression equal to a number, we can change the equation into its exponential form. However, to do this we need to make note that when a base is not given in a problem, it is understood to be the common logarithm with a base of 10.
So our logarithmic equation becomes
.
Now we need to solve for x. This will require solving a quadratic equation by factoring. Note: Most of the time solving by factoring will suffice. Very seldom will you need to solve a quadratic by another method.
So let’s solve for x.
Factoring and setting each term equal to zero results in
(x - 5)(x + 2) = 0
x - 5 = 0 or x + 2 = 0
x = 5 or x = -2
When solving logarithmic equations, we must ALWAYS check our answers. Logarithmic functions are not defined for negative values. Therefore we have to plug in our answers and make sure we are not taking the log of a negative number.
We’ll plug in x = 5 into the original equation.
We do not actually have to continue in the checking process as soon as we see that we are not taking the log of a negative number.
Now let’s plug in x = -2 into the original equation.
As soon as we see that we are taking the log of a negative number we know that x = -2 is NOT a solution.
So our solution is x = 5.
1. Solve
This problem is slightly different than the last example we worked. First of all, it involves the natural logarithm (link to exponents-e.doc). But you may also notice that there are log expressions on both sides of the equation. Our approach to this type of problem is to write each side as a single log expression. Once we do that, we can apply the property of logarithms that says
If then x = y.
Let’s get started.
First we’ll apply properties of logs and write the left side of the equation as a single expression using multiplication and write the right side with an exponent of 2 rather than a coefficient of 2.
Now we use the property of logs shown above to get
Factoring and setting each term equal to zero results in
x - 6 = 0 or x - 1 = 0
x = 6 or x = 1
If we check x = 6 in the original equation we do not have to take the log of a negative number, but plugging x = 1 into the original equation will cause us to take the log of a negative number and cannot be a solution.
Our answer to this problem is x = 6.
Examples | 718 | 3,089 | {"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} | 4.96875 | 5 | CC-MAIN-2022-05 | longest | en | 0.946979 |
https://web.pa.msu.edu/courses/1999fall/PHY471/set8.html | 1,591,463,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348517506.81/warc/CC-MAIN-20200606155701-20200606185701-00299.warc.gz | 585,402,197 | 4,497 | # Due Monday, 11/01/1999
## 2. [1+1+2+2+2+2] Unbound states of the finite square well
Consider the unbound states (E>0) of the finite square well . For a wave incident from the left, the solution to the Schrödinger equation can be written as .
a) Determine k and l.
b) Write down equations for all four boundary conditions for this problem.
c) Use the boundary conditions at x=+L to express C in terms of F and to express D in terms of F.
d) Use your results from c) and the boundary conditions at x=-L to eliminate A and to show that B can be expressed in terms of F as with .
e) Plot the transmission coefficient versus energy over a suitable range.
f) For which values of the energy is the transmission coefficient equal to 1?
## 3. [1+2+2+1+1] Unbound states of the square potential barrier
Consider a particle with mass m and energy E>V0 incident from the left on a potential .
a) Write down the solution of Schrödinger’s equation for each region.
b) State the boundary conditions that apply and obtain a sufficient number of equations to determine all unknown constants.
c) Solve for the transmission coefficient and plot it versus energy over a suitable range.
d) For which values of the energy is the transmission coefficient equal to 1?
e) Which value does the transmission coefficient approach for E>>V0? | 346 | 1,442 | {"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} | 2.875 | 3 | CC-MAIN-2020-24 | latest | en | 0.821715 |
https://www.homeworkmarket.com/content/part-you-will-submit-one-file-word-document%E2%80%A6 | 1,563,492,892,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525863.49/warc/CC-MAIN-20190718231656-20190719013656-00040.warc.gz | 745,950,220 | 40,350 | # Part A-You will submit one file, a Word document…
Part A (Items #1 and #12 are required but not graded)
You will submit one file, a Word document. Please limit each response to 250 words or less. Name the file in the following format: lastnamefirstinitialBTM8107-1.doc (example: smithbBTM8107-1.doc).
1. Briefly describe your area of research interest (1-3 sentences is sufficient).
2. List 4 variables that you might assess in a research project related to your research area. List one for each type of measurement scale: Nominal, ordinal, interval, and ratio. If you cannot think of a variable for each measurement scale, explain why the task is difficult.
3. Create one alternate hypothesis and its associated null hypothesis related to your research area.
4. Briefly describe whether you think your area of interest is more conducive to experimental or correlational research. What are the costs/benefits of each as it relates to your research area?
5. Reliability vs. Validity. Considering your area of research interest, discuss the importance of reliability and validity. Can you have one without the other? Why or why not?
6. Sample vs. Population. Considering your area of research interest, describe the difference between a sample and population. Why is it important to understand the difference between a sample and population in a statistics course?
7. Measures of Central Tendency. Below is a set of data that represent weight in pounds for a particular sample. Calculate the mean, median and mode. Which measure of central tendency best describes this data and why? You may use Excel, SPSS, some other software program, or a hand calculator for this problem.
110.00
117.00
120.00
118.00
104.00
100.00
107.00
115.00
115.00
115.00
114.00
100.00
117.00
115.00
103.00
105.00
110.00
115.00
250.00
275.00
8. Measures of Dispersion. For the data set above, calculate the range, the interquartile range, the variance, and the standard deviation. What do these measures tell you about the “spread” of the data?
9. Descriptive Statistics. Why is it important to perform basic descriptive statistics prior to conducting inferential statistical tests?
10. Statistical Significance. Revisit the hypotheses you created above in #5. If you conducted a statistical test based on these hypotheses and found a statistically significant result, what would that mean from both a statistical and practical standpoint? (Be sure to use the phrases “null hypothesis” and “effect size” in your answer).
11. Type I and Type II Error. The concept of Type I and Type II Error is critical and will come into play not only with each and every statistical test you perform, but when you are asked to conduct an a priori power analysis for your Dissertation Proposal. Considering your answer to #10, discuss the implications of making both a Type I and Type II error.
12. After completing Assignment #1, are there any areas of concern you have that you would like to share with your course instructor?
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### Research Methods & Basic Statistics, Entering Data, and Analysis
You will submit one file, a Word document. Please limit each response to 250 words or less. Name the file in the following format: lastnamefirstinitialBTM8107-1.doc (example: … | 782 | 3,438 | {"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} | 3.34375 | 3 | CC-MAIN-2019-30 | latest | en | 0.887775 |
https://puzzling.stackexchange.com/tags/graph-theory/hot | 1,575,711,843,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00468.warc.gz | 517,394,183 | 26,373 | # Tag Info
71
This is not impossible. Here is a possible solution:
57
You can't. Color them like a checkerboard - the top-left-front cell is black, and the ones adjacent to it are white, and the ones adjacent to those are black... Each prisoner in a white cell must move to a black cell, and each prisoner in a black cell must move to a white cell. But there are more of one color than the other, since there are an odd number ...
55
Explanation:
50
I started by assuming two adjacent states must be different colors. Therefore, I arbitrarily assigned blue to California and beige to Oregon. Nevada must therefore be a third color: green. Arizona must therefore be neither green nor blue: beige. Utah must therefore neither be green nor beige: blue. Idaho must therefore be neither beige, green, nor blue - ...
48
My most sincere apologies for this. Really.
45
Probably a bit simpler than what you were looking for. Looking at the other answers, I feel like there might be some details missing from your question.
44
It works when you use a really wide line:
36
It is impossible. Quite the same problem is "Seven Bridges of Königsberg", it was solved (proven) by Euler. Suppose you have drawn such a line and follow it from one room to another. Since you must use each door you must have a look at each room out of 5. What are these rooms? There will be 3 (at least) rooms you always go through - if you enter it you ...
36
This link explains why this puzzle is unsolvable: It also suggests a clever solution in "3 dimensions" (actually, a 2D solution transfomed into a 3D one).
34
Just for fun: Actually, there is solution, which formally satisfies all the rules. You just need to walk through a wall! Hard, but possible!
31
Option 1: Ask a mathematician to explain why the complete bipartite $K_{3,3}$ graph is non planar. Option 2: Put this drawing and a stack of cash in an envelope. Deliver the envelope to a competent engineer.
27
Yes! This is a classic graph theory problem. Anywhere where lines meet is called a vertex, and the degree of a vertex is the number of lines that meet there. Euler proved that as long as a graph has either 0 or 2 vertices of odd degree, and the graph is connected (consists of a single piece), then it can be drawn as you specified. Furthermore, if there are ...
25
Yes there is a solution with a very simple strategy: Start in (1,1). Always go the right most square that's unvisited I'll try to illustrate it. I checked it by hand on an 9x9 board and a very nice pattern emerges that makes it clear it works on any X by X board. \begin{array}{c|cccc} \ &1&2&3&4&5&6&7&8&9\\ \hline 1 &...
23
Similar to klm's solution, this one requires a little bit of lateral thinking but doesn't actually involve walking through a wall. Instead, you have to fold the corner of the piece of paper that the picture is drawn on, to form a bridge over a wall.
21
First of all, with some half-sibling incest, it's possible for a student to only have one grandpa. If that happens, then all 64 students have that grandpa as well. So, we can assume everyone has two grandpas. Let's say that a "grandpa triangle" is a set of three grandpas, where each pair have a common grandchild. If there is no grandpa triangle, then ...
21
Let $(x,y)$ be the square in row $x$, column $y$, so that a fantasy knight can move from $(x,y)$ to $(y,z)$. A closed tour is described by a cyclic sequence $$x_0,x_1,x_2,\ldots,x_{99^2-1},x_{99^2}=x_0,$$ where the knight moves from $(x_0,x_1)$ to $(x_1,x_2)$, then to $(x_2,x_3)$, and so on up to $(x_{99^2-1},x_0)$, then finally back to $(x_0,x_1)$. Each ...
19
You can draw a graph with six vertices, one for each room and one for the outside. Draw an edge to represent each door. The puzzle asks for an Eulerian path, which can be done if no more than two vertices have an odd number of edges coming in. In this graph the top two rooms, the middle bottom room, and the outside all have an odd number of edges coming ...
19
As atonement for my insolent lateral-thinking answer, I offer an optimality proof. If you keep repeating the correct code, the are six possible different orders: 1 abcdabcdabcd 2 abdcabdcabdc 3 acbdacbdacbd 4 acdbacdbacdb 5 adbcadbcadbc 6 adcbadcbadcb Each of the orders contains four possible codes. The orders are important, since after testing one code ...
18
Alternatively, going "through" a door need not necessarily be interpreted in the same way as one would in a house. This single, continuous line passes exactly once through each door, which is the constraint in the original question. Of course, it also conveniently side-steps making any other assumptions and takes certain liberties with the walls.
18
Here's one solution (not sure if it's unique): How I found it: by following the logic used to answer this similar question. How I found this specific solution:
16
Each of the four diagrams with letters but in each case These are, respectively, so this is a memorial to I expect it was constructed using graphs because
16
There are a few nodes that can be linked together immediately, giving us a good starting point: Most of those starting links are on the bottom half of the triangle, so that's where I started. Specifically, I was quickly able to fill in the bottom-right: Followed by the rest of the bottom: The next bit threw me briefly, but I managed to figure out the ...
15
Reiterate problem Bunny: a chess piece which moves like a bishop but only one square from its current position. It may also hop over another piece. Hop: a bunny hops when another piece is in a square touching the current square (no diagonals). In its turn it occupies the square 3 squares from its current square in the direction of the hopped piece. The ...
15
The state you have to end in will be Here's a possible trip you can take:
15
Number the nodes as follows: A valid solution is to check: To see that this works, first note that the fox can only go from a hole in an even-numbered level of the tree to a hole in an odd-numbered level of the tree and vice-versa. If the fox starts in an odd level, the sequence 4,2,5,2,1,3,6,3,7 guarantees the hunter will catch it. This works because the ...
15
First, I found all pairs that satisfy the criteria: Next, we can start making some deductions: That gives this new grid: Next, That resolves the rest of the path, but not the numbers: For the numbers, The final answer to the puzzle: And to double-check the result:
14
This is impossible Assume 2 houses are fully connected to 3 wells. There must be an 'inner' well which is surrounded by the connections to the other 2 wells. For the third house to be connected to the inner well, it must be in one of the spaces between the inner well and one of the outer ones. But this means it is seperated from the other outer well. ...
14
An observation is.. Using this observation, one of many possible solutions is (left image is front of cube, right image is inside view):
13
There are a finite number of states she can be in, where a state is determined by direction, road, and upcoming turn. Each state has a unique predecessor and successor; therefore, there are no branches in the directed graph of states as vertices and edges connecting each state to its successor. This means every state must be part of a cycle.
13
This is Here is a picture describing why. r = room p = pool g = garden a = armoury
Only top voted, non community-wiki answers of a minimum length are eligible | 1,816 | 7,503 | {"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} | 3.5625 | 4 | CC-MAIN-2019-51 | longest | en | 0.934237 |
https://www.albert.io/ie/sat-math/spotting-a-fire | 1,485,005,192,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281084.84/warc/CC-MAIN-20170116095121-00528-ip-10-171-10-70.ec2.internal.warc.gz | 867,619,743 | 20,260 | Free Version
Easy
# Spotting a Fire
SATMAT-EVSWJ1
A man in a tower which is $250$ feet above the ground spots a fire in the distance. The angle of depression to the fire from his location in the tower is ${ 6 }^{ \circ }$.
Which of the following is a correct expression for the horizontal distance between the tower and the fire?
A
$(250)\tan { \left( { 6 }^{ \circ } \right) }$
B
$(250)\tan { \left( { 84}^{ \circ } \right) }$
C
$\cfrac { 250 }{ \tan\left( { 84 }^{ \circ } \right) }$
D
$\sqrt { { 250 }^{ 2 }+{ 6 }^{ 2 } }$ | 182 | 537 | {"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} | 2.953125 | 3 | CC-MAIN-2017-04 | longest | en | 0.773026 |
https://www.easyelimu.com/qa/3596/object-placed-front-concave-mirror-length-calculate-distance | 1,718,799,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00829.warc.gz | 664,118,779 | 6,865 | # An object is placed 15cm in front of a concave mirror of focal length 10cm.calculate the image distance
845 views
An object is placed 15cm in front of a concave mirror of focal length 10cm.calculate the image distance
1/f=1/v+1/u
1/10=1/v+1/15
1/30=1/v
V=30cm | 86 | 263 | {"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-2024-26 | latest | en | 0.793057 |
https://tr.coursera.org/learn/sequence-models-tensorflow-gcp | 1,550,756,105,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00447.warc.gz | 703,742,865 | 121,886 | 4.6
87 ratings
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#### 100% online
Start instantly and learn at your own schedule.
#### Approx. 11 hours to complete
Suggested: 17 hours/week...
#### English
Subtitles: English
#### 100% online
Start instantly and learn at your own schedule.
#### Approx. 11 hours to complete
Suggested: 17 hours/week...
#### English
Subtitles: English
### Syllabus - What you will learn from this course
Week
1
4 hours to complete
## Working with Sequences
In this module, you’ll learn what a sequence is, see how you can prepare sequence data for modeling, and be introduced to some classical approaches to sequence modeling and practice applying them....
14 videos (Total 41 min), 1 reading, 4 quizzes
14 videos
Getting Started with Google Cloud Platform and Qwiklabs3m
Sequence data and models5m
From sequences to inputs2m
Modeling sequences with linear models2m
Lab intro: using linear models for sequences20s
Lab solution: using linear models for sequences7m
Modeling sequences with DNNs2m
Lab intro: using DNNs for sequences19s
Lab solution: using DNNs for sequences2m
Modeling sequences with CNNs3m
Lab intro: using CNNs for sequences19s
Lab solution: using CNNs for sequences3m
The variable-length problem4m
How to send course feedback10m
1 practice exercise
Working with Sequences
15 minutes to complete
## Recurrent Neural Networks
In this module, we introduce recurrent neural nets, explain how they address the variable-length sequence problem, explain how our traditional optimization procedure applies to RNNs, and review the limits of what RNNs can and can’t represent....
4 videos (Total 15 min), 1 quiz
4 videos
How RNNs represent the past4m
The limits of what RNNs can represent5m
1 practice exercise
Recurrent Neural Networks
4 hours to complete
## Dealing with Longer Sequences
In this module we dive deeper into RNNs. We’ll talk about LSTMs, Deep RNNs, working with real world data, and more....
14 videos (Total 62 min), 4 quizzes
14 videos
LSTMs and GRUs6m
RNNs in TensorFlow2m
Lab Intro: Time series prediction: end-to-end (rnn)45s
Lab Solution: Time series prediction: end-to-end (rnn)10m
Deep RNNs1m
Lab Intro: Time series prediction: end-to-end (rnn2)26s
Lab Solution: Time series prediction: end-to-end (rnn2)6m
Improving our Loss Function2m
Demo: Time series prediction: end-to-end (rnnN)3m
Working with Real Data10m
Lab Intro: Time Series Prediction - Temperature from Weather Data1m
Lab Solution: Time Series Prediction - Temperature from Weather Data11m
Summary1m
1 practice exercise
Dealing with Longer Sequences
Week
2
2 hours to complete
## Text Classification
In this module we look at different ways of working with text and how to create your own text classification models. ...
8 videos (Total 35 min), 2 quizzes
8 videos
Text Classification6m
Selecting a Model2m
Lab Intro: Text Classification47s
Lab Solution: Text Classification11m
Python vs Native TensorFlow4m
Demo: Text Classification with Native TensorFlow7m
Summary1m
1 practice exercise
Text Classification
1 hour to complete
## Reusable Embeddings
Labeled data for our classification models is expensive and precious. Here we will address how we can reuse pre-trained embeddings to make our models with TensorFlow Hub....
6 videos (Total 28 min), 2 quizzes
6 videos
Modern methods of making word embeddings8m
Introducing TensorFlow Hub1m
Lab Intro: Evaluating a pre-trained embedding from TensorFlow Hub24s
Lab Solution: TensorFlow Hub9m
Using TensorFlow Hub within an estimator1m
1 practice exercise
Reusable Embeddings
3 hours to complete
## Encoder-Decoder Models
In this module, we focus on a sequence-to-sequence model called the encoder-decoder network to solve tasks, such as Machine Translation, Text Summarization and Question Answering....
10 videos (Total 84 min), 3 quizzes
10 videos
Attention Networks4m
Training Encoder-Decoder Models with TensorFlow6m
Introducing Tensor2Tensor11m
Lab Intro: Cloud poetry: Training custom text models on Cloud ML Engine1m
Lab Solution: Cloud poetry: Training custom text models on Cloud ML Engine25m
AutoML Translation4m
Dialogflow6m
Lab Intro: Introducing Dialogflow54s
Lab Solution: Dialogflow13m
1 practice exercise
Encoder-Decoder Models
14 minutes to complete
## Summary
In this final module, we review what you have learned so far about sequence modeling for time-series and natural language data. ...
1 video (Total 4 min), 1 reading
1 video
4.6
12 Reviews
### Top Reviews
By MDFeb 3rd 2019
Very good.The explanation of the RNN was very good but the tensor2tensor was very hard.
## Instructor
We help millions of organizations empower their employees, serve their customers, and build what’s next for their businesses with innovative technology created in—and for—the cloud. Our products are engineered for security, reliability, and scalability, running the full stack from infrastructure to applications to devices and hardware. Our teams are dedicated to helping customers apply our technologies to create success....
This 5-course specialization focuses on advanced machine learning topics using Google Cloud Platform where you will get hands-on experience optimizing, deploying, and scaling production ML models of various types in hands-on labs. This specialization picks up where “Machine Learning on GCP” left off and teaches you how to build scalable, accurate, and production-ready models for structured data, image data, time-series, and natural language text. It ends with a course on building recommendation systems. Topics introduced in earlier courses are referenced in later courses, so it is recommended that you take the courses in exactly this order.... | 1,316 | 5,652 | {"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} | 2.875 | 3 | CC-MAIN-2019-09 | longest | en | 0.838456 |
https://whatisconvert.com/81-centimeters-in-nanometers | 1,713,951,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819089.82/warc/CC-MAIN-20240424080812-20240424110812-00231.warc.gz | 564,893,205 | 7,080 | ## Convert 81 Centimeters to Nanometers
To calculate 81 Centimeters to the corresponding value in Nanometers, multiply the quantity in Centimeters by 10000000 (conversion factor). In this case we should multiply 81 Centimeters by 10000000 to get the equivalent result in Nanometers:
81 Centimeters x 10000000 = 810000000 Nanometers
81 Centimeters is equivalent to 810000000 Nanometers.
## How to convert from Centimeters to Nanometers
The conversion factor from Centimeters to Nanometers is 10000000. To find out how many Centimeters in Nanometers, multiply by the conversion factor or use the Length converter above. Eighty-one Centimeters is equivalent to eight hundred ten million Nanometers.
## Definition of Centimeter
The centimeter (symbol: cm) is a unit of length in the metric system. It is also the base unit in the centimeter-gram-second system of units. The centimeter practical unit of length for many everyday measurements. A centimeter is equal to 0.01(or 1E-2) meter.
## Definition of Nanometer
A nanometer (sumbol: nm) is a unit of spatial measurement that is 10-9 meter, or one billionth of a meter. It is commonly used in nanotechnology, the building of extremely small machines. The SI prefix "nano" represents a factor of 10-9, or in exponential notation, 1E-9. So 1 nanometre = 10-9 metre.
## Using the Centimeters to Nanometers converter you can get answers to questions like the following:
• How many Nanometers are in 81 Centimeters?
• 81 Centimeters is equal to how many Nanometers?
• How to convert 81 Centimeters to Nanometers?
• How many is 81 Centimeters in Nanometers?
• What is 81 Centimeters in Nanometers?
• How much is 81 Centimeters in Nanometers?
• How many nm are in 81 cm?
• 81 cm is equal to how many nm?
• How to convert 81 cm to nm?
• How many is 81 cm in nm?
• What is 81 cm in nm?
• How much is 81 cm in nm? | 471 | 1,862 | {"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-2024-18 | latest | en | 0.871095 |
https://origin.geeksforgeeks.org/find-other-two-sides-and-angles-of-a-right-angle-triangle/?ref=lbp | 1,686,424,467,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00475.warc.gz | 493,008,047 | 45,716 | GFG App
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# Find other two sides and angles of a right angle triangle
Given one side of right angle triangle, check if there exists a right angle triangle possible with any other two sides of the triangle. If possible print length of the other two sides and all the angles of the triangle.
Examples:
Input : a = 12
Output : Sides are a = 12, b = 35, c = 37
Angles are A = 18.9246, B = 71.0754, C = 90
Explanation: a = 12, b = 35 and c = 37 form right
angle triangle because
12*12 + 35*35 = 37*37
Input : a = 6
Output : Sides are a = 6, b = 8, c = 10
Angles are A = 36.8699, B = 53.1301, C = 90
Approach to check if triangle exists and finding Sides
To solve this problem we first observe the Pythagoras equation. If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
This relationship is represented by the formula:
a*a + b*b = c*c
Case 1: a is an odd number: Given a, find b and c
c2 - b2 = a2
OR
c = (a2 + 1)/2;
b = (a2 - 1)/2;
Above solution works only for case when a is odd, because a2 + 1 is divisible by 2 only for odd a.
Case 2 : a is an even number: When c-b is 2 & c+b is (a2)/2
c-b = 2 & c+b = (a2)/2
Hence,
c = (a2)/4 + 1;
b = (a2)/4 - 1;
This works when a is even.
Approach to find Angles
First find all sides of triangle. Then Applied “SSS” rule that’s means law of cosine:
Below is the implementation of the above approach:
## C++
// C++ program to print all sides and angles of right // angle triangle given one side #include #include using namespace std; #define PI 3.1415926535 // Function to find angle A // Angle in front of side a double findAnglesA(double a, double b, double c) { // applied cosine rule double A = acos((b * b + c * c - a * a) / (2 * b * c)); // convert into degrees return A * 180 / PI; } // Function to find angle B // Angle in front of side b double findAnglesB(double a, double b, double c) { // applied cosine rule double B = acos((a * a + c * c - b * b) / (2 * a * c)); // convert into degrees and return return B * 180 / PI; } // Function to print all angles // of the right angled triangle void printAngles(int a, int b, int c) { double x = (double)a; double y = (double)b; double z = (double)c; // for calculate angle A double A = findAnglesA(x, y, z); // for calculate angle B double B = findAnglesB(x, y, z); cout << "Angles are A = " << A << ", B = " << B << ", C = " << 90 << endl; } // Function to find other two sides of the // right angled triangle void printOtherSides(int n) { int b,c; // if n is odd if (n & 1) { // case of n = 1 handled separately if (n == 1) cout << -1 << endl; else { b = (n*n-1)/2; c = (n*n+1)/2; cout << "Side b = " << b << ", Side c = " << c << endl; } } else { // case of n = 2 handled separately if (n == 2) cout << -1 << endl; else { b = n*n/4-1; c = n*n/4+1; cout << "Side b = " << b << ", Side c = " << c << endl; } } // Print angles of the triangle printAngles(n,b,c); } // Driver Program int main() { int a = 12; printOtherSides(a); return 0; }
## Java
// Java program to print all sides and angles of right // angle triangle given one side import java.io.*; class GFG { static double PI = 3.1415926535; // Function to find angle A // Angle in front of side a static double findAnglesA(double a, double b, double c) { // applied cosine rule double A = Math.acos((b * b + c * c - a * a) / (2 * b * c)); // convert into degrees return A * 180 / PI; } // Function to find angle B // Angle in front of side b static double findAnglesB(double a, double b, double c) { // applied cosine rule double B = Math.acos((a * a + c * c - b * b) / (2 * a * c)); // convert into degrees and return return B * 180 / PI; } // Function to print all angles // of the right angled triangle static void printAngles(int a, int b, int c) { double x = (double)a; double y = (double)b; double z = (double)c; // for calculate angle A double A = findAnglesA(x, y, z); // for calculate angle B double B = findAnglesB(x, y, z); System.out.println( "Angles are A = " + A + ", B = " + B + ", C = " + 90); } // Function to find other two sides of the // right angled triangle static void printOtherSides(int n) { int b=0,c=0; // if n is odd if ((n & 1)>0) { // case of n = 1 handled separately if (n == 1) System.out.println( -1); else { b = (n*n-1)/2; c = (n*n+1)/2; System.out.println( "Side b = " + b + ", Side c = " + c ); } } else { // case of n = 2 handled separately if (n == 2) System.out.println( -1); else { b = n*n/4-1; c = n*n/4+1; System.out.println( "Side b = " + b + ", Side c = " + c); } } // Print angles of the triangle printAngles(n,b,c); } // Driver Program public static void main (String[] args) { int a = 12; printOtherSides(a); } } // This code is contributed // by inder_verma..
## Python 3
# Python 3 program to print all # sides and angles of right # angle triangle given one side import math PI = 3.1415926535 # Function to find angle A # Angle in front of side a def findAnglesA( a, b, c): # applied cosine rule A = math.acos((b * b + c * c - a * a) / (2 * b * c)) # convert into degrees return A * 180 / PI # Function to find angle B # Angle in front of side b def findAnglesB(a, b, c): # applied cosine rule B = math.acos((a * a + c * c - b * b) / (2 * a * c)) # convert into degrees # and return return B * 180 / PI # Function to print all angles # of the right angled triangle def printAngles(a, b, c): x = a y = b z = c # for calculate angle A A = findAnglesA(x, y, z) # for calculate angle B B = findAnglesB(x, y, z) print("Angles are A = ", A, ", B = ", B , ", C = ", "90 ") # Function to find other two sides # of the right angled triangle def printOtherSides(n): # if n is odd if (n & 1) : # case of n = 1 handled # separately if (n == 1): print("-1") else: b = (n * n - 1) // 2 c = (n * n + 1) // 2 print("Side b = ", b, " Side c = ", c) else: # case of n = 2 handled # separately if (n == 2) : print("-1") else: b = n * n // 4 - 1; c = n * n // 4 + 1; print("Side b = " , b, ", Side c = " , c) # Print angles of the triangle printAngles(n, b, c) # Driver Code if __name__ == "__main__": a = 12 printOtherSides(a) # This code is contributed # by ChitraNayal
## C#
// C# program to print all sides // and angles of right angle // triangle given one side using System; class GFG { static double PI = 3.1415926535; // Function to find angle A // Angle in front of side a static double findAnglesA(double a, double b, double c) { // applied cosine rule double A = Math.Acos((b * b + c * c - a * a) / (2 * b * c)); // convert into degrees return A * 180 / PI; } // Function to find angle B // Angle in front of side b static double findAnglesB(double a, double b, double c) { // applied cosine rule double B = Math.Acos((a * a + c * c - b * b) / (2 * a * c)); // convert into degrees and return return B * 180 / PI; } // Function to print all angles // of the right angled triangle static void printAngles(int a, int b, int c) { double x = (double)a; double y = (double)b; double z = (double)c; // for calculate angle A double A = findAnglesA(x, y, z); // for calculate angle B double B = findAnglesB(x, y, z); Console.WriteLine( "Angles are A = " + A + ", B = " + B + ", C = " + 90); } // Function to find other two sides // of the right angled triangle static void printOtherSides(int n) { int b = 0, c = 0; // if n is odd if ((n & 1) > 0) { // case of n = 1 handled separately if (n == 1) Console.WriteLine( -1); else { b = (n * n - 1) / 2; c = (n * n + 1) / 2; Console.WriteLine( "Side b = " + b + ", Side c = " + c); } } else { // case of n = 2 handled separately if (n == 2) Console.WriteLine( -1); else { b = n * n / 4 - 1; c = n * n / 4 + 1; Console.WriteLine( "Side b = " + b + ", Side c = " + c); } } // Print angles of the triangle printAngles(n, b, c); } // Driver Code public static void Main () { int a = 12; printOtherSides(a); } } // This code is contributed // by inder_verma
## PHP
## Javascript
Output:
Side b = 35, Side c = 37
Angles are A = 18.9246, B = 71.0754, C = 90
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
My Personal Notes arrow_drop_up | 3,255 | 10,256 | {"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} | 4.65625 | 5 | CC-MAIN-2023-23 | latest | en | 0.770347 |
https://numberworld.info/26287 | 1,716,700,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00177.warc.gz | 369,942,266 | 3,791 | # Number 26287
### Properties of number 26287
Cross Sum:
Factorization:
Divisors:
1, 97, 271, 26287
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
66af
Base 32:
plf
sin(26287)
-0.96200890138826
cos(26287)
-0.27301808300873
tan(26287)
3.5236087323838
ln(26287)
10.176829799383
lg(26287)
4.4197410251407
sqrt(26287)
162.13266173107
Square(26287)
### Number Look Up
Look Up
26287 which is pronounced (twenty-six thousand two hundred eighty-seven) is a unique figure. The cross sum of 26287 is 25. If you factorisate the figure 26287 you will get these result 97 * 271. The figure 26287 has 4 divisors ( 1, 97, 271, 26287 ) whith a sum of 26656. The number 26287 is not a prime number. The number 26287 is not a fibonacci number. The figure 26287 is not a Bell Number. The number 26287 is not a Catalan Number. The convertion of 26287 to base 2 (Binary) is 110011010101111. The convertion of 26287 to base 3 (Ternary) is 1100001121. The convertion of 26287 to base 4 (Quaternary) is 12122233. The convertion of 26287 to base 5 (Quintal) is 1320122. The convertion of 26287 to base 8 (Octal) is 63257. The convertion of 26287 to base 16 (Hexadecimal) is 66af. The convertion of 26287 to base 32 is plf. The sine of 26287 is -0.96200890138826. The cosine of the figure 26287 is -0.27301808300873. The tangent of the figure 26287 is 3.5236087323838. The root of 26287 is 162.13266173107.
If you square 26287 you will get the following result 691006369. The natural logarithm of 26287 is 10.176829799383 and the decimal logarithm is 4.4197410251407. I hope that you now know that 26287 is unique number! | 609 | 1,752 | {"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 | 0.765889 |
https://www.weegy.com/?ConversationId=55474E1A&Link=i&ModeType=2 | 1,606,723,975,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00074.warc.gz | 897,517,036 | 14,432 | The area where one plate sinks into the asthenosphere underneath another plate is known as a subduction zonea transform boundaryan earthquake boundarya convection zone
The area where one plate sinks into the asthenosphere underneath another plate is known as : a subduction zone.
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sujaysen|Points 2711|
Question
Updated 1/31/2016 11:45:20 AM
Edited by jeifunk [1/31/2016 11:45:19 AM], Confirmed by jeifunk [1/31/2016 11:45:20 AM]
Rating
Questions asked by the same visitor
4y - 6y + 9y = -2 {-2/11} {-2/7} {2/7} {2/11}
Weegy: y = -2/7 User: (x - 3) - 2(x + 6) = -5 {-10} {-4} {8} Weegy: its -10. wait ill check again User: (x - 3) - 2(x + 6) = -5 {-10} {-4} {8} Weegy: its -10. wait ill check again User: 7x + 9 = 6x - 5 x = -14 x = -4 x = 4 x = 14 Weegy: 4-7x+10=15-6x-8 4+10=14 and 15-8=7 so now you have... 14-7x=7-6x - - - - - - - - - - - - - - - - - - - ? 7x+7x 14=7+x now use opposites to get -x= -7 which equals x=7 is your answer. User: 17x - 6 + 3x - 5 = x + 11 + 4x 20x - 11 = 5x + 11 20x + 1 = 5x + 11 9x = 16x Weegy: 20x-11=5x+11 User: An equation _____ has one solution. always sometimes never Weegy: It's either a trick question, [ or the person who wrote the question is guilty of fuzzy thinking. a linear (1st order) equation will have one solution a quadratic (2nd order) equation will have two solutions a cubic (3rd order) equation will have three solutions ... an nth order equation will have n solutions. An equation may have more than one solution, or the solution(s) may be complex, rather than real, but it will always have one solution - and sometimes more than one. Had the question been "An equation _____ has only one solution", the answer would have been #2, "sometimes". ] User: An equation _____ has one solution. always sometimes never Weegy: It's either a trick question, [ or the person who wrote the question is guilty of fuzzy thinking. a linear (1st order) equation will have one solution a quadratic (2nd order) equation will have two solutions a cubic (3rd order) equation will have three solutions ... an nth order equation will have n solutions. An equation may have more than one solution, or the solution(s) may be complex, rather than real, but it will always have one solution - and sometimes more than one. Had the question been "An equation _____ has only one solution", the answer would have been #2, "sometimes". ] User: Which of the following equations has the given solution set? Solution set: Ø -2m + 5 = ... (More)
Question
Updated 3/12/2014 3:28:22 AM
7x + 9 = 6x - 5
7x - 6x = -5 - 9
x = -14
9(x - 2) = 18
9x - 18 = 18
9x = 18 + 18
9x = 36
x = 36/9
x = 4
Simplify 8(3 - 2x). 24x - 16 16x - 24 24 - 16x User: Simplify 8(3 - 2x). 24x - 16 16x - 24 24 - 16x User: Simplify 11(2x + 3). 11x + 5 22x + 3 22x + 33 User: Simplify 3x + (-12x) - 5x. -14x -9x 10x 20x User: Simplify 2x + x. 2x² 3x 3x ²
Weegy: x = 11/7 User: Score: Simplify -2[9 - (x + 7)]. -32 - x -4 - x -4 + 2x -32 + 2x (More)
Question
Updated 6/4/2014 12:43:31 AM
8(3 - 2x) = 24 - 16x
Confirmed by andrewpallarca [6/4/2014 1:27:03 PM]
11(2x + 3) = 22x + 33
Confirmed by andrewpallarca [6/4/2014 1:27:14 PM]
3x + (-12x) - 5x = 3x - 12x - 5x = -9x - 5x = -14x
Confirmed by andrewpallarca [6/4/2014 1:28:03 PM]
2x + x = 3x
Confirmed by andrewpallarca [6/4/2014 1:28:05 PM]
-2[9 - (x + 7)]
= -2(9 - x - 7)
= -2(2 - x)
= -4 + 2x
Confirmed by andrewpallarca [6/4/2014 1:29:13 PM]
One side of a square is 3x + 2. Which expression represents the perimeter? 2(3x + 2) (3x + 2)² 4(3x + 2)
Weegy: 9x^4. If you are satisfied with my answer, please click 'Good' on the ratings. Thank you! User: If x represents the number, then which of the following equations could be used to solve the problem? 4x + 6 = 94 4(x + 6) = 94 x + 6 = 4(94) User: Dwayne's garden is triangle-shaped with two equal sides and a third side that is 4 ft more than the length of an equal side. If the perimeter is 49 ft, how long is each side? Which equation represents this? 2x = 49 2x + 4 = 49 3x = 49 3x + 4 = 49 Weegy: D)3x + 4 = 49 User: Thomas has \$6.35 in dimes and quarters. The number of dimes is three more than three times the number of quarters. How many quarters does he have? If q represents the number of quarters, then which of the following expressions represents the value of the number of dimes in cents? 3q + 3 6.35 - q 10(3q + 3) Weegy: a)3q + 3 User: After John worked at a job for 10 years, his salary doubled. If he started at \$x, his salary after 10 years is _____. \$x \$x+ 2 \$x- 2 \$2x Weegy: If his salary doubled after 10 years it is D. \$2x User: A man walks 10 miles in x hours. How far does he walk in one hour? Which expression represents this? 10 + x miles 10x miles 10/x miles User: Which of the following could represent consecutive even integers? x + 1, x + 2 x + 1, x + 3 2x, 2x + 1 Weegy: 3x - 4y = 30 User: The sum of three numbers is 69. If the second number is equal to the first diminished by 8, and the third number is 5 times the first. What are the numbers? If x represents the first number, then which of the following equations could be used to solve the problem? x = 6x - 8 69 = 6x - 8 69 = 7x - 8 User: The sum of two numbers is x. If one of the numbers is 12, what is the other? 12 -x x- 12 12 +x 12x User: Twice a certain number is tripled. The resulting number is _____. 2x+ 3 2x- 3 (2x)3 (2x) Weegy: If twice a certain number is tripled, the resulting number is C. (2x)3 User: The value in cents ... (More)
Question
Updated 6/26/2014 9:58:18 PM
4 + (27 - 12·2) ÷ 2
= 4 + (27 - 24) ÷ 2
= 4 + 3/2
= 4 + 1.5
= 5.5
Confirmed by jeifunk [6/26/2014 9:58:44 PM]
The coefficient for 3 · 2x is 6.
Confirmed by jeifunk [6/26/2014 9:58:57 PM]
In the expression (x^2 + 3), the base is x.
Confirmed by jeifunk [6/26/2014 9:59:05 PM]
42 + 6 ÷ 2
= 42 + 3
= 45
Confirmed by jeifunk [6/26/2014 10:02:36 PM]
y^5 for y = 1 is 1.
1^5 = 1
Confirmed by jeifunk [6/26/2014 10:02:41 PM]
Scientists have standardized a concise definition of life. True False User: Scientists distinguish whether an entity is living or nonliving by describing the _____ of life. characteristics value length quality
Weegy: characteristics. Scientists distinguish whether an entity is living or nonliving by describing the "characteristics" of life. User: It is often easier to determine if an object or organism possesses the characteristics of life rather than life itself. True False User: Select all that apply. Characteristics that are always present in living organisms are _____. complexity homeostasis metabolism heredity movement cells sensitivity reproduction (More)
Question
Updated 1/26/2016 12:00:59 PM
Scientists have standardized a concise definition of life. FALSE.
Confirmed by jeifunk [1/26/2016 12:14:24 PM]
It is often easier to determine if an object or organism possesses the characteristics of life rather than life itself. TRUE.
Confirmed by jeifunk [1/26/2016 12:14:35 PM]
Characteristics that are always present in a living organisms are cells, metabolism, reproduction, homeostasis, movement, and sensitivity.
Confirmed by jeifunk [1/26/2016 12:15:15 PM]
How many terms are in the expression shown below? x2 - 10xy + 3y + y2 - 1 2 3 4 5
Weegy: There are 3 terms. x, y, and xy.anything else? User: How many terms are in the expression shown below? x2 - 10xy + 3y + y2 - 1 2 3 4 5 Weegy: There are 3 terms. x, y, and xy.anything else? User: Simplify 5m - 13n - n + 4m. 9m - 14n 9m - 13 9m2+ 13n2 -5mn (More)
Question
Updated 5/13/2015 7:13:30 AM
There are 5 terms in the following expression x^2 - 10xy + 3y + y^2 - 1 .
Confirmed by Andrew. [5/13/2015 7:16:08 AM]
5m - 13n - n + 4m
= 9m - 13n - n
= 9m - 14n
Confirmed by Andrew. [5/13/2015 7:16:13 AM]
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# IMG_0005_NEW_0004 - its decomposition products phosphorus...
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Equilibrium Galculations Example 1, Continued.. 6 Define change to get to equilibrium and write expressions for equilibrium conc.s t. Lt.c.B. rautel 21il(s) Initial 2 1 0 Change -x -x +2x Equilibrium 2-x 1-x 0 +2x @ Substitute into expression for K and solve for x K_ DH!1^ ox n+il'ru Px)cru For an equation: m2 +bx+c =O ---:--- There are two roots: * - -bltlbz -4ac Hr(g) + trr equilibrium l"!t"., - * 6 Equilibrium Calculations Example I O Write balanced equation for reaction: Hr(s) + : 2HI (g) O Write equilibrium expression: I Find initial concentrations: e.g. : [Hz]o :2M, [Iz]o : I M, [HI]o : 0 M @ Find Q + Find direction for shift to equilibrium tHlu ffi = System will shift,o rn.[[email protected][L - lFzol Equilibrium Galculations (#; --:AExamP"' #o = o,,n \- ?rff gof PCl, is placed in a 500 ml reaction vessel and allowed to
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Unformatted text preview: its decomposition products phosphorus trichloride and chlorine at250 "C. ( P7ls + l0l> * 0b ': Given that Ko:78.3 O What are the equilibrium compositions in molll? O What is the percentage decomposition of . PCI.? (l.c t.imfl roG F: Pds t o{z .)1r= 9h-+rL o-3 o o lQ'ta;hu o-t-x Equilibrium Galculations Small Equilibrium Constants A mixture of 0.482 mol N2 and 0.933 mol 02 is placed in a vessel of volume 10.0 L and allowed to form N2O at a temperature for which K:2.0 x10-13 L mol-l. What is the equilibrium composition of the mixture? 2Nz(g)+Oz(g) : 2NzO(g) 2Nz(g) + Oz(g) :2NzO(e) g.r.(gL 0-01, o->x -x +7x .0til2-2.x o.o1l9-/ Z 8 [Nz] : lo.or+til-oyr 6: [N1of io,i : ffi^"* ^ : [Nr,FloJ = >tlo 77 Pr' "'9-X...
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Home >Technology >News >Opinion | The more things change, the more they’re invariant
# Opinion | The more things change, the more they’re invariant
The search for an invariant gives us a different perspective on a problem we’re faced with, and that new perspective often takes us closer to a solution
Remember the Hummer shuffle, which made an appearance in this column a couple of years ago? (“The fixed point of it all"). It’s simply this: On a pack of cards, turn over the top two. If they were all face-down to start with, the Hummer shuffle turns the top two cards face-up.
As I wrote then: “Now cut the deck however you choose; do the Hummer shuffle again—in fact, do both moves as many times as you like, in any order. When you stop, you will have a deck that has some unpredictable number of face-up cards. But even if that’s unpredictable, there is one feature of your card deck that has stayed invariant through all your shuffling and cutting: the number of face-up cards is even."
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That small but significant truth is the basis for various intriguing card tricks. With a roomful of people a few days ago, I even tried a slight variation: I turned only the top card face-up before doing the Hummer shuffle on the pack. Now, through any number of cuts and further Hummers, the number of face-up cards stayed odd—and that was the basis for the trick I played on them. I have to confess, though, that it left them singularly unimpressed.
Still, there’s a point here: About invariance. This is a principle that applies all the way from card tricks to logical puzzles and on into realms of mathematics and physics, and is of immense value every time. Given some objects, possibly mathematical, that we’re interested in and an operation we want to subject them to, we look for some property of the objects that doesn’t change once we apply the operation. With the cards I mentioned above, the operations are the Hummer shuffle and the cut. Do them once, twice or a hundred times, and the properties I mentioned—the oddness or the evenness of the number of face-up cards—is invariant.
So, let me throw some other invariants at you, to give you a flavour of why the idea is useful.
* Hold a tennis ball in your hand and rotate it. You will agree that neither its surface area nor its volume changes: They are invariant under such rotation.
* Take any two numbers and consider the difference between them. For example, 871 and 47: The difference is 824. Add the same quantity to both numbers—say 131, to produce 1,002 and 178—and the difference between the two new numbers remains 824: That difference is invariant under such mutual addition. Trivial, you think? But the study of number theory begins with apparently trivial fundamentals like this.
* Draw a triangle on a sheet of paper. Now change the lengths of its three sides any which way, drawing a new triangle each time. Whatever triangle you come up with, the sum of its angles is always 180° (degrees): That sum is invariant regardless of what triangle you draw. This is, of course, basic trigonometry.
* Fashion a teacup out of clay or plasticine. Now stretch and reshape it—but no tearing!—so it becomes doughnut-shaped. Both teacup and doughnut have one hole all the way through the clay, and no more: That one hole is invariant under any amount of stretching or reshaping you do to the clay. The study of properties that are invariant under this kind of reshaping is at the foundation of the entire mathematical field of topology.
Intriguingly, the invariant I mentioned about the angles of a triangle does not necessarily hold on three-dimensional surfaces—like teacups or doughnuts or even planets. Consider, for example, the triangle formed on the surface of the Earth by the 0° and 90° meridians (lines of longitude) and the Equator. All three are straight lines. The two meridians meet at the North Pole, and of course each meets the Equator. All three angles are 90°, meaning the triangle’s total is 270°. Not 180°, like you’ve always believed about triangles: That invariance is only for two-dimensional triangles.
The point, for someone like me who likes dipping into mathematics, is that invariance is about seeing things in different ways.
For example, give this puzzle a thought. I have a bag with 100 marbles, each with a random number on it. You remove two marbles at random, add their numbers, write that on a new marble, put it in the bag and throw away the two you took out. Now there are 99 marbles in the bag. Repeat until there’s only one marble left in the bag. What’s the number on it? (Stop reading here if you’d like a few minutes to think about this without hints.)
You might try solving this by writing down 100 numbers, then doing the two-out-one-in operation over and over… tedious at best. Then you might try it with just 10 numbers; probably still tedious.
But instead, you could ask: What about this set of numbers stays invariant under this particular two-out-one-in operation? I suspect you’ll get an answer to that if you work out what happens if I had only two numbers. But try it, nevertheless, with 100.
What can you do with 100 random numbers? Maybe you can add them all up. Hold on to that total. When you remove two, that figure in your mind decreases by their sum and the bag is down to 98 marbles. But then you put their sum back as a single number, going up to 99 marbles. Which means the total of the set of 99 is the same as the total of all 100. Aha! Are you on to something here? Do the operation again, and you realize the total of the resulting 98 numbers is the same as the total of 99, the same as the total of the original 100.
Indeed you’re on to something: The invariant here is the sum of all the numbers in the collection. And this means that when you’re down to one marble, voilà, the number on it is the total of the original 100.
The search for an invariant gives us a different perspective on a problem we’re faced with, and perspective often takes us closer to a solution. Put it another way: Invariance gives us a deeper understanding, and that’s why it applies in so many areas of mathematics.
In physics, invariance is embedded in the conservation laws, like the conservation of momentum. Momentum is the product of an object’s mass and its velocity. The law tells us that when two objects collide, the sum of their respective momentums stays the same. This will make immediate sense to you if you think of standing on a street and being hit by a speeding car. Slamming into you will slow the car down, certainly. It will also speed you up. You may even have time to calculate that the sum of the car’s momentum and yours before the collision remains the same after, given that the car has slowed and you are suddenly flying rapidly through the air. This applies in reverse, too. Imagine going out for a run, turning a blind corner and smacking hard into a stationary SUV. As you fall to the ground in agony, your momentum sinks to zero. But console yourself by remembering that you have in effect transferred all of your momentum to the SUV, thus moving it a tiny bit.
In either case, momentum is conserved. The sum of your momentum and the car’s isn’t changed by the collision. In other words, it’s invariant.
Similarly with angular momentum, which applies to objects that rotate. It’s defined as the momentum (as above) of the moving object, multiplied by the radius of the circle it traces as it rotates. The law explains why a figure skater doing a pirouette might start by stretching out her arms, then pull them in so she can whirl faster and faster. It explains why Venus travels faster on its path around the Sun than our Earth does, even though both planets are nearly the same size. Or why a spinning top stays upright. Or why, if you tie a stone to a string and whirl it around your head, it moves faster as you shorten the string.
In each case, again, angular momentum is conserved. Invariance, again.
Understanding such invariance about objects in motion helps us understand the working of our solar system, or the flight of an aircraft, or any number of other phenomena around us. It was crucial in planning the paths of Indian Space Research Organisation’s Chandrayaan and Mangalyaan missions. (Regular readers of this column will remember I used the analogy of the whirling stone to explain their orbits). This is why invariance is so fundamental, so invariably fascinating.
Though I don’t know if saying all this to the folks who were so indifferent to my card trick the other day would have changed anything. Invariantly unsmiling, all of them.
Once a computer scientist, Dilip D’Souza now lives in Mumbai and writes for his dinners. His Twitter handle is @DeathEndsFun | 1,962 | 8,867 | {"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} | 3.40625 | 3 | CC-MAIN-2021-04 | latest | en | 0.952534 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1512/2/s/a/ | 1,603,426,026,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00572.warc.gz | 800,137,578 | 44,124 | # Properties
Label 1512.2.s.a Level $1512$ Weight $2$ Character orbit 1512.s Analytic conductor $12.073$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1512 = 2^{3} \cdot 3^{3} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1512.s (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$12.0733807856$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{25}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -4 \zeta_{6} q^{5} + ( 3 - 2 \zeta_{6} ) q^{7} +O(q^{10})$$ $$q -4 \zeta_{6} q^{5} + ( 3 - 2 \zeta_{6} ) q^{7} + ( 2 - 2 \zeta_{6} ) q^{11} + 5 q^{13} + ( 6 - 6 \zeta_{6} ) q^{17} + 4 \zeta_{6} q^{19} + 6 \zeta_{6} q^{23} + ( -11 + 11 \zeta_{6} ) q^{25} -6 q^{29} + ( 7 - 7 \zeta_{6} ) q^{31} + ( -8 - 4 \zeta_{6} ) q^{35} -7 \zeta_{6} q^{37} -2 q^{41} -7 q^{43} -2 \zeta_{6} q^{47} + ( 5 - 8 \zeta_{6} ) q^{49} + ( 6 - 6 \zeta_{6} ) q^{53} -8 q^{55} + ( 6 - 6 \zeta_{6} ) q^{59} + 9 \zeta_{6} q^{61} -20 \zeta_{6} q^{65} + ( 7 - 7 \zeta_{6} ) q^{67} + 8 q^{71} + ( -10 + 10 \zeta_{6} ) q^{73} + ( 2 - 6 \zeta_{6} ) q^{77} -\zeta_{6} q^{79} -14 q^{83} -24 q^{85} + 12 \zeta_{6} q^{89} + ( 15 - 10 \zeta_{6} ) q^{91} + ( 16 - 16 \zeta_{6} ) q^{95} -15 q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 4q^{5} + 4q^{7} + O(q^{10})$$ $$2q - 4q^{5} + 4q^{7} + 2q^{11} + 10q^{13} + 6q^{17} + 4q^{19} + 6q^{23} - 11q^{25} - 12q^{29} + 7q^{31} - 20q^{35} - 7q^{37} - 4q^{41} - 14q^{43} - 2q^{47} + 2q^{49} + 6q^{53} - 16q^{55} + 6q^{59} + 9q^{61} - 20q^{65} + 7q^{67} + 16q^{71} - 10q^{73} - 2q^{77} - q^{79} - 28q^{83} - 48q^{85} + 12q^{89} + 20q^{91} + 16q^{95} - 30q^{97} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1512\mathbb{Z}\right)^\times$$.
$$n$$ $$757$$ $$785$$ $$1081$$ $$1135$$ $$\chi(n)$$ $$1$$ $$1$$ $$-\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
865.1
0.5 + 0.866025i 0.5 − 0.866025i
0 0 0 −2.00000 3.46410i 0 2.00000 1.73205i 0 0 0
1297.1 0 0 0 −2.00000 + 3.46410i 0 2.00000 + 1.73205i 0 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1512.2.s.a 2
3.b odd 2 1 1512.2.s.j yes 2
7.c even 3 1 inner 1512.2.s.a 2
21.h odd 6 1 1512.2.s.j yes 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
1512.2.s.a 2 1.a even 1 1 trivial
1512.2.s.a 2 7.c even 3 1 inner
1512.2.s.j yes 2 3.b odd 2 1
1512.2.s.j yes 2 21.h odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(1512, [\chi])$$:
$$T_{5}^{2} + 4 T_{5} + 16$$ $$T_{11}^{2} - 2 T_{11} + 4$$ $$T_{13} - 5$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ 1
$3$ 1
$5$ $$1 + 4 T + 11 T^{2} + 20 T^{3} + 25 T^{4}$$
$7$ $$1 - 4 T + 7 T^{2}$$
$11$ $$1 - 2 T - 7 T^{2} - 22 T^{3} + 121 T^{4}$$
$13$ $$( 1 - 5 T + 13 T^{2} )^{2}$$
$17$ $$1 - 6 T + 19 T^{2} - 102 T^{3} + 289 T^{4}$$
$19$ $$1 - 4 T - 3 T^{2} - 76 T^{3} + 361 T^{4}$$
$23$ $$1 - 6 T + 13 T^{2} - 138 T^{3} + 529 T^{4}$$
$29$ $$( 1 + 6 T + 29 T^{2} )^{2}$$
$31$ $$( 1 - 11 T + 31 T^{2} )( 1 + 4 T + 31 T^{2} )$$
$37$ $$1 + 7 T + 12 T^{2} + 259 T^{3} + 1369 T^{4}$$
$41$ $$( 1 + 2 T + 41 T^{2} )^{2}$$
$43$ $$( 1 + 7 T + 43 T^{2} )^{2}$$
$47$ $$1 + 2 T - 43 T^{2} + 94 T^{3} + 2209 T^{4}$$
$53$ $$1 - 6 T - 17 T^{2} - 318 T^{3} + 2809 T^{4}$$
$59$ $$1 - 6 T - 23 T^{2} - 354 T^{3} + 3481 T^{4}$$
$61$ $$1 - 9 T + 20 T^{2} - 549 T^{3} + 3721 T^{4}$$
$67$ $$1 - 7 T - 18 T^{2} - 469 T^{3} + 4489 T^{4}$$
$71$ $$( 1 - 8 T + 71 T^{2} )^{2}$$
$73$ $$( 1 - 7 T + 73 T^{2} )( 1 + 17 T + 73 T^{2} )$$
$79$ $$1 + T - 78 T^{2} + 79 T^{3} + 6241 T^{4}$$
$83$ $$( 1 + 14 T + 83 T^{2} )^{2}$$
$89$ $$1 - 12 T + 55 T^{2} - 1068 T^{3} + 7921 T^{4}$$
$97$ $$( 1 + 15 T + 97 T^{2} )^{2}$$ | 2,384 | 4,757 | {"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} | 2.703125 | 3 | CC-MAIN-2020-45 | latest | en | 0.298676 |
https://discourse.shapr3d.com/t/moving-a-simple-shape/1128 | 1,718,590,128,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00121.warc.gz | 175,552,462 | 7,415 | # Moving a simple shape
I’m hoping I’m just overlooking this simple function. If I draw a circle and want to simply move it a few mm in one direction, I thought I could just do this by selecting the circle and dragging it into position with the pencil (see 13:25 in this tutorial video: https://youtu.be/OFsbZikomSA). However, I don’t seem to be able to move the circle after drawing it. I have to use the free form transform tool which seems excessive. If I tap the circle and try to move it all that happens is another line is drawn.
How do I move a shape with the pencil by dragging it around?
Select the sketch with the pencil. Then you can simply drag it’s points to position them.
So it’s not possible to grab a circle shape and drag it at its current size to somewhere else in the workspace? I’ve figured out that I can change its radius and circumference by dragging the two points that appear when you tap it. Several of your tutorial videos though show the whole circle being moved as one to reposition it. That does seem much more intuitive. Has this changed?
If you lock the radius of the circle, then if you drag a point of the circle it will move the circle. Otherwise (since the circle is defined by two points) you will change the position of one point, and the circle will be recalculated from the points.
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Makes perfect sense. Thank you.
However woth the next release this is going to change, and when you drag a selected sketch, it will be moved, instead of being extruded.
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If I may just add a thought to this discussion:
… It would be very helpful if a circle could be selected [for Transform/Translate operations] by its centre.
… and, incidentally; the ability to select a line by its midpoint would also be convenient.
.
MichaelG.
[as a new user, I may have ‘missed a trick’ … if so, please educate me]
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Works for a simple circle b/c you can select/lock radius, but how do you simply move an entire extruded shape? I can see no easy way to do this other than ‘transform’ which is not a simple process. I move shapes, parts, etc. constantly and it seems there should be a way to intuitively grab and drag these shapes I’ve created. Am I missing something? Can’t find a tutorial for this?
There are 4 different ways to move an object:
• general transform, using the gizmo (arrows and rotation arrows)
• translate tool (by defining a start and end point of the translation)
• rotate tool (by defining an axis and an angle of the rotation)
• mirror tool (by defining the plane of the reflection)
There is no other way to move an object in Shapr. What you want is unfortunately under-defined, since you it would require moving an object in 3D space on a 2D screen.
Thanks - transform/translate works, but it seems very imprecise. How do I pick fixed coordinates to move the shape exactly? Or failing that, How can i grab a shape and hover with it until I can place it where I want it? (I do this in Sketchup rather easily)
Also, I find it’s a bit the same with drawing shapes. It would be ideal if shapr had a rectangle function instead of having to draw 4 individual lines. (?!)
And if I could snap an additional, smaller rectangle to the same starting coordinates, i could really make better use of the extrude function.
I realize Shapr is not Sketchup, but these things are easy in SU.
I like the app, but these are the deal breakers for me when it comes to buying a subscription. Thanks again for your help.
OK - my bad, I think I got it, you have to double tap screen and transform in 2D, and then you can pinpoint where to move shapes and then pull back to 3D to view. It’s certainly not intuitive, but works very well.
Figured it out thanks. See my last post. Thanks anyway
Not intuitive at all sadly. Still can’t figure out how to move sketch object on the ipad. - figured it out biggest problem is you cant see what sketches are combined. | 895 | 3,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.943562 |
https://math.stackexchange.com/questions/89272/possible-mistake-in-krauses-localization-theory | 1,624,065,890,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00169.warc.gz | 348,648,281 | 41,309 | # Possible mistake in Krause's “Localization Theory”
I'm a little confused about a proof in this note by H. Krause; see page 12, Prop. 3.5.1. Can you explain me how the $\beta_i$'s are determined? I can't catch how to choose them in order to coequalize both $\alpha_i'$ and $\alpha_i''$, and I also think there's a mistake in the criterion for a morphism to be in the saturation of $\Sigma$... I'll also welcome any other reference for this proposition.
• I think I'm able to prove it in a much simpler way than Krause's one; I'm just looking for a check on your experience, and considering this result turns out to be quite useful... Thanks! – fosco Dec 10 '11 at 13:34
For context, the question is about the following statement:
Suppose $\mathcal{C}$ is a category with small coproducts and that $\Sigma$ is a left multiplicative subset closed under taking coproducts, that is to say $\Sigma$ satisfies:
1. (Nontriviality and closure under composition): $\Sigma$ contains all identity morphisms and is closed under taking compositions.
2. (Ore condition): If $X'\; \xleftarrow{\sigma} \; X \; \xrightarrow{\alpha} \; Z$ is such that $\sigma \in \Sigma$ then there exists a commutative square
with $\sigma' \in \Sigma$.
Note: If in addition $\alpha \in \Sigma$ then the diagonal $\sigma'\alpha: X \to Y'$ is a morphism in $\Sigma$ by 1.
3. (Cancellation condition) If $\alpha,\beta: X \to Y$ are parallel morphisms for which there exists $\sigma: X' \to X$ in $\Sigma$ such that $\alpha\sigma = \beta\sigma$ then there exists $\sigma': Y \to Y'$ in $\Sigma$ such that $\sigma'\alpha = \sigma'\beta$
4. (Closure under coproducts): If $\sigma_i : X_i \to Y_i$ belong to $\Sigma$ then so does $\coprod \sigma_i : \coprod X_i \to \coprod Y_i$.
Then the category of left fractions $\mathcal{C}[\Sigma^{-1}]$ has coproducts and the quotient functor $q: \mathcal{C} \to \mathcal{C}[\Sigma^{-1}]$ preserves coproducts.
The thing that needs to be shown is that the canonical map $$\tag{3.5.1} \mathcal{C}[\Sigma^{-1}]\left( \coprod X_i, Y\right)\; \xrightarrow{f}\; \prod\mathcal{C}[\Sigma^{-1}]\left( X_i, Y \right)$$ is a bijection. Krause first establishes surjectivity of $f$ and the point of contention is its injectivity.
I'll first fill in the details in Krause's argument (which is perfectly okay) because I think that nothing but straightforward manipulations of fractions is needed and getting used to those is a necessity if one wants to understand any of this.
## 1. Constructing the $\beta_i$'s
Given two left fractions $\coprod X_i \;\xrightarrow{\alpha^\prime} \; Z^\prime \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^\prime}\; Y$ and $\coprod X_i \;\xrightarrow{\alpha^{\prime\prime}} \; Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y$ that are identified under $f$ we want to show that the fractions are equivalent.
To say that the images of these two fractions under $f$ are identified is to say that for each $i$ the fractions $X_i \;\xrightarrow{\alpha_{i}^\prime} \; Z^\prime \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^\prime}\; Y$ and $X_i \;\xrightarrow{\alpha_{i}^{\prime\prime}} \; Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y$ are equivalent.
Notice that the diagram $Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y \; \xrightarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime}}\;Z'$ appears in all the fractions for all $i$. Applying the Ore condition to this wedge we get a morphism $Y \; \xrightarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma}\; Z$ of the form $\sigma= \tau' \sigma' = \tau''\sigma'' \in \Sigma$.
The diagram
exhibits the fraction $X_{i} \; \xrightarrow{\tau'\alpha_{i}'} \; Z \; \xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma} \; Y$ as equivalent to $X_i \;\xrightarrow{\alpha_{i}^\prime} \; Z^\prime \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^\prime}\; Y$. Similarly, $X_{i} \; \xrightarrow{\tau''\alpha_{i}^{\prime\prime}} \; Z \; \xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma} \; Y$ is equivalent to $X_i \;\xrightarrow{\alpha_{i}^{\prime\prime}} \; Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y$.
At this point Krause replaces $\tau'\alpha_{i}^\prime$ by $\alpha_{i}^\prime$ and similarly for $\alpha_{i}^{\prime\prime}$ — this is justified by his sentence “We may assume that $Z^\prime = Z = Z^{\prime\prime}$ and $\sigma^\prime = \sigma = \sigma^{\prime\prime}$ since […]” — to avoid confusion we won't do this here.
Since equivalence of fractions is transitive (it is an equivalence relation — check this!), we conclude that the two fractions involving $Z$ are equivalent. But this means that we have a commutative diagram
From $\beta_{i}^\prime\sigma = \beta_{i}^{\prime\prime}\sigma$ we conclude with the cancellation condition that there is $\sigma_{i}^\prime: \tilde{Z}_i \to Z_i$ in $\Sigma$ such that $\sigma_{i}^\prime \beta_{i}^\prime = \sigma_{i}^{\prime}\beta_{i}^{\prime\prime}$. Now put $\beta_i = \sigma_{i}^\prime \beta_{i}^\prime = \sigma_{i}^{\prime}\beta_{i}^{\prime\prime}$ and observe that $\beta_{i} \tau'\alpha_{i}^\prime = \beta_{i}\tau^{\prime\prime}\alpha_{i}^{\prime\prime}$, as well as $\beta_i \sigma = \sigma_{i}^\prime (\beta_{i}^\prime \sigma) \in \Sigma$, as claimed by Krause (remembering the paragraph in italics above).
Added: To see that $\beta_i$ is indeed in the saturation of $\Sigma$ (see also the next section of the answer), we need only show that we can postcompose it with a morphism $\phi_i$ such that $\phi_i\beta_i \in \Sigma$. Here's one (not particularly elegant) way to do it:
Apply the Ore condition to the wedge $Z \; \xleftarrow{\sigma} \; Y \; \xrightarrow{\sigma_{i}^\prime \tilde{\sigma}_i} \; Z_i$ to get the commutative rectangle on the left below
The dotted diagonal arrow can be either one of $\beta_{i}^\prime$ or $\beta_{i}^{\prime\prime}$ and it doesn't make the diagram commutative. Nevertheless, $\tau_i \beta_i = \tau_i \sigma_{i}^\prime\beta_{i}^\prime$ and $\upsilon_i$ are two morphisms $Z \to \tilde{W}_i$ such that $\tau_i \beta_i\sigma = \upsilon_i \sigma$. Thus, by cancellation, we can find $\tau_{i}^\prime: \tilde{W}_i \to W_i$ in $\Sigma$ such that $\tau_{i}^\prime\tau_i\beta_i = \tau_{i}^\prime \upsilon \in \Sigma$ and thus $\phi_i = \tau_{i}^\prime \tau_i$ does what we want.
## 2. The criterion for a morphism to be in the saturation of $\Sigma$.
By definition, a morphism $\phi \in \mathcal{C}$ belongs to the saturation $\overline{\Sigma}$ of $\Sigma$ if and only if the localization functor $q: \mathcal{C} \to \mathcal{C}[\Sigma^{-1}]$ sends $\phi$ to an invertible morphism.
Claim. A morphism $\phi$ is in the saturation of $\Sigma$ if and only if there are morphisms $\phi'$ and $\phi''$ such that $\phi'' \phi \in \Sigma$ and $\phi \phi' \in \Sigma$.
Indeed, assume there are such $\phi'$ and $\phi''$. Then $(\phi''\phi)^{-1} \phi''$ is a left inverse of $\phi$ and $\phi'(\phi\phi')^{-1}$ is a right inverse of $\phi$ in $\mathcal{C}[\Sigma^{-1}]$, so $\phi$ is invertible, too.
Recall the explicit form of $q(\phi) = [\phi, 1]$. To say that $[\phi,1]$ has a left inverse is to say that there is a morphism $[\alpha,\sigma]$ such that $[\alpha,\sigma][\phi,1] \sim [1_A,1_A]$. Recalling the definition of the composition in the category of fractions we see that $[\alpha,\sigma][\phi,1] = [\alpha\phi,\sigma]$ and to say that the latter is equivalent to $[1,1]$ is to give a commutative diagram
telling us that $(\beta \alpha)\phi = \sigma^\prime \in \Sigma$ so that we can take $\phi^{\prime\prime} = \beta\alpha$. Similarly for the proof of existence of $\phi^\prime$.
## 3. End of the argument
With these arguments at hand, I think it is safe to refer back to Krause's notes, which I reproduce for the convenience of the reader — (3.5.1) is the same as the displayed equation above:
## 4. Alternative proof and some references
The classic reference for the calculus of fractions is Gabriel-Zisman, Calculus of Fractions and Homotopy Theory, but it is relatively terse. A detailed exposition can be found in Schubert, Categories, chapter 19 (at least in the German edition). For those who want to verify the basics I always recommend having pages 300 ff. of Lam's Lectures on modules and rings nearby because the arguments given there can be easily generalized from rings to categories by keeping track of source and target of the morphisms.
As for an alternative proof of the proposition you ask about, you can appeal to the fact that the morphism classes in $\mathcal{C}[\Sigma^{-1}]$ can be described as filtered colimits and keeping track of those. This is done in the proof of Theorem 19.2.8 of (the German edition of) Schubert's book. Filling in the details doesn't seem easier to me than Krause's argument, though, as the verification of all the naturality assertions involves precisely the same arguments, but hidden in a formalism. However, this is probably a matter of taste…
• Thanks a lot! I spent some time in reading your answer in order to give you a sensible comment. Maybe you reversed the order of the composition $\tau'\alpha',\tau''\alpha''$ when you composed with $\beta_i$; in the second equivalence of squares, $\beta_i'\tau_i'\alpha_i'=\tau_i'\alpha_i'$ is clearly a mistake. Finally I can't see where is a $\tau$ such that $\tau\beta_i\in \Sigma$. Thanks for your patience and complete answer! – fosco Dec 19 '11 at 18:23
• You're absolutely right about the reversal of the order and the other typo. I'll fix it and address your other question tomorrow. It's getting too late now. I'll ping you when it's done. – t.b. Dec 19 '11 at 21:21
• @tetrapharmakon: I've corrected a few typos and added an argument to the answer. I hope it's okay now. – t.b. Dec 20 '11 at 1:40
• Yeah, now it seems all right! Thanks! – fosco Dec 20 '11 at 5:57 | 2,976 | 9,919 | {"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} | 2.90625 | 3 | CC-MAIN-2021-25 | latest | en | 0.857946 |
https://www.aqua-calc.com/calculate/mole-to-volume-and-weight/substance/rubidium-blank-chloride-coma-and-blank-solid | 1,702,101,412,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00760.warc.gz | 707,754,148 | 8,056 | # Moles of Rubidium chloride, solid
## rubidium chloride, solid: convert moles to volume and weight
### Volume of 1 mole of Rubidium chloride, solid
centimeter³ 43.81 milliliter 43.81 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.19 inch³ 2.67 US fluid ounce 1.48 liter 0.04 US gallon 0.01 meter³ 4.38 × 10-5 US pint 0.09 metric cup 0.18 US quart 0.05 metric tablespoon 2.92 US tablespoon 2.96 metric teaspoon 8.76 US teaspoon 8.89
### Weight of 1 mole of Rubidium chloride, solid
carat 604.61 ounce 4.27 gram 120.92 pound 0.27 kilogram 0.12 tonne 0 milligram 120 921
### The entered amount of Rubidium chloride, solid in various units of amount of substance
centimole 100 micromole 1 000 000 decimole 10 millimole 1 000 gigamole 1 × 10-9 mole 1 kilogram-mole 0 nanomole 1 000 000 000 kilomole 0 picomole 1 000 000 000 000 megamole 1 × 10-6 pound-mole 0
#### Foods, Nutrients and Calories
CREAMY GERMAN STYLE QUARK, UPC: 854263004165 weigh(s) 159 grams per metric cup or 5.3 ounces per US cup, and contain(s) 67 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
1937 foods that contain Total sugar alcohols. List of these foods starting with the highest contents of Total sugar alcohols and the lowest contents of Total sugar alcohols
#### Gravels, Substances and Oils
Sand, Coral weighs 1 576 kg/m³ (98.38647 lb/ft³) with specific gravity of 1.576 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Soda ash, heavy (bulk) [Na2CO3] weighs 1 040 kg/m³ (64.92508 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-22, liquid (R22) with temperature in the range of -51.12°C (-60.016°F) to 71.12°C (160.016°F)
#### Weights and Measurements
A kilogram force meter (kgf-m) is a non-SI (non-System International) measurement unit of torque.
A force that acts upon an object can cause the acceleration of the object.
t/m³ to µg/pt conversion table, t/m³ to µg/pt unit converter or convert between all units of density measurement.
#### Calculators
Cubic equation calculator. Real and complex roots of cubic equations | 717 | 2,408 | {"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-2023-50 | latest | en | 0.59488 |
https://www.airmilescalculator.com/distance/dme-to-vko/ | 1,696,334,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00763.warc.gz | 669,756,516 | 12,449 | # How far is Moscow from Moscow?
The distance between Moscow (Moscow Domodedovo Airport) and Moscow (Vnukovo International Airport) is 28 miles / 46 kilometers / 25 nautical miles.
The driving distance from Moscow (DME) to Moscow (VKO) is 36 miles / 58 kilometers, and travel time by car is about 52 minutes.
28
Miles
46
Kilometers
25
Nautical miles
## Distance from Moscow to Moscow
There are several ways to calculate the distance from Moscow to Moscow. Here are two standard methods:
Vincenty's formula (applied above)
• 28.299 miles
• 45.543 kilometers
• 24.591 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 28.215 miles
• 45.408 kilometers
• 24.518 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Moscow to Moscow?
The estimated flight time from Moscow Domodedovo Airport to Vnukovo International Airport is 33 minutes.
## What is the time difference between Moscow and Moscow?
There is no time difference between Moscow and Moscow.
## Flight carbon footprint between Moscow Domodedovo Airport (DME) and Vnukovo International Airport (VKO)
On average, flying from Moscow to Moscow generates about 29 kg of CO2 per passenger, and 29 kilograms equals 64 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Moscow to Moscow
See the map of the shortest flight path between Moscow Domodedovo Airport (DME) and Vnukovo International Airport (VKO).
## Airport information
Origin Moscow Domodedovo Airport
City: Moscow
Country: Russia
IATA Code: DME
ICAO Code: UUDD
Coordinates: 55°24′31″N, 37°54′22″E
Destination Vnukovo International Airport
City: Moscow
Country: Russia
IATA Code: VKO
ICAO Code: UUWW
Coordinates: 55°35′29″N, 37°15′41″E | 511 | 2,059 | {"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-2023-40 | latest | en | 0.875472 |
http://www.math.sc.edu/course-descriptions-math-1xx-through-6xx | 1,493,402,694,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123046.75/warc/CC-MAIN-20170423031203-00108-ip-10-145-167-34.ec2.internal.warc.gz | 600,769,278 | 18,074 | College of Arts & SciencesDepartment of Mathematics
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Scilab manual >> Linear Algebra > ereduc
# ereduc
computes matrix column echelon form by qz transformations
### Calling Sequence
`[E,Q,Z [,stair [,rk]]]=ereduc(X,tol)`
### Arguments
X
m x n matrix with real entries.
tol
real positive scalar.
E
column echelon form matrix
Q
m x m unitary matrix
Z
n x n unitary matrix
stair
vector of indexes,
*
`ISTAIR(i) = + j` if the boundary element `E(i,j)` is a corner point.
*
`ISTAIR(i) = - j` if the boundary element `E(i,j)` is not a corner point.
`(i=1,...,M)`
rk
integer, estimated rank of the matrix
### Description
Given an `m x n` matrix `X` (not necessarily regular) the function ereduc computes a unitary transformed matrix `E=Q*X*Z` which is in column echelon form (trapezoidal form). Furthermore the rank of matrix `X` is determined.
### Examples
```X=[1 2 3;4 5 6]
[E,Q,Z ,stair ,rk]=ereduc(X,1.d-15)```
### Authors
Th.G.J. Beelen (Philips Glass Eindhoven). SLICOT
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# Double Refraction in Water Bottle
## Objective
See double refraction in water bottle.
## Introduction
Double refraction is a property of anisotropic material which offers different refractive indices for different orientations of E-field of light (birefringence). In such cases, often an incident beam splits in two parts on entering the medium. In this experiment it appears that a beam incident on water in a bottle splits into two on entering the water mass. But it is certainly not the case of double refraction in its true meaning.
## Procedure
1. Take water in the bottle up to about 3/4 of height. Put one or two drops of dettol in this water. Place the bottle on a horizontal surface.
2. Take a laser torch and let the laser fall on the water surface obliquely. It will refract in water and the path will be visible from the side.
3. Adjust the position of the laser so that it falls right at the junction of the glass surface and the water surface. Do fine adjustments and you will see two refracted beams!
Question: Are both the beams straight? Draw a diagram showing how the two beams appear.
## Discussion
The water surface near the glass surface gets curved due to surface tension. The angle of incidence for the part of the beam on this part of water surface is different from what the beam makes with the flat part of the water surface. This gives appearance of two refracted beams. | 320 | 1,477 | {"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-2020-45 | latest | en | 0.91892 |
https://en.wikipedia.org/wiki/Talk:H-theorem | 1,521,699,210,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647777.59/warc/CC-MAIN-20180322053608-20180322073608-00337.warc.gz | 558,693,906 | 28,056 | # Talk:H-theorem
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## Origin of the H formula ?
how did boltzmann come up with his definition for H? (Yosofun 06:42, 12 Apr 2005 (UTC))
## Fluctuation theorem
I'm cutting the following line:
The final resolution of the time reversibility paradox is found in the Fluctuation Theorem.
There is stuff which could be said about the Fluctuation Theorem as a modern generalisation of work on the prediction of entropy increase.
But the FT has nothing to say about the Arrow of time, because, like Boltzmann's H-theorem, the entropy increase it predicts so well arises because one assumes one can ignore correlations and similar 'hidden information' at times after the initial time. If you start with a non-equilibrium state and try to use the Fluctuation Theorem to predict backwards how that non-equilibrium state came to be, it will predict that it arose spontaneously as a fluctuation from an equilibium state -- because those are the only possibilities it knows about.
The cause of time asymmetry in the observed universe must be looked for elsewhere. Jheald 12:28, 21 October 2005 (UTC)
## Hi everybody!
While reading my personal copy of Tolman's book and I saw the chapter on H - theorem The Shannon information entropy, which is talked about in another book I have been reading by Feynman called lectures on computation contains a lot about Shannon's ideas. I find this subject interesting and I can help with the article with some inline citations if I can find something that Toleman said in his book inside of the article that needs an inline source
According to the history of entropy article Shannon information entropy is related to "H" and interestingly Shannon recognized his same formula by reading the same book I am reading now Toleman.
By the way Jheald, nice to see you have been working on this article! Also thanks so much for recommending the Penrose book, I have been studying it very intently. I am kinda stuck on chapter 12 but I have been skipping ahead and reading the later chapters as well.
Penrose seems to talk about just about everything in his book, but don't remember H theorem, and yes I have looked in the index but was wondering if it was called something else.
Hobojaks (talk) 07:00, 22 February 2009 (UTC)
Note that Tolman was written in 1938, though it's still an outstanding book. Shannon only widely introduced the idea of the Shannon entropy ten years later. Relationships between the two are touched on here in the Shannon entropy article. They are still the subject of some controversy. The link between the two ideas has been propounded most vigorously by E.T. Jaynes and his subsequent followers, from 1957 onwards; others still decry it, though as early as 1930 G.N. Lewis had written that "Gain in entropy always means loss of information, and nothing more".
Penrose touches on the idea of the H-theorem, and the related issue of Loschmidt's paradox in section 27.5 (page 696 in the UK hardback); though he doesn't mention either by name, essentially takes the idea of coarse graining as read, and quickly moves on to the gravitational and cosmological issues he's most interested in.
H.D. Zeh's The Physical Basis of the Direction of Time, now into its fifth edition, is often given as the standard overview reference in this area; though note that it's written at a professional level, not a popular one. Chapter 3 looks at thermodynamics, before the book gets into decoherence and then black hole and cosmological thermodynamics. Jheald (talk) 11:03, 22 February 2009 (UTC)
But I have to say it is a pretty hard book to digest in a short while. I get the feeling that I am going to be reading it for quite some time, because I find the section on quantum mechanics particularly difficult, and yet required to understand some of his most profound ideas.
I found an article on Boltzmann_entropy which I am starting to understand is related to H.
The citations I found from Tolman so far, don't need to go in the first paragraph if this makes the first paragraph seem to wordy, but I think the actual page numbers and wording might actually belong someplace at least in the article?
One thing that I have found in my experience to be true is that citations take more work than typing, so maybe the main contribution here is providing actual inline page numbers to a source for some of the material.
Um, I don't want to sound overbearing, and it's really good the way you're working through this; but can I suggest that you wait until you understand what is in the article first, so that you're confident you have a solid grasp of the big picture first, before you start trying to rewrite it?
While you're getting there, questions like "I don't understand this" or "This doesn't seem quite right to me" or "Why is the article saying that" are maybe more useful than trying to rewrite it as you're going along.
If you have access to a good university library, you also might find that some more modern "introduction to statistical mechanics" books may give a quicker, more accessible presentation. It's always good to have a look at what more than one source of exposition is saying (and the references would be good for the article, too!) Jheald (talk) 21:35, 24 February 2009 (UTC)
I'm not saying there isn't a lot wrong with this article, because there most definitely is. But I do think the best way to fix it is from the big picture first; i.e. to do some fairly comprehensive reading first, and work out what the most important things we want to say are first, rather than starting in on it piecemeal. Jheald (talk) 22:30, 24 February 2009 (UTC)
I agree to this methodology. However this suggests a more major undertaking than I had originally planned. I had at first thought to work on providing inline citations to the existing article with possibly some slight rewordings to bring the wikipedia text into agreement with those sources. That is OK though because the more I learn about H theorem the more interesting it becomes. Also the concept of the H theorem is deeper and more complicated than I first thought it would be! I especially appreciate your help in working through this. I am sure that there are both passive current and future readers of this discussion who are, or will be equally grateful for your contributions.
Hobojaks (talk) 21:08, 25 February 2009 (UTC)
### Some material for possible inclusion
In classical thermodynamics, the H-theorem was a thermodynamic principle first introduced by Boltzmann in 1872. The H-theorem demonstrates the tendency of the molecules of a system to approach their equilibrium Maxwell-Boltzmann distribution[1]. The quantity called H by Boltzmann can be regraded as an appropriate measure of the extent to which the condition of the system deviates from that corresponding to equilibrium[2]. This quantity H has a tendency to decrease with time to a minimum and thus for the system to approach its equilibrium condition.[3].
The H-theorem has been restated in quantum mechanical form.[4]
The decrease in the quantity H is related to the increase in the entropy[citation needed]. For example the approach to equilibrium of an ideal gas in an irreversible process, can be demonstrated using an approach based on the Boltzmann equation[citation needed].
It appears to predict an irreversible increase in entropy, despite microscopically reversible dynamics. This has led to much discussion.
ergodic hypothesis[5] —Preceding unsigned comment added by 76.191.171.210 (talk) 07:23, 24 February 2009 (UTC)
Hobojaks (talk) 19:41, 23 February 2009 (UTC)
### Problem list
The quantity H is defined as the integral over velocity space :
${\displaystyle \displaystyle H\ {\stackrel {\mathrm {def} }{=}}\ \int {P({\ln P})d^{3}v}=\left\langle {\ln P}\right\rangle }$ (1)
where P(v) is the probability. H is a forerunner of Shannon's information entropy.
(1) Exactly what is velocity space would it be better to call it phase space
• The integral isn't over phase space. That was an adaptation suggested by Gibbs some years later. (Tolman p. 165 and onwards). Boltzmann's original integral was over the 3-dimensional space of possible molecular velocities, not the 6N-dimensional phase space.
Actually I have managed to find somewhat of a source for this. But it is only in a problem set. The source is Fundamentals of statistical and thermal physics, Rieff 1965, The index lists two places were it talks about H theorem the first is in a problem set. These problems do define H as an integral containing a ${\displaystyle d^{3}}$ term. But it is multiplied by a distribution function, times the log of a distribution function, which is just a little different from a probability. So still no source for the current formula, but possibly just the slightest hope for it?Hobojaks (talk) 04:05, 27 February 2009 (UTC)
• On further consideration, it looks like this may well be wrong. Tolman presents Boltzmann as working in a 6-dimensional "μ-space". (Not the 6N-dimensional phase space, though).
On the other hand, the key input to the H-theorem is the time-reversibility of the transition coefficients. I don't think that depends on position, so it may not be impossible that Boltzmann first presented the derivation purely for the velocity distribution. We would need to check a reference that actually goes into the detail of the development of Boltzmann's thought to be sure - probably something by S. Brush. Jheald (talk) 22:12, 24 February 2009 (UTC)
I have been studying Toleman pg 45 as near as I can tell, "μ-space" is one of two flavors of phase space which he draws a distinction between. "...The phase space for any individual kind of element (molecule) is called the μ-space for that species of element." The other flavor of phase space is gama-space. According to Toleman the key distinction of "μ-space" is that if you switch two identical molecules this does not really count as two different states in "μ-space", but it would gama-space.Hobojaks (talk) 23:20, 25 February 2009 (UTC) I have discovered that in addition to page 45 the whole section 27 starting on page 74 is dedicated to explaining "μ-space". In section 27 it is made clear that "μ-space" is the phase space of a single molecule, and that a gamma-space for the whole system of many identical molecules can be made up by combining all the μ-spaces of the individual particles. He then explains a procedure for eliminating all the permutations that come about by switching two identical particles, adding that this concept will be of great importance when switching to quantum mechanics. μ-space in the case of a single molecule of a mono atomic gas would be six dimensional but in general the μ-space of a molecule would have 2f dimensions corresponding the the each of the generalized coordinates and generalized momentum for each degree of freedom p & q Hobojaks (talk) 08:43, 26 February 2009 (UTC)
1(a) Also a problem with the formula in question is that it says that P is the probability but it does not say the probability of what. I have found a citation in Rief Fundamentals of Statistical and thermal physics which may be relevant. This formula is for a limited special case of an ideal mono-atomic gas, and the problem tells the student to assume that for f they can use the equilibrium Maxwell distribution.
${\displaystyle H=\int d^{3}\mathbf {v} f\,\log \,f}$[6]
The student is then ask to show that H = -S/k where S is the entropy per unit volume of a mono-atomic ideal gas.
Is perhaps the student being guided down the path that Boltzmann once followed. I can't be sure of this, but I can make one observation. There needs to be a symbol for two different versions of H. One for the extensive property of the entire system and one for the intensive property. The H in the specialized Rieff-Boltzmann H theorem is the H per unit volume, and clearly not the H for the 'system' as a whole. This is a point that I struggled with considerably, and would like to get conformation that I am correct on?
In order to prevent future readers of this article from suffering the difficulty I have suffered I propose that we develop some notation to distinguish between H and H per unit volume in the article, and clearly indicate which one we are talking about each time we indicate a formula!!!!
Note that Toleman's formulas for H also use the notation were there is an f function, but where f is more generalized.
Perhaps a substitution were f is substituted for P and it is clearly stated that f is the equilibrium Maxwell distribution, and that the formula is specialized is in order. Note that the start of the article gives a very restricted definition for the H theorem, which is no longer true of the more general presentation the article currently gives.
(2) I can't find this exact formula for H anywhere in Tolman, who lists a number of different forms. It looks like there is a strange cube term on the differential operator, like it is the third derivative of v, which I guess is velocity space, the article really does not say.
• It's maybe an evolution of notation since Tolman's time, but the meaning is quite clear. The d3 v is being used as a shorthand for dv1 dv2 dv3.
(3) Shouldn't there be multiple integration symbols. ${\displaystyle \int ...\int }$
• Again, it's an evolution of notation. The integral is understood to be an integral over volume elements d3 v, rather than considering it a series of line integrals. The use of a single integral sign to indicate an integral over all the elements of a set is not unusual.
(4) Would it be better to have ${\displaystyle \delta p\,}$ and ${\displaystyle \delta q\,}$ indicating an inexact differentials of momentum and generalized coordinate? Especially in a variant of the formula that uses a summation rather than an integration?
• Boltzmann's integral wasn't an integral over phase space, it was an integral over velocity space.
(5) As near as I can tell there are at least three different versions of H theorem, at least in the context that Tolman uses the term. There is the original version of Boltzmann then there is the version of Gibbs which modifies Boltzmann's early version. Then there is the quantum mechanical version.
• Yes. This is something the article could perhaps make clearer. The basic structure of the argument is the same in each case however. The quantum mechanical version is (IMO) the one that is most stripped down to fundamentals, and easiest to see what is going on, so in my view it is the right choice that that is the one the article chooses to show how the H theorem comes about.
(6)There is another redefinition of entropy currently underway, that includes the concept that a black hole can have an entropy. Yet this new black hole including version of entropy is in contradiction with H theorem, at least if I understand correctly, but this is neglected in the article.
• See Black hole information paradox, and in particular recent discussions on the talk page there. There is sort of a connection - in both cases information you used to have about the system is being thrown away in your later description of the system, despite a deterministic evolution. With the H-theorem we know how that's happening. With the BH information paradox it is yet quite not so clear. It may be because the BH is not always being described in a fully quantum mechanical way, with a definitely finite number of states; it may also be related to information becoming unobservable in decoherence.
Zeh has written on this and it is covered in his book; but I don't think raising the issue of BH unitarity, which is still quite murky, is helpful in understanding the H-theorem, which is much simpler, and is well understood. It might be worth a mention in the "See also" section, if one could judiciously word the right covering phrase to make the connection. But anything else here would be a unnecessary diversion into confusion, when what we should be trying to do is make the main ideas and issues with the H theorem come over as clearly and simply as we can.
(7)Some older authors (including Tolman) equate H with the macroscopic entropy, it seems this important idea has taken a back seat to the idea of equating H to Shannon entropy? According to Tolman this association is one of the greatest achievements of physics.
• No, I think what Tolman is saying is quite right. The connection with Shannon entropy (for those who accept the connection) just makes the achievement even richer.
Gibbs's version of H (and its quantum analogue) - apart from a minus sign, and a completely irrelevant factor of kB - is essentially the standard, most general expression for microscopic entropy. I would say that the identification of this entropy at a microscopic level with classical thermodynamic entropy, so that one can write
${\displaystyle S=-k\sum p\log p}$
is definitely one of the greatest achievements of physics, without any doubt whatsoever.
If you go with Jaynes then Shannon's work makes Gibbs's formula even more of a fundamental intellectual achievement, because Shannon's work tells us that we can give a precise meaning to what is quite an abstract mathematics. Shannon shows that Gibbs's - Σ p ln p is the natural measure of information. So the H in the formula precisely connects information with classical thermodynamic entropy. The importance of H is precisely that it connects with both.
But classical thermodynamic entropy increases. How can we relate that to an amount of information, and to deterministic dynamics? This is where the H theorem comes in. It shows that if we approximate what's going on in the molecular collisions, or in quantum processes, making the processes probabilistic instead of deterministic because we can't keep track of the full details of the system - i.e. if the only information we can keep track of is in the macroscopic variables, then we systematically lose information; so if we use S to describe classical thermodynamic entropy, which is a state function of the macroscopic variables of the system, S will tend to increase.
So in short: the idea, that S given by the formula above, directly relates to classical thermodynamic entropy, is and remains absolutely fundamental.
The further idea, that that S can naturally be interpreted as the amount of Shannon information, which would be needed to define the microscopic system fully, but which is not conveyed by macroscopic variables alone, is the icing on the cake, which people can take or leave according to taste. Jheald (talk) 21:16, 24 February 2009 (UTC)
I think a lot of the hostility shown towards Shannon's information entropy is a gut feeling that it somehow dilutes Boltzmann's (and Gibbs's) great achievement! It doesn't, of course, as Shannon's contribution is only secondary and derivative :P Physchim62 (talk) 23:59, 25 February 2009 (UTC)
8 There are actually two classical forms of H theorem. The original old school Boltzmann version and the 1902 Gibbs version, based on a different quantity which is essentially the value of double bar H in Toleman's notation. This latter idea is a quantity
"which characterizes the condition of the ensemble of systems which we use as
appropriate for representing the continued behavior of the actual system of interest."(Toleman pg 166)
### Some important formula
It would seem to me that equations 47.5 to 47.9 of on page 135 - 136 of tolemans book in the section titled definition of the quantity H are very important. When I get chance I will start to transcribe these here for possible inclusion. —Preceding unsigned comment added by Hobojaks (talkcontribs) 21:14, 24 February 2009 (UTC)
They are all variants on ${\displaystyle H=\sum p_{i}\log p_{i}}$, which is really the key idea. Best to go with that, I think. Jheald (talk) 22:36, 24 February 2009 (UTC)
A good inline reference for that formula would be (Toleman 1938 pg 460 formula 104.7)
If you look closely in that version there is a double bar over the H, and it might be nice to have the double bar over the wikipedia version as well. Apparently in Toleman's thinking double bar H is a property of an ensemble where as H with no bar is a property of an individual system.
Hobojaks (talk) 20:46, 25 February 2009 (UTC)
Something like H or ${\displaystyle {\overline {\overline {H}}}}$? Physchim62 (talk) 12:29, 26 February 2009 (UTC)
${\displaystyle {\overline {\overline {H}}}=\sum p_{k}\log p_{k}}$[7]
This now looks exactly like the text from toleman's chapter The Quantum Mechanical H-theorem. Maybe someone can help me that understands the guidelines for inline citations to get the formant exactly correct for this important citation but then I propose that we include this line in the article.
It may also be wise to start with the 'quantum mechanical' formula as J heald has suggested using the rational that it is the simplest form of H theorem.
Not just that, but it gives a nice general notation based on transition probabilities between states, into which terms the classical versions can easily be cast as well. Jheald (talk) 11:26, 1 March 2009 (UTC)
The double bar tells us that we must constantly remember that this is not a property of a single system but an average property of an ensemble of systems.
I would use H, without the double bar, for the first time you are introducing the system. -- After all, it's only Tolman who uses the double bar, it's not "standard" notation. For other uses, we could than qualify H in difference ways, eg perhaps with a subscript, if we really want to emphasise that it is being calculated for a different probability distribution. Jheald (talk) 11:26, 1 March 2009 (UTC)
Also since we are contemplating including in this article boltzmann's original H quantity it will help to distinguish between the two quantities.
As a side note, Gibbs has a really nasty habit of stealing other peoples terminology, and introducing his own ideas, under the same name as another concept. Boltzmann's H-theorem, Clausisus's entropy, and Hamilton's vector are important on this list.
I have not really found anybody who has commented on the somewhat cosmic significance of the fact that in fact Boltzmann proved that the "true entropy" has to be a negative quantity. Sign switching was something that Gibbs delighted in, also switching the sign of the scalar part of the square of a quaternion. Hence V 'dot' V = ${\displaystyle -V^{2}}$ with Gibb's hermaphrodidical notation on the right, and Hamilton's in my point of view more elegant notation on the right. It would be interesting of a citation for this observation could be found, because I don't think it is OR on my part? It is in part what drives my interest in H - theorem. My other favorite book on thermodynamics relegates H -theorem to an apendex. Yet still a good source.Hobojaks (talk) 19:15, 26 February 2009 (UTC)
## Notes
1. ^ Toleman 1938 pg 134
2. ^ toleman 1938 p 134
3. ^ Toleman 1938 p 134
4. ^ Tolman 1938 chapter XII The Quantum Mechanical H-Theorem.
5. ^ Tolemam 1938 pg 65
6. ^ Rieff 1965 pg. 546
7. ^ Toleman 1938 pg 460 formula 104.7
Hobojaks (talk) 19:20, 26 February 2009 (UTC)
## Plan for a little editing.
1)Complete, I am changing the order of the quantum mechanical version of H theorem, with the other section.
As per discussion this is the more relevant definition of H theorem. This may require some cleanup.
(2)Complete, I am moving the discussion of Shannon entropy into the first quantum mechanical section of the article.
(3)Completed, I am pasting in the inline citation from Toleman for the first formula on entropy as suggested by J-heald.
(4)First stage now complete,These changes being made, my work will be done with the quantum mechanical section, and I also plan to leave the first paragraph of the article alone for a while to let these changes cool down for a while.
(5) I would then like to reconstruct the section on the original Boltzmann H theorem along the lines as it was developed by Toleman, providing the proper inline citations. With a note about the formula offered by Rief.
5(a) My first step will be to copy formulas 47.5 to 47.9 found in Toleman page 135.
Toleman is pretty good about tracing the historical development of H theorem. I plan on making it clear that this is Toleman's development of the concept, leaving any parts of Boltzman's original formulation, not covered by Toleman a bow for another Ulysses. The two developments are close but the extent that Boltzman's development differs from Toleman's so far I have only Toleman's account.
I now have all the fomulas i wanted to add to the article except for one that I am still working on the proper math tags for.
See the really ugly period at the start of the lower part of the formula that I used to get the equals signs to line up????
${\displaystyle \gamma \,}$
${\displaystyle H=\sum (n_{i}\log n_{i}-n_{i}\log \delta v_{\gamma })}$[1]
${\displaystyle .\,\,\,\,\,=\sum n_{i}\log n_{i}+const}$[2]
I plan on changing the formula current formula for H per unit volume, to make it more in line with the format and definitions of Reif with the appropriate inline citations. Or possibly doing away with it all together? But this would require more discussion. Can anybody provide an inline source citation for this formula?
(6)This will leave the article on H-theorem with still more work to do, but I feel like maybe it will be someone else's turn to take it from there for a while, and I plan on a cooling off period where I might participate in discussion, but not do any more editing of the article for a while. Hobojaks (talk) 20:36, 28 February 2009 (UTC)
No time now, but I'll try and comment in more detail on Monday. Jheald (talk) 11:28, 1 March 2009 (UTC)
## Icky notation
I found things here that look like this:
ναβ(pβ-pα)
How uncouth.
I changed that to this:
ναβ(pβ − pα)
Note: In non-TeX math notation,
• variables are to be italicized; parentheses and digits are NOT;
• a minus sign is not a stubby little hyphen;
• binary operators minus, plus, equals, etc. are preceded and followed by a space (don't worry about that in TeX; the software does it automatically); with binary operators I make the space non-breakable. (Minus and plus as unary operators are not followed by a space; thus +5 or −5 is correct.)
Also:
${\displaystyle \int f(x)\,dx=g(x)+{\text{constant}}\,}$
is correct, but
${\displaystyle \int f(x)\,dx=g(x)+constant\,}$
is wrong. "Constant" should be set in text mode, not left to get italicized as if the letters were variables. This also affects spacing in some contexts. Michael Hardy (talk) 18:54, 2 March 2009 (UTC)
I feel like I have been blessed by the math tag gods here!!!
You definitely made the formulas look a lot better. I will need study your techniques. I am still mastering math tags.
The double bar over the H however it seems to me is very important. Taking it off comes at the expense having an equation from a verifiable source.
The double bar shows that the quantity we are talking about is an average over an ensembly, and that it is different from the other quantity H used in other sections of the article.
## The "Criticism" section is complete nonsense
I am not sure who is the "source" being cited but the criticism clearly shows that its author has zero understanding for the proof presented in the article - and zero understanding for thermodynamics in general.
1) It's the very point of the proof that one divides the evolution into infinitesimal steps that are long enough that the "thermalization" occurs in the interval, and it produces nonzero probabilities for indistinguishable microstates. On the other hand, because this time can still be essentially arbitrarily short for macroscopic systems, the first-order perturbation theory becomes actually exact, so there is no error introduced by the first-order approximation of the Fermi's rule. At any rate, even if a higher-order corrected rule were used instead, one could still prove the inequality.
2) The derivation clearly talks about the probabilities of all possible microstates of the system. It makes no assumption about the number of "particles" or other degrees of freedom that describe the system. So in this sense, the derivation is absolutely exact and surely doesn't neglect any interactions. Only the "critic" is introducing some assumptions about the number of particles and the organization of the Hilbert space as a tensor product. The proof is doing nothing of the sort and is completely general in this sense.
3) It's completely absurd that an isolated system with many (N) particles will "sit in one microstate". Quite on the contrary, it will try to evolve into an equal mixture of completely all microstates with equal probabilities. That's what the proof shows - and what every sane person knows from the 2nd law of thermodynamics and everyday life, anyway.--Lumidek (talk) 09:27, 3 April 2009 (UTC)
I have to agree with the absurdity of this critique. Even if the case can be advanced for the potential invalidity of the H-theorem proof, it must be done under what most would consider rather extreme assumptions. Contrary to the criticism of its applicability, the H-theorem assumes rather mundane conditions, e.g. non interacting particles. Despite what the critic states, non-interaction represents most gases under "ordinary" conditions to an excellent degree for most applications.
While it is possible to prepare states that violate this theorem (e.g. Jaynes, 1971), they are systems which violate the assumptions of the theory, specifically with respect to the strength of potentials. This is not necessarily the case in N-body quantum systems, since even these systems may be treated as systems of N non interacting quantum particles, with successive perturbations (in the form of potentials) approaching the "true" system. The H-theorem will be perfectly valid for this system provided the potentials are sufficiently weak. It should be noted that a system can always be placed in a state that will violate the H-theorem, even in the absence of potentials. What is physically important is that the number of accessible states that will continue to violate the H-theorem as the system evolves decreases with exponential rapidity. For macroscopic systems this is essentially immediate, even if the initial system was H-theorem violating.
This section is out of place in this article, except perhaps as a note linking to another page, if the critic should like to create it, that dicusses scenarios in which the H-theorem is violated, which would necessarily be limited to those cases which strongly violate the assumptions supporting the theory. 68.19.49.85 (talk) 17:28, 14 October 2009 (UTC)
## Terribly written article!
1) If Boltzmann proved the H-theorem, why in the world does the article start off with the quantum version of it? First the classical treatment should be given, followed by the quantum version.
2) The article says to look up the Wikipedia entry on Shannon's information entropy in order to better understand the definition of H. This is ridiculous! The H-theorem article should not only be self-contained, but it should be able to explain the H-theorems's foundations better than any other article, since after all, it's about the H-theorem!
3) There are misplaced interjections, such as mentioning "a recent controversy called the Black hole information paradox." at the end of the Quantum mechanical H-theorem section, and a discussion about Loschmidt's paradox and molecular chaos at the end of the Boltzmann's H-theorem section. These should be discussed appropriately in the Analysis section or in their own section.
4) The Analysis section quotes various researchers without references and makes a personal attack on Gull in the paragraph "(It may be interesting ...". Overall it reads like an incoherent rant and rambling.
5) The article is full of sentences devoid of any information, like: "This has led to much discussion." "and thus a better feel for the meaning of H." " something must be wrong" " and therefore begs the question." (even linking to begs the question) "But this would not be very interesting." "With such an assumption, one moves firmly into the domain of predictive physics: if the assumption goes wrong, it may produce predictions which are systematically and reproducibly wrong." — Preceding unsigned comment added by Michael assis (talkcontribs) 20:32, 10 November 2012 (UTC)
## Isn't the introduction wrong?
″In classical statistical mechanics, the H-theorem, introduced by Ludwig Boltzmann in 1872, describes the tendency to increase in the quantity H (defined below) in a nearly-ideal gas of molecules.″
Shouldn't the H quantity increase? Also
"As this quantity H was meant to represent the entropy of thermodynamics, the H-theorem was an early demonstration of the power of statistical mechanics as it claimed to derive the second law of thermodynamics—a statement about fundamentally irreversible processes—from reversible microscopic mechanics."
If I'm not wrong it should be noted somewhere that H should be the negative entropy, because it leads to a lot of confusion otherwise (also considering that in many contexts the symbol H represents the common entropy). Correct me if I'm wrong, I'm not an expert on the subject.
~~maioenrico~~ — Preceding unsigned comment added by 79.45.212.156 (talk) 15:00, 13 December 2015 (UTC)
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1. ^ Toleman 1938 pg. 135 formula 47.7
2. ^ Toleman 1938 pg. 135 formula 47.7 | 7,906 | 35,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "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} | 2.796875 | 3 | CC-MAIN-2018-13 | latest | en | 0.949833 |
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https://www.finance-lib.com/financial-term-bin-transfer.html | 1,632,242,604,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.57/warc/CC-MAIN-20210921161350-20210921191350-00140.warc.gz | 785,298,423 | 6,130 | Financial Terms Bin transfer
# Definition of Bin transfer
## Bin transfer
A transaction to move inventory from one storage bin to another.
# Related Terms:
## Accumulation bin
A location in which components destined for the shop floor are
accumulated before delivery.
## Bin
A storage area, typically a subdivision of a single level of a storage rack.
## Binomial model
A method of pricing options or other equity derivatives in
which the probability over time of each possible price follows a binomial
distribution. The basic assumption is that prices can move to only two values
(one higher and one lower) over any short time period.
## Binomial option pricing model
An option pricing model in which the underlying asset can take on only two
possible, discrete values in the next time period for each value that it can take on in the preceding time period.
## Building a binomial tree
For a binomial option model: plotting the two
possible short-term price-changes values, and then the subsequent two values
each, and then the subsequent two values each, and so on over time, is known
as “building a binomial tree.” See binomial model.
## Combination matching
Also called horizon matching, a variation of multiperiod immunization and cash
flow matching in which a portfolio is created that is always duration matched and also cash-matched in the
first few years.
## Combination strategy
A strategy in which a put and with the same strike price and expiration are either both
bought or both sold. Related: Straddle
## Depository transfer check (DTC)
Check made out directly by a local bank to a particular firm or person.
## EFT (electronic funds transfer)
Funds which are electronically credited to your account (e.g. direct deposit), or electronically debited from your account on an ongoing basis (e.g. a pre-authorized monthly bill payment, or a monthly loan or mortgage payment). A wire transfer is a form of EFT.
## Electronic depository transfers
The transfer of funds between bank accounts through the Automated
Clearing House (ACH) system.
## Interplant transfer
The movement of inventory from one company location to
another, usually requiring a transfer transaction.
## negotiated transfer price
an intracompany charge for goods
or services set through a process of negotiation between
the selling and purchasing unit managers
## Official unrequited transfers
Include a variety of subsidies, military aid, voluntary cancellation of debt,
contributions to international organizations, indemnities imposed under peace treaties, technical assistance,
taxes, fines, etc.
## Private unrequited transfers
Refers to resident immigrant workers' remittances to their country of origin as
well as gifts, dowries, inheritances, prizes, charitable contributions, etc.
## Q ratio or Tobin's Q ratio
Market value of a firm's assets divided by replacement value of the firm's assets.
Quadratic programming Variant of linear programming whereby the equations are quadratic rather than linear.
## Robinson-Patman Act
a law that prohibits companies from pricing the same products at different amounts when those amounts do not reflect related cost differences
## Tobin's Q
Market value of assets divided by replacement value of assets. A Tobin's Q ratio greater than 1
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## Transfer agent
ndividual or institution appointed by a company to look after the transfer of securities.
## Transfer Payment
A grant or gift that is not payment for services rendered.
## Transfer price
The price at which one unit of a firm sells goods or services to another unit of the same firm.
## Transfer price
The price at which goods or services are bought and sold within divisions of the same organization, as opposed to an arm’s-length price at which sales may be made to an external customer.
## transfer price
an internal charge established for the exchange
of goods or services between organizational units
of the same company
## Transfer price
The price at which one part of a company sells a product or service to
another part of the same company.
## transfer time
the time consumed by moving products or
components from one place to another
## Transferable put right
An option issued by the firm to its shareholders to sell the firm one share of its
common stock at a fixed price (the strike price) within a stated period (the time to maturity). The put right is
"transferable" because it can be traded in the capital markets.
## Transferred-in cost
The cost that a product accumulates during its tenure in another
department that is earlier in the production process.
## two-bin system
an inventory ordering system in which two
containers (or stacks) of raw materials or parts are available
for use; when one container is depleted, the removal
of materials from the second container begins and a purchase
order is placed to refill the first container
## Two-bin system
A system in which parts are reordered when their supply in one
storage bin is exhausted, requiring usage from a backup bin until the replenishment
arrives.
## Unilateral transfers
Items in the current account of the balance of payments of a country's accounting books
that corresponds to gifts from foreigners or pension payments to foreign residents who once worked in the
country whose balance of payments is being considered.
## wire transfer
An electronic transmission of money from one place to another. For example, you might request that your bank transfer money from your bank account in Vancouver to the account of a relative in Quebec City. To do this, you would provide the relative’s name and account number, as well as the address of the bank in Quebec City. Your bank would then "wire" the funds, which would usually arrive within a couple of days.
## modified FIFO method (of process costing)
the method of cost assignment that uses FIFO to compute a cost per
equivalent unit but, in transferring units from a department,
the costs of the beginning inventory units and the
units started and completed are combined and averaged
## Trust Company
Organization usually combined with a commercial bank, which is engaged as a trustee for individuals or businesses in the administration of Trust funds, estates, custodial arrangements, stock transfer and registration, and other related services. | 1,280 | 6,388 | {"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-2021-39 | latest | en | 0.904859 |
https://metanumbers.com/1531896 | 1,643,328,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305317.17/warc/CC-MAIN-20220127223432-20220128013432-00236.warc.gz | 436,951,718 | 7,750 | # 1531896 (number)
1,531,896 (one million five hundred thirty-one thousand eight hundred ninety-six) is an even seven-digits composite number following 1531895 and preceding 1531897. In scientific notation, it is written as 1.531896 × 106. The sum of its digits is 33. It has a total of 7 prime factors and 64 positive divisors. There are 470,400 positive integers (up to 1531896) that are relatively prime to 1531896.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 33
• Digital Root 6
## Name
Short name 1 million 531 thousand 896 one million five hundred thirty-one thousand eight hundred ninety-six
## Notation
Scientific notation 1.531896 × 106 1.531896 × 106
## Prime Factorization of 1531896
Prime Factorization 23 × 3 × 29 × 31 × 71
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 382974 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 1,531,896 is 23 × 3 × 29 × 31 × 71. Since it has a total of 7 prime factors, 1,531,896 is a composite number.
## Divisors of 1531896
64 divisors
Even divisors 48 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 64 Total number of the positive divisors of n σ(n) 4.1472e+06 Sum of all the positive divisors of n s(n) 2.6153e+06 Sum of the proper positive divisors of n A(n) 64800 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1237.7 Returns the nth root of the product of n divisors H(n) 23.6404 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,531,896 can be divided by 64 positive divisors (out of which 48 are even, and 16 are odd). The sum of these divisors (counting 1,531,896) is 4,147,200, the average is 64,800.
## Other Arithmetic Functions (n = 1531896)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 470400 Total number of positive integers not greater than n that are coprime to n λ(n) 420 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 116128 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 470,400 positive integers (less than 1,531,896) that are coprime with 1,531,896. And there are approximately 116,128 prime numbers less than or equal to 1,531,896.
## Divisibility of 1531896
m n mod m 2 3 4 5 6 7 8 9 0 0 0 1 0 2 0 6
The number 1,531,896 is divisible by 2, 3, 4, 6 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (1531896)
Base System Value
2 Binary 101110101111111111000
3 Ternary 2212211100220
4 Quaternary 11311333320
5 Quinary 343010041
6 Senary 52500040
8 Octal 5657770
10 Decimal 1531896
12 Duodecimal 61a620
20 Vigesimal 9b9eg
36 Base36 wu0o
## Basic calculations (n = 1531896)
### Multiplication
n×y
n×2 3063792 4595688 6127584 7659480
### Division
n÷y
n÷2 765948 510632 382974 306379
### Exponentiation
ny
n2 2346705354816 3594908546221211136 5507026022322088454393856 8436191135491118014932130430976
### Nth Root
y√n
2√n 1237.7 115.277 35.1809 17.2602
## 1531896 as geometric shapes
### Circle
Diameter 3.06379e+06 9.62519e+06 7.37239e+12
### Sphere
Volume 1.50583e+19 2.94896e+13 9.62519e+06
### Square
Length = n
Perimeter 6.12758e+06 2.34671e+12 2.16643e+06
### Cube
Length = n
Surface area 1.40802e+13 3.59491e+18 2.65332e+06
### Equilateral Triangle
Length = n
Perimeter 4.59569e+06 1.01615e+12 1.32666e+06
### Triangular Pyramid
Length = n
Surface area 4.06461e+12 4.23664e+17 1.25079e+06
## Cryptographic Hash Functions
md5 4eb984565c650a4cecde7e1c24c5ffd2 521cd64b688cd7f0d351d3402c37a13d69712254 6f4b2ff3c0a4003c35ef1c83dbc265beecb72fb6482b384ce6495bc2dfda4b84 b656d1babff5382682eb4b449a975325629ea94057e09be49992b2b14df489f88fd12a9837af313b25aa9c8b8312083ab707c8998131d479c14a1a2226c28608 9f847b3e44680367773e2f626ed5aa0fcb527538 | 1,548 | 4,319 | {"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} | 3.484375 | 3 | CC-MAIN-2022-05 | latest | en | 0.807168 |
https://chrischona2015.org/how-many-oz-in-half-pint/ | 1,657,106,577,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00140.warc.gz | 228,531,031 | 5,041 | How countless Ounces in a Pint: convert a pint come ounces may sound choose a straightforward task, but it requires much more information for much more than one feasible answer.
You are watching: How many oz in half pint
The conventional measurement the a pint can vary depending upon the problem to be measured. It may also readjust with the usage of U.S. And also Imperial standards. It is among the most frequently referred methods of weight provided in day-to-day life. Pints the milk and beer space the most recognized units of volume.
However, how is that solitary pint converted when considering the range of ounces? Is the pint come be convert an royal pint or U.S. Sized pint? space you converting a dry pint to ounces or is the a fluid pint? every of those makes a far-ranging difference in the variety of ounces in the final conversion. Once, those determinants have been found the switch meets the adhering to standards.
## How numerous Ounces in a dry Pint (U.S.)?
The very first conversion will be through a dry pint. The volume uncovered in this dry pint is 18.62 ounces.
### How numerous Ounces in a liquid Pint (U.S.)?
To transform this pint to ounces is various in measurement to the dry pint. A U.S. Pint (Liquid) is same to 16 fluid ounces.
The use of liquid ounces measures a volume weight, unequal ounces which are a mass unit. The subtle distinction is the reason for the uneven variety of ounces that equal both liquid and a dry pint.
### How countless Ounces in imperial Pint?
The unified Kingdom renders use the the imperial pint measurement in ~ a 20% larger volume than the unit of measure up in the U.S. That renders the switch of an royal pint to same 20 ounces.
### How numerous Ounces in a Pint of ice Cream?
There are about 16 ounces in a pint of ice Cream.
### How plenty of Ounces in a Pint of Blueberries?
Blueberries are measured in a dry type making the size of a pint roughly 2 cups. The converts to a measure of 12 ounces.
Units of measurement because that fruit will count on the form of berry weighed. Even if it is you space using the tiny compact blueberry or a larger juicy strawberry will change the variety of ounces that the exact same pint will certainly weigh.
### How numerous Ounces in a Pint the Strawberries?
A larger berry prefer the strawberry will significantly require an ext fruit to make a pint. That may seem as if it renders strawberries room a more challenging item to transform to ounces. When it merely requires a little more thought to measure the weight.
Things to consider are the kind of the berries, whole, sliced, or even pureed. As soon as using whole strawberries, a pint is composed of 3.25 cup of fruit. Because that sliced, the conversion starts through 2 cup while a pint that pureed strawberry is same to 1.5 cups of berries.
1 cup of strawberries is same to 7.1 ounces.
The load of a pint of entirety strawberries is then equal to 23.075 ounces.
A pint the sliced strawberries converts come 14.2 ounces.
See more: Soulstorm Map Creator?: Dawn Of War Soulstorm Map Editor For Soulstorm
The pint the pureed strawberries then measures at 10.65 ounces.
1 cup equals 7.1 ounces
Number that Rows
Us liquid PintUs liquid OunceImperial Pintimperial Ounce | 730 | 3,251 | {"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-2022-27 | latest | en | 0.910086 |
https://la.mathworks.com/matlabcentral/cody/problems/591-create-a-patchwork-matrix/solutions/853583 | 1,606,615,288,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195967.34/warc/CC-MAIN-20201129004335-20201129034335-00203.warc.gz | 377,923,832 | 17,220 | Cody
# Problem 591. Create a patchwork matrix
Solution 853583
Submitted on 20 Mar 2016 by William
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
P = [1 0; 1 2]; m1 = eye(2); m2 = ones(2); M_correct = [1 0 0 0; 0 1 0 0; 1 0 1 1; 0 1 1 1]; assert(isequal(patchworkMatrix(P,m1,m2),M_correct))
2 Pass
P = 2-eye(4); m1 = eye(2); m2 = ones(2); M_correct = [1 0 1 1 1 1 1 1; 0 1 1 1 1 1 1 1; 1 1 1 0 1 1 1 1; 1 1 0 1 1 1 1 1; 1 1 1 1 1 0 1 1; 1 1 1 1 0 1 1 1; 1 1 1 1 1 1 1 0; 1 1 1 1 1 1 0 1]; assert(isequal(patchworkMatrix(P,m1,m2),M_correct))
3 Pass
P = [2 3 2 3]; m1 = 1; m2 = 2; m3 = 3; M_correct = [2 3 2 3]; assert(isequal(patchworkMatrix(P,m1,m2,m3),M_correct))
4 Pass
P = [6 5; 4 3; 2 1]; m1 = rand(2,3); m2 = rand(2,3); m3 = rand(2,3); m4 = rand(2,3); m5 = rand(2,3); m6 = rand(2,3); M_correct = [m6 m5; m4 m3; m2 m1]; assert(isequal(patchworkMatrix(P,m1,m2,m3,m4,m5,m6),M_correct))
5 Pass
P = zeros(2); m1 = rand(3,2); m2 = rand(3,2); m3 = rand(3,2); m4 = rand(3,2); m5 = rand(3,2); m6 = rand(3,2); M_correct = zeros(6,4); assert(isequal(patchworkMatrix(P,m1,m2,m3,m4,m5,m6),M_correct))
6 Pass
P = []; m = cell(100); assert(isempty(patchworkMatrix(P,m{:})))
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Start Hunting! | 656 | 1,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-50 | latest | en | 0.473299 |
https://nanoscale.blogspot.com/2021/01/idle-speculation-can-teach-physics.html?showComment=1609533081587 | 1,618,753,425,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00182.warc.gz | 509,490,312 | 16,595 | ## Friday, January 01, 2021
### Idle speculation can teach physics, vacuum energy edition
To start the new year, a silly anecdote ending in real science.
Back when I was in grad school, around 25 years ago, I was goofing around chatting with one of my fellow group members, explaining about my brilliant (ahem) vacuum energy extraction machine. See, I had read this paper by Robert L. Forward, which proposed an interesting idea, that one could use the Casimir effect to somehow "extract energy from the vacuum" - see here (pdf).
Fig from here.
(For those not in the know: the Casimir effect is an attractive (usually) interaction between conductors that grows rapidly at very small separations. The non-exotic explanation for the force is that it is a relativistic generalization of the van der Waals force. The exotic explanation for the force is that conducting boundaries interact with zero-point fluctuations of the electromagnetic field, so that "empty" space outside the region of the conductors has higher energy density. As explained in the wiki link and my previous post on the topic, the non-exotic explanation seemingly covers everything without needing exotic physics.)
Anyway, my (not serious) idea was, conceptually, to make a parallel plate structure where one plate is gold (e.g.) and the other is one of the high temperature superconductors. Those systems are rather lousy conductors in the normal state. So, the idea was, cool the system just below the superconducting transition. The one plate becomes superconducting, leading ideally to dramatically increased Casimir attraction between the plates. Let the plates get closer, doing work on some external load. Then warm the plates just slightly, so that the superconductivity goes away. The attraction should lessen, and the plate would spring back, doing less work of the opposite sign. It's not obvious that the energy required to switch the superconductivity is larger than the energy one could extract from running such a cycle. Of course, there has to be a catch (as Forward himself points out in the linked pdf above). In our conversation, I realized that the interactions between the plates would very likely modify the superconducting transition, probably in just the way needed to avoid extracting net energy through this process.
Fast forward to last week, when I randomly came upon this article. Researchers actually did an experiment using nanomechanical resonators to try to measure the impact of the Casimir interactions on the superconducting transition in (ultrasmooth, quench-condensed) lead films. They were not able to resolve anything (like a change in the transition temperature) in this first attempt, but it shows that techniques now exist to probe such tiny effects, and that idly throwing ideas around can sometimes stumble upon real physics.
#### 3 comments:
gilroy0 said...
I wasn't the group member (obviously) but I think I remember the discussion. I'm a little sad to hear that you don't need the exotic explanation. :)
Douglas Natelson said...
Yeah, me too :-) It was BB, btw.
Anonymous said...
This is a nice channel with loads of high-quality content on CMP.
https://www.youtube.com/c/SPICEmainz/videos | 693 | 3,238 | {"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-2021-17 | longest | en | 0.91556 |
https://fabricesalvaire.github.io/Patro/api/Patro/GeometryEngine/Bezier.html | 1,550,797,914,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247511573.67/warc/CC-MAIN-20190221233437-20190222015437-00027.warc.gz | 533,056,704 | 8,547 | # 10.1.4.1. Bezier¶
Module to implement Bézier curve.
For resources on Bézier curve see this section.
class Patro.GeometryEngine.Bezier.BezierMixin2D[source]
Mixin to implements 2D Bezier Curve.
LineInterpolationPrecision = 0.05
_logger = <Logger Patro.GeometryEngine.Bezier.BezierMixin2D (WARNING)>
_map_to_line(line)[source]
distance_to_point(point)[source]
interpolated_length(dt=None)[source]
Length of the curve obtained via line interpolation
length_at_t(t, cache=False)[source]
Compute the length of the curve at t.
non_parametric_curve(line)[source]
Return the non-parametric Bezier curve D(ti, di(t)) where di(t) is the distance of the curve from the baseline of the fat-line, ti is equally spaced in [0, 1].
split_at_two_t(t1, t2)[source]
t_at_length(length, precision=1e-06)[source]
Compute t for the given length. Length must lie in [0, curve length] range].
class Patro.GeometryEngine.Bezier.CubicBezier2D(p0, p1, p2, p3)[source]
Class to implements 2D Cubic Bezier Curve.
BASIS = array([[ 1, -3, 3, -1], [ 0, 3, -6, 3], [ 0, 0, 3, -3], [ 0, 0, 0, 1]])
INVERSE_BASIS = array([[1. , 1. , 1. , 1. ], [0. , 0.33333333, 0.66666667, 1. ], [0. , 0. , 0.33333333, 1. ], [0. , 0. , 0. , 1. ]])
InterpolationPrecision = 0.001
__init__(p0, p1, p2, p3)[source]
Initialize self. See help(type(self)) for accurate signature.
__repr__()[source]
Return repr(self).
static _clip_convex_hull(hull_top, hull_bottom, d_min, d_max)[source]
static _clip_convex_hull_part(part, top, threshold)[source]
Clip the bottom or top part of the convex hull.
part is a list of points, top is a boolean flag to indicate if it corresponds to the top part, threshold is d_min if top part else d_max.
_clipping_convex_hull()[source]
_d01()[source]
Return the distance between 0 and 1 quadratic aproximations
static _instersect_curve(curve1, curve2, t_min=0, t_max=1, u_min=0, u_max=1, old_delta_t=1, reverse=False, recursion=0, recursion_limit=32, t_limit=0.8, locations=[])[source]
Compute the intersection of two Bézier curves.
Code inspired from
_logger = <Logger Patro.GeometryEngine.Bezier.CubicMixin2D (WARNING)>
_q_point()[source]
Return the control point for mid-point quadratic approximation
_t_max()[source]
adaptive_length_approximation()[source]
area
Compute the area delimited by the curve and the segment across the start and stop point.
closest_point(point)[source]
Return the closest point on the curve to the given point.
For more details see this section.
fat_line()[source]
intersect_line(line)[source]
Find the intersections of the curve with a line.
is_flat_enough(flatness)[source]
Determines if a curve is sufficiently flat, meaning it appears as a straight line and has curve-time that is enough linear, as specified by the given flatness parameter.
For more details see this section.
length
mid_point_quadratic_approximation()[source]
point_at_t(t)[source]
q_length()[source]
Return the length of the mid-point quadratic approximation
split_at_t(t)[source]
Split the curve at given position
tangent1
tangent_at(t)[source]
to_spline()[source]
class Patro.GeometryEngine.Bezier.QuadraticBezier2D(p0, p1, p2)[source]
Class to implements 2D Quadratic Bezier Curve.
BASIS = array([[ 1, -2, 1], [ 0, 2, -2], [ 0, 0, 1]])
INVERSE_BASIS = array([[-2, 1, -2], [-1, -3, 1], [-1, -1, -2]])
__init__(p0, p1, p2)[source]
Initialize self. See help(type(self)) for accurate signature.
__repr__()[source]
Return repr(self).
_logger = <Logger Patro.GeometryEngine.Bezier.QuadraticBezier2D (WARNING)>
closest_point(point)[source]
Return the closest point on the curve to the given point.
For more details see this section.
fat_line()[source]
intersect_line(line)[source]
Find the intersections of the curve with a line.
For more details see this section.
length
Compute the length of the curve.
For more details see this section.
normal0
point_at_t(t)[source]
split_at_t(t)[source]
Split the curve at given position
tangent0
tangent1
tangent_at(t)[source]
to_cubic()[source]
Elevate the quadratic Bézier curve to a cubic Bézier cubic with the same shape.
For more details see this section. | 1,169 | 4,139 | {"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} | 2.71875 | 3 | CC-MAIN-2019-09 | longest | en | 0.520357 |
https://forums.ni.com:443/t5/LabVIEW/What-is-the-structure-of-solving-recursive-summation-in-labVIEW/m-p/3809951#M1075876 | 1,675,558,413,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00464.warc.gz | 282,845,782 | 32,625 | # LabVIEW
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## What is the structure of solving recursive summation in labVIEW
Hi,
Can anyone suggest the correct structure of how to solve recursive equations in LabVIEW?
For instance, I would like to calculate the coefficients of g matrix [g_0,g_1,...,g_n]
which can be defined by : please see the attached photo.
Note that g_0=1;
Ameer
Message 1 of 3
(1,483 Views)
## Re: What is the structure of solving recursive summation in labVIEW
You could use a recursive VI 🙂
Message 2 of 3
(1,481 Views)
## Re: What is the structure of solving recursive summation in labVIEW
The most straight-forward (but sometimes programmatically challenging) way to solve a recursive problem is, as gregoryj points out, to use a recursive algorithm. In LabVIEW, you do this by making your VI "reentrant".
However, another method that can be a little easier for some problems is to try to transform the recursive problem into an iterative one, as iteration (a.k.a. "looping") is something most languages (including LabVIEW) supports quite well.
Here your best tools are Pencil and Paper. Write down explicitly the formulas for g0, g1, g2, and g3. Notice that each formula can be expressed as a summation between another array, v, the elements of the previously-generated elements of g, and some coefficients that depend on the specific index of g being generated. The only "trick" is that you use the elements of the g array in the opposite order of the elements in the g array, but that's just an "indexing" problem. So an iterative solution to generating, say, N elements of g from N elements of v should be simple to code in LabVIEW once you "see" the iterative pattern.
Bob Schor
Message 3 of 3
(1,416 Views) | 427 | 1,755 | {"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-2023-06 | latest | en | 0.911205 |
https://www.transtutors.com/questions/a-body-of-mass-0-5-kg-travels-in-a-straight-line-with-velocity-v-ax-3-2-where-a-5-m--6794876.htm | 1,623,649,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00426.warc.gz | 950,631,955 | 14,658 | # A body of mass 0.5 kg travels in a straight line with velocity v =ax 3/2 where a = 5 m–1/2 s–1. What 1 answer below »
A body of mass 0.5 kg travels in a straight line with velocity v =ax 3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
KETHAMALA V
Ans:
Mass of the body,m = 0.5 kg
Velocity of the body, v=ax3/2
here, a = 5 m-1/2 s-1.
Initial velocity at x...
## Recent Questions in Physics
Looking for Something Else? Ask a Similar Question | 172 | 513 | {"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-2021-25 | latest | en | 0.902092 |
https://www.physicsforums.com/threads/err-real-easy-problem-but-whatd-he-or-i-do-wrong.90173/ | 1,702,100,280,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00267.warc.gz | 1,029,097,695 | 14,749 | # Err, real easy problem but what'd he(or I) do wrong
• schattenjaeger
#### schattenjaeger
the problem was that a guy wants to average a speed of 180 km/h while driving 4 laps around a race track.
after 2 laps, he's only averaged 150, so its asking what the average would have to be so that he does attain his goal.
So everybody would read the question, put 210, then check the answer in the back of the book and they would assume they are right.
But from the start i knew it wasnt what it seemed. Since If 1 lap=1 km, then going 180 km/h, it would take you 80 seconds to finish all 4 laps.
But, going 150, it would take you 48 seconds to finish 2 laps, and then going 210, it would take you roughly 35 seconds. which comes out to ~83 seconds, and not 80.
So The point is to find how fast you have to go to do a lap in 32, seconds, which comes out to 225 km/h.
When i first started explain it, no one would believe me, i felt like galileo or something
I got 210 myself, and upon close scrutinizing I can't find an error in either of our calculations, and all things considered, the inverse ability function probably made me wrong(what's the saying, you know you're a physics major when you can do vector calculus but don't remember long division?)
You're correct. Can you read my working ?
#### Attachments
• schatenjaegger.jpg
20.6 KB · Views: 393 | 349 | 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} | 3.28125 | 3 | CC-MAIN-2023-50 | latest | en | 0.969035 |
https://www.fllcasts.com/tutorials/195-ev3-robot-with-motors-in-opposite-directions-proportional-like-following-part-3 | 1,713,216,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00435.warc.gz | 718,330,695 | 23,419 | # EV3 Robot with Motors in Opposite Directions. Proportional like following. Part 3 ProPreview
Let's implement a more advanced program for this robot to learn how to use with motors in opposite directions - and this is to implement the Proportional Line Following algorithm.
• #195
• 13 Mar 2016
• 6:40
The algorithm controls the motors based on the sensors. But in this robot the motors are placed in opposite directions. We show how to modify the Proportional Line Following program to make it work for this robot.
We are also following not a black line, but a red line. Which is interesting because red should be a color close to white. This is how the color sensor should work. Or is it?
Find the complete short course at:
EV3 Robot with Motors in Opposite Directions. The Frankenstein.
A course on proportional smooth line following can be found at:
Proportional Line Following with LEGO Mindstorms EV3
### Proportional line following with the robot
Follow the line backwards. It shouldn`t be difficult at all. But you have to understand how the program works. Record a video, take a picture and share it with us in the comments section below.
#### Tasks description, submission and evaluation are available to subscribed users.
Subscribe now to access the full capacity and get feedback.
### English
In this video we are using our very strange Frankenstein robot, that has the 2 motors in opposite directions. And we are going to implement the proportional line following algorithm so that we can customize it to work with this very strange robot.
First I'll go to fllcasts.com site and I'll find the playlist for proportional line following. It's in the playlist section and it's right here. Proportional Line Following with LEGO Mindstorms EV3. Then in the third video, we have a complete final solution in the materials section. This here is the video. And we have the section for proportional smooth line following. When you download the program, you'll get something that looks like this. This is the program for proportional line following from this episode. And there are a few modifications that we must do in this program. First, we'll use the third sensor, so the sensor on port 3. In our case this is our right sensor.
What I'm trying to do now is to show you how to modify this program, so that it works for your robots. Because we received a number of questions on how we can modify this program for your constructions and for the way you build your robots and this is what I would like to show you today. Using our very strange robot. So I change the port for the color sensor, then I change the motors, our left motor in our case is C and the right is B. These are the 2 motors. So left motor - C, right motor - B. And the sensor is on port 3. These are the first 2 modifications that you must do on your program. Then we must detect the value of the color sensor. When I'm over the table, I can see that we detect a value of 60, that's the value that currently the sensor detects and when I move over the red line, I move the robot over the red line, when I'm over the red line, the value is 38. So over the table - 60, over the red line - 40. The value between 40 and 60 is 50. Third modification to do is to change the threshold value to 50. This one here and this one here.
This should be all. Let's now download, run our program and see how it behaves. Starting the program . It moves, but it seems that the robot is not detecting the line. So something strange is happening here. What's strange is that we must take into account that the motors of this robot. are placed in opposite directions, so whenever we want to use the right motor - B, we must actually in order to turn left with this motor, we must turn the motor back, because of its position. If we turn the motor forward the robot will turn right. And if we move the motor backward the robot will turn left. and that's different for this robot compared to the other robots that we built in which the motors are in the same direction. I'll change this in the program, so that we have the opposite calculated value passed to the motor for turning right and left.
We have our left motor - C, that's placed in the correct way, so whenever we move forward with this motor, the robot moves forward, but motor B is place in reversed. This means that we must reverse the value for this motor. I'll take a math block, I'll take the value that we calculate for motor right, and I'll subtract this from 0. What would the subtraction do? If the value for the right motor is let's say 10, then the result will be -10 and if the value for this motor is -10, then the result will be 10. This means that by subtracting this from 0 we actually reverse the value of motor right. I'll pass this, right here. I think we are ready to download and run.
The robot searches for the line, detects it and starts following it.
I hope you understood how to modify the proportional line following program for your robot. And how to use this program, even for very strange robots, where the motors are place in different directions. Find the instructions below for building this robot along with the program. If you find it difficult to follow the instructions in the video just download the programs, they're below in the materials section and directly use them. If they are not working just drop us a comment and we'll get back to you and try to see why they're not working. | 1,200 | 5,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.931488 |
https://forums.augi.com/showthread.php?176584-Multiply-a-value-in-mm-with-a-unit-less-value-(type-ul)-the-result-is-10-times-larger-why&s=91ff4c9b8e21131fba5d4df7f96076be | 1,701,985,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00503.warc.gz | 304,350,763 | 10,499 | See the top rated post in this thread. Click here
# Thread: Multiply a value in mm with a unit-less value (type ul) the result is 10 times larger, why?
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## Multiply a value in mm with a unit-less value (type ul) the result is 10 times larger, why?
Hi!
I'm using parameters in my design to speed up changes and run into a strange problem.
If I multiply a value in mm with a unitless value (type ul) the result is 10 times larger than it suppose to be. Why?
See my attachment for understanding.
Patrik
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## Re: Multiply a value in mm with a unit-less value (type ul) the result is 10 times larger, why?
Welcome to AUGI.
What happens if you add some more parenthesis? (PlateW - (((NoRows-1mm) * 8 ul) + 4 mm) * 4 ul) /AA
Also, instead of posting a Word file, please use the forum tools to post a picture (jpeg or png). That way users don't have to download anything. They can use the forum's image preview.
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## Re: Multiply a value in mm with a unit-less value (type ul) the result is 10 times larger, why?
Originally Posted by patrik.hertel801199
Hi!
I'm using parameters in my design to speed up changes and run into a strange problem.
If I multiply a value in mm with a unitless value (type ul) the result is 10 times larger than it suppose to be. Why?
See my attachment for understanding.
Patrik
Adist (mm) = Cdist (mm) * Arel (mm)
Try to modify the equation this way:
Adist (mm) = Cdist (mm) * Arel (mm) / 1 mm
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## Re: Multiply a value in mm with a unit-less value (type ul) the result is 10 times larger, why?
Originally Posted by patrik.hertel801199
Hi!
If I multiply a value in mm with a unitless value (type ul) the result is 10 times larger than it suppose to be. Why?
Patrik
Simple, Inventor internally calculates distances in cm (centimeters). That's 10mm:=)))
ul (unitless) means use native units, cm!
See VBA programming examples or some API manual.
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## Re: Multiply a value in mm with a unit-less value (type ul) the result is 10 times larger, why?
Originally Posted by patrik.hertel801199
Hi!
I'm using parameters in my design to speed up changes and run into a strange problem.
If I multiply a value in mm with a unitless value (type ul) the result is 10 times larger than it suppose to be. Why?
See my attachment for understanding.
Patrik
Hi Patrik
first of all , always make sure your unit settings are OK for all parameters. The red text in the entry-cells is red for a reason!!
example : you calculate Adist (a distance or length with unit mm) from multiplying Cdist and Arel (both distance or length) : the result actually has a unit of square mm. you could either devide the result by 1 mm (since mm^2/mm = mm), but i think it is more likely that one of your inputs (Arel) is not a distance but just a number (unitless)
cheers
rob sman
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• | 803 | 3,053 | {"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} | 3.296875 | 3 | CC-MAIN-2023-50 | latest | en | 0.769638 |
https://stats.stackexchange.com/questions/517498/bayes-by-backprop-applied-to-regression | 1,716,796,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00248.warc.gz | 460,563,388 | 41,049 | # Bayes by Backprop applied to Regression
I have been reading the Bayes by Backprop paper "Weight Uncertainty in Neural Networks" from 2015 (https://arxiv.org/pdf/1505.05424.pdf). I think I have a decent understanding of the content of the paper up until the part where it is applied to a regression problem. My issue is that to do the backpropagation we have to do computations involving the likelihood term $$P(\mathcal{D}|w)$$ where $$\mathcal{D}$$ is the data. This is based on the model specified by our network so in the classification case I believe this expression for any of our data points is just the probability predicted by the forward pass through the network.
After working on MNIST as a classification example in the paper (which I believe I understand assuming the above is correct) the authors move on to a regression example where they just generate some data from a simple function and attempt to fit it with their method and an ordinary neural network. They mention that the two networks are fit by minimising a conditional Gaussian loss and I'm not sure what is meant by this. The only thing that seems reasonable to me is that in the case of both networks the underlying model is that $$Y|_X$$ is Gaussian. I believe this would mean in the ordinary network we are treating the network output as an estimate for the mean of $$Y|_{X_i}$$ and therefore we use a squared error loss for training. In the Bayes by Backprop case then we would want $$P(\mathcal{D}|w)$$ to correspond to the likelihood for the target value so this would be the likelihood for a Gaussian with mean $$\hat{Y}_i$$ evaluated at the data $$(X_i, Y_i)$$, where $$\hat{Y}_i$$ is the value predicted by the network.
My first two questions are as follows:
• Is my understanding set out above correct?
• For the Bayes by Backprop regression above, we have no variance specified but we need to calculate the likelihood. How is this dealt with?
The results for this part are displayed by plotting the fits with confidence bands. I'm unsure on exactly how the confidence bands are being obtained. Firstly for the Bayesian approach, the result of fitting the network is a posterior probability distribution for each weight so the distribution for $$\hat{Y}|_X$$ is determined by these, however I'm not sure how practical it would be to use this to compute confidence bands due to fact that this distribution will generally have a complicated dependence on the weight distributions. Since prediction is achieved by sampling weights and averaging multiple predictions $$\hat{Y}$$ maybe it would make more sense to use these predictions to compute a confidence band too, but then we would need to make some sort of assumption on the posterior for $$\hat{Y}|_X$$ or justify using the central limit theorem. Secondly for the ordinary neural network there are also confidence regions on the plot. I'm not sure where these come from as my motivation for why this Bayesian approach was important is because on its own an ordinary neural network doesn't give us error estimates. Based on these points then:
• What exactly are the displayed confidence bands for the Bayes by Backprop network fit? Is it one of the two approaches I mentioned?
• How are confidence regions being obtained for the ordinary neural network? I was wondering if this is linked to the second of my first two questions since that involved specifying a variance for $$Y|_X$$ but that would be constant and the bands in the plot are not of constant width.
• This website works best if you ask one question at a time. Feel free to post each of these as separate questions. Mar 31, 2021 at 0:24
• They're all related to one subsection of the same paper, do they still need to be separated? They would probably each have to almost the full length to explain the context. Mar 31, 2021 at 0:43
• It’s helpful for getting faster answers. Several people may have the time or expertise to answer one part, but not the complete question. A partial answer to the uber-question doesn’t properly qualify as an answer here. Mar 31, 2021 at 0:49
I agree the authors didn't do a great job of clarifying this. My best guess is that the regression model looks like: $$y_i \sim \mathcal{N}(\mu=f(x_i), \sigma^2 = g(x_i))$$ where $$f$$ and $$g$$ are neural networks (or more precisely, probably two branches of the same neural network).
1. sample $$\theta$$ from the variational posterior,
2. given $$x$$, sample $$y$$ according to the above model,
4. compute the 25 and 75% percentile of the sampled $$y$$.
• I have seen other approaches where something like the network producing estimates for both $\mu$ and $\sigma^2$ is done, so I guess that seems likely. Apr 2, 2021 at 14:28
• My comment on the confidence bands was based on the fact that the distribution of $y$ for a given $x$ is also determined by this method (as well as the weights distributions). As a result I'm not sure why we would use sampling of $y$ to come up with a confidence region when we actually know the form of the distribution for $y$ so we don't need to do this empirically. Apr 2, 2021 at 14:34
• you know the conditional $p(y|x,\theta)$ but not the marginal $p(y|x)$. so either you integrate (intractable), or you just sample, which is by far easier Apr 2, 2021 at 16:37
• @EGN123 could you elaborate on that? where have you seen this? i'm looking for a way to apply bayes by backprop to regression and having estimates for both $\mu$ and $\sigma^2$ is exactly the outcome i want. Jul 1, 2021 at 17:10 | 1,281 | 5,517 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.948485 |
http://mathhelpforum.com/geometry/81088-find-4th-coordinate-parallelogram.html | 1,480,758,292,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00500-ip-10-31-129-80.ec2.internal.warc.gz | 173,491,032 | 10,103 | # Thread: find 4th coordinate of parallelogram
1. ## find 4th coordinate of parallelogram
How do I find the x and y values algebraically of the fourth coordinate of a parallelogram if I know the other 3 points?
TYIA
2. Originally Posted by hello
How do I find the x and y values algebraically of the fourth coordinate of a parallelogram if I know the other 3 points?
TYIA
Hi
If you know the coordinates of A, B, C and you want to find the coordinates of D such that ABCD is a parallelogram then you can use the vector relation
$\overrightarrow{AB} = \overrightarrow{DC}$ which leads to
$x_B - x_A = x_C - x_D$
$y_B - y_A = y_C - y_D$
from which you can find the coordinates of D
3. thanks for the quick response | 196 | 721 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "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} | 3.125 | 3 | CC-MAIN-2016-50 | longest | en | 0.821751 |
https://www.openlb.net/forum/topic/undesired-condensation-in-multicomponent-flows/ | 1,723,676,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00364.warc.gz | 713,544,816 | 13,770 | OpenLB – Open Source Lattice Boltzmann Code Forums on OpenLB General Topics Undesired condensation in multicomponent flows
Viewing 4 posts - 1 through 4 (of 4 total)
• Author
Posts
• #1964
lollipopcorn
Participant
Hi there,
When simulating multicomponent flows using SC method, I would like to change the interfacial tension between two fluids by changing G value. However, in the case of two fluids of density 1 and 0.8, unexpected condensation occurs when the G value is increased from 3 to 4. How to solve this problem?
Thanks a lot,
John
#2790
mgaedtke
Keymaster
Hey John,
what do you mean by condensation in combination with multiple components? What is your setup up and what is the effect you see there? Please provide some more information, so that I can reproduce your effect.
Best, Max
#2795
lollipopcorn
Participant
Dear Max,
Thanks for your reply. When I say condensation I was actually meaning the precipitation of one fluid in the other fluid. The setup is that fluid 1 is being injected from the left end, displacing fluid 2. (rectangle geometry, constant pressure at right end, bounceback at top and bottom, BouncbackRho1=0.7, BouncebackRho2=0.3). When I increase G from 3 to 4, after the start of the simulation, though most of fluid 1 is still in the left region, I see some little bubbles of fluid 1 precipitation in fluid 2 at up and bottom walls in the middle and right region. it feels like part of fluid one dissolves in fluid 2 and then precipitates at the top and bottom boundary walls.
Why does this happen? especially, if its due to the partial miscibility of two fluids, then, why does it happen when G is 4 but doesn’t when G is 3. since G is a measurement of repulsive forces between fluids, shouldn’t the fluids be “more immiscible” when G is 4?
Thanks,
John
#2799
mathias
Keymaster
Dear John,
What about the initial conditions? How close is the one fluid to 0. Did you try to plax with it like 0.01 or 0.001? Then G=4, the force seperating the fluids in stronger. So, the results seems all right.
Best
Mathias
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http://www.chegg.com/homework-help/abstract-algebra-3rd-edition-chapter-14.7-problem-6e-solution-9780471433347 | 1,475,104,468,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661768.10/warc/CC-MAIN-20160924173741-00147-ip-10-143-35-109.ec2.internal.warc.gz | 387,071,391 | 16,095 | Solutions
Abstract Algebra
# Abstract Algebra (3rd Edition) View more editions Solutions for Chapter 14.7 Problem 6E
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9780471433347ISBN-13: 0471433349ISBN: Authors: | 197 | 729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-40 | latest | en | 0.845208 |
https://brilliant.org/problems/should-i-plot-them/ | 1,532,040,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00508.warc.gz | 606,692,946 | 11,049 | # Should I plot them?
Calculus Level 5
Find the slope of the tangent to the curve $$y=\dfrac{1}{2x}+3$$ at the point where $$x=-1$$. Find the angle which this tangent makes with the curve $$y=2x^2+2$$, where $$x < 0$$. Submit your answer as the angle in degrees. You may use the fact that $$\arctan \left( \frac{1}{2} \right)\approx 26^\circ$$.
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http://www.readbag.com/seas-upenn-meam520-notes02-introrobotkinematics5 | 1,600,889,452,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212039.16/warc/CC-MAIN-20200923175652-20200923205652-00423.warc.gz | 204,152,321 | 15,391 | #### Read ROBOT GEOMETRY AND KINEMATICS text version
`V. Kumar5. Introduction to Robot Geometry and KinematicsThe goal of this chapter is to introduce the basic terminology and notation used in robot geometry and kinematics, and to discuss the methods used for the analysis and control of robot manipulators. The scope of this discussion will be limited, for the most part, to robots with planar geometry. The analysis of manipulators with three-dimensional geometry can be found in any robotics text1.5.1 Some definitions and examples We will use the term mechanical system to describe a system or a collection of rigid or flexible bodies that may be connected together by joints. A mechanism is a mechanical system that has the main purpose of transferring motion and/or forces from one or more sources to one or more outputs. A linkage is a mechanical system consisting of rigid bodies called links that are connected by either pin joints or sliding joints. In this section, we will consider mechanical systems consisting of rigid bodies, but we will also consider other types of joints. Degrees of freedom of a system The number of independent variables (or coordinates) required to completely specify the configuration of the mechanical system. While the above definition of the number of degrees of freedom is motivated by the need to describe or analyze a mechanical system, it also is very important for controlling or driving a mechanical system. It is also the number of independent inputs required to drive all the rigid bodies in the mechanical system. Examples: (a) A point on a plane has two degrees of freedom. A point in space has three degrees of freedom. (b) A pendulum restricted to swing in a plane has one degree of freedom.1Inparticular, two books offer an excellent treatment while keeping the mathematics at a very simple level: (a) Craig, J. J. Introduction to Robotics, Addison-Wesley, 1989; and (b) Paul, R., Robot Manipulators, Mathematics, Programming and Control, The MIT Press, Cambridge, 1981.-1- (c) A planar rigid body (or a lamina) has three degrees of freedom. There are two if you consider translations and an additional one when you include rotations. (d) The mechanical system consisting of two planar rigid bodies connected by a pin joint has four degrees of freedom. Specifying the position and orientation of the first rigid body requires three variables. Since the second one rotates relative to the first one, we need an additional variable to describe its motion. Thus, the total number of independent variables or the number of degrees of freedom is four. (e) A rigid body in three dimensions has six degrees of freedom. There are three translatory degrees of freedom. In addition, there are three different ways you can rotate a rigid body. For example, consider rotations about the x, y, and z axes. It turns out that any rigid body rotation can be accomplished by successive rotations about the x, y, and z axes. If the three angles of rotation are considered to be the variables that describe the rotation of the rigid body, it is evident there are three rotational degrees of freedom. (f) Two rigid bodies in three dimensions connected by a pin joint have seven degrees of freedom. Specifying the position and orientation of the first rigid body requires six variables. Since the second one rotates relative to the first one, we need an additional variable to describe its motion. Thus, the total number of independent variables or the number of degrees of freedom is seven. Kinematic chain A system of rigid bodies connected together by joints. A chain is called closed if it forms a closed loop. A chain that is not closed is called an open chain. Serial chain If each link of an open chain except the first and the last link is connected to two other links it is called a serial chain. An example of a serial chain can be seen in the schematic of the PUMA 560 series robot2, an industrial robot manufactured by Unimation Inc., shown in Figure 1. The trunk is bolted to a fixed table or the floor. The shoulder rotates about a vertical axis with respect to the trunk. The upper arm rotates about a horizontal axis with respect to the shoulder. This rotation is the shoulder joint rotation. The forearm rotates about a horizontal axis (the elbow) with respect to the upper arm. Finally, the wrist consists of an assembly of three rigid bodies with three2TheProgrammable Universal Machine for Assembly (PUMA) was developed in 1978 by Unimation Inc. using a set of specifications provided by General Motors.Robot Geometry and Kinematics-2-V. Kumar additional rotations. Thus the robot arm consists of seven rigid bodies (the first one is fixed) and six joints connecting the rigid bodies.Figure 1 The six degree-of-freedom PUMA 560 robot manipulator.Figure 2 The six degree-of-freedom T3 robot manipulator.Robot Geometry and Kinematics-3-V. Kumar Another schematic of an industrial robot arm, the T3 made by Cincinnati Milacron, is shown in Figure 2. Once again, it is possible to model it as a collection of seven rigid bodies (the first being fixed) connected by six joints3. Types of joints There are mainly four types of joints that are found in robot manipulators: · · · · Revolute, rotary or pin joint (R) Prismatic or sliding joint (P) Spherical or ball joint (S) Helical or screw joint (H)The revolute joint allows a rotation between the two connecting links. The best example of this is the hinge used to attach a door to the frame. The prismatic joint allows a pure translation between the two connecting links. The connection between a piston and a cylinder in an internal combustion engine or a compressor is via a prismatic joint. The spherical joint between two links allows the first link to rotate in all possible ways with respect to the second. The best example of this is seen in the human body. The shoulder and hip joints, called ball and socket joints, are spherical joints. The helical joint allows a helical motion between the two connecting bodies. A good example of this is the relative motion between a bolt and a nut. Planar chain All the links of a planar chain are constrained to move in or parallel to the same plane. A planar chain can only allow prismatic and revolute joints. In fact, the axes of the revolute joints must be perpendicular to the plane of the chain while the axes of the prismatic joints must be parallel to or lie in the plane of the chain. An example of a planar chain is shown in Figure 3. Almost all internal combustion engines use a slider crank mechanism. The high pressure of the expanding gases in the combustion chamber is used to translate the piston and the mechanism converts this translatory movement into the rotary movement of the crank. This mechanical system consists of three revolute joints and one prismatic joint. The example in Figure 3 is a planar, closed, kinematic chain. Examples of planar, serial chains are shown in Figure 4 and 5.3Thisis a convenient model. A more accurate kinematic model is required to model the coupling between the actuator that drives the elbow joint and the elbow joint.Robot Geometry and Kinematics-4-V. Kumar Connectivity of a joint The number of degrees of freedom of a rigid body connected to a fixed rigid body through the joint. The revolute, prismatic and helical joint have a connectivity 1. The spherical joint has a connectivity of 3. Sometimes one uses the term "degree of freedom of a joint" instead of the connectivity of a joint.Crank shaft Connecting rodr l PistonCylinderFxFigure 3 A schematic of a slider crank mechanismEND-EFFECTORLinkRLink JointRJointLinkRACTUATORSJointFigure 4 A schematic of a planar manipulator with three revolute jointsRobot Geometry and Kinematics-5-V. Kumar END-EFFECTOR R P ACTUATORS RFigure 5 A schematic of a planar manipulator with two revolute and one prismatic joints Mobility The mobility of a chain is the number of degrees of freedom of the chain. Most books will use the term "number of degrees of freedom" for the mobility. In a serial chain, the mobility of the chain is easily calculated. If there are n joints and joint i has a connectivity fi,M = fii =1nMost industrial robots have either revolute or prismatic joints (fi = 1) and therefore the mobility or the number of degrees of freedom of the robot arm is also equal to the number of joints. Sometimes, an n degree of freedom robot or a robot with mobility n is also called an n axis robot. Since a rigid body in space has six degrees of freedom, the most general robots are designed to have six joints. This way, the end effector or the link that is furthest away from the base can be made to assume any position or orientation (within some range). However, if the end effector needs to moved around in a plane, the robot need only have three degrees of freedom. Two examples4 of planar, three degree of freedom robots (technically, mobility three robots) are shown in Figures 4 and 5.4Notethat we do not count the opening and closing of the gripper as a degree of freedom. The gripper is usually completely open or completely shut and it is not continuously controlled as the other joints are. Also, the gripper freedom does not participate in the positioning and orienting of a part held by the gripper.Robot Geometry and Kinematics-6-V. Kumar When closed loops are present in the kinematic chain (that is, the chain is no longer serial, or even open), it is more difficult to determine the number of degrees of freedom or the mobility of the robot. But there is a simple formula that one can derive for this purpose. Let n be the number of moving links and let g be the number of joints, with fi being the connectivity of joint i. Each rigid body has six degrees if we consider spatial motions. If there were no joints, since there are n moving rigid bodies, the system would have 6n degrees of freedom. The effect of each joint is to constrain the relative motion of the two connecting bodies. If the joint has a connectivity fi, it imposes (6-fi) constraints on the relative motion. In other words, since there are fi different ways for one body to move relative to another, there (6-fi) different ways in which the body is constrained from moving relative to another. Therefore, the number of degrees of freedom or the mobility of a chain (including the special case of a serial chain) is given by:M = 6n - (6 - f i )i =1gor,M = 6(n - g ) + f ii =1g(2)END-EFFECTORACTUATORSFigure 6 A planar parallel manipulator.Robot Geometry and Kinematics-7-V. Kumar In the special case of planar motion, since each unconstrained rigid body has 3 degrees of freedom, this equation is modified to:M = 3(n - g ) + f ii =1g(3)Example 1 In Figures 4 and Figure 5, since n=g=3, Equation (3) reduces to the special case of Equation (1). And since f1 = f2 = f3 = 1, and M=g. Example 2 In the slider crank mechanism shown in Figure 3, n=3 andg=4. Since it is a planar mechanism we use Equation (2). All four joints have connectivity one: f1 = f2 = f3 = f4 = 1, and M=1. Example 3 Consider the parallel manipulator shown in Figure 6. Here, n = 7, g=9, and fi=1. According to Equation (3), M =3. There are correspondingly three actuators in the manipulator. Contrast this arrangement with the arrangement shown in Figures 4 and 5. The three actuators are mounted in parallel in Figure 6. In Figures 4 and 5, they are mounted "sequentially" in a serial fashion. The Stewart Platform The Stewart-Gough or the Stewart Platform5 device is a six degree of freedom (mobility six) kinematic chain with closed loops. The kinematic chain consists of a base and a moving platform each of which is a spatial hexagon. See Figure 7. Every vertex of the base hexagon is connected to one vertex of the moving platform hexagon by one leg. Similarly, every vertex of the moving hexagon is connected to a vertex of the base hexagon by a leg. There are six such legs. Each leg has is a serial chain consisting of two revolute joints with intersecting axes, a prismatic joint and a spherical joint. Typically the prismatic joints are actuated. The mobility of a Stewart Platform can be easily verified to be six. Each leg has three links and four joints. If we include the moving platform,n = 6 × 3+1 = 19.Stewart, "A Platform with Six Degrees of Freedom," The Institution of Mechanical Engineers, Proceedings 1965-66, Vol. 180 Part 1, No. 15, pages 371-386.5D.Robot Geometry and Kinematics-8-V. Kumar (a) A machine tool based on the Stewart Platform (Ingersoll Rand)6SEND EFFECTORLeg 5 Leg 4 Leg 3 Leg 6R PLeg 2Leg 1RBASELEG NO. i(b) A schematic showing the six legs (left) and the RRPS chain (right). Figure 7 The Stewart Platform6M.Valenti, "Machine Tools Get Smarter," Mechanical Engineering, Vol.117, No.11, November 1995.Robot Geometry and Kinematics-9-V. Kumar The connectivity of the revolute and the prismatic joint is one. The connectivity of the spherical joint is three. Since there are 6×2 revolute joints, 6 prismatic joints and 6 spherical joints, f i = 12 + 6 + 6 × 3 = 36i =1gAccording to Equation (3), M = 6(19 - 24) + 36 = 6 The Stewart Platform has actuators for all its six prismatic joints and it is therefore possible to control all six degrees of freedom.(a) The Adept 1850 Palletizer(b) side view (axes 2-4 are numbered)(c) top view (axes 2-4 are numbered) Figure 8 The Adept 1850 Palletizer There are four degrees of freedom in this SCARA manipulator. Joint 1 is a sliding joint that carries the manipulator arm up or down. Joints 2-4 are rotary joints with vertical axes.Robot Geometry and Kinematics-10-V. Kumar 5.2 Geometry of planar robot manipulators The mathematical modeling of spatial linkages is quite involved. It is useful to start with planar robots because the kinematics of planar mechanisms is generally much simpler to analyze. Also, planar examples illustrate the basic problems encountered in robot design, analysis and control without having to get too deeply involved in the mathematics. However, while the examples we will discuss will involve kinematic chains that are planar, all the definitions and ideas presented in this section are general and extend to the most general spatial mechanisms. We will start with the example of the planar manipulator with three revolute joints. The manipulator is called a planar 3R manipulator. While there may not be any three degree of freedom (d.o.f.) industrial robots with this geometry, the planar 3R geometry can be found in many robot manipulators. For example, the shoulder swivel, elbow extension, and pitch of the Cincinnati Milacron T3 robot (Figure 2) can be described as a planar 3R chain. Similarly, in a four d.o.f. SCARA manipulator (Figure 8), if we ignore the prismatic joint for lowering or raising the gripper, the other three joints form a planar 3R chain. Thus, it is instructive to study the planar 3R manipulator as an example. In order to specify the geometry of the planar 3R robot, we require three parameters, l1, l2, and l3. These are the three link lengths. In Figure 9, the three joint angles are labeled 1, 2, and 3. These are obviously variable. The precise definitions for the link lengths and joint angles are as follows. For each pair of adjacent axes we can define a common normal or the perpendicular between the axes. · · · The ith common normal is the perpendicular between the axes for joint i and joint i+1. The ith link length is the length of the ith common normal, or the distance between the axes for joint i and joint i+1. The ith joint angle is the angle between the (i-1)th common normal and ith common normal measured counter clockwise going from the (i-1)th common normal to the ith common normal.Note that there is some ambiguity as far as the link length of the most distal link and the joint angle of the most proximal link are concerned. We define the link length of the most distal link from the most distal joint axis to a reference point or a tool point on the end effector7. Generally, this is the center of the gripper or the end point of the tool. Since there is no zeroth common normal, we measure the first joint angle from a convenient reference line. Here, we have chosen this to be the x axis of a conveniently defined fixed coordinate system.7Thereference point is often called the tool center point (TCP).Robot Geometry and Kinematics-11-V. Kumar Another set of variables that is useful to define is the set of coordinates for the end effector. These coordinates define the position and orientation of the end effector. With a convenient choice of a reference point on the end effector, we can describe the position of the end effector using the coordinates of the reference point (x, y) and the orientation using the angle . The three end effector coordinates (x, y, ) completely specify the position and orientation of the end effector8.REFERENCE POINT(x, y)l33l2y l11 2xFigure 9 The joint variables and link lengths for a 3R planar manipulator As another example, consider the three d.o.f. cylindrical robot in Figure 10. If we ignore the lift freedom, the rotation of the base and the extension of the arm give us the two d.o.f. robot shown in Figure 11 that we can call the R-P manipulator. It consists of a revolute joint and prismatic joint as shown in the figure. 1, the base rotation, and d2, the arm extension, are the two joint variables. Note that there are no constant parameters such as the three link lengths in the 3R manipulator. The joint variable 1 is defined as before. Since there is no zeroth common normal,8Thedescription of the position and orientation of a three dimensional rigid body is significantly more complicated. For a spatial manipulator, a typical set of end effector coordinates would include three variables (x, y, z) for the position, and three Euler angles (, , ) for the orientation. -12V. KumarRobot Geometry and Kinematics we measure the joint angle from the x axis which we have chosen to be horizontal. d2 is defined as the distance from joint axis 1 to the reference point on the end effector. As in the previous example, the end effector coordinates are variables that completely specify the position and orientation of the end effector. In the figure, they are (x, y, ). Finally, we consider a Cartesian robot consisting of two prismatic joints at right angles. The P-P chain is found in x-y tables, plotters and milling machines. A schematic is shown in Figure 12. The simplest spatial manipulator is based on the P-P-P chain, which has a third prismatic joint. The three joint axes are mutually orthogonal. The Gantry robot in Figure 13 has this geometry. If you ignore the vertical up/down degree of freedom it is a P-P manipulator.Figure 10 The RT3300 cylindrical robot (Seiko)REFERENCE POINT (x, y)yd2 2 1 1xFigure 11 The joint variables and link lengths for a R-P planar manipulatorRobot Geometry and Kinematics-13-V. Kumar d2 d1Figure 12 The joint variables for a P-P planar manipulatorFigure 13 The G365 Gantry robot manipulator (CRS Robotics) on the left, and the Biomek 2000 Laboratory Automation Workstation (Beckman Coulter) on the right both have tooling mounted at the end of a P-P-P chain. The end effector of a manipulator that has only prismatic joints is constrained to remain in the same orientation. Thus, the end effector coordinates for the P-P manipulator only include the coordinates of the reference point on the end effector (x, y). In summary, in each case, we defined a set of constant parameters called link lengths (li) and set of joint variables or joint coordinates consisting of either joint angles (i) or displacementsRobot Geometry and Kinematics-14-V. Kumar (di). We also defined a set of variables called end effector coordinates. The link lengths are constant parameters that define the geometry of the manipulator. The joint variables define the configuration of the manipulator by specifying the position of each joint. The end effector coordinates define the position and orientation of the end effector. If the joint coordinates specify the configuration of the manipulator, they should also specify the position and orientation of the end effector. Thus one should expect to find an explicit dependence of the end effector coordinates on the joint coordinates. Although it may not be obvious, there is also a dependence of the joint coordinates on the end effector coordinates. The next subsection will address this dependence and analyse the kinematics of robot manipulators.5.3 Kinematic analysis of planar serial chains Kinematics is the study of motion. In this subsection, we will explore the relationship between joint movements and end effector movements. More precisely, we will try to develop equations that will make explicit the dependence of end effector coordinates on joint coordinates and vice versa. We will start with the example of the planar 3R manipulator. From basic trigonometry, the position and orientation of the end effector can be written in terms of the joint coordinates in the following way: = (1 + 2 + 3 )y = l1 sin 1 + l2 sin (1 + 2 ) + l3 sin (1 + 2 + 3 )x = l1 cos 1 + l2 cos(1 + 2 ) + l3 cos(1 + 2 + 3 )(4)Note that all the angles have been measured counter clockwise and the link lengths are assumed to be positive going from one joint axis to the immediately distal joint axis. Equation (4) is a set of three nonlinear equations9 that describe the relationship between end effector coordinates and joint coordinates. Notice that we have explicit equations for the end effector coordinates in terms of joint coordinates. However, to find the joint coordinates for a given set of end effector coordinates (x, y, ), one needs to solve the nonlinear equations for 1, 2, and 3. The kinematics of the planar R-P manipulator is easier to formulate. The equations are:9Thethird equation is linear but the system of equations is nonlinear.Robot Geometry and Kinematics-15-V. Kumar x = d 2 cos 1 y = d 2 sin 1 = 1 Again the end effector coordinates are explicitly given in terms of the joint coordinates. However, since the equations are simpler (than in (4)), you would expect the algebra involved in solving for the joint coordinates in terms of the end effector coordinates to be easier. Notice that in contrast to (4), now there are three equations in only two joint coordinates, 1, and d2. Thus, in general, we cannot solve for the joint coordinates for an arbitrary set of end effector coordinates. Said another way, the robot cannot, by moving its two joints, reach an arbitrary end effector position and orientation. Let us instead consider only the position of the end effector described by (x, y), the coordinates of the end effector tool point or reference point . We have only two equations: (5)x = d 2 cos 1 y = d 2 sin 1Given the end effector coordinates (x, y), the joint variables can be computed to be:d2 = + x2 + y2(6)(7) y 1 = tan -1 x Notice that we restricted d2 to positive values. A negative d2 may be physically achieved by allowing the end effector reference point to pass through the origin of the x-y coordinate system over to another quadrant. In this case, we obtain another solution:d2 = - x2 + y2 y 1 = tan -1 x(8)In both cases (7-8), the inverse tangent function is multivalued10. In particular, (9) tan(x) = tan(x + k), k=...-2, -1, 0, 1, 2, ... However, if we limit 1 to the range 0<1<2, there is a unique value of 1 that is consistent with the given (x, y) and the computed d2 (for which there are two choices).Appendix 1, we define another inverse tangent function called atan2 that takes two arguments, the sine and the cosine of an angle, and returns a unique angle in the range [0, 2).10InRobot Geometry and Kinematics-16-V. Kumar The existence of multiple solutions is typical when we solve nonlinear equations. As we will see later, this poses some interesting questions when we consider the control of robot manipulators. The planar Cartesian manipulator is trivial to analyze. The equations for kinematic analysis are:x = d2,y = d1(10)The simplicity of the kinematic equations makes the conversion from joint to end effector coordinates and back trivial. This is the reason why P-P chains are so popular in such automation equipment as robots, overhead cranes, and milling machines. Direct kinematics As seen earlier, there are two types of coordinates that are useful for describing the configuration of the system. If we focus our attention on the task and the end effector, we would prefer to use Cartesian coordinates or end effector coordinates. The set of all such coordinates is generally referred to as the Cartesian space or end effector space11. The other set of coordinates is the so called joint coordinates that is useful for describing the configuration of the mechanical lnkage. The set of all such coordinates is generally called the joint space. In robotics, it is often necessary to be able to "map" joint coordinates to end effector coordinates. This map or the procedure used to obtain end effector coordinates from joint coordinates is called direct kinematics. For example, for the 3-R manipulator, the procedure reduces to simply substituting the values for the joint angles in the equations = (1 + 2 + 3 )y = l1 sin 1 + l2 sin (1 + 2 ) + l3 sin (1 + 2 + 3 )x = l1 cos 1 + l2 cos(1 + 2 ) + l3 cos(1 + 2 + 3 )and determining the Cartesian coordinates, x, y, and . For the other examples of open chains discussed so far (R-P, P-P) the process is even simpler (since the equations are simpler). In fact, for all serial chains (spatial chains included), the direct kinematics procedure is fairly straight forward. On the other hand, the same procedure becomes more complicated if the mechanism contains one or more closed loops. In addition, the direct kinematics may yield more than one solution oreach member of this set is an n-tuple, we can think of it as a vector and the space is really a vector space. But we shall not need this abstraction here.11SinceRobot Geometry and Kinematics-17-V. Kumar no solution in such cases. For example, in the planar parallel manipulator in Figure 3, the joint positions or coordinates are the lengths of the three telescoping links (q1, q2, q3) and the end effector coordinates (x, y, ) are the position and orientation of the floating triangle. It can be shown that depending on the value of (q1, q2, q3), the number of (real) solutions for (x, y, ) can be anywhere from zero to six. For the Stewart Platform in Figure 4, this number has been shown to be anywhere from zero to forty.5.4 Inverse kinematics The analysis or procedure that is used to compute the joint coordinates for a given set of end effector coordinates is called inverse kinematics. Basically, this procedure involves solving a set of equations. However the equations are, in general, nonlinear and complex, and therefore, the inverse kinematics analysis can become quite involved. Also, as mentioned earlier, even if it is possible to solve the nonlinear equations, uniqueness is not guaranteed. There may not (and in general, will not) be a unique12 set of joint coordinates for the given end effector coordinates. We saw that for the R-P manipulator, the direct kinematics equations are:x = d 2 cos 1 y = d 2 sin 1(6)If we restrict the revolute joint to have a joint angle in the interval [0, 2), there are two solutions for the inverse kinematics: y x d 2 = x 2 + y 2 , 1 = atan 2 , , = ±1 d d 2 2 Here we have used the atan2 function in Appendix 1 to uniquely specify the joint angle 1. However, depending on the choice of , there are two solutions for d2, and therefore for 1. The inverse kinematics analysis for a planar 3-R manipulator appears to be complicated but we can derive analytical solutions. Recall that the direct kinematics equations (4) are: x = l1 cos 1 + l 2 cos(1 + 2 ) + l3 cos(1 + 2 + 3 ) = (1 + 2 + 3 )(4a) (4b) (4c)y = l1 sin 1 + l 2 sin (1 + 2 ) + l3 sin (1 + 2 + 3 )12Theonly case in which the analysis is trivial is the P-P manipulator. In this case, there is a unique solution for the inverse kinematics.Robot Geometry and Kinematics-18-V. Kumar We assume that we are given the Cartesian coordinates, x, y, and and we want to find analytical expressions for the joint angles 1, 2, and 3 in terms of the Cartesian coordinates. Substituting (4c) into (4a) and (4b) we can eliminate 3 so that we have two equations in 1 and 2: x - l3 cos = l1 cos 1 + l 2 cos(1 + 2 ) y - l3 sin = l1 sin 1 + l2 sin (1 + 2 ) (d) (e)where the unknowns have been grouped on the right hand side; the left hand side depends only on the end effector or Cartesian coordinates and are therefore known. Rename the left hand sides, x = x - l3 cos , y = y - l3 sin , for convenience. We regroup terms in (d) and (e), square both sides in each equation and add them:(x - l1 cos 1 )2 = (l 2 cos(1 + 2 ))2+( y - l1 sin 1 )2 = (l 2 sin (1 + 2 ))2After rearranging the terms we get a single nonlinear equation in 1:2 (- 2l1x) cos 1 + (- 2l1 y)sin 1 + (x2 + y2 + l12 - l2 ) = 0(f)Notice that we started with three nonlinear equations in three unknowns in (a-c). We reduced the problem to solving two nonlinear equations in two unknowns (d-e). And now we have simplified it further to solving a single nonlinear equation in one unknown (f). Equation (f) is of the type P cos + Q sin +R = 0 (g)Equations of this type can be solved using a simple substitution as shown in Appendix 2. There are two solutions for 1 given by: - x 2 + y 2 + l 2 - l 2 1 2 1 = + cos -1 2 2 2l1 x + y () (h)where, x - y = a tan 2 , 2 2 x 2 + y 2 x + y , and = ±1 .Robot Geometry and Kinematics-19-V. Kumar Note that there are two solutions for 1, one corresponding to =+1, the other corresponding to =-1. Substituting any one of these solutions back into Equations (d) and (e) gives us:x - l1 cos 1 l2 y - l1 sin 1 sin (1 + 2 ) = l2 cos(1 + 2 ) = This allows us to solve for 2 using the atan2 function in Appendix 1: y - l1 sin 1 x - l1 cos 1 - 1 2 = atan 2 , l2 l2 (i)Thus, for each solution for 1, there is one (unique) solution for 2. Finally, 3 can be easily determined from (c): 3 = - 1 - 2 (j)Equations (h-j) are the inverse kinematics solution for the 3-R manipulator. For a given end effector position and orientation, there are two different ways of reaching it, each corresponding to a different value of . These different configurations are shown in Figure 14.REFER ENC E P OINT(x,y) =+1 =-1Figure 14 The two inverse kinematics solutions for the 3R manipulator: "elbow-up" configuration (=+1) and the "elbow-down" configuration (= -1) Commanding a robot to move the end effector to a certain position and orientation is ambiguous because there are two configurations that the robot must choose from. From aRobot Geometry and Kinematics-20-V. Kumar practical point of view, if the joint limits are such that one configuration cannot be reached this ambiguity is automatically resolved13.5.5 Velocity analysisWhen controlling a robot to go from one position to another, it is not just enough to determine the joint and end effector coordinates of the target position. It may be necessary to continuously control the trajectory or the path taken by the robot as it moves toward the target position. This is essential to avoid obstacles in the workspace. More importantly, there are tasks where the trajectory of the end effector is critical. For example, when welding, it is necessary to maintain the tool at a desired orientation and a fixed distance away from the workpiece while moving uniformly14 along a desired path. Thus one needs to control the velocity of the end effector or the tool during the motion. Since the control action occurs at the joints, it is only possible to control the joint velocities. Therefore, there is a need to be able to take the desired end effector velocities and calculate from them the joint velocities. All this requires a more detailed kinematic analysis, one that addresses velocities or the rate of change of coordinates in contrast to the previous section where we only looked at positions or coordinates. Consider the 3R manipulator as an example. By differentiating Equation (4) with respect to time, it is possible to obtain equations that relate the the different velocities.& & & & & & & x = -l11s1 - l2 1 + 2 s12 - l3 1 + 2 + 3 s123 & & & & & & & y = l c +l + c +l + + c & & & & = 1 + 2 + 3(1 1 12(1()2 12))3(1(23 123))where we have used the short hand notation:s1 = sin 1, s12 = sin (1 + 2), s123 = sin (1 + 2 + 3) c1 = cos 1, c12 = cos (1 + 2), c123 = cos (1 + 2 + 3)& & i denotes the joint speed for the ith joint or the time derivative of the ith joint angles, and x , & & y , and are the time derivatives of the end effector coordinates. Rearranging the terms, we canwrite this equation in matrix form:is true of the human arm. If you consider planar movements, because the human elbow cannot be hyper extended, there is a unique solution for the inverse kinematics. Thus the central nervous system does not have to worry about which configuration to adopt for a reaching task. 14In some cases, a weaving motion is required and the trajectory of the tool is more complicated.13ThisRobot Geometry and Kinematics-21-V. Kumar & & x - (l1s1 + l2 s12 + l3s123 ) - (l2 s12 + l3 s123 ) - l3 s123 1 & & y = (l1c1 + l2c12 + l3c123 ) (l2c12 + l3c123 ) l3c123 2 & & 1 1 1 3 (11)The 3×3 matrix is called the Jacobian matrix15 and we will denote it by the symbol J. If you look at the elements of the matrix they express the rate of change of the end effector coordinates with respect to the joint coordinates: x 1 y J= 1 1 x 2 y 2 2 x 3 y 3 3 Given the rate at which the joints are changing, or the vector of joint velocities,& 1 & & q = 2 , & 3 using Equation (11), we can obtain expressions for the end effector velocities,& x & & p = y . & If the Jacobian matrix is non singular (its determinant is non zero and the matrix is invertible), then we can get the following expression for the joint velocities in terms of the end effector velocities:& & p = Jq,& & q = J -1 p(12)Thus if the task (for example, welding) is specified in terms of a desired end effector velocity, Equation (12) can be used to compute the desired joint velocity provided the Jacobian is non singular:15Thename Jacobian comes from the terminology used in multi-dimensional calculus.Robot Geometry and Kinematics-22-V. Kumar & 1 & 2 & 3 des=- (l1s1 + l2 s12 + l3 s123 ) - (l2 s12 + l3 s123 ) - l3 s123 (l1c1 + l2c12 + l3c123 ) (l2c12 + l3c123 ) l3c123 1 1 1 -1& x & y & desNaturally we want to determine the conditions under which the Jacobian becomes singular. This can be done by computing the determinant of J and setting it to zero. Fortunately, the expression for the determinant of the Jacobian, in this example, can be simplified using trigonometric identities to: |J| = l1 l2 sin 2 (13) This means that the Jacobian is singular only when 2 is either 0 or 180 degrees. Physically, this corresponds to the elbow being completely extended or completely flexed. Thus, as long we avoid going through this configuration, the robot will be able to follow any desired end effector velocity.5.6 Appendix5.6.1 The ambiguity in inverse trigonometric functions Inverse trigonometric functions have multiple values. Even within a 360 degree range they have two values. For example, if y = sin x the inverse sin function gives two values in a 360 degree interval: sin-1y = x, -x Of course we can add or subtract 2 from either of these solutions and obtain another solution. This is true of the inverse cosine and inverse tangent functions as well. If y = cos x, the inverse cosine function yields: cos-1y = x, -x Similarly, for the tangent function y = tan x,Robot Geometry and Kinematics-23-V. Kumar the inverse tangent function yields: tan-1y = x, +x This multiplicity is particularly troublesome in robot control where an ambiguity may mean that there is more than one way of reaching a desired position (see discussion on inverse kinematics). This problem is circumvented by defining the atan2 function which requires two arguments and returns a unique answer in a 360 range. The atan2 function takes as arguments the sine and cosine of a number and returns the number. Thus if , s = sin x; c = cos xthe atan2 function takes s and c as the argument and returns x: atan2 (s, c) = x The main idea is that the additional information provided by the second argument eliminates the ambiguity in solving for x. To see this consider the simple problem where we are given: 1 s=2; 3 c = 2and we are required to solve for x. If we use the inverse sine function and restrict the answer to be in the interval [0, 2), we get the result: 1 5 x = sin-1 2 = 6 , 6 3 Since we know the cosine to be 2 , we can quickly verify by taking cosines of both candidate solutions that the first solution is correct and the second one is incorrect. 3 3 5 cos 6 = 2 ; cos 6 = - 2 The atan2 function goes through a similar algorithm to figure out a unique solution in the range [0,2). 1 3 atan2 ( 2 , 2 ) = 6 The atan2 function is a standard function in most C, Pascal and Fortran compilers.Robot Geometry and Kinematics-24-V. Kumar 5.6.2 Solution of the nonlinear equation in (g) P cos + Q sin +R = 0 Define so that P cos = 2+Q2 P (g)andsin =Q P2+Q2Note that this is always possible. can be determined by using the atan2 function: Q P , = a tan 2 2 2 P2 + Q2 P +Q Now (g) can be rewritten as: cos cos + sin sin + or cos( - ) = -R P2 + Q2 R P2 + Q2 =0 This gives us two solutions for in terms of the known angle : , = ±1 = + cos 2 2 P +Q -1 -RRobot Geometry and Kinematics-25-V. Kumar `
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https://www.physicsforums.com/threads/linear-algebra-proofs.341484/ | 1,531,872,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00461.warc.gz | 949,015,786 | 13,035 | # Homework Help: Linear algebra proofs
1. Sep 29, 2009
### mafquestion
1) prove that for any five vectors (x1, ..., x5) in R3 there exist real numbers (c1, ...., c5), not all zero, so that BOTH
c1x1+c2x2+c3x3+c4x4+c5x5=0 AND c1+c2+c3+c4+c5=0
2)Let T:R5-->R5 be a linear transformation and x1, x2 & x3 be three non-zero vectors in R5 so that
T(x1)=x1
T(x2)=x1+x2
T(x3)=x2+x3
prove that {x1, x2, x3} are three linearly independent vectors.
any help would be greatly appreciated, thank you!
2. Sep 29, 2009
### aPhilosopher
I've thought up a proof for the first one but it might be too complicated. I'll try to think of a simpler one if somebody else doesn't.
As for the second, assume that you have a linear combination of the three equal to zero. Map it under the matrix and see if something cool happens. Then see if it happens again. There's probably a contradiction with the assumptions in there somewhere ;)
3. Sep 29, 2009
### Office_Shredder
Staff Emeritus
For question 1)
Extend a vector in R3 to one in R4 by adding a 1 in the fourth entry.
Last edited: Sep 29, 2009 | 341 | 1,089 | {"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} | 3.640625 | 4 | CC-MAIN-2018-30 | latest | en | 0.900149 |
https://www.physicsforums.com/threads/3-phase-synchronous-machine.422838/#post-2847979 | 1,643,174,674,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00537.warc.gz | 963,554,489 | 14,341 | # 3 Phase Synchronous Machine
## Homework Statement
A star-connected, 3-phase, 100kVA, 415V, 50Hz synchronous motor having a synchronous reactance of 1.2ohms and negligible resistance operates at its rated terminal voltage, rated current and per-phase field-induced emf of 228V. Determine the power factor.
The excitation is adjusted until the numerical value of the power factor is the same as initially but the sign of the reactive power is changed. Determine the new value of field induced emf.
## Homework Equations
Power = 3 Vp.Ip.PF.
Complex Power = Vl.Il/root3
V = Ef + jIaXs
Vl = Vp. root 3
Complex Power = S = P + jQ
## The Attempt at a Solution
I calculated the following(correctly, I think):
The synchronous angular velocity is 100.pi rad/s.
The line current is 46.37 Amps
IaXs = 55.65V
Phase Voltage = 239.6V
I can't seem to find the power factor though.
The answers are PF = 0.89, New value of field induced emf = 365V
As always any help would be greatly appreciated. Thank you | 272 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-05 | latest | en | 0.907688 |
http://www.fieldtriptoolbox.org/faq/what_is_the_idea_behind_statistical_inference_at_the_second-level | 1,527,210,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00544.warc.gz | 369,189,337 | 6,784 | A common approach for statistical inference is to split the statistics into two levels. At level 1 you compute a within-subject statistic which describes the effect size (i.e. no inference). At level 2 you do the inference, testing whether the effect is consistent over subjects.
Conventionally you would do the inference over the single subject between-condition difference in the averages. Here you replace the difference between the averages by another difference measure. Using the t-score as the measure of difference suppresses differences in the non-consistent parts of the data (e.g. channels that don't show an effect and time points in the baseline) and stresses the difference in consistent large-effect channels. The second level statistical inference thereby becomes more sensitive. Furthermore, the multiple comparison problem at the second level becomes easier to solve using randomization testing, because the randomization distribution will be less broadened by the uninteresting noisy parts of the data.
Some notes:
1. also in conventional statistics you do it in two steps: the first being the computation of the difference in a measure of central tendency (the mean), the second being the inference based on the between-subject consistency of these difference scores. The first step in conventional statistics (the computation of the mean) is often not recognized as such.
2. if you have different numbers of trials between subjects, it is better to carry z-scores to the second level. You can convert t- to z-scores.
3. instead of taking conventional t-scores to start with, you can use a Winsorized estimate of the mean and standard deviation to compute the t-scores at the first level. That makes them more rubust for outliers (common in EEG).
4. quite often beta values estimated with a GLM are used as the statistic to be carried on to the second level. This allows you to explain part of the uninteresting variance in the data with confound-regressors. | 389 | 1,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-22 | latest | en | 0.937175 |
https://www.physicsforums.com/threads/rgb-to-wavelenght-or-viceversa.81702/ | 1,670,626,664,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00631.warc.gz | 994,551,932 | 14,890 | # RGB to wavelenght or viceversa.
neurocomp2003
How would one convert RGB values to a wavelength value
OR
a wavelength value to RGB?
ie 180,180,60=?
Last edited:
Gold Member
I think that may be impossible. A blend of different colors of light does not have a unique wavelength. For example, you can't talk about the wavelength of the color brown because it is a mixture of light at different wavelengths.
neurocomp2003
but doesn't teh superposition mean that those different wavelengths...superimpose into one wave? with a wavelength because its a finite #?
Gold Member
No, waves of different frequencies do not combine into a single, normal wave. For example you can add two cosine waves of different frequencies. Suppose the average frequency of the two waves is m and that one wave has frequency (m+x) and the other (m-x). Then the function describing the amplitude of the wave is A=cos[(m+x)t] + cos[(m-x)t]. Using Euler's formula, that becomes
A = Re: e^[i(m+x)t] + e^[i(m-x)t]
= Re: {e^(imt)}*{e^(ixt) + e^(-ixt)}
= Re: {cos(mt) +i sin(mt)}*{cos(xt) + i sin(xt) + cos(-xt) + i sin(-xt)}
= Re: {cos(mt) +i sin(mt)}*{2cos(xt)}
= 2cos(mt)cos(xt)
Which is not a simple cosine wave (It looks like a cosine wave on top of a cosine wave for x<<m)
The eye contains three different light sensitive pigments which have maximum sensitivities at wavelengths corresponding roughly to red, green and blue. It is the relative extent to which these pigments are activated that determines what color is percieved. Each of these pigments can pick out of a complicated waveform that component of the the wave at the frequency to which it is most sensitive. No pure light could activate pigments that peak at different frequencies equally. This means that some colors can not be produced by pure light.
neurocomp2003
"It is the relative extent to which these pigments are activated that determines what color is percieved."
if so does this mean that the bipolar cells of the eye don't receive on/off signals from the cones&rods but REAL valued signals?
Gold Member
The frequency with which a neuron is activated determines the perceived strength of the signal. For example, to use random numbers, if a cell with one pigment is activated 10 times a second and a different one is activated fifteen times a second, the brain will interpret this as meaning the second color is more abundant. | 576 | 2,381 | {"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-2022-49 | latest | en | 0.919924 |
https://kmmiles.com/755-miles-in-km | 1,660,448,248,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00533.warc.gz | 338,196,211 | 6,674 | kmmiles.com
# 755 miles in km
## Result
755 miles equals 1214.795 km
You can also convert 755 km to miles.
## Conversion formula
Multiply the amount of miles by the conversion factor to get the result in km:
755 mi × 1.609 = 1214.795 km
## How to convert 755 miles to km?
The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:
1 mi = 1.609 km
To convert 755 miles into km we have to multiply 755 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:
1 mi → 1.609 km
755 mi → L(km)
Solve the above proportion to obtain the length L in km:
L(km) = 755 mi × 1.609 km
L(km) = 1214.795 km
The final result is:
755 mi → 1214.795 km
We conclude that 755 miles is equivalent to 1214.795 km:
755 miles = 1214.795 km
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case seven hundred fifty-five miles is approximately one thousand two hundred fourteen point seven nine five km:
755 miles ≅ 1214.795 km
## Conversion table
For quick reference purposes, below is the miles to kilometers conversion table:
miles (mi) kilometers (km)
756 miles 1216.404 km
757 miles 1218.013 km
758 miles 1219.622 km
759 miles 1221.231 km
760 miles 1222.84 km
761 miles 1224.449 km
762 miles 1226.058 km
763 miles 1227.667 km
764 miles 1229.276 km
765 miles 1230.885 km
## Units definitions
The units involved in this conversion are miles and kilometers. This is how they are defined:
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world. | 603 | 2,201 | {"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-2022-33 | latest | en | 0.830379 |
http://www.cardtrick.ca/trick/Count-Down.html | 1,601,374,875,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00191.warc.gz | 150,488,540 | 4,151 | ## Card Tricks
Card from the Box
A volunteer is asked politely to choose a card from a deck...
Aces Memory
Effect: Trick uses full or almost full card deck: one ace,...
Nice and Easy
Shuffle deck. Let someone choose any card, memorise it and...
Telepathy
EFFECT: The Magician spreads the cards in his hands and as...
Count On Me
This mathematical trick can be performed with any pack ...
From and shuffled deck of cards deal 6 piles of 5 cards ea...
Cardeenie Single
Effect: Showing the front side of your hand empty and then...
I Can't Believe They Don't Get It!
ffect: Good trick, easy to figure out, but it works on peo...
King of the Hill
Preparation: Before you start the trick place the four Kin...
The Lost Kings
Effect: Magician shows the four kings to the spectators. H...
## Count Down
Original Author: Unknown
Trick Description: You shuffle the deck several times and then ask a person to tell you when to STOP when they think you drop ten cards. You then count the correct amount of cards actually dropped. You then tell them to memorize the top card and put it back into the deck shuffling anyway they want to. You then fan all the cards out in a spiral towards the middle and pick their card out for them.
How Its Done
1. Shuffle the card deck a lot, memorizing the bottom card. Depending on the war you shuffle, you should be able to keep a certain card always on the bottom. I usually lead with the right hand first so that is always the bottom card no matter how many times you shuffle. If you lose the card, keep shuffling until you memorize the bottom card.
2. Since you now know what the bottom card really is, it is a simple job getting the other person to pick the card. When you drop cards, no matter how many you drop, you should count the DROPPED cards back to them and eventually the bottom card will be on top.
3. After they shuffle and give the cards back to you, throw the cards out in a pattern to confuse them, then show them their card.
(I like to confuse them even more by throwing out about 10-20 more than the one they actually pick and then say "Is this your card?". Since they will say no... dig through the stack and get the real one.
Next: Cut To It
Previous: Chased | 513 | 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 | 3 | CC-MAIN-2020-40 | latest | en | 0.935378 |
https://www.gurobi.com/documentation/9.5/refman/does_my_model_have_numeric.html | 1,702,289,937,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00487.warc.gz | 874,107,250 | 123,183 | Filter Content By
Version
## Does my model have numerical issues?
You can follow these steps to help determine whether a model is experiencing numerical issues:
1. Isolate the model for testing by exporting a model file and a parameter file. The easiest way to do this is to create a gurobi.env file in your working directory that contains the following line:
Record 1
Once you have done this, running your program will produce one recordingXYZ.grbr file for each Gurobi environment your program creates (where XYZ is a three-digit, 0-padded value that increases from 000). You can replay this recording file using gurobi_cl (e.g., gurobi_cl recording000.grbr). Consult this section for more information on recording files.
2. Using the Gurobi Interactive shell, run some simple Python code to read the model that the replay produces, and print the summary statistics:
m = read('gurobi.rew')
m.printStats()
The output will look like:
Statistics for model (null) :
Linear constraint matrix : 25050 Constrs, 15820 Vars, 94874 NZs
Variable types : 14836 Continuous, 984 Integer
Matrix coefficient range : [ 0.00099, 6e+06 ]
Objective coefficient range : [ 0.2, 65 ]
Variable bound range : [ 1, 5e+07 ]
RHS coefficient range : [ 1, 5e+07 ]
The range of numerical coefficients is one indication of potential numerical issues. As a very rough guideline, the ratio of the largest to the smallest coefficient should be less than ; smaller is better.
In this example, the matrix range is
3. If possible, re-solve the model using the same parameters and review the logs. With the Python shell, use code like the following:
m.read('gurobi.prm')
m.optimize()
Here are some examples of warning messages that suggest numerical issues:
Warning: Model contains large matrix coefficient range
Consider reformulating model or setting NumericFocus parameter
to avoid numerical issues.
Warning: Markowitz tolerance tightened to 0.5
Numeric error
Numerical trouble encountered
Restart crossover...
Sub-optimal termination
Warning: ... variables dropped from basis
Warning: unscaled primal violation = ... and residual = ...
Warning: unscaled dual violation = ... and residual = ...
4. When the optimize function completes, print solution statistics. With the Python shell, use code like the following:
m.printQuality()
which provides a summary of solution quality:
Solution quality statistics for model Unnamed :
Maximum violation:
Bound : 2.98023224e-08 (X234)
Constraint : 9.30786133e-04 (C5)
Integrality : 0.00000000e+00
Violations that are larger than the tolerances are another indication of numerical issues. Also, for a pure LP (without integer variables), print the condition number via the following Python command:
m.KappaExact
The condition number measures the potential for error in linear calculations; a large condition number, such as , is another indication of possible numerical issues, see this section for more details.
5. If changing parameters (e.g., Method or Seed) leads to a different optimization status (e.g., Infeasible instead of optimal), or if the optimal objective values changes, this is usually a sign of numerical issues. To further assess this you can tighten tolerances (to the order of or even ), and see if the behavior of the solver becomes consistent again. Note that tightening tolerances usually comes at the price of more computing time, and should not be considered as a solution for numerical issues.
Choose the evaluation license that fits you best, and start working with our Expert Team for technical guidance and support.
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##### Cloud Trial
Request free trial hours, so you can see how quickly and easily a model can be solved on the cloud. | 887 | 4,064 | {"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} | 2.53125 | 3 | CC-MAIN-2023-50 | longest | en | 0.734115 |
https://math.stackexchange.com/questions/876905/is-there-a-way-to-determine-the-matrix-of-lambdakt-given-the-matrix-of-t/877554#877554 | 1,718,786,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00704.warc.gz | 343,194,093 | 37,136 | # Is there a way to determine the matrix of $\Lambda^k(T)$ given the matrix of $T$?
Let $T$ be an endomorphism of a finite dimensional vector space $V$. Suppose that $(v_1,\ldots v_n)$ is an ordered basis of $V$. And let $[T]$ be the matrix of $T$ with respect to this basis.
Is there a way to compute the matrix of $\Lambda^k(T)$ with respect to the obvious basis given $[T]$? The 'obvious' basis is the $\binom{n}{k}$ $k$-wedge tuples from $\{v_1,\ldots,v_n\}$ with increasing indices.
There have been many questions on here about how to compute the characteristic polynomial and the coefficients $(-1)^k\text{tr}(\Lambda^k(T))$, but I haven't seen any interest in the matrix of $\Lambda^k(T)$ itself. I suspect it can be built from the minors of $[T]$, but I have no idea how to proceed.
• I recall reading about this calculation in books.google.com/books/about/… (the Advanced Calculus a Differential Forms Approach, by H.M. Edwards). Commented Jul 24, 2014 at 13:45
Let $T(v_i) = \sum_j a_{ji} \cdot v_j$, so that $(a_{ji})$ is the matrix of $T$ w.r.t. to $(v_1,\dotsc,v_n)$. The wedges $v_{i_1} \wedge \dotsc \wedge v_{i_k}$ with $i_1<\dotsc<i_k$ form a basis of $\Lambda^k(V)$. We have:
$$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} a_{j_1,i_1} \dotsc a_{j_k,i_j} \cdot v_{j_1} \wedge \dotsc \wedge v_{j_k}$$
If the indices $j_\ell$ are not pairwise distinct, the summand is zero. If not, we may choose a permutation to order the indices. The result is: $$\sum_{j_1<\dotsc<j_k} \sum_{\pi \in \Sigma_k} \mathrm{sgn}(\sigma) a_{j_{\pi(1)},i_1} \dotsc a_{j_{\pi(k)},i_j} \cdot v_{j_1} \wedge \dotsc \wedge v_{j_k}$$ Thus, the entry of the matrix of $\Lambda^k(T)$ in the row $(j_1<\dotsc<j_k)$ and the column $(i_1<\dotsc<i_k)$ is $$\sum_{\pi \in \Sigma_k} \mathrm{sgn}(\sigma) a_{j_{\pi(1)},i_1} \dotsc a_{j_{\pi(k)},i_j} .$$ This is nothing else than the determinant of the submatrix of $(a_{ji})$ corresponding to the rows $(j_1<\dotsc<j_k)$ and the columns $(i_1<\dotsc<i_k)$.
Conclusion: The entries of the matrix of $\Lambda^k(T)$ are the $k$-minors of the matrix of $T$.
I thought it might be worthwhile to add an explicit example. Consider $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ with $$[T] = \left[ \begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]$$ In particular, using the usual $e_1 = [ 1,0,0 ]^T=(1,0,0)$, $e_2 = (0,1,0)$, $e_3=(0,0,1)$ we have $$T(e_1) = (a,b,c), \ \ \ \ T(e_2) = (d,e,f),\ \ \ \ T(e_3) = (g,h,i).$$ Calculate the wedge products of these images \begin{align} T(e_1) \wedge T(e_2) &= (a,b,c) \wedge (d,e,f) \\ &= (ae-bd)e_1 \wedge e_2+(af-cd)e_1 \wedge e_3+(bf-ce)e_2 \wedge e_3 \end{align} Next, \begin{align} T(e_1) \wedge T(e_3) &= (a,b,c) \wedge (g,h,i) \\ &= (ah-gb)e_1 \wedge e_2+ (ai-cg)e_1 \wedge e_3+(bi-ch)e_2 \wedge e_3 \end{align} and \begin{align} T(e_2) \wedge T(e_3) &= (d,e,f) \wedge (g,h,i) \\ &= (dh-eg)e_1 \wedge e_2+(di-fg)e_1 \wedge e_3+(ei-fh)e_2 \wedge e_3 \end{align} The basis for $\Lambda^2 \mathbb{R}^3$ with increasing indices is $\{ e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \}$. Considering this the standard basis, we find in view of the calculations above and the definition $(\Lambda^2 T)(v,w) = T(v) \wedge T(w)$, $$[\Lambda^2 T] = \left[ \begin{array}{c|c|c} ae-bd & ah-gb & dh-eg \\ \hline af-cd & ai-cg & di-fg \\ \hline bf-ce & bi-ch & ei-fh \end{array}\right]$$ This is a matrix of minors of $[T]$. However, given our choice of basis, the minors are out of place relative to their usual assignment in the Laplace expansion. If we instead order the basis $\{ e_2 \wedge e_3, e_1 \wedge e_3, e_1 \wedge e_2 \}'$ then $$[\Lambda^2 T]' = \left[ \begin{array}{c|c|c} ei-fh & bi-ch & bf-ce \\ \hline di-fg & ai-cg & af-cd \\ \hline dh-eg & ah-gb & ae-bd \end{array}\right]$$ That's better. Now the minors are where they ought to be in my humble opinion. I suppose this is not terribly surprising, the strictly increasing basis is not natural for making the correspondence for vectors and one and two forms in $\mathbb{R}^3$.
• That was very illuminating. Thank you
– user123641
Commented Jul 25, 2014 at 4:18
• glad to help, I once made a sign-error in the 12-slot of the above. All the sudden I had students taking cell-phone pictures... Commented Jul 25, 2014 at 4:43 | 1,622 | 4,352 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.737096 |
https://learncplusplus.org/fr/category/learn-cpp/ | 1,638,155,120,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00121.warc.gz | 425,675,441 | 30,143 | ### Why You Should Know ELU Artificial Neural Net Functions
What is an Exponential Linear Unit or ELU? How can we use an ELU Activation Function? What do we need to know about activation functions? An Activation Function ( phi() ) also called as transfer function, or threshold function, determines the activation value ( a = phi(sum) ) from a given value (sum) from the Net Input Function . Net Input Function, here the sum is a sum of signals…
### What You Need To Know About C++ Gaussian Error Linear Units
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### What Is The SoftPlus Activation Function in C++ Neural Nets?
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What is a Binary Step Function? Should we use Binary Step Function or Heaviside Step Function? Are Binary Step functions and Heaviside Step functions the same thing? What is a Unit Step Function? Briefly, all these terms are same, let’s explain these terms. What is an activation function in C++ AI? An Activation Function ( phi() ) also called as transfer function, or threshold…
### This Is How To Make Artificial Neuron Models in C++
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### What Is The C++ Builder REST Debugger And How Do We Use It?
What is the REST Debugger? In RAD Studio there is a superbly useful tool called the REST Debugger. It comes included with RAD Studio C++ Builder. It is very useful to test REST Requests such as “Gets” and “Posts” to see what effect they have and…
### This Is How To Visualize Kinematics In Windows C++ Apps
What is kinematics? How can we use kinematics in programming? How can we draw kinematic drawings in C++? How can we simulate kinematic animations in C++ Builder? Let’s answer these questions. What does Kinematics mean? Kinematicsis a subfield of physics; it describes themotionof points, rigid bodies, and systems of these rigid body groups without considering the…
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What is basic_string in modern C++? How can I use basic_string in C++? Is std::basic string same as std::string? What is basic_string? The basic_string (std::basic_string and std::pmr::basic_string) is a class template thatstores and manipulates sequences…
### How To Make A New Windows VCL Static Library In C++
What do we mean by the term “static library”? What is a DLL? How can I create a new DLL dynamic library? Can I create a new DLL using the FMX framework? How can I create a function in a dynamic library? Where can I find a simple DLL example which uses C++…
### How To Automate Background Compiling C++ With TwineCompile
Learn how to use the TwineCompile SORTA Automatic Background Compile feature of TwineCompile. Install TwineCompile via the GetIt Package Manager for C++Builder with Update Subscription. TwineCompile lists its features as: Advanced compile system uses multi-threading technology and caching techniques to make C++ compiles 50x faster!Automatic background compiling engine ensures that files are… | 831 | 3,703 | {"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-2021-49 | latest | en | 0.768424 |
https://www.xarg.org/puzzle/codingame/divisibility-windows/ | 1,701,685,389,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100527.35/warc/CC-MAIN-20231204083733-20231204113733-00781.warc.gz | 1,196,588,878 | 5,197 | # Codingame Solution: Divisibility windows
## Goal
Given two numbers d and x, the aim is to find a and b the two numbers divisible by d around and closest to x, i.e.:
a is the largest multiple of d such that a < x,
b is the smallest multiple of d such that b > x.
Input
Line 1 : d
Line 2 : x
Output
Line 1 : a and b space-separated
Constraints
x is never divisible by d
d < x
2 <= d <= 65536
3 <= x <= 2147418111
## Solution
The problem statement asks to find $$a, b$$ such that $$a<x<b$$ and $$d | a,b$$. Since $$d<x$$ and $$d\not | x$$, it's enough to scale $$x$$ down by $$d$$ and the integer part of that result $$r=\left\lfloor\frac{x}{d}\right\rfloor$$ must be the maximum, such that $$a=r \cdot d$$.
For the same argument, we can say that $$s=\left\lceil\frac{x}{d}\right\rceil$$, such that $$b=s \cdot d$$ or even simpler $$b= a+ d$$
const d = +readline();
const x = +readline();
print([Math.floor(x / d) * d, Math.ceil(x / d) * d].join(" "))
« Back to problem overview | 314 | 987 | {"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": 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} | 3.875 | 4 | CC-MAIN-2023-50 | latest | en | 0.82662 |
https://www.nzqa.govt.nz/ncea/subjects/mathematics/clarifications/level-1/as91033/ | 1,686,363,615,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00647.warc.gz | 958,642,026 | 6,528 | # Level 1
## 91033: Apply knowledge of geometric representations in solving problems
Updated June 2014. This document has been updated to address new issues that have arisen from moderation.
### Solving problems
For the award of the standard, students must apply knowledge of geometric representationsin solving problems. To provide evidence, student responses need to clearly relate to the context of the problem. For example, if the context is a logo design, then students need to produce an actual logo rather than a series of unrelated loci and constructions.
The problem needs to provide sufficient scope for students to demonstrate and develop their own thinking. If there are parts to the problem, all of the parts need to contribute to the solution.
Students need to make their own decisions about what to do and how to solve problems. Where an assessment task has a series of instructions that lead students through a step or a sequence of steps towards the solution, it is likely that the opportunity for students to demonstrate all levels of thinking will be compromised.
### Expected evidence for Achieved
For the award of Achieved, the requirements include selecting and using a range of methods.
To be used as evidence, a ‘method’ must be at the appropriate curriculum level for the standard, for example, the evidence for bearings needs to be more than correctly using the bearings 000, 090, 180 and 270 and for two-dimensional representations of three-dimensional objects - the shapes being represented need to be more complex than a cuboid.
For a scale diagram to provide evidence, it is not appropriate for students to be given the scale.
While a grid may be provided, it is not appropriate for it to be numbered.
### Communicating solutions
At all levels there is a requirement relating to the communication of the geometric representations.
At Achieved level, the representations need to be correctly identified using appropriate geometrical terms, for example, a locus could be identified as the set of points less than 3 cm from A or a construction as an equilateral triangle on the side AB.
At Merit level, descriptions or instructions must allow the representations to be positioned correctly, for example when describing the construction of an equilateral triangle, the size, position and orientation would be required.
At Excellence level, accurate diagrams and descriptions or instructions for geometrical representations are required, and contextual aspects of the design(s) need to be discussed. | 481 | 2,540 | {"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-2023-23 | latest | en | 0.930687 |
http://www.indiabix.com/placement-papers/bsnl/3132 | 1,561,522,541,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00103.warc.gz | 254,450,569 | 11,296 | # Placement Papers - BSNL
## Why BSNL Placement Papers?
Learn and practice the placement papers of BSNL and find out how much you score before you appear for your next interview and written test.
## Where can I get BSNL Placement Papers with Answers?
IndiaBIX provides you lots of fully solved BSNL Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve BSNL Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### BSNL GE-JTO Recruitment Examination -IV Question Paper
Posted By : Vidya Rating : +15, -4
New Page 1
Paper : BSNL GE-JTO Recruitment Examination -IV Question Paper
NOTE : All Answer Are In BOLD Style.
1. When a piece of copper and another of germanium are cooled from room temperature to 800 K then the resistance of -
a) Each of them increases
b) Each of them decreases
c) Copper increases and germanium decreases
d) Copper decreases and germanium increases
2. When a signal of 10 mV at 75 MHz is to be measured then which of the following instrument can be used -
a) VTVM
b) Cathode ray oscilloscope
c) Moving iron voltmeter
d) Digital multimeter
3. When a sample of germanium and silicon having same impurity density are kept at room temperature then -
a) Both will have equal value of resistivity
b) Both will have equal value negative resistivity
c) Resistivity of germanium will be higher than that of silicon
d) Resistivity of silicon will be higher than that of germanium
4. When an RC driving point impedance function has zeros at s= -2 and s=-5 then the admissible poles for the function would be
a) s = 0; s = -6
b) s = 0; s = -3
c) s = 0; s = -1
d) s = -3; s = -4
5. For the n-type semiconductor with n = Np and p = n2/ND, the hole concentration will fall below the intrinsic value because some of the holes -
a) drop back to acceptor impurity states
b) drop to donor impurity states
c) Virtually leave the crystal
d) recombine with the electrons
6. The location of lighting arrestor is -
a) Near the transformer
b) Near the circuit breaker
c) Away from the transformer
d) None
7. Time constant of an RC circuit increases if the value of the resistance is -
a) Increased
b) Decreased
c) Neither a nor b
d) Both a and b
8. Telemetering is a method of
a) Counting pulses sent over long distances
b) Transmitting pictures from one place to another
c) Transmitting information concerning a process over a distance
d) None
9. When the gauge factor of a strain gauge is 2, stress is 1050 kg/cm2, Y = 2.1? 106 kg/cm2 and R is 100 ohms then the value of DR will be -
a) 2W
b) 3W
c) 4W
d) 1W
10. As the drain voltage is increased for a junction FET in the pinch off region then the drain current
a) Becomes zero
b) Abruptly decreases
c) Abruptly increases
d) Remains constant
11. One of the following, which is not a transducer in the true sense, is -
a) Thermocouple
b) Piezoelectric pick up
c) Photo-Voltaic cell
d) LCD
12. When a transistor is required to match a 100W signal source with a high impedance output circuit then the connection that would be used is -
a) Common base
b) Common collector
c) Common emitter
d) Emitter follower
13. In a JFET gates are always
a) forward biased
b) reverse biased
c) unbiased
d) none
14. The main factor which differentiate a DE MOSFET from an E only MOSFET is the absence of
a) insulated gate
b) electrons
c) channel
d) P-N junction
15. An SCR conducts appreciable current when
a) Anode and gate are both negative with respect to cathode
b) Anode and gate are both positive with respect to cathode
c) Anode is negative and gate is positive with respect to cathode
d) Gate is negative and anode is positive with respect to cathode
16. Silicon is not suitable for fabrication of light emitting diodes because it is -
a) An indirect band gap semiconductor
b) A direct band gap semiconductor
c) A wide band gap semiconductor
d) A narrow band gap semiconductor
17. An average responding rectifier type electronic ac voltmeter has its scale calibrated in terms of the rms value of a sine wave, when a square wave voltage of peak magnitude 100V is measured using this voltmeter then the reading indicated by the meter, will be
a) 111V
b) 100V
c) 90.09V
d) 70.7V
20. Which one of the following conditions for Z parameters would hold for a two port network containing linear bilateral passive circuit elements -
a) Z11 = Z22
b) Z12Z21 = Z11Z22
c) Z11Z12 = Z22Z21
d) Z12 = Z21
25. While calculating Rth, constant current sources in the circuit are -
a) replaced by opens
b) replaced by 'shorts'
c) treated in parallel with other voltage sources
d) converted into equivalent voltage sources
26. Maxwell's loop current method of solving electrical networks
a) uses branch currents
b) utilizes kirchhoff's voltage law
c) is confined to single-loop circuits
d) is a network reduction method
27. A transmission line of characteristic impedance Z0 = 50 ohms, phase velocity Vp = 2 x 108 m/s and length l = 1m is terminated by a load ZL= ( 30 - j 40 ) ohms. The input impedance of the line for a frequency of 100 MHz will be
a)(30 + j40 ) ohms
b)( 30 - j40 ) ohms
c)(50 + j40 ) ohms
d)(50 - j40 ) ohms
28. For an elliptically polarized wave incident on the interface of a dielectric at the Brewster angle then the reflected wave will be
a) Elliptically polarized
b) Linearly polarized
c) Right circularly polarized
d) Left circularly polarized
29. A yagi antenna has a driven antenna-
a) Only
b) With a reflector
c) With one or more directors
d) With a reflector and one or more directors
30. The number of lobes on each side of a 3l resonant antenna is -
a) 3 b) 6 c) 2 d) 1
31. The electric field intensity of a Hertizian dipole at a remote point varies as
Ans. 1/r
32. Radiation resistance of a half wave folded dipole is
a) 72 W b) 144W c) 288 W d) 216W
33. When a carrier wave is modulated at 100% it?s power is increased by -
a)100% b)150 % c)50% d)0%
34. On a clear sky day, the atmospheric radio noise is strongest -
a) During morning hours
b) Around mid-day
c) During nights
d) In the afternoon
35. TV broadcasting system in India is as per CCIR -
a) System B
b) System I
c) System M
d) System X
36. For the safety measurement of the internal resistance of a 25-0-25 mA meter, a laboratory multimeter whose sensitivity is equal to -
a) 1k ohm/volt can be used
b) 10 k ohm/volt can be used
c) 100 k ohm/volt can be used
d) 200 k ohm/volt can be used
37. In order to measure moisture in wood the most suitable method is -
a) Electrical conduction
b) Electrical - capacitive
d) Equilirium- moisture vs humidity
38. The flow rate of elctrically conducting liquid without any suspended practicle cannot be measured by
a) turbine flow meters
b) electromagnetic flow meters
c) ultrasonic flow meters
d) thermistor based heat loss flow meters
39. The most useful transducer for displacement sensing with excellent sensitivity, linearity and resolution is
a) an incremental encoder
b) an abosolute encoder
c) LVDT
d) a strain gauge
40. When variable reluctance type techometer has 150 teeth on the rotor & the counter records 13,500 pulses per second then the rotational speed will be-
a) 4800 rpm b) 5400 rpm c) 6000 rpm d) 7200 rpm.
41. On a voltage scale, zero dB m in a 600-ohm system could refer to
a) 1.732 V b) 1.0 V c) 0.7746 V d) 0.5V
42. One of the following devices which is required in addition in order to measure pressure using LVDT is
a) strain gauge b) pitot tube c) Bourden tube d) Rotameter
43. It is required to measure temperature in the range of 1300 deg C to 1500 deg C The most suitable thermocouple to be used as a transducer would be
a) chromel - constantan
b) Iron - constantan
c) chromel - alumel
d) platinum- rhodium
44. In a CSI if frequency of output voltage is f Hz, then frequency of input voltage to CSI is
a) f
b) 2 f
c) f/2
d) 3 f
46. Maximum value of charging resistance in an UJT is associated with-
a) peak point
b) valley point
c) any point between peak and valley
d) after the valley point
47. Thyristor A has rated gate current of 2A and thyristor B a rated gate current of 100 mA
a) A is a GTO and B is a conventional SCR
b) B is a GTO and A is a conventional SCR
c) B may operate as a transistor
d) none of the above
48. In a 3 phase full converter, the output voltage during overlap is equal to-
a) zero
b) source voltage
c) source voltage minus the inductance drop
d) average value of the conducting phase voltages
49. Mark old the correct statement for Cycloconverters
a) step-down Cycloconverter (CC) works on natural commutation
b) step up CC requires no forced commutation
c) load commutated CC works on line commutation
d) none of the above
50. In a 3 phase full converter if load current is I and ripple free, then average thyristor current is
Ans. b) 1/3(I)
51. In the RF amplifier stage cascade (CE-CB) amplifier is used because it gives
a) Large voltage gain
b) Low output impedance
c) Large isolation between the input and the output
d) None of the above
52. Silicon diode is less suited for low voltage rectifier operation because-
a) it can withstand high temperature
b) ensures low PIV of the diodes
c) ensures lower values of capacitance in the filter
d) reduces ripple content
53. An amplifier of class A is that in which
a) Base is biased to cut-off
b) Ic flows most of the time
c) Ie flows all the time
d) Vc often raises to Vcc
54. A transistor is in active region when-
a) Ib = bIc b) Ic=bIb c) Ic=Ie d) Ic=Ib
55. For coupling purposes in RF amplifier a buffer amplifier is used because it provides
56. A transistor has CE parameter as hie = 10kW, hre =20 x 10-4 , hse = 100, hoe = 25 ms. The hib for this transistor will be-
a) 100 W b) 99.01 W c) 5m W d) 101kW
a) 110 MHz b) 112 Hz c) 114 MHz d) 120 MHz
60. A dc to dc converter having an efficiency of 80% is delivering 16W to a load) If the converter is generating an output of 200V from an input source of 20V, then the current drawn from the source will be
a) 0.1A b) 0.5A c) 1.0A d) 10.0A | 2,832 | 10,050 | {"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} | 3.28125 | 3 | CC-MAIN-2019-26 | latest | en | 0.889191 |
https://carrageenan.info/answers-on-questions/how-many-strings-of-c9-lights-can-be-connected.html | 1,652,740,497,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00460.warc.gz | 210,863,735 | 13,021 | # How many strings of c9 lights can be connected?
## How many strands of lights can you put together?
Each string light boasts a wattage of 10 watts. So, you can safely connect no more than 10 of these string lights together.
## How many LED Christmas light strands can you connect?
In this case, most traditional incandescent Christmas mini lights only allow you to connect 4 or 5 sets end to end but with many LED mini light strings you can connect 40 to 50+ together depending on the light count. Consider your circuits: Most household circuits are 15 or 20 amps.
## Can you string too many Christmas lights together?
Because light strings have a maximum wattage capacity, which is why many string lights come with a little fuse just in case you connect too many together at once. The fuse is designed to blow so you don’t overload and damage your Christmas lights. This happens when your wattage exceeds the amp capacity of the circuit.
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## How many strings of lights do I need for a 9 foot tree?
A good rule of thumb is 100 lights for every 1.5 feet of tree.
How many strands do I need?
Tree Height Number of Lights
6 to 7 Feet 400 to 700
7.5 to 8.5 Feet 700 to 1,000
9 to 10 Feet 1,000 to 1,300
12 Feet 1,500 to 2,000
## How many lights can you daisy chain?
Divide 210 by the number of watts each string uses, and this will give you your number of strings you can successfully daisychain per outlet. For a 20-watt strand, that’s 10.5 string lights connected in series.
## How many Christmas lights should be on a circuit?
So you could, put 105 strings of those lights on a single circuit (1440÷12=105). However, a single circuit could only take 11 strings of large, incandescent bulbs that use 125 watts (1440÷125=11.5).
## How many LEDS can be on a 20 amp circuit?
You can put up to 40 lights (based on 50 watt light) on a 20 amp breaker.
## Can you string LED and regular Christmas lights together?
These light sets are manufactured in series and the same amount of current runs through all the light sets strung together. And the more lights that are strung together, the more current they “pull” or require. In other words: Do NOT run LED and incandescent light strings plugged into each other.
## What are three warning signs of an overloaded electrical circuit?
• Flickering, blinking, or dimming lights.
• Frequently tripped circuit breakers or blown fuses.
• Warm or discolored wall plates.
• Cracking, sizzling, or buzzing from receptacles.
• Burning odor coming from receptacles or wall switches.
• Mild shock or tingle from appliances, receptacles, or switches.
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## How many amps does a string of C9 Christmas lights use?
Incandescent Mini Light Sets use 3 amp fuses. Heavy Duty light sets 5V use 3 amp fuses. C7, C9 light sets use 5 amp fuses. G40, G50 light sets use 5 amp fuses.
## How many C9 lights do I need for a 7 foot tree?
Example: 7 ft tree with 45″ diameter 7 x 45 (divided by 2) = 157 lights Safety Tips: Do not connect mini lights end to end to C7 or C9 lights.
## How many feet of string lights do I need?
Simply buy string lights that are 2 to 6 feet longer than your linear measurement to create the swag of your choice. It’s always better to have more string length than you need because you can always shorten it by doubling it up on the ends.
## Which Christmas lights are the best?
The best holiday lights
• The best classic lights. Brizled LED Mini String Lights.
• The best battery-powered lights. KooPower Waterproof Battery Fairy Lights.
• The best large-bulb lights. Good Tidings Ceramic Multi-Colored Holiday Lights.
• The best smart lights. Twinkly Smart LED String Lights.
• The best solar-powered holiday lights.
## How much credit can i get on a credit card?How much credit can i get on a credit card?
Contents1 What is the average credit limit on a credit card?2 What is the average credit card limit UK?3 Can I get credit card with 10000 salary?4 How much should | 999 | 4,087 | {"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-2022-21 | latest | en | 0.901279 |
https://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/06/01/05/02/0003/ | 1,716,736,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00251.warc.gz | 233,504,609 | 9,484 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
HypergeometricPFQ
http://functions.wolfram.com/07.31.06.0018.01
Input Form
HypergeometricPFQ[{Subscript[a, 1], \[Ellipsis], Subscript[a, q + 1]}, {Subscript[b, 1], \[Ellipsis], Subscript[b, q]}, z] == (Product[Gamma[Subscript[b, k]], {k, 1, q}]/Product[Gamma[Subscript[a, k]], {k, 1, q + 1}]) Sum[GammaResidue[{{0}, {1 - Subscript[a, 1], \[Ellipsis], 1 - Subscript[a, q + 1]}}, {{}, {1 - Subscript[b, 1], \[Ellipsis], 1 - Subscript[b, q]}}, {1 - Subscript[a, k], 1, i}, -z], {k, 1, q + 1}, {i, 0, Infinity}] /; Abs[z] > 1 && ForAll[{j, k}, Element[{j, k}, Integers] && j != k && 1 <= j <= q + 1 && 1 <= k <= q + 1, !Element[Subscript[a, j] - Subscript[a, k], Integers]]
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[SubscriptBox["a", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["a", RowBox[List["q", "+", "1"]]]]], "}"]], ",", RowBox[List["{", RowBox[List[SubscriptBox["b", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["b", "q"]]], "}"]], ",", "z"]], "]"]], "\[Equal]", RowBox[List[FractionBox[RowBox[List[" ", RowBox[List[UnderoverscriptBox["\[Product]", RowBox[List["k", "=", "1"]], "q"], RowBox[List["Gamma", "[", SubscriptBox["b", "k"], "]"]]]]]], RowBox[List[UnderoverscriptBox["\[Product]", RowBox[List["k", "=", "1"]], RowBox[List["q", "+", "1"]]], RowBox[List["Gamma", "[", SubscriptBox["a", "k"], "]"]]]]], RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "1"]], RowBox[List["q", "+", "1"]]], RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["i", "=", "0"]], "\[Infinity]"], RowBox[List["GammaResidue", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List["{", "0", "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "-", SubscriptBox["a", "1"]]], ",", "\[Ellipsis]", ",", RowBox[List["1", "-", SubscriptBox["a", RowBox[List["q", "+", "1"]]]]]]], "}"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["{", "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "-", SubscriptBox["b", "1"]]], ",", "\[Ellipsis]", ",", RowBox[List["1", "-", SubscriptBox["b", "q"]]]]], "}"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "-", SubscriptBox["a", "k"]]], ",", "1", ",", "i"]], "}"]], ",", RowBox[List["-", "z"]]]], "]"]]]]]]]]]], "/;", RowBox[List[RowBox[List[RowBox[List["Abs", "[", "z", "]"]], ">", "1"]], "\[And]", RowBox[List[SubscriptBox["\[ForAll]", RowBox[List[RowBox[List["{", RowBox[List["j", ",", "k"]], "}"]], ",", RowBox[List[RowBox[List[RowBox[List["{", RowBox[List["j", ",", "k"]], "}"]], "\[Element]", "Integers"]], "\[And]", RowBox[List["j", "\[NotEqual]", "k"]], "\[And]", RowBox[List["1", "\[LessEqual]", "j", "\[LessEqual]", RowBox[List["q", "+", "1"]]]], "\[And]", RowBox[List["1", "\[LessEqual]", "k", "\[LessEqual]", RowBox[List["q", "+", "1"]]]]]]]]], RowBox[List["(", "\[InvisibleSpace]", RowBox[List["Not", "[", RowBox[List[RowBox[List[SubscriptBox["a", "j"], "-", SubscriptBox["a", "k"]]], "\[Element]", "Integers"]], "]"]], ")"]]]]]]]]]]
MathML Form
q + 1 F q ( a 1 , , a q + 1 ; b 1 , , b q ; z ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox[RowBox[List["q", "+", "1"]], TraditionalForm]], SubscriptBox["F", FormBox["q", TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[SubscriptBox["a", "1"], HypergeometricPFQ, Rule[Editable, True]], ",", TagBox["\[Ellipsis]", HypergeometricPFQ, Rule[Editable, True]], ",", TagBox[SubscriptBox["a", RowBox[List["q", "+", "1"]]], HypergeometricPFQ, Rule[Editable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ, Rule[Editable, True]], ";", TagBox[TagBox[RowBox[List[TagBox[SubscriptBox["b", "1"], HypergeometricPFQ, Rule[Editable, True]], ",", TagBox["\[Ellipsis]", HypergeometricPFQ, Rule[Editable, True]], ",", TagBox[SubscriptBox["b", "q"], HypergeometricPFQ, Rule[Editable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ, Rule[Editable, True]], ";", TagBox["z", HypergeometricPFQ, Rule[Editable, True]]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]], Rule[Editable, True]], HypergeometricPFQ] k = 1 q Γ ( b k ) k = 1 q + 1 Γ ( a k ) k = 1 q + 1 i = 0 ΓRes ( 0 ; 1 - a 1 , , 1 - a q + 1 ; ; 1 - b 1 , , 1 - b q ; 1 - a k , 1 , i ; - z ) /; "\[LeftBracketingBar]" z "\[RightBracketingBar]" > 1 { j , k } , { j , k } TagBox["\[DoubleStruckCapitalZ]", Function[Integers]] j k 1 j q + 1 1 k q + 1 ( a j - a k TagBox["\[DoubleStruckCapitalZ]", Function[Integers]] ) FormBox RowBox RowBox TagBox TagBox RowBox RowBox SubscriptBox FormBox RowBox q + 1 TraditionalForm SubscriptBox F FormBox q TraditionalForm RowBox ( RowBox TagBox TagBox RowBox TagBox SubscriptBox a 1 HypergeometricPFQ Rule Editable , TagBox HypergeometricPFQ Rule Editable , TagBox SubscriptBox a RowBox q + 1 HypergeometricPFQ Rule Editable InterpretTemplate Function SlotSequence 1 HypergeometricPFQ Rule Editable ; TagBox TagBox RowBox TagBox SubscriptBox b 1 HypergeometricPFQ Rule Editable , TagBox HypergeometricPFQ Rule Editable , TagBox SubscriptBox b q HypergeometricPFQ Rule Editable InterpretTemplate Function SlotSequence 1 HypergeometricPFQ Rule Editable ; TagBox z HypergeometricPFQ Rule Editable ) InterpretTemplate Function HypergeometricPFQ Slot 1 Slot 2 Slot 3 Rule Editable HypergeometricPFQ RowBox FractionBox RowBox UnderoverscriptBox RowBox k = 1 q RowBox Γ ( SubscriptBox b k ) RowBox UnderoverscriptBox RowBox k = 1 RowBox q + 1 RowBox Γ ( SubscriptBox a k ) RowBox UnderoverscriptBox RowBox k = 1 RowBox q + 1 ErrorBox RowBox UnderoverscriptBox RowBox i = 0 RowBox ΓRes ( RowBox RowBox RowBox GridBox RowBox 0 ; RowBox RowBox 1 - SubscriptBox a 1 , , RowBox RowBox 1 - SubscriptBox a RowBox q + 1 ; ; RowBox RowBox 1 - SubscriptBox b 1 , , RowBox RowBox 1 - SubscriptBox b q ; 1 - SubscriptBox a k , 1 , RowBox i ; RowBox - z ) /; RowBox RowBox RowBox z > 1 RowBox SubscriptBox RowBox RowBox { RowBox j , k } , RowBox RowBox RowBox { RowBox j , k } TagBox Function RowBox j k RowBox 1 j RowBox q + 1 RowBox 1 k RowBox q + 1 RowBox ( RowBox RowBox SubscriptBox a j - SubscriptBox a k TagBox Function ) TraditionalForm [/itex]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 2,224 | 6,767 | {"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} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.179474 |
https://brainmass.com/math/ordinary-differential-equations/solve-homogenous-order-ode-cauchy-euler-equation-27338 | 1,713,818,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00741.warc.gz | 126,486,713 | 6,899 | Purchase Solution
# Solve a homogenous 2nd order ODE.
Not what you're looking for?
Find y as a function of x if:
(x^2)(y'') + 19xy' +81y = x^2
y(1) = 9 y'(1) = -3
Hint: First assume that at least one solution to the corresponding homogeneous equation is of the form . You may have to use some other method to find the second solution to make a fundamental set of solutions. Then use one of the two methods to find a particular solution.
##### Solution Summary
A homogenous 2nd order ODE is solved. The expert solves a homogenous 2nd order ODE. A Cauchy-Euler equation is examined.
##### Solution Preview
This is a Cauchy-Euler equation. To solve that we let x=exp(z) and plug this into the equation. In general with this change of variable the equation:
x^2y''+ axy'+ by= f(x) becomes:
y''+ (a-1)y'+ by= f(exp(z)) ...
##### Probability Quiz
Some questions on probability
This quiz test you on how well you are familiar with solving quadratic inequalities.
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
##### Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them. | 332 | 1,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.878152 |
https://epolylearning.com/Heat-Transfer/Conduction-With-Heat-Generation/discuss/6915 | 1,675,759,590,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00706.warc.gz | 256,578,249 | 8,955 | # E-PolyLearning
## Discussion Forum
Que. If Q X is heat generated in at distance ‘x’, then heat conducted out at a distance (x + d x) will be a. Q X + 3d (Q X) d x /d x b. 2Q X + d (Q X) d x /d x c. d (Q X) d x /d x d. Q X + d (Q X) d x /d x
Correct Answer:Q X + d (Q X) d x /d x | 118 | 284 | {"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-2023-06 | latest | en | 0.562461 |
https://www.goconqr.com/flashcard/2808922/forces | 1,618,699,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464065.57/warc/CC-MAIN-20210417222733-20210418012733-00035.warc.gz | 896,286,746 | 10,634 | # Forces
Flashcards by Taylor Smith, updated more than 1 year ago
Created by Taylor Smith almost 6 years ago
8
1
### Description
Cambridge IGCSE Chemistry, Biology and Physics Flashcards on Forces, created by Taylor Smith on 05/25/2015.
## Resource summary
Question Answer Force A push or a pull Contact Force A force that is exerted directly on an object. When does friction occur? When 2 surfaces are pushed against each other so that their particles are in close contact. Effects of friction - Friction causes heat - Moving objects are slowed down - Surfaces wear Lubricant A substance that reduces friction between 2 moving surfaces Static friction The opposite frictional force that is at its maximum value and before the object starts to move Sliding friction The opposing frictional force that is slightly less than the static friction and the object moves at a constant speed. Newton's 3rd Law The mutual similar forces of action and reaction between 2 bodies are equal, opposite and collinear. Nature of Force - Forces may be attractive or repulsive. - Forces have both magnitude and direction. Forces having magnitude and direction (vector) If more than 1 force is applied, the combined effect of all the forces may keep the object stationary if the forces are balanced, or move the object if unbalanced. Resultant Force The single force that has the combined effect if all the forces operating on a body. Hooke's Law The extension of a spiral spring is directly proportional to the stretching force, provided that the spring is not stretched beyond its elastic limit Rotation The effect of a force when it is applied to a body which is free to move around a point. Fulcrum/Pivot The point about which rotation occurs. Moment The product of the force and the perpendicular distance of the line of action of the force form the pivot. Moment (Formula) T = Fd (T (tau) = moment, F = force, d = perpendicular distance) Law of Moments For a lever in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. A body is in equilibrium when: - The resultant force is 0. - The sum of clockwise moments equals the sum of anticlockwise moments. Centre of mass/gravity The point at which the whole weight of the body is considered to act. Position of centre of gravity - It determines whether a body topples or not. - A body topples when the vertical line through its centre of mass falls outside of its base and has a moment which topples the body. Stability of an object is increased by: 1. Lowering the centre of mass. 2. Increasing the are of the base. Stable equilibrium - The object returns to its initial position. - Centre of mass rises when displaced. - Weight has a moment around the point of contact (hence it rolling back). Unstable equilibrium - Object moves further away from its initial position when slightly displaced. - Centre of mass falls because it has a moment about the point of contact which increases displacement. Neutral equilibrium - The body remains in its new position after being displaced. - Weight acts through the point of contact, therefore, there is no moment to increase or decrease the displacement. Lever A device that can turn about a pivot. Effort The force applied to operate the lever. Load The resisting force. 1st class lever - A lever that has the fulcrum between the effort and the load (EFL) - E.g. Scissors, see-saw, pliers 2nd class lever - Load is between the effort and the fulcrum (ELF). - E.g. Wheelbarrow, nutcracker, bottle opener. 3rd class lever _ Effort is between the load and the fulcrum. - E.g. Elbow, sugar tongs. Resultant of a number of vectors The single vector that will have the same effect as all the original vectors acting together. Equilibriant of a number of forces - The single force that keeps all the other forces in equilibrium. - It's equal in magnitude but opposite direction to the resultant of the forces. Triangle Law If 3 forces are in equilibrium, they van be represented in both magnitude and direction by 3 sides of a triangle taken in order. Triangle Law Effect of force and mass on acceleration Acceleration is: 1) Directly proportional to the resultant force for a fixed mass. 2) Indirectly proportional the mass for a constant force. - Therefore, F = ma Newton's 2nd Law When a resultant force is applied to a body, it produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass of the object.
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Forces and their effects | 1,044 | 4,802 | {"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-2021-17 | latest | en | 0.90507 |
https://www.coursehero.com/file/6962092/mathlamaredutermsaspx-College-Algebra-Do-not-get-excited/ | 1,487,900,035,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171271.48/warc/CC-MAIN-20170219104611-00592-ip-10-171-10-108.ec2.internal.warc.gz | 793,297,402 | 22,015 | Alg_Complete
# Mathlamaredutermsaspx college algebra do not get
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Unformatted text preview: x + 10 gallons of the 40% solution once we’re done mixing the two. Here is the basic work equation for this problem. æ Amount of alcohol ö æ Amount of alcohol ö æ Amount of alcohol ö ç ÷+ç ÷=ç ÷ è in 50% Solution ø è in 35% Solution ø è in 40% Solution ø æ Volume of ö æ Volume of ö æ Volume of ö ÷ + ( 0.35 ) ç ÷ = ( 0.4 ) ç ÷ è 50% Solution ø è 35% Solution ø è 40% Solution ø ( 0.5 ) ç Now, plug in the volumes and solve for x. 0.5 x + 0.35 (10 ) = 0.4 ( x + 10 ) 0.5 x + 3.5 = 0.4 x + 4 0.1x = 0.5 0.5 x= = 5gallons 0.1 So, we need 5 gallons of the 50% solution to get a 40% solution. Example 11 We have a 40% acid solution and we want 75 liters of a 15% acid solution. How much water should we put into the 40% solution to do this? Solution Let x be the amount of water we need to add to the 40% solution. Now, we also don’t how much of the 40% solution we’ll need. However, since we know the final volume (75 liters) we will know that we will need 75 - x liters of the 40% solution. Here is the word equation for this problem. æ Amount o...
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## This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
Ask a homework question - tutors are online | 465 | 1,492 | {"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} | 4.09375 | 4 | CC-MAIN-2017-09 | longest | en | 0.877904 |
https://www.experts-exchange.com/questions/28358271/Scoring.html | 1,480,869,498,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00429-ip-10-31-129-80.ec2.internal.warc.gz | 961,495,431 | 29,711 | Solved
# Scoring
Posted on 2014-02-06
104 Views
Sir gowflow need little modification.
## 01.
On Button click analyze
if C29=”High” then
+1 @ Cell B49 & D49 & F49 & H49
## 02.
Find C22 Value from Range C2:C11 (+ or – 3 point leverage)
if True then find respective factor in range B39:B48 & Add +1 (if it is at Close Hits then put +1 @ Unity)else nothing next step
## 03.
Find C23 Value in Range C2:C11(+ or – 3 point leverage)
if true then find respective factor in range D39:D48 & Add +1(if it is at Close Hits then put +1 @ Unity )else nothing next step
## 04.
Find C22 value from range G2:G11(+ or – 3 point leverage)
if true then find respective factor in range F39:F48 & Add +1 (if it is at Point Hits then put +1 @ Unity )else nothing next step
## 05.
Find C23 value from range G2:G11(+ or – 3 point leverage)
if true then find respective factor in range H39:H48 & Add +1 (if it is at Point Hits then put +1 @ Unity )else nothing. Endsame way for LOW if you need detailed Step pls let me know I will post here.
Thanks
Scoring-G-V04.xlsm
0
Question by:itjockey
• 5
• 4
LVL 29
Expert Comment
All this should replace the previoous code or it is on top of the previous code ?
gowflow
0
LVL 29
Expert Comment
gowflow
0
LVL 8
Author Comment
0
LVL 8
Author Comment
it s your call, what ever you feel to do so? I just add one row Unity in Both High & Low process.
Thanks
0
LVL 29
Accepted Solution
gowflow earned 500 total points
try this
gowflow
Scoring-G-V05.xlsm
0
LVL 8
Author Closing Comment
Perfect
0
LVL 8
Author Comment
May I Post My Next Question?
Thanks
0
LVL 29
Expert Comment
yes
0
LVL 8
Author Comment
here it is Scoring.
Thanks
0
## Featured Post
This collection of functions covers all the normal rounding methods of just about any numeric value.
Since upgrading to Office 2013 or higher installing the Smart Indenter addin will fail. This article will explain how to install it so it will work regardless of the Office version installed.
This Micro Tutorial will demonstrate how to use longer labels with horizontal bar charts instead of the vertical column chart.
This Micro Tutorial demonstrates in Microsoft Excel how to consolidate your marketing data by creating an interactive charts using form controls. This creates cool drop-downs for viewers of your chart to choose from. | 649 | 2,343 | {"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-2016-50 | longest | en | 0.767471 |
http://www.opencm3.net/doc/help/gen_html/realgeometry/src/RealPoint.i3.html | 1,669,747,072,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00464.warc.gz | 85,711,011 | 2,956 | realgeometry/src/RealPoint.i3
Distributed only by permission.
Created on Sep 15 1988 by Jorge Stolfi
Contributed by Michel Dagenais (dagenais@vlsi.polymtl.ca), 1994.
```INTERFACE RealPoint;
```
Real pairs.
This interface implements the 2-dimensional vector space R^2. It defines the basic linear operations on real pairs (arrays with 2 REAL components). See R2x2.i3 and R2IO.i3 for additional operations.
Index: vectors; reals, vectors of; points; geometry; linear algebra; numerical routines
****************************************************************** WARNING: DO NOT EDIT THIS FILE. IT WAS GENERATED MECHANICALLY. See the Makefile for more details. ******************************************************************
Taken from the mg library distributed with DEC SRC Modula3 2.11. There was no makefile there explaining why it should not be edited. Simple cleaning and removing of dependencies with other mg modules performed.
```IMPORT Random;
TYPE
Axis = [0..1]; (* Coordinate selectors *)
Point = ARRAY Axis OF REAL; (* A point of R2 *)
T = Point;
CONST
Origin = T{0.0, 0.0};
Ones = T{1.0, 1.0};
PROCEDURE Unit(i: Axis): T;
(* The unit vector on the i-th axis. *)
PROCEDURE Equal(READONLY x, y: T): BOOLEAN;
(* Equality *)
(* TRUE if x = (0 0) *)
PROCEDURE Sum (READONLY x: T): REAL;
(* Sum of the coordinates x[i]. *)
(* Maximum of the x[i]. *)
(* Index i such that ABS(x[i]) is largest. *)
(* Minimum of the x[i]. *)
(* Sum of x[i]**2. *)
(* Sum of ABS(x[i]). *)
(* Euclidean length (sqrt of sum of x[i]**2). *)
(* Maximum of ABS(READONLY x[i]). *)
PROCEDURE L1Dist(READONLY x, y: T): REAL;
(* L1Norm(Sub(x,y)). *)
PROCEDURE Dist(READONLY x, y: T): REAL;
PROCEDURE L2Dist(READONLY x, y: T): REAL;
(* L2Norm(Sub(x,y)). *)
PROCEDURE L2DistSq(READONLY x, y: T): REAL;
(* SumSq(Sub(x,y)). *)
PROCEDURE LInfDist(READONLY x, y: T): REAL;
(* LInfNorm(Sub(x,y)). *)
PROCEDURE RelDist(READONLY x, y: T; eps: REAL := 1.0e-37): REAL;
(* Relative distance between two points, useful in convergence tests.
The /eps/ parameter specifies the magnitude of the (additive) noise
in each coordinate.
The value of RelDist is defined as max_i rdist(x[i], y[i]) where
|
| rdist(u,v) = max(|u-v|-eps, 0) / max(|u|, |v|, eps)
|
The result is between 0.0 and 2.0. A result below 2.0e-7 means the vectors
differ only at the roundoff error level. *)
PROCEDURE Dot (READONLY x, y: T): REAL;
(* Dot product of vectors x and y. *)
PROCEDURE Cos (READONLY x, y: T): REAL;
(* Co-sine of the angle between x and y. *)
PROCEDURE Det(READONLY p0, p1: T): REAL;
(* Determinant of the matrix with given points as rows *)
(* Vector sum x + y. *)
PROCEDURE Sub(READONLY x, y: T): T;
(* Vector difference x - y. *)
(* Negation of all coordinates. *)
PROCEDURE Scale(alpha: REAL; READONLY x: T): T;
(* Scalar multiplication alpha * x. *)
PROCEDURE Shift(READONLY x: T; delta: REAL): T;
(* Adds delta to every coordinate of x. *)
PROCEDURE Mix (READONLY x: T; alpha: REAL; READONLY y: T; beta: REAL): T;
(* Linear combination x * alpha + y * beta. *)
PROCEDURE Weigh (READONLY x, y: T): T;
(* Returns z such that z[i] = x[i] * y[i]. *)
TYPE Function = PROCEDURE (x: REAL): REAL;
PROCEDURE FMap (READONLY x: T; F: Function): T;
(* Returns z such that z[i] = F(x[i]). *)
(* Divides x by Length(x). An error if x = (0 0). *)
(* Cross product: Det(p, q, r, ...) = Dot(p, Cross(q, r, ...)) *)
PROCEDURE Throw(lo, hi: REAL; src: Random.T := NIL): T;
(* Returns a random vector with coordinates independently and
uniformly distributed between lo (inclusive) and hi (exclusive).
The "src" parameter is the source of randomness (See Random.def). *)
END RealPoint.
```
RealPoint's implementation is in:
procedure RealPoint.Equal is in:
procedure RealPoint.Max is in:
procedure RealPoint.Min is in:
procedure RealPoint.Sub is in:
procedure RealPoint.Minus is in:
procedure RealPoint.Scale is in:
```
``` | 1,116 | 3,936 | {"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} | 2.765625 | 3 | CC-MAIN-2022-49 | latest | en | 0.638625 |
https://testbook.com/question-answer/a-steel-plate-of-size-250-mm-150-mm-times--62972a6daedf0c66a2a76e1c | 1,680,118,615,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00649.warc.gz | 637,832,149 | 82,943 | # A steel plate of size 250 mm × 150 mm × 10mm with holes for two number of 16 mm diameter bolts having ultimate strength of 410 MPa, the design strength of plate in rupture of critical section is
This question was previously asked in
MPPSC AE CE 2017 Official Paper - II
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1. 336 kN
2. 382 kN
3. 365 kN
4. 280 kN
Option 1 : 336 kN
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## Detailed Solution
Concept:
Design strength of plate or tensile strength or tearing strength:
(i) Design strength based on gross area yielding:
$${P_T} = {A_g} × \frac{{{f_y}}}{{1.1}}$$
(ii) Design strength based on net area cracking(rupture):
$${P_T} = {A_{net}} × \frac{{0.9{f_u}}}{{1.25}}$$
Calculation:
Given, Diameter of bolt = 16 mm, diameter of hole = 16 + 2 = 18 mm
A= 150 × 10 = 1500 mm2, An = (150 - 18 × 2 ) × 10 = 1140 mm2, fu = 410 MPa
∵ Design strength of plate in rupture:,
⇒ $${P_T} = {A_{net}} × \frac{{0.9{f_u}}}{{1.25}}$$
$$0.9 × 1140 × 410 \over 1.25$$
PT = 336528 N = 336.528 kN ≈ 336 kN
So, the design strength of the critical section in rupture is 336 kN | 406 | 1,132 | {"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} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.75107 |
https://ju-sledovat.com/business/accounting/economic-order-quantity-and-inventory-588983id-p-1233xvx3 | 1,642,510,802,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00405.warc.gz | 408,425,975 | 15,844 | Home
# Average inventory EOQ calculator
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2. This simple Economic Order Quantity (EOQ) calculator can be used for computing the economic (optimal) quantity of goods or services a firm needs to order. The calculator also offers a visualization of the EOQ model in graphic form. To utilize this calculator, simply fill in all the fields below and then click the Calculate EOQ button
3. imize costs related to inventory, like holding and ordering costs. It is always good practice to reduce your costs as much as possible, to maximize your profits
4. EOQ = √(100000 ÷ 10) EOQ = √100000. EOQ = 312. Therefore, the economic order quantity is 312 units per order. Sources and more resources. NC State University - Economic Order Quantity (EOQ) Model: Inventory Management Models: A Tutorial - Part of a larger tutorial, this page details how to calculate the economic order quantity
5. Economic Order Quantity (EOQ) is derived from a formula that consists of annual demand, holding cost, and order cost. This formula aims at striking a balance between the amount you sell and the amount you spend to manage your inventory. Calculate Economic Order Quantity for your busines
6. imizes the total cost of ordering and holding the inventory. The calculator is used as follows
7. The cost of carrying inventory can be calculated by multiplying the cost of carrying a unit of inventory by the average number of units carried, usually for a year. If inventory is used at a steady pace, and restocked when empty, then the average number of units held would be the order size divided by 2. C=Carrying cost per unit of inventory
### EOQ (Economic Order Quantity) Calculator - Good Calculator
• Average Inventory = EOQ / 2 x \$12 First, we need to compute the Economic Order Quantity: EOQ = Square Root ((2 x Annual Demand x Order Cost) / Holding Cost) EOQ = 417 unit
• e just the right quantity to use when restocking inventory. Economic Order Quantity in inventory management is often used in supply chain and operations for the purposes of demand forecasting. It is often referred to in the industry as Reorder Quantity (or ROQ)
• Average Inventory Formula is used to calculate the mean value of Inventory at a certain point of time by taking the average of the Inventory at the beginning and at the end of the accounting period. It helps management to understand the Inventory, the business needs to hold during its daily course of business
• How to Calculate Average Inventory. Average inventory is used to estimate the amount of inventory that a business typically has on hand over a longer time period than just the last month. Since the inventory balance is calculated as of the end of the last business day of a month, it may vary considerably from the average amount over a longer time period, depending upon whether there was a.
• Average Inventory Formula = (Begining Inventory + End Inventory) / 2. Average (Avg) Inventory is the mean value of Inventory which is calculated at a certain point of time by taking the average of the Inventory at the beginning and at the end of the accounting period. See Also Average Inventory Period Calculator
• • Average inventory in EOQ model: Q* = Optimal order quantity (i.e., the EOQ) D = Annual demand, in units, for the inventory item; C o = Ordering cost per order; C h = Carrying or holding cost per unit per year; P = Purchase cost per unit of the inventory ite
### EOQ Calculator (Economic Order Quantity
• Annual demand is 20,000 units so the company will have to place 16 orders (= annual demand of 20,000 divided by order size of 1,265). Total ordering cost is hence \$6,400 (\$400 multiplied by 16). Average inventory held is 632.5 (= (0+1,265)/2) which means total annual carrying costs would be \$6,325 (i.e. 632.5 × \$10)
• imizing the total cost per order by setting the first-order derivative to zero. The components of the formula that make up the total cost per order are the cost of holding inventory and the cost of ordering that inventory
• The below table shows the calculation of the number of orders per year. A number of orders per year = Annual quantity demanded/ EOQ. So, the calculation of EOQ for numbers of orders per year is =2000/200 Therefore, a number of orders per year =1
• imizes the sum of ordering and holding costs related to raw materials or merchandise inventories.In other words, it is the optimal inventory size that should be ordered with the supplier to
• The average inventory will be in between the two extremes, i.e. 6000 units ((12000 + 0) ÷ 2). The annual holding cost will therefore be \$600 (6000 x \$0.1) if only one order is placed. Now if the number of orders is increased to two, the average inventory will reduce to half along with the annual holding cost
• imum total cost. Let's suppose that the business owner from our previous example has got a volume discount offer
http://www.driveyoursuccess.com This video explains how to calculate economic order quantity using the time-tested Wilson EOQ formula.The video provides a st.. Economic Order Quantity is Calculated as: Economic Order Quantity = √(2SD/H) EOQ = √2 (25 Crore) (1 Crore)/(10 Cr0re) EOQ = √5; EOQ = 2.2360; Hence the ideal order size is 2.2360 to meet customer demands and minimize costs. It is also the reordering point at which new inventory should be ordered. Explanation of Economic Order Quantity Formul
It costs the company \$5 per year to hold a pair of jeans in inventory, and the fixed cost to place an order is \$2. The EOQ formula is the square root of (2 x 1,000 pairs x \$2 order cost) / (\$5.. EOQ Calculator (Economic Order Quantity) Enter the carrying cost per unit per year, the fixed cost per order, and the demand of units per year into the EOQ calculator below. EOQ is used to determine the most cost efficient way for a company to order goods Economic Order Quantity is the level of inventory that minimizes the total inventory holding costs and ordering costs. It is one of the oldest classical production scheduling models. Economic order quantity refers to that number (quantity) ordered in a single purchase so that the accumulated costs of ordering and carrying costs are at the minimum level When calculating a three-month inventory average, the shoe company achieves the average by adding the current inventory of \$10,000 to the previous three months of inventory, recorded as \$9,000,.. Using the EOQ Calculator. The Excel EOQ calculator, available for download below, can be used to calculate the quantity of inventory units which minimizes the total cost of ordering and holding the inventory. The calculator is used as follows. Demand. Enter the unit demand for the product for the accounting period
### EOQ Calculator (Economic Order Quantity) - Captain Calculato
The Economic Order Quantity (EOQ) calculator to calculate the inventory cost of the product Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator Use the EOQ formula to start optimizing your inventory costs. What is Economic Order Quantity? Economic order quantity (EOQ), refers to the optimum amount of an item that should be ordered at any given point in time, such that the total annual cost of carrying and ordering that item is minimized. EOQ is also sometimes known as the optimum lot size
The basic Economic Order Quantity (EOQ) formula is as follows: and applied this to your entire inventory in the EOQ calculation. Your average inventory on a particular piece of software and on 80 lb. bags of concrete mix both came to \$10,000. The EOQ formula applied a \$500 storage cost to the average quantity of each of these items even. This script calculates the average inventory for a month in a sales business. This is done with 2 entries each month and the total is divided by 2. Enter the monthly beginning inventory and the monthly ending inventory. Click on Calculate and see the average inventory for that month. You may click on Clear Values and do another EOQ: Established EOQ (if you need to find this, visit EOQ Calculator) μ: Average item daily demand. z r: z value for confidence level (see below) σ: Standard deviation for daily demand. LT: Lead time . Solution. RP = Your RP is your review period, the number of days from one review to the next. TIL = TIL is your target inventory level. SS 3/6/2017 Self Assessment FINC 330 6380 Business Finance (2172) UMUC Learning Management System 2/7 Where where : Q* = the Economic Order Quantity C = carrying cost per unit S = total demand in units over the planning period O= ordering cost per order. Q* = (2*13,346 * \$10/\$4.01)^ (1/2) = 258 Step 2: Calculate the average inventory held during the year 258 / 2 = 12
### Economic Order Quantity (EOQ) : Formula and Calculato
• imum discount quantities above the EOQ level (i.e. 252 units, 500 units or 1000 units in the example above)
• The relationship between average inventory and Economic Order Quantity While you don't use the average inventory figure to directly calculate EOQ, your holding costs depend on average inventory. Large orders have lower ordering costs because fewer orders are made in the year, but it results in higher holding costs because there is more stock.
• The Economic Order Quantity is a balance between order acquisition costs and stock holding costs. The following elements need to be in place to calculate the EOQ: - annual demand is constant and known in advance, - the unit purchase price is constant, - all orders are delivered in one batch and at the same time
### EOQ Calculator Excel Double Entry Bookkeepin
• imizes total holding and ordering costs for the year. Even if all the assumptions don't hold exactly, the EOQ gives us a good indication of whether or not current order quantities are reasonable
• With this safety stock, you must, therefore, cover yourself on these uncertainties lead time and demand. Safety Stock with EOQ (Economic Order Quantity) Before going into detail on the different methods, I come back to Wilson's /EOQ formula (which you can find in another video). This Wilson formula will optimize the frequency of your order as well as the quantity to be ordered
• Wilson's formula is meant to calculate the economic order quantity (EOQ), which means that the more inventory you have, the more expensive it is.If you don't have stock, you don't have costs, but the more you increase your inventory, the more the cost of owning inventory will increase.This phenomenon is represented by the straight green line in the graph below
Q12 - The order cost per order of an inventory is Rs. 400 with an annual carrying cost of Rs. 10 per unit. The Economic Order Quantity (EOQ) for an annual demand of 2000 units is. 400; 440; 480; 500; Solution: Economic order quantity (EOQ) = = 400 units. Q 11 - The Economic Order Quantity (EOQ) is calculated as (2D*S/h)^1/2 (DS*/h)^1/2 (D*S. The Reorder Point Calculator is a small part of the advanced inventory management features you will find in Fishbowl Manufacturing and Fishbowl Warehouse. With Fishbowl, you can keep track of sales trends, seasonal changes in demand, supplier history, inventory levels and many other factors that can influence your Reorder Point The EOQ Inventory Formula by J. M. Cargal Qh QT CDu 2 2 0 Q T h CD h T h uu 2 2 2 2 0 T h Q T h uuk 2 2 2 Tu * 2CDh Q* 2CD h In this formula the order cost per unit time is CD/Q and Qh/2 is the average inventory cost per uni 2.2.1 Calculate the Economic Order Quantity (EOQ). [5] 2.2.2 Calculate the re-order point if the organisation does not keep safety (minimum level) inventory. [2] 2.2.3 Calculate the re-order point if the organisation has a policy to keep safety inventory. [2] 2.2.4 Calculate the safety inventory that should be kept by the organisation. [2 The aim behind the calculations of EOQ and ROL is to weigh up these; and other advantages and disadvantages and to find a suitable compromise level. EOQ. When determining how much to order at a time, an organisation will recognise that: as order quantity rises, average inventory rises and the total annual cost of holding inventory rise
The EOQ formula calculates the optimal number of units you should buy to keep enough inventory in stock while minimizing storage and production costs. Calculate it by finding the square root of two times the product quantity times the set-up costs, divided by the storage cost per unit If the system should calculate the economic order quantity (EOQ), select this check box. If the system should not calculate the EOQ, clear this check box. Note: This field, as well as the Based on Weighted Average Monthly Usage area, is displayed only if the Based on Weighted Average Monthly Usage option is selected in the Calculation Type area The economic production quantity model (also known as the EPQ model) determines the quantity a company or retailer should order to minimize the total inventory costs by balancing the inventory holding cost and average fixed ordering cost. The EPQ model was developed by E.W. Taft in 1918. This method is an extension of the economic order quantity model (also known as the EOQ model) Economic order quantity (EOQ) is an equation used to determine the ideal quantity of inventory to stock in your warehouse so that you don't spend too much on storage but also don't run out of products. MOQ is the amount of product a supplier or seller requires a purchaser to buy at one time. An MOQ can cause an ecommerce business to purchase more inventory than they need at one time. Average Inventory Period = Days In Period / Inventory Turnover. To calculate, first determine the inventory turnover rate during the period of time to be measured. Typical measurement periods are one year or one quarter but some companies may want to monitor more frequently. Inventory turnover can be figured a few different ways, but the.
### Economic Order Quantity Calculator - UltimateCalculators
calculating EOQ EX: Units of merchandise needed per year is 3,000 Ordering cost \$700 per order and carrying cost 2.5% of the unit costs, unit's price amounted to \$5,000 calculate EOQ EOQ = 183.3 units (rounded to nearest full upper unit to become 184 units (# of orders placed = 3,000 units ÷ 184 units = 16.30 order (rounded to nearest full upper number. Inventory is an important account to keep accurate on a company's financial statements. The average inventory is used in several financial ratios, such as the Cost Of Goods Sold. Analysts base part of their analysis on a company by using these financial ratios, which makes the need for accurate inventory volume. The economic order quantity for inventory management is the least amount of inventory the business can carry while continuing to meet demand. Keeping inventory levels low lends itself well to just-in-time inventory business models, a strategy of producing product only when an order is received The beginning and ending inventory balances at each period must be determined to calculate average inventory. Sum the inventory balances, and divide by the total number of periods. A person can use either the beginning or ending inventory balances for purposes of the calculation. These balances for each period must be summed and then divided by.
And for the average inventory to be coestimated in a way that provides an observable total system inventory, holding cost, service level, and a picture of what is happening to the overall system. Calculating individual parameters like EOQ without an appreciation for the systemwide does not make any sense 4. Average inventory. Average Inventory is the median value of inventory, over a defined time period. The Average Inventory ratio evens out seasonal fluctuations, effectively normalizing the data. It is an indicator of how fast inventory is selling, and the average volume kept on hand. A fluctuation may highlight issues with purchasing or sales. 5 Economic Order Quantity (EOQ) is a method of calculating the quantity of the stock that needs to be re-ordered by taking into consideration the demand for that particular item/product and your inventory holding costs. EOQ is the answer to the question; what quantity of stock should you reorder to replenish your inventory?
Calculate the EOQ. b. Determine the average level of inventory. (Note: Use a 365-day year to calculate daily usage.) c. Determine the reorder point. d. Indicate which of the following variables change if the firm does not hold the safety stock: (1) order cost, (2) carrying cost, (3) total inventory cost, (4) reorder point, (5) economic order. Let's say you own a smoke and vapor shop plus you need to order cartridges through the factory often, so that you can market it to your customers. The probl.. Total Inventory Cost Total Inventory cost is the total cost associated with ordering and carrying inventory, not including the actual cost of the inventory itself. It is important for companies to understand what factors influence the total cost they pay, so as to be able to minimize it Production Order Quantity (EOQ) Calculator. More about the Production Order Quantity for you to have a better understanding of the results provided by this solver. The production order quantity is a type of inventory policy that computes the order quantity \(POQ\) that minimizes the total annual inventory costs, that consists of the sum of the annual setup costs and the annual holding costs
### How to Best Use the Economic Order Quantity Calculator
1. The firm should maintain a Safety Stock of (40 x 3) 120 units. So, the Reorder Point will be 240 + 120 = 360 units. As such, the maximum inventory will be equal to EOQ plus the Safety Stock, i.e., 800 units + 120 = 920 units. Illustration 4: Calculate the reorder point from the following particulars: Annual Demand — 1, 04,000 units
2. i = Carrying cost, per dollar, of average inventory c = Unit cost at last purchase, maintained by the application The value in the EOQ maximum number of stockouts per year field on the EOQ tab of the Data -> Parts Items -> Setup -> Options screen is used to calculate when and how many of each part you should order to optimize management of.
3. imum level. In other words, the quantity that is ordered at one time should be so, which will
4. The company's reputation and profitability are always at stake if it has too much or too little inventory on hand. 43% of business owners have considered overbuying inventory as a challenge, whereas 36% said the same for underbuying inventory. To provide a solution to this, in 1913, Ford W. Harris designed a model of Economic Order Quantity (EOQ) through an explanation
5. The cost of each unit is \$99, and the inventory carrying cost is \$9 per unit per year The average ordering cost is \$29 per order. It takes about 5 days for an order to arrive, and the demand for 1 week is 119 units
6. imum quantity to attain each price break 5. ick the order quantity that has the lowest TA
7. ing the restocking level. Where: = average demand during the reorder period plus thereplenishment lead time (if there is a delay getting new products in). SS = safety stock. This is a cushion of inventory held to mitigate the.
### How can we determine the average inventory level (from EOQ
• Calculate and illustrate average total (with and without safety stocks) and pipeline inventory as well as EOQ. List components of Inventory carrying costs. Good Inventory Management Inventory decisions which incorporate the: The right _____ The right _____ At the right _____ _____ When is a refrigerator not a refrigerator
• imize cost of inventory
• The EOQ is a useful measure as it takes into account all costs associated with inventory as well as the average inventory level. The figure you calculate is the optimum inventory level for your.
Average Inventory Ending Cost - what is average cost? Bond Price - Know Face Value of Bond Economic Order Quantity - Minimize total holding Costs; Employee's State Insurance Rate - For India Calculations can be difficult and to calculate average inventory ending cost use this free online business calculator. Average Inventory. optimal average cost backorder (b) optimal average cost (\$) Reducing Setup Costs Sensitivity Analysis T a(T) EOQ Calculator opt. average cost opt. order interval opt. order quantity optimal order quantity Average Cost Cost Reduction Order Quantity Holding Cost Setup Cost Total EPQ Calculator a(T)/a(T^*) 0.05 \$42.50 0.50 1.00 2.13 0.10 \$25.00 0. Calculating EOQ and inventory costs Inventory Management: Safety Stock, Re-Order, EOQ Inventory management EQQ, average inventory, order per year, and average daily demand Calculating EOQ, annual holding costs and ordering costs Calculating EOQ and Reorder point EOQ Model Manufacturing and Services of Southeastern Bell Inventory Polic Economic Order Quantity and Inventory Control System EOQ, Reorder point and average inventory Calculating the EOQ and various inventory costs Import Beer product line: Method for optimal ordering quantities Economic Order Quantity Rish Corporation: calculate economic ordering quantity, average inventory and more... Economic Order Quantity - EOQ An Economic Order Quantity (EOQ) is an inventory-related evaluation to determine the optimum order quantity which a company should use to ensure that Inventory is not overstocked whilst at the same time maintaining sufficient stock to prevent a stock-out. The objective therefore is to minimise the combined costs of acquiring and carrying inventory
Economic order quantity (EOQ) For businesses that do not use just in time (JIT) inventory management systems , there is an optimum order quantity for inventory items, known as the EOQ. The aim of the EOQ model is to minimise the total cost of holding and ordering inventory CALCULATE the Economic Order Quantity from the following information. Also state the number of orders to be placed in a year. Consumption of materials per annum : 10,000 kg. Order placing cost per order : ` 50 Cost per kg. of raw materials : ` 2 Storage costs : 8% on average inventory SOLUTION EOQ = 2A O C ×× A = Units consumed during yea
It's a tool for keeping tabs on the inventory level of your items so that you do not run out of stock. Whenever the item count reaches the reorder point, you get automatically notified and you know it is the right time to create a purchase order with your supplier The well-known Economic Order Quantity (EOQ) model assumes that the demand and ordering cost is constant over time, and there is no lead-time for each order delivered. (Harris, 1913) The purpose of the EOQ model is to help the manufacturer determine the best order size that could minimize the total inventory holding cost and ordering cost Inventory costs are the costs of keeping stock of your goods expressed as a percentage of the inventory value. Capital, warehousing, taxes, depreciation are some of the costs included in the total annual inventory cost. This page shows the Total annual inventory cost formula to calculate the total annual inventory cost
### Online Calculator of Average Inventory Ending Cos
Calculation of Economic Order Quantity (EOQ)! The quantity to be ordered at one time is known as 'ordering quantity' and should be determined with good care. If it is small, a number of orders will have to be placed in a year involving costs in terms of clerical labour, material handling, etc If the actual order quantity is the economic order quantity in a problem that meets the assumptions of the economic order quantity, the average amount of inventory on hand: A) is smaller than the holding cost per unit. B) is zero. C) is one-half of the economic order quantity. D) is affected by the amount of product cost Economic Order Quantity (EOQ): Model descriptionI The EOQ model is a simple deterministic model that illustrates the trade-o s between ordering and inventory costs. Consider a single warehouse facing constant demand for a single item. The warehouse orders from the supplier, who is assumed to have an unlimited quantity of the product We know that the average inventory level is Q/2, where Q is the order quantity. If we order Q* (the EOQ) units each time, the value of the average inventory can be computed by multiplying the average inventory by the unit purchase cost, P. That is: Average Monetary Value of Inventory = P x (Q*/2) 40 What is EOQ (economic order quantity)? Also referred to as 'optimum lot size,' the economic order quantity, or EOQ, is a calculation designed to find the optimal order quantity for businesses to minimize logistics costs, warehousing space, stockouts, and overstock costs. Why you should be calculating EOQ. Calculating the EOQ for your.
Finding the EOQ, Total Cost, and TBOFor the bird feeders in Example 12.2, calculate the EOQ and its total annual cycle-inventory cost.How frequently will orders be placed if the EOQ is used?Using the formulas for EOQ and annual cost, we getFigure 12.6 shows that the total annual cost is much less than the \$3,033 costof the current policy of. For calculating the carrying cost, we need to find average inventory first. Average inventory is opening stock plus purchase divided by 2. We have no opening stock so that average inventory would be 895/2 = 447.5 units. So, the total carrying cost is (447.5 * 5) \$2,237.5. Total inventory cost for company at EOQ is the ordering cost plus holding. Economic Order Quantity. Economic Order Quantity (EOQ) is the level of inventory that minimizes the total cost of holding and ordering inventory over a period of time. Usually the time period is one year. The total cost of inventory is the sum of the purchase, ordering and holding costs. As a formula Let Q be the Economic Order Quantity (EOQ), R be the demand per unit time in units, (From Fig. 4.1, if total quantity is OB i.e., Q, then average inventory=Q/2) Putting expressions of (2) ad (3) in (1) The objective is to determine the quantity to order which minimizes the total annual inventory cost average inventory average inventory vs. Calculating EOQ OM2013 - 13 19 1. Determine ordering costs (not necessarily easy) 2. Determine holding costs (not necessarily easy) 3. Calculate EOQ Determining reorder point in EOQ-model- R = dL - OM2013 - 13 20 Inventory Time Lot size Q Reorder point R Lead time L demand during lead time usage rate d.
Moving average: in one site they had that example with column 1 going from jan to december showing the actual sales and in column 2 was the moving average forecast. in that example they were taking the average of 3 months. on one site the guy was entering the formula (average of jan, feb and march)in the april cell of column 2 The simplicity of the model resides in its ability to calculate such optimal quantity only considering demand, and the ordering and holding costs. This application calculates the EOQ given an annual demand estimate, together with the total yearly orders and the total annual cost The reorder point (ROP) is the level of inventory which triggers an action to replenish that particular inventory stock. It is a minimum amount of an item which a firm holds in stock, such that, when stock falls to this amount, the item must be reordered. It is normally calculated as the forecast usage during the replenishment lead time plus safety stock Total annual holding cost = Average inventory (EOQ/2) x holding cost per unit of inventory. Total annual ordering cost = Number of orders x cost of placing an order. There is also a formula that allows us to calculate the Total Annual Costs (TAC) i.e. the total of purchasing costs, holding costs and ordering costs STEP 1: Find out carrying cost per unit, stock-out cost per unit, economic order quantity (EOQ) and reorder level. STEP 2: Using historical data, assign different probabilities to occurrence of difference demand levels. STEP 3: Select a safety stock level and find out expected stock out costs and carrying costs using the following formulas
### Inventory Turnover Calculator - Good Calculator
1. 7 Solving for optimal Q and R Start with a Q0 value and iterate until the Q values converge Q0=EOQ Q1 R1 Q R2 Q3 R3 R4 2 1 2 1 Remember: To find Q, you need n(R) = L(z) Lookup for z in the Normal tables Example - Rainbow Colors Rainbow Colors paint store uses a (Q,R) inventory system to control its stock levels
2. imize both inventory and carrying costs. Here's the formula: EOQ = square root of (2 x demand x ordering costs) / carrying costs
3. In accounting, the inventory turnover (also referred to as inventory turns or stock turnover), is the number of times the inventory is sold or consumed during a given time period, typically a year. Inventory turnover is typically measured either at the SKU (Stock-Keeping Unit) level, or averaged out at a more aggregate level. Numerically, the inventory turnover is frequently defined as the.
4. Calculating the reorder point, safety stock and average order quantity over time can help you avoid hoarding excess inventory and free up your working capital. Let's say you recently expanded your business and a couple of your new products have been selling like hot cakes
5. Q* = Optimal number of pieces per order (EOQ) D = Annual demand in units for the Inventory item S = Setup or ordering cost for each order H = Holding or carrying cost per unit per year Annual holding cost = (Average inventory level) x (Holding cost per unit per year) Order quantity 2 = (Holding cost per unit per year
Inventory Formula (Table of Contents) Inventory Formula; Examples of Inventory Formula (With Excel Template) Inventory Formula Calculator; Inventory Formula. Inventory, in very simple terms, is basically products, goods, raw material which are not utilized by the business and expected to be used Inventory Management also known as stock management is a crucial part of working capital management. EOQ is one of the most prominent models used widely for effective inventory management. EOQ calculates the ordering quantity of inventory using inputs of carrying cost, ordering cost, annual usage of the said inventory
### Calculate average Average Inventory? Yahoo Answer
Since the average daily sales for the Ghost are 2 (as calculated earlier on this page), that means the safety stock for Ghost is about 14 x 2 = 28. [We've also written a more detailed article on safety stock, if you'd like to calculate it based on lead-time demand.] Putting the reorder point formula togethe For the calculation of the EOQ, buffer inventory is irrelevant. If asked to calculate the total annual holding cost, then it is the average inventory (EOQ/2) plus the buffer inventory that is multiplied by the holding cost per unit Qo=√ [(2DS) ÷H], where Qo = Economic Order Quantity, D = annual demand (9600 crates/year), S = ordering cost (\$28), and H = the annual carrying cost (35% of purchase price/crate which is \$3.50). Ordering in smaller quantities, like the EOQ model suggests, helps to reduce cost by reducing the holding cost of the inventory There are tools to help you calculate and reduce the cost of carrying inventory. Many businesses use formulas to find the exact amounts they spend on various inventory costs. One of the most helpful formulas in business management is inventory turnover ratio , which is the cost of goods sold divided by average inventory
The Economic Order Quantity model (EOQ) is a mathematical model used to calculate the quantity of inventory to order from a supplier each time that an order is made. The aim of the model is to identify the order quantity for any item of inventory that minimizes total annual inventory costs Safety inventory is the average amount of inventory. the Economic Order Quantity (EOQ) model. This optimal order interval for this model is given by T⁄ = r 2K h‚: (5) The optimal order quantity for this model is known as the EOQ and is given by Q⁄ = ‚T⁄ = r 2K‚ h: (6) Finally, the optimal average cost is given by a⁄ · a(T⁄) = p 2Kh‚ (7) 1.2 Sensitivity Analysis The basic formula for the reorder point is to multiply the average daily usage rate for an inventory item by the lead time in days to replenish it. For example, ABC International uses an average of 25 units of its green widget every day, and the number of days it takes for the supplier to replenish inventory is four days Academia.edu is a platform for academics to share research papers Models of inventory management contain different parameters. An issue is observable in the classical models which can be related to the determination of the quantity of the economic order and the quantity of the economic production. In these models, the parameters like setup and holding costs and also the rate of demands are fixed. This matter causes the quantity of the economic ordering in.
### EOQ Calculator Economic Order Quantity Formula
a. Calculate the EOQ. b. Determine the average level of inventory. (Note: Use a 365-day year to calculate daily usage.) c. Determine the reorder point. d. Indicate which of the following variables change if the firm does not hold the safety stock: (1) order cost, (2) carrying cost, (3) total inventory cost, (4) reorder point, (5) economic order. Average inventory. 3. DP. d._____ Holding cost. 4. SS. e._____ Variance in lead time. Extra amount of inventory beyond that needed to meet average demand during lead time. 10. H (Use the EOQ and ROP formulas to answer this question.) Which variables could you change in each equation if you wanted to reduce inventory costs in your. Economic order quantity (EOQ) When determining how much to order at a time, an organisation will recognise that: as order quantity rises, average stock rises and the total annual cost of holding stock rises. as order quantity rises, the number of orders decreases and the total annual re-order costs decrease.. The economic order quantity (EOQ) is the order quantity which minimises the total. How to Get Rid of Average Costing in QuickBooks Reports. QuickBooks Enterprise can be set up to use the average cost of items to compute values or first in first out. FIFO provides a method of calculating the total worth of your inventory. With FIFO enabled, average cost is turned off and your inventory calculations.
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[–]United States 80 points81 points (24 children)
If you put \$40K down on a \$200,000 house (20% down), you'll have a \$160,000 mortgage. On a 15-year loan at the current 3.5% average rate (bankrate.com mortage calculator), you'll pay \$1143 per month for your mortgage. About \$450/month of that will be interest during first year, so you'll pay about \$5400 in interest. That's deductible from your gross income at your highest tax rate (28% at your salary), so you take off a bit over \$1500 off your taxes. You'll pay property taxes (and deduct them), but that is a net increase in taxes since you deduct the amount from your taxable income (i.e.: it's a deduction, not a credit). You'll end up saving a few hundred a year on taxes -- maybe a grand??? Maybe...
Meanwhile, you are paying \$19,440/yr as interest on \$270K of student loans at 7.2%.
So take on another big loan plus property taxes plus maintenance costs to save \$1500 in income taxes while you keep paying over \$19K/yr in interest on your student loans? Thanks but no thanks! Just paying off \$21K in student loans will save you \$1500/yr without also paying property taxes and upkeep on the house.
Pay down the loan.
[–] 8 points9 points (1 child)
A very important note you assume the OP already has itemized deductions equal to the standard deduction because you only get the mortgage benefit through itemization. You might not even get the full benefit of the interest and each year the standard deduction is going up reducing the potential tax benefit
[–] 4 points5 points (0 children)
If OP is in an expensive state in terms of income taxes, (likely, considering that income figure), then itemizing is a certainty just from the state income tax alone.
[–] 6 points7 points (0 children)
This response is absolutely brilliant. I'd also say that 7.2% guaranteed money is better than you'll see anywhere else right now.
[–] 1 point2 points (15 children)
What about the notion that payments to rent are gone forever, whereas the principal payments on his mortgage he "keeps" as part of his net worth (and recoups when he sells the home)? I feel like that should factor into your final equation but haven't done the final math on it. The accountant might be saying it for the wrong reason, but it might make perfect economic sense to buy a home rather than rent.
[–] 49 points50 points (6 children)
Also gone forever: interest, buying costs, selling costs, maintenance, insurance, opportunity cost, etc.
[–]United States 9 points10 points (2 children)
There's truth that there are advantages to buying rather than renting, but even those come with significant caveats. For example, you need to put down roots to make a mortgage worthwhile. You build up ownership slowly at first because so much of each payment goes to interest. When you sell you must either go "For Sale by Owner" or expect to fork over 6% of the sales price to a realtor. On a \$200K home, that's \$12,000! Add in maintenance and property taxes and the benefits weaken more. However, if you bite the bullet and commit to staying put for a long time the benefits outway the costs -- but if a tempting out-of-state job opportunity arises you'll need to pass.
I think home ownership for some people is GREAT, but for many, many others it really isn't all it's made out to be. I've been in the same place for many years and will be mortgage free in 2 more years (at which time I'll be 48). Of course the house will be about due for a new roof about that time, the water heater is getting quite old, and one bathroom needs a serious overhaul (fixtures, floor, wall coverings, vent fan, lighting...). On the plus side, no one got to say "no" when we decided to convert our daughter's door into a TARDIS and no one scowls about my laser cutter in the third bedroom.
Bottom line: home ownership is not the win-win no-brainer that people often believe it is.
[–] 1 point2 points (0 children)
More people need to fully understand what you just said. Way too many people rushing out to buy a home because it's the "logical thing to do" without doing any of the math or thinking about the downsides.
[–] 0 points1 point (0 children)
There are several major issues that aren't being considered here. One is inflation and low interest rates. We are still at pretty historic lows in terms of interest, which means more principal is paid off in those first few years than with higher interest rate loans. So it is possible to gain decent equity over time, at a faster rate.
The second is appreciation. Not speculation, but appreciation due to inflation. Housing follows inflation. So a 200K home will appreciate at roughly 3% a year in urban areas, and most cities at about 5-7% (local inflation).
If he puts 40K down 7.2% towards his loans0, that's roughly \$2900 in interest "saved" per year. However, on a 200K house, he is gaining value at 3% (low end) to 5-7% (high end, NYC, SF, etc). 3% on the FULL 200K price is 6K per year he gains, versus "saving" 3K/year in interest. If he sells in the first 2-3 years, he basically loses that appreciation to fee's, but after that time appreciation really kicks in.
Add in tax savings on interest payments, higher principal paydowns in the earlier years due to low interest rates, depreciation write offs, inflation and the fact that we're just coming out of a housing cycle, buying would probably be a pretty good decision on his part.
I own half a dozen rentals right now, and all those deductions add up! My interest rates plus maintenance and other fees are all below the inflation for the city, so I'm being paid to borrow money! It's wonderful.
[–] -2 points-1 points (4 children)
Usually only after five or ten years. You pay most of the interest first; the bank gets their share first. Then you get to build equity.
The bigger the down payment, the better off you are. Saving for 5 years renting to double or triple your down payment can let you halve your mortgage payment and halve what you pay to the bank in interest. And then you can potentially only have a mortgage for 15 years instead of 30.
[–] 1 point2 points (1 child)
What? A larger down payment is a huge opportunity cost for any investments you would have otherwise paid off. Every dollar the OP puts into his debt is 7.2% guaranteed returns versus a 3.5% interest he saves by adding to his down payment.
[–] -1 points0 points (0 children)
Of course he he should pay off the debts first before considering to save for a down payment. I just forgot to say it.
[–] -3 points-2 points (1 child)
You're preaching to the choir, but all of this needs to be considered in the analysis. Everyone yelling "Fire your accountant!" is skipping some of the harder questions/math.
[–]Wiki Contributor 2 points3 points (0 children)
No, it really doesn't. His accountant gave him advice that would not be in his benefit for at least 5+ years, if not longer depending on his situation. For his accountant to throw out a blanket statement of "you'll save money on your taxes by purchasing a home" is flat out ignorant and misleading.
[–] 0 points1 point (2 children)
If our OP pays the smallest down payment possible, 3.5%, the amount that he can pay toward his student loan will barely differ at all.
Given the OP's situation, if he gets an interest only mortgage at 3.5% down, he would essentially be swapping a rent payment for an interest and property tax payment. The big difference is that one is tax deductible, and one is not. We don't know how big interest and property taxes are likely to be compared to rent, but it is probably going to be cheaper after accounting for taxes.
[–]United States 0 points1 point (1 child)
Nope. Nope. Nope... and no thanks.
You're correct that OP might come out slightly ahead, but the financial risks for that slight reward are terrible. He'd be locking himself into a loan with very few exit options. With 3.5% down on a \$200K house he'd owe \$193,000. A 30-year fixed Interest-Only loan averages around 5% now, so he'd pay \$9650 per year in interest. We'll make a guess at \$4000 per year in taxes and assume 1% annual maintenance costs for another \$2000 -- and maybe \$1000 for homeowners insurance. With the deductions we should end up somewhere in the \$1K/month range (very rough estimate).
In return, he needs to commit to living in the house for a very long time for this to be theoretically worthwhile. Otherwise, he sells and the realtor takes 6% or \$12,000 (at a \$200,000 selling price). Since you've suggested interest-only, he'll spend several thousand out-of-pocket to exit. In addition to that downside, if the economy hits another soft spot and home prices fall just a little, he's underwater on his mortgage. Next, if he loses his job, he's locked in with this mortgage at a time when he may need to move quickly to pursue another opportunity. Finally, in addition to the normal estimation of costs for maintenance, he runs the risk that a large repair could crop up that wasn't anticipated.
When all is said and done, you get an incremental reward on an outsized risk if you can create enough tax advantage to make it feasible at all. If you're taking a significant risk, you better be expecting a significant reward. This amount of debt for such a small return makes no sense.
Hammering down the student loans is a guaranteed return -- and a big one at 7.2% interest rates. Pay down the loan.
[–] -1 points0 points (0 children)
So the assumptions I am making are that:
1. It is a non-recourse state (a large number of states are) for mortgages. In other words, if the prices of the house drops, it is the banks problem, and if the price of the house goes up, he gets all of the gains.
2. He is in the highest tax bracket possible. Marginal tax rates, Federal - State - City, (again, depending on where OP lives. By the numbers, I am guessing CA/NY, but I am guessing heavily here), are often in the 50% range. So after the deductions, we are looking at costs far closer to 600 then 1000. That should make it several hundred dollars a month cheaper then rent.
3. Since the point of comparison is against renting, there are no point in discussing a 30 year fixed. 5/1 ARM is far cheaper, and still offers a more stable payment then renting. Yes, the interest rate can go up, but so can rents. 5/1 ARM is still in the 3% range, which helps the math here a lot. In any event, our OP will in a much better position financially when it comes time to refinance in 5 years.
4. Maintenance can often be deferred until after the student loans are paid off. Obviously depends on what is broken, but that does help the math here by a bit.
Doing the math, on a 400K place that rents for 2K a month, our OP would have to pay 16K a year in interest+tax. After taxes, that is 700 a month. Budget another 200 a month for repairs, brings us up to 900 a month, post tax. Which means that OP saves 1100 a month. If we assume that the price of the house simply keeps pace with 2% inflation, that means that it will gain 8K a year in value. So OP saves 13K in rent a year + 8K a year in home value gains, he will make realtor fees back in a year.
TL;DR: The name of the game isn't investing, it is to structure your payments in a way that the government pays for half of your rent.
[–] -2 points-1 points (1 child)
He should also be able to get tax deductions from the interest he is paying on (private?) student loans. So while that is a large principle, he isn't doing terrible by not paying it off as fast as possible. Also, by buying a home he is saving on rent, getting more tax deductions, and investing in his future.
That being said, those tax deductions could be going away if they are not renewed by the government. So his tax saving plan would become a tax burden.
[–] 1 point2 points (0 children)
Nope, income way too high for that.
[–] 39 points40 points (29 children)
You accountant shouldn't be your accountant any more. Without numbers, He's telling you to pay the bank interest so you can right it off and get a credit for 35% of the total interest paid.
If you get a mortgage and pay the bank 10,000 this year in interest, you will save \$3,500 in taxes. How about this. You mail me a check for \$10,000 and I'll give you \$5,000 for your taxes.
There are other ways to earn tax breaks that don't put you into further debt. Seriously consider firing this gentleman.
[–] 1 point2 points (1 child)
If you get a mortgage and pay the bank 10,000 this year in interest, you will save \$3,500 in taxes.
Not even, since that doesn't factor in the difference between itemizing and taking the standard deduction
[–] 2 points3 points (0 children)
Since OP makes \$175k, there's a chance he itemizes already just because of state taxes and charitable giving. So each dollar of mortgage interest could represent savings at the marginal rate.
[–] 4 points5 points (3 children)
While I agree with the sentiment generally, this analysis doesn't take into account the fact that OP is paying rent right now. If he is already paying \$1500 a month in rent, and instead pays \$1600/month for his mortgage (with ~\$500 of that being for interest, just to pick a number), he effectively gets \$175 off his taxes per month (assuming he already has enough other deductions to itemize, which I assume his accountant knows and we don't), all for sending the bank, effectively, an extra \$100 per month.
And that's not taking into account that the \$1500/month sent into rent is gone forever, whereas the amount he puts into the principal of the mortgage (in my example, \$1000 a month) he recoups when he sells the house, even without assuming any appreciation of the home.
[–] 6 points7 points (0 children)
As YoYo mentioned, dont forget interest, buying costs, selling costs, maintenance, insurance, opportunity cost, etc, all that are lost forever. Owning a home is not always a net gain, especially if the savings from rent can go towards investments like a stock index with a 7% annual return or (even better) that debt he has that has a guaranteed 7.2% return on whatever he pays off.
[–] 0 points1 point (0 children)
But how would he deal with the additional \$100k in debt if he ever had an emergency?
[–] -2 points-1 points (0 children)
Your numbers are way way off.
[–] 2 points3 points (8 children)
You are assuming that he pays nothing for rent now. If his mortgage payment is the same as his current rent payment then he won't be paying any more than he is now but he will get a tax deduction for some of that monthly payment.
[–] 5 points6 points (4 children)
But he gets the added risk of a house. Any thing goes wrong and he is several thousand in the hole.
[–] 0 points1 point (3 children)
Depends on the state, some states are non-recourse for mortgages.
[–] 1 point2 points (2 children)
I'm talking about roof leaking, heater goes out, bug infestation. Stuff that can cost big bucks that come out of nowhere. While it is unlikely that they pop up immediately, it is possible to pay over \$10k in repairs the first year just because the roof developed a leak.
[–] 0 points1 point (1 child)
For someone at an income level of 172K, 10K is not going to make a huge difference of one way or another. As long as OP figures in the correct expected value for repairs, it will turn out more or less fine.
[–] 0 points1 point (0 children)
But his questions is about coming out ahead in the short term.
The reason this house came to his attention was to save money on his loans via tax itemization. A \$10k bill would not help that.
[–] -5 points-4 points (2 children)
That's nothing like what the accountant suggested and also very unlikely. Almost always buying a house involves payments more than rent, so using an unrealistic scenario to make the accountant's advice actually work is a weird curiosity, but still doesn't make it good advice.
Paying extra in interest to get a tax deduction COSTS extra. It only makes sense if the net cost (expense less the deduction) is somehow better for you.
[–] 2 points3 points (1 child)
Almost always buying a house involves payments more than rent, so using an unrealistic scenario to make the accountant's advice actually work is a weird curiosity, but still doesn't make it good advice.
I would like to see some data to back this up. If it almost always costs more to buy a house than to rent then most landlords are losing money.
I'm also not suggesting that this is a good idea. Buying a house just for a tax break can be a very bad idea. It doesn't mean you won't save money over renting though.
[–] -3 points-2 points (0 children)
No. Most people rent a more modest apartment than the house they ultimately buy. Most landlords are not renting the exact same place to people who buy it from them. People do buy up, down and sideways when they go from renting to buying, but more often than not, it's to step UP in housing - both in cost and in place.
[–] 0 points1 point (5 children)
your accountant shouldn't be giving you investment or money management advice. He should be explaining to you in vivid detail the implications of your current, actual spending habits on your taxes.
[–] 3 points4 points (3 children)
My accountant told me to get married. It wasn't good advice for him because my husband is now an accountant, so I don't need him anymore :)
[–] 2 points3 points (2 children)
Well...unless he was that accountant. And that was a very weird marriage proposal
[–] 0 points1 point (0 children)
He was not that accountant
[–] 0 points1 point (0 children)
It could have been epic. I am imagining a romance novel around that now.
Harelquin, "Let me do your numbers"
[–] 1 point2 points (0 children)
Only if he is not a very good accountant. A good accountant should always be advising you on the future and ways to improve your financial return given the accounting and tax implications of different behaviors and actions. Indeed, planning is the biggest reason for most individuals to use accountants, not just helping file a form reporting history. By then it is too late to do anything about it.
[–] -1 points0 points (6 children)
You're ignoring his rent, which is likely FAR more than the mortgage on the equivalent house would be.
[–] 0 points1 point (5 children)
This is a fair point, but for me, paying rent makes up for the incidental costs associated with homeownership.
There is also the benefit of not being tied down to a place, or adding to the pile of debt OP already owes.
I guess the real truth is that "saving on taxes" is not a reason to buy a house. Buy a house when you have the money and are established enough financial, professional, and interpersonally to say "I want to live here for the next 7 years"
[–] 0 points1 point (4 children)
You think you aren't paying for your landlord's "incidental costs associated with homeownership" through your rent?
[–] 0 points1 point (1 child)
Yes, these other costs will be factored into your rent, but you're far less likely to need a few thousand dollars at the drop of a hat when you're a renter. As with a lot of rent vs buy stuff, it's a bit of a lifestyle question - would you rather pay a bit more a month & not have to worry if something breaks, or pay a bit less a month & deal with all that crap yourself?
[–] 0 points1 point (0 children)
"A bit less"
lol
[–] -1 points0 points (1 child)
Rental property is completely different than personal residence.
[–] 0 points1 point (0 children)
As someone who owns both, you're right. My renters pay a shitton more in rent than they would for a mortgage on the same property. It's how real estate investors make money.
[–] -2 points-1 points (0 children)
Thank you!
[–] 9 points10 points (1 child)
If you buy a modest home, that means your total interest paid will not be very high, so I can't see you saving a lot of money by itemizing (unless you have a bunch of other deductions you can take if you do itemize)
Given the fact your loan is at 7.2%, any down payment/closing costs towards purchasing a house will cost you a lot of interest on your student loan.
[–] 1 point2 points (0 children)
It really depends on how much money it will save. In some areas buying a similar house over renting can save you hundreds of dollars a month that could be going to the student loans. A 30 year mortgage will be mostly interest regardless of the amount and you need to consider the property tax deduction as well.
On the other hand I would never buy a house unless I knew I would want to live there for at least 5 to 10 years. You could also consider buying a cheaper house now to live in and save money and then rent it out later.
[–] 12 points13 points (0 children)
Say you buy a \$100k home; you pay \$20k down and borrow \$80k on a 30 year note at 4.5%. Your monthly payment will be about \$405 (it will be higher because of escrow for property tax and insurance). The first year you will pay about \$3500 in interest. You can deduct that plus property taxes (let's say \$2000, but this will vary wildly by region). So that gives you \$5500 in deductions. The standard deduction for 2013 is \$6100, so you still have to come up with \$600 in other deductions before you begin to see any benefit. Then the benefit is only on marginal dollars over the standard deduction.
So, unless you are already itemizing or close to itemizing, there really won't be much tax benefit from a modest house. Meanwhile paying that \$20k toward your student loans saves you \$1440 a year guaranteed. You really won't get that return anywhere else.
Now, if you want to buy a house and it fits in your budget, it may not be much different cost from renting if you can get some tax benefit because of you high income. I just wouldn't buy only for taxes. Don't forget that owning a house opens you up for lots of little and big emergency expenses over and above normal mortgage+taxes+insurance+utilities. Not to mention initial expenses like closing costs, moving expense, furniture, etc. I own a house and am glad that I do, but don't do it as a tax avoidance vehicle. There are better things to do to lower your tax bill like max out retirement contributions. Even buying an investment property is likely to lower your tax bill more than buying a house to live in if you don't pay cash (depreciation expense makes it a loss on paper even with a positive cash flow).
[–]Wiki Contributor 10 points11 points (0 children)
You need to find another accountant.
[–] 19 points20 points (10 children)
He suggests I buy a modest home, states the mortgage won't be much higher than my rent and I will save a lot on taxes.
Buying a home "just to save on taxes" is a silly idea. Buy a home if you are ready and want to buy a home, not before.
[–] 0 points1 point (0 children)
But look at all that mortgage interest you get to deduct... /s
[–] 1 point2 points (0 children)
Med school?
[–] 1 point2 points (0 children)
Here's the math I can glean from the facts you provide:
Student loans: 240K left at 7.2%. Monthly payments of around \$3000 I'm guessing. Let's say you put your \$40K towards these loans.... you will save a total of ~\$30K in interest payments over the life of the loan! Not an insignificant amount at all.
House: This is more complicated, but we need to figure out how much you can save. Others have done the math, and it depends on the size of your mortgage, but odds are you won't be able to save more than ~\$1500-\$2000 in income taxes, and the number will decline every year. So this is nice but not significant savings. You will build up equity, (~\$7000 for the first year would be a reasonable guess if it's a \$200K house). But with taxes and generally paying more money towards this additional debt, it will also slow down your student loan payments, and you'll be paying even more interest on your student loans (which remember, are substantial and have a high interest rate).
All this being said, if you plan on taking a long time to pay off these loans, buying the house does make financial sense. Another key factor is the amount you plan on spending on the house. Since we have no info on how fast you want to pay down the loans or how much you want to spend on the house, I won't even try to speculate when the break even point is.
[–] 4 points5 points (1 child)
You're accountant is an idiot. You're paying a dickload in interest on your student loan debt every year. Thats making a significantly larger dent in your income instead of the marginal tax incentive you might gain.
[–] 6 points7 points (0 children)
You're accountant is an idiot.
Ouch.
[–] 0 points1 point (1 child)
[–] 2 points3 points (0 children)
Maybe throwing the accountant on OP's debt will help somewhat, too?
[–] 0 points1 point (0 children)
Define 'much higher'.
If you actually want a house and expect to be in the same area for awhile (>=5 years) then I might go for it, although the level of debt would be concerning to me. Whether or not you'd actually save money though depends on the monthly mortgage payment, taxes, and maintenance costs/fees vs. your rent payment.
It doesn't make much sense to spend more than you would otherwise to 'save' money on taxes when you're still coming out behind. You're only saving money on your marginal tax rate for anything above the standard deduction amount, which is pretty generous to begin with.
It's like the old JC Penny sales advertisements, "The more you spend, the more you save!", sure but you're still spending more so the additional 10/20% off is irrelevant unless you were going to buy that stuff to begin with.
[–] 0 points1 point (0 children)
I like how a bunch of people's immediate advice is to fire the accountant. As if they have access to the information your accountant has. Chances are his advice is good within the confines of saving money on taxes.
The question then is do you want to own a house and is buying a house what you would do if the money were a wash? If not, why not? Are you going to move within the next 5 years? Do you must not want the responsibility of dealing with the costs (in both time and money) that ownership entails?
[–] 0 points1 point (0 children)
Live frugally on almost nothing (\$50k per year would be just fine anywhere in America, even in SF or NY) and get rid of your loan in the next two years. Do not take on more debt.
Get rid of your student debt, on which you're paying an insane amount of interest, and then consider buying a house.
Your accountant's advice is terrible in case anything happens. Get injured? Lose your job? How will you pay off your \$400k+ debt? He's setting you up for a disaster.
[–] 0 points1 point (0 children)
The cost of owning a home vs the taxes you save on the interest will never work out in your favor. For example, I paid \$8k in interest last year. That equated to a savings of about \$1600 on my taxes which still means the interest cost me \$6400. So the bank makes \$6400 off of me rather than \$8,000 - would you call that a win?
Stay with renting. You can predict your monthly expenses (vs having unexpected expenses pop up as happens often in home ownership). Pay down your debt hardcore. Once you're under \$50-70k in student debt, then consider buying a house if you know you are going to live in it for 5-10 years minimum
[–] -1 points0 points (0 children)
Are you paying this guy for his advice?
[–] 0 points1 point (0 children)
One has nothing to do with the other. Find a new accountant.
[–] 0 points1 point (0 children)
Get a new accountant.
[–] 0 points1 point (0 children)
If a whole bunch of things he says are correct and you take on an additional \$100,000 in debt you may save several hundred dollars per year.
This is nuts. Your 'Accountant' needs to be fired immediately. | 6,724 | 27,960 | {"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-2017-43 | latest | en | 0.956513 |
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Discussion in 'Stability' started by Eugene Mak, Mar 13, 2014.
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### peterengSenior Member
I think we need to clarify the use of "CoG". the CoG only refers to the centroid of the masses involved. The CoG of the waterplane and volume does not exist or the term is being used incorrectly. The correct term is the centroid (or geometric centre) for these positions. And yes as Tansl states this problem is not simple. Plus the answer you seek needs to be stated so you can either develop a method to approximate this or you need to do it accurately using FE. The main problem as I see it is the catenary load is non linear so depending on your cable geometry this will make the problem easier or harder. ie the further out you place the anchors the more the catenary becomes significant, the more difficult the probem becomes to solve manually. Regards Peter S
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### jehardimanSenior Member
Do not confuse a reference axis related to the body, to the global axis system. TANSL wishes to look only at the body, not how the body is moving in the global axis system. Are you looking to minimize the pitch/roll motion of the buoy, or to minimize the off station x/y location of the buoy? Based upon cross-coupling, these two are different. Let me get back to the office on Monday and I will check were to look in Berteaux.
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### TANSLSenior Member
Of course, the movements of the object always depend on the degrees of freedom we give it. This is obvious and would not need to explain. But any movement of the object can be reduced to a combination of a translation and a rotation. This is also obvious. The rotation can always be decomposed into three rotations around the three principal axes of the object, X axis, Y axis and Z axis Those axes, always pass through the center of gravity of the float at any times. This is what, may be not so obvious, but, in my opinion, is how the movement works.
(..... this is the plane that the bouy trims around : that is not possible)
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### peterengSenior Member
Hi Eugene - If you publish a drawing with all the required geometry and weights etc I'll consider running it as a FE model for your info. I can establish its stability condition if this is what you want. If you have a thrust on the turbine this can be included as well. You state that you may use synthetic cable for the anchors. Chains are used as a damper. If you use synthetic that is close to neutral bouyancy how do you pretension the system and maintain tension in the system
(due to tide?)? I've been involved in rubber bungy stability of marina pontoons and it works very well. I assume you will have tidal variation to account for again this is what the chains are used for. Cheers Peter S
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### jehardimanSenior Member
Try this paper, you can read it from that site, but you cannot print it. (...though whatever is on the screen is effectively yours anyway...)
https://dspace.mit.edu/handle/1721.1/46545
I looked in Buoy Engineering, and while the theory is there, there is not a clean worked example for a moored spar buoy. Perhaps a better text, which has a worked example for a moored structure, is Advanced Dynamics of Marine Structures by Hooft. The worked example is chapter 9 but it presumes a decent knowledge of hydromechanics. For a basic introduction to floating structure dynamics with lots of worked examples I would recommend Dynamics of Marine Vehicles by Bhattacharyya. These two books, are part of the Wiley-Intersience series (like Beryeaux) and one builds on the other.
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### Eugene MakJunior Member
Hi Guys, thanks a lot for your help so far. That’s quite amazing experience being able to draw on knowledge across the globe.
We have run quite a few tests in a home-made wind tunnel with the bottom part dropped out and having a rain water tub underneath. The definite conclusion is “the closer anchor attached to the surface, the less angle of a heel”.
https://dl.dropboxusercontent.com/u/41354924/March 2014/13.50, Test 3.MOV
Hi Eugene - If you publish a drawing with all the required geometry and weights etc I'll consider running it as a FE model for your info. I can establish its stability condition if this is what you want. If you have a thrust on the turbine this can be included as well. by PETERENG
Yes please. Attached is “Buoy Schematics”. That’s was our initial shot. One thing we would change is a padeye would have to move up, as close to the surface as possible. I will send the thrust value as a side force in N shortly.
I am not that strong at maths and softwares. My personal preference is hands-on model testing and observations. The typical fisherman buoy (attached) has to stay upright all the time so it can be detected from afar and has to be constructed on a shoe lace budget. That’s pretty much about us...
You state that you may use synthetic cable for the anchors. Chains are used as a damper. If you use synthetic that is close to neutral bouyancy how do you pretension the system and maintain tension in the system (due to tide?)? by PETERENG
Why do I have to pretension the system? I am not worried about it “going for a walk” in the large envelope, the only thing I am concerned is reducing a pitch/ heel. On the other hand I am going to test this prototype in Baltic, where the tide virtually non-existent.
Thanks a lot
Eugene
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### TANSLSenior Member
I think that some, including myself, did not understand the nature of the buoy which you spoke and, therefore, what we have said does not apply to this case.
Cheers
8. Joined: Aug 2004
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Location: Port Orchard, Washington, USA
### jehardimanSenior Member
If that's the size it is, there is already a 60cm model out there.
http://www.wpi.edu/Pubs/E-project/A.../unrestricted/MQP_Final_Report_03_09_2011.pdf
FWIW, that much free surface in the water ballast is going to be a problem. You need to calculate a max/min draft based on expected water density changes and ballast for that making the water ballast tank (if any) very small with minimial free surface.
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# Creating a new data structure while filtering its data origin.
P: n/a Hi everyone. I'm trying to work with very simple data structures but I'm stuck in the very first steps. If someone has the luxury of a few minutes and can give an advice how to resolve this, I'll really appreciate it. 1- I have a list of tuples like this: lista= [(162, 141, 3), (162, 141, 3), (162, 141, 3), (168, 141, 2), (168, 141, 2), (168, 141, 2), (201, 141, 1), (213, 141, 1), (203, 141, 1), (562, 142, 4), (562, 142, 4), (562, 142, 4), (568, 142, 2), (568, 142, 2), (568, 142, 2), (501, 142, 1), (513, 142, 1), (503, 142, 1)] and I want to end with a dict like this: {141: {1: [203, 213, 201], 2: [168, ], 3: [162, ]}, 142: {1: [503, 513, 501], 2: [568, ], 4: [562, ]}} the logic of the final output: a) the outer dict's key is a set() of the 2rd value of the input. b) the inner dict's key is a set() of the 3th value for tuples which 3rd value equals a). c) the inner list will be fill up with the 1st value of every tuple which 3rd value equals b) and its 2rd value equals a). So far, the only thing it seems I can achieve is the first part: outer_dict = dict([(x,dict()) for x in set(row[1] for row in lista)]) >From then on, I'm starting to get tired after several successful failures (I tried with itertools, with straight loops ...) and I don't know which can be the easier way to get that final output. Thanks in advance. Mar 28 '07 #1
5 Replies
P: n/a mm***@fibertel.com.ar wrote: Hi everyone. I'm trying to work with very simple data structures but I'm stuck in the very first steps. If someone has the luxury of a few minutes and can give an advice how to resolve this, I'll really appreciate it. 1- I have a list of tuples like this: lista= [(162, 141, 3), (162, 141, 3), (162, 141, 3), (168, 141, 2), (168, 141, 2), (168, 141, 2), (201, 141, 1), (213, 141, 1), (203, 141, 1), (562, 142, 4), (562, 142, 4), (562, 142, 4), (568, 142, 2), (568, 142, 2), (568, 142, 2), (501, 142, 1), (513, 142, 1), (503, 142, 1)] and I want to end with a dict like this: {141: {1: [203, 213, 201], 2: [168, ], 3: [162, ]}, 142: {1: [503, 513, 501], 2: [568, ], 4: [562, ]}} the logic of the final output: a) the outer dict's key is a set() of the 2rd value of the input. b) the inner dict's key is a set() of the 3th value for tuples which 3rd value equals a). c) the inner list will be fill up with the 1st value of every tuple which 3rd value equals b) and its 2rd value equals a). So far, the only thing it seems I can achieve is the first part: outer_dict = dict([(x,dict()) for x in set(row[1] for row in lista)]) >>From then on, I'm starting to get tired after several successful failures (I tried with itertools, with straight loops ...) and I don't know which can be the easier way to get that final output. Thanks in advance. d={} for a, b, c in lista: if d.has_key(b): if d[b].has_key(c): if a not in d[b][c]: d[b][c].append(a) else: d[b][c]=[a] else: d[b]={c:[a]} print d -Larry Bates Mar 28 '07 #2
P: n/a mm***@fibertel.com.ar wrote: Hi everyone. I'm trying to work with very simple data structures but I'm stuck in the very first steps. If someone has the luxury of a few minutes and can give an advice how to resolve this, I'll really appreciate it. 1- I have a list of tuples like this: lista= [(162, 141, 3), (162, 141, 3), (162, 141, 3), (168, 141, 2), (168, 141, 2), (168, 141, 2), (201, 141, 1), (213, 141, 1), (203, 141, 1), (562, 142, 4), (562, 142, 4), (562, 142, 4), (568, 142, 2), (568, 142, 2), (568, 142, 2), (501, 142, 1), (513, 142, 1), (503, 142, 1)] and I want to end with a dict like this: {141: {1: [203, 213, 201], 2: [168, ], 3: [162, ]}, 142: {1: [503, 513, 501], 2: [568, ], 4: [562, ]}} the logic of the final output: a) the outer dict's key is a set() of the 2rd value of the input. b) the inner dict's key is a set() of the 3th value for tuples which 3rd value equals a). c) the inner list will be fill up with the 1st value of every tuple which 3rd value equals b) and its 2rd value equals a). So far, the only thing it seems I can achieve is the first part: outer_dict = dict([(x,dict()) for x in set(row[1] for row in lista)]) >>From then on, I'm starting to get tired after several successful failures (I tried with itertools, with straight loops ...) and I don't know which can be the easier way to get that final output. Thanks in advance. d={} for a, b, c in lista: if d.has_key(b): if d[b].has_key(c): if a not in d[b][c]: d[b][c].append(a) else: d[b][c]=[a] else: d[b]={c:[a]} print d -Larry Bates Mar 28 '07 #3
P: n/a On Mar 28, 4:44 pm,
P: n/a On Mar 28, 1:44 pm,
P: n/a mm***@fibertel.com.ar wrote: Hi everyone. I'm trying to work with very simple data structures but I'm stuck in the very first steps. If someone has the luxury of a few minutes and can give an advice how to resolve this, I'll really appreciate it. 1- I have a list of tuples like this: lista= [(162, 141, 3), (162, 141, 3), (162, 141, 3), (168, 141, 2), (168, 141, 2), (168, 141, 2), (201, 141, 1), (213, 141, 1), (203, 141, 1), (562, 142, 4), (562, 142, 4), (562, 142, 4), (568, 142, 2), (568, 142, 2), (568, 142, 2), (501, 142, 1), (513, 142, 1), (503, 142, 1)] and I want to end with a dict like this: {141: {1: [203, 213, 201], 2: [168, ], 3: [162, ]}, 142: {1: [503, 513, 501], 2: [568, ], 4: [562, ]}} the logic of the final output: a) the outer dict's key is a set() of the 2rd value of the input. b) the inner dict's key is a set() of the 3th value for tuples which 3rd value equals a). c) the inner list will be fill up with the 1st value of every tuple which 3rd value equals b) and its 2rd value equals a). So far, the only thing it seems I can achieve is the first part: outer_dict = dict([(x,dict()) for x in set(row[1] for row in lista)]) >>From then on, I'm starting to get tired after several successful failures (I tried with itertools, with straight loops ...) and I don't know which can be the easier way to get that final output. Thanks in advance. lista, wanted = ... result = {} for a,b,c in lista: inner = result.setdefault(b, {}).setdefault(c, []) if a not in inner: inner.insert(0, a) # I had used append, but ... print result == wanted --Scott David Daniels sc***********@acm.org Mar 30 '07 #6
### This discussion thread is closed
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https://statisticasoftware.wordpress.com/2013/04/25/root-cause-analysis/ | 1,563,420,784,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.64/warc/CC-MAIN-20190718022001-20190718044001-00315.warc.gz | 534,035,025 | 16,666 | Root Cause Analysis
The term root cause analysis is commonly used in manufacturing to summarize the activities involved in determining the variables or factors that impact the final quality or yield of the respective processes. For example, if a particular pattern of defects emerges in the manufacture of silicon chips, engineers will pursue various methods and strategies for root cause analysis to determine the ultimate causes of those patterns of quality problems.
One method of root cause analysis is variable screening (also called feature selection), where analytic tools are used to find the variables most highly associated with the quality issue. Interaction terms between these variables can also be part of the root cause. The analysis will yield a list of variables that are the best predictors of the quality issues. These variables can be explored further to gain additional insight.
Design terms
The columns of the design matrix (design terms) for interaction effects are created as follows:
Continuous-by-continuous predictor interactions – A single column is created in the design matrix for each product of the continuous predictor columns.
Continuous-by-categorical predictor interactions – First, the number of unique values (classes) in the categorical predictor is determined. As many columns as there are unique values in the categorical predictors are generated. For each column j of the k columns (unique values), a 1 is generated if the respective observation belongs to class j, and a 0 otherwise. Each column (with the 0/1 indicator codes) is then multiplied by the continuous predictor variable. Hence, for continuous-by-categorical predictor interactions, the program will generate as many columns in the design matrix as there are unique values in the categorical predictor.
Categorical-by-categorical predictor interactions – The unique combinations of groups or classes are enumerated into a single column in the design matrix. For example, the interaction between two categorical predictors with two unique values (classes) each would result in a single column with (2*2 =) 4 values. Note that these coded columns in the design matrix are technically “confounded” with the main effects. In other words, if one of the categorical predictors is strongly related to the dependent variable in the analysis, it is likely that some of the interactions with other categorical predictors will show strong relationships with the dependent variable as well.
Higher-order interactions (e.g., three-way interactions) are created accordingly, i.e., they are generated as the products of continuous and categorical predictors following the rules outlined above. For example, a three-way interaction column would be generated by multiplying a two-way interaction with another effect.
Variable Screening
Options for variable screening enable you to screen predictor variables for regression and classification problems as well as the methods that can be used to find the predictors that are important. In general, predictor statistics can be computed by the respective method, and then predictors can be ranked based on the method-specific measure of predictor importance. The following methods may be appropriate:
Linear model. A linear fit model using stepwise selection of predictors is a simple approach to the regression problem. Predictor importance is computed by ranking the p-values for each predictor effect. For tied p-values, the rankings are based on the ranking of the F-values. For classification tasks, a stepwise linear discriminant function analysis can be used. Predictor importance is computed by ranking the values of the Wilks’ lambda statistics for each predictor.
Classification and regression trees. For classification and regression trees, the standard rankings for predictor importance are used.
Boosted trees. For boosted trees models (stochastic gradient boosting), the standard rankings for predictor importance are used.
MARSplines. For multivariate regression splines (MARSplines), rankings are computed based on the number of times that each predictor was used (referenced) in a basis function. The more frequently a predictor was used (referenced by a basis function), the greater is its importance.
Neural networks. For neural networks, the final importance rankings for the predictors is computed by averaging the importance rankings for each predictor over a set of networks. | 805 | 4,448 | {"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-2019-30 | latest | en | 0.875299 |
https://matharguments180.blogspot.com/2014/08/219-how-far-out-can-you-go.html | 1,516,771,875,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893397.98/warc/CC-MAIN-20180124050449-20180124070449-00371.warc.gz | 724,537,681 | 17,588 | Saturday, August 16, 2014
219: How far out can you go?
A series of identical blocks.
A series of identical rectangular blocks is stacked out at their balancing points from the top down. You can show with a simple center of mass calculation the total "stick-out" distance; that is, the horizontal distance from the back of the bottom block to the back of the top block is
d = ½(1 + ½ + ⅓ + ¼ + ...)
Thanks to for finding these. | 109 | 430 | {"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-2018-05 | longest | en | 0.917178 |
http://www.enotes.com/homework-help/let-f-x-x-2-2x-9-x-rr-determine-set-values-real-454389 | 1,462,431,913,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.68/warc/CC-MAIN-20160428161525-00208-ip-10-239-7-51.ec2.internal.warc.gz | 502,594,061 | 11,524 | # Let `f(x) = x^2+2x+9; x in RR` Determine the set of values of a real constant `lambda` for which the equation `f(x)=lambda` ; that f(x) has no real solution for x.
Posted on
`f(x) = x^2+2x+9`
`x^2+2x+9 = lambda`
`x^2+2x+9-lambda = 0`
If the above quadratic function does not have real roots then the discriminant(Delta) should be less than 0.
`Delta = 4-4xx1xx(9-lambda)<0`
`4<4(9-lambda)`
`lambda<8`
So for `f(x) = lambda` where there is no solutions `lambda<8`
Sources: | 177 | 483 | {"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-2016-18 | longest | en | 0.695545 |
http://mathoverflow.net/feeds/question/88741 | 1,371,645,769,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708766848/warc/CC-MAIN-20130516125246-00002-ip-10-60-113-184.ec2.internal.warc.gz | 166,278,498 | 3,334 | Methods for determining domains of influence - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:42:50Z http://mathoverflow.net/feeds/question/88741 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence Methods for determining domains of influence Igor Khavkine 2012-02-17T17:30:14Z 2012-02-28T10:54:20Z <p>Given a hyperbolic PDE, the <em>domain of influence</em> of a spacetime point $x$, say $I_x$ though $x$ could be replaced by any set, can be defined in two ways. Lets call one of them <em>geometric</em> ($I_x^G$) and the other <em>analytical</em> ($I_x^A$). In Lorentzian geometry, the geometric domain of influence consists of the interior of the cone of null geodesics emanating from $x$ (let me not bother about whether to include the boundary in the definition of $I_x^G$ or not). In general, a similar definition can be given using characteristic cones instead of null cones. The analytical domain of influence can be defined as the set of all spacetime points $y$ such that for every neighborhood neighborhood $O$ of $x$ there exist two solutions $u_1$ and $u_2$ satisfying the condition $u_1(x')=u_2(x')$, for all $x'$ on a Cauchy surface passing through $x$ except for $x'\in O$, and also the condition $u_1(y)\ne u_2(y)$. The latter one is the definition used in Lax's book on Hyperbolic PDEs.</p> <p>Similar defintions can be given for the geometric and analytical domain of dependence, say $D_S^G$ and $D_S^A$. Such a definition should capture the desired equality $D_K = I_{S\setminus K}$, for a Cauchy surface $S$ and $K\subset S$ (once again, being sloppy with boundaries). I know that energy methods can be used to establish that $D^G_K \subseteq D^A_K$ (the analytical domain of dependence is at least as large as the geometric one). Hence, by duality, the same methods establish $I_K^A \subseteq I_K^G$ (that the geometric domain of influence is at least as large as the analytical one).</p> <p>My question is about the reverse inclusion, $I_K^G \subseteq I_K^A$ or by duality $D_K^A \subseteq D_K^G$. Maybe it's too much to ask for the analytical and geometric definitions to coincide. But when they do, what methods are used to establish that? When they don't what methods can identify the obstruction? Except briefly in Lax's book, I don't know what references discuss this problem explicitly, so those would also be appreciated!</p> http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence/89107#89107 Answer by Stefan Waldmann for Methods for determining domains of influence Stefan Waldmann 2012-02-21T14:24:50Z 2012-02-21T14:24:50Z <p>I'm not quite sure if this is really the situation you are interested in, but in the book of Bär, Ginoux, and Pfäffle: Wave Equations on Lorentzian Manifolds and Quantization. ESI Lectures in Mathematics and Physics, European Mathematical Society, 2007, they discuss in quite some detail the Cauchy problem for hyperbolic linear wave equations on, and that is the catch, globally hyperbolic spacetimes. I guess that one should require something like that since otherwise you can at best hope for some local statements. But in their situation, I'm pretty sure to remember correctly, they have statements like the one you are looking for (don't they?). In any case, this is maybe a too special situation for you, but the book is nevertheless very nice. Unlike many other texts on hyperbolic PDE, it emphasizes the geometry very much.</p> http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence/89419#89419 Answer by Willie Wong for Methods for determining domains of influence Willie Wong 2012-02-24T16:39:27Z 2012-02-24T16:39:27Z <p>I highly doubt the result you actually asked for is true. </p> <p>Consider the <strong>linear wave equation</strong> on $(1+3)$ Minkowski space. The <em>analytic domain of influence</em> of a point $x$ as Lax defined it, which morally says that $y$ is in the analytic domain only if one can find perturbations in arbitrary small neighborhoods of $x$ that change $y$ (if I interpret your question statement correctly), actually consists of only the null cone emanating from $x$ and nothing more, since strong Huygen's principle holds. </p> <p>The same is true for the linear wave equation on $(1+(2k+1))$ Minkowski spaces. </p> <p>The opposite conclusion can be drawn on $(1+2k)$ dimensional Minkowski spaces, where the Green's function have support inside the cone. </p> <p>For more general situations, you may want to consult the classical result of Atiyah-Bott-Garding on the existence of <a href="http://en.wikipedia.org/wiki/Petrovsky_lacuna" rel="nofollow">Petrowsky lacunae</a>. For any linear hyperbolic equation that admits a lacuna, the analytic domain of influence cannot cover the entirety of the geometric one. </p> <p>But for the result that you seem to actually want, where you should replace the analytic domain of dependence by a suitable "convex" envelope of it, I don't know if such a result is proven anywhere, but my guess is that, at least for the "local" version one can approach it using some sort of geometric optics construction. </p> <p>For possible references (I haven't actually finished reading either, so they may not contain what you want), maybe you want to look at Michael Beals' book on propagation of singularities (sorry, the title escapes me at the moment) or Rauch's notes on <em>Hyperbolic PDEs and Geometric Optics</em> which I think you can find floating around on the internet. </p> | 1,454 | 5,585 | {"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} | 3.296875 | 3 | CC-MAIN-2013-20 | latest | en | 0.83297 |
http://nrich.maths.org/public/leg.php?code=5005&cl=2&cldcmpid=6968 | 1,503,263,513,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106990.33/warc/CC-MAIN-20170820204359-20170820224359-00356.warc.gz | 310,369,691 | 10,391 | # Search by Topic
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Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws? | 2,332 | 9,642 | {"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} | 4.15625 | 4 | CC-MAIN-2017-34 | latest | en | 0.87664 |
https://chemistry.stackexchange.com/questions/170475/diffusion-migration-and-einstein-equation/172296 | 1,718,805,241,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00471.warc.gz | 136,175,106 | 41,718 | # Diffusion, Migration and Einstein Equation
In the textbook: Electrochemical Systems by Newman and Alyea, 3rd edition, chapter 11.9: Moderately Dilute Solutions, equation for the mole flux of the component $$i$$ is given by: $$N_i = - \frac {u_i c_i} {z_i F} \nabla \bar\mu_i\ + c_i v \tag {1}$$
where $$u_i$$ is the ionic mobility, $$c_i$$ is the concentration, $$\bar \mu_i$$ is the electrochemical potential and $$v$$ is the velocity of the streaming fluid (assuming a low concentration of the component $$i$$ as situation becomes more complicated at higher concentrations).
One thing to note is that I wrote ionic mobility $$u_i$$ a bit differently compared to the textbook. I defined it as a terminal velocity of the ion in the unitary electric field, so its measuring unit is $$[\frac {m^2}{Vs}]$$ in my case. In the textbook they defined it also as a terminal velocity in the unitary electric field, but divided by $$z_i F$$. This is something to keep in mind to avoid confusion.
Basically, equation (1) states that the mole flux of the component has three contributions: diffusion, migration and convection. Diffusion and migration are accounted for in the first term (electrochemical potential gradient) and convection is accounted for in the second term.
My problem is with the first term $$-\frac {u_i c_i} {z_i F} \nabla \mu_i$$ which I know how to derive and I will do this here as I don't understand something about the equation (1).
To start, we define diffusion mole flux from as the 1st Fick's law: $$N_{i,dif} = -\frac {D_i c_i} {RT} \nabla \mu_i \tag {2}$$
where $$D_i$$ is the diffusion coefficient and $$\mu_i$$ is the chemical potential of the component $$i$$
Migration mole flux is given by: $$N_{i,mig} = -u_i c_i \nabla \phi \tag {3}$$
where $$\phi$$ is the electric potential.
If there is no convection, we can write total mole flux of the component as: $$N_i = -(\frac {D_i c_i} {RT} \nabla \mu_i + u_i c_i \nabla \phi) \tag {4}$$
We recall the definition of electrochemical potential $$\bar \mu_i$$: $$\bar \mu_i = \mu_i + z_i F \phi \tag {5}$$
If we assume that system is in thermodynamic equilibrium than electrochemical potential gradient and net mole flux vanish: $$\nabla \bar \mu_i = 0 \tag {6}$$ $$N_i = 0 \tag {7}$$
From equations (5) and (6), we can write: $$\frac {\nabla \mu_i}{\nabla \phi} = -z_i F \tag {8}$$
From equations (4) and (7), we can write: $$\frac {\nabla \mu_i}{\nabla \phi} = - \frac {u_i RT}{D_i} \tag {9}$$
By equating (8) and (9), we get the Einstein equation, which relates diffusion coefficient $$D_i$$ and ionic mobility $$u_i$$ in the thermodynamic equilibrium: $$D_i = \frac {u_i RT}{z_i F} \tag {10}$$
If we know substitute the equation (10) into equation (4), we can write: $$N_i = -u_i c_i(\frac {1}{z_i F} \nabla \mu_i + \nabla \phi) = - \frac {u_i c_i} {z_i F} \nabla \bar\mu_i\ \tag {11}$$
After the derivation of the first term in the equation (1), I can get to my point. Namely, that the first term in the equation must be equal to zero since it is impossible to write this term in that form (via gradient of the electrochemical potential) without using Einstein equation. We saw earlier that Einstein equation is only valid in the thermodynamic equilibrium and therefore equation (6) applies.
• I would check another book containing the same topic for the same derivation. Have you checked Bard and Faulkner?
– ACR
Commented Jan 9, 2023 at 3:39
• I didn't, but I will. Commented Jan 9, 2023 at 7:38
• @AChem Just checked Bard and Faulkner, Chapter 4.2: Migration. Basically, they explained nothing. They stated that the connection is given by the Einstein equation, but without any derivation. Commented Jan 10, 2023 at 2:10
• Dario, We will to check the original literature now. You don't stop or rest until you find a satisfying answer. This is the true essense of science. This requires going deeper and deeper in terms of literature search. Visit a library and check other electrochemistry textbooks. Also, ask your instructor.
– ACR
Commented Jan 10, 2023 at 2:46
• @AChem Yes, I'll definetly ask my supervisor. Internet provides some idea about the derivation from random walk which might give a good explanation. Commented Jan 10, 2023 at 4:30
Before writing, I understand your derivation, but I do not agree with Eq. (8). The gradient of a scalar field is a vector, so it is not correct to divide a vector quantity like $$\underline{\nabla}\mu_i$$ with $$\underline{\nabla}\phi$$.
Indeed the flux of species $$j$$, in a $$1$$-dimensional system, can be written as $$N_j = -D_j\dfrac{dC_j}{dx} - u_j C_j \dfrac{d\phi}{dx} \tag{1}$$
Here is a somewhat informal derivation of your Eq. (10). Imagine the charged species $$j$$ being subjected to two forces, the drag force and the electrical force, but both being balanced. By Newton's 2nd law we have \begin{align} F_D &= F_E \\ 6 \pi \mu r_j v_D &= z_j e E \\ \dfrac{v_D}{E} &= \dfrac{z_j e}{6 \pi \mu r_j} \\ u_j &= \dfrac{k_BT}{6 \pi \mu r_j}\dfrac{z_j e}{k_BT} \hspace{1 cm} \left(D_j = \dfrac{k_BT}{6 \pi \mu r_j} \hspace{0.5 cm} \text{(Einstein-Stokes law)}\right) \\ u_j &= D_j\dfrac{z_j e N_A}{N_Ak_B T} \\ u_j &= D_j\dfrac{z_j F}{RT} \tag{2} \\ \end{align} This is the Einstein-Smoluchowski equation. Combining Eqs. (1) and (2) we have $$N_j = -D_j\dfrac{dC_j}{dx} - \dfrac{F}{RT} z_jD_jC_j \dfrac{d\phi}{dx} \tag{3}$$ If you couple Eq. (3) with the convection part, i.e. $$C_j\mathbf{v}$$, the result is the famous Nernst-Planck equation.
Eq. (2) was derived by considering that the charged species $$j$$ is in an equilibrium between its forces. Although I didn't, you can derive Einstein-Stokes law by doing the same, but considering: (1) the balance between the chemical force $$d\mu_j/dx$$ and the drag force, and (2) comparing the derived expression with Fick's law. Thus, how come this relation, obtained by an equilibrium between forces, is combined with the expression to describe the motion of species in $$N_j$$?. The easiest answer is the following: so we can reduce the number of phenomenological and constant parameters, to describe the motion, from two to one.
Nernst-Planck equation is a good approximation to the molar flux, if the concentrations are sufficiently low. This means that you can neglect ion-ion interactions in the medium. This is the reason why you are able to change the activity by the concentration in the chemical part of $$\tilde{\mu}_j$$. It also assumes that the flux can be written as a linear combination of driving forces, and the phenomenological coefficients, are typically considered as constants. But indeed, coupling it with Eq. (2) goes much further. If you want to be more precise, Eq. (1) should be written $$N_j = -D_j^\infty\dfrac{dC_j}{dx} - u_j^\infty C_j \dfrac{d\phi}{dx} \tag{4}$$ So you need two phenomenological parameters to evaluate the flux. However, sometimes it is hard to obtain both coefficients, $$D_j^\infty$$ and $$u_j^\infty$$, so we go around with this by applying Eq. (2). I will give you an example of my own. I am modeling lithium-sulfur batteries, and there are publications where $$\ce{Li+}$$ (and other sulfur species) are as high as $$4-5$$ mol/dm$$^3$$, and other authors and myself still use Eq. (3).
• Thank you for the detailed answer. Yes, you are correct that it isn't very rigorous to write vectors in fractions although I think it leads to correct results nevertheless. It's similar to when scientists treat differentials algebraically, it's wrong mathematically, but it still delivers correct result. Commented Mar 21, 2023 at 19:14
• Problem with my question is that it merely proves that Einstein equation must be valid in thermodynamic equilibrium and not that it isn't valid outside of it (which I claim in the question). Basically the same derivation is carried on the wikipedia. The thing is that to prove general validity of Einstein equation, one needs to go beyond macroscopic thermodynamics and electrochemistry. Einstein derived it using the concept of random walk which is a mathematical idea quite different from laws of thermodynamics and electrochemistry. Commented Mar 21, 2023 at 19:21
• It is interesting that you model lithium-sulfur batteries. I research hydrogen fuel cell air electrodes (SOFC/SOEC). Commented Mar 21, 2023 at 19:26 | 2,327 | 8,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 47, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.916155 |
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# Qabalistic & Thelemic Proofs Section I
## The Order & Value
of the English Alphabet
Not long after my initial introduction to Liber Al vel Legis in 1993 e.v., I
came across the first volume of The Key of It All: Book One: The Eastern
Mysteries, by David Allen Hulse, in which was promised a second
volume, Book Two : The Western Mysteries. Not yet published, this book
was purported to contain a Magickal alphanumeric code for English. The
idea seized my imagination; for the concept of an English Cabala had
chapter two of Liber AL vel Legis:
"Thou shalt obtain the order & value of the English Alphabet;
thou shalt find new symbols to attribute them unto."
The Advent of Book One supplied me with some of the necessary tools to
explore the above command. Utilizing certain Yoga techniques, I began
to probe the Silent Pool of Knowledge for an answer to the riddle, which
cameupon me in a flash of revelation while meditating upon the 11th
Path of the Tree of Life, the Fool of the Tarot - the Secret Cipher instantly
and enigmatically appearing in my mind's eye as 1=A=0. AHA!
While the incredulous are welcome to their opinions on this matter, it
was at this moment that I "heard" a soft yet quite firm voice, identifiable
as neither male or female, which ordered me to get a pen and paper and
record what was about to be "transmitted" to me, the result being
presented in the form of the short document which I have entitled LIBER
A vel FOLLIS.
After applying the process of Gematria with this new Cipher, extended
serially to yield A=0 through Z=25, I was astounded by the numeric
correlations with various Qabalistic and Thelemic Words and Numbers.
Assuming that the code to be expounded in Book Two of The Key of It All
must be one and the same, I found myself waiting anxiously for its
publication.
Surprisingly, it contained a serial English value of A=1, the code already
used by modern Numerologists. Hulse cited numerous Qabalistic and
Thelemic correlations which are undeniably convincing, leading me to
hypothesize that the true value of the English Alphabet might in fact be
a binary code, A=1 and 0. The similarity between this binary code with
the dual value of the Fool of the Tarot and the language of our modern
computers, 1/0, not to mention the esoteric symbology of the circle and
the line, the yoni and the lingham, seemed far more than coincidental,
and I entertained the notion that perhaps I should write Mr. Hulse to see
what this thoughts might be.
As it turned out, I got my chance to show him the results of my work in
person a few short months after my first discoveries. I had only very
recently met (under strange circumstances which I won't go into here) a
new acquaintance. After a visit to my home he noted my all too apparent
fascination with Magick, and suggested that I might be interested to
meet a friend of his with a similar calling, David Hulse. Needless to say, I
was floored by this synchronicity. My friend soon arranged a meeting,
and so I prepared my notes and waited for the appointed time to arrive.
To make a long story short, David and I began exchanging our ideas; and
though he was initially skeptical of my code, he finally conceded the
radical validity of A=0 as a unique Magickal system. He would often cite
a meaningful numerological correlation in A=1, to which I would counter
with a reciprocal association in A=0, much to our mutual amusement
and amazement. David likened these codes to a fisherman's net, which
we were using to capture shimmering messages of Knowledge, and
suggested that my code had effectively tightened the skein of this net. I
likened the A=0 code to a wild card, the Joker, or Fool if you will, which
added an air of unpredictability; since A=0 words possess
disproportionately smaller numeric values than the A=1 code,
depending upon how many times the letter A appears in them.
David often referred to A=0 as The Shadow to A=1 (though
Qabalistically it might be seen the other way around), which led me to
label these two as the LVX and NOX Cabalas (Latin words meaning Light
and Darkness). We finally agreed that work with the NOX or A=0 code
had just begun, and so it was that the responsibility for expounding it fell
upon my shoulders. David's occupation with the LVX or A=1 code had
already resulted in a hitherto unpublished work alluded to in The Key of
It All, a copy of which was given to me in an act of unrestrained trust and
esteem. It being improper to prematurely divulge the contents of this
Magickal Work, we will therefore respectfully touch upon the A=1 code
only when absolutely necessary.
Ultimately, both are important and valid systems which mutually support
each other, and so they could be best rendered as 1=A=0 for my NOX
cipher and 0=A=1 for the LUX code, equations which reveal both the
order and the value of the English Alphabet for each. Interestingly
enough, 1=A=0 mirrors the Name of the Gnostic Godhead, IAO, while
0=A=1 mirrors its inverse formula OAI (vide Crowley's Liber Stellae
Rubae).
Here then is the order & value of the English Alphabet as revealed by the
Cipher 1=A=0.
## The Order & Value of the English Alphabet
Armed with this serial order & value, 1=A=0 through 26=Z=25 (with 1
signifying the order and 0 being the value of the first letter A), we may
now test it by obtaining the additive values of specific Qabalistic and
Thelemic Words, Names, and Numbers. If this is a valid order & value,
then the numerological correlations and metaphors should carry
immediate conviction, while retaining subtle and direct simplicity, all in
accordance with the conditions delineated earlier. For this reason, we will
only list the strongest of correlations here. Many more are given in LIBER
A vel FOLLIS and THE ENGLISH CABALA - 111, both of which contain an
extensive Word/Number dictionary similar to the Sepher Sephiroth in
Crowley's 777 And Other Qabalistic Writings .
It should be noted that numerologists have been assigning the English
Alphabet a serial value of A = 1 through Z = 26 for many years. This
system does have its merits and champions (see Michael Freedman's
Gematria: Magic/Magick and Fr.Nigris' English Gematria), and may be
seen to work surprisingly well in tandem with A = 0. This is suggestive of
a true binary code (1/0) for the English Alphabet. For example, modern
English Cabalists see a critical importance in there being 26 letters in the
English Alphabet because the Hebrew God Name YHVH (Yahweh or
Jehovah) adds to 26 in gematria. The strength of the LVX value seems to
be verified by the correlation since GOD = 26 in A = 1. An interesting
and meaningful interdependence between A = 1 and A = 0 appears
when the following equation is taken into account:
Z = 26 in A = 1
Z = 25 in A = 0
GOD = 26 in A = 1 (G 7 + O 15 + D 4 = 26)
MAN = 25 in A = 0 (M 12 + A 0 + N 13 + 25)
Thus, the final letter of completion in the English Alphabet, the letter Z,
would serve as a metaphor of both God and Man when comparing the
results of A = 1 and A = 0. It is important to note that 25 has a square
root of 5, the traditional esoteric number of Man, symbolized by the fivepointed star - the Pentagram.
We can take this idea even further as follows:
Z 26 - Z 25 = 1 (One)
Z 26 + Z 25 = 51
I AM ONE = 51 in A = 0.
In this way, our Sacred Cipher A = 0 goes a step further than the A = 1
code by demonstrating an essential occult Truth - God and Man are One.
Again, the greatest proponent of the A = 1 system to date is the author
of the two volume masterworks The Key of It All : Book One : The Eastern
Mysteries & Book Two : The Western Mysteries , David Allen Hulse. In his
Book Two, Hulse gives compelling Qabalistic and Thelemic correlations
for this code. The reader is encouraged to review this work, which, as
mentioned earlier, played a pivotal role in the discovery of our Sacred
Cipher. After all, it was Hulse who inspired our terms LVX and NOX for the
Grand Binary Code when he casually remarked that A = 0 appeared to
be the shadow to his code.
And so, the proper reading of 1 = A = 0 shows both the order and the
value for the English Alphabet, the order of A being first or ONE and the
value of A being NOTHING or ZERO, and it is upon this order & value that
the Magick of THE ENGLISH CABALA - 111 is formulated.
"The Perfect and the Perfect are one Perfect and not two; nay, are
none!"
Liber AL vel Legis - the threefold Book of Law I:12 :45
The student is also encouraged to compare our Sacred Cipher to other
systems, most notably Linda Falorio's English Qabalah, or the New Aeon
English Qabala of the Hermetic Order of QBLH and Jake Stratton-Kent's
English Qaballa, both of which are derived from Jim Lees' seminal work in
the 70's. While these elaborate systems certainly yield interesting and
convincing permutations which undoubtedly work for those who have
much time and effort invested in them, it is our belief that none can
approach the crystalline clarity of our Cipher 1=A=0. The originators of
these codes have not labored in vain, however; for these systems
certainly do work for their creators. Ultimately, they all demonstrate the
unequivocal power of our Magickal Art of Gematria through contrast and
example.
The following web pages represent a brief rundown of the most
compelling Qabalistic and Thelemic proofs of our order & value for the
English Alphabet 1 = A = 0.
This ends Section I of our Qabalistic and Thelemic Proofs. Clickable Links
to Sections II, III, IV, V, VI and VII are as follows:
II. The Flaming Sword
III. Numbers & Words - Introduction
IV. Numbers & Words - Correlations
V. The Grand Puzzle
VI. The Three Keys
VII. New Symbols | 2,446 | 9,660 | {"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} | 2.875 | 3 | CC-MAIN-2019-30 | latest | en | 0.939373 |
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Algorithms
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The Algorithmic Machine That means I know how to follow a sequence of steps.
Control Structures Sequence Selection (or decision) Repetition
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Sequence Control Structure 1 - 9 1 - 9 1 - 9 1 - 9 STOP 1 - 9 Move Forward Move Backward Turn Right Turn Left Stop/Pause
Selection Control Structure ENDIF YES IF NO ??? ???
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Repetition Control Structure NEXT ??? ??? FOR #
A First Program START 5 4 STOP 1 END STOP 1
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A Square Program START 5 4 STOP 1 STOP 1 5 4 STOP 1 STOP 1 5 4 STOP 1 STOP 1 END 5 4 STOP 1 STOP 1
A Better Square Program START 5 4 STOP 1 STOP 1 END FOR 4 NEXT
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Are 5 squares 5x as long? START END 5 4 STOP 1 STOP 1 FOR 4 NEXT FOR 5 NEXT 5x
Autonomous Operation
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“Infinite” Repetition NEXT FOR 1
Selection Control Structure YES IF SL ON NO Do ??? if the left sensor is ON Do ??? if the left sensor is OFF
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Final Program: The Wanderer START 1 END IF FOR ∞ IF SL OFF IF SR OFF END IF YES YES NO NO YES IF SL ON 3 2 END IF NO NEXT 3 2 END IF IF SR ON YES NO
From Flowchart to Code BEGIN
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https://math.stackexchange.com/questions/tagged/open-map?sort=frequent&pageSize=50 | 1,555,643,661,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526966.26/warc/CC-MAIN-20190419021416-20190419042441-00044.warc.gz | 475,729,429 | 23,059 | # Questions tagged [open-map]
In topology, an open map is a function between two topological spaces which maps open sets to open sets. That is, a function f : X → Y is open if for any open set U in X, the image f(U) is open in Y. A map may be open, closed, both, or neither; in particular, an open map need not be closed and vice versa.
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### Open maps which are not continuous
What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of ...
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### continuous image of a locally compact space is locally compact
Is continuous image of a locally compact space is locally compact? Let $X$ be locally compact(l.c.).Let $f:X\to Y$ is continuous and surjective. A space $X$ is locally compact if for each $x\in X$ ...
Quotient map from $X$ to $Y$ is continuous and surjective with a property : $f^{-1}(U)$ is open in $X$ iff $U$ is open in $Y$. But when it is open map? What condition need? | 280 | 1,058 | {"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} | 2.625 | 3 | CC-MAIN-2019-18 | latest | en | 0.922535 |
https://republicofsouthossetia.org/question/equation-of-a-circle-mastery-test-2-the-equation-y2-2-2y-1-0-is-the-general-form-of-the-equation-16309686-19/ | 1,638,255,820,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00288.warc.gz | 578,332,759 | 13,358 | ## Equation of a Circle: Mastery Test 2 The equation x² + y2 – 2x + 2y – 1 = 0 is the general form of the equation of a circl
Question
Equation of a Circle: Mastery Test
2
The equation x² + y2 – 2x + 2y – 1 = 0 is the general form of
the equation of a circle. What is the standard form of the
equation?
(x – 1)2 – (y – 1)2 = 1
(x – 1)2 + (y + 1)2 = 3
© (x + 1)2 + (y – 1)2 = 4
(x – 1)² + (y + 1)2 = 1
in progress 0
2 months 2021-09-23T22:58:11+00:00 1 Answer 0 views 0
(x-1)^2+(y+1)^2=3
Step-by-step explanation:
x² + y2 – 2x + 2y – 1 = 0
add the 1 to get it to the other side of the equation
x² + y2 – 2x + 2y = +1
group the x’s and y’s
(x² -2x) + (y2+2y) = +1
then you’ll complete the square on the -2x and + 2y. that just means divide by two and then raise it to the 2nd power.
so (-2/2)^2 and (+2/2)^2
(x²-2x+1)+(y2+2y+1) = 1+1+1
you add the one’s to the other side because whatever is done to one side must be done to the other
you’ll then need to factor again.
(x-1)^2+(y+1)^2=3
to factor it take one of your squared x’s, the sign of the middle term within the parentheses , then the square root of the last term within the parentheses. remeber to put your ^2 (raised to the 2nd power) outside of the parentheses when you finish. | 487 | 1,255 | {"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} | 4.40625 | 4 | CC-MAIN-2021-49 | latest | en | 0.869162 |
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# econ1010a_2011s_may_27_ans.doc
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YORK UNIVERSITY Department of Economics ECON1010A: Principles of Macroeconomics Midterm #2: May 27, 2011 Duration - 1 hour Aids Allowed Non-programmable calculators only Time Allowed: 60 Minutes The total marks for this test are 50. The test is divided into two parts: Part I - Problem format - is worth 37 marks (37 of the total mark of 50) Part II – Explanations (8 marks) and Multiple choice (5 marks) (13 of the total mark of 50) Show your work where applicable. Please use pen instead of pencil. Print your name and student number clearlyon the front of the exam and on any loose pages. Name: . (FamilyName) (Given Name) Student #: . There are 6 pages to the exam. - 1/6 - ECON1010A: Midterm Test #2: May 27, 2011 Name: _______________________ Part I: Place your answers (and work where necessary) in the space provided. Clearlylabel all axes, curves, and points. 1. MacroModel (12 Marks) An economy has the following set of macroeconomic equations. Consumption: C = 230 + 0.84Yd Net Taxes T = 0.25Y Investment: I = 50 + 0.2Y Exports X = 570 Government Spending: G = 520 Imports IM = 20 + 0.08Y a) What is the Consumption equation as a function of Y (not Yd)? (1 mark) 1 mark: C = 230 + 0.63Y from something like 230 + 0.84(Y – 0.25Y) b) What is the Aggregate Expenditure equation? (2 marks) 1 mark: correct Autonomous part = 1,350 from AE as sum of above C + I + G + X - IM 1 mark: correct Marginal part = 0.75 from AE = 1,350 + 0.75Y c) Calculate the value of Autonomous Spending Multiplier. (1 mark) 1 mark: = 4 from something like 1/(1 – 0.75) d) Calculate the value of Equilibrium Income. (1 mark) 1 mark: Y = 5,400 from AE = Y or 1,350/(1 – 0.75) e)Calculate Consumption at equilibrium. (1 mark) 1 mark: C =3,632 from 230 + 0.63*5,400 or 180 + 0.84*Yd (they’d have to calculate, see below) f) Calculate Net Exports at Equilibrium. (1 mark) 1 mark: X – IM = 118 from 570 – (20 + 0.08*5,400), must be positive g) Calculate the government budget surplus (+)/deficit (-) at equilibrium? 1 mark: = +830 from something like 0.25*5,400 – 520 h)Prove that your equilibrium income is correct. (1 mark) 1 mark: Show C + I + G + X – IM = Y, i.e. 3,632 + 1,130 + 520 + 118 = 5,400 i) Draw a diagram showing equilibrium Y given the AE function. Be sure to label numerically Autonomous Expenditure, the slope of AE, and equilibrium Y. (3 marks) 1 mark: AE as a straight line intersecting the vertical axis above 0. Must designate AE and Y (or better, real AE and real Y) as the axes 1 mark: Equilibrium Y at intersection of AE and 45 (equality) line 1 mark: AE intercept = 1,350, and equilibrium Y = 5,400 - 2/6 - ECON1010A: Midterm Test #2: May 27, 2011 Name: _______________________ 2. AE/Y and Fiscal Policy (12 marks) The following are a country’s equations for Consumption, Taxes, and Aggregate Expenditure (C, I, G, X and IM are included in the AE function). C = 350 + 0.8Yd T = -30 + 0.2Y AE = 1,600 + 0.6Y a) What is equilibriumGDP? (1 mark) 1 mark Y = 4,000 from Y = 1,600 + 0.6Y or 1,600/(1 – 0.6) b) What is Disposable Income at equilibrium GDP? (1 mark) 1 mark = 3,230 from 4,000 – (-30 + 0.2*4,000) or something for Y - T c) What are Personal Savings at equilibriumGDP? (1 mark) 1 mark: 296 from 3,230 – (350 + 0.8*3,230) or some numbers indicating Yd - C d) What is the change in Y caused by an increase in Exports of 12 (+12)? (1 mark) 1 mark: = +30 from +12/(1 – 0.6) or 12*2.5 (must be positive) e) What is the change in inventories at Y = 3,600? (2 marks) 1 mark: calculate AE = 3,760 from 1,600 + 0.6*3,600 1 mark: Change Inventories = 3,600 – 3,760 = -160 from Yo – Aeo (must be negative) f) What is the amount of the fixed tax multiplier? (1 mark) 1 mark: = -2 from –0.8*2.5 or –0.8/(1 – 0.6) (must be negative) g) What is the change in Government Spending necessary to increase equilibriumGDP by 30 ceteris paribus? (1 mark) 1 mark: = +12 from 30/(1/(1-0.6)) or 30/2.5 h) What is the change in Taxes given an increase in equilibriumGDP of 30? (1 mark) 1 mark: +6 from 0.2*Change Y (or they could do this as difference between two Y’s) i) What is the change in the Government budget (surplus (+)/deficit (-) given an increase in Government Spending = +30? (1 mark) 1 mark: = -6 from 0.2 (30) – 12 (must be negative) j)What is the change in equilibriumGDP given an increase in Fixed taxes of 15 (+15) ceteris paribus? (1 mark) 1 mark: =-30 from –2* (+15) or –0.8*(15)*2.5(must be negative) k) What is the change in equilibriumGDP given an increase in Fixed Transfer Payments of 15 (+15) ceteris paribus? (1 mark) 1 mark: =+30 from –2* (-15) or –0.8*(-15)*2.5(must be negative) - 3/6 - ECON1010A: Midterm Test #2: May 27, 2011 Name: _______________________ 3. Money Supply (13 marks) The following conditions hold in Canada’s banking system for all parts of this question: deposits are demand deposits; banks are chartered banks; there are neither excess reserves nor excess circulation; and banks loan, rather than buy securities with, excess reserves. A) Assume that the only items on the
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Subtraction
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Animal Math
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Animal Math
Searching for a worksheet that practices addition and subtraction? This Animal Math printable is perfect for beginning math stars!
Math
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Subtraction Color by Number
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Subtraction Color by Number
Someone help this elephant! He's lost all his color, and he doesn't know how to get it back. Use subtraction to come to the elephant's rescue!
Math
Worksheet
Subtract & Color
Worksheet
Subtract & Color
This cute math worksheet's fun subtraction practice for your young learner who'll get to color-in the correct answers.
Math
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Subtraction Facts: Cupcake Subtraction
Worksheet
Subtraction Facts: Cupcake Subtraction
Want to soothe those subtraction woes? Cupcakes to the rescue! This neat, sweet worksheet will guide your kid through subtraction facts.
Math
Worksheet
Worksheet
Math
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Complete the Inverse Operation
Worksheet
Complete the Inverse Operation
Your child will develop the skills necessary to recognize patterns by completing inverse operations as well make a connection between addition and subtraction.
Math
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Race to 120
Worksheet
Race to 120
Build number sense and familiarity with numbers 1–120 with this fun game! Students will roll a die and move that many more spaces. The first player to reach 120 wins!
Math
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Hanukkah Number Line Problems
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Hanukkah Number Line Problems
Take these Hanukkah themed word problems for a spin!
Math
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Fact Family Practice: Icy Fact Families
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Fact Family Practice: Icy Fact Families
In this fact family practice worksheet, kids show that they understand the inverse relationship between addition and subtraction.
Math
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Solve the Story Problem
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Solve the Story Problem
Practice solving different types of subtraction word problems with this worksheet. Plenty of space is included for students to show their thinking!
Math
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At the Grocery Store: Addition and Subtraction
Worksheet
At the Grocery Store: Addition and Subtraction
Math
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Subtraction Fact Families
Worksheet
Subtraction Fact Families
This solve-and-check worksheet encourages kids to solve subtraction problems, and then check their answers with addition.
Math
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Math Maze: Single-Digit Addition and Subtraction
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Math Maze: Single-Digit Addition and Subtraction
Math
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Thanksgiving Subtraction #1
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Thanksgiving Subtraction #1
Children practice subtraction within 20 in this fall-themed math worksheet.
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Subtracting with Stars
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Subtracting with Stars
Need to add some sparkle to your child's subtraction practice? Let the stars guide your math whiz as he tackles 12 problems amidst the stars.
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Math Story Problems
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Math Story Problems
Build great math skills with these story problems. Use the picture to help you visualize each problem.
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Hanukkah Number Line Subtraction
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Hanukkah Number Line Subtraction
Practice using the number line to count and subtract numbers up to nine in this Hanukkah themed worksheet.
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Even in nature, solving story problems is important. How many apples will be left on a tree if seven are picked?
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Make it Fair With Equal Groups
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Make it Fair With Equal Groups
Kids love it when things are fair! Help students make equal groups of toys while exploring how the equals sign means “the same as.” Students will draw the missing objects and write the missing factors to balance the equations.
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Math Jam #4
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Math Jam #4
Build your child's confidence in knowing math facts, so that he gets nothing but net. Take a shot at addition and subtraction problems!
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Math
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Underwater Math
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Underwater Math
Encourage your first grader to reach down deep and take the plunge into math practice.
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Balance Each Side
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Balance Each Side
This worksheet helps students understand how an equals sign means “the same as.” By filling in missing factors, students gain practice seeing how multiple number combinations can make the same number. | 888 | 4,262 | {"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} | 2.828125 | 3 | CC-MAIN-2021-04 | latest | en | 0.858115 |
https://cs.stackexchange.com/questions/10538/bit-what-is-the-intuition-behind-a-binary-indexed-tree-and-how-was-it-thought-a/95954 | 1,657,104,079,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00756.warc.gz | 230,520,346 | 66,773 | # BIT: What is the intuition behind a binary indexed tree and how was it thought about?
A binary indexed tree has very less or relatively no literature as compared to other data structures. The only place where it is taught is the topcoder tutorial. Although the tutorial is complete in all the explanations, I cannot understand the intuition behind such a tree? How was it invented? What is the actual proof of its correctness?
• An article on Wikipedia claims that these are called Fenwick trees. Mar 16, 2013 at 19:32
• @DavidHarkness- Peter Fenwick invented the data structure, so they're sometimes called Fenwick trees. In his original paper (found at citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.14.8917), he referred to them as binary indexed trees. The two terms are often used interchangeably. Mar 18, 2013 at 15:56
• The following answer conveys a very nice "visual" intuition of binary indexed trees cs.stackexchange.com/questions/42811/…. May 20, 2015 at 20:40
• I know how you feel, the first time I read the topcoder article, it just seemed like magic. May 3, 2017 at 11:46
Intuitively, you can think of a binary indexed tree as a compressed representation of a binary tree that is itself an optimization of a standard array representation. This answer goes into one possible derivation.
Let's suppose, for example, that you want to store cumulative frequencies for a total of 7 different elements. You could start off by writing out seven buckets into which the numbers will be distributed:
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
1 2 3 4 5 6 7
Now, let's suppose that the cumulative frequencies look something like this:
[ 5 ] [ 6 ] [14 ] [25 ] [77 ] [105] [105]
1 2 3 4 5 6 7
Using this version of the array, you can increment the cumulative frequency of any element by increasing the value of the number stored at that spot, then incrementing the frequencies of everything that come afterwards. For example, to increase the cumulative frequency of 3 by 7, we could add 7 to each element in the array at or after position 3, as shown here:
[ 5 ] [ 6 ] [21 ] [32 ] [84 ] [112] [112]
1 2 3 4 5 6 7
The problem with this is that it takes O(n) time to do this, which is pretty slow if n is large.
One way that we can think about improving this operation would be to change what we store in the buckets. Rather than storing the cumulative frequency up to the given point, you can instead think of just storing the amount that the current frequency has increased relative to the previous bucket. For example, in our case, we would rewrite the above buckets as follows:
Before:
[ 5 ] [ 6 ] [21 ] [32 ] [84 ] [112] [112]
1 2 3 4 5 6 7
After:
[ +5] [ +1] [+15] [+11] [+52] [+28] [ +0]
1 2 3 4 5 6 7
Now, we can increment the frequency within a bucket in time O(1) by just adding the appropriate amount to that bucket. However, the total cost of doing a lookup now becomes O(n), since we have to recompute the total in the bucket by summing up the values in all smaller buckets.
The first major insight we need to get from here to a binary indexed tree is the following: rather than continuously recomputing the sum of the array elements that precede a particular element, what if we were to precompute the total sum of all the elements before specific points in the sequence? If we could do that, then we could figure out the cumulative sum at a point by just summing up the right combination of these precomputed sums.
One way to do this is to change the representation from being an array of buckets to being a binary tree of nodes. Each node will be annotated with a value that represents the cumulative sum of all the nodes to the left of that given node. For example, suppose we construct the following binary tree from these nodes:
4
/ \
2 6
/ \ / \
1 3 5 7
Now, we can augment each node by storing the cumulative sum of all the values including that node and its left subtree. For example, given our values, we would store the following:
Before:
[ +5] [ +1] [+15] [+11] [+52] [+28] [ +0]
1 2 3 4 5 6 7
After:
4
[+32]
/ \
2 6
[ +6] [+80]
/ \ / \
1 3 5 7
[ +5] [+15] [+52] [ +0]
Given this tree structure, it's easy to determine the cumulative sum up to a point. The idea is the following: we maintain a counter, initially 0, then do a normal binary search up until we find the node in question. As we do so, we also the following: any time that we move right, we also add in the current value to the counter.
For example, suppose we want to look up the sum for 3. To do so, we do the following:
• Start at the root (4). Counter is 0.
• Go left to node (2). Counter is 0.
• Go right to node (3). Counter is 0 + 6 = 6.
• Find node (3). Counter is 6 + 15 = 21.
You could imagine also running this process in reverse: starting at a given node, initialize the counter to that node's value, then walk up the tree to the root. Any time you follow a right child link upward, add in the value at the node you arrive at. For example, to find the frequency for 3, we could do the following:
• Start at node (3). Counter is 15.
• Go upward to node (2). Counter is 15 + 6 = 21.
• Go upward to node (4). Counter is 21.
To increment the frequency of a node (and, implicitly, the frequencies of all nodes that come after it), we need to update the set of nodes in the tree that include that node in its left subtree. To do this, we do the following: increment the frequency for that node, then start walking up to the root of the tree. Any time you follow a link that takes you up as a left child, increment the frequency of the node you encounter by adding in the current value.
For example, to increment the frequency of node 1 by five, we would do the following:
4
[+32]
/ \
2 6
[ +6] [+80]
/ \ / \
> 1 3 5 7
[ +5] [+15] [+52] [ +0]
Starting at node 1, increment its frequency by 5 to get
4
[+32]
/ \
2 6
[ +6] [+80]
/ \ / \
> 1 3 5 7
[+10] [+15] [+52] [ +0]
Now, go to its parent:
4
[+32]
/ \
> 2 6
[ +6] [+80]
/ \ / \
1 3 5 7
[+10] [+15] [+52] [ +0]
We followed a left child link upward, so we increment this node's frequency as well:
4
[+32]
/ \
> 2 6
[+11] [+80]
/ \ / \
1 3 5 7
[+10] [+15] [+52] [ +0]
We now go to its parent:
> 4
[+32]
/ \
2 6
[+11] [+80]
/ \ / \
1 3 5 7
[+10] [+15] [+52] [ +0]
That was a left child link, so we increment this node as well:
4
[+37]
/ \
2 6
[+11] [+80]
/ \ / \
1 3 5 7
[+10] [+15] [+52] [ +0]
And now we're done!
The final step is to convert from this to a binary indexed tree, and this is where we get to do some fun things with binary numbers. Let's rewrite each bucket index in this tree in binary:
100
[+37]
/ \
010 110
[+11] [+80]
/ \ / \
001 011 101 111
[+10] [+15] [+52] [ +0]
Here, we can make a very, very cool observation. Take any of these binary numbers and find the very last 1 that was set in the number, then drop that bit off, along with all the bits that come after it. You are now left with the following:
(empty)
[+37]
/ \
0 1
[+11] [+80]
/ \ / \
00 01 10 11
[+10] [+15] [+52] [ +0]
Here is a really, really cool observation: if you treat 0 to mean "left" and 1 to mean "right," the remaining bits on each number spell out exactly how to start at the root and then walk down to that number. For example, node 5 has binary pattern 101. The last 1 is the final bit, so we drop that to get 10. Indeed, if you start at the root, go right (1), then go left (0), you end up at node 5!
The reason that this is significant is that our lookup and update operations depend on the access path from the node back up to the root and whether we're following left or right child links. For example, during a lookup, we just care about the right links we follow. During an update, we just care about the left links we follow. This binary indexed tree does all of this super efficiently by just using the bits in the index.
The key trick is the following property of this perfect binary tree:
Given node n, the next node on the access path back up to the root in which we go right is given by taking the binary representation of n and removing the last 1.
For example, take a look at the access path for node 7, which is 111. The nodes on the access path to the root that we take that involve following a right pointer upward is
• Node 7: 111
• Node 6: 110
• Node 4: 100
All of these are right links. If we take the access path for node 3, which is 011, and look at the nodes where we go right, we get
• Node 3: 011
• Node 2: 010
• (Node 4: 100, which follows a left link)
This means that we can very, very efficiently compute the cumulative sum up to a node as follows:
• Write out node n in binary.
• Set the counter to 0.
• Repeat the following while n ≠ 0:
• Add in the value at node n.
• Clear the rightmost 1 bit from n.
Similarly, let's think about how we would do an update step. To do this, we would want to follow the access path back up to the root, updating all nodes where we followed a left link upward. We can do this by essentially doing the above algorithm, but switching all 1's to 0's and 0's to 1's.
The final step in the binary indexed tree is to note that because of this bitwise trickery, we don't even need to have the tree stored explicitly anymore. We can just store all the nodes in an array of length n, then use the bitwise twiddling techniques to navigate the tree implicitly. In fact, that's exactly what the bitwise indexed tree does - it stores the nodes in an array, then uses these bitwise tricks to efficiently simulate walking upward in this tree.
Hope this helps!
• Feb 8, 2015 at 19:42
• This is by far the best explanation I've read on the topic so far, among all the sources I've found on the Internet. Well done ! Apr 14, 2015 at 19:01
• How did Fenwick get this smart? May 3, 2017 at 14:31
• This is a very great explanation but suffers from the same problem as every other explanation as well as Fenwick's own paper, doesn't provide a proof! Jan 16, 2018 at 8:05
• Could you describe what does "to the left of that given node" mean precisely? Aug 4, 2018 at 22:07
I think that the original paper by Fenwick is much clearer. The answer above by @templatetypedef requires some "very cool observations" about the indexing of a perfect binary tree, which are confusing and magical to me.
Fenwick simply said that the responsibility range of every node in the interrogation tree would be according to its last set bit:
E.g. as the last set bit of 6==00110 is a "2-bit" it will be responsible for a range of 2 nodes. For 12==01100, it is a "4-bit", so it will be responsible for a range of 4 nodes.
So when querying F(12)==F(01100), we strip the bits one-by-one, getting F(9:12) + F(1:8). This is not nearly a rigorous proof, but I think that's it's more obvious when put so simply on the numbers axis and not on a perfect binary tree, what are the responsibilities of each node, and why is the query cost equals to the number of set bits.
If this is still unclear the paper is very recommended.
• i think you are both right. but intuition is a matter beyond correctness and a right solution solution can also be extremely "wrong" in intuition, so to speak. What I would like to add on is about the last set bits, aka the least significant bit. When I read the paper, Fenwick said instead the branching ratio depends on the number of trailing zeros. That's as far as I can see in the paper. What I can infer is that not only the branching ratio determines the number of attached nodes, but also the amount of partial sum. say 8, (1000), it can hold 8 nodes, including itself, hence 8 partial sum. Mar 7 at 23:11 | 3,307 | 12,248 | {"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} | 3.71875 | 4 | CC-MAIN-2022-27 | longest | en | 0.944144 |
https://www.coursehero.com/file/6776608/alignment/ | 1,529,508,974,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00144.warc.gz | 784,656,720 | 589,764 | alignment
alignment - Pairwise Sequence Alignment using Dynamic...
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Page 1 Pairwise Sequence Alignment using Dynamic Programmin g What is sequence alignment? Given two sequences of letters, and a scoring scheme for evaluating matching letters, find the optimal pairing of letters from one sequence to letters of the other sequence. Align: THIS IS A RATHER LONGER SENTENCE THAN THE NEXT. THIS IS A SHORT SENTENCE. THIS IS A RATHER LONGER SENTENCE THAN THE NEXT. THIS IS A ######SHORT## SENTENCE##############. OR THIS IS A SHORT#########SENTENCE##############.
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Page 2 Aligning biological sequences DNA (4 letter alphabet + gap) TTGACAC TTTACAC Proteins (20 letter alphabet + gap) RKVA--GMAKPNM RKIAVAAASKPAV Statement of Problem Given 2 sequences scoring system for evaluating match(or mismatch) of two characters penalty function for gaps in sequences Produce Optimal pairing of sequences that retains the order of characters in each sequence, perhaps introducing gaps, such that the total score is optimal.
Page 3 Why align sequences? Lots of sequences with unknown structure and function. A few sequences with known structure and function. If they align, they are similar, maybe due to common descent. If they are similar, then they might have same structure or function. If one of them has known structure/function, then alignment to the other yields insight about how the structure or function works. Multiple alignment Pairwise alignment (two at a time) is much easier than multiple alignment (N at a time). This is a rather longer sentence than the next. This is a short sentence. This is the next sentence. Rather long is the next concept. Rather longer than what is the next concept.
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Page 4 Drawing alignments Exact Matches OK, Inexact Costly, Gaps cheap. This is a rather longer sentence than the next. This is a ############# sentence##############. Exact Matches OK, Inexact Costly, Gaps cheap. This is a *rather longer*sentence than the next. This is a s###h####o###rtsentence##############. Exact Matches OK, Inexact Moderate, Gaps cheap. This is a rather longer sentence than ########the next#########. This is a ##short###### sentence###############################. Exact Matches cheap, Inexact cheap, Gaps expensive. This is a rather longer sentence than the next. This is a short sentence.###################### Multiple Alignment (NP-hard) This is a rather longer sentence than ########the next#########. This is a short######## sentence####################3##########. This is ######################################the next sentence. ##########Rather long is ########the############# next concept#. ##########Rather longer #########than what is the next concept#.
Page 5 There used to be dot matrices. Put one sequence along the top row of a matrix. Put the other sequence along the left column of the matrix. • Plot a dot everytime there is a match between an element of row sequence and an element of the column sequence. • Diagonal lines indicate areas of match.
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Page 6 Problems with dot matrices Rely on visual analysis Difficult to find optimal alignments Need scoring schemes more sophisticated that “identical match” Difficult to estimate significance of alignments Gaps The thing that makes alignment hard is the possibility that gaps are introduced in one sequence (corresponding to a shortening of the protein chain, for example).
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https://www.edplace.com/worksheet_info/maths/keystage4/year10/topic/31/5677/understand-place-value | 1,579,563,883,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00105.warc.gz | 840,394,927 | 27,241 | # Understand Place Value
In this worksheet, students will calculate the number value represented by specific digits, using knowledge of place value in both larger whole numbers and decimals.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: Pearson Edexcel, OCR, Eduqas, AQA
Curriculum topic: Number, Number Operations and Integers
Curriculum subtopic: Structure and Calculation, Whole Number Theory
Difficulty level:
### QUESTION 1 of 10
When using place value, it is useful to use a table that looks something like this:
Thousands Hundreds Tens Units . Tenths Hundredths Thousandths x 1000 x 100 x 10 x 1 . ÷ 10 ÷ 100 ÷ 1000 8 7 6 5 . 4 3 2
If we want to know what magnitude each of the numbers represents, we need to look at which column it lies in.
e.g. The number 7 in the table is in the 'Hundreds' column.
This means we need to multiply by 100 to find its true value.
7 x 100 = 700
This means that the digit 7 in this table represents the number 700 in reality.
e.g. The number 2 in the table is in the 'Thousandths' column.
This means we need to divide by 1000 to find its true value.
2 ÷ 1000 = 0.002
This means that the digit 2 represents 0.002.
In this activity, you will calculate the number value represented by specific digits using your knowledge of place value in both larger whole numbers and decimals.
Let's get going!
In the number 347.218, what does the digit 4 represent?
In the number 347.218, what does the digit 3 represent?
3
30
300
In the number 347.218, which place value column is the 7 in?
Hundreds
Tens
Units
Tenths
In the number 347.218, what does the digit 2 represent?
2
0.2
20
In the number 347.218, which column is the 8 in?
Tenths
Hundredths
Thousandths
Consider the number 281.326.
Then complete the sentence below.
Tenths
Hundredths
Thousandths
In the number 281.326 what does the digit 8 represent?
In the number 381.326, what does the digit 2 represent?
2
0.02
0.0002
Look at the number 281.336.
Now complete the sentence below.
2
0.02
0.0002
For the number 281.356, match each digit with its correct place value column.
## Column B
2
Tenths
8
Hundredths
1
Thousandths
3
Units
5
Hundreds
6
Tens
• Question 1
In the number 347.218, what does the digit 4 represent?
40
EDDIE SAYS
4 is in the tens column. 4 x 10 = 40 The common mistake here is to just think '4'. Remember, place value is important here. It would only be 4 if it was placed in the units or ones column, so where the 7 is in the question number. Let's try another...
• Question 2
In the number 347.218, what does the digit 3 represent?
300
EDDIE SAYS
How did you do here? 3 is in the hundreds column: 3 x 100 Did you get that one?
• Question 3
In the number 347.218, which place value column is the 7 in?
Units
EDDIE SAYS
Hopefully, you're getting the hang of this with each question! The 7 is the first digit to the left of the decimal places. This means it in the units column. Did you know that the 'units' column can be referred to as the 'ones' column too?
• Question 4
In the number 347.218, what does the digit 2 represent?
0.2
EDDIE SAYS
The 2 is the first number to the right of the decimal place, this means it's in the tenths column. To get the answer, you have to think - what is 2 tenths as a decimal? The calculation to complete it: 2 ÷ 10 = 0.2
• Question 5
In the number 347.218, which column is the 8 in?
Thousandths
EDDIE SAYS
If we look at the columns to the right of the decimal point, they're called tenths, hundredths and thousandths. 8 is in the third column after the decimal, so it's in the thousandths column. Refer back to the table in the intro to check this if you are at all unsure or would like to see this represented visually.
• Question 6
Consider the number 281.326.
Then complete the sentence below.
EDDIE SAYS
1 is in the units column: 1 x 1 Good effort!
• Question 7
In the number 281.326 what does the digit 8 represent?
80
EDDIE SAYS
8 is in the tens column, this means there are 8 lots of 10: 8 x 10 How did you find that one?
• Question 8
In the number 381.326, what does the digit 2 represent?
0.02
EDDIE SAYS
2 is in the hundredths column (the second to the right after the decimal). If it's to the right, don't forget we need to divide: 2 ÷ 100 = 0.02
• Question 9
Look at the number 281.336.
Now complete the sentence below.
EDDIE SAYS
Are these challenges getting easier? 2 is in the hundreds column, this means we have 2 lots of 100: 2 x 100
• Question 10
For the number 281.356, match each digit with its correct place value column.
## Column B
2
Hundreds
8
Tens
1
Units
3
Tenths
5
Hundredths
6
Thousandths
EDDIE SAYS
Remember the difference between hundreds and hundredths? Starting at the left, the order is: Hundreds, tens, units, (decimal point), tenths, hundredths, thousandths. Refer back to the intro for a final time to check this out. Do you feel that this place value table is clear in your head now? Well done, that’s another activity completed!
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Get started | 1,546 | 5,502 | {"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} | 4.625 | 5 | CC-MAIN-2020-05 | longest | en | 0.794456 |
https://support.ilovegrowingmarijuana.com/t/what-wattage-hps-for-a-3-5-by-3-5-room/9309 | 1,601,422,934,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00760.warc.gz | 637,409,887 | 7,010 | # What wattage (HPS) for a 3.5 by 3.5 room?
one all ways like to upgrade - considering a 400 watt would a 150 be to small ?
1 Like
Okay this is how i understand it 50 watts per square foot for mh/hps your just over 12 feet if my math is right now heat will be and issue so hope you have thought about air cooled hood of some kind …
3x3 400 watt
4x4 600 watt
5x5 1000 wattts
But lets ask the big guys
@Majiktoker @latewood
Hope i got this right lol
2 Likes
I have a 4ftx4ft tent and I have a 600w in it
2 Likes
I have my old setup 3x3 400 watt mh/hps but went to LEDs
1 Like
No it would be 3x3=9 then 9x50=450 watts
2 Likes
3.5x3.5 12.25
I just went to 600 watt more light better then less but your right
Plus more heat my way lol will need to be air cooled haha
1 Like
I’m thinkin about just addin an LED to mine, like a 600w one
2 Likes
That’s 3.5ftx3.5ft=12.25sqft (grow space)
12.25sqftx50w(per sqft)=612.5 total watts
It’s easy math Lol
2 Likes
Ive got two 600 equivalent 275 each true watts
And one 300 equivalent 135 true watts
In my 2x4x7 tent and my setup works great just to small
3 Likes
Perfect you have 685 true watts ️
Hats more than what you need for that space 2x4=8
8x50=400 your in the clear I think buddy Good on lighting, now Im on to learn and master heating and ventilation systems
2 Likes
1 Like
Id put 1000 watts in there if i could keep it cool but im crazy like that push it to the max
1 Like
Do you have heat issues with the 600 watter ?
Thanks ! very help-full chart
1 Like
Well I would if I didn’t have a big a\$\$ fan blowin into the tent
My Temps are hanging around 77-80 an that’s with only 75% power, I haven’t tried full power of the 600w. I’m waitin till I can get my fans and filter before I crank it all the way
1 Like
Yea - one all ways needs the extra room… Be nice to have two rooms one for veg and one for flower
Thanks to all the answers !! you guy’s are a great help - maybe 400 with no heat issues or 600 with a bit of extra heat - (fans needed - could use 2) but running cost would not be that great
1 Like
To be honest your going to need fans with 400watt or the 600watt but id go big because youll end up wanting more trust me im about
To by my third tent because ive out grown the others lol
1 Like
You are right in thinking about it !! Thanks !!! Now to find a bulb and reflector. Need to figure the “cfm’s” for fans. I forget the formula for cfm’s
1 Like | 759 | 2,471 | {"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-2020-40 | latest | en | 0.900046 |
https://www.kodytools.com/units/tension/from/gfpmicrom/to/mnpft | 1,716,120,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00733.warc.gz | 760,473,493 | 17,278 | # Gram Force/Micrometer to Millinewton/Foot Converter
1 Gram Force/Micrometer = 2989066.92 Millinewton/Foot
## One Gram Force/Micrometer is Equal to How Many Millinewton/Foot?
The answer is one Gram Force/Micrometer is equal to 2989066.92 Millinewton/Foot and that means we can also write it as 1 Gram Force/Micrometer = 2989066.92 Millinewton/Foot. Feel free to use our online unit conversion calculator to convert the unit from Gram Force/Micrometer to Millinewton/Foot. Just simply enter value 1 in Gram Force/Micrometer and see the result in Millinewton/Foot.
Manually converting Gram Force/Micrometer to Millinewton/Foot can be time-consuming,especially when you don’t have enough knowledge about Surface Tension units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Gram Force/Micrometer to Millinewton/Foot converter tool to get the job done as soon as possible.
We have so many online tools available to convert Gram Force/Micrometer to Millinewton/Foot, but not every online tool gives an accurate result and that is why we have created this online Gram Force/Micrometer to Millinewton/Foot converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Gram Force/Micrometer to Millinewton/Foot (gf/μm to mN/ft)
By using our Gram Force/Micrometer to Millinewton/Foot conversion tool, you know that one Gram Force/Micrometer is equivalent to 2989066.92 Millinewton/Foot. Hence, to convert Gram Force/Micrometer to Millinewton/Foot, we just need to multiply the number by 2989066.92. We are going to use very simple Gram Force/Micrometer to Millinewton/Foot conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gram Force/Micrometer} = 1 \times 2989066.92 = \text{2989066.92 Millinewton/Foot}$$
## What Unit of Measure is Gram Force/Micrometer?
Gram force per micrometer is a unit of measurement for surface tension. Surface tension is considered to be equal to one gram force per micrometer if the force along a line of one micrometer length, where the force is parallel to the surface but perpendicular to the line is equal to one gram force.
## What is the Symbol of Gram Force/Micrometer?
The symbol of Gram Force/Micrometer is gf/μm. This means you can also write one Gram Force/Micrometer as 1 gf/μm.
## What Unit of Measure is Millinewton/Foot?
Millinewton per foot is a unit of measurement for surface tension. Surface tension is considered to be equal to one millinewton per foot if the force along a line of one foot length, where the force is parallel to the surface but perpendicular to the line is equal to one millinewton.
## What is the Symbol of Millinewton/Foot?
The symbol of Millinewton/Foot is mN/ft. This means you can also write one Millinewton/Foot as 1 mN/ft.
## How to Use Gram Force/Micrometer to Millinewton/Foot Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Gram Force/Micrometer and in the first input field, enter a value.
• From the second dropdown, select Millinewton/Foot.
• Instantly, the tool will convert the value from Gram Force/Micrometer to Millinewton/Foot and display the result in the second input field.
## Example of Gram Force/Micrometer to Millinewton/Foot Converter Tool
Gram Force/Micrometer
1
Millinewton/Foot
2989066.92
# Gram Force/Micrometer to Millinewton/Foot Conversion Table
Gram Force/Micrometer [gf/μm]Millinewton/Foot [mN/ft]Description
1 Gram Force/Micrometer2989066.92 Millinewton/Foot1 Gram Force/Micrometer = 2989066.92 Millinewton/Foot
2 Gram Force/Micrometer5978133.84 Millinewton/Foot2 Gram Force/Micrometer = 5978133.84 Millinewton/Foot
3 Gram Force/Micrometer8967200.76 Millinewton/Foot3 Gram Force/Micrometer = 8967200.76 Millinewton/Foot
4 Gram Force/Micrometer11956267.68 Millinewton/Foot4 Gram Force/Micrometer = 11956267.68 Millinewton/Foot
5 Gram Force/Micrometer14945334.6 Millinewton/Foot5 Gram Force/Micrometer = 14945334.6 Millinewton/Foot
6 Gram Force/Micrometer17934401.52 Millinewton/Foot6 Gram Force/Micrometer = 17934401.52 Millinewton/Foot
7 Gram Force/Micrometer20923468.44 Millinewton/Foot7 Gram Force/Micrometer = 20923468.44 Millinewton/Foot
8 Gram Force/Micrometer23912535.36 Millinewton/Foot8 Gram Force/Micrometer = 23912535.36 Millinewton/Foot
9 Gram Force/Micrometer26901602.28 Millinewton/Foot9 Gram Force/Micrometer = 26901602.28 Millinewton/Foot
10 Gram Force/Micrometer29890669.2 Millinewton/Foot10 Gram Force/Micrometer = 29890669.2 Millinewton/Foot
100 Gram Force/Micrometer298906692 Millinewton/Foot100 Gram Force/Micrometer = 298906692 Millinewton/Foot
1000 Gram Force/Micrometer2989066920 Millinewton/Foot1000 Gram Force/Micrometer = 2989066920 Millinewton/Foot
# Gram Force/Micrometer to Other Units Conversion Table
ConversionDescription
1 Gram Force/Micrometer = 9806.65 Newton/Meter1 Gram Force/Micrometer in Newton/Meter is equal to 9806.65
1 Gram Force/Micrometer = 98.07 Newton/Centimeter1 Gram Force/Micrometer in Newton/Centimeter is equal to 98.07
1 Gram Force/Micrometer = 9.81 Newton/Millimeter1 Gram Force/Micrometer in Newton/Millimeter is equal to 9.81
1 Gram Force/Micrometer = 0.00980665 Newton/Micrometer1 Gram Force/Micrometer in Newton/Micrometer is equal to 0.00980665
1 Gram Force/Micrometer = 0.00000980665 Newton/Nanometer1 Gram Force/Micrometer in Newton/Nanometer is equal to 0.00000980665
1 Gram Force/Micrometer = 8967.2 Newton/Yard1 Gram Force/Micrometer in Newton/Yard is equal to 8967.2
1 Gram Force/Micrometer = 2989.07 Newton/Foot1 Gram Force/Micrometer in Newton/Foot is equal to 2989.07
1 Gram Force/Micrometer = 249.09 Newton/Inch1 Gram Force/Micrometer in Newton/Inch is equal to 249.09
1 Gram Force/Micrometer = 9.81 Kilonewton/Meter1 Gram Force/Micrometer in Kilonewton/Meter is equal to 9.81
1 Gram Force/Micrometer = 0.0980665 Kilonewton/Centimeter1 Gram Force/Micrometer in Kilonewton/Centimeter is equal to 0.0980665
1 Gram Force/Micrometer = 0.00980665 Kilonewton/Millimeter1 Gram Force/Micrometer in Kilonewton/Millimeter is equal to 0.00980665
1 Gram Force/Micrometer = 0.00000980665 Kilonewton/Micrometer1 Gram Force/Micrometer in Kilonewton/Micrometer is equal to 0.00000980665
1 Gram Force/Micrometer = 9.80665e-9 Kilonewton/Nanometer1 Gram Force/Micrometer in Kilonewton/Nanometer is equal to 9.80665e-9
1 Gram Force/Micrometer = 8.97 Kilonewton/Yard1 Gram Force/Micrometer in Kilonewton/Yard is equal to 8.97
1 Gram Force/Micrometer = 2.99 Kilonewton/Foot1 Gram Force/Micrometer in Kilonewton/Foot is equal to 2.99
1 Gram Force/Micrometer = 0.24908891 Kilonewton/Inch1 Gram Force/Micrometer in Kilonewton/Inch is equal to 0.24908891
1 Gram Force/Micrometer = 9806650 Millinewton/Meter1 Gram Force/Micrometer in Millinewton/Meter is equal to 9806650
1 Gram Force/Micrometer = 98066.5 Millinewton/Centimeter1 Gram Force/Micrometer in Millinewton/Centimeter is equal to 98066.5
1 Gram Force/Micrometer = 9806.65 Millinewton/Millimeter1 Gram Force/Micrometer in Millinewton/Millimeter is equal to 9806.65
1 Gram Force/Micrometer = 9.81 Millinewton/Micrometer1 Gram Force/Micrometer in Millinewton/Micrometer is equal to 9.81
1 Gram Force/Micrometer = 0.00980665 Millinewton/Nanometer1 Gram Force/Micrometer in Millinewton/Nanometer is equal to 0.00980665
1 Gram Force/Micrometer = 8967200.76 Millinewton/Yard1 Gram Force/Micrometer in Millinewton/Yard is equal to 8967200.76
1 Gram Force/Micrometer = 2989066.92 Millinewton/Foot1 Gram Force/Micrometer in Millinewton/Foot is equal to 2989066.92
1 Gram Force/Micrometer = 249088.91 Millinewton/Inch1 Gram Force/Micrometer in Millinewton/Inch is equal to 249088.91
1 Gram Force/Micrometer = 9806650000 Micronewton/Meter1 Gram Force/Micrometer in Micronewton/Meter is equal to 9806650000
1 Gram Force/Micrometer = 98066500 Micronewton/Centimeter1 Gram Force/Micrometer in Micronewton/Centimeter is equal to 98066500
1 Gram Force/Micrometer = 9806650 Micronewton/Millimeter1 Gram Force/Micrometer in Micronewton/Millimeter is equal to 9806650
1 Gram Force/Micrometer = 9806.65 Micronewton/Micrometer1 Gram Force/Micrometer in Micronewton/Micrometer is equal to 9806.65
1 Gram Force/Micrometer = 9.81 Micronewton/Nanometer1 Gram Force/Micrometer in Micronewton/Nanometer is equal to 9.81
1 Gram Force/Micrometer = 8967200760 Micronewton/Yard1 Gram Force/Micrometer in Micronewton/Yard is equal to 8967200760
1 Gram Force/Micrometer = 2989066920 Micronewton/Foot1 Gram Force/Micrometer in Micronewton/Foot is equal to 2989066920
1 Gram Force/Micrometer = 249088910 Micronewton/Inch1 Gram Force/Micrometer in Micronewton/Inch is equal to 249088910
1 Gram Force/Micrometer = 980665000 Dyne/Meter1 Gram Force/Micrometer in Dyne/Meter is equal to 980665000
1 Gram Force/Micrometer = 9806650 Dyne/Centimeter1 Gram Force/Micrometer in Dyne/Centimeter is equal to 9806650
1 Gram Force/Micrometer = 980665 Dyne/Millimeter1 Gram Force/Micrometer in Dyne/Millimeter is equal to 980665
1 Gram Force/Micrometer = 980.66 Dyne/Micrometer1 Gram Force/Micrometer in Dyne/Micrometer is equal to 980.66
1 Gram Force/Micrometer = 0.980665 Dyne/Nanometer1 Gram Force/Micrometer in Dyne/Nanometer is equal to 0.980665
1 Gram Force/Micrometer = 896720076 Dyne/Yard1 Gram Force/Micrometer in Dyne/Yard is equal to 896720076
1 Gram Force/Micrometer = 298906692 Dyne/Foot1 Gram Force/Micrometer in Dyne/Foot is equal to 298906692
1 Gram Force/Micrometer = 24908891 Dyne/Inch1 Gram Force/Micrometer in Dyne/Inch is equal to 24908891
1 Gram Force/Micrometer = 1000000 Gram Force/Meter1 Gram Force/Micrometer in Gram Force/Meter is equal to 1000000
1 Gram Force/Micrometer = 10000 Gram Force/Centimeter1 Gram Force/Micrometer in Gram Force/Centimeter is equal to 10000
1 Gram Force/Micrometer = 1000 Gram Force/Millimeter1 Gram Force/Micrometer in Gram Force/Millimeter is equal to 1000
1 Gram Force/Micrometer = 0.001 Gram Force/Nanometer1 Gram Force/Micrometer in Gram Force/Nanometer is equal to 0.001
1 Gram Force/Micrometer = 914400 Gram Force/Yard1 Gram Force/Micrometer in Gram Force/Yard is equal to 914400
1 Gram Force/Micrometer = 304800 Gram Force/Foot1 Gram Force/Micrometer in Gram Force/Foot is equal to 304800
1 Gram Force/Micrometer = 25400 Gram Force/Inch1 Gram Force/Micrometer in Gram Force/Inch is equal to 25400
1 Gram Force/Micrometer = 1000 Kilogram Force/Meter1 Gram Force/Micrometer in Kilogram Force/Meter is equal to 1000
1 Gram Force/Micrometer = 10 Kilogram Force/Centimeter1 Gram Force/Micrometer in Kilogram Force/Centimeter is equal to 10
1 Gram Force/Micrometer = 1 Kilogram Force/Millimeter1 Gram Force/Micrometer in Kilogram Force/Millimeter is equal to 1
1 Gram Force/Micrometer = 0.001 Kilogram Force/Micrometer1 Gram Force/Micrometer in Kilogram Force/Micrometer is equal to 0.001
1 Gram Force/Micrometer = 0.000001 Kilogram Force/Nanometer1 Gram Force/Micrometer in Kilogram Force/Nanometer is equal to 0.000001
1 Gram Force/Micrometer = 914.4 Kilogram Force/Yard1 Gram Force/Micrometer in Kilogram Force/Yard is equal to 914.4
1 Gram Force/Micrometer = 304.8 Kilogram Force/Foot1 Gram Force/Micrometer in Kilogram Force/Foot is equal to 304.8
1 Gram Force/Micrometer = 25.4 Kilogram Force/Inch1 Gram Force/Micrometer in Kilogram Force/Inch is equal to 25.4
1 Gram Force/Micrometer = 1000000 Pond/Meter1 Gram Force/Micrometer in Pond/Meter is equal to 1000000
1 Gram Force/Micrometer = 10000 Pond/Centimeter1 Gram Force/Micrometer in Pond/Centimeter is equal to 10000
1 Gram Force/Micrometer = 1000 Pond/Millimeter1 Gram Force/Micrometer in Pond/Millimeter is equal to 1000
1 Gram Force/Micrometer = 1 Pond/Micrometer1 Gram Force/Micrometer in Pond/Micrometer is equal to 1
1 Gram Force/Micrometer = 0.001 Pond/Nanometer1 Gram Force/Micrometer in Pond/Nanometer is equal to 0.001
1 Gram Force/Micrometer = 914400 Pond/Yard1 Gram Force/Micrometer in Pond/Yard is equal to 914400
1 Gram Force/Micrometer = 304800 Pond/Foot1 Gram Force/Micrometer in Pond/Foot is equal to 304800
1 Gram Force/Micrometer = 25400 Pond/Inch1 Gram Force/Micrometer in Pond/Inch is equal to 25400
1 Gram Force/Micrometer = 1000 Kilopond/Meter1 Gram Force/Micrometer in Kilopond/Meter is equal to 1000
1 Gram Force/Micrometer = 10 Kilopond/Centimeter1 Gram Force/Micrometer in Kilopond/Centimeter is equal to 10
1 Gram Force/Micrometer = 1 Kilopond/Millimeter1 Gram Force/Micrometer in Kilopond/Millimeter is equal to 1
1 Gram Force/Micrometer = 0.001 Kilopond/Micrometer1 Gram Force/Micrometer in Kilopond/Micrometer is equal to 0.001
1 Gram Force/Micrometer = 0.000001 Kilopond/Nanometer1 Gram Force/Micrometer in Kilopond/Nanometer is equal to 0.000001
1 Gram Force/Micrometer = 914.4 Kilopond/Yard1 Gram Force/Micrometer in Kilopond/Yard is equal to 914.4
1 Gram Force/Micrometer = 304.8 Kilopond/Foot1 Gram Force/Micrometer in Kilopond/Foot is equal to 304.8
1 Gram Force/Micrometer = 25.4 Kilopond/Inch1 Gram Force/Micrometer in Kilopond/Inch is equal to 25.4
1 Gram Force/Micrometer = 2204.62 Pound Force/Meter1 Gram Force/Micrometer in Pound Force/Meter is equal to 2204.62
1 Gram Force/Micrometer = 22.05 Pound Force/Centimeter1 Gram Force/Micrometer in Pound Force/Centimeter is equal to 22.05
1 Gram Force/Micrometer = 2.2 Pound Force/Millimeter1 Gram Force/Micrometer in Pound Force/Millimeter is equal to 2.2
1 Gram Force/Micrometer = 0.0022046226218488 Pound Force/Micrometer1 Gram Force/Micrometer in Pound Force/Micrometer is equal to 0.0022046226218488
1 Gram Force/Micrometer = 0.0000022046226218488 Pound Force/Nanometer1 Gram Force/Micrometer in Pound Force/Nanometer is equal to 0.0000022046226218488
1 Gram Force/Micrometer = 2015.91 Pound Force/Yard1 Gram Force/Micrometer in Pound Force/Yard is equal to 2015.91
1 Gram Force/Micrometer = 671.97 Pound Force/Foot1 Gram Force/Micrometer in Pound Force/Foot is equal to 671.97
1 Gram Force/Micrometer = 56 Pound Force/Inch1 Gram Force/Micrometer in Pound Force/Inch is equal to 56
1 Gram Force/Micrometer = 35273.96 Ounce Force/Meter1 Gram Force/Micrometer in Ounce Force/Meter is equal to 35273.96
1 Gram Force/Micrometer = 352.74 Ounce Force/Centimeter1 Gram Force/Micrometer in Ounce Force/Centimeter is equal to 352.74
1 Gram Force/Micrometer = 35.27 Ounce Force/Millimeter1 Gram Force/Micrometer in Ounce Force/Millimeter is equal to 35.27
1 Gram Force/Micrometer = 0.0352739621 Ounce Force/Micrometer1 Gram Force/Micrometer in Ounce Force/Micrometer is equal to 0.0352739621
1 Gram Force/Micrometer = 0.0000352739621 Ounce Force/Nanometer1 Gram Force/Micrometer in Ounce Force/Nanometer is equal to 0.0000352739621
1 Gram Force/Micrometer = 32254.51 Ounce Force/Yard1 Gram Force/Micrometer in Ounce Force/Yard is equal to 32254.51
1 Gram Force/Micrometer = 10751.5 Ounce Force/Foot1 Gram Force/Micrometer in Ounce Force/Foot is equal to 10751.5
1 Gram Force/Micrometer = 895.96 Ounce Force/Inch1 Gram Force/Micrometer in Ounce Force/Inch is equal to 895.96
1 Gram Force/Micrometer = 70931.61 Poundal/Meter1 Gram Force/Micrometer in Poundal/Meter is equal to 70931.61
1 Gram Force/Micrometer = 709.32 Poundal/Centimeter1 Gram Force/Micrometer in Poundal/Centimeter is equal to 709.32
1 Gram Force/Micrometer = 70.93 Poundal/Millimeter1 Gram Force/Micrometer in Poundal/Millimeter is equal to 70.93
1 Gram Force/Micrometer = 0.070931611876605 Poundal/Micrometer1 Gram Force/Micrometer in Poundal/Micrometer is equal to 0.070931611876605
1 Gram Force/Micrometer = 0.000070931611876605 Poundal/Nanometer1 Gram Force/Micrometer in Poundal/Nanometer is equal to 0.000070931611876605
1 Gram Force/Micrometer = 64859.87 Poundal/Yard1 Gram Force/Micrometer in Poundal/Yard is equal to 64859.87
1 Gram Force/Micrometer = 21619.96 Poundal/Foot1 Gram Force/Micrometer in Poundal/Foot is equal to 21619.96
1 Gram Force/Micrometer = 1801.66 Poundal/Inch1 Gram Force/Micrometer in Poundal/Inch is equal to 1801.66
1 Gram Force/Micrometer = 98066500000 Erg/Square Meter1 Gram Force/Micrometer in Erg/Square Meter is equal to 98066500000
1 Gram Force/Micrometer = 9806650 Erg/Square Centimeter1 Gram Force/Micrometer in Erg/Square Centimeter is equal to 9806650 | 5,007 | 16,181 | {"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} | 3.34375 | 3 | CC-MAIN-2024-22 | latest | en | 0.841363 |
https://metanumbers.com/635213 | 1,642,907,246,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00195.warc.gz | 422,020,566 | 7,382 | # 635213 (number)
635,213 (six hundred thirty-five thousand two hundred thirteen) is an odd six-digits composite number following 635212 and preceding 635214. In scientific notation, it is written as 6.35213 × 105. The sum of its digits is 20. It has a total of 2 prime factors and 4 positive divisors. There are 619,680 positive integers (up to 635213) that are relatively prime to 635213.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 20
• Digital Root 2
## Name
Short name 635 thousand 213 six hundred thirty-five thousand two hundred thirteen
## Notation
Scientific notation 6.35213 × 105 635.213 × 103
## Prime Factorization of 635213
Prime Factorization 41 × 15493
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 635213 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 635,213 is 41 × 15493. Since it has a total of 2 prime factors, 635,213 is a composite number.
## Divisors of 635213
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 650748 Sum of all the positive divisors of n s(n) 15535 Sum of the proper positive divisors of n A(n) 162687 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 797.003 Returns the nth root of the product of n divisors H(n) 3.90451 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 635,213 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 635,213) is 650,748, the average is 162,687.
## Other Arithmetic Functions (n = 635213)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 619680 Total number of positive integers not greater than n that are coprime to n λ(n) 154920 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 51605 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 619,680 positive integers (less than 635,213) that are coprime with 635,213. And there are approximately 51,605 prime numbers less than or equal to 635,213.
## Divisibility of 635213
m n mod m 2 3 4 5 6 7 8 9 1 2 1 3 5 5 5 2
635,213 is not divisible by any number less than or equal to 9.
## Classification of 635213
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (635213)
Base System Value
2 Binary 10011011000101001101
3 Ternary 1012021100102
4 Quaternary 2123011031
5 Quinary 130311323
6 Senary 21340445
8 Octal 2330515
10 Decimal 635213
12 Duodecimal 267725
20 Vigesimal 3j80d
36 Base36 dm4t
## Basic calculations (n = 635213)
### Multiplication
n×y
n×2 1270426 1905639 2540852 3176065
### Division
n÷y
n÷2 317606 211738 158803 127043
### Exponentiation
ny
n2 403495555369 256305622212608597 162808663202537744726161 103418179378873608440738907293
### Nth Root
y√n
2√n 797.003 85.962 28.2312 14.4738
## 635213 as geometric shapes
### Circle
Diameter 1.27043e+06 3.99116e+06 1.26762e+12
### Sphere
Volume 1.07361e+18 5.07047e+12 3.99116e+06
### Square
Length = n
Perimeter 2.54085e+06 4.03496e+11 898327
### Cube
Length = n
Surface area 2.42097e+12 2.56306e+17 1.10022e+06
### Equilateral Triangle
Length = n
Perimeter 1.90564e+06 1.74719e+11 550111
### Triangular Pyramid
Length = n
Surface area 6.98875e+11 3.02059e+16 518649
## Cryptographic Hash Functions
md5 b92bbbf31bac78d160a160db0a42cdd3 4e3134b326db9840194ff16a7b673587cf0916c9 df7afdd3b14b1929ef1b688792c74d2f238c64f0b542844c774b86230edc42ab 6596a7e329c60b43b4b931139dc9ff15e1a12c36158e3e00df1e9c639fb0f861202f7617b0c304bbd7a4d2c6430bcee269fc717a2d232220affe7bc34850fbae a4de0d542c8f44902cd249cc7fecc31dccddd8d2 | 1,477 | 4,275 | {"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} | 3.734375 | 4 | CC-MAIN-2022-05 | latest | en | 0.810387 |
https://discuss.datasciencedojo.com/t/how-to-find-minimum-and-maximum-value-of-array/963 | 1,686,220,875,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00434.warc.gz | 236,666,585 | 4,905 | # How to find minimum and maximum value of array?
My goal is to find the maximum and minimum values of an array. My teacher told me about NumPy’s `amin` and `amax` methods and give me the code below:
I’m confused about the difference between Numpy’s min and amin() functions. Do they help us find the minimum value of the array? and same for max and amax(). Can you provide me with solutions that help me in determining the minimum and maximum values?
Hello @safa, the `numpy.amin()` function is similar to `numpy.min()` function, but it always returns a scalar value by computing the minimum value over the entire input array, regardless of the axis parameter. In short, the key difference between `numpy.min()` and `numpy.amin()` is that `numpy.min()` returns the minimum value along a specified axis, while `numpy.amin()` returns the minimum value over the entire array. The same is true for `numpy.max()` and `numpy.amax()`.
Note: `axis` is the parameter by which you can specify which dimension in the array you want the minimum or maximum from.
I hope this helps clear your confusion! | 245 | 1,095 | {"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} | 3.078125 | 3 | CC-MAIN-2023-23 | latest | en | 0.807209 |
https://www.haskell.org/hoogle/?hoogle=foldr&start=41 | 1,435,772,852,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095183.13/warc/CC-MAIN-20150627031815-00032-ip-10-179-60-89.ec2.internal.warc.gz | 912,007,091 | 5,452 | # foldr
foldr, applied to a binary operator, a starting value (typically the right-identity of the operator), and a list, reduces the list using the binary operator, from right to left: > foldr f z [x1, x2, ..., xn] == x1 `f` (x2 `f` ... (xn `f` z)...)
foldr, applied to a binary operator, a starting value (typically the right-identity of the operator), and a packed string, reduces the packed string using the binary operator, from right to left.
O(n) foldr, applied to a binary operator, a starting value (typically the right-identity of the operator), and a Text, reduces the Text using the binary operator, from right to left. Subject to fusion.
O(n). Fold the elements in the set using the given right-associative binary operator, such that foldr f z == foldr f z . toAscList. For example, > toAscList set = foldr (:) [] set
foldr, applied to a binary operator, a starting value (typically the right-identity of the operator), and a ByteString, reduces the ByteString using the binary operator, from right to left.
O(n). Fold the values in the map using the given right-associative binary operator, such that foldr f z == foldr f z . elems. For example, > elems map = foldr (:) [] map > let f a len = len + (length a) > foldr f 0 (fromList [(5,"a"), (3,"bbb")]) == 4
O(n). Fold the values in the map using the given right-associative binary operator, such that foldr f z == foldr f z . elems. For example, > elems map = foldr (:) [] map > let f a len = len + (length a) > foldr f 0 (fromList [(5,"a"), (3,"bbb")]) == 4
O(n). Fold the elements in the set using the given right-associative binary operator, such that foldr f z == foldr f z . toAscList. For example, > toAscList set = foldr (:) [] set
foldr1 is a variant of foldr that has no starting value argument, and thus must be applied to non-empty lists.
Fold over the elements of a structure, associating to the right, but strictly.
Monadic fold over the elements of a structure, associating to the right, i.e. from right to left.
'foldr\'' is a strict variant of foldr
O(n). A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
'foldr\'' is like foldr, but strict in the accumulator.
O(n). A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
O(n). A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
O(n). A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
foldr1 is a variant of foldr that has no starting value argument, and thus must be applied to non-empty ByteStrings
O(n) A variant of foldr that has no starting value argument, and thus must be applied to a non-empty Text. Subject to fusion.
foldr1 is a variant of foldr that has no starting value argument, and thus must be applied to non-empty ByteStrings
foldr1 is a variant of foldr that has no starting value argument, and thus must be applied to non-empty ByteStrings An exception will be thrown in the case of an empty ByteString.
A strict variant of foldr1
'foldr1\'' is a variant of foldr1, but is strict in the accumulator.
Consume the chunks of a lazy ByteString with a natural right fold.
Consume the chunks of a lazy Text with a natural right fold.
foldrWithIndex is a version of foldr that also provides access to the index of each element.
O(n). Fold the keys and values in the map using the given right-associative binary operator, such that foldrWithKey f z == foldr (uncurry f) z . toAscList. For example, > keys map = foldrWithKey (\k x ks -> k:ks) [] map > let f k a result = result ++ "(" ++ (show k) ++ ":" ++ a ++ ")" > foldrWithKey f "Map: " (fromList [(5,"a"), (3,"b")]) == "Map: (5:a)(3:b)"
O(n). Fold the keys and values in the map using the given right-associative binary operator, such that foldrWithKey f z == foldr (uncurry f) z . toAscList. For example, > keys map = foldrWithKey (\k x ks -> k:ks) [] map > let f k a result = result ++ "(" ++ (show k) ++ ":" ++ a ++ ")" > foldrWithKey f "Map: " (fromList [(5,"a"), (3,"b")]) == "Map: (5:a)(3:b)"
O(n). A strict version of foldrWithKey. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
O(n). A strict version of foldrWithKey. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
The unfoldr function is a `dual' to foldr: while foldr reduces a list to a summary value, unfoldr builds a list from a seed value. The function takes the element and returns Nothing if it is done producing the list or returns Just (a,b), in which case, a is a prepended to the list and b is used as the next element in a recursive call. For example, > iterate f == unfoldr (\x -> Just (x, f x)) In some cases, unfoldr can undo a foldr operation: > unfoldr f' (foldr f z xs) == xs if the following holds: > f' (f x y) = Just (x,y) > f' z = Nothing A simple use of unfoldr: > unfoldr (\b -> if b == 0 then Nothing else Just (b, b-1)) 10 > [10,9,8,7,6,5,4,3,2,1]
Create a Builder that encodes a sequence generated from a seed value using a BoundedPrim for each sequence element.
Encode a list of values represented as an unfoldr with a FixedPrim.
O(n) The unfoldr function is analogous to the List 'unfoldr'. unfoldr builds a ByteString from a seed value. The function takes the element and returns Nothing if it is done producing the ByteString or returns Just (a,b), in which case, a is a prepending to the ByteString and b is used as the next element in a recursive call.
O(n), unfoldr function is analogous to the List 'unfoldr'. unfoldr builds a ByteString from a seed value. The function takes the element and returns Nothing if it is done producing the ByteString or returns Just (a,b), in which case, a is the next character in the string, and b is the seed value for further production. Examples: > unfoldr (\x -> if x <= '9' then Just (x, succ x) else Nothing) '0' == "0123456789"
O(n), unfoldr function is analogous to the List unfoldr. unfoldr builds a Text from a seed value. The function takes the element and returns Nothing if it is done producing the Text, otherwise Just (a,b). In this case, a is the next Char in the string, and b is the seed value for further production. Performs replacement on invalid scalar values.
O(n), unfoldr function is analogous to the List unfoldr. unfoldr builds a Text from a seed value. The function takes the element and returns Nothing if it is done producing the Text, otherwise Just (a,b). In this case, a is the next Char in the string, and b is the seed value for further production. Subject to fusion. Performs replacement on invalid scalar values.
O(n) The unfoldr function is analogous to the List 'unfoldr'. unfoldr builds a ByteString from a seed value. The function takes the element and returns Nothing if it is done producing the ByteString or returns Just (a,b), in which case, a is a prepending to the ByteString and b is used as the next element in a recursive call.
O(n), unfoldr function is analogous to the List 'unfoldr'. unfoldr builds a ByteString from a seed value. The function takes the element and returns Nothing if it is done producing the ByteString or returns Just (a,b), in which case, a is the next byte in the string, and b is the seed value for further production. Examples: > unfoldr (\x -> if x <= 5 then Just (x, x + 1) else Nothing) 0 > == pack [0, 1, 2, 3, 4, 5]
Builds a sequence from a seed value. Takes time linear in the number of generated elements. WARNING: If the number of generated elements is infinite, this method will not terminate.
O(n) Like unfoldr, unfoldrN builds a ByteString from a seed value. However, the length of the result is limited by the first argument to unfoldrN. This function is more efficient than unfoldr when the maximum length of the result is known. The following equation relates unfoldrN and unfoldr: > unfoldrN n f s == take n (unfoldr f s)
O(n) Like unfoldr, unfoldrN builds a Text from a seed value. However, the length of the result should be limited by the first argument to unfoldrN. This function is more efficient than unfoldr when the maximum length of the result is known and correct, otherwise its performance is similar to unfoldr. Subject to fusion. Performs replacement on invalid scalar values.
O(n) Like unfoldr, unfoldrN builds a ByteString from a seed value. However, the length of the result is limited by the first argument to unfoldrN. This function is more efficient than unfoldr when the maximum length of the result is known. The following equation relates unfoldrN and unfoldr: > fst (unfoldrN n f s) == take n (unfoldr f s)
O(n) Like unfoldr, unfoldrN builds a Text from a seed value. However, the length of the result should be limited by the first argument to unfoldrN. This function is more efficient than unfoldr when the maximum length of the result is known and correct, otherwise its performance is similar to unfoldr. Performs replacement on invalid scalar values. | 2,346 | 9,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2015-27 | longest | en | 0.84243 |
https://socratic.org/questions/how-do-solve-the-following-linear-system-4x-2y-2-x-3y-2 | 1,716,219,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00090.warc.gz | 487,090,739 | 6,103 | # How do solve the following linear system?: 4x-2y=2 , x-3y=-2 ?
Oct 7, 2016
$\text{solution : x=1 and y=1 }$
#### Explanation:
$\text{given :}$
$4 x - 2 y = 2 \text{ (1)}$
$x - 3 y = - 2 \text{ (2)}$
$\text{expand both side of equation (2) by 4 }$
$4 \left(x - 3 y\right) = - 2 \cdot 4$
$4 x - 12 y = - 8$
$\textcolor{red}{4 x} = 12 y - 8 \text{ (3)}$
$\text{rearrange (1)}$
$\textcolor{red}{4 x} = 2 y + 2 \text{ (4)}$
$\text{(3) and (4) are equal.}$
$12 y - 8 = 2 y + 2$
$12 y - 2 y = 2 + 8$
$10 y = 10$
$y = 1$
$\text{now use (1) or (2)}$
$4 x - 2 \cdot 1 = 2$
$4 x - 2 = 2$
$4 x = 2 + 2$
$4 x = 4$
$x = 1$ | 311 | 633 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "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} | 4.53125 | 5 | CC-MAIN-2024-22 | latest | en | 0.383081 |
https://riskwiki.vosesoftware.com/Numberofsamplestogetaspecifics.php | 1,723,456,748,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00159.warc.gz | 388,593,334 | 20,385 | Number of hypergeometric samples to get a specific number of successes | Vose Software
# Number of hypergeometric samples to get a specific number of successes
Consider the situation where we are sampling without replacement from a population M with D items with the characteristic of interest until we have s items with the required characteristic. The distribution of the number of failures we will have before the s success can be easily calculated in the same manner as we developed the Negative Binomial distribution. The probability of observing (s-1) successes in (x+s-1) trials (i.e. x failures) is given by direct application of the Hypergeometric distribution:
The probability p of then observing a success in the next trial (the (s+x)th trial), is simply the number of D items remaining (=D-(s-1)) divided by the size of the population remaining (= M-(s+x-1)):
and the probability of having exactly x failures up to the sth success, where trials are stopped at the sth success, is then the product of these two probabilities:
This is the probability mass function for the Inverse Hypergeometric distribution InvHypergeo(s,D,M) and is analogous to the Negative Binomial distribution for the binomial process and the Gamma distribution for the Poisson process. So:
n = s + VoseInvHypergeo(s,D,M)
For a population M that is large compared to s, the Inverse Hypergeometric distribution is closely approximated by the Negative Binomial:
InvHypergeo(s,D,M) » NegBin(s,D/M)
and if the probability D/M is very small:
InvHypergeo(s,D,M) » Gamma(s,M/D)
The four figures below show examples of the Inverse Hypergeometric distribution. In the first figure you can see the probability mass function of the number of failures before getting 4 successes when drawing samples from a population 50 in which 5 individuals have the characteristic you are interested in. We leave to you the task to explain in words the figures 2 - 4.
An Inverse Hypergeometric distribution shifted k units along the domain is sometimes called a Negative Hypergeometric distribution. | 443 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.906299 |
https://numbersworksheet.com/adding-subtracting-poitive-and-negative-numbers-worksheet/ | 1,709,075,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00588.warc.gz | 431,414,639 | 14,810 | # Adding Subtracting Poitive And Negative Numbers Worksheet
The Bad Amounts Worksheet is a great way to commence educating your kids the thought of unfavorable amounts. A poor number is any amount which is lower than absolutely no. It might be extra or subtracted. The minus sign suggests the unfavorable quantity. You may also publish negative phone numbers in parentheses. Listed below can be a worksheet to acquire started out. This worksheet has an array of unfavorable amounts from -10 to 10. Adding Subtracting Poitive And Negative Numbers Worksheet.
Negative amounts are a lot as their worth is lower than absolutely no
A negative amount carries a importance under absolutely nothing. It can be expressed with a quantity series in 2 methods: with the positive amount created as the initially digit, and with the unfavorable number published as being the last digit. A confident quantity is created using a as well as sign ( ) prior to it, however it is recommended to create it this way. It is assumed to be a positive number if the number is not written with a plus sign.
## These are depicted with a minus indication
In ancient Greece, unfavorable numbers were not applied. These folks were dismissed, since their mathematics was based on geometrical methods. When European scholars started translating ancient Arabic text messages from North Africa, they got to acknowledge adverse figures and appreciated them. Right now, bad numbers are displayed with a minus sign. To learn more about the history and origins of bad phone numbers, read through this post. Then, consider these cases to see how bad numbers have advanced as time passes.
## They are often included or subtracted
Positive numbers and negative numbers are easy to add and subtract because the sign of the numbers is the same, as you might already know. Negative numbers, on the other hand, have a larger absolute value, but they are closer to than positive numbers are. They can still be added and subtracted just like positive ones, although these numbers have some special rules for arithmetic. You can even add and subtract negative figures utilizing a variety range and use a similar guidelines for subtraction and addition as you may do for good figures.
## These are represented from a number in parentheses
A negative quantity is displayed with a variety covered in parentheses. The unfavorable signal is transformed into its binary equivalent, along with the two’s go with is kept in a similar spot in storage. Sometimes a negative number is represented by a positive number, though the result is always negative. In such cases, the parentheses should be incorporated. If you have any questions about the meaning of negative numbers, you should consult a book on math.
## They may be separated with a optimistic variety
Unfavorable numbers may be divided and multiplied like beneficial amounts. They can also be divided by other bad numbers. However, they are not equal to one another. At the first try you grow a negative amount from a good number, you will definately get zero as a result. To make the perfect solution, you need to select which indicator your solution must have. It really is quicker to keep in mind a poor quantity when it is printed in brackets. | 626 | 3,260 | {"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} | 4.28125 | 4 | CC-MAIN-2024-10 | latest | en | 0.966733 |
https://netlib.sandia.gov/slatec/src/mgsbv.f | 1,716,544,728,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00034.warc.gz | 364,832,588 | 3,447 | *DECK MGSBV SUBROUTINE MGSBV (M, N, A, IA, NIV, IFLAG, S, P, IP, INHOMO, V, W, + WCND) C***BEGIN PROLOGUE MGSBV C***SUBSIDIARY C***PURPOSE Subsidiary to BVSUP C***LIBRARY SLATEC C***TYPE SINGLE PRECISION (MGSBV-S, DMGSBV-D) C***AUTHOR Watts, H. A., (SNLA) C***DESCRIPTION C C ********************************************************************** C Orthogonalize a set of N real vectors and determine their rank C C ********************************************************************** C INPUT C ********************************************************************** C M = Dimension of vectors C N = No. of vectors C A = Array whose first N cols contain the vectors C IA = First dimension of array A (col length) C NIV = Number of independent vectors needed C INHOMO = 1 Corresponds to having a non-zero particular solution C V = Particular solution vector (not included in the pivoting) C INDPVT = 1 Means pivoting will not be used C C ********************************************************************** C OUTPUT C ********************************************************************** C NIV = No. of linear independent vectors in input set C A = Matrix whose first NIV cols. contain NIV orthogonal vectors C which span the vector space determined by the input vectors C IFLAG C = 0 success C = 1 incorrect input C = 2 rank of new vectors less than N C P = Decomposition matrix. P is upper triangular and C (old vectors) = (new vectors) * P. C The old vectors will be reordered due to pivoting C The dimension of p must be .GE. N*(N+1)/2. C ( N*(2*N+1) when N .NE. NFCC ) C IP = Pivoting vector. The dimension of IP must be .GE. N. C ( 2*N when N .NE. NFCC ) C S = Square of norms of incoming vectors C V = Vector which is orthogonal to the vectors of A C W = Orthogonalization information for the vector V C WCND = Worst case (smallest) norm decrement value of the C vectors being orthogonalized (represents a test C for linear dependence of the vectors) C ********************************************************************** C C***SEE ALSO BVSUP C***ROUTINES CALLED PRVEC, SDOT C***COMMON BLOCKS ML18JR, ML5MCO C***REVISION HISTORY (YYMMDD) C 750601 DATE WRITTEN C 890531 Changed all specific intrinsics to generic. (WRB) C 890831 Modified array declarations. (WRB) C 890921 Realigned order of variables in certain COMMON blocks. C (WRB) C 891214 Prologue converted to Version 4.0 format. (BAB) C 900328 Added TYPE section. (WRB) C 910722 Updated AUTHOR section. (ALS) C***END PROLOGUE MGSBV C DIMENSION A(IA,*),V(*),W(*),P(*),IP(*),S(*) C C COMMON /ML18JR/ AE,RE,TOL,NXPTS,NIC,NOPG,MXNON,NDISK,NTAPE,NEQ, 1 INDPVT,INTEG,NPS,NTP,NEQIVP,NUMORT,NFCC, 2 ICOCO C COMMON /ML5MCO/ URO,SRU,EPS,SQOVFL,TWOU,FOURU,LPAR C C***FIRST EXECUTABLE STATEMENT MGSBV IF(M .GT. 0 .AND. N .GT. 0 .AND. IA .GE. M) GO TO 10 IFLAG=1 RETURN C 10 JP=0 IFLAG=0 NP1=N+1 Y=0.0 M2=M/2 C C CALCULATE SQUARE OF NORMS OF INCOMING VECTORS AND SEARCH FOR C VECTOR WITH LARGEST MAGNITUDE C J=0 DO 30 I=1,N VL=SDOT(M,A(1,I),1,A(1,I),1) S(I)=VL IF (N .EQ. NFCC) GO TO 25 J=2*I-1 P(J)=VL IP(J)=J 25 J=J+1 P(J)=VL IP(J)=J IF(VL .LE. Y) GO TO 30 Y=VL IX=I 30 CONTINUE IF (INDPVT .NE. 1) GO TO 33 IX=1 Y=P(1) 33 LIX=IX IF (N .NE. NFCC) LIX=2*IX-1 P(LIX)=P(1) S(NP1)=0. IF (INHOMO .EQ. 1) S(NP1)=SDOT(M,V,1,V,1) WCND=1. NIVN=NIV NIV=0 C IF(Y .EQ. 0.0) GO TO 170 C ********************************************************************** DO 140 NR=1,N IF (NIVN .EQ. NIV) GO TO 150 NIV=NR IF(IX .EQ. NR) GO TO 80 C C PIVOTING OF COLUMNS OF P MATRIX C NN=N LIX=IX LR=NR IF (N .EQ. NFCC) GO TO 40 NN=NFCC LIX=2*IX-1 LR=2*NR-1 40 IF(NR .EQ. 1) GO TO 60 KD=LIX-LR KJ=LR NRM1=LR-1 DO 50 J=1,NRM1 PSAVE=P(KJ) JK=KJ+KD P(KJ)=P(JK) P(JK)=PSAVE 50 KJ=KJ+NN-J JY=JK+NMNR JZ=JY-KD P(JY)=P(JZ) 60 IZ=IP(LIX) IP(LIX)=IP(LR) IP(LR)=IZ SV=S(IX) S(IX)=S(NR) S(NR)=SV IF (N .EQ. NFCC) GO TO 69 IF (NR .EQ. 1) GO TO 67 KJ=LR+1 DO 65 K=1,NRM1 PSAVE=P(KJ) JK=KJ+KD P(KJ)=P(JK) P(JK)=PSAVE 65 KJ=KJ+NFCC-K 67 IZ=IP(LIX+1) IP(LIX+1)=IP(LR+1) IP(LR+1)=IZ C C PIVOTING OF COLUMNS OF VECTORS C 69 DO 70 L=1,M T=A(L,IX) A(L,IX)=A(L,NR) 70 A(L,NR)=T C C CALCULATE P(NR,NR) AS NORM SQUARED OF PIVOTAL VECTOR C 80 JP=JP+1 P(JP)=Y RY=1.0/Y NMNR=N-NR IF (N .EQ. NFCC) GO TO 85 NMNR=NFCC-(2*NR-1) JP=JP+1 P(JP)=0. KP=JP+NMNR P(KP)=Y 85 IF(NR .EQ. N .OR. NIVN .EQ. NIV) GO TO 125 C C CALCULATE ORTHOGONAL PROJECTION VECTORS AND SEARCH FOR LARGEST NORM C Y=0.0 IP1=NR+1 IX=IP1 C **************************************** DO 120 J=IP1,N DOT=SDOT(M,A(1,NR),1,A(1,J),1) JP=JP+1 JQ=JP+NMNR IF (N .NE. NFCC) JQ=JQ+NMNR-1 P(JQ)=P(JP)-DOT*(DOT*RY) P(JP)=DOT*RY DO 90 I = 1,M 90 A(I,J)=A(I,J)-P(JP)*A(I,NR) IF (N .EQ. NFCC) GO TO 99 KP=JP+NMNR JP=JP+1 PJP=RY*PRVEC(M,A(1,NR),A(1,J)) P(JP)=PJP P(KP)=-PJP KP=KP+1 P(KP)=RY*DOT DO 95 K=1,M2 L=M2+K A(K,J)=A(K,J)-PJP*A(L,NR) 95 A(L,J)=A(L,J)+PJP*A(K,NR) P(JQ)=P(JQ)-PJP*(PJP/RY) C C TEST FOR CANCELLATION IN RECURRENCE RELATION C 99 IF(P(JQ) .GT. S(J)*SRU) GO TO 100 P(JQ)=SDOT(M,A(1,J),1,A(1,J),1) 100 IF(P(JQ) .LE. Y) GO TO 120 Y=P(JQ) IX=J 120 CONTINUE IF (N .NE. NFCC) JP=KP C **************************************** IF(INDPVT .EQ. 1) IX=IP1 C C RECOMPUTE NORM SQUARED OF PIVOTAL VECTOR WITH SCALAR PRODUCT C Y=SDOT(M,A(1,IX),1,A(1,IX),1) IF(Y .LE. EPS*S(IX)) GO TO 170 WCND=MIN(WCND,Y/S(IX)) C C COMPUTE ORTHOGONAL PROJECTION OF PARTICULAR SOLUTION C 125 IF(INHOMO .NE. 1) GO TO 140 LR=NR IF (N .NE. NFCC) LR=2*NR-1 W(LR)=SDOT(M,A(1,NR),1,V,1)*RY DO 130 I=1,M 130 V(I)=V(I)-W(LR)*A(I,NR) IF (N .EQ. NFCC) GO TO 140 LR=2*NR W(LR)=RY*PRVEC(M,V,A(1,NR)) DO 135 K=1,M2 L=M2+K V(K)=V(K)+W(LR)*A(L,NR) 135 V(L)=V(L)-W(LR)*A(K,NR) 140 CONTINUE C ********************************************************************** C C TEST FOR LINEAR DEPENDENCE OF PARTICULAR SOLUTION C 150 IF(INHOMO .NE. 1) RETURN IF ((N .GT. 1) .AND. (S(NP1) .LT. 1.0)) RETURN VNORM=SDOT(M,V,1,V,1) IF (S(NP1) .NE. 0.) WCND=MIN(WCND,VNORM/S(NP1)) IF(VNORM .GE. EPS*S(NP1)) RETURN 170 IFLAG=2 WCND=EPS RETURN END | 2,309 | 5,948 | {"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-2024-22 | latest | en | 0.685088 |
https://www.thoughtco.com/calculate-limiting-reactant-and-theoretical-yield-609565 | 1,679,836,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00372.warc.gz | 1,140,430,062 | 39,399 | # How to Calculate Limiting Reactant and Theoretical Yield
The limiting reactant of a reaction is the reactant that would run out first if all the reactants were to be reacted together. Once the limiting reactant is completely consumed, the reaction would cease to progress. The theoretic yield of a reaction is the amount of products produced when the limiting reactant runs out. This worked example chemistry problem shows how to determine the limiting reactant and calculate the theoretical yield of a chemical reaction.
## Limiting Reactant and Theoretical Yield Problem
You are given the following reaction:
2 H2(g) + O2(g) → 2 H2O(l)
Calculate:
a. the stoichiometric ratio of moles H2 to moles O2
b. the actual moles H2 to moles O2 when 1.50 mol H2 is mixed with 1.00 mol O2
c. the limiting reactant (H2 or O2) for the mixture in part (b)
d. the theoretical yield, in moles, of H2O for the mixture in part (b)
Solution
a. The stoichiometric ratio is given by using the coefficients of the balanced equation. The coefficients are the numbers listed before each formula. This equation is already balanced, so refer to the tutorial on balancing equations if you need further help:
2 mol H2 / mol O2
b. The actual ratio refers to the number of moles actually provided for the reaction. This may or may not be the same as the stoichiometric ratio. In this case, it is different:
1.50 mol H2 / 1.00 mol O2 = 1.50 mol H2 / mol O2
c. Note that the actual ratio of smaller than the required or stoichiometric ratio, which means there is insufficient H2 to react with all of the O2 that has been provided. The 'insufficient' component (H2) is the limiting reactant. Another way to put it is to say that O2 is in excess. When the reaction has proceeded to completion, all of the H2 will have been consumed, leaving some O2 and the product, H2O.
d. Theoretical yield is based on the calculation using the amount of limiting reactant, 1.50 mol H2. Given that 2 mol H2 forms 2 mol H2O, we get:
theoretical yield H2O = 1.50 mol H2 x 2 mol H2O / 2 mol H2
theoretical yield H2O = 1.50 mol H2O
Note that the only requirement for performing this calculation is knowing the amount of the limiting reactant and the ratio of the amount of limiting reactant to the amount of product.
Answers
a. 2 mol H2 / mol O2
b. 1.50 mol H2 / mol O2
c. H2
d. 1.50 mol H2O
## Tips for Working This Type of Problem
• The most important point to remember is that you are dealing with the molar ratio between the reactants and products. If you are given a value in grams, you need to convert it to moles. If you're asked to supply a number in grams, you convert back from the moles used in the calculation.
• The limiting reactant isn't automatically the one with the smallest number of moles. For example, say you have 1.0 moles of hydrogen and 0.9 moles of oxygen in the reaction to make water. If you didn't look at the stoichiometric ratio between the reactants, you might choose oxygen as the limiting reactant, yet hydrogen and oxygen react in a 2:1 ratio, so you'd actually expend the hydrogen much sooner than you'd use up the oxygen.
• When you're asked to give quantities, watch the number of significant figures. They always matter in chemistry!
Format
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Your Citation
Helmenstine, Anne Marie, Ph.D. "How to Calculate Limiting Reactant and Theoretical Yield." ThoughtCo, Jul. 29, 2021, thoughtco.com/calculate-limiting-reactant-and-theoretical-yield-609565. Helmenstine, Anne Marie, Ph.D. (2021, July 29). How to Calculate Limiting Reactant and Theoretical Yield. Retrieved from https://www.thoughtco.com/calculate-limiting-reactant-and-theoretical-yield-609565 Helmenstine, Anne Marie, Ph.D. "How to Calculate Limiting Reactant and Theoretical Yield." ThoughtCo. https://www.thoughtco.com/calculate-limiting-reactant-and-theoretical-yield-609565 (accessed March 26, 2023). | 1,001 | 3,884 | {"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} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.918294 |
https://mrelementarymath.com/math-books-that-will-change-your/ | 1,653,418,595,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00265.warc.gz | 466,753,464 | 32,998 | Take the stress out of implementing math centers in your classroom...
# Math Books That Will Change Your Teaching Blog Hop
Today I am participating in the Math Books That Will Change Your Teaching Blog Hop organized by Brandi of The Research Based Classroom. Take this journey with me and 11 other excellent educators as we discuss books that have changed our teaching. Each blogger will be giving away a copy of the book they reviewed, for FREE!!! Simply enter the giveaway on each blog.
I am very happy to participate in this blog hop because I have an opportunity to share one of my favorite books. It was a difficult decision to pick just one book to share because I have a variety of math books that I enjoy.
I would like to introduce……..
Why I Selected this Book?
A couple of months ago, I facilitated a professional development that highlighted the focus math standards for each grade from Kdg. – 5th. While preparing for the PD, I realized the importance of various standards and how they build upon one another. Additionally, I thought about how each grade has fluency facts that should be mastered before moving to the next grade. It was like a light bulb went off in my head and I really thought about the importance of teaching addition & subtraction facts utilizing a systematic approach. This book is an EXCELLENT resource to help orchestrate such a task . In fact, it focuses on helping students understand groups of facts and then building on that understanding with additional sets of facts versus fact memorization.
AHA MOMENT….
I really connected with this book as I’ve often thought of ways that I could make teaching addition facts easier and more efficient. When I came across the suggested teaching sequence below, I was in love. Teaching in this strategic manner really supports student learning and builds conceptual understanding. The suggested teaching sequence for addition facts included:
• First, teaching +1 and +2 addition facts
• Then, teaching +0 addition facts
• Next, teaching +10 addition facts
• Then, teaching the doubles facts
• Lastly, teaching the facts that make 10 (ie. 6 + 4, 7 + 3, etc.)
Interesting Features from the Book
Each chapter contains the following components:
• Big Ideas for the foundational facts
• A literature link- Yes, this book incorporates literature for each of the foundational facts.
• Games & Activities that align with each of the foundational facts.
• Ways to build automaticity through targeted practice.
• Connections to subtraction
This book also contains a study guide for professional learning communities. That section of the book has questions that align with each chapter for the book study.
Two of my Favorite Activities from the Book
1) Hop the Line is an activity in which students add 1, add 2, subtract 1 or subtract 2 until they reach the end of the line. The students use a spinner to determine how many numbers to jump on the line. If the student spins a number that moves them off the number line, they lose a turn. The object of the game is to be the first person to get to 20. The book has a worksheet with a number line but I created my own number line so the game would be interactive.
2) Ladders is a game that provides students with practice on near-doubles facts. To play, two students take turns spinning the Ladders Spinner and recording the sum on the ladder. The sum must be recorded in order from the least number (on the bottom of the ladder) to the greatest number (on the top of the ladder). A player will lose a turn, if they spin a sum that doesn’t have a place on the ladder. The object of the game is to be the first player to complete the ladder with the sums in order.
This game is great because it can be played with many addition facts as well as multiplication facts. The students see the importance of being strategic with the placement of the sum.
Final Words
One of the great things about being involved in this wonderful blog hop is an opportunity for you, the reader to win a FREE copy of the book. Yes, a FREE copy shipped to your house! You can enter the raffle by completing the Rafflecopter Giveaway. A winner will be selected on Saturday, August 30th. The winner will be announced via my blog and Facebook Fan Page.
a Rafflecopter giveaway
Be sure to hop along our blog hop to the next wonderful mathematician Brandi, (the blog hop sponsor) by clicking on the link below. Brandi will review the book, Number Sense Routines. Be sure to check out her post!
19 | 994 | 4,522 | {"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-2022-21 | longest | en | 0.958552 |
http://chimera.labs.oreilly.com/books/1234000001642/ch04.html | 1,516,772,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893397.98/warc/CC-MAIN-20180124050449-20180124070449-00303.warc.gz | 61,242,533 | 7,272 | # Chapter 4. Logic and Recursion
## Étude 4-1: Using case
Change the area/3 function that you wrote in Étude 3-2 so that it uses a case instead of pattern matching. Use a guard on the function definition to ensure that the numeric arguments are both greater than zero.
See a suggested solution in Appendix A.
## Étude 4-2: Recursion
This is a typical exercise for recursion: finding the greatest common divisor (GCD) of two integers. Instead of giving Euclid’s method, we’ll do this with a method devised by Edsger W. Dijkstra. The neat part about Dijkstra’s method is that you don’t need to do any division to find the result. Here is the method.
To find the GCD of integers M and N:
• If M and N are equal, the result is M.
• If M is greater than N, the result is the GCD of M - N and N
• Otherwise M must be less than N, and the result is the GCD of M and N - M.
Write a function gcd/2 in a module named Dijkstra that implements the algorithm. This program is an ideal place to use Elixir’s cond construct. Here is some sample output.
iex(1)> c("dijkstra.ex")
[Dijkstra]
iex(2)> Dijkstra.gcd(2, 8)
2
iex(3)> Dijkstra.gcd(14, 21)
7
iex(4)> Dijkstra.gcd(125, 46)
1
iex(5)> Dijkstra.gcd(120, 36)
12
See a suggested solution in Appendix A. You can also use guards with multiple clauses to solve this étude; the solution for that approach is in Appendix A. In general, use of multiple clauses is considered more in the spirit of Elixir.
The next two exercises involve writing code to raise a number to an integer power (like 2.53 or 4-2) and finding the nth root of a number, such as the cube root of 1728 or the fifth root of 3.2.
These capabilities already exist with the :math.pow/2 function, so you may wonder why I’m asking you to re-invent the wheel. The reason is not to replace :math.pow/2, but to experiment with recursion by writing functions that can be expressed quite nicely that way.
## Étude 4-3: Non-Tail Recursive Functions
Create a module named Powers (no relation to Francis Gary Powers), and write a function named raise/2 which takes parameters x and n and returns the value of xn.
Here’s the information you need to know to write the function:
• Any number to the power 0 equals 1.
• Any number to the power 1 is that number itself — that stops the recursion.
• When n is positive, xn is equal to x times x(n - 1) — there’s your recursion.
• When n is negative, xn is equal to 1.0 / x-n
Note that this algorithm is not tail recursive.
The Elixir kernel already has a raise/2 function, so your function will cause a conflict unless you add this after the defmodule:
import Kernel, except: [raise: 2]
Here is some sample output.
iex(1)> c("powers.ex")
$Powers$
iex(2)> Powers.raise(5,1)
5
iex(3)> Powers.raise(2,3)
8
iex(4)> Powers.raise(1.2, 3)
1.728
iex(5)> Powers.raise(2, 0)
1
iex(6)> Powers.raise(2, -3)
0.125
If you try raising 0 to a negative power, you will get an error message. For now, just let the error occur. You will learn more about error handling in Chapter 10.
See a suggested solution in Appendix A.
## Étude 4-4: Tail Recursion with an Accumulator
Practice the "accumulator trick." Rewrite the raise/2 function for n greater than zero so that it calls a helper function raise/3 This new function has x, n, and an accumulator as its parameters.
Your raise/2 function will return 1 when n is equal to 0, and will return 1.0 / raise(x, -n) when n is less than zero.
When n is greater than zero, raise/2 will call raise/3 with arguments x, n, and 1 as the accumulator.
The raise/3 function will return the accumulator when n equals 0 (this will stop the recursion).
Otherwise, recursively call raise/3 with x, n - 1, and x times the accumulator as its arguments.
The raise/3 function is tail recursive.
Because the kernel also contains a raise/3 function, you have to change your import as follows:
import Kernel, except: [raise: 2, raise: 3]
See a suggested solution in Appendix A.
## Étude 4-5: Recursion with a Helper Function
In this exercise, you will add a function nth_root/2 to the Powers module. This new function finds the nth root of a number, where n is an integer. For example, nth_root(36, 2) will calculate the square root of 36, and nth_root(1.728, 3) will return the cube root of 1.728.
The algorithm used here is the Newton-Raphson method for calculating roots. (See http://en.wikipedia.org/wiki/Newton%27s_method for details).
You will need a helper function nth_root/3, whose parameters are x, n, and an approximation to the result, which we will call a. nth_root/3 works as follows:
• Calculate f as (an - x)
• Calculate f_prime as n * a(n - 1)
• Calculate your next approximation (call it next) as a - f / f_prime
• Calculate the change in value (call it change) as the absolute value of next - a
• If the change is less than some limit (say, 1.0e-8), stop the recursion and return next; that’s as close to the root as you are going to get.
• Otherwise, call the nth_root/3 function again with x, n, and next as its arguments.
For your first approximation, use x / 2.0. Thus, your nth_root/2 function will simply be this:
nth_root(x, n) → nth_root(x, n, x / 2.0)
Use IO.puts to show each new approximation as you calculate it. If your argument name is estimate, you would do something like this:
IO.puts("Current guess is #{estimate}")
Here is some sample output.
iex(1)> c("powers.ex")
$Powers$
iex(2)> Powers.nth_root(27, 3)
Current guess is 13.5
Current guess is 9.049382716049383
Current guess is 6.142823558176272
Current guess is 4.333725614685509
Current guess is 3.3683535855517652
Current guess is 3.038813723595138
Current guess is 3.0004936436555805
Current guess is 3.000000081210202
Current guess is 3.000000000000002
3.0
See a suggested solution in Appendix A. | 1,592 | 5,778 | {"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} | 4.21875 | 4 | CC-MAIN-2018-05 | latest | en | 0.85592 |
https://www.jiskha.com/display.cgi?id=1371955010 | 1,516,753,025,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892802.73/warc/CC-MAIN-20180123231023-20180124011023-00009.warc.gz | 915,327,525 | 3,683 | # math check
posted by .
Consider the approximately normal population of heights of male college students with mean ì = 69 inches and standard deviation of ó = 4.6 inches. A random sample of 25 heights is obtained.
(b) Find the proportion of male college students whose height is greater than 69 inches. (Give your answer correct to four decimal places.)
.8212
(e) Find P(x > 70). (Give your answer correct to four decimal places.)
0.8212 .
(f) Find P(x < 67). (Give your answer correct to four decimal places.)
.8211
• math check -
Don't need to waste your time on working this, already to late.
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11. Assume that the population of heights of male college students is approximately normally distributed with mean m of 72.15 inches and standard deviation s of 6.39 inches. A random sample of 96 heights is obtained. Show all work. …
9. ### math help
Consider the approximately normal population of heights of male college students with mean ì = 69 inches and standard deviation of ó = 4.6 inches. A random sample of 25 heights is obtained (e) Find P(x > 70). (Give your answer …
10. ### Math
Consider the approximately normal population of heights of male college students with mean ì = 69 inches and standard deviation of ó = 4.6 inches. A random sample of 25 heights is obtained. (b) Find the proportion of male college …
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