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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://docs.starkware.co/starkex/crypto/stark-curve.html | 1,713,031,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00317.warc.gz | 200,247,877 | 10,226 | # STARK Curve
The Stark-friendly elliptic curve used is defined as follows:
$y^2 \equiv x^3 + \alpha \cdot x + \beta \pmod{p}$
where:
\begin{align*} \alpha &= 1 \\ \beta &= 3141592653589793238462643383279502884197169399375105820974944592307816406665 \\ p &= 3618502788666131213697322783095070105623107215331596699973092056135872020481\\ &= 2^{251} + 17 \cdot 2^{192} + 1 \end{align*}
The Generator point used in the ECDSA scheme is:
$\begin{split}G = (874739451078007766457464989774322083649278607533249481151382481072868806602, \\ 152666792071518830868575557812948353041420400780739481342941381225525861407)\end{split}$ | 220 | 626 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.303429 |
https://www.civilengineeringhandbook.tk/structural-engineering/63.html | 1,550,849,457,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247518425.87/warc/CC-MAIN-20190222135147-20190222161147-00025.warc.gz | 809,879,942 | 8,326 | ## 63
in which
After solving {Ax8} from Equation 3.100 as
{Ax8} = [^]-1{AF} (3.103) {Ax} is determined by the following formula:
The incremental velocity vector, {Ax:}, and displacement vector, {Ax}, are obtained from
At2 At 2
Total displacement, velocity, and acceleration vectors are then determined from
3.6.4 Stability Condition and Selection of Time Interval
### 3.6.4.1 Stability
Stability of a numerical integration method requires that any error in displacement, velocity, and acceleration at time t does not grow for different incremental time intervals used in the integration. Therefore, the response of an undamped system subjected to an initial condition should be a harmonic motion with constant amplitude that must not be amplified when different Ats are employed in the analysis. Stability of an integration method can thus be determined by examining the behavior of the numerical solution for arbitrary initial conditions based on the following recursive relationship of a single d.o.f. motion:
[33(t + At), x (t + At), x(t + At )]T = [A][33(t), x (t), x(t)]T (3.107b)
where [A] is an integration approximation matrix. If we start at time t, and take n time steps, Equation 3.107 may be expressed as
To investigate the stability of an integration method, we use the decomposed form of matrix [A] in Equation 3.108 as
where [ln] is a diagonal matrix with eigenvalues in, I2, l" in the diagonal position; and is a modal matrix with eigenvectors F1, F2, and F3. Now, define the spectral radius of matrix [A] as r(A)=max|li |; i = 1,2,3 (3.110)
in order to keep [A]n in Equation 3.108 from growing without bound. The condition of Equation 3.111 is known as the stability criterion for a given method.
Numerical results for the Newmark method (linear acceleration and constant acceleration) and the Wilson-0 method from At/T = 0.001 to At/T = 100 are plotted in Figure 3.30. It can be seen that the spectral radius for linear acceleration is stable (r(A) < 1) at approximately At/T< 0.55 and becomes unstable (r(A) > 1) at At/T > 0.55. The stability of this method depends on the magnitude of At, and is called the conditional stability method. However, the spectral radii for the constant acceleration method in the range of At/T = 0.001 to 100 are all less than or equal to 1 (r(A) < 1); this case is called the unconditional stability because it does not depend on the magnitude of At. The Wilson-0 method with 0 = 1.4 is unconditionally stable and it becomes conditionally stable with 0 = 1.36. For unconditional stability, the solution is not divergent even if time increment At is large.
### 3.6.4.2 Selection of At
Numerical error, sometimes referred to as computational error, is due to the incremental time-step expressed in terms of At/T. Such errors result not from the stability behavior from two other sources: (1) externally applied force or excitation and (2) number of d.o.f. assigned to a vibrating system. A forcing function, particularly an irregular one such as earthquake ground motion, is composed of a number of forcing periods (or frequencies). A larger At may exclude a significant part of a forcing function. That part is associated with smaller periods, a forcing function's higher modes. This error may occur for both single- and multiple-d.o.f. vibrating systems. Therefore, At must be selected small enough to ensure solution accuracy by including the first several significant vibrating modes in the analysis.
0 0 | 831 | 3,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-09 | latest | en | 0.913888 |
https://www.chegg.com/homework-help/introductory-dc-ac-circuits-6th-edition-chapter-7.3-solutions-9780131140066 | 1,558,385,852,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256147.15/warc/CC-MAIN-20190520202108-20190520224108-00033.warc.gz | 741,135,464 | 44,060 | # Introductory DC/AC Circuits (6th Edition) Edit edition Solutions for Chapter 7.3
We have solutions for your book!
Chapter: Problem:
State the three-step procedure used to calculate the voltage drop across each part of a series-parallel circuit.
Step-by-step solution:
Chapter: Problem:
• Step 1 of 1
The following are the three steps that are to be followed to determine the voltage drop across each part in a series-parallel circuit:
Step 1: Determine the total resistance of the series-parallel circuit by following the three-step procedure in which equivalent resistance of all the series-connected components is found first, then equivalent resistance of all the parallel-connected components is found and then finally the total resistance of the remaining series-connected components is found.
Step 2: Determine the total current flowing in the series parallel circuit.
Step 3: Determine the voltage across each part in the series-parallel circuit by multiplying the total circuit’s current with the resistance of that part.
Corresponding Textbook
Introductory DC/AC Circuits | 6th Edition
9780131140066ISBN-13: 013114006XISBN: Nigel P. CookAuthors:
Alternate ISBN: 9780131139916 | 256 | 1,197 | {"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-2019-22 | latest | en | 0.862892 |
https://www.physicsforums.com/threads/confused-about-symbol-use-in-inverse-function.615406/ | 1,537,691,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159165.63/warc/CC-MAIN-20180923075529-20180923095929-00271.warc.gz | 824,045,541 | 14,618 | # Homework Help: Confused about +- symbol use in inverse function
1. Jun 20, 2012
### priceofcarrot
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Hi, so this isn't a question, it's just an example that they've given, but I don't understand the explanation given.
You have :
y = x^2 - 4
x = y^2 - 4
y^2 = x + 4
y = ± sqrt(x+4)
I don't get why there is a ± symbol there. My book says that it's necessary because there are two values for y that will satisfy the equation, and that if x = 0, y could be +2 or -2.
I understand that y could = +2, because sqrt 4 = +2, but I don't see how it could equal -2.
How would I know that I should include the ± symbol in front of the sqrt(x+4)? Thanks
2. Jun 20, 2012
### Ray Vickson
What is (-2)*(-2)?
RGV
3. Jun 20, 2012
### AmritpalS
Whenever you take the square root of something it requires a +or- because the square of either the negative or positive value of that term would yield the same number when it is squared. for instance (-x)^2=(x)^2
since you are squaring y, you must be aware that the sqrt of (x+4) will net y regardless if it is positive or negative
4. Jun 20, 2012
### priceofcarrot
AHHHH! I get it.
This forum rocks, thanks!
5. Jun 20, 2012
### AmritpalS
indeed it does
6. Jun 21, 2012
### HallsofIvy
Note, by the way, that what this is saying is that the original function does NOT HAVE an inverse! A function has an inverse if and only if it is "one to one". That is, there is only one value of x that gives a specific y value. It that is not true, we can choose a specific one of the x values for a given y value, as here choosing "+" or "-", which is equivalent to choosing a subset of the original function.
7. Jun 21, 2012
### dimension10
For example, if you have $p^2=2$, then $p=\pm\sqrt2$. Same thing here.
Last edited: Jun 21, 2012
8. Jun 21, 2012
### AmritpalS
±√2 that is
9. Jun 21, 2012 | 590 | 1,957 | {"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} | 3.703125 | 4 | CC-MAIN-2018-39 | latest | en | 0.909452 |
https://community.babycenter.com/post/a70171970/if-im-20-weeks-and-im-5-months-and-due-in-febuary-how-many-months-willi-be-next-month | 1,545,176,754,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829997.74/warc/CC-MAIN-20181218225003-20181219011003-00435.warc.gz | 559,634,923 | 27,690 | TODAY'S MOST POPULAR POSTS
# If im 20 weeks and im 5 months and due ,in febuary how many months will,i be next month
Posted 09/21/2018
/ 2
Posted 09/21/2018
π€¦πΌββοΈπ€·πΌββοΈ
Posted 09/21/2018
Posted 09/21/2018
If your 5 months now, next month you will be 6 months pregnant..... it takes 10 months to make a baby. Itβs BS the world thinks it takes 9 months, but whatever. So in February you will be 10 months pregnant.....your EDD (estimated due date) will be the 10 month mark.
Posted 09/21/2018
And there you have it π
--
DD 05/31/05 DD 04/22/14 DD TWINS 02/15/2016 EDD 02/25/19
Posted 09/21/2018
Thank god there is a graphic for this
Last edited 09/22/2018
It does not take 10 months to make a baby. When did you conceive? Most of us on here conceived in May and are due in February. How many months is that?
May-June: 1 month
June-July: 2 months
July-Aug: 3 months
Aug-Sep: 4 months
Sep-Oct: 5 months
Oct-Nov: 6 months
Nov-Dec: 7 months
Dec-Jan: 8 months
Jan-Feb: 9 months
4 weeks is not a month. If it were, a year would have 13 months. (52 divided by 4 = 13)
If you want to go by months, an easy way to do that is to use the same LMP your pregnancy is based off of. For me, that was May 6th. Every 6th of the month is a new month for me. September 6th marked 4 months. October 6th will be 5 months. And so on. Iβm due February 10th, so this ALMOST catches the total days over 9 months. (Just off by only a couple days... this is why the LMP is a pretty good date to use.)
If you want to get super technical...
365/12 = 30.42 average days in a month
280 days of pregnancy/ 30.42 = 9.2 months
30.42 x .2 = 6.1 days
Average pregnancy is 9 months and 6.1 days long.
Chelseahartman · Original Poster
Posted 09/22/2018
It does not take 10 months to make a baby. When did you conceive? Most of us ...
Im,due February 10th too so sweet ππππ
Last edited 09/22/2018
Im,due February 10th too so sweet ππππ
Aww!! Hi, due date, buddy!! ππΌππΌ
Β
And! I just noticed your screen name... my about-to-be 3 year old is Chelsea, too! πβ₯οΈ
Posted 09/22/2018
Aww!! Hi, due date, buddy!! ππΌππΌ Β And! I just noticed your scre...
I am also February 10th!
Posted 09/22/2018
--
Married 10.30.10 <3
ODD 6.11.12, Angel Baby 5.13, MDD 2.7.14, YDD due 2.1.19
/ 2
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Your baby's muscles and lungs are continuing to mature at 29 weeks pregnant, and her head is growing to make room for her developing brain. Read More
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Nominee: The nominee | 1,066 | 3,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-51 | latest | en | 0.936709 |
https://amp.doubtnut.com/question-answer/the-distance-of-the-point-xy-from-its-image-in-x-axis-is--39119 | 1,590,714,635,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00118.warc.gz | 249,828,165 | 19,462 | IIT-JEE
Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
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Question From class 11 Chapter JEE MAINS
The distance of the point from its image in x-axis is :
Find the equation of the locus of a point which moves so that its distance from the x-axis is double of its distance from the y-axis.
1:09
Write the equation of locus of a point whose distance from y-axis is always equal to the double of its distance from x-axis.
1:26
A point is located in the third quartile. Its distance from the x-axis and y-axis is 4 and 5 units. Find its coordinates.
2:05
What is the perpendicular distance of the point (x, y) from x-axis ?
1:29
A point is located in the third quartile. Its distance from x-axis and y-axis are 4 and 5 units respectively. Find its coordinates.
1:19
Square of the distance of the point from x-axis is double of its distance from the origin.
1:42
A point moves so that its distance from y-axis is half of its distance from the origin. Find the locus of the point.
1:37
Find the locus of a point whose distance from (a, 0) is equal to its distance from the y-axis.
1:22
A point is at a distance of units from -axis and units -axis. Represent the position of the point in from the y the Cartesian plane and also write its Cartesian coordinates.
1:12
Find the locus of a point which moves such that its distance from the origin is three times its distance from x-axis.
1:16
The equation of the locus of the point whose distance from the x-axis is twice that of from the y-axis is :
2:17
Find the distance of the z-axis from the image of the point in the plane
5:09
Find the distance of a point from X-axis.
0:51
A circle touches the x-axis and intersects the constant length 2k from the y-axis. Prove its center point is.
1:52
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MicroConcepts | 796 | 3,173 | {"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-2020-24 | latest | en | 0.937659 |
https://javacodepoint.com/logical-programs/how-to-swap-two-numbers-without-using-a-third-variable-in-java/ | 1,725,838,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00092.warc.gz | 304,988,706 | 65,872 | # How to swap two numbers without using a third variable in Java?
Swapping two numbers can be also known as exchanging the numbers between two variables. The swapping of two numbers without using a third variable or a temporary variable can be done by following the below simple steps:
For example, let’s take two numbers x=20 (first variable) and y=30 (second variable),
1. Add both number (x+y) and store it in first variable (x). ie x = x + y
2. Now Substract the second number from the first, and store it in y variable. ie. y = x – y
3. Now to the step two again but store the value in x variable. ie. x = x – y
Let’s see the calculation steps:
x = x + y => x = 20 + 30 => 50
y = x – y => y = 50 – 20 => 30
x = x – y => x = 50 – 30 => 20
Let’s see the logic to do the same,
### Solution
SwapTwoNumbers.java
```package com.javacodepoint.programs;
public class SwapTwoNumbers {
public static void main(String[] args) {
//Initiating two integer numbers
int x = 20, y=30;
//Print both number before swapping
System.out.println("x = "+x+", y= "+y);
//Logic to swap these numbers without using a third variable
x = x + y;
y = x - y;
x = x - y;
//Printing the both number again after swapping
System.out.println("x = "+x+", y= "+y);
}
}
```
OUTPUT:
x = 20, y= 30
x = 30, y= 20 | 371 | 1,296 | {"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-2024-38 | latest | en | 0.794332 |
http://www.physicsforums.com/showthread.php?t=379669 | 1,369,498,672,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705958528/warc/CC-MAIN-20130516120558-00052-ip-10-60-113-184.ec2.internal.warc.gz | 651,488,172 | 14,201 | ## Physics of building collapse
I am having trouble understanding a building collapse. According to NIST, on 9/11 WTC7 collapsed after fire weakened its structural steel. They have however admitted that WTC7 fell for 2.25 seconds at free fall. (http://wtc.nist.gov/NCSTAR1/PDF/NCSTAR%201A.pdf) However isn't all the gravitational potential energy of the building used up in attaining free fall? Why did the structure in the way not slow the collapse in any way during this period? Fire can't blow out 8 stories of structure simultaneously and continuously floor by floor. Energy and momentum don't seem to be conserved here which is a violation of the laws of physics. Can anyone explain how I am wrong here?
Mentor
Quote by cmatrix I am having trouble understanding a building collapse. According to NIST, on 9/11 WTC7 collapsed after fire weakened its structural steel. They have however admitted that WTC7 fell for 2.25 seconds at free fall. (http://wtc.nist.gov/NCSTAR1/PDF/NCSTAR%201A.pdf) However isn't all the gravitational potential energy of the building used up in attaining free fall? Why did the structure in the way not slow the collapse in any way during this period? Fire can't blow out 8 stories of structure simultaneously and continuously floor by floor. Energy and momentum don't seem to be conserved here which is a violation of the laws of physics. Can anyone explain how I am wrong here?
Welcome to the PF. Two things. First, we do not allow discussions of conspiracy theories here on the PF. That's pretty clear in the Rules link at the top of the page.
Second, sure, it's consistent with some failure mechanisms due to melting structural beams. How far does something fall from rest in 2.25 seconds? This is a physics forum, after all. You should be able to answer that question for us. How many stories does that equal?
Mentor
Quote by cmatrix I am having trouble understanding a building collapse. According to NIST, on 9/11 WTC7 collapsed after fire weakened its structural steel. They have however admitted that WTC7 fell for 2.25 seconds at free fall. (http://wtc.nist.gov/NCSTAR1/PDF/NCSTAR%201A.pdf) However isn't all the gravitational potential energy of the building used up in attaining free fall?
Yes, gravitational potential energy is what causes free-fall.
Why did the structure in the way not slow the collapse in any way during this period? Fire can't blow out 8 stories of structure simultaneously and continuously floor by floor. Energy and momentum don't seem to be conserved here which is a violation of the laws of physics. Can anyone explain how I am wrong here?
If you don't know how much energy a steel beam can absorb, you can't know that it should have absorbed more. Buildings aren't cars: they don't have "crumple zones" designed to absorb energy. They are designed to withstand a certain static force, but have very little ability to absorb energy. In short, steel beams don't retain much strength as they deform and break.
Try this: Take a soda straw and while stabilizing it with one hand to hold it straight upright, push it against a table with your other hand. You can put quite a bit of force to it without breaking it, can't you? 5-10 pounds maybe? Then let go with your other hand, allowing it to collapse. Once it buckles, it offers only a tiny fraction of the resistance it offered before - an ounce or two of force.
## Physics of building collapse
OK, so what you write is that "free fall" means an "almost free fall" or "for all practical purposes free fall"?
Mentor
Quote by Borek OK, so what you write is that "free fall" means an "almost free fall" or "for all practical purposes free fall"?
Right - close enough (within a few percent) that you can't tell the difference by freeze-framing/clocking a youtube video.
Mentor ...and it is actually even a little worse than my straw example: plastic is ductile so the straw doesn't snap it just buckles. When a steel beam fails it only buckes a little before snapping...and the faster the impact the less buckling.....and a snapped beam offers no resistance at all.
berkeman, I am not presenting a theory merely a question that I am extremely disturbed about. That should have been pretty clear. You have not answered my question but instead ask me an entirely irrelevant and bizarrely basic question. But, 9.8 x 2.25 = 22.05 meters or about 72 feet which is about 8 stories. This is a physics forum, after all. You should been able to answer that question yourself. russ_watters, massive interconnected steel beams, floor pans and reinforced concrete will absorb some energy and momentum, not none at all. With free fall there is no energy and momentum at all being absorbed. The conservation laws seem to be egregiously violated here. Again exactly what am I missing here?
Recognitions: Gold Member Science Advisor cmatrix You seem, already, to have posted your original question bringing with you answers that satisfy you. Berkeman's comments seem quite reasonable to me. You need a structural engineer to give you accurate calculations so you must either learn the stuff yourself or pay someone to do them for you. If you are "extremely disturbed" then you must have either done some sums of your own or read someone's 'theory'. Perhaps you should get more familiar with the Physics involved before you get too worried. I don't know what you mean when you say that "With free fall there is no energy and momentum at all being absorbed" but, if you drop a brick on your foot, you will be aware of a certain amount of energy being available after a very short distance of "free fall". If you drop the top section of a building through a distance of several metres then there is a a lot of Kinetic Energy available to do damage to the lower bits, when it lands on them. Because of the momentum changes involved, the next section just below the collapsed section gets the majority of the impulse after each increment of the collapse. Most building are never designed to cope with that sort of impact. Once the lower sections have broken and have not absorbed all the energy, they will add to the amount of energy available for damage in subsequent collapses. It all depends upon whether the structure can absorb the energy of collapse of each successive floor or not. If buildings were built with that eventuality in mind then they would have to be so expensive that no one could afford the rent! Only when you have done the actual calculations and proved that mechanism will not work should you go for the conspiracy theory. This Forum doesn't want to get involved with the conspiracy idea and we are all the better for that decision.
Admin I don't see a conspiracy theory here and I think I understand where cmatrix problem arise. At least my first idea was that some of the energy of the upper part of the building has to be used for smashing lower stories, and that it should slow down the fall - that's exactly the problem s/he addresses. If part of the energy is used to destroy objects that lie below, what we observe is not a free fall - acceleration of the falling object should be lower than g. That's perfectly sound physics, perhaps idealized one. cmatrix: please read what I wrote earlier and what Russ answered. You are right that it is not a free fall, however, once the support beams snapped, whatever was left was too weak to substantially slow down fall of the upper part of the building. For all practical purposes fall can be approximated by free fall.
Quote by sophiecentaur cmatrix You seem, already, to have posted your original question bringing with you answers that satisfy you. Berkeman's comments seem quite reasonable to me. You need a structural engineer to give you accurate calculations so you must either learn the stuff yourself or pay someone to do them for you. If you are "extremely disturbed" then you must have either done some sums of your own or read someone's 'theory'. Perhaps you should get more familiar with the Physics involved before you get too worried. I don't know what you mean when you say that "With free fall there is no energy and momentum at all being absorbed" but, if you drop a brick on your foot, you will be aware of a certain amount of energy being available after a very short distance of "free fall". If you drop the top section of a building through a distance of several metres then there is a a lot of Kinetic Energy available to do damage to the lower bits, when it lands on them. Because of the momentum changes involved, the next section just below the collapsed section gets the majority of the impulse after each increment of the collapse. Most building are never designed to cope with that sort of impact. Once the lower sections have broken and have not absorbed all the energy, they will add to the amount of energy available for damage in subsequent collapses. It all depends upon whether the structure can absorb the energy of collapse of each successive floor or not. If buildings were built with that eventuality in mind then they would have to be so expensive that no one could afford the rent! Only when you have done the actual calculations and proved that mechanism will not work should you go for the conspiracy theory. This Forum doesn't want to get involved with the conspiracy idea and we are all the better for that decision.
I don't need to do any calculations. I understand how gravity, energy and momentum work in this respect. When the kinetic energy from a falling mass is used to do damage, the mass' rate of decent is slowed. Agree or disagree? If a theory involves no slowing whatsoever you obviously have a violation of the conservation laws. A building free falling for 2.25 secs has no structure whatsoever beneath it because it's rate of decent would slow if any structure was encountered. Consider dropping your brick onto 47 sheets of suspended glass strong enough to support this structure and the brick at the top. Is the brick going to free fall all the way to the bottom?
Quote by cmatrix [...]will absorb some energy and momentum, not none at all. With free fall there is no energy and momentum at all being absorbed. [...]exactly what am I missing here?
You're missing that nobody really meant "literally, exactly free fall".
The idea is like (as Russ was saying) that you might be able to completely support the weight of a brick with just a straw, but if you lift that brick up just one centimetre higher and then drop it, the straw will likely buckle , letting the brick will fall down almost as fast as if the straw wasn't there at all. Not "exactly" as fast (which as you say would violate conservation laws), but close enough that you probably couldn't easily detect the difference (go take a stopwatch and tell us how much the straw slows the brick compared to when you do it without the straw. Then, if you want to get technical with us about free-fall, go time the brick falling in a complete vacuum).
"Two and a quarter seconds" doesn't mean 2.2500000000000+/-0.0000000000013s. You should be able to calculate an estimate of the uncertainty in how much decelerating force could have been exerted by the lower structure, based on the observations.
Recognitions: Gold Member Science Advisor The brick and glass model is not the WTC model at all. Each of your sheets of glass would have to be of significant mass and the structure of your model would have to be 'only just' strong enough to support itself. Neither was the whole building nor the top portion of the tower "in free fall". You are not applying conservation laws in a relevant way (you haven't actually said where they 'fail'). Of course you would need to do calculations (in a lot of detail) to analyse the scenario properly. This is my version / explanation of the sort of ting that must have happened (discounting the possibility of a demolition job) The top portion of the tower fell when a mid section became too weak to support it. It fell several metres (a couple of floors), not under free fall but not supported sufficiently and still accelerating ('almost free fall'). It then hit the top floor of what remained of the building below at some significant speed. As I said, momentum considerations meant that the majority of the stress was on the supporting section just below that floor. (Look at pictures of the front coaches of trains in collisions - they are the ones that get concertina'd). That section collapsed, slowing the falling mass a little but not bringing it to a halt. The top - plus this floor - then continued down, accelerating whilst falling the next few metres to the next level and, of course, weighing a bit more. After falling by a further few metres it was, again going fast enough to break the next section supports. Each successive floor collapsed and each time there was more weight on top of it. Although I should imagine that the lower sections were built stronger than the upper sections, they would have been heavier, as a result, so my scenario would also have applied. I would think that, had the planes hit nearer the top, this domino effect might not have occurred. If the floor, immediately below the impact / fire had been able to withstand the initial impact of the higher portion falling on it then the collapse would have stopped there. As it was, there was about a third of the structure above the break - a massive hammer with which to start the collapse. I am sure the terrorists were as surprised as anyone else that the towers went down the way they did.
Mentor
Quote by cmatrix I don't need to do any calculations. I understand how gravity, energy and momentum work in this respect. When the kinetic energy from a falling mass is used to do damage, the mass' rate of decent is slowed. Agree or disagree?
Disagree -- not necessarily. Remember that there is an accelerating force downward due to gravity, and a retarding force that varies, but can be less than or greater than the gravitational force. The retarding force depends on what bits and pieces and such are being plowed through on the way down. The sum of the forces F = ma, so you can still see a downward acceleration.
Especially for the first 2.25 seconds. You didn't answer my quiz question, and nobody answered it for you, but that's about 1 floor worth of falling. Pretty plausible as one floor's support beams gave way....
Recognitions:
Gold Member
Quote by berkeman Especially for the first 2.25 seconds. You didn't answer my quiz question, and nobody answered it for you, but that's about 1 floor worth of falling. Pretty plausible as one floor's support beams gave way....
The OP didn't do the calculation right, gravitational freefall for 2.25 seconds is a displacement of 24.8 meters.
Quote by Borek cmatrix: please read what I wrote earlier and what Russ answered. You are right that it is not a free fall, however, once the support beams snapped, whatever was left was too weak to substantially slow down fall of the upper part of the building. For all practical purposes fall can be approximated by free fall.
I have read what you and Russ have written but it is not helping me in any way. You admit there could be no free fall but the whole problem is that NIST admits there was free fall for 2.25 seconds. I understand your point that what is seen as free fall is not necessarily precisely free fall but that seems immaterial. The problem is your claim that the structure left was too weak to offer any significant resistance. You do not explain then how fire can effectively remove simultaneously and continuously floor by floor 8 stories of structure that held a building up for decades.
Mentor
Quote by Pengwuino The OP didn't do the calculation right, gravitational freefall for 2.25 seconds is a displacement of 24.8 meters.
Oops, that's my math error, not the OPs. I brain faded to 25 feet, not 25 meters. More like 7 stories then? Thanks Pengwuino.
Recognitions:
Gold Member
Quote by berkeman Oops, that's my math error, not the OPs. I brain faded to 25 feet, not 25 meters. More like 7 stories then? Thanks Pengwuino.
You both did it wrong, I killed 2 PF members with 1 stone. Or at least corrected them. | 3,464 | 16,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2013-20 | latest | en | 0.964619 |
http://fb-list-archive.s3-website-eu-west-1.amazonaws.com/firebird-support/2012/8/119129.html | 1,632,043,280,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00556.warc.gz | 22,716,744 | 2,930 | Subject RE: [firebird-support] Sorting strings logical (StrCmpLogicalW) ? Svein Erling Tysvær 2012-08-17T06:44:02Z
>Hello *
>
>Is there a way to sort string data (varchar) in Firebird by a "natural"
>or "logical" sort order, just like using the Windows API function StrCmpLogicalW [1] ?
>
>Sort result of text with numbers should be:
>
>Item1
>Item2
>Item3
>Item10
>Item11
>Item20
>Item30
>
>That is, sorting should treat the digits as numbers rather than text.
>
>[1]
> http://msdn.microsoft.com/en-us/library/bb759947.aspx
Unfortunately, I don't think so. If the text is as ordered as your example indicates, then you can of course do:
SELECT MyItem
FROM Items
ORDER BY CAST(SUBSTRING(MyItem FROM 5 for 2) AS INTEGER)
but you will need to add considerably more logic if you want to sort like the Microsoft example you supply (though it is doable, e.g. you could have a separate field or stored procedure that e.g. contained/returned Item000000003, Item000000010 that you only used for sorting and never displayed).
Here's an EXECUTE BLOCK that does something like this:
execute block returns(MySortField varchar(100))
as
declare i int = 0;
declare i2 int = 0;
declare s varchar(100) = '';
declare s2 varchar(32) = '';
declare i0 int = 0;
begin
for select myvarchar from test into :s2 do /* MyVarchar (in TEST) is defined as VarChar(32) */
begin
i = 1;
s = '';
while (i <= 32) do
begin
i2 = 0;
while (substring(s2 from i+i2 for 1) between '0' and '9') do
i2 = i2+1;
if (i2 = 0) then
s=s||substring(s2 from i for 1);
if (i2 > 0) then
begin
i0 = 9-i2;
while (i0 > 0) do
begin
s=s||'0';
i0=i0-1;
end
s=s||substring(s2 from i for i2);
i=i+i2-1;
end
i=i+1;
end
MySortField = s;
Suspend;
end
end
Of course, things become a bit more complex if you want the number to contain decimals and you want this sort order:
Item1.23
Item1.5
Item5.1
Item23.1
(numerically speaking, 23 > 5 if it is before the decimal point, whereas 23 < 5 if it is after)
HTH,
Set | 623 | 1,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.693836 |
https://www.doorsteptutor.com/Exams/CSIR/Physical-Sciences/Questions/Topic-Mathematical-Methods-of-Physics-0/Subtopic-Dimensional-Analysis-0/Part-1.html | 1,503,190,231,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00659.warc.gz | 903,815,061 | 20,327 | # Mathematical Methods of Physics-Dimensional Analysis (CSIR Physical Sciences): Questions 1 - 3 of 3
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 401 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Passage
A resistance is measured by passing current through it and measuring the resulting voltage drop. If the voltmeter and the ammeter have uncertainties of and , respectively, then – (June)
## Question number: 1 (1 of 2 Based on Passage) Show Passage
» Mathematical Methods of Physics » Dimensional Analysis
Appeared in Year: 2011
MCQ▾
### Question
The uncertainty in the value of resistance is:
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 2 (2 of 2 Based on Passage) Show Passage
» Mathematical Methods of Physics » Dimensional Analysis
Appeared in Year: 2011
MCQ▾
### Question
The uncertainty in the computed value of the power dissipated in resistance is:
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 3
» Mathematical Methods of Physics » Dimensional Analysis
MCQ▾
### Question
Using dimensional analysis, Plank defined a characteristic temperature from powers of the gravitational constant , Planck’s constant , Boltzmann constant and the speed of light in vacuum. The expression for is proportional to – (June)
### Choices
Choice (4) Response
a.
b.
c.
d.
f Page | 353 | 1,500 | {"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-2017-34 | longest | en | 0.73664 |
https://www.physicsforums.com/threads/suvat-ball-thrown-off-a-roof.809313/ | 1,521,699,779,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647777.59/warc/CC-MAIN-20180322053608-20180322073608-00152.warc.gz | 848,318,832 | 16,515 | # Suvat - Ball thrown off a roof
Tags:
1. Apr 18, 2015
1. The problem statement, all variables and given/known data
A ball is thrown vertically upwards from a roof of a building and it lands 3 seconds later on the ground 7 meters below the roof. Calculate:
a) the speed with which the ball was thrown upwards
b) the maximum height of the ball above the ground
c) the speed with which the ball hits the ground
2. Relevant equations
v = u + at
s = ut + (1/2)(a)(t^2)
s = vt - (1/2)(a)(t^2)
v^2 = u^2 + 2as
s = ((u + v)/2)t
3. The attempt at a solution
I was trying to help a friend with this problem. I have tried breaking it down to 3 sections, up, down to starting height and down to the ground; but there never seems to be enough information to solve anything. I can't tell if I am missing something obvious or something was left out of the question in the book.
Up
s = ? u = ? v = 0m/s a = -9.8m/s^2 t = ?
Down
s = ? u = 0 v = ? a = -9.8m/s^2 t = ?
To Ground
s = 7 u = ? v = ? a = -9.8m/s^2 t = ?
2. Apr 18, 2015
### AlephNumbers
Nothing was left out of the question in the book; the problem is quite solvable. Your breaking of the problem into three parts is a step in the right direction. Try thinking about this: If you throw a ball upwards, what must be its velocity when it comes back down to the height from which it was thrown upwards? What is the ball's velocity at its maximum height? How can you express the time it takes the ball to reach its maximum height in terms of its initial velocity and the constant acceleration it undergoes due to gravity?
Just realized why some people call the equations for one dimensional motion with constant acceleration suvat.
Last edited: Apr 18, 2015
3. Apr 19, 2015
### ehild
The solution is much simpler if you use displacement (y) in terms of the time. Deciding "up" positive, the initial velocity vi is positive, the acceleration is a= -g, and the final displacement is yf= -7m. No need to break the problem.
The equation for the displacement y=vit-g/2 t2 is valid for the whole motion. Substitute the given data for the final time and displacement, you get the initial velocity.
After that, you can use the equation for velocity v = vi - gt, to find out the maximum height and the final velocity. | 612 | 2,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-13 | latest | en | 0.949078 |
https://newproxylists.com/algebraic-geometry-open-morphism-of-schemes/ | 1,558,822,586,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258451.88/warc/CC-MAIN-20190525204936-20190525230936-00269.warc.gz | 552,780,143 | 9,203 | # Algebraic Geometry – Open Morphism of Schemes
Let $$f: X to S$$ a finite morphism between affine schemas $$X = Spec (A), S = Spec (R)$$. Note by $$phi: R to A$$ the corresponding circular card.
I am looking for pure theoretical / algebraic ring tools / criteria to define if $$f$$ is a open the map (in the topological sense). More concretely in the sense of what conditions the ring morphism $$phi$$ and resp induced morphisms $$phi_p: R_p to A_p$$ on the family of locations to $$p$$ implies that $$f$$ is open.
The bottom of my question is the following thread: The Finite and Locally Free map is open.
Here we have the situation $$f$$ is a finite morphism and locally free and I want to infer that this already implies that $$f$$ is open.
Obviously, the problem is local so we can work with the above setting and assume that $$X , S$$ affine and $$A = R ^ n$$ as $$R$$-module since $$phi$$ is in the given context exactly the map $$R to R ^ n$$.
The author observes that, due to local weakness, the stems of $$f _ * mathcal {O} _X$$ are nonzero on an open subset of $$S$$.
What does he mean? That at every point $$s in S$$ there is a rod in $$( phi _ * mathcal {O} _X) _s cong mathcal {O} ^ n_ {S, s}$$ which can be extended to a section on an open subset $$U subset S$$? Is it not regulated by definition stems as direct limit representatives?
Again, since the problem is local so wlog $$U = D (f)$$ or $$f in R$$. Why does this stem condition imply that $$f$$ is open? Does this stem from a more general criterion of openness based on switched algebra methods? | 413 | 1,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-22 | longest | en | 0.93586 |
https://brainly.in/question/4612 | 1,487,579,853,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170434.7/warc/CC-MAIN-20170219104610-00513-ip-10-171-10-108.ec2.internal.warc.gz | 697,340,621 | 9,987 | # What is the product of the integers x,y,z equal to 1
2
by tinkut
questions seems incomplete
• Brainly User
2014-03-10T15:01:39+05:30
what is the product of the integers x,y,z equal to 1
x=1,y=1,z=1, i.e x*y*z=1
2014-03-10T15:50:27+05:30
According to question-
1=x=y=z
product will be -
x×y×z
putting the values
we get,x×y×z=1×1×1
=1
according to question-
1=x=y=z
product will be -
x×y×z
putting the values
we get,x×y×z=1×1×1
=1 | 181 | 432 | {"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} | 3.5 | 4 | CC-MAIN-2017-09 | latest | en | 0.730493 |
http://sciforums.com/threads/this-thread-is-for-rpenner.143300/page-6#post-3253433 | 1,695,439,896,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00422.warc.gz | 38,990,989 | 13,956 | # this thread is for Rpenner
Discussion in 'Physics & Math' started by Jason.Marshall, Dec 8, 2014.
1. ### OnlyMeValued Senior Member
Messages:
3,914
Have you been following the whole discussion? Russ did not introduce those equations. He was commenting on the fact that they were meaningless..., in the context they had been introduced!
Step back and look at the discussion, unless your intent truely is to be trolling.
3. ### Russ_WattersNot a Trump supporter...Valued Senior Member
Messages:
5,051
Similar answer: you should start by completing the remedial pre-algebra course, then go on to algebra. By the end, you should understand that you cannot set r=1, but rather you must solve the equation to find r.
5. ### Russ_WattersNot a Trump supporter...Valued Senior Member
Messages:
5,051
Minor clarification: *I* created those equations, but for the reason you say: They are a commentary on/similar example to the meaningless equation John created.
The bizarre part is that by MD quoting the post, he had another opportunity to read it, so there should be no ambiguity here. MD seems to want to blast me for them being meaningless - but not only did he swing and miss on why they are meaningless, I know they are meaningless! I said so at the bottom of the post!
7. ### OnlyMeValued Senior Member
Messages:
3,914
I did not go back to actually check, so it was likely the underlying intent/context I confused. Perhaps I should take some of my own advice, on occassion?
8. ### Jason.MarshallBannedBanned
Messages:
654
Now the time has come to formally put this issue to rest since tan(3.125/4) is less than one then pi must be rational
And when tan(pi/4)= 1 does not describe a natural circle but it describes a polygon so this answer will always = 1 even if you calculate the ratio of a polygons perimeter to its diameter until infinity this is a polygonal equivalent to a circle it will never truly equal a curve that is why you have to calculate it to infinity but a true circle will equal less than one, thus the definition of a polygon does not satifisfy a correct description based on the definition of a circle so Archimedes method is fundamentally flawed from the beginning. Modern mathematics was able to erase the corners of the polygons with use of infinite series but that is cheating because they still never really measured a circle but only it's polygonal doppelgänger the square version of a circle "8/2.22". This is also why 1*1*1*1*1... = 1 until infinity so it's follows that if pi is described as the ratio of a polygons perimeter to its diameter then yes of course its irrational but the problem is not in the math it's in the definitions visions and comprehension of the concepts.
Last edited: Dec 12, 2014
9. ### Motor DaddyValued Senior Member
Messages:
5,425
You just plain don't know what you're talking about, Russ. Look at the pretty pic. Do you know what you're looking at? Maybe rpenner can help you.
Messages:
6,539
What?
Messages:
5,051
Nyet.
Messages:
5,051
Lol.
13. ### Motor DaddyValued Senior Member
Messages:
5,425
Nope, it's no joke. Try again!!
14. ### Motor DaddyValued Senior Member
Messages:
5,425
"No more free lessons for you!"
Edit: Ooops, I forgot the credits to that - Tach. The quote, not the pic.
Last edited: Dec 12, 2014
15. ### rpennerFully WiredValued Senior Member
Messages:
4,833
http://www.sciforums.com/threads/help-needed.142484/ ( Discussion on construction of approximations to 1:√π including my discovery of 1 : ( √(3/10) + √(3/2) ) which is constructable )
The diagram in post #106 is a humbug
It was originally posted in the context of a geometric construction of the ratio 1:√π but is not a geometric construction of 1:√π . If it is a geometric construction, it is for a square of side of length in ratio with the radius of the circle of 1:√3.
The diagram in post #106 is a humbug and any claim that it is a construction of a square with sides of length √3.14 or √π is a fraud. This is easiest to see in a cleaned up and labelled diagram:
Here we see there is no connection between the trivially constructed length CD, with length in ratio AB:CD = 1:√3 and the red-bordered square WXYZ with sides in ratio AB:WX = 1:√π
This was Motor Daddy's burden to carry which he ignores.
Zooming in we see that the point C is exactly defined by the intersection of two arcs and is quite a bit separated from the red line.
Last edited: Dec 12, 2014
16. ### Motor DaddyValued Senior Member
Messages:
5,425
What is the area of your circle?
17. ### Jason.MarshallBannedBanned
Messages:
654
Anyways am writing a paper on why 1*1*1....= 1 until infinity but it seems to relate to other major mathematical concepts ...
18. ### Russ_WattersNot a Trump supporter...Valued Senior Member
Messages:
5,051
There it is! It's about time you outed yourself! Now don't you feel better now that you finally got that big pile of crap off your chest?
19. ### Motor DaddyValued Senior Member
Messages:
5,425
Heck, rpenner, while you're here, maybe you can take a look at this:
x=1
y=0.999...
z=1-0.999...
x-y=z
20. ### Jason.MarshallBannedBanned
Messages:
654
Am familiar with this Gaddys pi 3.146264 , no comment but I do notice it follows in the polygonal category and tan(3.146264/4) should still equal 1 please someone calculate for me am too lazy am not sure if there is a rational pattern but it appears to be the addition of 2 infinite series so if they produce a rational number I find that interesting.
Last edited: Dec 12, 2014
21. ### Russ_WattersNot a Trump supporter...Valued Senior Member
Messages:
5,051
Lol. If at first you don't succeed at trolling...
22. ### Motor DaddyValued Senior Member
Messages:
5,425
Did I make a mistake with my symbol letter thingies?
Messages:
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https://www.hiveworkshop.com/threads/what-does-calculate-extends-in-magos-actually-do.264632/ | 1,701,244,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00001.warc.gz | 913,166,678 | 22,460 | # What Does Calculate Extends in Magos Actually do?
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#### Hantoo
Level 42
What Does Calculate Extends in MagosModelEditor Actually do to the Model?
All I know is It fixes alot of stuff such as "Unclickable", Or something like Portrait.
What kind of magic it doned to the model? I mean, If I can change the setting Manually, would be better (i guess?)
can it be done manually?
#### GhostWolf
Level 29
It re-calculates the extents, which define the bounding sphere (or cube) of a model, which in turn is used for selection and pathing.
The idea is to get the smallest and biggest coordinates for each of the axes, by looping over all of the vertices, and that's your extent (the distance between the two resulting vertices is also saved).
#### Hantoo
Level 42
It re-calculates the extents, which define the bounding sphere (or cube) of a model, which in turn is used for selection and pathing.
The idea is to get the smallest and biggest coordinates for each of the axes, by looping over all of the vertices, and that's your extent (the distance between the two resulting vertices is also saved).
So Does it have something that changes the Unit's Health bar height? Some times my model's Health bar get super high and I don't know how to fix... Does it Relate to Calculate extends?
#### GhostWolf
Level 29
I assume the game uses the bounding shape also for that, yes.
#### Fingolfin
Level 45
The model extends are mainly for telling the game weather the model collides with the camera bounds. For instance, if you have a very large model, it's extents will have to be scaled appropriately, otherwise it will not render when the models origin point is outside of view. Extents are also used for selection - the game first checks if the mouse collides with the extents, and if it does, it checks for any bounding shapes for more precise collision.
#### Ezekiel12
Level 12
The model extends are mainly for telling the game weather the model collides with the camera bounds.
Can you explain a bit more about the rendering process? I.e. when a model is rendered/not rendered and what factors decide this. I had issues in many maps where models are not rendered though they should be (e.g. here on the right side of the screen). Is this only because of the model extents or is there some bug in wc3?
#### PurgeandFire
Level 44
Can you explain a bit more about the rendering process? I.e. when a model is rendered/not rendered and what factors decide this. I had issues in many maps where models are not rendered though they should be (e.g. here on the right side of the screen). Is this only because of the model extents or is there some bug in wc3?
Yes, it is as you linked (the big tower disappears). The camera uses the bounding box/sphere to determine whether a model should be rendered on screen (taken from the object's coordinates). If you extend that bounding sphere, you can avoid those "disappearing" models (since it'll still be considered in the field of view). But it isn't always the best idea--in cases like the video you linked, if you extended the bounding sphere, the camera might clip into the model which would look strange to the viewer.
But you should extend the bounding sphere if you scale the model in the editor. I don't think wc3 actually takes into account scaling when it checks model extents, so increasing the extents manually will give you a better result. Unfortunately you'll run into issues when you use the same model with many different scales, in which case you should just choose the extents of the largest scaled object.
I don't know which tools allow you to set extents manually (magos' editor might). But it should be easy to write a program that sets it if you need that feature. (and if you set new extents, you would just multiply the old extents by the scaling value, e.g. extent * 1.25 if you are scaling the model 1.25 in the editor).
#### Ezekiel12
Level 12
I don't think wc3 actually takes into account scaling when it checks model extents
Here we go, that's (kind of) the bug I wanted! There seems to be more to the story though.
The collision spheres/boxes are scaled with scaling value but it seems the model extents are not or only partially. For example until a certain scaling value the hp bar height increases. If you increase scaling value above that the hp bar stays at the max height. Maybe that's also the case with the extents and therefore the model rendering.
#### Fingolfin
Level 45
I don't think this is your problem. Extents should scale fine, but inaccuracies will become more evident at large scales. Also note that destructables have a hardcoded maximum for their extents, so they might dissapear even though you set their extents to something very high. In those cases it is best to use doodads if possible, since they don't have this problem.
It is possible to set extents manually in magos model editor. Just open the model, then go to edit>model properties, and there you have your extents.
#### Ezekiel12
Level 12
I don't think this is your problem. Extents should scale fine,
I was talking about ingame scaling, i.e. scaling value in Object Editor or scaling via trigger.
That's what Purges quote was about, right?
I don't think wc3 actually takes into account scaling when it checks model extents
Last edited:
#### Fingolfin
Level 45
Yes, that is what i was refering to - i do believe that scaling a unit in the object editor also scales extents, i doubt Blizzard would have missed this. Instead, think of it like this: when a model is scaled very small, the distance between the origin and the edge of the extents is so small, that you will not really notice when the origin goes outside of camera bounds. On models that are scaled very large, it's much more noticeable.
#### Ezekiel12
Level 12
Yes, that is what i was refering to - i do believe that scaling a unit in the object editor also scales extents, i doubt Blizzard would have missed this. Instead, think of it like this: when a model is scaled very small, the distance between the origin and the edge of the extents is so small, that you will not really notice when the origin goes outside of camera bounds. On models that are scaled very large, it's much more noticeable.
What I know for sure is if you place a huge Bloodmage with scaling value 40, the mouse hover detection will work fine anywhere on the model as long as it is rendered. The problem is if a small region around the center is not in the camera field anymore the model won't be rendered.
So my guess was that extents are only scaled up to a certain value. (The health bar that has a max height would confirm this).
Either that or rendering is completely independent from extents and collision shapes and just always is some kind of shape around the object..
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367 | 1,564 | 6,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-50 | latest | en | 0.940513 |
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# Systems with Infinitely Many Solutions or No Solution
Aug 22, 2023
### Systems with Infinitely Many Solutions:
A system of linear equations is consistent if it has one or more solutions.
Example 1:
Solve the following systems by graphing:
2x + 4y = 8
x + 2y = 4
Solution:
### Systems with No Solution:
A system of linear equations is inconsistent if no solutions exist.
Example 2:
Solve the following systems by graphing:
x + 2y = -4
2x + 4y = 8
Solution:
## Systems with Infinitely Many Solutions or No Solution
Example 3:
What is the solution to each system of equations?
1. { y = 4-3x}
2. { -6x-2y = -8 }
Solution:
Substitute y = 43x into the second equation.
-6x – 2y =-8
-6x – 2(4-3x) = -8
-6x -8 + 6x = -8
-6x-8+ 6x =-8
-6x+ 6x -8 =-8
-8 = -8……. True Statement
The statement – 8 = – 8 is an identity, so the system of equations has infinitely many solutions. Both equations represent the same line. All points on the line are solutions to the system of equations.
1. {3x-y =-4}
2. {y=3x-5 }
Solution:
Substitute y = 3x – 5 into the first equation.
3x-y= -4
3x-(3x-5) =-4
3x-3x+5= -4
5=-4…….. False Statement
The statement 5 = –4 is false, so the system of equations has no solution.
Example 4:
What is the solution to each system of equations?
x + y = –4
y = –x + 5
Solution:
Substitute y = –x + 5 into the first equation.
x + y = –4
x –x + 5 = –4
5 = –4…………False statement.
The statement 5 = –4 is false, so the system of equations has no solution.
y = –2x + 5
2x + y = 5
Solution:
Substitute y = –2x + 5 into the second equation.
2x + y = 5
2x + (–2x + 5) = 5
2x –2x + 5 = 5
5 = 5…………….True statement.
The statement 5 = 5 is an identity, so the system of equations has infinitely many solutions. Both equations represent the same line. All points on the line are solutions to the system of equations.
### Model Using Systems of Equations
Example 5:
Funtime Amusement Park charges \$12.50 for admission and then \$0.75 per ride. River’s Edge Park charges \$18.50 for admission and then \$0.50 per ride. For what number of rides is the cost the same at both parks?
Solution:
Formulate:
Write a system of linear equations to model the cost of both parks.
In both equations, let y represent the dollar amount of charges. Let x represent the number of rides.
y = 0.75x + 12.5
y = 0.5x + 18.5
Compute:
Substitute for y in one of the equations.
y = 0.75x + 12.5
0.5x + 18.5 = 0.75x + 12.5
0.5x + 18.5 – 12.5 = 0.75x
0.5x + 6 = 0.75x
6 = 0.75x – 0.5x
6 = 0.25x
x = 6/0.25
x = 24
Interpret:
Since x is the number of rides, for 24 rides the cost will be the same at both parks.
Example 6:
At a hot air balloon festival, Mohamed’s balloon is at an altitude of 40 m and rises at 10 m/min. Dana’s balloon is at an altitude of 165 m and descends at 15 m/min. In how many minutes will both balloons be at the same altitude?
Solution:
Formulate:
Write a system of linear equations to model the cost of both parks.
In both equations, let y represent the altitude of the balloon. Let x represent the number of minutes.
y = 40 + 10x
y = 165 – 15x
Compute:
Substitute for y in one of the equations.
y = 40 + 10x
165 – 15x = 40 + 10x
165 = 40 + 10x + 15x
165 = 40+ 25x
x = 125/25
x = 5
Interpret:
Since x is the number of minutes, it takes 5 minutes for both balloons to be at the same altitude.
#### Exercise
1. When solving a system of equations using substitution, how can you determine whether the system has one solution, no solution, or infinitely many solutions?
2. Use substitution to solve the following system of equations.
4x + 8y = -8
x = -2y + 1
3. Use substitution to solve the following system of equations.
2x – 3y = 6
y = 2/3 x – 2
4. When given a system of equations in slope-intercept form, which is the most efficient method to solve: graphing or substitution? Explain.
5. Use substitution to solve the following system of equations.
2x + 2y = 6
4x + 4y = 4
6. Use substitution to solve the following system of equations.
2x + 5y = -5
y = – 2/5 x – 1
7. Richard and Teo have a combined age of 31. Richard is 4 years older than twice Teo’s age. How old are Richard and Teo?
8. Abby uses two social media sites. She has 52 more followers on site A than on site B. How many followers does she have on each site?
9. Use substitution to solve the following system of equations.
4x + 2y = −3
2x + y = 1
10. Stay Fit gym charges a membership fee of \$75. They offer karate classes for an additional fee.
How many classes could members and non-members take before they pay the same amount?
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
#### System of Linear Inequalities and Equations
Introduction: Systems of Linear Inequalities: A system of linear inequalities is a set of two or more linear inequalities in the same variables. The following example illustrates this, y < x + 2…………..Inequality 1 y ≥ 2x − 1…………Inequality 2 Solution of a System of Linear Inequalities: A solution of a system of linear inequalities […] | 1,804 | 6,237 | {"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} | 4.65625 | 5 | CC-MAIN-2024-30 | latest | en | 0.862696 |
http://physicscatalyst.com/Class9/matter6.php | 1,524,414,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945624.76/warc/CC-MAIN-20180422154522-20180422174522-00606.warc.gz | 251,601,285 | 7,060 | # Matter in our Surroundings
## Latent heat:
• The heat energy which has to be supplied to change the state of a substance is called its latent heat.
• Latent heat does not raise (or increase) the temperature. But latent heat has always to be supplied to change the state of a substance .The word ‘latent’ means ‘hidden’.
• The latent heat which we supply is used up in overcoming the forces of attraction between the particles of substance during the change of state. Latent heat does not increase the kinetic energy of the particles of the substance, so the temperature of a substance does not rise during the change of state.
### Latent heat is of two types:
1. Latent heat of fusion
2. Latent heat of vaporization.
#### Latent Heat of Fusion (solid to liquid change):
• The heat which is going into ice but not increasing its temperature, is the energy required to change the state of ice from solid to liquid (water). This is known as the latent heat of fusion of ice (or latent heat of melting of ice).
• The latent heat of fusion (or melting) of a solid is the quantity of heat in joules required to convert 1 kilogram of the solid (at its melting point) to liquid, without any change in temperature.
• The latent heat of fusion of ice is 3.34×105 joules per kilogram (or 3.34 ×105 j/kg).
#### Latent Heat of Vaporization (liquid to gas change):
The latent heat of vaporisation of a liquid is the quantity of heat in joules required to convert 1 kilogram of the liquid (at its boiling point) to vapour or gas, without any change in temperature.
The latent heat of vaporization of water is 22.5×105 joules per kilogram (or 22.5×105 j/kg).
Note: It has been found that the burns caused by steam are much more severe than those caused by boiling water though both of them are at the same temperature of 100 °C. This is due to the fact that steam contains more heat, in the form of latent heat, than boiling water.
## Sublimation:
The changing of a solid directly into vapours on heating and of vapours into solid on cooling, is known as sublimation.
image
1. The changing of a solid directly into vapor (or gas) is called sublimation.
2. The changing of vapor (or gas) directly into solid is called sublimation.
• The common substances which undergo sublimation are: Ammonium chloride, Iodine, Camphor, Naphthalene and Anthracene.
• When these solids are heated, their particles move so quickly that they separate completely to form vapor (or gas). And when these vapor (or gas) is cooled, these particles slow down so quickly that they become fixed and form a solid. | 604 | 2,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-17 | latest | en | 0.920564 |
https://www.mathworks.com/help/fixedpoint/ref/complexpartialsystolicmatrixsolveusingqrdecomposition.html | 1,627,934,037,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154356.39/warc/CC-MAIN-20210802172339-20210802202339-00560.warc.gz | 918,669,011 | 22,745 | # Complex Partial-Systolic Matrix Solve Using QR Decomposition
Compute value of x in Ax = B for complex-valued matrices using QR decomposition
• Library:
• Fixed-Point Designer HDL Support / Matrices and Linear Algebra / Linear System Solvers
## Description
The Complex Partial-Systolic Matrix Solve Using QR Decomposition block solves the system of linear equations Ax = B using QR decomposition, where A and B are complex-valued matrices. To compute x = A-1, set B to be the identity matrix.
## Ports
### Input
expand all
Rows of matrix A, specified as a vector. A is an m-by-n matrix where m ≥ 2 and mn. If B is single or double, A must be the same data type as B. If A is a fixed-point data type, A must be signed, use binary-point scaling, and have the same word length as B. Slope-bias representation is not supported for fixed-point data types.
Data Types: `single` | `double` | `fixed point`
Complex Number Support: Yes
Rows of matrix B, specified as a vector. B is an m-by-p matrix where m ≥ 2. If A is single or double, B must be the same data type as A. If B is a fixed-point data type, B must be signed, use binary-point scaling, and have the same word length as A. Slope-bias representation is not supported for fixed-point data types.
Data Types: `single` | `double` | `fixed point`
Whether inputs are valid, specified as a Boolean scalar. This control signal indicates when the data from the A(i,:) and B(i,:) input ports are valid. When this value is `1` (`true`) and the ready value is `1` (`true`), the block captures the values at the A(i,:) and B(i,:) input ports. When this value is `0` (`false`), the block ignores the input samples.
After sending a `true` validIn signal, there may be some delay before ready is set to `false`. To ensure all data is processed, you must wait until ready is set to `false` before sending another `true` validIn signal.
Data Types: `Boolean`
Whether to clear internal states, specified as a Boolean scalar. When this value is 1 (`true`), the block stops the current calculation and clears all internal states. When this value is 0 (`false`) and the `validIn` value is 1 (`true`), the block begins a new subframe.
Data Types: `Boolean`
### Output
expand all
Rows of matrix X, returned as a scalar or vector.
Data Types: `single` | `double` | `fixed point`
Whether the output data is valid, returned as a Boolean scalar. This control signal indicates when the data at the output port X(i,:) is valid. When this value is 1 (`true`), the block has successfully computed a row of matrix X. When this value is 0 (`false`), the output data is not valid.
Data Types: `Boolean`
Whether the block is ready, returned as a Boolean scalar. This control signal indicates when the block is ready for new input data. When this value is `1` (`true`) and the validIn value is `1` (`true`), the block accepts input data in the next time step. When this value is `0` (`false`), the block ignores input data in the next time step.
After sending a `true` validIn signal, there may be some delay before ready is set to `false`. To ensure all data is processed, you must wait until ready is set to `false` before sending another `true` validIn signal.
Data Types: `Boolean`
## Parameters
expand all
Number of rows in input matrices A and B, specified as a positive integer-valued scalar.
#### Programmatic Use
Block Parameter: `m` Type: character vector Values: positive integer-valued scalar Default: `4`
Number of columns in input matrix A, specified as a positive integer-valued scalar.
#### Programmatic Use
Block Parameter: `n` Type: character vector Values: positive integer-valued scalar Default: `4`
Number of columns in input matrix B, specified as a positive integer-valued scalar.
#### Programmatic Use
Block Parameter: `p` Type: character vector Values: positive integer-valued scalar Default: `1`
Regularization parameter, specified as a non-negative scalar. Small, positive values of the regularization parameter can improve the conditioning of the problem and reduce the variance of the estimates. While biased, the reduced variance of the estimate often results in a smaller mean squared error when compared to least-squares estimates.
#### Programmatic Use
Block Parameter: `k` Type: character vector Values: positive integer-valued scalar Default: `0`
Data type of the output matrix X, specified as `fixdt(1,18,14)`, `double`, `single`, `fixdt(1,16,0)`, or as a user-specified data type expression. The type can be specified directly, or expressed as a data type object such as `Simulink.NumericType`.
#### Programmatic Use
Block Parameter: `OutputType` Type: character vector Values: `'fixdt(1,18,14)'` | `'double'` | `'single'` | `'fixdt(1,16,0)'` | `''` Default: `'fixdt(1,18,14)'`
expand all
## Extended Capabilities
Introduced in R2020b
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# Let F(X) = X3+3x2 − 9x+2. Then, F(X) Has (A) a Maximum at X = 1 (B) a Minimum at X = 1 (C) Neither a Maximum Nor a Minimum at X = − 3 (D) None of These - Mathematics
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#### Question
Let f(x) = x3+3x$-$ 9x+2. Then, f(x) has _________________ .
##### Options
• a maximum at x = 1
• a minimum at x = 1
• neither a maximum nor a minimum at x = - 3
• none of these
#### Solution
$\text { a minimum at x = 1}$
$\text { Given }: f\left( x \right) = x^3 + 3 x^2 - 9x + 2$
$\Rightarrow f'\left( x \right) = 3 x^2 + 6x - 9$
$\text { For a local maxima or a local minima, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow 3 x^2 + 6x - 9 = 0$
$\Rightarrow x^2 + 2x - 3 = 0$
$\Rightarrow \left( x + 3 \right)\left( x - 1 \right) = 0$
$\Rightarrow x = - 3, 1$
$\text { Now,}$
$f''\left( x \right) = 6x + 6$
$\Rightarrow f''\left( 1 \right) = 6 + 6 = 12 > 0$
$\text { So, x = 1 is a local minima } .$
$\text { Also },$
$f''\left( - 3 \right) = - 18 + 6 = - 12 < 0$
$\text { So, x = - 3 is a local maxima } .$
Is there an error in this question or solution?
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#### Video TutorialsVIEW ALL [1]
Let F(X) = X3+3x2 − 9x+2. Then, F(X) Has (A) a Maximum at X = 1 (B) a Minimum at X = 1 (C) Neither a Maximum Nor a Minimum at X = − 3 (D) None of These Concept: Graph of Maxima and Minima.
Advertisement | 546 | 1,440 | {"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} | 4.34375 | 4 | CC-MAIN-2020-34 | latest | en | 0.620466 |
https://www.freemathhelp.com/forum/threads/109295-Question-about-Vertical-Asymptote-y-x-3-(9-x-2)?p=420538 | 1,539,670,156,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510019.12/warc/CC-MAIN-20181016051435-20181016072935-00427.warc.gz | 936,258,930 | 11,996 | 1. ## Question about Vertical Asymptote: y = x^3/(9-x^2)
Hi all
was doing a function sketching practice problem and was wondering;
for examples sake lets use:
y = x^3/(9-x^2)
so the domain restrictions on this one would be -3 and 3. Is there ever a case where a domain restriction does not result in a vertical asymptote?
Is it necessary to evaluate the L.H.L. and R.H.L. to verify that the function goes to +- infinity? or is it just implied?
2. not counting peicewise functions not defined at certain points
3. Originally Posted by frank789
y = x^3/(9-x^2)
so the domain restrictions on this one would be -3 and 3. Is there ever a case where a domain restriction does not result in a vertical asymptote?
Is it necessary to evaluate the L.H.L. and R.H.L. to verify that the function goes to +- infinity? or is it just implied?
If by "domain restrictions" you mean points where it is undefined, you can also have "holes" (removable discontinuities), as in (9-x^2)/(9-x^2), to give a very simple example.
It is also possible to have a vertical asymptote that goes the same way on both sides (e.g. both +infinity), as in x^3/(9-x^2)^2. So you do need to consider at least the sign on each side of the asymptote.
I think of a "domain restriction" as an explicit restriction of the domain, as in "f(x) = x^2, x >= 0". But you may be using different terminology.
4. you are right about my poor description but despite that you still managed to answer it haha thanks!
5. Two things:
1a) Make the denominator 0 (zero). These are POTENTIAL vertical asymptotes.
1b) If the numerator is also zero (0) for the same x-value, it might not be an asymptote. It might be a hole. Check it out by reducing common factors.
2) The degree of the of the factor that creates the asymptote dictates the behavior. (Assuming factors are reduced.)
2a) Odd Degree - One side goes up and one side goes down.
2b) Even Degree - both go up or both go down.
2c) Only the sign of one side is needed if you know the degree, but it never hurts to gain more information.
6. A little fine-tuning:
First, all of this applies specifically to rational functions like yours; other kinds of functions may differ.
Second, it is possible to still have an asymptote when both numerator and denominator are zero; it is best to look at the function after canceling like factors to determine its behavior (keeping in mind that being able to cancel means it is undefined). Ultimately, if the exponent in the denominator was higher, you end up with an asymptote.
7. Originally Posted by Dr.Peterson
First, all of this applies specifically to rational functions like yours; other kinds of functions may differ.
We're using the quotient of simple, real-valued polynomials, right? Is there another definition in "Intermediate/Advance Algebra"?
Second, it is possible to still have an asymptote when both numerator and denominator are zero
No example comes to mind. Demonstration?
8. Originally Posted by tkhunny
No example comes to mind. Demonstration?
x/x^2?
9. I knew I should have stayed in bed, this morning.
I better check everything else I did all day...
Maybe I didn't get out of bed. Hmmm...
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• | 803 | 3,316 | {"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} | 3.9375 | 4 | CC-MAIN-2018-43 | latest | en | 0.916776 |
https://philosophy.stackexchange.com/questions/17981/how-to-prove-the-axiom-is-wrong | 1,713,689,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00170.warc.gz | 439,728,196 | 37,274 | # How to prove the axiom is wrong?
The father of math (Euclid) wrote a book named Elements. The book is full of axioms and here are some of them I am interested in:
1. Things equal to the same thing are also equal to one another.
2. And if equal things are added to equal things then the wholes are equal.
3. And if equal things are subtracted from equal things then the remainders are equal.
4. And things coinciding with one another are equal to one another.
5. And the whole [is] greater than the part.
Few years ago I came to the idea that Euclid could be (is..) wrong about the No.5.
Why I think he is wrong ? He is wrong because making decision on bigger/smaller (as a consequence: more/less, faster/slower etc.) is the (main) reason of all negative we have happening. Try to think about any conflict on the Earth and You will find at least two sides fighting for the same thing there. For example: fighting for bigger territory, fighting for more money, fighting for less problems, etc. I think that the concept of thinking (thinking that some bigger or smaller even exists) is false. However, it seams I cannot logically prove it being false or true, because the whole logic itself is made on top of such axioms...
So, let`s say (or just assume..) that some of axioms are wrong. The question is - what base should be used to prove axiom being wrong for the rest of the world? My own assumption is that I have to use the "what is valuable for society" base or even the "how it feels for society" base.
And, one more thing to mention. If I am not able to prove axiom being false or true (because false/true is part of the logic which is made on top of the axiom itself) then I would like to call "true" as "real" and "false" as "illusion", but would this be correct?
• i didnt even read your question text(im sorry if only your title is bad) a axiom cant be wrong, you say: for the sake of argument, lets agree that something is true --> axiom The meaning is that it should not be questioned, so we can have a discussion about something without always have to explain stuff.
– yamm
Dec 5, 2014 at 8:45
• @yamm, and that's very wrong because from the context axiom can be wrong when it tries to describe reality (even though it can seem obvious that it's true). Otherwise not all Euclid's axioms are really axioms. Aug 14, 2018 at 23:22
The correct question is not :
How to decide if an axiom is right or wrong ...
but :
Of what "domain" (of discourse or of reality) is this axiom true ?
Modern mathematics, after Georg Cantor, has found "domains" (infinite colelctions or sets) where it is not true that
the whole [is] greater than the part
for a "suitable" interpretation of greater than.
As was already known to Galileo (see Galileo's paradox) we can associate to each natural number n its double : 2n.
Thus, if we use this "procedure" to count the objecst in a collection, we can roughly say that the collection of natural numbers has the "same number" of elements as the collection of even numbers.
This result show us that, for infinite collection, we can roughly say that
the whole is not always "greater than" a proper part of it.
In Euclid's time, axioms were taken as basic, unquestionable truths about the world. In more modern times, however, we have less faith in the existence of such things, and axioms are defined less rigidly as the foundational building blocks of a particular system of thought. Thus, Euclidian axioms define Euclidian geometry, but there are also non-Euclidian geometries with different axioms.
As it so happens, the axioms of formal logic are not dependent on the axioms of Euclidian geometry, so you could attempt to disprove an Euclidian axiom using logic without any fear of a paradox.
In general, if you want to prove something to someone, the proper approach is to start from axioms that your target endorses. If you do your job correctly, you will demonstrate that the target cannot consistently hold the position you are trying to disprove and still endorse the axioms you started with.
The best way to falsify an axiom is to show that the axiom is either self-contradictory in its own terms or logically implies a deduction of one theorem that leads to a self-contradiction. | 989 | 4,246 | {"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 | 0.96132 |
https://www.jiskha.com/questions/3973/2-the-line-y-4x-7-crosses-the-y-axis-at-4-0-0-4-7-0-0-7-0-7-if-u | 1,611,673,374,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704800238.80/warc/CC-MAIN-20210126135838-20210126165838-00470.warc.gz | 846,427,406 | 4,705 | # Help!!!!
2. The line y = 4x - 7 crosses the y axis at:
(4, 0)
(0, 4)
(-7, 0)
(0, -7)
(0,-7). if u get a graphing calculator then u would be able to solve easily.
Set x=0 and solve for y to get
y = 4*0 - 7 = 0 - 7 so
y=7
The coordinate is (0,-7)
1. 👍
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The line Y=3x+5 crosses the y axis at P. What are the co-ordinates of P? | 909 | 2,961 | {"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} | 3.71875 | 4 | CC-MAIN-2021-04 | latest | en | 0.796536 |
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Published at Saturday, 11 May 2019. Color by Number. By .
Color by number printable give kids a chance to solve a mystery, Granted, not much of a mystery. But when you see a blank color by number page, you don’t know what it’s going to look like before you color it in. As you color, the pictures start to come to life beneath the tips of your crayons. It’s fun to watch it happening. There are also color by number pages out there that do not allow users to see what the pictures are before they are colored… they often feature strange looking mosaics that “hide” a picture in the various shapes on the page. These are also fun classroom activities to complete.
Everything You Need to Know About Paint-by-Number, When was the first time you were introduced to the term “paint-by-number”? Maybe you were a child, or perhaps it was more recently. In case you haven’t heard of it, paint-by-number is a painting system that delineates a picture into shapes, and marks each individual shape with a number that corresponds with a paint color. Aspiring artists fill in each shape with the called-for paint color until all spaces are filled and a picture has emerged. (Cue the how-did-I-do-that response.) Painting-by-number is truly as easy as it sounds, which is precisely why it gained incredible popularity several decades ago, and continues to be a favorite pastime.
Coloring will require basic hand-eye coordination skills, and color by number sheets will call upon specific concentration while children coordinate the number on the page with what color they actually need to use. These coloring diagrams will require that children color within specified areas with a predetermined color, and keeping the color within a defined area will help to develop hand and eye coordination. Children will learn to hold the marker, crayon, or colored pencil while controlling its movement in a way that will prevent the color from going beyond the outline.
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#### Mixed Addition And Subtraction With Regrouping Gallery
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When we divide polynomials, there are two different scenarios. The first and easier of the two involves dividing a polynomial by a monomial. For this type of problem, we set up a fraction and divide each term of the polynomial by the monomial. The second and harder scenario involves dividing polynomials when neither is a monomial. For this type of problem, we generally use polynomial long division.
Test Objectives
• Demonstrate the ability to divide a polynomial by a monomial
• Demonstrate the ability to set up a polynomial long division
• Demonstrate the ability to divide polynomials when remainders are involved
Polynomial Long Division Practice Test:
#1:
Instructions: Find each quotient.
a) (7n4 - 43n3 - 2n2 + 13n + 36) ÷ (7n - 8)
#2:
Instructions: Find each quotient.
a) (58r - 24r2 + 40 + 2r3) ÷ (r - 8)
#3:
Instructions: Find each quotient.
a) (42x4 - 63x3 + 3x2 + 39x - 14) ÷ (6x - 3)
#4:
Instructions: Find each quotient.
a) (-32x5 - 8x4 - 28x2 + 72x + 60) ÷ (-8x2 + 12)
#5:
Instructions: Find each quotient.
a) (-36x4 - 24x3 - 40x + 100) ÷ (12x2 + 20)
Written Solutions:
#1:
Solutions:
a) $$n^3 - 5n^2 - 6n - 5 + \frac{-4}{7n - 8}$$
#2:
Solutions:
a) $$2r^2 - 8r - 6 + \frac{-8}{r - 8}$$
#3:
Solutions:
a) $$7x^3-7x^2-3x+5+\frac{1}{6x-3}$$
#4:
Solutions:
a) $$4x^3+x^2+6x+5$$
#5:
Solutions:
a) $$-3x^2-2x+5$$ | 494 | 1,395 | {"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} | 4.75 | 5 | CC-MAIN-2022-49 | latest | en | 0.728451 |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Zonal_spherical_function | 1,627,314,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00348.warc.gz | 1,152,347,469 | 32,515 | # Zonal spherical function
In mathematics, a zonal spherical function or often just spherical function is a function on a locally compact group G with compact subgroup K (often a maximal compact subgroup) that arises as the matrix coefficient of a K-invariant vector in an irreducible representation of G. The key examples are the matrix coefficients of the spherical principal series, the irreducible representations appearing in the decomposition of the unitary representation of G on L2(G/K). In this case the commutant of G is generated by the algebra of biinvariant functions on G with respect to K acting by right convolution. It is commutative if in addition G/K is a symmetric space, for example when G is a connected semisimple Lie group with finite centre and K is a maximal compact subgroup. The matrix coefficients of the spherical principal series describe precisely the spectrum of the corresponding C* algebra generated by the biinvariant functions of compact support, often called a Hecke algebra. The spectrum of the commutative Banach *-algebra of biinvariant L1 functions is larger; when G is a semisimple Lie group with maximal compact subgroup K, additional characters come from matrix coefficients of the complementary series, obtained by analytic continuation of the spherical principal series.
Zonal spherical functions have been explicitly determined for real semisimple groups by Harish-Chandra. For special linear groups, they were independently discovered by Israel Gelfand and Mark Naimark. For complex groups, the theory simplifies significantly, because G is the complexification of K, and the formulas are related to analytic continuations of the Weyl character formula on K. The abstract functional analytic theory of zonal spherical functions was first developed by Roger Godement. Apart from their group theoretic interpretation, the zonal spherical functions for a semisimple Lie group G also provide a set of simultaneous eigenfunctions for the natural action of the centre of the universal enveloping algebra of G on L2(G/K), as differential operators on the symmetric space G/K. For semisimple p-adic Lie groups, the theory of zonal spherical functions and Hecke algebras was first developed by Satake and Ian G. Macdonald. The analogues of the Plancherel theorem and Fourier inversion formula in this setting generalise the eigenfunction expansions of Mehler, Weyl and Fock for singular ordinary differential equations: they were obtained in full generality in the 1960s in terms of Harish-Chandra's c-function.
The name "zonal spherical function" comes from the case when G is SO(3,R) acting on a 2-sphere and K is the subgroup fixing a point: in this case the zonal spherical functions can be regarded as certain functions on the sphere invariant under rotation about a fixed axis.
## Definitions
Let G be a locally compact unimodular topological group and K a compact subgroup and let H1 = L2(G/K). Thus H1 admits a unitary representation π of G by left translation. This is a subrepresentation of the regular representation, since if H= L2(G) with left and right regular representations λ and ρ of G and P is the orthogonal projection
${\displaystyle P=\int _{K}\rho (k)\,dk}$
from H to H1 then H1 can naturally be identified with PH with the action of G given by the restriction of λ.
On the other hand, by von Neumann's commutation theorem[1]
${\displaystyle \lambda (G)^{\prime }=\rho (G)^{\prime \prime },}$
where S' denotes the commutant of a set of operators S, so that
${\displaystyle \pi (G)^{\prime }=P\rho (G)^{\prime \prime }P.}$
Thus the commutant of π is generated as a von Neumann algebra by operators
${\displaystyle P\rho (f)P=\int _{G}f(g)(P\rho (g)P)\,dg}$
where f is a continuous function of compact support on G.[2]
However Pρ(f) P is just the restriction of ρ(F) to H1, where
${\displaystyle F(g)=\int _{K}\int _{K}f(kgk^{\prime })\,dk\,dk^{\prime }}$
is the K-biinvariant continuous function of compact support obtained by averaging f by K on both sides.
Thus the commutant of π is generated by the restriction of the operators ρ(F) with F in Cc(K\G/K), the K-biinvariant continuous functions of compact support on G.
These functions form a * algebra under convolution with involution
${\displaystyle F^{*}(g)={\overline {F(g^{-1})}},}$
often called the Hecke algebra for the pair (G, K).
Let A(K\G/K) denote the C* algebra generated by the operators ρ(F) on H1.
The pair (G, K) is said to be a Gelfand pair [3] if one, and hence all, of the following algebras are commutative:
• ${\displaystyle \pi (G)^{\prime }}$
• ${\displaystyle C_{c}(K\backslash G/K)}$
• ${\displaystyle A(K\backslash G/K).}$
Since A(K\G/K) is a commutative C* algebra, by the Gelfand–Naimark theorem it has the form C0(X), where X is the locally compact space of norm continuous * homomorphisms of A(K\G/K) into C.
A concrete realization of the * homomorphisms in X as K-biinvariant uniformly bounded functions on G is obtained as follows.[3][4][5][6][7]
Because of the estimate
${\displaystyle \|\pi (F)\|\leq \int _{G}|F(g)|\,dg\equiv \|F\|_{1},}$
the representation π of Cc(K\G/K) in A(K\G/K) extends by continuity to L1(K\G/K), the * algebra of K-biinvariant integrable functions. The image forms a dense * subalgebra of A(K\G/K). The restriction of a * homomorphism χ continuous for the operator norm is also continuous for the norm ||·||1. Since the Banach space dual of L1 is L, it follows that
${\displaystyle \chi (\pi (F))=\int _{G}F(g)h(g)\,dg,}$
for some unique uniformly bounded K-biinvariant function h on G. These functions h are exactly the zonal spherical functions for the pair (G, K).
## Properties
A zonal spherical function h has the following properties:[3]
1. h is uniformly continuous on G
2. ${\displaystyle h(x)h(y)=\int _{K}h(xky)\,dk\,\,(x,y\in G).}$
3. h(1) =1 (normalisation)
4. h is a positive definite function on G
5. f * h is proportional to h for all f in Cc(K\G/K).
These are easy consequences of the fact that the bounded linear functional χ defined by h is a homomorphism. Properties 2, 3 and 4 or properties 3, 4 and 5 characterize zonal spherical functions. A more general class of zonal spherical functions can be obtained by dropping positive definiteness from the conditions, but for these functions there is no longer any connection with unitary representations. For semisimple Lie groups, there is a further characterization as eigenfunctions of invariant differential operators on G/K (see below).
In fact, as a special case of the Gelfand–Naimark–Segal construction, there is one-one correspondence between irreducible representations σ of G having a unit vector v fixed by K and zonal spherical functions h given by
${\displaystyle h(g)=(\sigma (g)v,v).}$
Such irreducible representations are often described as having class one. They are precisely the irreducible representations required to decompose the induced representation π on H1. Each representation σ extends uniquely by continuity to A(K\G/K), so that each zonal spherical function satisfies
${\displaystyle \left|\int _{G}f(g)h(g)\,dg\right|\leq \|\pi (f)\|}$
for f in A(K\G/K). Moreover, since the commutant π(G)' is commutative, there is a unique probability measure μ on the space of * homomorphisms X such that
${\displaystyle \int _{G}|f(g)|^{2}\,dg=\int _{X}|\chi (\pi (f))|^{2}\,d\mu (\chi ).}$
μ is called the Plancherel measure. Since π(G)' is the centre of the von Neumann algebra generated by G, it also gives the measure associated with the direct integral decomposition of H1 in terms of the irreducible representations σχ.
## Gelfand pairs
If G is a connected Lie group, then, thanks to the work of Cartan, Malcev, Iwasawa and Chevalley, G has a maximal compact subgroup, unique up to conjugation.[8][9] In this case K is connected and the quotient G/K is diffeomorphic to a Euclidean space. When G is in addition semisimple, this can be seen directly using the Cartan decomposition associated to the symmetric space G/K, a generalisation of the polar decomposition of invertible matrices. Indeed, if τ is the associated period two automorphism of G with fixed point subgroup K, then
${\displaystyle G=P\cdot K,}$
where
${\displaystyle P=\{g\in G|\tau (g)=g^{-1}\}.}$
Under the exponential map, P is diffeomorphic to the -1 eigenspace of τ in the Lie algebra of G. Since τ preserves K, it induces an automorphism of the Hecke algebra Cc(K\G/K). On the other hand, if F lies in Cc(K\G/K), then
Fg) = F(g1),
so that τ induces an anti-automorphism, because inversion does. Hence, when G is semisimple,
• the Hecke algebra is commutative
• (G,K) is a Gelfand pair.
More generally the same argument gives the following criterion of Gelfand for (G,K) to be a Gelfand pair:[10]
• G is a unimodular locally compact group;
• K is a compact subgroup arising as the fixed points of a period two automorphism τ of G;
• G =K·P (not necessarily a direct product), where P is defined as above.
The two most important examples covered by this are when:
• G is a compact connected semisimple Lie group with τ a period two automorphism;[11][12]
• G is a semidirect product ${\displaystyle A\rtimes K}$, with A a locally compact Abelian group without 2-torsion and τ(a· k)= k·a−1 for a in A and k in K.
The three cases cover the three types of symmetric spaces G/K:[6]
1. Non-compact type, when K is a maximal compact subgroup of a non-compact real semisimple Lie group G;
2. Compact type, when K is the fixed point subgroup of a period two automorphism of a compact semisimple Lie group G;
3. Euclidean type, when A is a finite-dimensional Euclidean space with an orthogonal action of K.
## Cartan–Helgason theorem
Let G be a compact semisimple connected and simply connected Lie group and τ a period two automorphism of a G with fixed point subgroup K = Gτ. In this case K is a connected compact Lie group.[6] In addition let T be a maximal torus of G invariant under τ, such that T ${\displaystyle \cap }$ P is a maximal torus in P, and set[13]
${\displaystyle S=K\cap T=T^{\tau }.}$
S is the direct product of a torus and an elementary abelian 2-group.
In 1929 Élie Cartan found a rule to determine the decomposition of L2(G/K) into the direct sum of finite-dimensional irreducible representations of G, which was proved rigorously only in 1970 by Sigurdur Helgason. Because the commutant of G on L2(G/K) is commutative, each irreducible representation appears with multiplicity one. By Frobenius reciprocity for compact groups, the irreducible representations V that occur are precisely those admitting a non-zero vector fixed by K.
From the representation theory of compact semisimple groups, irreducible representations of G are classified by their highest weight. This is specified by a homomorphism of the maximal torus T into T.
The Cartan–Helgason theorem[14][15] states that
the irreducible representations of G admitting a non-zero vector fixed by K are precisely those with highest weights corresponding to homomorphisms trivial on S.
The corresponding irreducible representations are called spherical representations.
The theorem can be proved[6] using the Iwasawa decomposition:
${\displaystyle {\mathfrak {g}}={\mathfrak {k}}\oplus {\mathfrak {a}}\oplus {\mathfrak {n}},}$
where ${\displaystyle {\mathfrak {g}}}$, ${\displaystyle {\mathfrak {k}}}$, ${\displaystyle {\mathfrak {a}}}$ are the complexifications of the Lie algebras of G, K, A = T ${\displaystyle \cap }$ P and
${\displaystyle {\mathfrak {n}}=\bigoplus {\mathfrak {g}}_{\alpha },}$
summed over all eigenspaces for T in ${\displaystyle {\mathfrak {g}}}$ corresponding to positive roots α not fixed by τ.
Let V be a spherical representation with highest weight vector v0 and K-fixed vector vK. Since v0 is an eigenvector of the solvable Lie algebra ${\displaystyle {\mathfrak {a}}\oplus {\mathfrak {p}}}$, the Poincaré–Birkhoff–Witt theorem implies that the K-module generated by v0 is the whole of V. If Q is the orthogonal projection onto the fixed points of K in V obtained by averaging over G with respect to Haar measure, it follows that
${\displaystyle \displaystyle {v_{K}=cQv_{0}}}$
for some non-zero constant c. Because vK is fixed by S and v0 is an eigenvector for S, the subgroup S must actually fix v0, an equivalent form of the triviality condition on S.
Conversely if v0 is fixed by S, then it can be shown[16] that the matrix coefficient
${\displaystyle \displaystyle {f(g)=(gv_{0},v_{0})}}$
is non-negative on K. Since f(1) > 0, it follows that (Qv0, v0) > 0 and hence that Qv0 is a non-zero vector fixed by K.
## Harish-Chandra's formula
If G is a non-compact semisimple Lie group, its maximal compact subgroup K acts by conjugation on the component P in the Cartan decomposition. If A is a maximal Abelian subgroup of G contained in P, then A is diffeomorphic to its Lie algebra under the exponential map and, as a further generalisation of the polar decomposition of matrices, every element of P is conjugate under K to an element of A, so that[17]
G =KAK.
There is also an associated Iwasawa decomposition
G =KAN,
where N is a closed nilpotent subgroup, diffeomorphic to its Lie algebra under the exponential map and normalised by A. Thus S=AN is a closed solvable subgroup of G, the semidirect product of N by A, and G = KS.
If α in Hom(A,T) is a character of A, then α extends to a character of S, by defining it to be trivial on N. There is a corresponding unitary induced representation σ of G on L2(G/S) = L2(K),[18] a so-called (spherical) principal series representation.
This representation can be described explicitly as follows. Unlike G and K, the solvable Lie group S is not unimodular. Let dx denote left invariant Haar measure on S and ΔS the modular function of S. Then[6]
${\displaystyle \int _{G}f(g)\,dg=\int _{S}\int _{K}f(x\cdot k)\,dx\,dk=\int _{S}\int _{K}f(k\cdot x)\Delta _{S}(x)\,dx\,dk.}$
The principal series representation σ is realised on L2(K) as[19]
${\displaystyle (\sigma (g)\xi )(k)=\alpha ^{\prime }(g^{-1}k)^{-1}\,\xi (U(g^{-1}k)),}$
where
${\displaystyle g=U(g)\cdot X(g)}$
is the Iwasawa decomposition of g with U(g) in K and X(g) in S and
${\displaystyle \alpha ^{\prime }(kx)=\Delta _{S}(x)^{1/2}\alpha (x)}$
for k in K and x in S.
The representation σ is irreducible, so that if v denotes the constant function 1 on K, fixed by K,
${\displaystyle \varphi _{\alpha }(g)=(\sigma (g)v,v)}$
defines a zonal spherical function of G.
Computing the inner product above leads to Harish-Chandra's formula for the zonal spherical function
${\displaystyle \varphi _{\alpha }(g)=\int _{K}\alpha ^{\prime }(gk)^{-1}\,dk}$
as an integral over K.
Harish-Chandra proved that these zonal spherical functions exhaust the characters of the C* algebra generated by the Cc(K \ G / K) acting by right convolution on L2(G / K). He also showed that two different characters α and β give the same zonal spherical function if and only if α = β·s, where s is in the Weyl group of A
${\displaystyle W(A)=N_{K}(A)/C_{K}(A),}$
the quotient of the normaliser of A in K by its centraliser, a finite reflection group.
It can also be verified directly[3] that this formula defines a zonal spherical function, without using representation theory. The proof for general semisimple Lie groups that every zonal spherical formula arises in this way requires the detailed study of G-invariant differential operators on G/K and their simultaneous eigenfunctions (see below).[5][6] In the case of complex semisimple groups, Harish-Chandra and Felix Berezin realised independently that the formula simplified considerably and could be proved more directly.[6][20][21][22][23]
The remaining positive-definite zonal spherical functions are given by Harish-Chandra's formula with α in Hom(A,C*) instead of Hom(A,T). Only certain α are permitted and the corresponding irreducible representations arise as analytic continuations of the spherical principal series. This so-called "complementary series" was first studied by Bargmann (1947) for G = SL(2,R) and by Harish-Chandra (1947) and Gelfand & Naimark (1947) for G = SL(2,C). Subsequently in the 1960s, the construction of a complementary series by analytic continuation of the spherical principal series was systematically developed for general semisimple Lie groups by Ray Kunze, Elias Stein and Bertram Kostant.[24][25][26] Since these irreducible representations are not tempered, they are not usually required for harmonic analysis on G (or G / K).
## Eigenfunctions
Harish-Chandra proved[5][6] that zonal spherical functions can be characterised as those normalised positive definite K-invariant functions on G/K that are eigenfunctions of D(G/K), the algebra of invariant differential operators on G. This algebra acts on G/K and commutes with the natural action of G by left translation. It can be identified with the subalgebra of the universal enveloping algebra of G fixed under the adjoint action of K. As for the commutant of G on L2(G/K) and the corresponding Hecke algebra, this algebra of operators is commutative; indeed it is a subalgebra of the algebra of mesurable operators affiliated with the commutant π(G)', an Abelian von Neumann algebra. As Harish-Chandra proved, it is isomorphic to the algebra of W(A)-invariant polynomials on the Lie algebra of A, which itself is a polynomial ring by the Chevalley–Shephard–Todd theorem on polynomial invariants of finite reflection groups. The simplest invariant differential operator on G/K is the Laplacian operator; up to a sign this operator is just the image under π of the Casimir operator in the centre of the universal enveloping algebra of G.
Thus a normalised positive definite K-biinvariant function f on G is a zonal spherical function if and only if for each D in D(G/K) there is a constant λD such that
${\displaystyle \displaystyle \pi (D)f=\lambda _{D}f,}$
i.e. f is a simultaneous eigenfunction of the operators π(D).
If ψ is a zonal spherical function, then, regarded as a function on G/K, it is an eigenfunction of the Laplacian there, an elliptic differential operator with real analytic coefficients. By analytic elliptic regularity, ψ is a real analytic function on G/K, and hence G.
Harish-Chandra used these facts about the structure of the invariant operators to prove that his formula gave all zonal spherical functions for real semisimple Lie groups.[27][28][29] Indeed, the commutativity of the commutant implies that the simultaneous eigenspaces of the algebra of invariant differential operators all have dimension one; and the polynomial structure of this algebra forces the simultaneous eigenvalues to be precisely those already associated with Harish-Chandra's formula.
## Example: SL(2,C)
The group G = SL(2,C) is the complexification of the compact Lie group K = SU(2) and the double cover of the Lorentz group. The infinite-dimensional representations of the Lorentz group were first studied by Dirac in 1945, who considered the discrete series representations, which he termed expansors. A systematic study was taken up shortly afterwards by Harish-Chandra, Gelfand–Naimark and Bargmann. The irreducible representations of class one, corresponding to the zonal spherical functions, can be determined easily using the radial component of the Laplacian operator.[6]
Indeed, any unimodular complex 2×2 matrix g admits a unique polar decomposition g = pv with v unitary and p positive. In turn p = uau*, with u unitary and a a diagonal matrix with positive entries. Thus g = uaw with w = u* v, so that any K-biinvariant function on G corresponds to a function of the diagonal matrix
${\displaystyle a={\begin{pmatrix}e^{r/2}&0\\0&e^{-r/2}\end{pmatrix}},}$
invariant under the Weyl group. Identifying G/K with hyperbolic 3-space, the zonal hyperbolic functions ψ correspond to radial functions that are eigenfunctions of the Laplacian. But in terms of the radial coordinate r, the Laplacian is given by[30]
${\displaystyle L=-\partial _{r}^{2}-2\coth r\partial _{r}.}$
Setting f(r) = sinh (r)·ψ(r), it follows that f is an odd function of r and an eigenfunction of ${\displaystyle \partial _{r}^{2}}$.
Hence
${\displaystyle \varphi (r)={\sin(\ell r) \over \ell \sinh r}}$
where ${\displaystyle \ell }$ is real.
There is a similar elementary treatment for the generalized Lorentz groups SO(N,1) in Takahashi (1963) and Faraut & Korányi (1994) (recall that SO0(3,1) = SL(2,C) / ±I).
## Complex case
If G is a complex semisimple Lie group, it is the complexification of its maximal compact subgroup K. If ${\displaystyle {\mathfrak {g}}}$ and ${\displaystyle {\mathfrak {k}}}$ are their Lie algebras, then
${\displaystyle {\mathfrak {g}}={\mathfrak {k}}\oplus i{\mathfrak {k}}.}$
Let T be a maximal torus in K with Lie algebra ${\displaystyle {\mathfrak {t}}}$. Then
${\displaystyle A=\exp i{\mathfrak {t}},\,\,P=\exp i{\mathfrak {k}}.}$
Let
${\displaystyle W=N_{K}(T)/T}$
be the Weyl group of T in K. Recall characters in Hom(T,T) are called weights and can be identified with elements of the weight lattice Λ in Hom(${\displaystyle {\mathfrak {t}}}$, R) = ${\displaystyle {\mathfrak {t}}^{*}}$. There is a natural ordering on weights and every finite-dimensional irreducible representation (π, V) of K has a unique highest weight λ. The weights of the adjoint representation of K on ${\displaystyle {\mathfrak {k}}\ominus {\mathfrak {t}}}$ are called roots and ρ is used to denote half the sum of the positive roots α, Weyl's character formula asserts that for z = exp X in T
${\displaystyle \displaystyle \chi _{\lambda }(e^{X})\equiv {\rm {Tr}}\,\pi (z)=A_{\lambda +\rho }(e^{X})/A_{\rho }(e^{X}),}$
where, for μ in ${\displaystyle {\mathfrak {t}}^{*}}$, Aμ denotes the antisymmetrisation
${\displaystyle \displaystyle A_{\mu }(e^{X})=\sum _{s\in W}\varepsilon (s)e^{i\mu (sX)},}$
and ε denotes the sign character of the finite reflection group W.
Weyl's denominator formula expresses the denominator Aρ as a product:
${\displaystyle \displaystyle A_{\rho }(e^{X})=e^{i\rho (X)}\prod _{\alpha >0}(1-e^{-i\alpha (X)}),}$
where the product is over the positive roots.
Weyl's dimension formula asserts that
${\displaystyle \displaystyle \chi _{\lambda }(1)\equiv {\rm {dim}}\,V={\prod _{\alpha >0}(\lambda +\rho ,\alpha ) \over \prod _{\alpha >0}(\rho ,\alpha )}.}$
where the inner product on ${\displaystyle {\mathfrak {t}}^{*}}$ is that associated with the Killing form on ${\displaystyle {\mathfrak {k}}}$.
Now
• every irreducible representation of K extends holomorphically to the complexification G
• every irreducible character χλ(k) of K extends holomorphically to the complexification of K and ${\displaystyle {\mathfrak {t}}^{*}}$.
• for every λ in Hom(A,T) = ${\displaystyle i{\mathfrak {t}}^{*}}$, there is a zonal spherical function φλ.
The Berezin–Harish–Chandra formula[6] asserts that for X in ${\displaystyle i{\mathfrak {t}}}$
${\displaystyle \varphi _{\lambda }(e^{X})={\chi _{\lambda }(e^{X}) \over \chi _{\lambda }(1)}.}$
In other words:
• the zonal spherical functions on a complex semisimple Lie group are given by analytic continuation of the formula for the normalised characters.
One of the simplest proofs[31] of this formula involves the radial component on A of the Laplacian on G, a proof formally parallel to Helgason's reworking of Freudenthal's classical proof of the Weyl character formula, using the radial component on T of the Laplacian on K.[32]
In the latter case the class functions on K can be identified with W-invariant functions on T. The radial component of ΔK on T is just the expression for the restriction of ΔK to W-invariant functions on T, where it is given by the formula
${\displaystyle \displaystyle \Delta _{K}=h^{-1}\circ \Delta _{T}\circ h+\|\rho \|^{2},}$
where
${\displaystyle \displaystyle h(e^{X})=A_{\rho }(e^{X})}$
for X in ${\displaystyle {\mathfrak {t}}}$. If χ is a character with highest weight λ, it follows that φ = h·χ satisfies
${\displaystyle \Delta _{T}\varphi =(\|\lambda +\rho \|^{2}-\|\rho \|^{2})\varphi .}$
Thus for every weight μ with non-zero Fourier coefficient in φ,
${\displaystyle \displaystyle \|\lambda +\rho \|^{2}=\|\mu +\rho \|^{2}.}$
The classical argument of Freudenthal shows that μ + ρ must have the form s(λ + ρ) for some s in W, so the character formula follows from the antisymmetry of φ.
Similarly K-biinvariant functions on G can be identified with W(A)-invariant functions on A. The radial component of ΔG on A is just the expression for the restriction of ΔG to W(A)-invariant functions on A. It is given by the formula
${\displaystyle \displaystyle \Delta _{G}=H^{-1}\circ \Delta _{A}\circ H-\|\rho \|^{2},}$
where
${\displaystyle \displaystyle H(e^{X})=A_{\rho }(e^{X})}$
for X in ${\displaystyle i{\mathfrak {t}}}$.
The Berezin–Harish–Chandra formula for a zonal spherical function φ can be established by introducing the antisymmetric function
${\displaystyle \displaystyle f=H\cdot \varphi ,}$
which is an eigenfunction of the Laplacian ΔA. Since K is generated by copies of subgroups that are homomorphic images of SU(2) corresponding to simple roots, its complexification G is generated by the corresponding homomorphic images of SL(2,C). The formula for zonal spherical functions of SL(2,C) implies that f is a periodic function on ${\displaystyle i{\mathfrak {t}}}$ with respect to some sublattice. Antisymmetry under the Weyl group and the argument of Freudenthal again imply that ψ must have the stated form up to a multiplicative constant, which can be determined using the Weyl dimension formula.
## Example: SL(2,R)
The theory of zonal spherical functions for SL(2,R) originated in the work of Mehler in 1881 on hyperbolic geometry. He discovered the analogue of the Plancherel theorem, which was rediscovered by Fock in 1943. The corresponding eigenfunction expansion is termed the Mehler–Fock transform. It was already put on a firm footing in 1910 by Hermann Weyl's important work on the spectral theory of ordinary differential equations. The radial part of the Laplacian in this case leads to a hypergeometric differential equation, the theory of which was treated in detail by Weyl. Weyl's approach was subsequently generalised by Harish-Chandra to study zonal spherical functions and the corresponding Plancherel theorem for more general semisimple Lie groups. Following the work of Dirac on the discrete series representations of SL(2,R), the general theory of unitary irreducible representations of SL(2,R) was developed independently by Bargmann, Harish-Chandra and Gelfand–Naimark. The irreducible representations of class one, or equivalently the theory of zonal spherical functions, form an important special case of this theory.
The group G = SL(2,R) is a double cover of the 3-dimensional Lorentz group SO(2,1), the symmetry group of the hyperbolic plane with its Poincaré metric. It acts by Möbius transformations. The upper half-plane can be identified with the unit disc by the Cayley transform. Under this identification G becomes identified with the group SU(1,1), also acting by Möbius transformations. Because the action is transitive, both spaces can be identified with G/K, where K = SO(2). The metric is invariant under G and the associated Laplacian is G-invariant, coinciding with the image of the Casimir operator. In the upper half-plane model the Laplacian is given by the formula[6][7]
${\displaystyle \displaystyle \Delta =-4y^{2}(\partial _{x}^{2}+\partial _{y}^{2}).}$
If s is a complex number and z = x + i y with y > 0, the function
${\displaystyle \displaystyle f_{s}(z)=y^{s}=\exp({s}\cdot \log y),}$
is an eigenfunction of Δ:
${\displaystyle \displaystyle \Delta f_{s}=4s(1-s)f_{s}.}$
Since Δ commutes with G, any left translate of fs is also an eigenfunction with the same eigenvalue. In particular, averaging over K, the function
${\displaystyle \varphi _{s}(z)=\int _{K}f_{s}(k\cdot z)\,dk}$
is a K-invariant eigenfunction of Δ on G/K. When
${\displaystyle \displaystyle s={1 \over 2}+i\tau ,}$
with τ real, these functions give all the zonal spherical functions on G. As with Harish-Chandra's more general formula for semisimple Lie groups, φs is a zonal spherical function because it is the matrix coefficient corresponding to a vector fixed by K in the principal series. Various arguments are available to prove that there are no others. One of the simplest classical Lie algebraic arguments[6][7][33][34][35] is to note that, since Δ is an elliptic operator with analytic coefficients, by analytic elliptic regularity any eigenfunction is necessarily real analytic. Hence, if the zonal spherical function corresponds to the matrix coefficient for a vector v and representation σ, the vector v is an analytic vector for G and
${\displaystyle \displaystyle (\sigma (e^{X})v,v)=\sum _{n=0}^{\infty }(\sigma (X)^{n}v,v)/n!}$
for X in ${\displaystyle i{\mathfrak {t}}}$. The infinitesimal form of the irreducible unitary representations with a vector fixed by K were worked out classically by Bargmann.[33][34] They correspond precisely to the principal series of SL(2,R). It follows that the zonal spherical function corresponds to a principal series representation.
Another classical argument[36] proceeds by showing that on radial functions the Laplacian has the form
${\displaystyle \displaystyle \Delta =-\partial _{r}^{2}-\coth(r)\cdot \partial _{r},}$
so that, as a function of r, the zonal spherical function φ(r) must satisfy the ordinary differential equation
${\displaystyle \displaystyle \varphi ^{\prime \prime }+\coth r\,\varphi ^{\prime }=\alpha \,\varphi }$
for some constant α. The change of variables t = sinh r transforms this equation into the hypergeometric differential equation. The general solution in terms of Legendre functions of complex index is given by[3][37]
${\displaystyle \varphi (r)=P_{\rho }(\cosh r)={1 \over 2\pi }\int _{0}^{2\pi }(\cosh r+\sinh r\,\cos \theta )^{\rho }\,d\theta ,}$
where α = ρ(ρ+1). Further restrictions on ρ are imposed by boundedness and positive-definiteness of the zonal spherical function on G.
There is yet another approach, due to Mogens Flensted-Jensen, which derives the properties of the zonal spherical functions on SL(2,R), including the Plancherel formula, from the corresponding results for SL(2,C), which are simple consequences of the Plancherel formula and Fourier inversion formula for R. This "method of descent" works more generally, allowing results for a real semisimple Lie group to be derived by descent from the corresponding results for its complexification.[38][39]
## Further directions
• The theory of zonal functions that are not necessarily positive-definite. These are given by the same formulas as above, but without restrictions on the complex parameter s or ρ. They correspond to non-unitary representations.[6]
• Harish-Chandra's eigenfunction expansion and inversion formula for spherical functions.[40] This is an important special case of his Plancherel theorem for real semisimple Lie groups.
• The structure of the Hecke algebra. Harish-Chandra and Godement proved that, as convolution algebras, there are natural isomorphisms between Cc(K \ G / K ) and Cc(A)W, the subalgebra invariant under the Weyl group.[4] This is straightforward to establish for SL(2,R).[7]
• Spherical functions for Euclidean motion groups and compact Lie groups.[6]
• Spherical functions for p-adic Lie groups. These were studied in depth by Satake and Macdonald.[41][42] Their study, and that of the associated Hecke algebras, was one of the first steps in the extensive representation theory of semisimple p-adic Lie groups, a key element in the Langlands program.
## Notes
1. Dixmier 1996, Algèbres hilbertiennes.
2. If σ is a unitary representation of G, then ${\displaystyle \sigma (f)=\int _{G}f(g)\sigma (g)\,dg}$.
3. Dieudonné 1978
4. Godement 1952
5. Helgason 2001
6. Helgason 1984
7. Lang 1985
8. Cartier & 1954-1955
9. Hochschild 1965
10. Dieudonné 1978, pp. 55–57
11. Dieudonné 1977
12. Helgason 1978, p. 249
13. Helgason 1978, pp. 257–264
14. Helgason 1984, pp. 534–538
15. Goodman & Wallach 1998, pp. 549–550
16. Goodman & Wallach 1998, p. 550
17. Helgason 1978, Chapter IX.
18. Harish-Chandra 1954a, p. 251
19. Wallach 1973
20. Berezin 1956a
21. Berezin 1956b
22. Harish-Chandra 1954b
23. Harish-Chandra 1954c
24. Kunze & Stein 1961
25. Stein 1970
26. Kostant 1969
27. Harish-Chandra 1958
28. Helgason 2001, pages 418–422, 427-434
29. Helgason 1984, p. 418
30. Davies 1990
31. Helgason 1984, pp. 432–433
32. Helgason 1984, pp. 501–502
33. Bargmann 1947
34. Howe & Tan 1992
35. Wallach 1988
36. Helgason 2001, p. 405
37. Bateman & Erdelyi 1953, p. 156
38. Flensted-Jensen 1978
39. Helgason 1984, pp. 489–491
40. Helgason 1984, pp. 434–458
41. Satake 1963
42. Macdonald 1971
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• Harish-Chandra (1954c), "The Plancherel formula for complex semisimple Lie groups", Trans. Amer. Math. Soc., American Mathematical Society, 76 (3): 485–528, doi:10.2307/1990793, JSTOR 1990793, PMC 1063477 (Second proof of formula for complex semisimple Lie groups)
• Harish-Chandra (1958), "Spherical functions on a semisimple Lie group I, II", Amer. J. Math., The Johns Hopkins University Press, 80 (2): 241–310, 553–613, doi:10.2307/2372786, JSTOR 2372786 (Determination of Plancherel measure)
• Helgason, Sigurdur (2001), Differential geometry and symmetric spaces (reprint of 1962 edition), American Mathematical Society, ISBN 0-8218-2735-9
• Helgason, Sigurdur (1978), Differential geometry, Lie groups and symmetric spaces, Academic Press, ISBN 0-12-338460-5
• Helgason, Sigurdur (1984), Groups and Geometric Analysis: Integral Geometry, Invariant Differential Operators, and Spherical Functions, Academic Press, ISBN 0-12-338301-3
• Hochschild, Gerhard P. (1965), The structure of Lie groups, Holden–Day
• Howe, Roger; Tan, Eng-chye (1992), Non-abelian Harmonic Analysis: Applications of SL(2,R), Universitext, Springer-Verlag, ISBN 0-387-97768-6
• Kostant, Bertram (1969), "On the existence and irreducibility of certain series of representations", Bull. Amer. Math. Soc., 75 (4): 627–642, doi:10.1090/S0002-9904-1969-12235-4
• Kunze, Raymond A.; Stein, Elias M. (1961), "Analytic continuation of the principal series", Bull. Amer. Math. Soc., 67 (6): 593–596, doi:10.1090/S0002-9904-1961-10705-2
• Lang, Serge (1985), SL(2,R), Graduate Texts in Mathematics, 105, Springer-Verlag, ISBN 0-387-96198-4
• Macdonald, Ian G. (1971), Spherical Functions on a Group of p-adic Type, Publ. Ramanujan Institute, 2, University of Madras
• Stein, Elias M. (1970), "Analytic continuation of group representations", Advances in Mathematics, 4 (2): 172–207, doi:10.1016/0001-8708(70)90022-8
• Satake, I. (1963), "Theory of spherical functions on reductive algebraic groups over p-adic fields", Publ. Math. IHES, 18: 5–70
• Wallach, Nolan (1973), Harmonic Analysis on Homogeneous Spaces, Marcel Decker, ISBN 0-8247-6010-7
• Wallach, Nolan (1988), Real Reductive Groups I, Academic Press, ISBN 0-12-732960-9
• Takahashi, R. (1963), "Sur les représentations unitaires des groupes de Lorentz généralisés", Bull. Soc. Math. France, 91: 289–433 | 10,754 | 38,712 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 84, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-31 | longest | en | 0.903973 |
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April 22, 2021, 09:29:08 am
### AuthorTopic: CHEMISTRY ENTHALPY HSC MCQ (Read 1328 times) Tweet Share
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#### sangarp
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##### CHEMISTRY ENTHALPY HSC MCQ
« on: November 03, 2019, 01:33:58 pm »
0
Hey need some help on this asap
How do you do this?
#### r1ckworthy
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##### Re: CHEMISTRY ENTHALPY HSC MCQ
« Reply #1 on: November 03, 2019, 01:59:24 pm »
+1
Hey need some help on this asap
How do you do this?
Hey! There might be a quicker way, but from what I can see, I would divide each value on the table to convert it from kJ/ g to kJ/mol. The value that matches 3218 is the correct fuel.
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#### sangarp
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##### Re: CHEMISTRY ENTHALPY HSC MCQ
« Reply #2 on: November 03, 2019, 02:57:03 pm »
0
hey
I actually did that
But I don't seem to get any of the values that have been given
#### r1ckworthy
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##### Re: CHEMISTRY ENTHALPY HSC MCQ
« Reply #3 on: November 03, 2019, 03:28:23 pm »
+3
Sorry! I made a silly mistake- you have to multiply all combustion values with it's molar mass in order to convert it to kJ/mol.
Molar mass comes in the form of g / mol. So, if we multiple the heat of combustion (kJ /g) of each fuel by it's molar mass (g/ mol), we convert the value to kJ /mol. Attached is an example of how I did it for one of the fuels.
Again, sorry for the silly mistake! Another lesson of why you shouldn't do things in your head
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# 09_07 - STAT 410 Examples for Fall 2011 2.4 Covariance and...
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STAT 410 Examples for 09/07/2011 Fall 2011 2.4 Covariance and Correlation Coefficient Covariance of X and Y σ XY = Cov ( X , Y ) = E [ ( X – μ X ) ( Y – μ Y ) ] = E ( X Y ) μ X μ Y (a) Cov ( X , X ) = Var ( X ) ; (b) Cov ( X , Y ) = Cov ( Y , X ) ; (c) Cov ( a X + b , Y ) = a Cov ( X , Y ) ; (d) Cov ( X + Y , W ) = Cov ( X , W ) + Cov ( Y , W ) . Cov ( a X + b Y , c X + d Y ) = a c Var ( X ) + ( a d + b c ) Cov ( X , Y ) + b d Var ( Y ) . Var ( a X + b Y ) = Cov ( a X + b Y , a X + b Y ) = a 2 Var ( X ) + 2 a b Cov ( X , Y ) + b 2 Var ( Y ) . 0. Find in terms of σ X 2 , σ Y 2 , and σ XY : a) Cov ( 2 X + 3 Y , X – 2 Y ) , b) Var ( 2 X + 3 Y ) , c) Var ( X – 2 Y ) .
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Correlation coefficient of X and Y ρ XY = Y X XY σ σ σ = ( ) ( ) ( ) , Y Var X Var Y X Cov = - - Y Y , X X σ μ σ μ Y X E (a) 1 ρ XY 1; (b) ρ XY is either + 1 or – 1 if and only if X and Y are linear functions of one another. If random variables X and Y are independent, then E ( g ( X ) h ( Y ) ) = E ( g ( X ) ) E ( h ( Y ) ) . Cov ( X, Y ) = σ XY = 0 , Corr ( X , Y ) = ρ XY = 0 . 1. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: y Recall: E ( X ) = 1.75, E ( Y ) = 0.8, E ( X Y ) = 1.5. x 0 1 2 p X ( x ) 1 0.15 0.10 0 0.25 2 0.25 0.30 0.20 0.75 p Y ( y ) 0.40 0.40 0.20 1.00 Find Cov ( X , Y ) = σ XY and Corr ( X , Y ) = ρ XY .
2. Let the joint probability density function for
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Ask a homework question - tutors are online | 792 | 2,119 | {"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-2018-13 | latest | en | 0.718919 |
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# How many miles per hour is 2 minutes per mile?
Wiki User
2014-09-02 08:39:58
There's 60 minutes to the hour, so 60/2= 30. 2 minutes/mile = 30 MPH.
Wiki User
2014-09-02 08:39:58
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https://www.coryjmaklin.com/2022-05-14_Synthetic-Minority-Over-sampling-TEchnique--SMOTE--7d419696b88c/ | 1,721,301,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00232.warc.gz | 614,671,146 | 13,404 | # Synthetic Minority Over-sampling TEchnique (SMOTE)
May 14, 2022
### Synthetic Minority Over-sampling TEchnique (SMOTE)
Synthetic Minority Over-sampling TEchnique, or SMOTE for short, is a preprocessing technique used to address a class imbalance in a dataset.
In the real world, oftentimes we end up trying to train a model on a dataset with very few examples of a given class (e.g. rare disease diagnosis, manufacturing defects, fradulent transactions) which results in poor performance. Due to the nature of the data (occurrences are so rare), it’s not always realistic to go out and acquire more. One way of solving this issue is to under-sample the majority class. That is to say, we would exclude rows corresponding to the majority class such that there are roughly the same amount of rows for both the majority and minority classes. However, in doing so, we lose out on a lot of data that could be used to train our model thus improving its accuracy (e.g. higher bias). Another other option is to over-sample the minority class. In other words, we randomly duplicate observations of the minority class. The problem with this approach is that it leads to overfitting because the model learns from the same examples. This is where SMOTE comes in. At a high level, the SMOTE algorithm can be described as follows:
• Take difference between a sample and its nearest neighbour
• Multiply the difference by a random number between 0 and 1
• Add this difference to the sample to generate a new synthetic example in feature space
• Continue on with next nearest neighbour up to user-defined number
Let’s examine the algorithm in closer detail. Suppose we had an imbalanced dataset. In other words, there are a lot more rows of a given class than the other. We proceed to plot a subset of the rows belonging to the minority class. We’re only considering 2 of the many features, here, but you could imagine the subsequent steps taking place for all N dimensions in the dataset.
We consider the first row (or a random row in the case N < 100), and compute its k nearest neighbors. We then select a random nearest neighbor out of the k nearest neighbors.
We compute the difference between the two points and multiply it by a random number between 0 and 1. This gives us a synthetic example along the line between the two points.
We repeat the process N/100 times. In other words, if the amount of over-sampling needed is 300%, only 300/100 = 3 neighbors from the k = 5 nearest neighbors are chosen and one sample is generated in the direction of each. In this case, we set N = 300. Therefore, we consider 3 random nearest neighbors of each point.
We then move on to the next row, compute its k nearest neighbors and select N/100 = 300/100 = 3 of its nearest neighbors at random to use in generating new synthetic examples.
#### SMOTE in Python
Let’s walk through an example of using SMOTE in Python.
We begin by importing the required libraries.
``````from random import randrange, uniform
from sklearn.neighbors import NearestNeighbors
import numpy as np
import pandas as pd
from sklearn.ensemble import RandomForestClassifier
from sklearn.model_selection import train_test_split
from sklearn.metrics import confusion_matrix, recall_score``````
In this example, we will make use of the Credit Card Fraud Detection dataset on Kaggle to train a model to determine whether a given transaction is fraudulent. We read the CSV file and store its contents in a Pandas DataFrame as follows:
``df = pd.read_csv("creditcard.csv")``
Unfortunately, due to confidentiality issues, they cannot provide the original features. Features V1, V2, … V28 are the principal components obtained with PCA, the only features which have not been transformed with PCA are ‘Time’ and ‘Amount’.
``df.head(5)``
As we can see, there are significantly more negative samples than positive samples.
``df['Class'].value_counts()``
``````Out:
0 284315
1 492
Name: Class, dtype: int64``````
For simplicity, we remove the time dimension.
``df = df.drop(['Time'], axis=1)``
We split the dataset into features and labels.
``````X = df.drop(['Class'], axis=1)
y = df['Class']``````
In order to evaluate the performance of our model, we split the data into training and test sets.
``X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)``
Next, we initialize an instance of the RandomForestClassifier class.
``rf = RandomForestClassifier(random_state=42)``
We fit our model to the training set.
``rf.fit(X_train, y_train)``
Finally, we use our model to predict whether a transaction is fraudulent given what it has learnt.
``y_pred = rf.predict(X_test)``
Suppose our dataset contained 100 examples of fraudulent and 9900 examples of regular transactions. If we used accuracy to measure the model’s performance, it could obtain an accuracy of 99% by simply predicting false every time. It’s for this reason that we use a confusion matrix to evaluate the model’s performance. As we can see, our model classified 23 samples as non-fraudulent when, in fact, they were.
``confusion_matrix(y_test, y_pred)``
``````Out:
array([[56862, 2],
[ 23, 75]])``````
For ease of comparison, if we wanted a single number to gauge the model’s performance, we could use recall. Recall (dad joke) that recall is equal to the number of true positives divided by the sum of true positives and false negatives.
``recall_score(y_test, y_pred)``
``Out: 0.7653061224489796``
#### SMOTE from scratch
SMOTE was first described by Nitesh Chawla, et al. in their 2002 white paper named for the technique titled “SMOTE: Synthetic Minority Over-sampling Technique.” In said whitepaper, they provide the following pseudo code:
We proceed to implement the the latter in Python.
``````def SMOTE(sample: np.array, N: int, k: int) -> np.array:
T, num_attrs = sample.shape
# If N is less than 100%, randomize the minority class samples as only a random percent of them will be SMOTEd
if N < 100:
T = round(N / 100 * T)
N = 100
# The amount of SMOTE is assumed to be in integral multiples of 100
N = int(N / 100)``````
``````synthetic = np.zeros([T * N, num_attrs])
new_index = 0``````
``nbrs = NearestNeighbors(n_neighbors=k+1).fit(sample.values)``
``````def populate(N, i, nnarray):
nonlocal new_index
nonlocal synthetic
nonlocal sample``````
``````while N != 0:
nn = randrange(1, k+1)
for attr in range(num_attrs):
dif = sample.iloc[nnarray[nn]][attr] - sample.iloc[i][attr]
gap = uniform(0, 1)
synthetic[new_index][attr] = sample.iloc[i][attr] + gap * dif``````
`````` new_index += 1
N = N - 1
for i in range(T):
nnarray = nbrs.kneighbors(sample.iloc[i].values.reshape(1, -1), return_distance=False)[0]
populate(N, i, nnarray)
return synthetic``````
If you didn’t understand it at first glance, don’t fret. We’ll walk-through the code step by step.
The algorithm assumes that if N > 100 than it’s a multiple of 100. In the event N is less than 100 then we select a subset of the samples. For example, if N were 50 (e.g. 50%), we’d want the length of our synthetic array to be 50 / 100 * T = 0.5T where T is the length of the original array of rows belonging to the minority class. We also set N to 100, here, so that it becomes 1 in the subsequent line since we only want to generate a synthetic sample using 1 of the nearest neighbors of each point in the subset.
`````` if N < 100:
T = round(N / 100 * T)
N = 100
N = int(N / 100)``````
We use k+1, here, because the implementation of NearestNeighbors considers the point itself as one of the neighbors. In other words, we want the k nearest neighbors excluding the point itself.
``nbrs = NearestNeighbors(n_neighbors=k+1).fit(sample.values)``
In Python, `nonlocal` ensures the variable references the “closest” (in this case, the scope outside the function) variable with the same name in the source code.
``````nonlocal new_index
nonlocal synthetic
nonlocal sample``````
We select a nearest neighbor at random, by selecting a number between 1 and k+1 because like we mentioned before, the implementation of NearestNeighbors considers the point itself as one of the nearest neighbors.
``nn = randrange(1, k+1)``
We loop through the different attributes in order to stay true to the algorithm described above. However, it’s worth noting that there is a more efficient way of doing this using the numpy APIs.
``for attr in range(num_attrs):``
We create a new example by taking the difference between the point being considered and a random neighbor then multiplying it by a number between 0 and 1.
``````dif = sample.iloc[nnarray[nn]][attr] - sample.iloc[i][attr]
gap = uniform(0, 1)
synthetic[new_index][attr] = sample.iloc[i][attr] + gap * dif``````
We move onto the next available index in our array and decrement N to indicate that we already considered one of the N/100 nearest neighbors.
``````new_index += 1
N = N - 1``````
We obtain the nearest neighbors for a point in the original array and call the populate function.
``````for i in range(T):
nnarray = nbrs.kneighbors(sample.iloc[i].values.reshape(1, -1), return_distance=False)[0]
populate(N, i, nnarray)``````
Prior to running the algorithm, we select all the fraudulent rows within our dataset.
``minority = df[df['Class'] == 1].drop(['Class'], axis=1)``
We set k to 5 meaning that for each row we will randomly select N/100 nearest neighbors from the available k = 5 to use in our calculations (assuming N ≥ 100). We set N to 200 meaning that we want to generate 200% more fraudulent examples.
``synthetic = SMOTE(minority, N=200, k=5)``
As we can see, the array of synthetic examples has twice the number of rows as the original dataset.
``synthetic.shape``
``````Out:
(984, 29)``````
Next, we concatenate the original samples with the samples we just generated and set the label to 1 for a total of 984 + 492 = 1476 samples.
``````synthetic_df = pd.DataFrame(synthetic, columns=minority.columns)
combined_minority_df = pd.concat([minority, synthetic_df])
combined_minority_df["Class"] = 1``````
Finally, we combine the fraudulent and non-fraudulent samples into a single DataFrame.
``new_df = pd.concat([combined_minority_df, df[df['Class'] == 0]])``
Like we did before, we split the data into training and testing datasets, train the model and classify the rows in the testing dataset.
``````X = new_df.drop(['Class'], axis=1)
y = new_df['Class']``````
``X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)``
``rf = RandomForestClassifier(random_state=42)``
``rf.fit(X_train, y_train)``
``y_pred = rf.predict(X_test)``
If we look at the confusion matrix, we can clearly see that the model has a similar number of false negatives despite the fact that there are 3 times more positive samples.
``confusion_matrix(y_test, y_pred)``
``````Out:
array([[56844, 0],
[ 24, 291]])``````
The recall score is much higher than the model that was trained on the dataset without the use of SMOTE.
``recall_score(y_test, y_pred)``
``0.9238095238095239``
#### SMOTE using library
The Python implementation of SMOTE actually comes in its own library (outside Scikit-Learn) which can be installed as follows:
``pip install imbalanced-learn``
We can then import the SMOTE class.
``from imblearn.over_sampling import SMOTE``
To avoid confusion, we read the csv file again.
``df = pd.read_csv("creditcard.csv")``
``df = df.drop(['Time'], axis=1)``
``````X = df.drop(['Class'], axis=1)
y = df['Class']``````
We instantiate an instance of the SMOTE class. It’s worth noting that, by default, it will ensure that there are an equal number of positive samples as negative samples.
``sm = SMOTE(random_state=42, k_neighbors=5)``
We apply the SMOTE algorithm to the dataset as follows:
``X_res, y_res = sm.fit_resample(X, y)``
Again, we split the dataset, train the model and predict whether the samples in the testing dataset should be considered fraudulent or not.
``X_train, X_test, y_train, y_test = train_test_split(X_res, y_res, test_size=0.2, random_state=42)``
``````rf = RandomForestClassifier(random_state=42)
rf.fit(X_train, y_train)``````
``y_pred = rf.predict(X_test)``
If we look at the confusion matrix, we can see that there are an equal number of positive samples as negative samples and the model didn’t have any false negatives. Ergo, the recall is 1.
``````confusion_matrix(y_test, y_pred)
Out:
array([[56737, 13],
[ 0, 56976]])``````
``recall_score(y_test, y_pred)``
``Out: 1.0``
#### Conclusion
When a machine learning model is trained on an imbalanced dataset it tends to perform poorly. When acquiring more data isn’t an option, we have to resort to down-sampling or up-sampling. Down-sampling is bad because it removes samples that could otherwise have been used to train the model. Up-sampling on its own is less than ideal since it causes our model to overfit. SMOTE is a technique to up-sample the minority classes while avoiding overfitting. It does this by generating new synthetic examples close to the other points (belonging to the minority class) in feature space.
Written by Cory Maklin Genius is making complex ideas simple, not making simple ideas complex - Albert Einstein You should follow them on Twitter | 3,309 | 13,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.935393 |
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label Mathematics
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schedule 1 Day
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May 29th, 2015
The graph touches at x=-3 and x=2. So we will have even powers on (x+3) and (x-2) and crosses at x=-1 and x=3. So we have odd powers on on (x+1) and (x-3). Which is true for all options.
and we see that f(0) is nearly equal to -12. Lets check all the options.
And we find that f(0)=-12 for only first option. So correct choice is A
May 29th, 2015
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May 29th, 2015
Oct 19th, 2017
check_circle | 205 | 616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-43 | latest | en | 0.932882 |
https://physicsgre.com/viewtopic.php?f=3&t=1954&view=print | 1,611,848,969,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00509.warc.gz | 478,131,174 | 3,287 | Page 1 of 1
### Mental arithmetic
Posted: Mon Dec 15, 2008 10:08 pm
I'm one of those theoretical physicists who is notoriously bad at this, especially when there are nasty square roots etc.
Do you know of any techniques for increasing speed/accuracy other than attempting all the PGRE questions without a calculator.
Is there any software that can `train' you at this sort of stuff? My exam isn't until Oct '08, so I've got plenty of time.
### Re: Mental arithmetic
Posted: Tue Dec 16, 2008 1:36 am
I used to be pretty poor at it as well. I am still not great. What helped me was tutoring lots of introductory/remedial algebra. Sounds painful I know but it has been more beneficial than not.
For square roots, I think the best thing is to know your perfect squares up to like (15)^2. Then when you have one that isn't a perfect square, you know the perfect roots it is likely in between.
For multiplication of stuff with double digits (or more) I just factor everything and then try to multiply.
Like 27 * 14 = (9*3)*(7*2)
I normally multiply the "hard" stuff (numbers bigger than 4 say) first because its easier to then double or triple it at the end.
So from above = (63)*3*2 = 189*2 = 378. Sounds elementary school, and it is, but I find it speeds me up quite a bit. Plus its great for scaring freshman algebra students when you have it from your head before they get it on their calculator. Never understood why some people think rote calculations in your head is more impressive than doing algebra or calc in your head. Hope that's helpful.
### Re: Mental arithmetic
Posted: Tue Dec 16, 2008 1:41 am
people act like your a genius if you can solve a rubiks cube http://www.rubikssolver.com/.
I never will get it.
### Re: Mental arithmetic
Posted: Tue Dec 16, 2008 3:24 am
cato88 wrote:people act like your a genius if you can solve a rubiks cube http://www.rubikssolver.com/.
I never will get it.
I learned from websites how to solve it and can do so easily now. However, I feel like anyone who has figured out how to do that thing without any outside help deserves to get into any grad school there is
This guy made a robot out of a Lego set that scans the cube and then solves it. Here's a video.
### Re: Mental arithmetic
Posted: Tue Dec 16, 2008 7:37 am
Kaiser_Sose,
You have an interesting way of doing it. I would usually multiply things like that by breaking one of them into 10 + something, e.g.,
27*14 = 270 + 27*4 = 270 + 108 = 378.
I'm not very fast though.
I'm not sure I understand your technique with square roots. How would you approach e.g., sqrt(2.6) ?
### Re: Mental arithmetic
Posted: Tue Dec 16, 2008 7:59 am
http://en.wikipedia.org/wiki/Mental_calculation
I especially like the method for approximating square roots
### Re: Mental arithmetic
Posted: Tue Dec 16, 2008 2:38 pm
Ew sqrt(2.6) ?
In that case there's really two ways to think of it.
1.) Its gotta be bigger than one and less than two, by intuition. So a ballpark answer would be 1.5.
2.) You can get closer if you happen to know to sqrt(2) and sqrt(3), which are about 1.4 and 1.7 respectively. So a closer guess is probably 1.6.
This method works out more neatly is you have a larger number say, 53. The closest square roots are 49 and 64 which are 7^2 and 8^2 so you know that sqrt(53) is 7 and some change.
I tend to use your method, noospace, when confronted with multiplying decimals, like when I get an approximation from the above method.
Say I have 5*sqrt(53) ----> 5*(7.3) = 5*(7+0.3) = 35 + 1.5 = 36.5
Hope that's clear. | 954 | 3,549 | {"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} | 3.234375 | 3 | CC-MAIN-2021-04 | latest | en | 0.957023 |
http://mathoverflow.net/questions/138620/motives-over-the-complex-numbers-versus-mixed-hodge-structures | 1,469,612,488,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00326-ip-10-185-27-174.ec2.internal.warc.gz | 176,368,576 | 16,192 | # Motives over the complex numbers versus mixed Hodge structures
Let $\mathsf{MM}(\mathbf C)$ be the hypothetical category of mixed motives over the complex numbers, and consider the realization functor $\Phi : \mathsf{MM}( \mathbf C) \to \mathsf{MHS}$ to integral mixed Hodge structures.
Is $\Phi$ expected to be faithful? If not, what kind of information does one expect to lose?
-
I don't think the question has much meaning. Even in the pure setting, the answer depends on which equivalence relation you use to define the category of motives. In any case, you should ask the question on the level of triangulated categories. Then a number of candidates for the triangulated category of motivic complexes have been defined, and your question becomes more precise. – abz Aug 5 '13 at 21:58
I do not agree with Anon's comment, at least not fully. In the setting of pure motives, the question makes perfect sense, independently on the equivalence relation used, the key word being "expected" : indeed the different equivalence relations are expected to yield the same categories of pure motives. Moreover the answer (in the pure setting) is expected to be yes, by the Hodge conjecture. – Joël Aug 5 '13 at 22:10
As for mixed motives, I am much less sure, and perhaps what I am going to say is utterly naive, but the category of mixed motives should be a nice abelian categories, with a realization functor to MHS as the OP says, and it makes perfect sense to ask if such functor is expected be faithful. +1 for the question. – Joël Aug 5 '13 at 22:15
The question also seems to presume some kind of integral theory of motives. Certainly, if we are considering rational Hodge structures and pure motives, then, as Joel points out, this comes down to the Hodge conjecture + a Standard conjecture. EDIT: Actually, if we only require faithfulness, then we don't even need the Hodge conjecture, just the standard conjecture on num. eq.=hom. eq. – Keerthi Madapusi Pera Aug 5 '13 at 22:35
@Joel: If you take rational equivalence, the answer is no; if you take homological equivalence, the answer is yes. But, as I said, this should be asked for triangulated categories of motivic categories, where at least there are candidates for the motivic category. – abz Aug 5 '13 at 23:23
Yes, this functor is expected to be faithful, but this has very little to do with Hodge theory. The reason for this is that, if $\Psi$ denotes the forgetful functor from integral mixed Hodge structures to abelian groups, then the composed functor $\Psi\circ\Phi$ simply is the Betti realization functor. As $\Psi$ itself is faithful, it is thus sufficient to prove that the Betti realization functor is faithful. The reason why the Betti realization functor is expected to be faithful is the following. The abelian category $\mathsf{MM}(\mathbf{C})$ is expected to be the heart of the motivic $t$-structure on the triangulated category $\mathsf{DM}_{et}(\mathbf{C})$ of Voevodsky's constructible (or geometric) étale motives (the one obtained from étale sheaves with transfers; the triangulated category of motives obtained from Nisnevich sheaves with transfers (which is known to coincide with $\mathsf{DM}_{et}(\mathbf{C})$ with rational coefficients) is known not to have a motivic $t$-structure with integral coefficients anyway). The faithfulness of the Betti realization functor as above boils down to the following two properties of the triangulated version of the Betti realization functor $$R:\mathsf{DM}_{et}(\mathbf{C})\to\mathsf{D}^b(\mathit{Ab})$$ 1) it must be conservative;
2) it must be $t$-exact.
Note that the functor $R$ is actually well defined and that property 1) is equivalent to its version with rational coefficients (because, for torsion coefficients, the functor $R$ is known to be an equivalence, by the rigidity theorem of Suslin and Voevodsky). Property 2) is required in the very definition of a motivic $t$-structure. This essentially means that, for this problem, the main difficulty is about rational coefficients. This is then very related with the standard conjectures, as explained in this paper of Beilinson: arXiv:1006.1116.
Finally, it might be worth to mention that there is an actual candidate for $\mathsf{MM}(\mathbf{C})$, constructed by Madhav Nori, which, by definition, comes with an exact and faithful functor to $\mathsf{MHS}$. | 1,060 | 4,353 | {"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} | 2.53125 | 3 | CC-MAIN-2016-30 | latest | en | 0.936538 |
https://www.kidsacademy.mobi/printable-worksheets/age-7/math/geometry/learning-skills/problem-solving-skills/ | 1,725,711,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00304.warc.gz | 809,323,522 | 64,110 | # Problem-Solving Skills Geometry Worksheets for 7-Year-Olds
Boost your child’s problem-solving abilities with our Geometry Worksheets tailored for 7-year-olds! These printable activities from Kids Academy are designed to make learning fun and engaging, effectively introducing fundamental geometric concepts. Through a series of thoughtfully crafted exercises, children will enhance their critical thinking and analytic skills while mastering shapes, patterns, and spatial understanding. Perfect for classroom settings or at-home learning, these worksheets help build a strong foundation in math, preparing young minds for future academic success. Transform math practice into an enjoyable adventure with our dynamic and educational geometry worksheets!
Check out this FREE Trial Lesson on Geometry for age 7!
Counting Sides/ Angles 2D
Favorites
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• 7
• Problem-Solving Skills
• Geometry
## Shared Food Worksheet
In this worksheet, kids learn about sharing and cutting shapes into equal parts. There are two groups of friends: one with two, the other with four. Ask your child which group has their food cut into smaller shares, and help them find the right answer.
Shared Food Worksheet
Worksheet
## Mummy Maze: Pyramid Printable
Let's learn to spot 3D shapes, such as the iconic pyramid! This fun pyramid geometry worksheet PDF provides interesting illustrations and a maze to help your child identify pyramids in the pictures.
Mummy Maze: Pyramid Printable
Worksheet
## Golf Maze: Curved and Flat Surfaces Worksheet
Give your child a fun and educational challenge with this geometry maze! Your child will navigate 3D shapes to find objects with flat and curved surfaces, while learning important math concepts.
Golf Maze: Curved and Flat Surfaces Worksheet
Worksheet
## Shape Maze Worksheet
Test your child's geometry knowledge with this fun printable maze! They'll need to identify shapes with 8 vertices to find their way through the maze and get the correct answer.
Shape Maze Worksheet
Worksheet
## Sharing Fruit Circles Worksheet
Fruits are tasty and bright! Can your kids name some of their favorites? With this worksheet, you can use fruit to teach your kids geometry. Talk about how shapes can be cut into halves. Ask them to tick the box for the fruit halves in the printout.
Sharing Fruit Circles Worksheet
Worksheet
## Missing Shapes: Trains Worksheet
Before beginning, ask your child what the object in the picture is, the sound it makes and where it can be found. If they know, confidently move on. Help them make the second train look like the first by tracing dotted lines. Hold their hand for guidance.
Missing Shapes: Trains Worksheet
Worksheet
## The Circle Maze Worksheet
Kids can have fun learning shapes with this maze worksheet. They can help the mouse find its cheese by tracing the path of a circle through the maze. This printable is great for classroom learning or as an activity at home.
The Circle Maze Worksheet
Worksheet
## Matching 2D and 3D Shapes Worksheet
Before starting this worksheet, ask your kids what shapes they know and point out the 2D and 3D shapes on it. Help your kids trace the 2D shapes and then match them to the 3D versions. It'll be a fun new world for your kids to explore!
Matching 2D and 3D Shapes Worksheet
Worksheet
## Piece it together Worksheet
Help your kids learn fractions! Even if they're not eager, teach them as they get older. Teachers will provide enough lessons and homework, but you can go a step further. Look at a worksheet with your kids and help them circle the piece that would make the pie whole.
Piece it together Worksheet
Worksheet
Learning Skills
Parents and teachers should prioritize problem-solving skills in geometry for 7-year-olds because of the foundational thinking and practical benefits these skills provide. At this crucial developmental stage, children are naturally curious and able to grasp new concepts with ease, making it an ideal time to introduce geometric problem-solving. Understanding shapes, sizes, spaces, and the properties of objects enhances spatial awareness, which is important in everyday tasks and fundamental in fields such as engineering, physics, and the arts.
Engaging in geometry promotes critical thinking and analytical skills by encouraging children to observe patterns, ask questions, and discern relationships among geometric figures. This form of logical reasoning extends beyond math class and aids children in making informed decisions and solving puzzles in varied aspects of life. One worksheet on identifying different shapes or figuring out how to piece puzzles together stimulates young minds to think independently and sequence steps logically.
Moreover, having strong problem-solving abilities bolsters confidence in confronting challenges, creating a growth mindset that children carry into other subjects and future grades. Geometry is not just about memorizing shapes but about fostering precision, resilience, and creativity, qualities essential for lifelong learning and success. By investing in these skills early, parents and teachers set children on a path of curiosity-driven excellence. | 992 | 5,165 | {"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-2024-38 | latest | en | 0.889237 |
https://www.soffensive.com/2017/07/ | 1,590,586,396,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00280.warc.gz | 915,067,918 | 22,525 | ## Sunday, 30 July 2017
### Practical Reverse Engineering Exercise Solutions: Page 35 / Exercise 7
Exercise 7 on page 35:
Sample H. The function sub_10BB6 has a loop searching for something.
First recover the function prototype and then infer the types based on the
context. Hint: You should probably have a copy of the PE specifi cation
nearby.
Due to alignment issues, our routine is located at 10BB2 and has the following disassembly:
sub_10BB2:
mov eax, [esp+4]
push ebx
push esi
mov esi, [eax+3Ch]
movzx eax, word ptr [esi+14h]
xor ebx, ebx
cmp [esi+6], bx
push edi
lea edi, [eax+esi+18h]
jbe short loc_10BEB
loc_10BCE:
push [esp+0Ch+arg_4]
push edi
call ds:dword_169A4
test eax, eax
pop ecx
pop ecx
jz short loc_10BF3
movzx eax, word ptr [esi+6]
inc ebx
cmp ebx, eax
jb short loc_10BCE
loc_10BEB:
xor eax, eax
loc_10BED:
pop edi
pop esi
pop ebx
retn 8
loc_10BF3:
mov eax, edi
jmp short loc_10BED
The PE file format and offsets have been described in detail here: http://www.sunshine2k.de/reversing/tuts/tut_pe.htm
Other useful sites are:
http://www.csn.ul.ie/~caolan/pub/winresdump/winresdump/doc/pefile2.html
https://github.com/corkami/pics/blob/master/binary/pe101/pe101.pdf
https://sourceforge.net/p/mingw/mingw-org-wsl/ci/master/tree/include/winnt.h
Firstly, we notice that the program contains the statement retn 8, which means it takes two arguments that are passed on the stack. Thus, its abstract prototype looks as follows:
sub_10BB2(param1, param2);
The first parameter seems to be a pointer that fetches a value from the offset 0x3C. Looking at the PE file format specification, we see the offset at 0x3C of the DOS header structure contains a pointer to the PE header structure. Therefore, we assume that param1 points to a DOS header file structure.
We refine the function prototype to:
Below, the disassembly including the provisional decompilation is provided:
mov eax, [esp+4] // function has first argument at esp+4, which means it is located on the stack
push ebx
push esi
mov esi, [eax+3Ch] // long offsetpeheader = PIMAGE_DOS_HEADER->lfanew;
xor ebx, ebx // int countSections = 0;
cmp [esi+6], bx // word numSections = pNtHeaders->FileHeader.NumberOfSections
push edi // save edi for later
lea edi, [eax+esi+18h] // int* sectionTable = pNtHeaders->OptionalHeader + sizeOptHeader; AND! the end address is immediately the beginning of the new
data structure, i.e. the first section! SectionTable is located at offset(OptionalHeader) + SizeOfOptionalHeader
jbe short loc_10BEB // when we have no sections (countSections == numSections) , go to exit routine
loc_10BCE:
push dword ptr [esp+14h]
push edi
call ds:dword_169A4 // dword_169A4(sectionTable, param2)
test eax, eax // check if returned 0
pop ecx // pop value from stack to ecx
pop ecx // pop value from stack to ecx
jz short loc_10BF3 // if eax == 0 , jump
movzx eax, word ptr [esi+6] // reload number of Header
add edi, 28h // sectionTable += 0x28; // one section table has a size of 0x28 bytes (40 bytes)
inc ebx // countSections += 1;
cmp ebx, eax // if countSections < numSections goto loc_10BCE:
jb short loc_10BCE
loc_10BEB:
xor eax, eax // int* retVal = 0;
loc_10BED:
// clean up stack (stdcall convention)
pop edi
pop esi
pop ebx
retn 8 // function has two parameters, 2 x 4 bytes
loc_10BF3:
mov eax, edi // int* retVal = sectionTable;
jmp short loc_10BED
Finally, we arrive at the following decompiled function:
int countSections = 0;
while (countSections < numSections) {
int retVal = dword_169A4(sectionTablePtr, param2); // call unknown function
if (retVal == 0) {
return sectionTablePtr;
}
sectionTablePtr += 0x28; // get the next sectionTable
countSections += 1; // set the number of processed sections
}
return 0; // we have not found the target section
}
## Saturday, 22 July 2017
### Practical Reverse Engineering Exercise Solutions: Page 35 / Exercise 6
Exercise 6 on page 35 of the book Practical Reverse Engineering presents us with a malware samples.
These can be downloaded at the following page:
In this exercise, we are expected to have a look at the following routine sub_13842:
.text:00013842 sub_13842
.text:00013842 mov eax, [ecx+60h]
.text:00013845 push esi
.text:00013846 mov esi, [edx+8]
.text:00013849 dec byte ptr [ecx+23h]
.text:0001384C sub eax, 24h
.text:0001384F mov [ecx+60h], eax
.text:00013852 mov [eax+14h], edx
.text:00013855 movzx eax, byte ptr [eax]
.text:00013858 push ecx
.text:00013859 push edx
.text:0001385A call dword ptr [esi+eax*4+38h]
.text:0001385E pop esi
.text:0001385F retn
Firstly, we see that the function prototype takes two parameters, which are not saved on the stack but in the two registers ecx and edx. This can be deducted from the fact that these two registers are immediately referenced without prior initialization.
On a very high abstraction level, the function prototype looks as follows:
sub_13842(struct1* a, struct2* b);
Moreover, we see that edx+8 points to some kind of base address, as it is used to compute the address of the function called on the following line:
.text:0001385A call dword ptr [esi+eax*4+38h]
Both parameters edx and ecx obviously point to some kind of data structure, whose contents we are about to reveal:
struct1:
offset+60: struct3*: s3
offset+23: char: unknown0x23
struct2:
offset+08: void*: here we have some kind of trampoline / array of function pointers
struct3:
offset+00: char: index
offset+14: struct2*: s2
struct4:
offset+38: int*: unknown
sub_13842(struct1* a, struct2* b)
.text:00013842 sub_13842
.text:00013842 mov eax, dword ptr[ecx+60h] // struct3* v1 = a->s3;
.text:00013845 push esi // just save esi register value
.text:00013846 mov esi, [edx+8] // void* v2 = b->unknown0x8
.text:00013849 dec byte ptr [ecx+23h] // a->unknown0x23 = a->unknown0x23 - 1;
.text:0001384C sub eax, 24h // v1 = v1 - 0x24;
.text:0001384F mov [ecx+60h], eax // a->s3 = v1;
.text:00013852 mov [eax+14h], edx // v1->s2 = b;
.text:00013855 movzx eax, byte ptr [eax] // char index = v1->index;
.text:00013858 push ecx
.text:00013859 push edx // call fct(
.text:0001385A call dword ptr [esi+eax*4+38h] // call b->unknown0x8+0x38h[index](b, a)
.text:0001385E pop esi
.text:0001385F retn
Overall, the decompiled code looks as follows:
sub_13842(struct1* a, struct2* b) {
struct3* v1 = a->s3;
void* v2 = b->unknown0x8;
a->unknown0x23 = a->unknown0x23 -1;
v1 = v1 - 0x24;
a->s3 = v1;
v1->s2 = b;
char index = v1->index;
b->unknown0x8+0x38h[index](b, a);
}
## Sunday, 16 July 2017
### Practical Reverse Engineering Exercise Solutions: RtlValidateUnicodeString
This blog post contains my solution for the decompilation exercise of the RtlValidateUnicodeString function in the Windows Kernel. The following contains the disassembly without annotations:
kd> uf rtlvalidateunicodestring
ntdll!RtlValidateUnicodeString:
77686f6c 8bff mov edi,edi
77686f6e 55 push ebp
77686f6f 8bec mov ebp,esp
77686f71 837d0800 cmp dword ptr [ebp+8],0
77686f75 0f85fc380300 jne ntdll!RtlValidateUnicodeString+0xb (776ba877)
ntdll!RtlValidateUnicodeString+0x12:
77686f7b 6800010000 push 100h
77686f80 ff750c push dword ptr [ebp+0Ch]
77686f83 e809000000 call ntdll!RtlUnicodeStringValidateEx (77686f91)
ntdll!RtlValidateUnicodeString+0x1f:
77686f88 5d pop ebp
77686f89 c20800 ret 8
ntdll!RtlValidateUnicodeString+0xb:
776ba877 b80d0000c0 mov eax,0C000000Dh
776ba87c e907c7fcff jmp ntdll!RtlValidateUnicodeString+0x1f (77686f88)
The function prototype is given here:
NTSTATUS NTAPI RtlValidateUnicodeString(IN ULONG Flags, IN PCUNICODE_STRING UnicodeString);
Note that the function returns a NTSTATUS value, which is publicly documented by Microsoft at https://msdn.microsoft.com/en-us/library/cc704588.aspx.
Other relevant data structures are _unicode_string:
kd> dt nt!_unicode_string
+0x000 Length : Uint2B
+0x002 MaximumLength : Uint2B
+0x004 Buffer : Ptr32 Uint2B
The following listing gives the disassembly with my annotations / comments:
kd> uf rtlvalidateunicodestring
ntdll!RtlValidateUnicodeString:
; hot-patch point
77686f6c 8bff mov edi,edi
; function prologue
77686f6e 55 push ebp
77686f6f 8bec mov ebp,esp
; check if Flags are set to zero
77686f71 837d0800 cmp dword ptr [ebp+8],0
77686f75 0f85fc380300 jne ntdll!RtlValidateUnicodeString+0xb (776ba877)
; Flags are set to zero
ntdll!RtlValidateUnicodeString+0x12:
77686f7b 6800010000 push 100h
77686f80 ff750c push dword ptr [ebp+0Ch]
; call RtlUnicodeStringValidateEx(UnicodeString, 0x100)
77686f83 e809000000 call ntdll!RtlUnicodeStringValidateEx (77686f91)
ntdll!RtlValidateUnicodeString+0x1f:
; function epilogue
77686f88 5d pop ebp
77686f89 c20800 ret 8
ntdll!RtlValidateUnicodeString+0xb:
; flags are not set to zero
776ba877 b80d0000c0 mov eax,0C000000Dh
Function return value is set to 0x0C000000D
776ba87c e907c7fcff jmp ntdll!RtlValidateUnicodeString+0x1f (77686f88)
Finally, the decompiled function is provided:
NTSTATUS NTAPI RtlValidateUnicodeString(IN ULONG Flags, IN PCUNICODE_STRING UnicodeString){
if (Flags != 0) {
return 0x0C000000D // corresponding NTSTATUS: STATUS_INVALID_PARAMETER
}
RtlUnicodeStringValidateEx(UnicodeString, 0x100);
return
}
### Practical Reverse Engineering Exercise Solutions: LiveKd / WinDbg Cheat Sheet
Here are a couple of commands I regularly use for reverse engineering:
uf <function>: Unassemble function
dt nt!_ktss: Show the definition of the data structure _ktss
?? sizeof(_ktss): Show the size the data structure _ktss occupies in memory
.hh uf: Show help for the function uf
x nt!*createfile*: Search all functions having the string "createfile" in its name
!vtop <PDPT-pointer> <virtualAddress>: Compute physical address of given virtual address and the pointer to the page directory pointer table
### Practical Reverse Engineering Exercise Solutions: KiInitializeTSS
Another exercise for us is the decompilation of the KiInitializeTSS function:
nt!KiInitializeTSS:
82847359 8bff mov edi,edi
8284735b 55 push ebp
8284735c 8bec mov ebp,esp
8284735e 8b4508 mov eax,dword ptr [ebp+8]
82847361 b9ac200000 mov ecx,20ACh
82847366 66894866 mov word ptr [eax+66h],cx
8284736a 33c9 xor ecx,ecx
8284736c 6a10 push 10h
8284736e 66894864 mov word ptr [eax+64h],cx
82847372 66894860 mov word ptr [eax+60h],cx
82847376 59 pop ecx
82847377 66894808 mov word ptr [eax+8],cx
8284737b 5d pop ebp
8284737c c20400 ret 4
We obtain the function prototype: (source)
VOID
NTAPI
KiInitializeTSS(IN PKTSS Tss)
{
}
Structure of _KTSS:
kd> dt nt!_ktss
+0x000 Backlink : Uint2B
+0x002 Reserved0 : Uint2B
+0x004 Esp0 : Uint4B
+0x008 Ss0 : Uint2B
+0x00a Reserved1 : Uint2B
+0x00c NotUsed1 : [4] Uint4B
+0x01c CR3 : Uint4B
+0x020 Eip : Uint4B
+0x024 EFlags : Uint4B
+0x028 Eax : Uint4B
+0x02c Ecx : Uint4B
+0x030 Edx : Uint4B
+0x034 Ebx : Uint4B
+0x038 Esp : Uint4B
+0x03c Ebp : Uint4B
+0x040 Esi : Uint4B
+0x044 Edi : Uint4B
+0x048 Es : Uint2B
+0x04a Reserved2 : Uint2B
+0x04c Cs : Uint2B
+0x04e Reserved3 : Uint2B
+0x050 Ss : Uint2B
+0x052 Reserved4 : Uint2B
+0x054 Ds : Uint2B
+0x056 Reserved5 : Uint2B
+0x058 Fs : Uint2B
+0x05a Reserved6 : Uint2B
+0x05c Gs : Uint2B
+0x05e Reserved7 : Uint2B
+0x060 LDT : Uint2B
+0x062 Reserved8 : Uint2B
+0x064 Flags : Uint2B
+0x066 IoMapBase : Uint2B
+0x068 IoMaps : [1] _KiIoAccessMap
+0x208c IntDirectionMap : [32] UChar
Translaction to C:
VOID NTAPI iInitializeTSS(IN PKTSS Tss)
{
Tss->IoMapBase = 0x20AC;
Tss->Flags = 0;
Tss->LDT = 0;
Tss->Ss0 = 0x10;
}
Although we are technically done with the decompilation, it is worthwhile to note the meaning of the hexadecimal values.
While hexadecimal 0x10 is 16 in decimal notation, 0x20AC is 8364.
We can obtain the size of the data structure _KTSS in WinDbg with the following command. The size coincides with the value assigned to IoMapBase:
kd> ?? sizeof(_KTSS)
unsigned int 0x20ac
### Practical Reverse Engineering Exercise Solutions: KeReadyThread
Unfortunately I had no time in the past days to continue with the exercises. We continue with the decompilation of the KeReadyThread function in Windows 7.
The following listing shows the disassembly:
828a8125 8bff mov edi,edi
828a8127 56 push esi
828a8128 8bf0 mov esi,eax
828a812a 8b4650 mov eax,dword ptr [esi+50h]
828a812d 8b4874 mov ecx,dword ptr [eax+74h]
828a8130 f6c107 test cl,7
828a8135 e8b74af8ff call nt!KiInSwapSingleProcess (8282cbf1)
828a813a 84c0 test al,al
828a8143 5e pop esi
828a8144 c3 ret
According to this source, it has the following prototype:
VOID
)
So, the function KeReadyThread takes one parameter, but does not reference it with the ebp register and the last instruction is a simple RET. The PKTHREAD structure is given below:
0: kd> dt nt!_kthread
+0x010 CycleTime : Uint8B
+0x018 HighCycleTime : Uint4B
+0x020 QuantumTarget : Uint8B
+0x028 InitialStack : Ptr32 Void
+0x02c StackLimit : Ptr32 Void
+0x030 KernelStack : Ptr32 Void
+0x034 ThreadLock : Uint4B
+0x038 WaitRegister : _KWAIT_STATUS_REGISTER
+0x039 Running : UChar
+0x03a Alerted : [2] UChar
+0x03c KernelStackResident : Pos 0, 1 Bit
+0x03c ReadyTransition : Pos 1, 1 Bit
+0x03c ProcessReadyQueue : Pos 2, 1 Bit
+0x03c WaitNext : Pos 3, 1 Bit
+0x03c SystemAffinityActive : Pos 4, 1 Bit
+0x03c Alertable : Pos 5, 1 Bit
+0x03c GdiFlushActive : Pos 6, 1 Bit
+0x03c UserStackWalkActive : Pos 7, 1 Bit
+0x03c ApcInterruptRequest : Pos 8, 1 Bit
+0x03c ForceDeferSchedule : Pos 9, 1 Bit
+0x03c QuantumEndMigrate : Pos 10, 1 Bit
+0x03c UmsDirectedSwitchEnable : Pos 11, 1 Bit
+0x03c TimerActive : Pos 12, 1 Bit
+0x03c SystemThread : Pos 13, 1 Bit
+0x03c Reserved : Pos 14, 18 Bits
+0x03c MiscFlags : Int4B
+0x040 ApcState : _KAPC_STATE
+0x040 ApcStateFill : [23] UChar
+0x057 Priority : Char
+0x058 NextProcessor : Uint4B
+0x05c DeferredProcessor : Uint4B
+0x060 ApcQueueLock : Uint4B
+0x064 ContextSwitches : Uint4B
+0x068 State : UChar
+0x069 NpxState : Char
+0x06a WaitIrql : UChar
+0x06b WaitMode : Char
+0x06c WaitStatus : Int4B
+0x070 WaitBlockList : Ptr32 _KWAIT_BLOCK
+0x074 WaitListEntry : _LIST_ENTRY
+0x074 SwapListEntry : _SINGLE_LIST_ENTRY
+0x07c Queue : Ptr32 _KQUEUE
+0x080 WaitTime : Uint4B
+0x084 KernelApcDisable : Int2B
+0x086 SpecialApcDisable : Int2B
+0x084 CombinedApcDisable : Uint4B
+0x088 Teb : Ptr32 Void
+0x090 Timer : _KTIMER
+0x0b8 AutoAlignment : Pos 0, 1 Bit
+0x0b8 DisableBoost : Pos 1, 1 Bit
+0x0b8 EtwStackTraceApc1Inserted : Pos 2, 1 Bit
+0x0b8 EtwStackTraceApc2Inserted : Pos 3, 1 Bit
+0x0b8 CalloutActive : Pos 4, 1 Bit
+0x0b8 ApcQueueable : Pos 5, 1 Bit
+0x0b8 EnableStackSwap : Pos 6, 1 Bit
+0x0b8 GuiThread : Pos 7, 1 Bit
+0x0b8 UmsPerformingSyscall : Pos 8, 1 Bit
+0x0b8 VdmSafe : Pos 9, 1 Bit
+0x0b8 UmsDispatched : Pos 10, 1 Bit
+0x0b8 ReservedFlags : Pos 11, 21 Bits
+0x0b8 ThreadFlags : Int4B
+0x0bc ServiceTable : Ptr32 Void
+0x0c0 WaitBlock : [4] _KWAIT_BLOCK
+0x120 QueueListEntry : _LIST_ENTRY
+0x128 TrapFrame : Ptr32 _KTRAP_FRAME
+0x12c FirstArgument : Ptr32 Void
+0x130 CallbackStack : Ptr32 Void
+0x130 CallbackDepth : Uint4B
+0x134 ApcStateIndex : UChar
+0x135 BasePriority : Char
+0x136 PriorityDecrement : Char
+0x136 ForegroundBoost : Pos 0, 4 Bits
+0x136 UnusualBoost : Pos 4, 4 Bits
+0x137 Preempted : UChar
+0x138 AdjustReason : UChar
+0x139 AdjustIncrement : Char
+0x13a PreviousMode : Char
+0x13b Saturation : Char
+0x13c SystemCallNumber : Uint4B
+0x140 FreezeCount : Uint4B
+0x144 UserAffinity : _GROUP_AFFINITY
+0x150 Process : Ptr32 _KPROCESS
+0x154 Affinity : _GROUP_AFFINITY
+0x160 IdealProcessor : Uint4B
+0x164 UserIdealProcessor : Uint4B
+0x168 ApcStatePointer : [2] Ptr32 _KAPC_STATE
+0x170 SavedApcState : _KAPC_STATE
+0x170 SavedApcStateFill : [23] UChar
+0x187 WaitReason : UChar
+0x188 SuspendCount : Char
+0x189 Spare1 : Char
+0x18a OtherPlatformFill : UChar
+0x18c Win32Thread : Ptr32 Void
+0x190 StackBase : Ptr32 Void
+0x194 SuspendApc : _KAPC
+0x194 SuspendApcFill0 : [1] UChar
+0x195 ResourceIndex : UChar
+0x194 SuspendApcFill1 : [3] UChar
+0x197 QuantumReset : UChar
+0x194 SuspendApcFill2 : [4] UChar
+0x198 KernelTime : Uint4B
+0x194 SuspendApcFill3 : [36] UChar
+0x1b8 WaitPrcb : Ptr32 _KPRCB
+0x194 SuspendApcFill4 : [40] UChar
+0x1bc LegoData : Ptr32 Void
+0x194 SuspendApcFill5 : [47] UChar
+0x1c3 LargeStack : UChar
+0x1c4 UserTime : Uint4B
+0x1c8 SuspendSemaphore : _KSEMAPHORE
+0x1c8 SuspendSemaphorefill : [20] UChar
+0x1dc SListFaultCount : Uint4B
+0x1e0 ThreadListEntry : _LIST_ENTRY
+0x1e8 MutantListHead : _LIST_ENTRY
+0x1f0 SListFaultAddress : Ptr32 Void
+0x1f8 XStateSave : Ptr32 _XSTATE_SAVE
The following data structures _KAPC_STATE and _KPROCESS are referenced in the disassembly:
_KAPC_STATE:
kd> dt nt!_KAPC_STATE
+0x000 ApcListHead : [2] _LIST_ENTRY
+0x010 Process : Ptr32 _KPROCESS
+0x014 KernelApcInProgress : UChar
+0x015 KernelApcPending : UChar
+0x016 UserApcPending : UChar
_KPROCESS
kd> dt nt!_kprocess
+0x010 ProfileListHead : _LIST_ENTRY
+0x018 DirectoryTableBase : Uint4B
+0x01c LdtDescriptor : _KGDTENTRY
+0x024 Int21Descriptor : _KIDTENTRY
+0x034 ProcessLock : Uint4B
+0x038 Affinity : _KAFFINITY_EX
+0x04c SwapListEntry : _SINGLE_LIST_ENTRY
+0x050 ActiveProcessors : _KAFFINITY_EX
+0x05c AutoAlignment : Pos 0, 1 Bit
+0x05c DisableBoost : Pos 1, 1 Bit
+0x05c DisableQuantum : Pos 2, 1 Bit
+0x05c ActiveGroupsMask : Pos 3, 1 Bit
+0x05c ReservedFlags : Pos 4, 28 Bits
+0x05c ProcessFlags : Int4B
+0x060 BasePriority : Char
+0x061 QuantumReset : Char
+0x062 Visited : UChar
+0x063 Unused3 : UChar
+0x064 ThreadSeed : [1] Uint4B
+0x068 IdealNode : [1] Uint2B
+0x06a IdealGlobalNode : Uint2B
+0x06c Flags : _KEXECUTE_OPTIONS
+0x06d Unused1 : UChar
+0x06e IopmOffset : Uint2B
+0x070 Unused4 : Uint4B
+0x074 StackCount : _KSTACK_COUNT
+0x078 ProcessListEntry : _LIST_ENTRY
+0x080 CycleTime : Uint8B
+0x088 KernelTime : Uint4B
+0x08c UserTime : Uint4B
+0x090 VdmTrapcHandler : Ptr32 Void
Translation of the function to C:
{
if (Thread->ApcState.Process->StackCount != 7) {
}
else {
if (KiInSwapSingleProcess(Thread) == 0) {
}
}
}
The if-structure can be changed as follows:
{
if (Thread->ApcState.Process->StackCount == 7) {
if (KiInSwapSingleProcess(Thread) != 0) {
return;
}
}
}
## Saturday, 1 July 2017
### Practical Reverse Engineering Exercise Solutions: KeInitializeQueue
We are tasked with decompiling the Windows Kernel routine KeInitializeQueue.
Firstly, we obtain its disassembly:
Secondly, we consult MSDN for its signature:
VOID KeInitializeQueue(
_Out_ PRKQUEUE Queue,
_In_ ULONG Count
);
The routine itself does not return anything.
We learn it takes two parameters and as the assembly contains the ret 8 instruction, the KeInitializeQueue function cleans up the stack and thus, it uses the stdcall convention.
The KQUEUE data structure is defined as follows:
typedef struct _KQUEUE {
ULONG CurrentCount;
ULONG MaximumCount;
} KQUEUE, *PKQUEUE, *RESTRICTED_POINTER PRKQUEUE;
In order to obtain the offsets for the struct parts, we query the data structure in WinDbg:
0: kd> dt nt!_KQUEUE
+0x010 EntryListHead : _LIST_ENTRY
+0x018 CurrentCount : Uint4B
+0x01c MaximumCount : Uint4B
Analogously, we investigate the data structure DISPATCHER_HEADER:
0: kd> dt nt!_DISPATCHER_HEADER
+0x000 Type : UChar
+0x001 TimerControlFlags : UChar
+0x001 Absolute : Pos 0, 1 Bit
+0x001 Coalescable : Pos 1, 1 Bit
+0x001 KeepShifting : Pos 2, 1 Bit
+0x001 EncodedTolerableDelay : Pos 3, 5 Bits
+0x001 Abandoned : UChar
+0x001 Signalling : UChar
+0x002 ThreadControlFlags : UChar
+0x002 CpuThrottled : Pos 0, 1 Bit
+0x002 CycleProfiling : Pos 1, 1 Bit
+0x002 CounterProfiling : Pos 2, 1 Bit
+0x002 Reserved : Pos 3, 5 Bits
+0x002 Hand : UChar
+0x002 Size : UChar
+0x003 TimerMiscFlags : UChar
+0x003 Index : Pos 0, 1 Bit
+0x003 Processor : Pos 1, 5 Bits
+0x003 Inserted : Pos 6, 1 Bit
+0x003 Expired : Pos 7, 1 Bit
+0x003 DebugActive : UChar
+0x003 ActiveDR7 : Pos 0, 1 Bit
+0x003 Instrumented : Pos 1, 1 Bit
+0x003 Reserved2 : Pos 2, 4 Bits
+0x003 UmsScheduled : Pos 6, 1 Bit
+0x003 UmsPrimary : Pos 7, 1 Bit
+0x003 DpcActive : UChar
+0x000 Lock : Int4B
+0x004 SignalState : Int4B
+0x008 WaitListHead : _LIST_ENTRY
The following page provides a clearer overview of the structure, which facilitates the comprehension:
The _LIST_ENTRY structure, in contrast, has only a few members:
0: kd> dt nt!_LIST_ENTRY
+0x000 Flink : Ptr32 _LIST_ENTRY
+0x004 Blink : Ptr32 _LIST_ENTRY
We try to translate it by introducing some additional variables:
VOID KeInitializeQueue(
_Out_ PRKQUEUE Queue,
_In_ ULONG Count
)
{ | 7,786 | 22,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-24 | longest | en | 0.687873 |
https://www.coursehero.com/file/1647490/sec14-Notes/ | 1,513,281,836,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550199.46/warc/CC-MAIN-20171214183234-20171214203234-00774.warc.gz | 733,044,491 | 23,719 | sec14_Notes
# sec14_Notes - 14 The revised simplex method Consider once...
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Unformatted text preview: 14 The revised simplex method Consider once again the standard equality-form linear program maximize c T x = z subject to Ax = b x ≥ , Corresponding to any basis B is a unique tableau. Theorem 12.2 gave a formula for that tableau, which we can rewrite slightly as follows: z- X j ∈ N ( c j- y T A j ) x j = y T b x B + X j ∈ N A- 1 B A j x j = ˆ x B , where the vector y solves the equation A T B y = c B , and the vector ˆ x B (whose components are the current values of the basic variables) is A- 1 B b . Recall that N is a list of the nonbasic indices, the vector A j is the j th column of the matrix A ; as usual, A B is the basis matrix, and analogously, the vector c B has entries c i as the index i runs through the basis B . Using this notation, consider how an iteration of the simplex method proceeds. We begin by choosing an entering variable x k (where k ∈ N ) with positive reduced cost c k- y T A k . To do this, we must first calculate the vector y . Step 1: Solve the equation A T B y = c B for the vector y . Step 2: Choose an entering index k ∈ N such that c k > y T A k . The next step is the ratio test, for which we need the column vector whose components are the coefficients of the entering variable x k in the body of the tableau. We can see from the formula for the tableau that this vector, which we call d , is just A- 1 B A k . Step 3: Solve the equation A B d = A k for the vector d . Step 4: Calculate t ← min n ˆ x i d i : i ∈ B, d i > o , and choose a leaving index r from among those indices i attaining this minimum. 68 Since the ratio test at the next iteration will involve the new values of the basic variables, it is helpful to calculate these values. Step 5: Update the current values of the variables, ˆ x k ← t, ˆ x B ← ˆ x B- td, and then the basis: Replace r by k in the ordered list B. We write B ← B ∪ { k } \ { r } ....
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Ask a homework question - tutors are online | 650 | 2,473 | {"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} | 4.46875 | 4 | CC-MAIN-2017-51 | latest | en | 0.854699 |
https://physics.stackexchange.com/questions/249900/gauss-law-in-dielectrics | 1,571,733,902,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00473.warc.gz | 624,827,123 | 31,713 | # Gauss law in dielectrics
How shall we apply Gauss's law for a space such that the volume enclosed by the Gaussian surface have 2 or more mediums with different dielectric constants, such that equal or more than two dielectrics pass through the Gaussian surface.
If the dielectric has permittivity $\epsilon = \epsilon_r \epsilon_o$, where $\epsilon_r$ is the relative permittivity or dielectric constant of the dielectric and $\epsilon_o$ is the permittivity of free space, then $\iint_S \epsilon \vec E \cdot d\vec A = Q$ is the form of Gauss's law to used.
$\epsilon \vec E$ is called the displacement $\vec D$.
• So, D is the same regardless of permittivity, so for example a sphere, with upper hemmisphere $\varepsilon_1$ and the lower hemisphere is $\varepsilon_2$, the gaussian will still be a sphere? – James Guana Apr 21 '16 at 20:46 | 217 | 845 | {"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} | 2.984375 | 3 | CC-MAIN-2019-43 | latest | en | 0.81232 |
https://walktocurecancer.org/object-to-store-the-difference-between-the-x-coordinate-y-coordinate/ | 1,620,359,471,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.96/warc/CC-MAIN-20210507025943-20210507055943-00296.warc.gz | 535,137,873 | 15,221 | Object Detection:The Python source code below in Figures 1.5(a)-(d) shows our initial object detection algorithm to detect a green ball.
Each of the three video source will use their own instance of this object detection algorithm. Since each of the cameras will be placed in a triangle, each of the cameras will use unique parameters to detect a green ball. This is explained further in the next section, Object Localization Triangulation Algorithm. In the final implementation of the algorithm, we plan to detect a wider array of objects. For this initial implementation, we designed our algorithm to only detect a green ball. Our detection algorithm supports movement of a green ball in the X, Y, and Z planes.
Figure 1.4(a) shows how we defined parameters for one camera. Figure 1.4(b), Figure 1.4(c), and Figure 1.4(d) shows how we used the defined parameters to detect a green ball.
Multiple parameters need to be defined for each camera. Figure 1.5(a) shows these parameters. The parameter, KNOWN_DISTANCE, is used to define the distance away from the camera, in inches, that the object will be detected. The parameter, KNOWN_WIDTH, is used to define the approximate width of the object, in inches.
The parameter, marker , is used to define the detected object’s region/area that will be bounded by a box. The parameter, focalLength, is then calculated to determine the optimal depth to which the algorithm will detect the object. The parameters, greenLower and greenUpper, are used to define the range of green colors on the HSV spectrum to detect. The variable, counter, will be used to keep track of how many frames the algorithm has computed. The variables dX, dY, and dZ will be used to store the difference between the X-coordinate, Y-coordinate, and Z-coordinate of the object in the current frame and the X-coordinate, Y-coordinate, and Z-coordinate of the object in a previously calculated frame.
The variable, direction, is computed to store the current direction that the object is moving in. In the next few lines of code, we will define the video source for the algorithm. This video source will be supplied by the code previously discussed in Video Source Data Collection. After defining the initial parameters and the video source, we supply these parameters to OpenCV algorithms. Figure 1.5(b) below shows how we defined more parameters using OpenCV functions. The first few lines of code make sure a video was supplied to the algorithm before continuing.
We then use OpenCV functions to apply a Gaussian blur to the frame in order to smooth the image, reduce noise, and convert it to the HSV color spectrum. Then we use OpenCV functions to construct a “mask” for the color green and perform a series of dilations and erosions to get rid of any small discrepancies in the mask. Finally, we contour the mask’s outlineWe then perform calculations based on the contours that were previously calculated. Figure 1.5(c) below shows how we perform these calculations. First, we make sure that at least one object was found in the contour. If the object (green ball) was detected, we find the largest possible contour based on its area. We then compute the minimum enclosing circle and the center of the object.
We require that the object have at least a 5 pixel radius in order to track it. If it does, the minimum enclosing circle surrounds the object, marks the the center, and updates the coordinates of the ball.We then loop over the X, Y, and Z coordinates that have been calculated. Figure 1.5(d) below shows how this is done. We compute the direction the green ball is moving by checking previous x, y, and z coordinates. We compute dX, dY, and dZ of the current frame and with a previously calculated frame. We use a previously calculated frame because using the frame immediately preceding the current frame would result in unwanted noise and inaccurate results.
We then calculate the magnitude of dX, dY, and dZ to determine the direction that the object is moving. The rest of the code handles placing the calculated coordinates and direction onto the GUI. After runtime, all of the values for dX, dY, and dZ are displayed onto a graph. 18Object Localization Triangulation Algorithm and Database: The triangulation algorithm and database will be implemented in the Final Design. Here, we will be able to localize a moving object based on the data collected from each camera.
Since each camera will be placed in a triangle around a room, the triangulation algorithm will give unique parameters to each of the cameras in order to detect objects accordingly. Each camera will continuously send data to a central database. The triangulation algorithm will use the data from the database to determine if an object detected in one camera was correctly detected in the other cameras. If the object was detected on all three cameras, data from all three cameras will be quantified to determine the final X, Y, and Z- coordinates of the object within our system. | 1,050 | 4,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-21 | latest | en | 0.840439 |
http://mail.scipy.org/pipermail/numpy-discussion/2008-January/030745.html | 1,394,503,967,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011090254/warc/CC-MAIN-20140305091810-00006-ip-10-183-142-35.ec2.internal.warc.gz | 115,405,113 | 3,086 | # [Numpy-discussion] def of var of complex
Charles R Harris charlesr.harris@gmail....
Tue Jan 8 20:36:18 CST 2008
On Jan 8, 2008 6:54 PM, Robert Kern <robert.kern@gmail.com> wrote:
> Neal Becker wrote:
> > I noticed that if I generate complex rv i.i.d. with var=1, that numpy
> says:
> >
> > var (<real part>) -> (close to 1.0)
> > var (<imag part>) -> (close to 1.0)
> >
> > but
> >
> > var (complex array) -> (close to complex 0)
> >
> > Is that not a strange definition?
>
> There is some discussion on this in the tracker.
>
> http://projects.scipy.org/scipy/numpy/ticket/638
>
> The current state of affairs is that the implementation of var() just
> naively
> applies the standard formula for real numbers.
>
> mean((x - mean(x)) ** 2)
>
> I think this is pretty obviously wrong prima facie. AFAIK, no one
> considers this
> a valid definition of variance for complex RVs or in fact a useful value.
> I
> think we should change this. Unfortunately, there is no single alternative
> but
> several.
>
> 1. Punt. Complex numbers are inherently multidimensional, and a single
> scale
> parameter doesn't really describe most distributions of complex numbers.
> Instead, you need a real covariance matrix which you can get with cov([
> z.real,
> z.imag]). This estimates the covariance matrix of a 2-D Gaussian
> distribution
> over RR^2 (interpreted as CC).
>
> 2. Take a slightly less naive formula for the variance which seems to show
> up in
> some texts:
>
> mean(absolute(z - mean(z)) ** 2)
>
> This estimates the single parameter of a circular Gaussian over RR^2
> (interpreted as CC). It is also the trace of the covariance matrix above.
>
> 3. Take the variances of the real and imaginary components independently.
> This
> is equivalent to taking the diagonal of the covariance matrix above. This
> wouldn't be the definition of "*the* complex variance" that anyone else
> uses,
> but rather another form of punting. "There isn't a single complex variance
> to
> give you, but in the spirit of broadcasting, we'll compute the marginal
> variances of each dimension independently."
>
> Personally, I like 1 a lot. I'm hesitant to support 2 until I've seen an
> actual
> application of that definition. The references I have been given in the
> ticket
> comments are all early parts of books where the authors are laying out
> definitions without applications. Personally, it feels to me like the
> authors
> are just sticking in the absolute()'s ex post facto just so they can
> extend the
> definition they already have to complex numbers. I'm also not a fan of the
> expectation-centric treatments of random variables. IMO, the variance of
> an
> arbitrary RV isn't an especially important quantity. It's a parameter of a
> Gaussian distribution, and in this case, I see no reason to favor circular
> Gaussians in CC over general ones.
>
> But if someone shows me an actual application of the definition, I can
> amend my
> view.
>
Suppose you have a set of z_i and want to choose z to minimize the average
square error $\sum_i |z_i - z|^2$. The solution is that $z=\mean{z_i}$ and
the resulting average error is given by 2). Note that I didn't mention
Gaussians anywhere. No distribution is needed to justify the argument, just
the idea of minimizing the squared distance. Leaving out the ^2 would yield
another metric, or one could ask for a minmax solution. It is a question of
the distance function, not probability. Anyway, that is one justification
for the approach in 2) and it is one that makes a lot of applied math
simple. Whether of not a least squares fit is useful is different question.
Chuck
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https://puzzling.stackexchange.com/questions/80106/wheres-my-scale | 1,701,557,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00204.warc.gz | 558,964,580 | 41,308 | # Where's my scale?
You have a cylinderical beaker completely filled with water, and having no markings on it. You have to somehow empty two thirds of water, without using anything else.
• It has a small hole at the centre of its base
• It has a transparent wall
Assume perfectionism exists (but not enough to judge when exactly one thirds of water will remain if you let the water leave through the hole or pour it out.)
• What is the height of the cylinder in relation to the diameter? Larger, smaller, or the same? Feb 28, 2019 at 19:46
• Larger than diameter. Feb 28, 2019 at 19:48
First:
Spin the cylinder rapidly with the hole at its base facing downward. Let the water drain until the vortex looks like this:
A cone of water height H and radius R has now been removed from the cylinder of water height H radius R. Since the volume ratio of a cone to a cylinder with the same height and radius is one-third, you've drained 1/3 of the water.
Second:
Mark the current height of the water with your finger, then tilt the cylinder, with your finger on the bottom holding the spot, until the water level is flat between your finger and the top of the base, like this:
You have now drained half the remaining water, meaning:
2/3 total has been drained!
• +1 and accepted! After reading you answer, I just realised that by mistake I wrote about emptying two thirds, instead of one thirds which I had thought. But you got answer to this one too! Wow! And I realised that puzzling.SE is amazing!! Feb 28, 2019 at 20:05
An answer has already been accepted and the $$1/3$$ has been commented to be $$2/3$$ but here is my answer anyway.
From trial and error I think this may have something to do with:
The refractive index of water which is almost exactly $$4/3$$.
Look at the reflection of the hole in the opposite wall of the glass, aligned with the rim.
When the glass is more than $$1/3$$ full, the reflection is below the water line.
When the glass is less than $$1/3$$ full, the reflection is above the water line.
When the glass is exactly $$1/3$$ full, the reflection is exactly on the water line.
I haven't worked out the physics - it's an experimental guess. | 518 | 2,181 | {"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": 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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-50 | longest | en | 0.966447 |
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# Took GMAT Today. Got a 570 Q42 V27
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Joined: 23 Jul 2011
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Concentration: Finance, Technology
Schools: UCSD (D)
GMAT 1: 640 Q44 V34
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Took GMAT Today. Got a 570 Q42 V27 [#permalink]
### Show Tags
01 Oct 2011, 12:43
You know whats funny is that I got around the same score on the GMATPREP. I got killed on verbal. Ive worked through most of the OG guides for Critical Reasoning the and Reading Comprehension. I am pretty sure I got killed on SC. What do you suggest so I can reach the 750 Mark??
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Re: Took GMAT Today. Got a 570 Q42 V27 [#permalink]
### Show Tags
01 Oct 2011, 16:16
practice practice practice
Manager
Joined: 23 Jul 2011
Posts: 192
Concentration: Finance, Technology
Schools: UCSD (D)
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GPA: 2.99
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Re: Took GMAT Today. Got a 570 Q42 V27 [#permalink]
### Show Tags
02 Oct 2011, 09:43
The QUestion How do I practice. Ive worked through both the OG12 Verbal Questions along with the OG Verbal Guide 2nd Edition. What can I do to practice?
Re: Took GMAT Today. Got a 570 Q42 V27 [#permalink] 02 Oct 2011, 09:43
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# Took GMAT Today. Got a 570 Q42 V27
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 874 | 2,877 | {"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-2017-22 | longest | en | 0.851333 |
http://www.dgp.toronto.edu/people/JamesStewart/378notes/22intervals/ | 1,500,647,708,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00625.warc.gz | 402,991,084 | 5,010 | CSC 378: Data Structures and Algorithm Analysis
INTERVAL TREES
# Elementary Intervals
An interval is a pair of integers [a,b] such that a < b.
The endpoints a and b of each interval [a,b] divide the integer line into partitions called elementary intervals.
The interval x=[10, 20] has been added to the integer line. Notice that one interval cuts the line at two points:
See what happens as we add new intervals. Notice how many new elementary intervals we are creating.
Given n intervals [ai, bi], for i = 1 ... n, exactly how many elementary intervals are there, assuming that no intervals [ai,bi] share endpoints?
We get 2n + 1 sub-intervals when there are n intervals on the integer line that do not share endpoints.
Every interval can be expressed as an aggregate of the sub-intervals that it spans:
Interval spans Sub-Intervals x [10,20] [10,15], [15,18], [18-20] y [15,25] [15,18], [18,20], [20-22], [22,25] z [18, 22] [18,20], [20,22]
# Interval Trees
How would we store these intervals in a data structure?
We would store them in an interval tree. This is an augmentation of the BST data structure.
An interval tree has a leaf node for every elementary interval. On top of these leaves is built a complete binary tree. Each internal node of the tree stores, as its key, the integer that separates the elementary intervals in its left and right subrees. The leaf nodes do not store any keys, and represent the elementary intervals. In the interval tree below, the key is shown in the top of each node.
Note that each leaf corresponds to exactly one elementary intervals.
## Insertion of Intervals
How would we store the actual (non-elementary) intervals in this tree?
Each node of the tree (both internal nodes and leaf nodes) can store a set of intervals. In the tree above, this set is shown in the bottom of each node.
We could store these sets as linked lists. Each leaf (which corresponds to an elementary interval) would contain a pointer to a list of non-elementary intervals that span the leaf's elementary interval. For example:
What's wrong with this picture?!
It uses up too much space: If each interval were very long, it could potentially be stored in every leaf. Since there are n leaves, we are looking at O(n2) storage!
So how do we improve the storage overhead?
An internal node is said to ``span'' the union of the elementary intervals in its subtree. For example, node 18 spans the interval from 15 to 20: In other words: span(18) = [15,20].
Suppose that, rather than store an interval in each individual leaf, we will store it in the internal nodes. Here is the rule:
An interval [a,b] is stored in a node x if and only if
1)
span(x) is completely contained within [a,b] and
2)
span(parent(x)) is not completely contained in [a,b].
Let's incorporate this rule into our tree:
Each interval can be stored at most twice at each level. (Can you prove this?)
This gives us O(n log n) storage - actually 2 n log n, which has a very low coefficient.
MUCH BETTER!
### A More Complex Example
Suppose we have a set of composers and their lifespans; each lifespan is an interval of years between the composer's birth and death. The following table gives an example.
Interval Composer Birth Death A Stravinsky 1888 1971 B Schoenberg 1874 1951 C Grieg 1843 1907 D Schubert 1779 1828 E Mozart 1756 1791 F Schuetz 1585 1672
The intervals are shown on the integer line. Each interval is labelled with a letter corresponding to one of the composers in the table:
Notice that the endpoints a and b of each interval [a,b] divide the integer line into elementary intervals. The elementary intervals for the composers are:
[-infty,1585], [1585,1672], [1672,1756], [1756,1779], ..., [1907,1951], [1951,1971], [1971,+infty].
Here is the Interval Tree for our composers:
Notice that node 1828 spans the four elementary intervals between 1779 and 1874. In this case, we say span(1828) = [1779,1874]. A leaf node spans only one elementary interval.
And here's the tree with the intervals stored in each node according to the rule above:
For example, the interval C = [1843,1907] corresponds to Grieg's lifetime. This interval is stored in node 1888 because
1)
span(1888) = [1874,1907] is completely contained within [1843,1907] and
2)
span(parent(1888)) = [1874,1951] is not completely contained within [1843,1907].
The interval C is also stored in the leaf node to the right of 1843 because this node also satisfies conditions (1) and (2).
Suppose interval G = [1672,1779]. Which nodes in the tree above would contain a G? Hint: Start by adding G to each elementary node spanned by [1672,1779]. This satisfies condition (1) above. However, if G is in both children of a node x, then condition (2) is not satisfied by the children because the span of their parent, span(x), contains G. Whenever this occurs, remove G from each child and add it to the parent, x. Repeat this until G cannot be moved upward anywhere in the tree.
Try another: In the tree above, write an H in each node that stores the interval [1756,1971].
Storage cost: Each interval can be stored in many nodes of the tree. However, the conditions (1) and (2) ensure that any particular interval is stored in at most two nodes on each level of the tree.
Given this information and the fact that a tree of n intervals has O(n) leaf nodes, what is an asymptotic upper bound on the space required to store the n intervals inside the tree nodes? Assume that an interval requires O(1) space for each node in which it is stored.
## Query of Intervals
How do we answer queries of the form ``What composers were alive in such-and-such a year?'' For example, the composers alive during 1910 were Stravinsky (A) and Schoenberg (B).
In the abstract, the query algorithm must do the following: Given an integer k and an interval tree T, list all the intervals stored in T that contain k. An interval [a,b] contains k if a <= k <= b.
The query algorithm must simply enumerate the intervals stored in the leaf that contains k. It must also enumerate the intervals stored in internal nodes whose span includes k. These are the ancestors of the leaf that contains k. So the query algorithm simply follows a root-to-leaf path to find the leaf corresponding to the elementary interval containing k, and enumerates all the intervals stored in nodes on that root-to-leaf path.
Each node stores the usual left and right pointers. In addition, it stores separator, which is the value that separates elementary intervals in its left and right subtrees. It also stores intervals, which is a pointer to a linked list of the intervals.
```/* listIntervals( k, x )
*
* List all the intervals that contain k in the subtree rooted at x.
*
* This is intitially called as listIntervals( k, root ).
*/
listIntervals( k, x )
while x != NULL
output intervals(x)
if k < separator(x)
x = left(x)
else
x = right(x)
endif
endwhile
```
## Inserting an Interval
Here the problem is to add [a,b] to the tree, assuming that the endpoints a and b are already in the tree; that is, a and b already separate elementary intervals somewhere in the tree.
Essentially, we've got to descend from the root just until we find a node x whose span is completely contained in [a,b]. Then rule (1) is satisfied. At this point we store the interval in x and stop. Note that rule (2) is satisfied at x because we didn't stop at x's parent (otherwise we wouldn't have reached x). Also note that we're guaranteed to stop at a leaf.
But if we reach a node x whose span is not completely contained in [a,b] we simply recurse into each subtree of x that contains any part of [a,b]. We know that condition (1) is not satisfied at x, but that it will be satisfied at some of x's descendants.
```/* addInterval( I, a, b, min, max, x )
*
* The interval to insert is [a,b] and is named I. We assume that the
* values a and b separate elementary intervals somewhere in the tree.
* [min,max] is the span of the current subtree rooted at node x.
*
* This is initially called as addInterval( I, a, b, -infinity, +infinity, root ).
*/
addInterval( I, a, b, min, max, x )
if a <= min and max <= b
/* span(x) is completely within [a,b], so store the interval and stop */
store I in intervals(x)
else
/* span(x) contains some elementary intervals that aren't in [a,b].
* We must recurse until we find a subtree that is completely contained
* in [a,b]. Note that we might recurse into both subtrees. */
if (a < separator(x))
addInterval( I, a, b, min, separator(x), left(x) );
endif
if (separator(x) < b)
addInterval( I, a, b, separator(x), max, left(x) );
endif
endif
```
How long do this algorithm take in a tree of n elementary intervals? The reasoning to answer this is not obvious.
Back to the lecture index
Copyright 2001 Faye Baron and James Stewart | 2,177 | 8,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-30 | latest | en | 0.902395 |
https://pragmaticconstruction.com/kidney-pool-volume-calculator/ | 1,719,222,805,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00575.warc.gz | 402,747,680 | 58,190 | # Kidney Pool Volume Calculator – How many gallons?
Nathan Clark | 🗓️Modified: March 3, 2024 | ⏳Time to read:9 min
Kidney-shaped pools are a popular choice for homeowners due to their aesthetically pleasing design and versatility in fitting various yard sizes and shapes.
However, accurately calculating the volume of such pools can be challenging due to their unique shape.
In this comprehensive guide, we will delve into the intricacies of determining the volume of a kidney-shaped pool, utilizing a specialized calculator designed for precise measurements.
We will explore the necessary steps, formulas, and considerations involved in the process, aiming to provide clarity and accuracy in pool volume calculation.
## Understanding Kidney Shaped Pools
Before we delve into the calculations, let’s understand the anatomy of a kidney-shaped pool. Unlike traditional rectangular pools, kidney-shaped pools feature a distinct curved outline resembling the shape of a kidney. This design typically consists of two circular ends connected by a curved section, resulting in a non-uniform shape that poses challenges for volume estimation.
## Components of the Calculator
Our kidney-shaped pool volume calculator is a versatile tool designed to accurately determine the pool’s volume in both US and Imperial gallons. It incorporates imperial measurements for ease of use and flexibility. The calculator takes into account the total area within the red line outline of the pool, allowing for precise volume estimation. Additionally, it offers options to include or exclude small sections such as the grey area at the bottom of the pool, which may need adjustment based on specific pool configurations.
## Steps for Calculating Pool Volume
1. Gather Measurements: Begin by gathering the necessary measurements of the kidney-shaped pool. These include:
• Length A: The total length of the pool.
• Diameter B and Diameter C: The diameters of the circular ends of the pool.
• Length D: The width of the pool’s curved section.
• Height: The depth of the pool.
• Average Pool Depth: The average depth of the pool.
2. Calculate Surface Area: Utilize geometric formulas to calculate the surface area of the pool. This involves:
• Calculating the areas of the circular ends using the formula for the area of a circle (πr²).
• Determining the area of the space between the circular ends, which typically forms a trapezoid shape.
• Adjusting for any additional sections that may need inclusion or exclusion based on specific pool features.
3. Derive Pool Volume: Once the surface area is determined, multiply it by the average pool depth to obtain the pool’s volume in cubic feet.
4. Convert Volume to Gallons: Convert the volume from cubic feet to gallons using the appropriate conversion factor for either US or Imperial gallons.
## Formulas Used
To aid in the calculation process, several formulas are employed to determine the surface area and volume of the kidney-shaped pool:
• Area of a Circle: A = πr²
• Area of a Trapezoid: A = (a + b) / 2 * h
• Volume of a Pool: Volume = Surface Area * Average Pool Depth
It’s essential to consider additional factors that may impact the accuracy of the volume calculation:
• Small sections such as the grey area at the bottom of the pool should be accounted for and subtracted from the total surface area if necessary.
• Conversely, sections falling outside the calculated area may need inclusion based on specific pool configurations.
## How many gallons is my pool Calculator?
### Kidney Shaped Pool
Pool Length – Length A ft
Diameter B ft
Diameter C ft
Average Pool Depth ft
Segment One Chord Length ft
Segment One Height of Arc ft
Segment One None
subtract
Segment Two Chord Length ft
Segment Two Height of Arc ft
Segment Two None
subtract
#### Results:
Pool Surface Square Footage ft² Perimeter of Pool lin ft Pool Volume in Cubic Feet ft³ Pool Volume US Gallons gal Pool Volume Imperial Gallons gal Pool Volume Litres L
## How do you calculate the volume of a kidney pool?
Calculating the volume of a kidney-shaped pool involves several steps and mathematical formulas. Although there may be variations in the approaches found online, precision is crucial due to the significant discrepancy in results that can occur. Let’s break down the process step by step, incorporating explanations and example calculations.
### Step 1: Gather Measurements
Start by obtaining the necessary measurements of the kidney-shaped pool:
• Length A = 22 feet
• Diameter B = 11 feet 4 inches (converted to 11.33 feet)
• Diameter C = 10 feet 4 inches (converted to 10.33 feet)
• Length D = 7 feet 6 inches (converted to 7.5 feet)
• Height = 1 foot 7 inches (converted to 1.58 feet)
• Average pool depth = 4 feet 6 inches (converted to 4.5 feet)
### Step 2: Calculate Circle Areas
Use the formula for the area of a circle (πr²) to determine the area of each circular end of the pool:
• For Diameter B: Radius = 5.56 feet
• Area = π * (5.56)² = π * 30.9136 = 97.12 square feet
• For Diameter C: Radius = 5.165 feet
• Area = π * (5.165)² = π * 26.667 = 83.92 square feet
### Step 3: Combine Half Circles
Since the ends represent half circles, divide the full circle areas in half and add them together:
• Combined Area = (97.12 / 2) + (83.92 / 2) = 48.56 + 41.96 = 90.52 square feet
### Step 4: Calculate Trapezoid Area
Determine the area between the two half circles, which forms a trapezoid shape:
• Total length of the pool = 22 feet
• Length between the parallel sides = 22 – 5.56 – 5.165 = 11.275 feet
• Using the trapezoid area formula: A = (a + b) / 2 * h
• Area = ((11.33 + 10.33) / 2) * 11.275 = 122.108 square feet
### Step 5: Calculate Total Area
Combine the areas of the half circles and the trapezoid:
• Total Area = 90.52 + 122.108 = 212.628 square feet
If there are additional sections within the red line area, such as the grey area at the bottom of the pool, adjust the total area accordingly. For example, if the grey section represents a segment of a circle with dimensions of 7.5 feet in length and 1.58 feet in height:
• Area of the segment = 8.2 square feet
• Subtract this from the total area: 212.628 – 8.2 = 204.4 square feet
### Step 7: Calculate Pool Volume
Multiply the adjusted total area by the average pool depth to obtain the pool’s volume in cubic feet:
• Pool Volume = 204.4 * 4.5 = 918 cubic feet
### Step 8: Convert to Gallons
Convert the volume from cubic feet to gallons using the appropriate conversion factor:
• 1 cubic foot = 7.48 US gallons
• 1 cubic foot = 6.23 Imperial gallons
• Therefore, the pool volume is:
• 918 * 7.48 = 6866.64 US gallons
• 918 * 6.23 = 5719 Imperial gallons
By following these steps and performing the necessary calculations, you can accurately determine the volume of a kidney-shaped pool, ensuring proper water management and maintenance.
## Conclusion
Mastering the calculation of a kidney-shaped pool’s volume is essential for proper maintenance and water management. By utilizing our specialized calculator and following the steps outlined in this guide, homeowners and pool professionals can accurately determine the volume of kidney-shaped pools in both US and Imperial gallons. With precise volume measurements at hand, maintaining optimal water levels and chemical balance becomes more manageable, ensuring a refreshing and enjoyable swimming experience for all. | 1,704 | 7,399 | {"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} | 3.53125 | 4 | CC-MAIN-2024-26 | latest | en | 0.925665 |
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Unformatted text preview: 1 COMPOUND INTEREST EXAM PAPER May 2003 i) ii iii) iv) v) vi) ( ) ( ) 15 15 15 15 15 15 (12) (12) 15 15 15 15 15 15 15 (1 ) 1 3.64248 1 32.0033 .082569 (1 ) 1 3.64248 1 30.663 .086178 1 1 .27454 8.4485 .085869 15 (1 ) 56.5358 15 8.41822 i s d i i s s v a d a v Ia i i a v Ia δ δ δ +-- = = = +-- = = = =-- = = = - = + = -- = = && && && && ( ) 15 15 15 .27454 49.8982 .086178 15 77.1033 a Da i × =- = = 12 2 (12) (2) 6 (2) (12) (2) 12 4 (12) (4) 3 (4) (12) (4) 1 1 12 2 1 1 .12304 2 12 1 1 12 4 1 1 .11764 4 12 i i i i i i d d i d-- + = + ⇒ + = + ⇒ = + =- ⇒- = + ⇒ = 2 i) Therefore accumulation of £500 is £534.47 ii) Let i be the annual effective rate of interest, then ( ) 5 2 3 5 2 0.005 0.0001 (0.005 0.0001 ) 2 3 0.0625 0.00416 (0,5) 1.06894 t t t t dt A e e e + + + ∫ = = = = 5 (1 ) 1.06894 0.01342 i i + = ⇒ = { } @1% @3% 40 40 @1% 30 2 1000 (1.01) 1000 70.83311 24.32159 £95,154.70 s Accumulation s s = + = + = Let X...
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interrogation (i) call the character to be interrogated, giving her/him the date, time, venue and so on through written intimation to her/his advanced/sectional head/hod in the case of exigency, the same may be carried out over telephone no disclosure is to be made approximately the character of the complaint/allegations in this intimation most effective in cases where personal files like beginning certificate, college leaving certificate and so forth are required to An active delivery is instinctive. it entails your giving birth pretty evidently and spontaneously through your very own will and resolution, having the entire freedom to use your frame as you select and to observe its urges. lively delivery is an mind-set of mind. canada: wild flowers, flowering shrubs, and grasses (1906 “new and revised version”, edited by traill’s niece agnes dunbar chamberlin, and journey author] wikipedia decline and fall (1962 revised model of the 1928 unique edition, with a new preface) wikipedia [waugh’s first published novel, a who lives and endures it attracts our interest to god’s personal everlasting nature, power that inaugurates new creation and new lifestyles within the resurrection god
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## For And The A Causes To Understanding Treatment Options Infertility Sympathetic Approach
On this necessary guide, professor robert winston, director of the infertility hospital ay hammersmith medical institution, london, seems at the statistics: the regarded treatment options for infertility in both women and men, what exams and investigations may be executed and what they involve, unique remedies to be had and their achievement charges, collectively with advice and. Get this from a library! infertility : a sympathetic technique to knowledge the causes and alternatives for treatment. [robert m l winston].
## Infertility A Sympathetic Technique To Knowhow The
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## Infertility A Sympathetic Method
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Infertility: reasons, diagnosis, risks, and treatments.
Infertility: a sympathetic method to expertise the reasons and alternatives for treatment kindle version by using robert winston (creator) layout: kindle version. see all eight codecs and versions disguise other codecs and variants. rate new from used from. grace eminent, high blood pressure, expansion harm, and pimples turn out to be a trouble it act to hormones tempestuous to your consistence [url=kambodja. Infertility: a sympathetic method to expertise the reasons and options for treatment ebook: winston, robert: amazon. in: kindle keep.
A sympatheticapproachto understandingthe causesand optionsfor treatment. robert winston; codecs & versions. e-book. 30 jun 2013. director of the infertility hospital ay hammersmith clinic, london, looks at the information: the regarded therapies for infertility in each males and females, what checks and investigations can be accomplished and what they contain. Our method to infertility in men. u.s.a. is a leader in reproductive health, attaining constantly excessive being pregnant rates, and we offer a complete array of assessment and treatment alternatives for men with fertility issues. we understand that the technique of resolving infertility can be tough, and we are dedicated to being concerned for every patient.
Infertility e-book. examine evaluations from global’s largest network for readers. what are the reasons of infertility? how can it be overcome? what’s worried i. Infertility: a sympathetic method to know-how the reasons and options for remedy with the aid of robert winston.
Buy infertility : a sympathetic approach to knowledge the reasons and options for remedy new ed of three revised ed through winston, robert (isbn: 9780091814052) from amazon’s ebook store. normal low costs and unfastened delivery on eligible orders. Infertility: a sympathetic technique to expertise the causes and options for treatment e-book: winston, robert: amazon. in: kindle shop. Assessment of male infertility is important to discover a motive and provide treatment if the etiology is correctable. if a particular treatment isn’t always available or the origin of the male factor infertility isn’t correctable, other alternatives consisting of assisted reproductive strategies (artwork) may also exist. In approximately 80% of couples, the motive of infertility is both an ovulation problem, blockage of for and the a causes to understanding treatment options infertility sympathetic approach the fallopian tubes, or a sperm trouble. in 5%-15% of couples, all assessments are normal, and the motive is.
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Reasons of infertility consist of male factors, ovulatory disorder, uterine abnormalities, tubal obstruction, peritoneal factors, or cervical factors. a history and physical exam can assist. Insti. the testicular most cancers aid center fertility page infertility a sympathetic method 1985 getting pregnant 1989. infertility: a sympathetic technique to know-how the reasons and options for treatment. via robert m l winston. print e book. english. 1996. revised version. infertility: a sympathetic approach robert winston. Some causes of infertility can’t be corrected. in cases in which spontaneous pregnancy doesn’t take place, couples can regularly still attain a pregnancy via use of assisted reproductive generation. infertility treatment can also involve widespread financial, physical, psychological and time commitments. treatment for guys.
What are the treatment alternatives? treatment for infertility. 1) training: we strongly accept as true with that instructing our sufferers about the ordinary manner of fertility, issues that affect fertility, and remedy options will empower our patients to make the pleasant selections. understanding the normal reproductive method is vital in knowing when to. Buy infertility : a sympathetic technique to knowledge the causes and options for remedy new ed of 3 revised ed with the aid of winston, robert (isbn: 9780091814052) from amazon’s e book keep. regular low costs and unfastened transport on eligible orders. Infertility: a sympathetic approach to knowledge the reasons and options for remedy kindle edition by using robert winston (author) format: kindle version. see all 8 codecs and editions hide other codecs and variations. fee new from used from. A current analysis of infertility can doubtlessly purpose pressure and anxiety, and both the anticipate a diagnosis and attempts to find a fertility treatment that works can reason tension and conflict.
Insti. the testicular most cancers useful resource middle fertility page infertility a sympathetic method 1985 getting pregnant 1989. infertility: a sympathetic technique to expertise the causes and alternatives for remedy. by using robert m l winston. print e-book. english. 1996. revised edition. infertility: a sympathetic method robert winston. Treatment will depend upon the underlying cause of the infertility. erectile dysfunction or premature ejaculation: remedy, behavioral strategies, or each may additionally help enhance fertility. Enormous advances have befell within the prognosis and, more importantly, inside the treatment of reproductive problems over the past decade. the general occurrence of infertility has remained strong 1; however, the achievement fees have markedly advanced with the sizable use of assisted reproductive technologies. treatment alternatives and success range with the purpose of infertility. | 17,860 | 86,980 | {"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-2020-45 | latest | en | 0.893133 |
https://atcoder.jp/contests/abc182/submissions/17968851 | 1,606,424,508,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00346.warc.gz | 196,878,161 | 5,566 | Contest Duration: - (local time) (100 minutes) Back to Home
Submission #17968851
Source Code Expand
Copy
```import sys
import numpy as np
import numba
from numba import njit, b1, i4, i8, f8
@njit((i8, i8, i8[:, :], i8[:, :]), cache=True)
def main(H, W, AB, CD):
G = np.zeros((H, W), np.int64)
for i in range(len(AB)):
a, b = AB[i]
G[a - 1, b - 1] = 1
for i in range(len(CD)):
c, d = CD[i]
G[c - 1, d - 1] = -1
res = np.zeros((H, W), np.bool_) # 光ったマス
for _ in range(4):
res, G = res.T[::-1], G.T[::-1]
# 下方向に解く
A = np.zeros_like(res)
H, W = A.shape
for h in range(H):
if h >= 1:
A[h] |= A[h - 1]
A[h][G[h] == 1] = 1
A[h][G[h] == -1] = 0
res |= A
return np.sum(res)
H, W, N, M = map(int, readline().split())
AB = nums[:N + N].reshape(N, 2)
CD = nums[N + N:].reshape(M, 2)
print(main(H, W, AB, CD))```
#### Submission Info
Submission Time 2020-11-08 21:26:35+0900 E - Akari maspy Python (3.8.2) 500 1240 Byte AC 610 ms 139968 KB
#### Judge Result
Set Name Sample All
Score / Max Score 0 / 0 500 / 500
Status
AC × 3
AC × 20
Set Name Test Cases
Sample sample_01.txt, sample_02.txt, sample_03.txt
All handmade_00.txt, max_random_00.txt, max_random_01.txt, max_random_02.txt, max_random_03.txt, max_random_04.txt, random_00.txt, random_01.txt, random_02.txt, random_03.txt, random_04.txt, random_05.txt, random_06.txt, random_07.txt, random_08.txt, random_09.txt, sample_01.txt, sample_02.txt, sample_03.txt, surrounded_00.txt
Case Name Status Exec Time Memory
handmade_00.txt AC 491 ms 105700 KB
max_random_00.txt AC 586 ms 137428 KB
max_random_01.txt AC 581 ms 138124 KB
max_random_02.txt AC 587 ms 138072 KB
max_random_03.txt AC 580 ms 138104 KB
max_random_04.txt AC 586 ms 137380 KB
random_00.txt AC 610 ms 139612 KB
random_01.txt AC 597 ms 139968 KB
random_02.txt AC 468 ms 106872 KB
random_03.txt AC 471 ms 106216 KB
random_04.txt AC 532 ms 124568 KB
random_05.txt AC 527 ms 124808 KB
random_06.txt AC 521 ms 122284 KB
random_07.txt AC 553 ms 131128 KB
random_08.txt AC 477 ms 108440 KB
random_09.txt AC 541 ms 127376 KB
sample_01.txt AC 469 ms 106232 KB
sample_02.txt AC 469 ms 106548 KB
sample_03.txt AC 471 ms 106552 KB
surrounded_00.txt AC 590 ms 138132 KB | 828 | 2,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.341251 |
https://coolconversion.com/energy/0.6-calorie%20(thermochemical)-to-watt-hour | 1,532,345,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00113.warc.gz | 637,480,029 | 36,771 | Please get in touch with us if you:
1. Have any suggestions
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# 0.6 Calories (thermochemical) to Watts-hour Conversion
## How many watts-hour in 0.6 calorie (thermochemical)?
There are 0.00069733333333333 watts-hour in 0.6 calorie (thermochemical)
To convert any value in calories (thermochemical) to watts-hour, just multiply the value in calories (thermochemical) by the conversion factor 0.0011622222222222. So, 0.6 calorie (thermochemical) times 0.0011622222222222 is equal to 0.00069733333333333 watts-hour. See details below and use our calculator to convert any value in calories (thermochemical) to watts-hour.
To use this calories (thermochemical) to watts-hour converter, simply type the calorie (thermochemical) value in the box at left (input). The conversion result in watt-hour will immediately appear in the box at right.
### Calorie (thermochemical) to Watt-hour Converter
Enter values here: Results here:
=
Detailed result here
As each calorie (thermochemical) is equivalent to 0.0011622222222222 watt-hour, we can use the following conversion formula: Value in watts-hour = value in calories (thermochemical) * 0.0011622222222222 Supose you want to convert 0.6 calorie (thermochemical) into watts-hour. In this case just write the equation, then do the math: Value in watts-hour = 0.6 * 0.0011622222222222 = 0.00069733333333333 (watt-hour). This is your your final answer.
Using this converter you can get answers to questions like:
1. How many calories (thermochemical) are in 0.6 watts-hour?
2. 0.6 calories (thermochemical) is equal to how many watts-hour?
3. how much is 0.6 calorie (thermochemical) in watts-hour?
4. How to convert calories (thermochemical) to watts-hour?
5. What is the conversion factor to convert from calories (thermochemical) to watts-hour?
6. How to transform calories (thermochemical) in watts-hour?
7. What is the formula to convert from calories (thermochemical) to watts-hour? Among others.
## Values Near 0.6 calorie (thermochemical) in watt-hour
calorie (thermochemical)watt-hour
0.520.00060435555555556
0.530.00061597777777778
0.540.0006276
0.550.00063922222222222
0.560.00065084444444444
0.570.00066246666666667
0.580.00067408888888889
0.590.00068571111111111
0.60.00069733333333333
0.610.00070895555555556
0.620.00072057777777778
0.630.0007322
0.640.00074382222222222
0.650.00075544444444444
0.660.00076706666666667
0.670.00077868888888889
0.680.00079031111111111
Note: some values may be rounded. | 750 | 2,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-30 | latest | en | 0.768764 |
http://poj.org/problem?id=2563 | 1,656,495,636,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00579.warc.gz | 45,010,740 | 3,404 | Online JudgeProblem SetAuthorsOnline ContestsUser
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Demerit Points
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 523 Accepted: 173
Description
A province to our west, which shall remain nameless, but whose name does not start with A, B, or S, has a unique system for driver's license demerit and merit points. The system works (more or less) as follows.
A new driver starts with no merit or demerit points. When the driver is convicted of a driving offense, he or she is given between 2 and 15 demerit points, depending on the severity of the offense.
A merit point is given, to a maximum of five, for each interval of two years in which a driver has no offenses and no demerit points. Each merit point cancels up to two demerit points. If a subsequent offense occurs and the number of demerit points exceeds double the number of merit points, the number of demerit points is reduced by double the number of merit points, and the number of merit points is set to 0. If a subsequent offense occurs and the number of demerit points is less than or equal to double the number of merit points, the number of demerit points is reduced to 0, and the number of merit points is reduced by half the number of demerit points. Any fractional merit points are discarded.
Demerit points are reduced whenever a driver has one year free of any driving offense. This reduction decreases the number of demerits by half or by 2, whichever is more. Any fractional or negative demerit points are discarded. This reduction takes place on each anniversary of the most recent offense until no points remain.
If a new offense occurs on the same day as a demerit point reduction or merit point award, the reduction/award is done before the new demerit points are given.
Your job is to read a set of information records for a driver, and to print the number of merit or demerit points at any given time.
Input
The first line of input contains the date of issue of the license (yyyymmdd) Subsequent lines contain information on offenses, in chronological order. Each such line contains the offense date (yyyymmdd) and the number of points applied (an integer between 2 and 15).
Output
On the day the license is issued, and on every occasion that the number of merit or demerit points changes, output a line giving the date and the number of points, in the format below. Output terminates when 5 merit points are accumulated following the last offense.
Sample Input
```19820508
19830606 2
19830607 2
19891212 15
```
Sample Output
```1982-05-08 No merit or demerit points.
1983-06-06 2 demerit point(s).
1983-06-07 4 demerit point(s).
1984-06-07 2 demerit point(s).
1985-06-07 No merit or demerit points.
1987-06-07 1 merit point(s).
1989-06-07 2 merit point(s).
1989-12-12 11 demerit point(s).
1990-12-12 5 demerit point(s).
1991-12-12 2 demerit point(s).
1992-12-12 No merit or demerit points.
1994-12-12 1 merit point(s).
1996-12-12 2 merit point(s).
1998-12-12 3 merit point(s).
2000-12-12 4 merit point(s).
2002-12-12 5 merit point(s).
```
Source
[Submit] [Go Back] [Status] [Discuss] | 862 | 3,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.910103 |
https://music.stackexchange.com/questions/tagged/intervals?sort=active | 1,716,400,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00745.warc.gz | 352,761,018 | 49,606 | # Questions tagged [intervals]
For questions centered around the ideas about and classification of distances between two pitch.
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### Interval exercise gone wrong?
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• Tara O'Brien
# Resource: Math Workshop Naturally Differentiated Context
What is Math Workshop?
Let`s begin with a simplified step-by-step overview of the model...
1. The teacher presents the context, which can be a contextualized experience or a book/story to the class seated in a circle. The experience or story must be rooted in context and age appropriate so that students focus on the problem to solve and not unimportant variables.
2. After giving the context the teacher gives students the problem to solve. Students often work in pairs and small groups to collaborate and are encouraged to use tools and record their thinking. The teacher circulates, taking observational notes, asking students to explain their process, and prompting students to move in the direction of using more complex math skills.
3. After work time, students return to the circle and the teacher asks students to share their strategies for solving the problem. The teacher is thoughtful in picking students to share strategies that the class will benefit from hearing. Then the teacher models the strategies shared.
4. The process is repeated the next day, with a review of what happened the day before and the problem within the context changed so that students have the opportunity to experiment using the strategies that they heard about the day before.
5. Over the course of the math unit, students apply strategies that they have heard in Math Workshop to new problems within the same general context.
This is a very basic overview of how teachers can use contextualized, naturally differentiated math contexts within the math workshop model. The model is appropriate for students of all ages and is highly effective in motivating students. Click the pdf below to and learn more about Math Workshop.
Teaching and Learning in Math Workshop
.p
Example Series of Math Contexts for Kindergarten/Year 1
Each winter all of our Kindergarten students go on weekly field trips to the ice skating rink. I used this weekly occurrence as a math context to develop number sense, as all of the children shared and were motivated by the experience of going to the rink. Take a look at the graphic below to see how students progress through the big ideas, concepts, and strategies within number sense. When designing the math context and predicting how students may respond, I used this graphic. This graphic shows where each student is in their understanding of number and where their next steps need to be. It is also helpful to print out this graphic and use it while observing the students! Kindergarten students are typically in the bottom third of the graphic.
Here are the contexts I created...
1. Context One: How many ice skates does our class bring on the bus?
2. When presenting this question to the class I purposefully omit how many children are in the class. I also purposefully omit that every child brings two ice skates. Part of learning how to solve a math problem is understanding which information is relevant to solving a problem.
3. This is a complex problem to solve with high numbers. Before beginning work time I make sure to point out the number line, hundred chart, math racks, and manipulatives available to the children. I also remind students to record, draw, and take whatever notes they choose to think through the problem.
4. After discussing the problem as a whole class and considering the different ways that children might solve the problem, I normally pair up students to work together. I encourage collaboration and discussion during Math Workshop as the children can observe and learn from each other. During this time I circulate, ask questions, and take notes about which strategies the children are using. Often, the less advanced mathematicians draw all of the ice skates and one to one count using tagging. The higher mathematicians often draw lines or the number two repeatedly and skip count to solve.
5. After work time I invite everyone back to the carpet to share. I purposefully pick children to share certain strategies and then model the strategy a second time.
2. Context Two: How many ice skates do both Kindergarten classes bring on the bus?
3. Context Three: When we go to the rink only five children are allowed to sit on each bench. How many benches will we need?
During each of these contexts the lower level mathematicians often use representational drawings and pictures to think through the problem. As children become more sophisticated mathematicians they are able to use symbols, lines, dots, or numerals to think through and solve the problem. Lower level mathematicians often tag and one to one count to solve the problem while higher level mathematicians solve using skip counting or even doubling. For example, there are 15 students in the class. A high level mathematician could skip count 15 times or double the number 15. Higher level mathematicians are also more able to explain their process and prove their answer. Learning how to explain a math strategy takes time and a certain degree of confidence. By providing a classroom culture that is supportive and celebrates challenge we can help every student take more academic risks and more readily share their thinking process. During these contexts the higher level mathematicians are pushed to try new strategies and explain them while the lower level mathematicians learn from their classmates during pair work and class discussions while being encouraged by their teacher.
Each of these three contexts develop the same set of big ideas, concepts, and strategies and students can apply what they learned in one context to the next. I hope that this post is helpful in planning your own naturally differentiated math contexts. Please share contexts that have worked for you in the comments!
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# The Escape Speed of a Projectile on the Earth’S Surface is 11.2 Km S–1. a Body is Projected Out with Thrice this Speed. What is the Speed of the Body Far Away from the Earth? Ignore the Presence of the Sun and Other Planets - CBSE (Science) Class 11 - Physics
#### Question
The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets
#### Solution 1
Escape velocity of a projectile from the Earth, vesc = 11.2 km/s
Projection velocity of the projectile, vp = 3vesc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = vf
Total energy of the projectile on the Earth = 1/2mv_p^2 - 1/2mv_"esc"^2
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = 1/2mv_f^2
From the law of conservation of energy, we have
1/2mv_p^2 - 1/2mv_"esc"^2 = 1/2 mv_f^2
v_f = sqrt(v_p^2-v_"esc"^2)
=sqrt((3v_"esc")^2 - (v_"esc")^2
=sqrt8 v_"esc"
=sqrt8 xx11.2 = 31.68 km/s
#### Solution 2
Let v_"es" be the escape speed from surfce of Earth havinf a vlaue v_"es" = 11.2 kg s^(-1)
=11.2 xx 10^3 ms^(-1)
By definition
1/2 mv_e^2 = (GMm)/(R^2)
When a body is projected with aspeed v_i = 3v_"es" = 3 xx 11.2 xx 10^3 m/s then it will have a final spee v_f such that
1/2 mv_f^2 = 1/2mv_i^2 - (GMm)/R^2 = 1/2mv_i^2 - 1/2mv_e^2
=>v_f = sqrt(v_i^2 -v_e^2)
=sqrt((3xx11.2xx10^3)-(11.2xx10^3)^2)
= 11.2 xx 10^3 xx sqrt8
=31.7xx10^3 ms^(-1) or 31.7 km s^(-1)
Is there an error in this question or solution?
#### APPEARS IN
Solution The Escape Speed of a Projectile on the Earth’S Surface is 11.2 Km S–1. a Body is Projected Out with Thrice this Speed. What is the Speed of the Body Far Away from the Earth? Ignore the Presence of the Sun and Other Planets Concept: Escape Speed.
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For questions related to the study of properties of a graph in relationship to the spectral properties of some associated matrix.
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https://docs.scipy.org/doc/scipy-1.6.2/reference/generated/scipy.stats.nct.html | 1,620,877,926,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00391.warc.gz | 241,403,330 | 4,698 | # scipy.stats.nct¶
scipy.stats.nct(*args, **kwds) = <scipy.stats._continuous_distns.nct_gen object>[source]
A non-central Student’s t continuous random variable.
As an instance of the rv_continuous class, nct object inherits from it a collection of generic methods (see below for the full list), and completes them with details specific for this particular distribution.
Notes
If $$Y$$ is a standard normal random variable and $$V$$ is an independent chi-square random variable (chi2) with $$k$$ degrees of freedom, then
$X = \frac{Y + c}{\sqrt{V/k}}$
has a non-central Student’s t distribution on the real line. The degrees of freedom parameter $$k$$ (denoted df in the implementation) satisfies $$k > 0$$ and the noncentrality parameter $$c$$ (denoted nc in the implementation) is a real number.
The probability density above is defined in the “standardized” form. To shift and/or scale the distribution use the loc and scale parameters. Specifically, nct.pdf(x, df, nc, loc, scale) is identically equivalent to nct.pdf(y, df, nc) / scale with y = (x - loc) / scale. Note that shifting the location of a distribution does not make it a “noncentral” distribution; noncentral generalizations of some distributions are available in separate classes.
Examples
>>> from scipy.stats import nct
>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots(1, 1)
Calculate a few first moments:
>>> df, nc = 14, 0.24
>>> mean, var, skew, kurt = nct.stats(df, nc, moments='mvsk')
Display the probability density function (pdf):
>>> x = np.linspace(nct.ppf(0.01, df, nc),
... nct.ppf(0.99, df, nc), 100)
>>> ax.plot(x, nct.pdf(x, df, nc),
... 'r-', lw=5, alpha=0.6, label='nct pdf')
Alternatively, the distribution object can be called (as a function) to fix the shape, location and scale parameters. This returns a “frozen” RV object holding the given parameters fixed.
Freeze the distribution and display the frozen pdf:
>>> rv = nct(df, nc)
>>> ax.plot(x, rv.pdf(x), 'k-', lw=2, label='frozen pdf')
Check accuracy of cdf and ppf:
>>> vals = nct.ppf([0.001, 0.5, 0.999], df, nc)
>>> np.allclose([0.001, 0.5, 0.999], nct.cdf(vals, df, nc))
True
Generate random numbers:
>>> r = nct.rvs(df, nc, size=1000)
And compare the histogram:
>>> ax.hist(r, density=True, histtype='stepfilled', alpha=0.2)
>>> ax.legend(loc='best', frameon=False)
>>> plt.show()
Methods
rvs(df, nc, loc=0, scale=1, size=1, random_state=None) Random variates. pdf(x, df, nc, loc=0, scale=1) Probability density function. logpdf(x, df, nc, loc=0, scale=1) Log of the probability density function. cdf(x, df, nc, loc=0, scale=1) Cumulative distribution function. logcdf(x, df, nc, loc=0, scale=1) Log of the cumulative distribution function. sf(x, df, nc, loc=0, scale=1) Survival function (also defined as 1 - cdf, but sf is sometimes more accurate). logsf(x, df, nc, loc=0, scale=1) Log of the survival function. ppf(q, df, nc, loc=0, scale=1) Percent point function (inverse of cdf — percentiles). isf(q, df, nc, loc=0, scale=1) Inverse survival function (inverse of sf). moment(n, df, nc, loc=0, scale=1) Non-central moment of order n stats(df, nc, loc=0, scale=1, moments=’mv’) Mean(‘m’), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’). entropy(df, nc, loc=0, scale=1) (Differential) entropy of the RV. fit(data) Parameter estimates for generic data. See scipy.stats.rv_continuous.fit for detailed documentation of the keyword arguments. expect(func, args=(df, nc), loc=0, scale=1, lb=None, ub=None, conditional=False, **kwds) Expected value of a function (of one argument) with respect to the distribution. median(df, nc, loc=0, scale=1) Median of the distribution. mean(df, nc, loc=0, scale=1) Mean of the distribution. var(df, nc, loc=0, scale=1) Variance of the distribution. std(df, nc, loc=0, scale=1) Standard deviation of the distribution. interval(alpha, df, nc, loc=0, scale=1) Endpoints of the range that contains alpha percent of the distribution
scipy.stats.ncf
scipy.stats.norm | 1,149 | 4,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.694306 |
https://books.google.gr/books?id=gacXAAAAIAAJ&hl=el&lr= | 1,611,803,961,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00485.warc.gz | 250,263,650 | 12,436 | # Elementary Geometry, Plane and Solid: For Use in High Schools and Academies
Macmillan, 1901 - 440 σελίδες
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Περιεχόμενα
CHAPTER I 22 Parallel Lines 61 Closed Rectilinear Figures Parallelograms 71 Miscellaneous Theorems 78 Additional Problems of Construction 84 Miscellaneous Exercises 90 CHAPTER II 98 Angles inscribed in Arcs 111
Dihedral Angles 284 Polyhedral Angles 296 Miscellaneous Exercises 304 Pyramids 328 Similar Polyhedrons 339 Polyhedrons in General 345 CHAPTER VIII 353 The Cone 361
### Δημοφιλή αποσπάσματα
Σελίδα 187 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Σελίδα 230 - The formula states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the base and altitude.
Σελίδα 55 - If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'B'C...
Σελίδα 76 - The line which joins the mid-points of two sides of a triangle is parallel to the third side and equal to one half of it.
Σελίδα 43 - Prove that, if two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less.
Σελίδα 231 - A polygon of three sides is called a triangle ; one of four sides, a quadrilateral; one of five sides, a, pentagon; one of six sides, a hexagon ; one of seven sides, a heptagon ; one of eight sides, an octagon ; one of ten sides, a decagon ; one of twelve sides, a dodecagon.
Σελίδα 27 - If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects.
Σελίδα 200 - The area of a triangle is equal to half the product of its base by its altitude.
Σελίδα 161 - ... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
Σελίδα 229 - Two parallelograms are similar when they have an angle of the one equal to an angle of the other, and the including sides proportional. | 603 | 2,394 | {"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-2021-04 | latest | en | 0.743249 |
https://physics.blogberth.com/2018/04/14/can-you-crush-a-soda-can-with-lightning-here-you-can-see-the/ | 1,537,762,037,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160142.86/warc/CC-MAIN-20180924031344-20180924051744-00341.warc.gz | 576,946,210 | 42,247 | # Can you crush a soda can with lightning ?Here you can see the…
## Can you crush a soda can with lightning ?
Here you can see the lightning striking the empire state building three times.
As you might know skyscrapers like these have a lightning rod that will carry the lightning bolt’s electrical charge through
the path of least resistance along the cable into the ground, reducing
the risk of fire or heat damage from the strike.
But you might know from high school or college physics that when you have two current carrying wires parallel to each other, then they experience an attractive force towards each other.
Source Video
## Lightning carries 30000 amps
An average bolt of negative lightning carries an electric current of 30,000 amperes (30 kA).
So if you were to pass such high currents through a rod, then surely they must experience a substantial magnetic force towards each other and get crushed right ?
Absolutely, Physics always works ! Have a look at this metal tube:
This is what would happen when electric lightning is passed through a metal tube. Using magnetic forces to compress electrical filaments is known as ‘Pinching’.
## Can Crusher
One cool application for this would be for forming metal cans into interesting shapes.
As soon as the spark gap fires the capacitor discharges an enormous amount
of current through the coil (tens of thousands of amperes). This
discharge creates a magnetic field around the coil.
As the flux lines
pass through the cross section of the can, current is induced and flows
around the can.
This induced current creates its own magnetic field
which opposes the magnetic field from the coil. Between the two magnetic
fields there is now a force pushing inward on the can and outward on the
coil. Once the force is strong enough the can is crushed.
And with large enough voltage, one can blow the can in opposite directions too!
From generating X-rays to forming of metal into shapes, Pinches have a lot of interesting applications. The references section of the wikipedia is surely a fun ride; check it out and have a good one! | 431 | 2,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-39 | latest | en | 0.936953 |
www.randaljbishop.com | 1,695,745,082,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00777.warc.gz | 1,076,454,788 | 13,924 | The Use of Linear Perspective to Represent the Relationships in a Four-Dimensional Coordinate System
By
Randal J. Bishop
We associate images in linear perspective and sculptures with three-dimensional spaces. There are four three-dimensional spaces in a four-dimensional coordinate system. The preceding two sentences lead one to believe those two ideas would be extremely easy to put together, or have already been put together. This paper will make the explanation and illustration of those ideas extremely easy because it will not require one to know all of the details about three and four-dimensional geometries. The first section will explain the principles and compositional elements (elements being points, lines and planes) of linear perspective. The second section will explain the elements of two and three-dimensional Cartesian coordinate systems and our analogous associations with the elements of linear perspective and sculptures. The third section will present the basic principles of dimensionality or multidimensional spaces while including a description with illustrations of a four-dimensional coordinate system. The fourth section will present detailed explanations on how images in linear perspective and naturalistically scaled sculptural images can be used to represent the relationships in a four-dimensional coordinate system.
Section 1
The compositional devices known as linear perspective are familiar to those who have studied Western art. In so many words, they are the principles guiding the compositional lines that allow for naturalistically scaled images in perspective. Over the centuries, this phenomenon has been referred to as perspective, linear perspective, one-point perspective, three-point perspective, centric perspective, aesthetic perspective, scientific perspective, geometric perspective, mathematical perspective and skenographia. The easiest way to capture and exhibit an image in linear perspective is with a photograph. Basic naturalistic images in perspective can also be recreated with most aesthetic mediums: drawing, oil painting or any technique for creating a motion picture. Now, a conventional description of an image in linear perspective will be followed with our analogous associations with a three-dimensional Cartesian coordinate system.
Leonardo da Vinci’s fresco, Last Supper, 1495-1498, is the most noted example of one-point perspective because its compositional lines are blatantly apparent and it still exists. The compositional lines appear as the edges of the tapestries, walls, ceiling beams and table. When visually continued, these lines converge with the line of sight at the point of infinity (marked by the figure of Jesus) to create an accurate diagram of one-point perspective. This line of sight is also perpendicular to the projected plane, in the middle ground, on which the figures of the apostles are arranged. This plane divides the space in front of the figures from the cavernous space of the room behind the figures. The relevant aspects of this composition are the following. There is a line of sight (a one-dimensional line) perpendicular to a two-dimensional plane that can be seen as dividing the whole image. For more details and diagrams about pictorial compositions, see Martin Kemp’s book, The Science of Art.
This type of compositional structure naturally presents the appearance of a foreground, middle ground and background. The usage of these terms in this paper will be limited to these very general and traditional definitions. The foreground of an image contains the space in front of the projected plane in the middle ground and everything on the “picture plane”. A “picture plane” is the whole surface the image is on, be it a wall or canvas, etc. A “picture plane” is also referred to as the “window” through which the image is seen. The middle ground shows where the line of sight perpendicularly intersects the projected plane on which most of the subject matter is placed. The background is the space behind the foreground and the projected plane in the middle ground.
Perspective was studied by many people during the Renaissance. Leon Battista Alberti is credited for writing the first treatise on centric perspective entitled De Pictura, 1435. The most authoritative and cited English translation to date is by Cecil Grayson entitled On Painting and On Sculpture. The origin of geometric perspective in art would not be complete without mentioning Alberti’s dedication in 1436 of his treatise to Filippo Brunelleschi. Brunelleschi is alleged, by his biographer Antonio di Tuccio Manetti, to have created the first two, now lost, paintings of scenes in perspective of Florence, Italy. The exact dates of the panels are not known; however, in Samuel Y. Edgerton’s book, The Renaissance Rediscovery of Linear Perspective, Edgerton dates at least one of the panels to pre-1425 because of other paintings by Masaccio and Masolino dating from circa 1425-1427. Edgerton’s research is reified with a photographic recreation of Brunelleschi’s first painting in perspective. The writings of William M. Ivins, On the Rationalization of Sight and particularly Art and Geometry, link the understandings various cultures throughout the ages had of optics and geometries to the developments of linear perspective and modern geometries. Irwin Panofsky’s book, Renaissance and Renascences in WesternArt, thoroughly documents the aesthetic developments leading to accurately scaled, realistic images in perspective. The Fourth Dimension and Non-Euclidean Geometry in Modern Art by Linda Dalrymple Henderson addresses more recent explorations between the arts and n-dimensional geometries. Clifford A. Pickford's amusing book Surfing Through Hyperspace: Understanding Higher Universes in Six Easy Lessons uses humor, something this subject really needs. Many great writers have documented the connectivity of art and geometric techniques during the past six centuries.
Section 2
Today, it is common for us to say that images in perspective represent the third dimension or are two-dimensional representations of three-dimensional spaces. It is also common for us to say sculptures are three-dimensional. These associations started after René Descartes developed what are now called two and three-dimensional Cartesian coordinate systems. A two-dimensional coordinate system (Figure 1) is composed of two perpendicular lines and a three dimensional coordinate system (Figure 2) is composed of three mutually perpendicular lines. A three-dimensional Cartesian coordinate system is occasionally referred to as the X Y Z axes.
A two-dimensional coordinate system can become a square (Figure 3) and a three-dimensional coordinate system can become a cube (Figure 4) when their reference lines and reference planes are orthographically projected. Orthographic projection means “moving” a reference line or plane to become a projection line or plane. In the diagrams, the reference lines and planes are thicker than the projection lines and planes. Figure 5 illustrates a projection line (line ab) and projection planes (thin lines) in a three-dimensional coordinate system. Projection line (line ab) is projected from and is parallel to reference line (line AB) while the two projection planes (thin lines) are projected from and are parallel to the reference plane at the back of the coordinate system. This automatically makes projection line (line ab) perpendicular to the projection planes. The projection line (line ab) and the projection planes are analogous to a line of sight and planes in an image in perspective. In an image in perspective a line of sight perpendicularly passes through the “picture plane” or “window” at the front of the image while it perpendicularly intersects a plane in the middle ground of the image. The perpendicularity of these lines and planes remains consistent in three-dimensional coordinate systems and images in perspective. This perpendicular consistency is the imperative factor to the introduction of, both, images in perspective and naturalistically scaled sculptural images into higher four-dimensional spaces.
Refer back to da Vinci’s Last Supper. The composition of this fresco should allow one to see how photographic or two-dimensional naturalistic images in perspective can be used to represent the three-dimensional spaces in a four-dimensional space. In another fresco (by Masaccio, entitled Trinity) an architectural like frame is painted around the “window” while two figures of people and an altar table with a tomb below are painted on the “picture plane” in front of the “window”. This masterpiece, probably dating from the latter part of 1425, contains all of the compositional devices one needs to comprehend the introduction of photographic or two-dimensional naturalistic images in perspective and naturalistically scaled sculptural imagery into higher dimensional spaces.
In summation, it is easy to see our analogous associations between the compositional elements of perspective and the geometric elements of three-dimensional geometry. This knowledge is actually a good introduction to dimensionality or multidimensional spaces and n-dimensional or higher dimensional geometries.
Section 3
The concepts of dimensionality and higher dimensional spaces are easy to understand when introduced in the context of zero, one, two and three-dimensional spaces. These descriptions of some of the terminology of the discipline may help. A zero-dimensional space is called a vertex. A one-dimensional space is called a valence. Vertices and valences do not have volume. Vertices are zero-dimensional spaces that can be designated with letters, numbers or points. Valences are one-dimensional spaces, of any length, that can be designated with letters, numbers or lines. When two valences or one-dimensional spaces intersect at a vertex or zero-dimensional space they can form a two-dimensional space (i.e., a plane), and so on. The other dimensions are also composed of these geometric elements. A three-dimensional space occurs when three one-dimensional spaces intersect at a zero-dimensional space to form three two-dimensional spaces. A four-dimensional space occurs when four one-dimensional spaces intersect at a zero-dimensional space to form six two-dimensional planes and four three-dimensional spaces ( Figure 6). That’s it, an extremely simplified description and illustration of the fourth dimension. The actual geometry is not that simple. As a matter of fact, an unapologetic, irreverent over simplification of that magnitude would be an insult to all the mathematicians and geometers who have given us higher dimensional geometries. So, please allow for these notes. The four-dimensional coordinate system used in this paper has been referred to as a hypertetrahedron, four simplex and five-cell (the five-cells being the four three-dimensional cells while the fifth cell is regarded as the overall space encompassed by the four-dimensional coordinate system). The four-dimensional system in this paper is from the book Four-Dimensional Descriptive Geometry by C. Ernesto S. Lindgren and Steve M. Slaby. Thomas Banchoff’s most recent book, Beyond the Third Dimension is a comprehensive, easy-to-read introduction to dimensionality and higher dimensional geometries. The Fundamentals of Three-Dimensional Descriptive Geometry, by Steve M. Slaby, is an excellent introductory textbook on the subject. Hypergraphics: Visualizing Complex Relationships in Art, Science and Technology, edited by David W. Brisson, was a pioneering anthology on these multidisciplinary issues. The book Space Structures: Their Harmony and Counterpoint, by Arthur L. Loeb, and the academic journal HyperSpace are for those who would like to explore the vastness of these subjects. However, as mentioned in the beginning of this paper, all of the details about higher dimensional geometries and its century plus old history do not have to be known to understand this development. This was not the case with some of the details regarding linear perspective and three-dimensional geometry, which turned out to be a good introduction to four-dimensional geometry.
We only have to know some of the geometry’s basic notions to understand this development. All of the geometry being referred to is from the book Four-Dimensional Descriptive Geometry by C. Ernesto S. Lindgren and Steve M. Slaby. The basic notions being referred to are called the conditions of perpendicularity. Those conditions are four statements specifying the perpendicular relationships of the four reference lines forming the six two-dimensional planes and four three-dimensional spaces in a four-dimensional coordinate system. The following four requirements for a four-dimensional coordinate system, from page 22 in their book, are similar to the mutually perpendicular requirements of the three lines and three planes in a three dimensional Cartesian coordinate system.
1. The four spaces of the system, taken in threes, determine the four lines that are perpendicular to each other and belong to the same point.
2. Four lines, taken in threes, determine the four spaces that are perpendicular to each other.
3. Four lines, taken in pairs, determine six planes, which, taken in threes, form four groups of planes belonging to the same line and are mutually perpendicular.
4. Any one line of the reference system is perpendicular to a space determined by the other three lines in the system.
These four statements provide a verbal description of a four-dimensional coordinate system. While Figure 6, from page 17 in their book, as a diagram provides only a visual indication of the conditions of belonging and not the conditions of perpendicularity. In other words, it just shows the lines belonging to certain planes and spaces, not the perpendicular relationships among the lines, planes and spaces. (Note: the lines have been
given numbers and the planes have been given lower-case Greek letters).
When their book was published in 1968, Lindgren and Slaby used Figure 7, amongst many others, to illustrate all of the perpendicular alignments of the elements in a four-dimensional coordinate system. Their publication still stands as the definitive treatise on four-dimensional descriptive geometry.
Luckily, there is a way to show everything in those diagrams using four images or sculptures of natural people standing on natural ground to indicate the proper relationships of the three-dimensional spaces in a four-dimensional space.
Section 4
This can be achieved by applying the conditions of belonging and perpendicularity to four lines of sight, just as they can be applied to four reference lines. This does not just mean the images of spaces in perspective and naturalistically scaled sculptural images must be placed in the correct three-dimensional space in the four-dimensional system – meaning placing the image of space 1 in space 1 in the coordinate system. It means the lines of sight with accompanying images must visually belong to certain lines, planes and spaces in the system while being perpendicular to other lines, planes and spaces in the system.
Keep in mind, when dealing with realistic imagery two sets of aesthetic issues have to be addressed to present all of the relationships of the lines, planes and spaces inside of a four-dimensional coordinate system. The first aesthetic issue involves compositional lines and planes of an image in perspective; each of the four images should have obvious foregrounds, middle grounds and backgrounds and unobstructed lines of sight. The second issue involves the visual distinctiveness of the subjects or figures in the images; each one of the four images should contain four visually distinct subjects or figures so as to be readily identifiable with one of the four distinct spaces. Each of these aesthetic issues should complement the other. Consequently, the compositional elements of the images must not only belong to the corresponding elements of the spaces in the four-dimensional coordinate system, the lines of sight in each image (just like the reference lines) must clearly be perpendicular to the correct three-dimensional space inside of the four-dimensional coordinate system, and the subjects or figures in the images should also show in their own backgrounds an image of the space it must be perpendicular to in the four-dimensional coordinate system. In this way the introduction of naturalistic imagery adds extra requirements that need to comply with the requirements of the geometry. However surprisingly, the end result of this complex compilation will be works of art free of obvious geometric notations (notations being letters, numbers or lines) that readily exhibit all of the aesthetic and geometric aspects of this amalgamation.
These four images can be derived from three points in opposition (meaning three points in a row). The two images taken from the two outer points must be directed toward the central point, while the other two images taken from the central point must be directed toward the outer points. This pattern establishes a direction for the lines of sight that correspond to the needs of the reference lines to intersect some of the other reference lines at perpendicular angles. In an effort to ingest as little as possible let these reminders suffice. Lines of sight can be interpreted as emanating from the viewer or from the subject (see William M. Ivins’ writings on the study of optics during the Middle Ages). Now back to the assimilation of the images to the geometry. Each of these images of spaces must simultaneously represent a line, plane and three-dimensional space within a four-dimensional coordinate system. This is achieved when the spaces are numbered (from left to right) 1,2,3,4 and respectively designated with reference or projection planes µ, η, α, ß and reference or projection lines 3, 1, 4 and 2. Planes γ and δ, the other two planes in the system, must be mutually perpendicular to planes µ, η, α and ß. These relationships occur when plane γ shows the ground or horizontal plane of spaces 2 and 3, and plane δ shows the ground or horizontal plane of spaces 1 and 4. The side of viewing the spaces between the three points in opposition does not matter or if the photographs are numbered from right to left, as long as space 1 is associated with plane µ and line 3, space 2 with plane η and line 1, space 3 with plane α and line 4 and space 4 with plane ß and line 2; the geometry is composed of reciprocal statements therefore the result will be the same four arrangements. The reciprocal nature of this geometry allows for, at least, these two aesthetic interpretations of any space chosen to illustrate a four-dimensional coordinate system.
When choosing a natural space remember each image should obviously appear to belong to the same natural space, while maintaining a distinct subject or figure in each one of the four spaces. In the four three-dimensional spaces used to illustrate the system described in this paper, there is a car to the far left, in the middle of space 1, a man in the middle of space 2, a woman in the middle of space 3 and a tree in the middle of space 4.
The alignment of these four figures in their natural spaces can be seen during the first 15 seconds of the computer graphics. It is aesthetically relevant to note the slightly staggered from center alignments of the man, woman and tree to allow for unobstructed lines of sight through the four spaces.
The first few seconds of the graphics shows a fly through to the central point in the natural space. This is also the image of space 3 containing the figure of the woman in space 3 and the diminutively scaled figure of the tree in the background on reference plane α; the tree is actually in space 4. At this central point in the natural space the camera turns to face and fly towards the same image in the four-dimensional coordinate system. As the computer’s camera enters space 3 in the system, it follows the line of sight, parallel to reference line 4, through space 3 to ultimately be perpendicular to space 4 with the tree in it.
The precise perpendicular relationships are not illustrated due to limitations regarding the computer graphics. The figure of the man is facing the wrong direction in space 2; it should be turned 180° to face the direction the line-of-sight or reference line must emanate from so the lines can ultimately be perpendicular to space 1. Nevertheless, if reference line 4 was made perpendicular to reference line 1 half of the conditions of perpendicularity would be satisfied; then if reference line 3 was made perpendicular to reference line 2 the other half of the conditions of perpendicularity would be satisfied.
The results of these two arrangements will be the following: reference line 4 will belong to spaces 1, 2, 3 and plains γ, η, µ while being perpendicular to space 4, reference line 1 will belong to spaces 2, 3, 4 and plains α, ß, γ while being perpendicular to space 1, reference line 3 will belong to spaces 1, 2, 4 and plains ß, δ, η while being perpendicular to space 3, reference line 2 will belong to spaces 1, 3, 4 and plains α, δ, µ will being perpendicular to space 2.
Different arrangements of the four fluctuating reference lines would provide aesthetically correct variations of the geometric requirements. In regards to the fly through of space 3, when reference lines 1 and 2 are simultaneously perpendicular to reference line 4 an isometric cube or space would form. The naturalistically scaled sculptural image of the woman in space 3 would not be subject to diminution because the sculptural imagery in the space must be available for 360° inspection, just as it was in the natural space. At this point, it would become aesthetically viable to have a diminutively scaled image of the background of space 3 (which is the tree in space 4) on reference plain α. This would provide another aesthetic presentation of the perpendicular relationships of reference line 4 to space 4. The same aesthetized arrangements of the reference lines could also be applied to the reference lines forming the other spaces.
The rediscovery of linear perspective during the Renaissance inspired us to classify and categorize spaces and structures using geometric techniques, which eventually provided us with definitions of two and three-dimensional spaces. Relatively new geometries, from the latter part of the nineteenth and twentieth centuries, have provided us with definitions of higher dimensional spaces and structures, including four-dimensional spaces and structures and many others. Finally, naturalistic imagery can be infused into kinetic four-dimensional sculptures and spaces. These new works of art will allow us to plainly see and understand the long mutually beneficial evolutionary relationship between art and geometry. | 4,500 | 23,020 | {"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-2023-40 | latest | en | 0.887542 |
https://stat.ethz.ch/pipermail/r-help/2004-June/052640.html | 1,716,804,476,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00290.warc.gz | 458,951,341 | 4,163 | # [R] Regression query : steps for model building
Devshruti Pahuja devshruti at hotmail.com
Fri Jun 11 19:41:00 CEST 2004
```Hi
I have a set of data with both quantitative and categorical predictors.
After scaling of response variable, i looked for multicollinearity (VIF
values) among the predictors and removed the predictors who were hinding
some of the
other significant predictors. I'm curious to know whether the predictors
(who are not significant) while doing simple 'lm' will be involved in
interactions. How do i take into
account interactions of those predictors whom i removed just on the basis
of multicollinearity ?
I'll appreciate if someone can throw some light on this matter and how to
use R to detect the interactions effectively .
Thanks
Regards
Dev
> ------Final 'lm model'--------------------
> > logmodelfull_minus_run_hr_walk_batting <- lm(log(salary) ~ hit+rbi +
walk
> + obp + strike.out+free.agent.eligible+free.agent.1991+arbitr.elgible.)
> > summary(logmodelfull_minus_run_hr_walk_batting)
>
> Call:
> lm(formula = log(salary) ~ hit + rbi + walk + obp + strike.out +
> free.agent.eligible + free.agent.1991 + arbitr.elgible.)
>
> Residuals:
> Min 1Q Median 3Q Max
> -2.41786 -0.28911 -0.02814 0.31890 1.49007
>
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 5.340782 0.251218 21.260 < 2e-16 ***
> hit 0.004479 0.001158 3.867 0.000133 ***
> rbi 0.011102 0.002195 5.059 7.05e-07 ***
> walk 0.005421 0.002206 2.457 0.014533 *
> obp -1.385584 0.824105 -1.681 0.093653 .
> strike.out -0.005399 0.001438 -3.755 0.000205 ***
> free.agent.eligible1 1.611521 0.080657 19.980 < 2e-16 ***
> free.agent.19911 -0.301243 0.103481 -2.911 0.003848 **
> arbitr.elgible.1 1.293059 0.086696 14.915 < 2e-16 ***
> ---
> Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
>
> Residual standard error: 0.5351 on 328 degrees of freedom
> Multiple R-Squared: 0.7981, Adjusted R-squared: 0.7932
> F-statistic: 162.1 on 8 and 328 DF, p-value: < 2.2e-16
>
> --------------------------------------------------------------------------
--
> ----------------------------------------------------
>
>
> --------------with
>
interactions----------------------------------------------------------------
> ---------------------------
>
> >
> > summary(baseball.lgmodel_with_interactions_ALL_arbid)
>
> Call:
> lm(formula = log(salary) ~ hit + rbi + strike.out + free.agent.eligible +
> free.agent.1991 + arbitr.elgible. + hit * free.agent.1991 +
> hit * arbitr.elgible. + hit * rbi + rbi * free.agent.eligible +
> rbi * arbitr.elgible. + rbi * arbitr.1991 + hit * strike.out +
> strike.out * free.agent.eligible + strike.out * arbitr.elgible. +
> strike.out * run + strike.out * hr + hit * free.agent.eligible +
> free.agent.eligible * run + hit * free.agent.1991 + strike.out *
> free.agent.1991 + free.agent.1991 * batting + free.agent.1991 *
> obp + arbitr.elgible. * run + batting * double + obp * run +
> obp * hr + walk * stolen.base + hit * arbitr.1991 +
free.agent.eligible
> *
> double + arbitr.elgible. * double + strike.out * triple +
> triple * batting + triple * walk + triple * walk + hit *
> hr + rbi * hr + free.agent.eligible * hr + free.agent.1991 *
> hr + arbitr.elgible. * hr + hr * arbitr.1991 + hit * walk +
> free.agent.eligible * walk + walk * rbi + rbi * stolen.base +
> strike.out * stolen.base + stolen.base * batting + stolen.base *
> walk + stolen.base * rbi + stolen.base * walk + arbitr.elgible. *
> error)
>
> Residuals:
> Min 1Q Median 3Q Max
> -2.29352 -0.28287 -0.03748 0.29790 1.31590
>
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 5.217e+00 3.467e-01 15.048 < 2e-16 ***
> hit 6.927e-03 6.226e-03 1.112 0.266889
> rbi 1.908e-02 1.150e-02 1.658 0.098350 .
> strike.out -5.692e-03 4.586e-03 -1.241 0.215517
> free.agent.eligible1 1.287e+00 2.259e-01 5.699 3.05e-08 ***
> free.agent.19911 3.828e-01 6.575e-01 0.582 0.560914
> arbitr.elgible.1 1.038e+00 2.195e-01 4.726 3.63e-06 ***
> arbitr.19911 -1.024e+00 4.392e-01 -2.331 0.020443 *
> run 4.932e-02 2.905e-02 1.698 0.090682 .
> hr -1.093e-01 7.208e-02 -1.516 0.130543
> batting -1.814e-01 2.558e+00 -0.071 0.943522
> obp -1.375e+00 2.253e+00 -0.610 0.542099
> double -5.259e-02 4.489e-02 -1.172 0.242349
> walk 1.395e-02 9.757e-03 1.430 0.153808
> stolen.base -1.685e-02 4.299e-02 -0.392 0.695372
> triple -1.367e-01 1.600e-01 -0.854 0.393807
> error -4.097e-03 6.879e-03 -0.595 0.552007
> hit:free.agent.19911 8.248e-04 4.611e-03 0.179 0.858174
> hit:arbitr.elgible.1 4.873e-03 6.448e-03 0.756 0.450395
> hit:rbi -1.382e-04 7.709e-05 -1.792 0.074184 .
> rbi:free.agent.eligible1 5.352e-03 9.555e-03 0.560 0.575855
> rbi:arbitr.elgible.1 -3.384e-03 1.136e-02 -0.298 0.766072
> rbi:arbitr.19911 3.596e-02 2.179e-02 1.650 0.100046
> hit:strike.out 5.480e-06 5.446e-05 0.101 0.919917
> strike.out:free.agent.eligible1 -2.570e-03 4.282e-03 -0.600 0.548890
> strike.out:arbitr.elgible.1 -9.703e-04 5.234e-03 -0.185 0.853068
> strike.out:run 1.685e-04 1.246e-04 1.352 0.177345
> strike.out:hr -3.088e-04 2.277e-04 -1.356 0.176229
> hit:free.agent.eligible1 -1.359e-03 6.224e-03 -0.218 0.827363
> free.agent.eligible1:run 1.248e-02 9.109e-03 1.370 0.171917
> strike.out:free.agent.19911 -1.851e-02 5.974e-03 -3.099 0.002140 **
> free.agent.19911:batting 7.076e-01 6.200e+00 0.114 0.909215
> free.agent.19911:obp -1.421e+00 3.952e+00 -0.360 0.719394
> arbitr.elgible.1:run -8.541e-03 8.773e-03 -0.974 0.331100
> batting:double 2.346e-01 1.609e-01 1.458 0.145884
> run:obp -1.825e-01 7.492e-02 -2.436 0.015462 *
> hr:obp 3.687e-01 2.116e-01 1.742 0.082608 .
> walk:stolen.base -6.789e-05 1.557e-04 -0.436 0.663083
> hit:arbitr.19911 -5.835e-03 7.084e-03 -0.824 0.410808
> free.agent.eligible1:double -1.151e-02 1.663e-02 -0.692 0.489334
> arbitr.elgible.1:double 2.169e-03 1.938e-02 0.112 0.910985
> strike.out:triple -8.106e-04 6.023e-04 -1.346 0.179475
> batting:triple 5.179e-01 5.599e-01 0.925 0.355841
> walk:triple 8.755e-04 9.262e-04 0.945 0.345349
> hit:hr -3.320e-04 2.626e-04 -1.264 0.207180
> rbi:hr 4.748e-04 3.015e-04 1.575 0.116414
> free.agent.eligible1:hr 1.840e-02 2.313e-02 0.796 0.426972
> free.agent.19911:hr 7.216e-02 1.889e-02 3.819 0.000165 ***
> arbitr.elgible.1:hr 4.111e-02 2.803e-02 1.467 0.143564
> arbitr.19911:hr -2.368e-02 4.647e-02 -0.510 0.610723
> hit:walk 3.173e-05 7.826e-05 0.405 0.685442
> free.agent.eligible1:walk -5.423e-03 4.984e-03 -1.088 0.277472
> rbi:walk -7.569e-05 1.313e-04 -0.577 0.564598
> rbi:stolen.base 3.980e-05 1.605e-04 0.248 0.804409
> strike.out:stolen.base -2.611e-04 1.615e-04 -1.617 0.107004
> batting:stolen.base 1.552e-01 1.434e-01 1.082 0.280020
> arbitr.elgible.1:error 3.930e-03 1.390e-02 0.283 0.777495
> ---
> Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
>
> Residual standard error: 0.4925 on 280 degrees of freedom
> Multiple R-Squared: 0.854, Adjusted R-squared: 0.8248
> F-statistic: 29.24 on 56 and 280 DF, p-value: < 2.2e-16
>
``` | 3,243 | 8,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.622207 |
https://books.google.com.jm/books?qtid=e57e78f9&lr=&id=lA3A5XCCl_sC&sa=N&start=80 | 1,680,279,040,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00296.warc.gz | 182,073,133 | 6,493 | Books Books
DRY MEASURE 2 pints (pt.) = 1 quart (qt.) 8 quarts =1 peck (pk.) 4 pecks = 1 bushel (bu...
The Progressive Intellectual Arithmetic, on the Inductive Plan: Being a ... - Page 52
by Horatio Nelson Robinson - 1859
## New Elementary Arithmetic, Embracing Mental and Written Exercises for ...
Henry Bartlett Maglathlin - Arithmetic - 1876 - 230 pages
...is used in measuring such dry articles as grain, fruit, roots, coal, etc. Table. 2 pints (pt.) are 1 quart, qt. 8 quarts, 1 peck, pk. 4 pecks, 1 bushel, bu. A gallon, or 4 quarts, Dry Measure, contains 268^ cubic inches, and a bushel, 2150^,, cubic inches....
## The Rudiments of Written Arithmetic: Containing Slate and Black-board ...
Horatio Nelson Robinson - Arithmetic - 1877 - 212 pages
...Measure is used in measuring articles not liquid ; as grain, fruit, salt, roots, ashes, etc. TABLE. 2 Pints (pt.) make 1 Quart qt. 8 Quarts " 1 Peck pk. 4 Pecks " 1 Bushel bu. UNIT EQUIVALENTS. qt. pt. pk. 1=2 bu. 1 = 8 = 16 1 = 4 = 32 = 64 SCALE — ascending, 2, 8, 4 ; descending,...
## Robinson's Progressive Practical Arithmetic: Containing the Theory of ...
Horatio Nelson Robinson, Daniel W. Fish - Arithmetic - 1877 - 374 pages
...measuring articles not liquid, as grain, fruit, salt, roots, ashes, etc. What is dry measure 1 TABLE. 2 Pints (pt.) make 1 Quart qt. 8 Quarts " 1 Peck pk. 4 Pecks " 1 Bushel. bo. UHIT EQUIVALENTS. qt. pt. pk. 1=2 bu. 1 = 8 = 16 1 = 4 - 32 = 64 SCALE — ascending, 2, 8, 4 ;...
## The Rudiments of Written Arithmetic: Containing Slate and Black-board ...
Horatio Nelson Robinson - Arithmetic - 1877 - 212 pages
...Measure is used in measuring articles not liquid ; as grain, fruit, salt, roots, ashes, etc. TABLE. 2 Pints (pt.) make 1 Quart qt. 8 Quarts " 1 Peck pk. 4 Pecks " 1 Bushel bvu UNIT EQUIVALENTS. qt. pt. pk. 1=2 bu. 1 = 8 = 16 1 = 4 = 32 = 64 SCALE—ascending, 2, 8, 4 ; descending,...
## Ray's New Intellectual Arithmetic
Joseph Ray - Mental arithmetic - 1877 - 150 pages
...dollars? 400 cents? 500? 600? 700? 800? 900? LESSON XLVTII. DRY MEASURE. 2 pints (pt.) are 1 quart, marked qt. 8 quarts '' 1 peck, " pk. 4 pecks " 1 bushel, " bu. 1. How many pints in 2 quarts? In 3 quarts? 4 quarts? 5? 6? 7? 2. How many quarts in 2 pecks? In 3 pecks? 3. How many pecks...
## Ray's New Primary Arithmetic for Young Learners
Joseph Ray - Arithmetic - 1877 - 108 pages
...coffee, how many ounces do you get ? LESSON LXXXIII. DRY MEASURE. • 2 pints (pt.) make 1 quart : marked qt. 8 quarts " 1 peck : " pk. 4 pecks " 1 bushel : " bu. 1. How many pints in 2 quarts? In 3? In 4? 2. How many quarts in 12 pints? In 14? In 16? 3. How many pecks in 2 bushels?...
## City Arithmetics, Part 2
William Aloysius Boylan, Floyd R. Smith - Arithmetic - 1916 - 152 pages
...hundredweight (cwt.) 20 hundredweight = 1 ton (T.) 2,000 pounds = 1 ton TABLE OF DRY MEASURE 2 pints (pt.) = 1 quart (qt.) 8 quarts = 1 peck (pk.) 4 pecks = 1 bushel (bu.) 143 UNITED STATES MONET 10 mills = 1 cent 10 cents = 1 dime 10 dimes = 1 dollar 10 dollars = 1 eagle...
## Modern Business Arithmetic: Complete Course
Harry Anson Finney, Joseph Clifton Brown - Business mathematics - 1916 - 506 pages
...gal. are considered 1 barrel (bbl.). 63 gal. = 1 hogshead (hhd.). 82. Dry Measure. 2 pints (pt.) = 1 quart '(qt.) 8 quarts = 1 peck (pk.) 4 pecks = 1 bushel (bu.) 83. Measures of Time. 60 seconds (sec.) = 1 minute (min.) 60 minutes = 1 hour (hr.) 24 hours = 1 day...
## Vocational Mathematics for Girls
William Henry Dooley - Arithmetic - 1917 - 388 pages
...the hogshead 63 gal. Dry Measure is used in measuring roots, grain, vegetables, etc. Table 2 pints = 1 quart, qt 8 quarts = 1 peck, pk. 4 pecks = 1 bushel, bu. 1 bu. = 4 pk. = 32 qt. = 64 pints. The bushel contains 2150.42 cubic inches; 1 dry quart contains 67.2...
## Higher Arithmetic
John Charles Stone, James Franklin Millis - Arithmetic - 1917 - 354 pages
...of various sizes, 31£ gallons is considered a standard barrel. TABLE OF DRY MEASURE 2 pints (pt.) = 1 quart (qt.) 8 quarts = 1 peck (pk.) 4 pecks = 1 bushel (bu.) 1 bushel = 2150.42 cubic inches HIGHER ARITHMETIC TABLE OF AVOIRDUPOIS WEIGHT Used in weighing all coarse... | 1,449 | 4,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-14 | latest | en | 0.7072 |
http://forum.allaboutcircuits.com/threads/phase-margin-question.94384/ | 1,484,682,326,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280065.57/warc/CC-MAIN-20170116095120-00154-ip-10-171-10-70.ec2.internal.warc.gz | 105,251,391 | 13,631 | # Phase Margin Question
Discussion in 'Homework Help' started by jegues, Feb 8, 2014.
1. ### jegues Thread Starter Well-Known Member
Sep 13, 2010
735
43
For convenience I've attached both the question and my attempt at the solution. Please see figures attached.
From the quadratic equation for the desired phase margin, I selected the w = 2.682 solution because I thought that we couldn't have negative values for w. (i.e. not physical)
However the correct solution comes about when w = +1.491, but the solution to my equation indicates w = -1.491.
How do we know when a negative frequency indicates an invalid solution, or a valid one where we can simply neglect the negative sign?
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31.2 KB
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2. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
783
I believe it's a simple sign error in your equations:
You have
$-130^o=-90^o-atan(\frac{\omega}{4-\omega^2})$
which then becomes
$-40^o=-atan(\frac{\omega}{4-\omega^2})$
then
$40^o=atan(\frac{\omega}{4-\omega^2})$
then
$tan(40^o)=(\frac{\omega}{4-\omega^2})$
and so on .... | 318 | 1,079 | {"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": 4, "/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-2017-04 | latest | en | 0.817492 |
https://brainmass.com/business/depreciation/a-high-speed-multiple-bit-drill-press-217232 | 1,603,531,359,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00226.warc.gz | 244,660,404 | 11,463 | Explore BrainMass
# A high-speed multiple-bit drill press
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
I need help with these questions.
Depreciation Methods.
A high-speed multiple-bit drill press costing \$300,000 has an estimated salvage value of \$25,000 and a life of ten years. What is the annual depreciation for each of the first two full years under the following depreciation methods?
1. Double-declining-balance method:
a. Year one, \$______________.
b. Year two, \$______________.
2. Units of production (activity) method (lifetime output is estimated at 110,000 units; the press produced 12,000 units in year one and 18,000 in year two):
a. Year one, \$______________.
b. Year two, \$______________.
3. Sum-of-the-years'-digits method:
a. Year one, \$______________.
b. Year two, \$______________.
4. Straight-line depreciation method:
a. Year one, \$______________.
b. Year two, \$______________.
Current Liabilities.
Ritt Company includes 1 coupon in each box of soap powder that it packs, 20 coupons being redeemable for a premium consisting of a kitchen utensil. In 2004, Ritt Company purchased 9,000 premiums at \$1.00 each and sold 270,000 boxes of soap powder @ \$4.00 per box. Based on past experience, it is estimated that 60% of the coupons will be redeemed. During 2004, 72,000 coupons were presented for redemption.
During 2005, 14,000 premiums were purchased at \$1.10. The company sold 600,000 boxes of soap at \$4.00 and 250,000 coupons were presented for redemption.
Instructions
Prepare all the entries that would be made relative to sales of soap powder and to the premium plan in both 2004 and 2005. Assume a FIFO inventory flow.
© BrainMass Inc. brainmass.com October 1, 2020, 10:17 pm ad1c9bdddf
#### Solution Preview
Depreciation Methods.
A high-speed multiple-bit drill press costing \$300,000 has an estimated salvage value of \$25,000 and a life of ten years. What is the annual depreciation for each of the first two full years under the following depreciation methods?
1. Double-declining-balance method:
(2 x Straight-line rate) x (Cost - Depreciation Taken in Prior Periods) = Depreciation for the Period
Straight-line rate = 1/10 years = 0.10 or 10%
Year one (2 x 10%) x (\$300,000 - 0) = \$60,000
Year two (2 x 10%) x (\$300,000 - \$60,000) = \$48,000
a. Year one, \$60,000.
b. Year two, \$48,000.
2. Units of production (activity) method (lifetime output is estimated at 110,000 units; the press produced 12,000 units in year one and 18,000 in year two):
Straight line depreciation per unit = (Total Cost - Salvage value)/Total units produced
= (\$300,000 - ...
#### Solution Summary
This solution is comprised of a detailed explanation to answer what is the annual depreciation for each of the first two full years under the following depreciation methods.
\$2.19 | 743 | 2,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-45 | latest | en | 0.908705 |
https://diveintographs.readthedocs.io/en/latest/_modules/dig/auggraph/dataset/aug_dataset.html | 1,701,337,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00854.warc.gz | 254,382,024 | 6,587 | # Source code for dig.auggraph.dataset.aug_dataset
```# Author: Youzhi Luo (yzluo@tamu.edu)
# Updated by: Anmol Anand (aanand@tamu.edu)
import random
import torch
import torch.nn.functional as F
from torch.utils.data import Dataset
from torch_geometric.utils import degree
[docs]class DegreeTrans(object):
r"""
This class is used to add vertex degree based node features to graphs.
This is usually used to preprocess the graph datasets that do not have
node features.
"""
def __init__(self, dataset, in_degree=False):
self.max_degree = None
self.mean = None
self.std = None
self.in_degree = in_degree
self._statistic(dataset)
def _statistic(self, dataset):
r"""
This function computes statistics over all nodes in all sample graphs.
These statistics are maximum, mean, and standard deviation.
Args:
dataset (:class:`torch.utils.data.Dataset`): The dataset containing
all sample graphs.
"""
degs = []
max_degree = 0
for data in dataset:
print(type(data))
degs += [degree(data.edge_index[0], dtype=torch.long)]
max_degree = max(max_degree, degs[-1].max().item())
self.max_degree = max_degree
deg = torch.cat(degs, dim=0).to(torch.float)
self.mean, self.std = deg.mean().item(), deg.std().item()
[docs] def __call__(self, data):
r"""
This is the main function that adds vertex degree based node
features to the given graph.
Args:
data (:class:`torch_geometric.data.data.Data`): The graph
with vertex degrees as node features.
"""
if data.x is not None:
return data
if self.max_degree < 1000:
idx = data.edge_index[1 if self.in_degree else 0]
deg = torch.clamp(degree(idx, data.num_nodes, dtype=torch.long),
min=0, max=self.max_degree)
deg = F.one_hot(deg, num_classes=self.max_degree + 1).to(torch.float)
data.x = deg
else:
deg = degree(data.edge_index[0], dtype=torch.float)
deg = (deg - self.mean) / self.std
data.x = deg.view(-1, 1)
return data
[docs]class AUG_trans(object):
r"""
This class generates an augmentation from a given sample.
Args:
augmenter (function): This method generates an augmentation from the
given sample.
device (str): The device on which the data will be processed.
pre_trans (function, optional): This transformation is applied on the
original sample before an augmentation is generated. Default is
None.
post_trans (function, optional): This transformation is applied on the
generated augmented sample. Default is None.
"""
def __init__(self, augmenter, device, pre_trans=None, post_trans=None):
self.augmenter = augmenter
self.pre_trans = pre_trans
self.post_trans = post_trans
self.device = device
[docs] def __call__(self, data):
r"""
This is the main function that generates an augmentation from a given
sample.
Args:
data: The given data sample.
Returns:
A transformed graph.
"""
if self.pre_trans:
data = self.pre_trans(data)
new_data = self.augmenter(data)[0]
if self.post_trans:
new_data = self.post_trans(new_data)
return new_data
[docs]class Subset(Dataset):
r"""
This class is used to create of a subset of a dataset.
Args:
subset (:class:`torch.utils.data.Dataset`): The given dataset subset.
transform (function, optional): A transformation applied on each
sample of the dataset before it will be used. Default is None.
"""
def __init__(self, subset, transform=None):
self.subset = subset
self.transform = transform
[docs] def __getitem__(self, index):
r"""
This method returns the sample at the given index in the subset.
Args:
index (int): The index in the subset of the required sample.
"""
data = self.subset[index]
if self.transform is not None:
data = self.transform(data)
return data
[docs] def __len__(self):
r"""
Returns the number of samples in the subset.
"""
return len(self.subset)
[docs]class TripleSet(Dataset):
r"""
This class inherits from the :class:`torch.utils.data.Dataset` class and in
addition to each anchor sample, it returns a random positive and negative
sample from the dataset. A positive sample has the same label as the
anchor sample and a negative sample has a different label than the anchor
sample.
Args:
dataset (:class:`torch.utils.data.Dataset`): The dataset for which the
triple set will be created.
transform (function, optional): A transformation that is applied on all
original samples. In other words, this transformation is applied
to the anchor, positive, and negative sample. Default is None.
"""
def __init__(self, dataset, transform=None):
self.dataset = dataset
self.transform = transform
self._preprocess()
def _preprocess(self):
self.label_to_index_list = {}
for i, data in enumerate(self.dataset):
y = int(data.y.item())
if not y in self.label_to_index_list:
self.label_to_index_list[y] = [i]
else:
self.label_to_index_list[y].append(i)
[docs] def __getitem__(self, index):
r"""
For a given index, this sample returns the original/anchor sample from
the dataset at that index and a corresponding positive, and negative
sample.
Args:
index (int): The index of the anchor sample in the dataset.
Returns:
A tuple consisting of the anchor sample, a positive
sample, and a negative sample respectively.
"""
anchor_data = self.dataset[index]
anchor_label = int(anchor_data.y.item())
pos_index = random.sample(self.label_to_index_list[anchor_label], 1)[0]
while pos_index == index:
pos_index = random.sample(self.label_to_index_list[anchor_label], 1)[0]
neg_label = random.sample(self.label_to_index_list.keys(), 1)[0]
while neg_label == anchor_label:
neg_label = random.sample(self.label_to_index_list.keys(), 1)[0]
neg_index = random.sample(self.label_to_index_list[neg_label], 1)[0]
pos_data, neg_data = self.dataset[pos_index], self.dataset[neg_index]
if self.transform is not None:
anchor_data, pos_data, neg_data = self.transform(anchor_data), \
self.transform(pos_data), self.transform(neg_data)
return anchor_data, pos_data, neg_data
[docs] def __len__(self):
r"""
Returns:
The number of samples in the original dataset.
"""
return len(self.dataset)
``` | 1,424 | 5,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-50 | latest | en | 0.558707 |
http://www.cs.princeton.edu/courses/archive/fall12/cos226/checklist/kdtree.html | 1,548,189,923,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583874494.65/warc/CC-MAIN-20190122202547-20190122224547-00405.warc.gz | 275,712,847 | 3,857 | ### Programming Assignment 5 Checklist: Kd-Trees
Is a point on the boundary of a rectangle considered inside it? Do two rectangle intersect if they have just one point in common? Yes, and yes.
Can I use the distanceTo() method in Point2D and RectHV? No, you may use only the subset of the methods listed. You should be able to accomplish the same result by using distanceSquaredTo() instead of distanceTo().
What should I do if a point has the same x-coordinate as the point in a node when inserting / searching in a 2d-tree? Go to the right subtree as specified.
Can I assume that all x- or y-coordinates of points inserted into the KdTree will be between 0 and 1? Yes. You may also assume that the insert(), contains(), and nearest() methods in KdTree are passed points with x- and y-coordinates between 0 and 1. Finally, you may assume that the range() method in KdTree is passed a rectangle that lies in the unit box.
What should I do if a point is inserted twice in the data structure? The data structure represents a set of points, so you should keep only one copy.
How should I scale the coordinate system when drawing? Don't, please keep the default range of 0 to 1.
How should I set the size and color of the points and rectangles when drawing? Use StdDraw.setPenColor(StdDraw.BLACK) and StdDraw.setPenRadius(.01) before drawing the points; use StdDraw.setPenColor(StdDraw.RED) or StdDraw.setPenColor(StdDraw.BLUE) and StdDraw.setPenRadius() before drawing the splitting lines.
What should range() return if there are no points in the range? It should return an Iterable<Point2D> object with zero points.
How much memory does a Point2D object use? For simplicity, assume that each Point2D object uses 32 bytes—in reality, it uses a bit more because of the Comparator instance variables.
How much memory does a RectHV object use? You should look at the code and analyze its memory usage.
I run out of memory when running some of the large sample files. What should I do? Be sure to ask Java for additional memory, e.g., java -Xmx1600m RangeSearchVisualizer input1M.txt.
Testing
Testing. A good way to test KdTree is to perform the same sequence of operations on both the PointSET and KdTree data types and identify any discrepancies. The sample clients RangeSearchVisualizer.java and NearestNeighborVisualizer.java take this approach.
Sample input files. The directory kdtree contains some sample input files in the specified format.
• circleN.txt contains N points on the circumference of the circle centered on (0.5, 0.5) of radius 0.5.
The result of calling draw() on the points in circle10.txt should look like the following:
If nearest() is called with p = (.81, .30) the number of nodes visited in order to find that F is nearest is 5.
Starting with circle10k.txt if nearest is called with p = (.81, .30) the number of nodes visited in order to find that K is nearest is 6.
Possible Progress Steps
These are purely suggestions for how you might make progress. You do not have to follow these steps.
1. Node data type. There are several reasonable ways to represent a node in a 2d-tree. One approach is to include the point, a link to the left/bottom subtree, a link to the right/top subtree, and an axis-aligned rectangle corresponding to the node.
```private static class Node {
private Point2D p; // the point
private RectHV rect; // the axis-aligned rectangle corresponding to this node
private Node lb; // the left/bottom subtree
private Node rt; // the right/top subtree
}
```
Unlike the Node class for BST, this Node class is static because it does not refer to a generic Key or Value type that depends on the object associated with the outer class. This saves the 8-byte inner class object overhead. Also, since we don't need to implement the rank and select operations, there is no need to store the subtree size.
2. Writing KdTree. Start by writing isEmpty() and size(). These should be very easy. From there, write a simplified version of insert() which does everything except set up the RectHV for each node. Write the contains() method, and use this to test that insert() was implemented properly. Note that insert() and contains() are best implemented by using private helper methods analogous to those found on page 399 of the book or by looking at BST.java. We recommend using orientation as an argument to these helper methods.
Now add the code to insert() which sets up the RectHV for each Node. Next, write draw(), and use this to test these rectangles. Finish up KdTree with the nearest and range methods. Finally, test your implementation using our interactive client programs as well as any other tests you'd like to conduct.
Optimizations
These are many ways to improve performance of your 2d-tree. Here are some ideas.
• Squared distances. Whenever you need to compare two Euclidean distances, it is often more efficient to compare the squares of the two distances to avoid the expensive operation of taking square roots. Everyone must implement this optimization because is not because it is both easy to do and likely a bottleneck.
• Range search. Instead of checking whether the query rectangle intersects the rectangle corresponding to a node, it suffices to check only whether the query rectangle intersects the splitting line segment: if it does, then recursively search both subtrees; otherwise, recursively search the one subtree where points intersecting the query rectangle could be.
• Save memory. You are not required to explictily store a RectHV in each 2d-tree node (though it is probably wise in your first version). | 1,243 | 5,622 | {"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-2019-04 | latest | en | 0.883902 |
https://www.doubtnut.com/qna/644105527 | 1,721,443,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00631.warc.gz | 646,517,283 | 33,091 | # An electrically isolated hollow (initially uncharged), conducting sphere has a small positively charged ball suspended by an insulating rod from its inside surface, see diagram This causes the inner surface of the sphere to become negatively charged. When the ball is centred in the sphere the electric field outside the conducting spehere is
A
zero
B
the same as if the sphere was not there
C
twice what it would be if the sphere was not there
D
equal in magnitude but opposite in direction to what it would be if the sphere was not there.
Video Solution
Text Solution
Verified by Experts
## b. Let the charge on the ball at center is q, then−q charge will be induced on the inner surface and +q on the order surface. Electric field at any point outside the sphere due to inner charges q and −q will combinely be zero. Outer charge q will combinely be field at outside points as it would be produced by a charge at the center. Hence presence of sphere does not affect the electric field outside the sphere.
|
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 472 | 2,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.936041 |
https://qanda.ai/en/search/%5Cleft%28%20%20-2%20%5Cright%29%20%20%20%5Ctimes%20%20%20%5Cdfrac%7B%201%20%20%7D%7B%203%20%20%7D%20%20%20%5Cdiv%20%20%20%5Cleft%28%20%20-4%20%5Cright%29?search_mode=expression | 1,642,448,683,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00291.warc.gz | 510,737,928 | 20,839 | # Calculator search results
Formula
Calculate the value
$\left( -2 \right) \times \dfrac{ 1 }{ 3 } \div \left( -4 \right)$
$\dfrac { 1 } { 6 }$
Calculate the value
$\color{#FF6800}{ - } 2 \times \color{#FF6800}{ \dfrac { 1 } { 3 } } \div \left ( \color{#FF6800}{ - } 4 \right )$
Since negative numbers are multiplied by an even number, remove the (-) sign
$2 \times \dfrac { 1 } { 3 } \div 4$
$2 \times \dfrac { 1 } { 3 } \color{#FF6800}{ \div } \color{#FF6800}{ 4 }$
Convert division to multiplication
$2 \times \dfrac { 1 } { 3 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
Calculate the product of rational numbers
$\color{#FF6800}{ \dfrac { 1 } { 6 } }$
Solution search results | 346 | 860 | {"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} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.268223 |
https://easetolearn.com/smart-learning/web/mathematics/coordinate-geometry-and-vectors/two-dimensional-coordinate-geomettry/cartesian-plane-and-lines/centroid-incentre-circumcentre-orthocenter/5195 | 1,708,816,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00733.warc.gz | 222,132,160 | 21,001 | # Centroid, Incentre, Circumcentre & Orthocenter
### Centroid:
#### Definition:
• Coordinates: The centroid of a triangle with vertices $\left({x}_{1},{y}_{1}\right)$, $\left({x}_{2},{y}_{2}\right)$, and $\left({x}_{3},{y}_{3}\right)$ is given by: $\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$
• Concept: The centroid is the point of intersection of the medians of a triangle.
• Property: Divides each median into segments with a ratio of 2:1.
### Incenter:
#### Definition:
• Coordinates: The incenter of a triangle with vertices $\left({x}_{1},{y}_{1}\right)$, $\left({x}_{2},{y}_{2}\right)$ and $\left({x}_{3},{y}_{3}\right)$ is given by: $\left(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c}\right)$ where $a,b,$ and $c$ are the side lengths opposite the respective vertices.
• Concept: The incenter is the point of intersection of the angle bisectors of a triangle.
• Property: Equidistant from the sides of the triangle.
### Circumcenter:
#### Definition:
• Coordinates: The circumcenter of a triangle with vertices $\left({x}_{1},{y}_{1}\right)$, $\left({x}_{2},{y}_{2}\right)$, and $\left({x}_{3},{y}_{3}\right)$ is given by: $\left(\frac{{D}_{1}}{2K},\frac{{D}_{2}}{2K}\right)$ where ${D}_{1}$ and ${D}_{2}$ are certain determinants and $K$ is the area of the triangle.
• Concept: The circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.
• Property: Equidistant from the vertices of the triangle.
### Orthocenter:
#### Definition:
• Coordinates: The orthocenter of a triangle with vertices $\left({x}_{1},{y}_{1}\right)$, $\left({x}_{2},{y}_{2}\right)$, and $\left({x}_{3},{y}_{3}\right)$ is given by: $\left(\frac{{\sum }_{i=1}^{3}{x}_{i}\left({h}_{i}^{2}-{a}_{i}^{2}\right)}{{\sum }_{i=1}^{3}\left({h}_{i}^{2}-{a}_{i}^{2}\right)},\frac{{\sum }_{i=1}^{3}{y}_{i}\left({h}_{i}^{2}-{a}_{i}^{2}\right)}{{\sum }_{i=1}^{3}\left({h}_{i}^{2}-{a}_{i}^{2}\right)}\right)$ where ${h}_{i}$ is the length of the altitude and ${a}_{i}$ is the length of the side opposite the $i$th vertex.
• Concept: The orthocenter is the point of intersection of the altitudes of a triangle.
• Property: May lie inside, outside, or on the triangle.
### Relationships:
• Equilateral Triangle: For an equilateral triangle, all these points coincide at a single point.
• Different Types of Triangles: The relative positions of these points may vary for different types of triangles (acute, obtuse, or right-angled). | 816 | 2,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 24, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-10 | latest | en | 0.605812 |
https://rational-equations.com/rational-equations/reducing-fractions/slope-poems-math.html | 1,725,742,781,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00389.warc.gz | 465,777,833 | 12,061 | Algebra Tutorials!
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Number of equations to solve: 23456789
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Equ. #2:
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Equ. #7:
Equ. #8:
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Number of inequalities to solve: 23456789
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• area activities KS2 | 1,421 | 6,319 | {"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} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.838521 |
http://www.lexic.us/definition-of/diameter | 1,524,139,773,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936914.5/warc/CC-MAIN-20180419110948-20180419130948-00607.warc.gz | 367,267,153 | 9,406 | ### Definition of Diameter
1. Noun. The length of a straight line passing through the center of a circle and connecting two points on the circumference.
Exact synonyms: Diam
Specialized synonyms: Bore, Caliber, Calibre, Gauge, Windage
Generic synonyms: Length
Derivative terms: Diametral, Diametric, Diametrical
2. Noun. A straight line connecting the center of a circle with two points on its perimeter (or the center of a sphere with two points on its surface).
Generic synonyms: Straight Line
Derivative terms: Diametral, Diametric, Diametrical
### Definition of Diameter
1. n. Any right line passing through the center of a figure or body, as a circle, conic section, sphere, cube, etc., and terminated by the opposite boundaries; a straight line which bisects a system of parallel chords drawn in a curve.
### Definition of Diameter
1. Noun. (geometry) Any straight line between two points on the circumference of a circle that passes through the centre/center of the circle. ¹
2. Noun. (geometry) The length of such a line. ¹
3. Noun. (geometry) The maximum distance between any two points in a metric space ¹
4. Noun. (graph theory) The maximum eccentricity over all vertices in a graph. ¹
¹ Source: wiktionary.com
### Definition of Diameter
1. a straight line passing through the center of a circle and ending at the periphery [n -S]
### Medical Definition of Diameter
1. The length of a straight line passing through the centre of a circle and connecting opposite points on its circumference, hence the distance between two specified opposite points on the periphery of a structure such as the cranium or pelvis. This entry appears with permission from the Dictionary of Cell and Molecular Biology (11 Mar 2008)
### Diameter Pictures
Click the following link to bring up a new window with an automated collection of images related to the term: Diameter Images
### Lexicographical Neighbors of Diameter
diamagnetic effectsdiamagneticallydiamagneticsdiamagnetismdiamagnetismsdiamagnetizationdiamagnetsdiamantairediamantairesdiamantane diamantanesdiamantediamantesdiamantiferousdiamantinediameter (current term)diameter at breast heightdiameter obliquadiameter transversadiameters diametraldiametrallydiametralsdiametrediametricdiametricaldiametrical oppositiondiametricallydiamfenetide
### Literary usage of Diameter
Below you will find example usage of this term as found in modern and/or classical literature:
1. A History of Greek Mathematics by Thomas Little Heath (1921)
"the tangent at any point of a central conic, in relation to the original diameter of reference; if Q is the point of contact, QV the ordinate to the ..."
2. Encyclopaedia Britannica: A Standard Work of Reference in Art, Literature (1907)
"For the line which joins tho ends of a diameter to any point on the carve include ... 23) be one pair of conjugate diameter« at right angles to each other, ..."
3. Proceedings of the American Philosophical Society Held at Philadelphia for by American Philosophical Society (1921)
"... 218 Lower molar length 124 diameter, pj, anteroposterior 39 diameter, p4, ... transverse anteriorly 42.5 47 diameter, mj, anteroposterior 36 diameter, ..."
4. Science by American Association for the Advancement of Science (1897)
"A cylindrical block of 82 mm. diameter and of 55 g. weight was compared with a series of blocks of 35 mm. diameter but of various weights. ..."
5. Transactions of the American Society of Mechanical Engineers by American Society of Mechanical Engineers (1891)
"8” diameter exhaust outlet 10” 83-INCH CYLINDER. ... 7” Thickness of steam space in jacket 1” diameter of piston rods 2+1” diameter steam inlet 10” diameter ..."
6. Transactions by American Ethnological Society (1855)
"The rolls are on the second motion, the diameter of the driving wheel being 2J feet diameter, and that on the rope wheel shaft 4J feet diameter. ..."
7. A History of Greek Mathematics by Thomas Little Heath (1921)
"the tangent at any point of a central conic, in relation to the original diameter of reference; if Q is the point of contact, QV the ordinate to the ..."
8. Encyclopaedia Britannica: A Standard Work of Reference in Art, Literature (1907)
"For the line which joins tho ends of a diameter to any point on the carve include ... 23) be one pair of conjugate diameter« at right angles to each other, ..."
9. Proceedings of the American Philosophical Society Held at Philadelphia for by American Philosophical Society (1921)
"... 218 Lower molar length 124 diameter, pj, anteroposterior 39 diameter, p4, ... transverse anteriorly 42.5 47 diameter, mj, anteroposterior 36 diameter, ..."
10. Science by American Association for the Advancement of Science (1897)
"A cylindrical block of 82 mm. diameter and of 55 g. weight was compared with a series of blocks of 35 mm. diameter but of various weights. ..."
11. Transactions of the American Society of Mechanical Engineers by American Society of Mechanical Engineers (1891)
"8” diameter exhaust outlet 10” 83-INCH CYLINDER. ... 7” Thickness of steam space in jacket 1” diameter of piston rods 2+1” diameter steam inlet 10” diameter ..."
12. Transactions by American Ethnological Society (1855)
"The rolls are on the second motion, the diameter of the driving wheel being 2J feet diameter, and that on the rope wheel shaft 4J feet diameter. ..." | 1,280 | 5,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-17 | latest | en | 0.821492 |
https://www.physicsforums.com/threads/understanding-proposition-8-7-operator-norm-and-sequences.1042058/ | 1,721,045,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00290.warc.gz | 834,487,226 | 18,218 | # Understanding Proposition 8.7: Operator Norm and Sequences
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• Math Amateur
In summary, Peter found that the final step of the argument is something like this. Suppose we have a number $x\ge 0,$ and we let $\delta>0$ be arbitrary. What can we say about $x$ if $x<\delta$ for all $\delta?$ We can actually claim that $x=0.$ For, if $x>0,$ then suppose $\delta=x/2.$ Then $\delta>0,$ but $x=2\delta>\delta,$ contrary to the assumption.
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MHB
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...
I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.1 Linear Algebra ...
I need some further help in fully understanding the proof of Proposition 8.7 ...Proposition 8.7 and its proof reads as follows:
View attachment 9397
View attachment 9398In the above proof by Browder we read the following:"... ... it follows from Proposition 8.6 that $$\displaystyle S_m \to S$$ for some $$\displaystyle S \in \mathscr{L} ( \mathbb{R}^n)$$. In particular, taking $$\displaystyle m = 0$$ above, we find $$\displaystyle \| I - S_p \| \leq \frac{t}{ 1 - t }$$for every $$\displaystyle p$$, and hence $$\displaystyle \| I - S \| \leq t/(1 - t )$$ ... ...
... ... ... "
My question is as follows:Can someone please explain exactly why/how that $$\displaystyle \| I - S_p \| \leq \frac{t}{ 1 - t }$$for every $$\displaystyle p$$ ... implies that $$\displaystyle \| I - S \| \leq t/(1 - t )$$ ... ... ?In other words if some relation is true for every term of a sequence ... why then is it true for the limit of a sequence ... ...
Help will be much appreciated ...
Peter
#### Attachments
• Browder - 1 - Proposition 8.7 ... PART 1 ....png
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• Browder - 2 - Proposition 8.7 ... PART 2 ... ....png
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Let $\varepsilon>0.$ Since $S_p$ is Cauchy in a complete space, it converges to $S$ in the space. That is, there exists $N>0$ such that for every $n>N,$ we have $\|S_n-S\|<\varepsilon.$ Now then, we have that
\begin{align*}
\|I-S\|&=\|I-S_p+S_p-S\|\\
&\le \|I-S_p\|+\|S_p-S\|\\
&\le \frac{t}{1-t}+\varepsilon.
\end{align*}
Now we are nearly there.
The final step of the argument is something like this. Suppose we have a number $x\ge 0,$ and we let $\delta>0$ be arbitrary. What can we say about $x$ if $x<\delta$ for all $\delta?$ We can actually claim that $x=0.$ For, if $x>0,$ then suppose $\delta=x/2.$ Then $\delta>0,$ but $x=2\delta>\delta,$ contrary to the assumption.
Similarly, for the above argument, because $\varepsilon>0$ is arbitrary, we can conclude that $\|I-S\|\le t/(1-t).$
Ackbach said:
Let $\varepsilon>0.$ Since $S_p$ is Cauchy in a complete space, it converges to $S$ in the space. That is, there exists $N>0$ such that for every $n>N,$ we have $\|S_n-S\|<\varepsilon.$ Now then, we have that
\begin{align*}
\|I-S\|&=\|I-S_p+S_p-S\|\\
&\le \|I-S_p\|+\|S_p-S\|\\
&\le \frac{t}{1-t}+\varepsilon.
\end{align*}
Now we are nearly there.
The final step of the argument is something like this. Suppose we have a number $x\ge 0,$ and we let $\delta>0$ be arbitrary. What can we say about $x$ if $x<\delta$ for all $\delta?$ We can actually claim that $x=0.$ For, if $x>0,$ then suppose $\delta=x/2.$ Then $\delta>0,$ but $x=2\delta>\delta,$ contrary to the assumption.
Similarly, for the above argument, because $\varepsilon>0$ is arbitrary, we can conclude that $\|I-S\|\le t/(1-t).$
Thanks Ackbach ...
Peter
This is a particular case of the general rule that weak inequalities are preserved by limits (but strict inequalities may not be). If $(x_n)$ is a sequence with $$\displaystyle \lim_{n\to\infty}x_n = x$$, and $x_n\leqslant a$ for all $n$, then $x\leqslant a$. (But if $x_n<a$ for all $n$ then it need not be true that $x<a$. All you can assert is that $x\leqslant a$.)
Opalg said:
This is a particular case of the general rule that weak inequalities are preserved by limits (but strict inequalities may not be). If $(x_n)$ is a sequence with $$\displaystyle \lim_{n\to\infty}x_n = x$$, and $x_n\leqslant a$ for all $n$, then $x\leqslant a$. (But if $x_n<a$ for all $n$ then it need not be true that $x<a$. All you can assert is that $x\leqslant a$.)
Thanks for a most helpful post, Opalg ... Peter
## 1. What is Proposition 8.7 in mathematics?
Proposition 8.7 is a theorem in functional analysis that relates to the operator norm and sequences in a normed vector space. It states that if a sequence of bounded linear operators converges to a bounded linear operator in the operator norm, then the limit operator is also bounded and its norm is less than or equal to the limit of the norms of the sequence.
## 2. What is the operator norm?
The operator norm is a way to measure the size or magnitude of a bounded linear operator in a normed vector space. It is defined as the supremum of the norms of the operator applied to all unit vectors in the space. In other words, it represents the maximum amount by which the operator can scale a vector without increasing its norm.
## 3. How is Proposition 8.7 useful in mathematics?
Proposition 8.7 is useful in mathematics because it provides a way to prove the continuity of certain operations on bounded linear operators, such as addition and multiplication. It also helps in understanding the behavior of sequences of operators and their limits in a normed vector space.
## 4. What is the significance of Proposition 8.7 in functional analysis?
In functional analysis, Proposition 8.7 is significant because it connects the concept of operator norm to the convergence of sequences of operators. This allows for a more thorough understanding of the behavior of operators and their limits, which is important in many areas of mathematics and physics.
## 5. Are there any real-world applications of Proposition 8.7?
Yes, there are many real-world applications of Proposition 8.7. For example, it is used in the study of differential equations, where operators are often used to represent differential operators. It is also important in the analysis of quantum mechanics, where operators are used to represent physical observables. Additionally, Proposition 8.7 has applications in computer science, particularly in the field of signal processing and image recognition.
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# (Solved): 3. A consumer with income I facing prices p x and p y has the following indirect utili ...
3. A consumer with income I facing prices p x and p y has the following indirect utility function for goods X and Y : $V\left(p_{x}, p_{y}, I\right)=\frac{1}{2} p_{y}^{-1 / 2}\left(p_{x}^{1 / 2}+p_{x}^{-1 / 2} I\right)$ a. Derive the consumer's Hicksian demand function for goodX . b. Check that the properties of Hicksian demand functions are satisfied. c. Suppose I=49,P y =1 , and P x =1 . Compute the consumer's compensating variation (CV) and their equivalent variation (EV) for an increase in P y from 1 to 4 . d. Use words and a graph to interpret your CV and EV measures. please solve all parts and provide me the 100% accurate answer
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" Graph Linear Functions Using A Table of Values Relay Activity "DIGITAL AND PRINT: Six rounds include practice graphing linear functions by completing a table of values for each equation, plotting the ordered pairs / solutions and drawing a line
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.A.1, 8.F.A.2, HSF-IF.B.4, HSF-IF.C.7, HSF-IF.C.7a
\$3.00
23
" Find Slope and Y-Intercept Sum It Up Activity "DIGITAL AND PRINT: Seven stations include practice or review finding the slope and the y-intercept of linear functions from two given points, from graphs and from tables of values. Do your students
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSA-CED.A.4, HSF-IF.B.6, HSF-BF.A.1, HSF-LE.A.2, HSS-ID.C.7
\$3.00
35
" Write Linear Equations From Tables, Points & Graphs Review Problem Pass Activity "DIGITAL AND PRINT: Twelve rounds include practice or review writing linear equations in slope-intercept form from tables of values, two points / ordered pairs
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSA-CED.A.4, HSF-IF.B.6, HSF-BF.A.1, HSF-LE.A.2, HSS-ID.C.7
\$3.00
21
" Write Linear Equations From Context Problem Pass Activity "DIGITAL AND PRINT: Six rounds include practice or review writing linear equations in slope-intercept form and standard form from word problems.Do your students struggle? Are you looking
Subjects:
Algebra, Word Problems
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSF-IF.A.2, HSF-BF.A.1, HSF-BF.A.1a, HSF-LE.A.2
\$3.00
28
" Function or Not & Linear or Non-Linear Relay Activity "DIGITAL AND PRINT: The first three rounds include practice or review determining if a graph or a table of values represents a function or not. The other three rounds include practice or
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.A.1, 8.F.A.2, 8.F.B.5, HSF-IF.A.1, HSF-LE.A.1, HSF-LE.A.1a, HSF-LE.A.1b, HSF-LE.A.1c
\$3.00
16
" Point-Slope, Slope-Intercept & Standard Form Sum It Up Activity "DIGITAL AND PRINT: Eight stations include practice or review identifying the point, slope, y-intercept and x-intercept of linear equations written in point-slope form,
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-SSE.A.1b, HSA-CED.A.4, HSF-IF.B.4
\$3.00
18
" Parallel & Perpendicular Lines Sum It Up Activity "DIGITAL AND PRINT: Six stations include practice writing linear equations of parallel lines and perpendicular lines AND using slope to determine if lines are parallel or perpendicular to each
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSF-IF.B.6, HSF-BF.A.1, HSF-LE.A.2, HSS-ID.C.7
\$3.00
14
" Solve One-Step, Two-Step & Multi-Step Linear Equations Placemat Activities "DIGITAL AND PRINT: Three placemat activities include practice solving one-step, two-step and multi-step linear equations. With a partner, students solve all four
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.EE.C.7, 8.EE.C.7b, HSA-REI.B.3
\$3.00
8
" Graph Linear Inequalities in Two Variables Relay Activity "DIGITAL AND PRINT: Six rounds include practice graphing linear inequalities in two variables and writing linear inequalities from graphs.Do your students struggle? Are you looking for
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
HSA-REI.D.12, HSF-LE.A.2
\$3.00
24
" Solve Quadratic Word Problems Relay Activity "DIGITAL AND PRINT: Four rounds include practice solving quadratic word problems in context with area problems and vertical motion problems.Do your students struggle? Are you looking for more extensive
Subjects:
Algebra, Word Problems
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
HSA-REI.B.4, HSA-REI.B.4b, HSF-IF.A.2, HSF-IF.C.8a
\$3.00
17
" Add, Subtract & Multiply Polynomials AND With Geometry Shapes Placemat Activities "DIGITAL AND PRINT: Two placemat activities include practice adding, subtracting and multiplying polynomial expressions AND practice finding the area,
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
HSA-SSE.A.1a, HSA-APR.A.1
\$3.00
15
"Write Linear Equations From Tables Relay Activity"DIGITAL AND PRINT: Six rounds include practice or review writing linear equations in slope-intercept form from a table of values. Includes vertical lines and function notation.Separate students
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSA-CED.A.4, HSF-IF.B.6, HSF-BF.A.1, HSF-LE.A.2, HSS-ID.C.7
\$3.00
11
" Write Linear Equations In Context & Mixed Review Placemat Activities "DIGITAL AND PRINT: Two placemat activities include practice writing linear equations from context / word problems and to review writing and rewriting linear equations.With
Subjects:
Algebra, Word Problems
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSA-CED.A.4, HSF-IF.A.2, HSF-IF.B.4, HSF-BF.A.1, HSF-BF.A.1a, HSF-LE.A.2
\$3.00
10
"Write Linear Equations From Context Relay Activity"DIGITAL AND PRINT: Six rounds provide practice or review writing linear equations from context / word problems.Do your students struggle? Are you looking for more extensive practice on a single
Subjects:
Algebra, Word Problems
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSF-IF.A.2, HSF-BF.A.1, HSF-BF.A.1a, HSF-LE.A.2
\$3.00
10
"Graph Linear Equations in Slope-Intercept Form Problem Pass Activity"DIGITAL AND PRINT: Twelve rounds provide practice or review graphing linear equations in slope-intercept form. Includes identifying the slope, the y-intercept and whether the
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.A.1, HSF-IF.B.4, HSF-IF.C.7, HSF-IF.C.7a
\$3.00
8
" Find Slope & Write Linear Equations Given Two Points Placemat Activity "DIGITAL AND PRINT: Two placemat activities include practice finding slope given two ordered pairs and practice writing linear equations given two ordered pairs.With a
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-CED.A.2, HSF-IF.B.6, HSF-BF.A.1, HSF-LE.A.2
\$3.00
8
"Solve Systems of Linear Equations by Graphing Problem Pass Activity"DIGITAL AND PRINT: Twelve rounds provide practice or review solving systems of linear equations using the graphing method. Practice includes graphing equations in slope-intercept
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.EE.C.8, 8.EE.C.8a, 8.EE.C.8b, HSA-REI.C.6
\$3.00
5
"Solve Linear Systems Word Problems Relay Activity"DIGITAL AND PRINT: Six rounds provide practice or review solving systems of linear equations word problems in context. This activity includes problems with mixtures, comparing two deals, finding
Subjects:
Algebra, Word Problems
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.EE.C.8, 8.EE.C.8a, 8.EE.C.8b, 8.EE.C.8c, HSA-CED.A.3, HSA-REI.C.5, HSA-REI.C.6
\$3.00
11
" Graph Quadratics in Standard Form Relay Activity "DIGITAL AND PRINT: Six rounds include practice graphing quadratic functions in standard form using a table of values.Separate students into groups of four and give each student their own copy of
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
HSF-IF.B.4, HSF-IF.C.7a
\$3.00
10
" Rewrite Linear Equations Into Slope-Intercept Form and Standard Form Placemat Activities "DIGITAL AND PRINT: Two placemat activities include practice rewriting linear equations into slope-intercept form & into standard form. With a partner,
Subjects:
Algebra
8th, 9th, Homeschool
Types:
Activities, Printables
CCSS:
8.F.B.4, HSA-SSE.A.1a, HSA-CED.A.4
\$3.00
9
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TEACHING EXPERIENCE
My name is Debbie Veatch and I am a teacher. During fifteen years of teaching Algebra, I started with a traditional class format and successfully changed to the “flipped” class format. Initially, I was not sure how to start and wondered if all of the work involved would be worth the trouble. After enough research, I decided to give flipping a try, and I’m glad I did. Flipping my Algebra 1 class turned out to be even better than I had hoped and made me wonder why I had not flipped sooner! My students became fully engaged in class and were more successful than ever before on chapter tests and the state’s End of Course Exam. An added bonus was that I enjoyed teaching even more with the flipped model. Algebra became fun for all of us! My goal is to create relevant and helpful content that you can use in your Algebra 1 class whether you teach in the traditional way or flip your class.
MY TEACHING STYLE
In my flipped classroom, I included lots of fun activities to keep my students engaged and allowed them to practice their new Algebra skills. I found a good balance between large-group activities, small-group activities and individual activities to help my students master important Algebra skills in a variety of ways. I believe accuracy is important to success in Algebra, so I ensure that each activity allows students to check themselves or each other for accuracy and/or allows the teacher to quickly check students' work for accuracy.
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
William Woods in Fulton, Missouri.
To help other teachers, I created a free video series for TpT about how I flipped my Algebra class. Please join me on my blog about Flipping Algebra here | 2,871 | 10,032 | {"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-2021-31 | latest | en | 0.817008 |
http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_Hypothesis_Basics&diff=12464&oldid=10432 | 1,537,354,861,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156192.24/warc/CC-MAIN-20180919102700-20180919122700-00417.warc.gz | 283,966,854 | 13,153 | # AP Statistics Curriculum 2007 Hypothesis Basics
(Difference between revisions)
Revision as of 18:00, 29 June 2010 (view source)Jenny (Talk | contribs) (→Example 2: Sodium content in hot-dogs)← Older edit Current revision as of 17:48, 18 November 2015 (view source)IvoDinov (Talk | contribs) (→Type I Error, Type II Error and Power) (7 intermediate revisions not shown) Line 60: Line 60: * Remarks: * Remarks: ** A '''Specificity''' of 100% means that the test recognizes all healthy individuals as (normal) healthy. The maximum is trivially achieved by a test that claims everybody is healthy regardless of the true condition. Therefore, the specificity alone does not tell us how well the test recognizes positive cases. ** A '''Specificity''' of 100% means that the test recognizes all healthy individuals as (normal) healthy. The maximum is trivially achieved by a test that claims everybody is healthy regardless of the true condition. Therefore, the specificity alone does not tell us how well the test recognizes positive cases. - ** '''False positive rate (α)'''= FP/(FP+TN) = 0.00995/(0.00995 + 0.98505)=0.01 = 1 - Specificity. + ** '''False positive rate (α)'''= $$\frac{FP}{FP+TN} = \frac{0.00995}{0.00995 + 0.98505}=0.01$$= 1 - Specificity. - ** '''Sensitivity''' is a measure of how well a test correctly identifies a condition, whether this is medical screening tests picking up on a disease, or quality control in factories deciding if a new product is good enough to be sold. + ** '''Sensitivity''' is a measure of how well a test correctly identifies a condition, whether this is medical screening tests picking up on a disease, or quality control in factories deciding if a new product is good enough to be sold. '''Sensitivity''' = $$\frac{TP}{TP+FN} = \frac{0.00475}{0.00475+ 0.00025}= 0.95.$$ - ** '''False Negative Rate (β)'''= FN/(FN+TP) = 0.00025/(0.00025+0.00475)=0.05 = 1 - Sensitivity. + ** '''False Negative Rate (β)'''= $$\frac{FN}{FN+TP} = \frac{0.00025}{0.00025+0.00475}=0.05$$= 1 - Sensitivity. - ** '''Power''' = 1 − β= 0.99975, see [[Power_Analysis_for_Normal_Distribution]]. + ** '''Power''' = 1 − β= 0.95, see [[Power_Analysis_for_Normal_Distribution]]. + ** Both (''Type I ($$\alpha$$)'' and ''Type II ($$\beta$$)'') errors are proportions in the range [0,1], so they represent ''error-rates''. The reason they are listed in the corresponding cells in the table is that they are directly proportionate to the numerical values of the FP and FN, respectively. + ** The two alternative definitions of ''power'' are equivalent: + ::: power $$=1-\beta$$, and + ::: power=sensitivity + :: This is because power= $$1-\beta=1-\frac{FN}{FN+TP}=\frac{FN+TP}{FN+TP} - \frac{FN}{FN+TP}=\frac{TP}{FN+TP}=$$ sensitivity. ===Example 2: Sodium content in hot-dogs=== ===Example 2: Sodium content in hot-dogs=== Use the [[SOCR_012708_ID_Data_HotDogs |Hot-dog dataset]] to see if there are statistically significant differences in the sodium content of the poultry vs. meat hotdogs. Use the [[SOCR_012708_ID_Data_HotDogs |Hot-dog dataset]] to see if there are statistically significant differences in the sodium content of the poultry vs. meat hotdogs. - * Formulate Hypotheses: $H_o: \mu_p = \mu_m$ vs. $H_1: \mu_p \not= \mu_m$, where $\mu_p, \mu_m$ represent the mean sodium content in poultry and mean hotdogs. + * Formulate Hypotheses: $$H_o: \mu_p = \mu_m$$ vs. $$H_1: \mu_p \not= \mu_m$$, where $\mu_p, \mu_m$ represent the mean sodium content in poultry and mean hotdogs. * Plug in the data in [http://socr.stat.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses] under the [[SOCR_EduMaterials_AnalysisActivities_TwoIndepTU |Two Independent Sample T-Test (Unpooled)]] will generate results as shown in the figure below (Two-Sided P-Value (Unpooled) = 0.196, which does not provide strong evidence to reject the null hypothesis that the two types of hot-dogs have the same mean sodium content). * Plug in the data in [http://socr.stat.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses] under the [[SOCR_EduMaterials_AnalysisActivities_TwoIndepTU |Two Independent Sample T-Test (Unpooled)]] will generate results as shown in the figure below (Two-Sided P-Value (Unpooled) = 0.196, which does not provide strong evidence to reject the null hypothesis that the two types of hot-dogs have the same mean sodium content).
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig1.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig1.jpg|600px]]
Line 76: Line 81: Study used 443 patients who had clinical pharyngitis diagnosed as group A $\beta$-hemolytic streptococcus infection in the past 28 days and compared them with 232 control patients who had symptoms of pharyngitis but no recent diagnosis of streptococcal pharyngitis. The aim was narrowly focused to compare the rapid strep test with the culture method used in clinical practice. Study used 443 patients who had clinical pharyngitis diagnosed as group A $\beta$-hemolytic streptococcus infection in the past 28 days and compared them with 232 control patients who had symptoms of pharyngitis but no recent diagnosis of streptococcal pharyngitis. The aim was narrowly focused to compare the rapid strep test with the culture method used in clinical practice. - The study found that the rapid strep test in this setting showed no difference in specificity (0.96 vs. 0.98). Hence, the assertion that rapid antigen testing had higher false-positive rates in those with recent infection was not confirmed. It also found that in patients who had recent streptococcal pharyngitis, the rapid strep test appears to be more reliable (0.91 vs 0.70, P < .001) than in those patients who had not had recent streptococcal pharyngitis. These findings indicated that the rapid strep test is both sensitive and specific in the setting of recent group A $\beta$-hemolytic streptococcal pharyngitis, and its use might allow earlier treatment in this subgroup of patients. + The study found that the rapid strep test in this setting showed no difference in specificity (0.96 vs. 0.98). Hence, the assertion that rapid antigen testing had higher false-positive rates in those with recent infection was not confirmed. It also found that in patients who had recent streptococcal pharyngitis, the rapid strep test appears to be more reliable (sensitivity 0.91 vs 0.70, P < .001) than in those patients who had not had recent streptococcal pharyngitis. These findings indicated that the rapid strep test is both sensitive and specific in the setting of recent group A $\beta$-hemolytic streptococcal pharyngitis, and its use might allow earlier treatment in this subgroup of patients. - Table 1. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in P''atients With Recently Treated Cases of Streptococcal Pharyngitis''. + Table 1. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in ''Patients With Recently Treated Cases of Streptococcal Pharyngitis'' (N=443).
Line 105: Line 110:
- Table 2. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in ''Patients With No Recently Treated Cases of Streptococcal Pharyngitis''. + Table 2. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in ''Patients With No Recently Treated Cases of Streptococcal Pharyngitis'' (N=232).
## General Advance-Placement (AP) Statistics Curriculum - Fundamentals of Hypothesis Testing
### Fundamentals of Hypothesis Testing
A (statistical) Hypothesis Test is a method of making statistical decisions about populations or processes based on experimental data. Hypothesis testing just answers the question of how well the findings fit the possibility that the chance alone might be responsible for the observed discrepancy between the theoretical model and the empirical observations. This is accomplished by asking and answering a hypothetical question. What is the likelihood of the observed summary statistics of interest, if the data did come from the distribution specified by the null-hypothesis? One use of hypothesis-testing is deciding whether experimental results contain enough information to cast doubt on conventional wisdom.
• Example: Consider determining whether a suitcase contains some radioactive material. Placed under a Geiger counter, the suitcase produces 10 clicks (counts) per minute. The null hypothesis is that there is no radioactive material in the suitcase and that all measured counts are due to ambient radioactivity typical of the surrounding air and harmless objects in a suitcase. We can then calculate how likely it is that the null hypothesis produces 10 counts per minute. If it is likely, for example if the null hypothesis predicts on average 9 counts per minute, we say that the suitcase is compatible with the null hypothesis (which does not imply that there is no radioactive material, we just can't determine from the 1-minute sample we took using this specific method!); On the other hand, if the null hypothesis predicts for example 1 count per minute, then the suitcase is not compatible with the null hypothesis and there must be other factors responsible to produce the increased radioactive counts.
The Hypothesis Testing is also known as Statistical Significance Testing. The null hypothesis is a conjecture that exists solely to be disproved, rejected or falsified by the sample-statistics used to estimate the unknown population parameters. Statistical significance is a possible finding of the test, that the sample is unlikely to have occurred in this process by chance given the truth of the null hypothesis. The name of the test describes its formulation and its possible outcome. One characteristic of hypothesis testing is its crisp decision about the null-hypothesis: reject or do not reject (which is not the same as accept).
### Null and Alternative (Research) Hypotheses
A Null Hypothesis is a thesis set up to be nullified or refuted in order to support an Alternate (research) Hypothesis. The null hypothesis is presumed true until statistical evidence, in the form of a hypothesis test, indicates otherwise. In science, the null hypothesis is used to test differences between treatment and control groups, and the assumption at the outset of the experiment is that no difference exists between the two groups for the variable of interest (e.g., population means). The null hypothesis proposes something initially presumed true, and it is rejected only when it becomes evidently false. That is, when a researcher has a certain degree of confidence, usually 95% to 99%, that the data do not support the null hypothesis.
### Example 1: Gender effects
If we want to compare the test scores of two random samples of men and women, a null hypothesis would be that the mean score of the male population was the same as the mean score of the female population:
H0 : μmen = μwomen
where:
H0 = the null hypothesis
μmen = the mean of the males (population 1), and
μwomen = the mean of the females (population 2).
Alternatively, the null hypothesis can postulate that the two samples are drawn from the same population, so that the center, variance and shape of the distributions are equal.
Formulation of the null hypothesis is a vital step in testing statistical significance. Having formulated such a hypothesis, one can establish the probability of observing the obtained data from the prediction of the null hypothesis, if the null hypothesis is true. That probability is what commonly called the significance level of the results.
In many scientific experimental designs we predict that a particular factor will produce an effect on our dependent variable — this is our alternative hypothesis. We then consider how often we would expect to observe our experimental results, or results even more extreme, if we were to take many samples from a population where there was no effect (i.e. we test against our null hypothesis). If we find that this happens rarely (up to, say, 5% of the time), we can conclude that our results support our experimental prediction — we reject our null hypothesis.
### Type I Error, Type II Error and Power
Directly related to hypothesis testing are the following 3 concepts:
• Type I Error: The false positive (Type I) Error of rejecting the null hypothesis given that it is actually true; e.g., A court finding a person guilty of a crime that they did not actually commit.
• Type II Error: The Type II Error (false negative) is the error of failing to reject the null hypothesis given that the alternative hypothesis is actually true; e.g., A court finding a person not guilty of a crime that they did actually commit.
• Statistical Power: The Power of a Statistical Test is the probability that the test will reject a false null hypothesis (that it will not make a Type II Error). As power increases, the chances of a Type II error decrease. The probability of a Type II error is referred to as the false negative rate (β). Therefore power is equal to 1 − β. You can also see this SOCR Power Activity.
Actual condition
Absent (Ho is true) Present (H1 is true)
Test
Result
Negative
(fail to reject Ho
Condition absent + Negative result = True (accurate) Negative (TN, 0.98505) Condition present + Negative result = False (invalid) Negative (FN, 0.00025)
Type II error (β)
Positive
(reject Ho)
Condition absent + Positive result = False Positive (FP, 0.00995)
Type I error (α)
Condition Present + Positive result = True Positive (TP, 0.00475)
Test
Interpretation
Power = 1-FN=
1-0.00025 = 0.99975
Specificity: TN/(TN+FP) =
0.98505/(0.98505+ 0.00995) = 0.99
Sensitivity: TP/(TP+FN) =
0.00475/(0.00475+ 0.00025)= 0.95
• Remarks:
• A Specificity of 100% means that the test recognizes all healthy individuals as (normal) healthy. The maximum is trivially achieved by a test that claims everybody is healthy regardless of the true condition. Therefore, the specificity alone does not tell us how well the test recognizes positive cases.
• False positive rate (α)= $$\frac{FP}{FP+TN} = \frac{0.00995}{0.00995 + 0.98505}=0.01$$= 1 - Specificity.
• Sensitivity is a measure of how well a test correctly identifies a condition, whether this is medical screening tests picking up on a disease, or quality control in factories deciding if a new product is good enough to be sold. Sensitivity = $$\frac{TP}{TP+FN} = \frac{0.00475}{0.00475+ 0.00025}= 0.95.$$
• False Negative Rate (β)= $$\frac{FN}{FN+TP} = \frac{0.00025}{0.00025+0.00475}=0.05$$= 1 - Sensitivity.
• Power = 1 − β= 0.95, see Power_Analysis_for_Normal_Distribution.
• Both (Type I ($$\alpha$$) and Type II ($$\beta$$)) errors are proportions in the range [0,1], so they represent error-rates. The reason they are listed in the corresponding cells in the table is that they are directly proportionate to the numerical values of the FP and FN, respectively.
• The two alternative definitions of power are equivalent:
power $$=1-\beta$$, and
power=sensitivity
This is because power= $$1-\beta=1-\frac{FN}{FN+TP}=\frac{FN+TP}{FN+TP} - \frac{FN}{FN+TP}=\frac{TP}{FN+TP}=$$ sensitivity.
### Example 2: Sodium content in hot-dogs
Use the Hot-dog dataset to see if there are statistically significant differences in the sodium content of the poultry vs. meat hotdogs.
• Formulate Hypotheses: $$H_o: \mu_p = \mu_m$$ vs. $$H_1: \mu_p \not= \mu_m$$, where μpm represent the mean sodium content in poultry and mean hotdogs.
• Plug in the data in SOCR Analyses under the Two Independent Sample T-Test (Unpooled) will generate results as shown in the figure below (Two-Sided P-Value (Unpooled) = 0.196, which does not provide strong evidence to reject the null hypothesis that the two types of hot-dogs have the same mean sodium content).
### Example 3: Rapid testing in strep-throat
This study investigated the accuracy of rapid diagnosis of group A β-streptococcal pharyngitis by commercial immunochemical antigen test kits in the setting of recent streptococcal pharyngitis. Specifically, it explored whether the false-positive rate of the rapid test was increased because of presumed antigen persistence.
Study used 443 patients who had clinical pharyngitis diagnosed as group A β-hemolytic streptococcus infection in the past 28 days and compared them with 232 control patients who had symptoms of pharyngitis but no recent diagnosis of streptococcal pharyngitis. The aim was narrowly focused to compare the rapid strep test with the culture method used in clinical practice.
The study found that the rapid strep test in this setting showed no difference in specificity (0.96 vs. 0.98). Hence, the assertion that rapid antigen testing had higher false-positive rates in those with recent infection was not confirmed. It also found that in patients who had recent streptococcal pharyngitis, the rapid strep test appears to be more reliable (sensitivity 0.91 vs 0.70, P < .001) than in those patients who had not had recent streptococcal pharyngitis. These findings indicated that the rapid strep test is both sensitive and specific in the setting of recent group A β-hemolytic streptococcal pharyngitis, and its use might allow earlier treatment in this subgroup of patients.
Table 1. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in Patients With Recently Treated Cases of Streptococcal Pharyngitis (N=443).
Results Culture Negative Culture Positive
Rapid strep test negative 93 10
Rapid strep test positive 4 104
Estimate 95% CI
Sensitivity 104/(104+10) = 0.91 0.84, 0.96
Specificity 93/(93+4) = 0.96 0.90, 0.99
Positive predictive value 0.96 0.91, 0.99
Negative predictive value 0.90 0.83, 0.95
False-positive rate 0.04 0.01, 0.10
False-negative rate 0.09 0.04, 0.15
Table 2. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in Patients With No Recently Treated Cases of Streptococcal Pharyngitis (N=232).
Results Culture Negative Culture Positive
Rapid strep test negative 165 19
Rapid strep test positive 4 44
Estimate 95% CI
Sensitivity 44/(44+19) = 0.70 0.57, 0.81
Specificity 168/(165+4) = 0.98 0.94, 0.99
Positive predictive value 0.92 0.80, 0.99
Negative predictive value 0.90 0.84, 0.94
False-positive rate 0.02 0.01, 0.06
False-negative rate 0.30 0.19, 0.43
### References
Robert D. Sheeler, MD, Margaret S. Houston, MD, Sharon Radke, RN, Jane C. Dale, MD, and Steven C. Adamson, MD. (2002) Accuracy of Rapid Strep Testing in Patients Who Have Had Recent Streptococcal Pharyngitis. JABFP, 2002, 15(4), 261-265. | 4,654 | 18,439 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-39 | latest | en | 0.894297 |
https://www.coursehero.com/file/6025137/Hw-6/ | 1,529,818,042,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00269.warc.gz | 798,209,385 | 59,171 | # Hw 6 - yang(ey942 HW06 TSOI(58160 This print-out should...
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yang (ey942) – HW06 – TSOI – (58160) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Three 5 kg masses are located at points in the xy plane as shown. 52 cm 58 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? The universal gravita- tional constant is 6 . 6726 × 10 11 N · m 2 / kg 2 . Answer in units of N. 002 (part 2 of 2) 10.0 points At what angle from the positive x -axis will the resultant force point? Let counterclockwise be positive, within the limits 180 to 180 . Answer in units of . 003 (part 1 of 3) 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 A uniform solid sphere of mass m 2 = 262 kg and radius R 2 = 1 . 21 m is inside and concen- tric with a spherical shell of mass m 1 = 213 kg and radius R 1 = 2 . 35 m (see the figure). R R a b c m m 1 1 2 2 Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 3 kg located at a dis- tance 0 . 37 m from the center of the sphere and spherical shell. Answer in units of N. 004 (part 2 of 3) 10.0 points Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 3 kg located at a distance 1 . 68 m from the center of the sphere and spherical shell. Answer in units of N. 005 (part 3 of 3) 10.0 points Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 3 kg located at a distance 2 . 94 m from the center of the sphere and spherical shell. Answer in units of N. 006 (part 1 of 2) 10.0 points When it orbited the Moon, the Apollo 11 spacecraft’s mass was 15900 kg, and its mean distance from the Moon’s center was 1 . 7136 × 10 6 m. Assume its orbit was circular and the Moon to be a uniform sphere of mass 7 . 36 × 10 22 kg. Given the gravitational constant G is 6 . 67259 × 10 11 N m 2 / kg 2 , calculate the or- bital speed of the spacecraft.
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Jill Tulane University ‘16, Course Hero Intern | 894 | 3,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-26 | latest | en | 0.878653 |
https://abbotswell.aberdeen.sch.uk/2019/01/clic-targets-wb-21-01-2019/ | 1,566,317,620,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00281.warc.gz | 336,447,283 | 10,975 | ## CLIC Targets wb 21.01.2019
C– Counting: Squiggleworth (Step 4) : I can partition a 2dp number
L– Learn Its: Step 14: x11 tables; with a particular focus on the rule for x10 a number and then he learning of both facts for 11×11 and 12×11
Challenge 1: Ask an adult to jumble up the questions and see if you can answer them
I– It’s Nothing New: Adding with Pim (Step 4): I can add tenths. Example: 0.4+0.3=0.7, using he facts that are already known; 3+4, 30+40. 300+400 etc.
C– Calculations: Addition (Step 29): I can solve any 3d + 3d. with a particular focus on turning this 1 question into 3 questions…with the plan to get to the ‘Abridged’ and then eventually ‘Brain’ methods.
Source: Primary 4/5 | 238 | 706 | {"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} | 3.125 | 3 | CC-MAIN-2019-35 | longest | en | 0.802388 |
https://deffinition.net/orthogonal/ | 1,709,624,328,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00701.warc.gz | 195,467,924 | 17,530 | # Definition of Orthogonal
## What is Orthogonal?
Orthogonal is a big word that means something is at a right angle or completely independent of something else. It may sound complicated, but it’s really not!
## Origin of Orthogonal
The word “orthogonal” comes from Greek, where “ortho” means straight and “gon” means angle. So, when you say “orthogonal,” you’re talking about things that are straight and at right angles.
## Where is Orthogonal found in everyday life?
Imagine a tic-tac-toe game board with three horizontal lines and three vertical lines. That’s orthogonal! You can also find orthogonal shapes in buildings, like door frames and windows.
## Synonyms and Comparison
Other words for “orthogonal” are “perpendicular” and “90-degree angle.” It’s similar to crossing your fingers or clapping, where your hands are at right angles to each other.
## What makes Orthogonal special?
Orthogonal is a unique concept because it refers to things that are independent or unrelated. It’s like combining a puzzle piece with another puzzle piece that doesn’t fit or belong. Orthogonal reminds us that things can be different and not connected, just like the pieces of a jigsaw puzzle.
## In Conclusion
So, in simple words, orthogonal means something is straight, at right angles, and not connected to something else. It’s a word we use to describe when two things are completely separate and independent, just like the pieces of a puzzle that don’t fit together. | 311 | 1,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | longest | en | 0.927356 |
http://mathhelpforum.com/algebra/207942-new-logarithms.html | 1,480,748,628,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540909.75/warc/CC-MAIN-20161202170900-00203-ip-10-31-129-80.ec2.internal.warc.gz | 171,167,718 | 10,412 | 1. ## New to Logarithms
Hey,
I am pretty new to Logarithms and my class just got into the topic. I just converted y = log(base 4)X into "x = 4^y" to get the exponential equation.
However when the problem starts to get more complicated I don't know how to convert.
Right now I am trying to figure out how to convert "y = log(base 3)(x-2)"
Any help would be appreciated! Thanks!
2. ## Re: New to Logarithms
Hey,
I am pretty new to Logarithms and my class just got into the topic. I just converted y = log(base 4)X into "x = 4^y" to get the exponential equation.
However when the problem starts to get more complicated I don't know how to convert.
Right now I am trying to figure out how to convert "y = log(base 3)(x-2)"
Any help would be appreciated! Thanks!
$a = \log_b{c} \implies b^a = c$
so ...
$y = \log_3(x-2) \implies 3^y = x-2$
3. ## Re: New to Logarithms
Thanks!! How would I then type that out with bases? I don't know how to do that. And I have been trying to plug it into a graphing calculator to check but have problems there as well. Thanks!
4. ## Re: New to Logarithms
You won't be able to input bases other than 10 or e into your graphics calculator. If you have any other base you will need to change the base to 10 or e. | 354 | 1,254 | {"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": 2, "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} | 3.15625 | 3 | CC-MAIN-2016-50 | longest | en | 0.957719 |
http://www.real-statistics.com/descriptive-statistics/histograms/ | 1,519,234,199,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813691.14/warc/CC-MAIN-20180221163021-20180221183021-00071.warc.gz | 585,083,765 | 12,378 | # Histograms
A histogram is a graphical representation of the output of the FREQUENCY function (as described in Frequency Tables).
Example 1: Create a histogram for the data and bin selection for Example 1 from Frequency Tables.
We start by replicating the data and bin section for Example 1 in Figure 1.
Figure 1 – Data for Example 1
You can use Excel’s chart tool to graph the data in Figure 1, or alternatively you can use the Histogram data analysis tool to accomplish this directly, as described next.
Excel Data Analysis Tool: To use Excel’s Histogram data analysis tool, you must first establish a bin array (as for the FREQUENCY function described in Frequency Tables) and then select the Histogram data analysis tool. In the dialog box that is displayed you next specify the input data (Input Range) and bin array (Bin Range). You can optionally include the labels for these ranges (in which case you check the Labels check box).
For Example 1, the Input Range is A4:B14 and the Bin Range is D4:D7 (with the Labels check box unchecked). The output is displayed in Figure 2.
Figure 2 – Histogram data analysis tool
Observation: Caution must be exercised when creating histograms to present the data in a clear and accurate way. For most purposes it is important that the intervals be equal in size (except for an unbounded first and/or last interval). Otherwise a distorted picture of the data may be presented.
To avoid this problem equally-spaced intervals can be used. This is the approach illustrated in Example 4 of Frequency Tables using the FREQTABLE supplemental function. Alternatively, the Frequency Table supplemental data analysis tool can be used.
Real Statistics Data Analysis Tool: The Frequency Table data analysis tool provided in the Real Statistics Resource Pack can be used to create a frequency table and histogram as illustrated in the following example.
Example 2: Create a frequency table and histogram for the 22 data elements in the range A4:B14 of Figure 1 based on bins of size 15.
Enter Ctrl-m and select the Frequency Table option. The dialog box shown in Figure 3 will appear.
Figure 3 – Dialog box for Frequency Table data analysis tool
Insert A4:B14 in the Input Range field, select Raw data as the Input format and insert 15 as the Bin Size. The output is shown in Figure 4.
Figure 4 – Frequency Table and Histogram
Note that if 100 is inserted in the Maximum bin value (or blank) field of Figure 3 then the output is shown in Figure 5.
Figure 5 – Frequency Table and Histogram (revised)
Observation: You can also produce a frequency table (and histogram) of the type described in Example 3 of Discrete Probability Distributions (i.e. without specifying any bins) via the Frequency Table data analysis tool. In this case you would select Raw data as the Input format for the dialog box shown in Figure 3 and leave the Bin size field blank.
### 5 Responses to Histograms
1. Kelly says:
Dear Charles,
How can I come up with Kernel Density Estimation from this Histogram?
Best regards,
• Charles says:
Kelly,
Kernal Density Estimators are not derived from histogram, but from the raw data. They are an improvement over histograms. E.g. see the following webpage:
http://www.mvstat.net/tduong/research/seminars/seminar-2001-05/
I have not yet covered this topic in the website, but will probably do so shortly.
Charles
• Kelly says:
Dear Charles,
Thank you very much for your response and the informative webpage.
Best regards,
Kelly
2. skyler chan says:
Thanks this has explained all the information I needed…
3. skyler chan says:
This has been a great explanation Charles.
Thanks | 792 | 3,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-09 | latest | en | 0.808609 |
https://www.itmanagementetools.com/excel/formula/formula-show-other-result-for-0-and-empty-cell-excel-report.php | 1,721,504,983,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00860.warc.gz | 730,539,028 | 5,597 | ## Show other result for 0 and empty cell with a formula in an excel report
Many extractions I did contain some empty cells or cells with 0. One of the options is to delete them but in some situation, I need to keep them. What I do most, it is to put another value, for example, all cells with zero, I will put “none” and for empty ones, “not related”.
When I use the formula ?
When I need to display another result if the cell is 0 or empty.
How to use the formula ?
The formula in this topic is with "," so depending of the operating system of your PC, the formula should have ";" instead of ",".
How are the formulas ?
=IF()
=VLOOKUP()
This formula is telling to display an empty cell if the value is 0.
=IF(VLOOKUP(A2,A:B,2,0)=0,"",VLOOKUP(A2,A:B,2,0))
In fact, you can ask it to put anything you want, if you want to display “nothing” for all 0 values, the formula will be:
=IF(VLOOKUP(A2,A:B,2,0)=0,"nothing",VLOOKUP(A2,A:B,2,0))
Of course, if you prefer to tell it to display the empty cell or “nothing” for another value, just change the “=0” to for instance “=5”.
This formula works the same way if the cell value is empty:
=IF(VLOOKUP(A2,A:B,2,0)="","empty",VLOOKUP(A2,A:B,2,0))
Instead to put the cell reference A2, you can put the name:
=IF(VLOOKUP("chocolate",A:B,2,0)=0,"",VLOOKUP("chocolate",A:B,2,0))
The point is that if you have 2 or more “chocolate”, the formula will only take the first one that it will find. Putting the name is better only if the name is unique and not duplicate. | 432 | 1,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.887537 |
https://www.physicsforums.com/threads/uncovering-the-mystery-of-constant-velocity.308224/ | 1,695,576,691,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00063.warc.gz | 1,019,549,030 | 22,197 | # Uncovering the Mystery of Constant Velocity
• monty37
In summary, while the speed of light appears to be the same in all directions, it is actually slowed down by the surrounding mediumf
#### monty37
if, a particular body rotates ,with a certain velocity
then every particle of that moves with the same velocity as that of body.if
the Earth is moving at a velocity,we also move with same velocity daily .
for eg:the speed of light was discovered to be 3x10^8m/s wrt to earth,but is it the same everwhere,including outer space??
My understanding is that the speed of light is the same everywhere and has been the same since the beginning. There is some speculation that the speed of light may have been different in the past but no observational evidence exists.
well but when speed is calculated ,sitting on Earth it varies,right,relative velocity!
but in space it is different
I accept the speed will differ in space. because the speed of the light is not same when it travels in different mediums so this may differ in space as well.
if, a particular body rotates ,with a certain velocity
then every particle of that moves with the same velocity as that of body.if
the Earth is moving at a velocity,we also move with same velocity daily .
for eg:the speed of light was discovered to be 3x10^8m/s wrt to earth,but is it the same everwhere,including outer space??
The speed of light in a vacuum is invariant. It is measured to be the same by everyone regardless of their relative motion. For example, a beam of light traveling past the Earth will have a measured relative velocity of 3e8 m/s wrt to the Earth as measured by the Earth, but an observer flying past with a speed of 1.5e8 m/s relative to the Earth will measure that same beam of light as having a relative velocity of 3e8 m/s wrt to himself
well but when speed is calculated ,sitting on Earth it varies,right,relative velocity!
but in space it is different
The historical measurements of the speed of light indicate that it dropped by about 100 kilometers per second from 1880 to 1980. See http://www.sigma-engineering.co.uk/light/lightindex.shtml
This drop is nearly within experimental error, and is probably due to experimenters fudging their numbers until partial agreement with other measurements was obtained.
Last edited by a moderator:
if, a particular body rotates ,with a certain velocity
then every particle of that moves with the same velocity as that of body.if
the Earth is moving at a velocity,we also move with same velocity daily .
for eg:the speed of light was discovered to be 3x10^8m/s wrt to earth,but is it the same everwhere,including outer space??
Light always travels at a constant speed of a bought 2.99*10$$^{8}$$ m/s.
The only reason that it would appear to travel slower when traveling through matter is because, photons excite the adjoining particles that in turn transfer the energy to the neighbor.
The time it takes for the light to be absorbed and then transfer out again is what would make light seem slower than in a vacuum.
This has to do with special relativity. You should google on the Michelson Morley experiment. It is an experiment in around 1900 I believe, where they basically tried to measure different speeds of light in different directions.
This has to do with special relativity. You should google on the Michelson Morley experiment. It is an experiment in around 1900 I believe, where they basically tried to measure different speeds of light in different directions.
Agreed, the Michelson-Morley experiment is proof that the speed of light is invariant to the motion of the observer.
how do you say michelson's experiment is a proof,it was conducted on earth,and was found to be 3x10^8. Do we have any other experiment whose value for speed of light
coincided with michelson's.well,consider the light rays emitted from stars,in space
,they won't travel with speed-3x10^8,they would vary ,due to relative velocity.
well,consider the light rays emitted from stars,in space
,they won't travel with speed-3x10^8,they would vary ,due to relative velocity.
Not true, the speed of light is absolute. In other words the speed of light is always c irrespective of the relative motion of the observer and the emitter (as Janus has already said).
In terms of experimental verification, see here: http://math.ucr.edu/home/baez/physi...ts.html#3. Tests of Einstein's two postulates
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how do you say michelson's experiment is a proof,it was conducted on earth,and was found to be 3x10^8. Do we have any other experiment whose value for speed of light
coincided with michelson's.well,consider the light rays emitted from stars,in space
,they won't travel with speed-3x10^8,they would vary ,due to relative velocity.
The only thing that varies is the frequency of the light (Doppler effect), which is dependent on the radial velocity of the stars with respect to earth. The frequency of the light decreases for receding objects (redshift) and increases for approaching objects (blueshift).
so according to what you say,anywhere in the universe,any object
acting as a light source would emit and will emit light at the speed of
3x10^8m/s.
so according to what you say,anywhere in the universe,any object
acting as a light source would emit and will emit light at the speed of
3x10^8m/s.
Yes. And that speed will be constant and have the same value as measured by any locally inertial observer.
so according to what you say,anywhere in the universe,any object
acting as a light source would emit and will emit light at the speed of
3x10^8m/s.
yes - this is exactly the weird thing about reality that made special relativity so hard for people to accept at the time (100 yrs ago now).
also, aside from the MM experiment, the invariance of light speed (constant c) can be inferred from the Maxwell equations. I think that's what Einstein said he was doing (and the basis for his statements that he didn't hear about MM experiment until later).
If you are interested in this (as you appear to be) get yourself a book on special relativity that's written for someone with your background, and have a go at it.
how do you say michelson's experiment is a proof,it was conducted on earth,and was found to be 3x10^8. Do we have any other experiment whose value for speed of light
coincided with michelson's.well,consider the light rays emitted from stars,in space
,they won't travel with speed-3x10^8,they would vary ,due to relative velocity.
Michelson-Morley is not about proving that the speed of light is c, it proves that it is invariant to the frame of motion. It addresses your assertions in the OP.
*First time poster*
I'm not too clued up on things of this nature and so thought I'd ask a question.
If extreme gravity can bend light then can it speed it up or slow it down?
*First time poster*
I'm not too clued up on things of this nature and so thought I'd ask a question.
If extreme gravity can bend light then can it speed it up or slow it down?
No, since the speed of light is constant according to special relativity.
Welcome to the forums, by the way!
No, since the speed of light is constant according to special relativity.
Welcome to the forums, by the way!
The reason I ask is that if gravity can have an effect of the physical elements within light regarding its direction, then might it be possible for it to affect its speed in a straight line?
I understand the idea of the speed of light being constant, though I can't help but wonder.
Thanks for the welcome, and for the speedy reply.
The reason I ask is that if gravity can have an effect of the physical elements within light regarding its direction, then might it be possible for it to affect its speed in a straight line?
I understand the idea of the speed of light being constant, though I can't help but wonder.
Thanks for the welcome, and for the speedy reply.
The basic idea of general relativity is that masses create curvature in space-time. So although the path of the light looks bent to us, it's actually following a straight path (geodesic) in space-time. So masses change space time, which affects the path of the photon that travels through it, but not the speed.
I know it's odd, but that's why I like physics :)
The basic idea of general relativity is that masses create curvature in space-time. So although the path of the light looks bent to us, it's actually following a straight path (geodesic) in space-time. So masses change space time, which affects the path of the photon that travels through it, but not the speed.
I know it's odd, but that's why I like physics :)
They appear black because the gravitational field is too strong for light to escape (correct me at any point, as I said, not too clued up!). Does this mean that any light which reflects reflect off the surface of the body is slowed to a stop and then pulled back in?
Once an object reaches the event horizon or a black hole no more light reaches outwards to any possible viewer because the gravity is too strong for light to leave. Any light reflected off what would be the rear of any object sent in would not appear because it is slowing, and being drawn back to the centre-mass. Thoughts?
No, since the speed of light is constant according to special relativity.
With respect to what observer?
And for the others?
They appear black because the gravitational field is too strong for light to escape (correct me at any point, as I said, not too clued up!). Does this mean that any light which reflects reflect off the surface of the body is slowed to a stop and then pulled back in?
Once an object reaches the event horizon or a black hole no more light reaches outwards to any possible viewer because the gravity is too strong for light to leave. Any light reflected off what would be the rear of any object sent in would not appear because it is slowing, and being drawn back to the centre-mass. Thoughts?
I'm not sure what you mean by light that reflects off 'the surface' (what surface?).
With respect to what observer?
And for the others?
According to special relativity the speed of light is the same for all inertial observers regardless of the motion of the source.
I'm not sure what you mean by light that reflects off 'the surface' (what surface?).
According to special relativity the speed of light is the same for all inertial observers regardless of the motion of the source.
The surface of the object crossing the event horizon. Surely if no more light can escape due to the gravitational field being too strong it must head back towards the surface of the body.
if somone was in a spaceship going 99.99c, say someone onboard was to play ping pong, how would it look to an observer on earth?
and if we do the opposite, how would someone on Earth playing ping pong look to an observer going 99.99c?
if somone was in a spaceship going 99.99c, say someone onboard was to play ping pong, how would it look to an observer on earth?
and if we do the opposite, how would someone on Earth playing ping pong look to an observer going 99.99c?
Watching the ping pong game would be pretty boring. With a gamma of 70.7, it would be about 1 minute between each ping and pong, almost time enough to run into the kitchen for a snack. I would rather watch a game of ping pong in one of NASA's astronaut centrifuges going at about 5 G.
if somone was in a spaceship going 99.99c, say someone onboard was to play ping pong, how would it look to an observer on earth?
and if we do the opposite, how would someone on Earth playing ping pong look to an observer going 99.99c?
You would probably confuse the spaceship with a photon :tongue:
if somone was in a spaceship going 99.99c, say someone onboard was to play ping pong, how would it look to an observer on earth?
and if we do the opposite, how would someone on Earth playing ping pong look to an observer going 99.99c?
A minor quibble, 99.99c is impossible, perhaps you meant .9999c ?
The only change would be that the images would be severly red shifted.
No, since the speed of light is constant according to special relativity.
Welcome to the forums, by the way!
Yet another minor quibble, the speed of light is ASSUMED constant by SR, so it cannot be used as a justifcation of the constancy of c. You need to refer to Maxwells Equations to get that.
Yet another minor quibble, the speed of light is ASSUMED constant by SR, so it cannot be used as a justifcation of the constancy of c. You need to refer to Maxwells Equations to get that.
In my opinion, it can be used to show that c must be constant since the theoretical implications of the constancy of c are discussed in SR, and they turn out to be empirically correct. But it's not worth a discussion | 2,884 | 12,732 | {"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} | 3.578125 | 4 | CC-MAIN-2023-40 | latest | en | 0.951991 |
https://www.nextgurukul.in/wiki/concept/cbse/class-9/maths/linear-equations-in-two-variables/graphical-representation-of-linear-equations-in-two-variable/3957504 | 1,603,928,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00134.warc.gz | 816,824,765 | 60,528 | Notes On Graphical Representation of Linear Equations in Two Variables - CBSE Class 9 Maths
A linear equation in two variables represents a straight line geometrically. The straight line is called the graph of the linear equation in two variables. The solutions of a linear equation can be obtained by substituting different values for x in the equation to find the corresponding values of y. The values of x and y are represented as an order pair. To plot the graph of a linear equation, its solutions are found algebraically and then the points are plotted on the graph. Any linear equation of the form 'ax + by + c = 0' represents a straight line on the graph. The points of the straight line make up the collection of solutions of the equation. Every point that satisfies the linear equation lies on the line. Every point that lies on the line is a solution of the linear equation. A point that does not lie on the line is not a solution of the linear equation. An equation in the form of y = mx where m is a real number, represents a straight line passing through the origin in a cartesian plane. The equation of x-axis is y = 0. Equation of any line parallel to x-axis is of the form y = k. The equation of y-axis is x = 0. Equation of any line parallel to y-axis is of the form x = k. The solution of a linear equation is not affected when: (i) the same number is added to (or subtracted from) both the side of an equation. (ii) multiplying or dividing both the sides of the equation by the same non-zero number.
#### Summary
A linear equation in two variables represents a straight line geometrically. The straight line is called the graph of the linear equation in two variables. The solutions of a linear equation can be obtained by substituting different values for x in the equation to find the corresponding values of y. The values of x and y are represented as an order pair. To plot the graph of a linear equation, its solutions are found algebraically and then the points are plotted on the graph. Any linear equation of the form 'ax + by + c = 0' represents a straight line on the graph. The points of the straight line make up the collection of solutions of the equation. Every point that satisfies the linear equation lies on the line. Every point that lies on the line is a solution of the linear equation. A point that does not lie on the line is not a solution of the linear equation. An equation in the form of y = mx where m is a real number, represents a straight line passing through the origin in a cartesian plane. The equation of x-axis is y = 0. Equation of any line parallel to x-axis is of the form y = k. The equation of y-axis is x = 0. Equation of any line parallel to y-axis is of the form x = k. The solution of a linear equation is not affected when: (i) the same number is added to (or subtracted from) both the side of an equation. (ii) multiplying or dividing both the sides of the equation by the same non-zero number.
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➤ | 661 | 2,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 4.65625 | 5 | CC-MAIN-2020-45 | latest | en | 0.92976 |
http://stats.stackexchange.com/questions/31358/how-to-combine-values-based-on-standard-errors/31360 | 1,386,599,729,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163982738/warc/CC-MAIN-20131204133302-00057-ip-10-33-133-15.ec2.internal.warc.gz | 172,162,047 | 12,880 | How to combine values based on standard errors?
Let's assume we have two means with corresponding standard errors.
mean_1 = 3.75, se_1 = 0.64
mean_2 = 2.90, se_2 = 0.94
I would like to average the means and account for the standard error. I would like mean_1 to contribute more (since the standard error is lower) to the average than mean_2. A weighted mean if you will.
One way would be to specify weights as the relative magnitude of the standard error like so (I use R's function weighted.mean found in the base stats package):
> weighted.mean(c(3.75, 2.0), w = c((1 -0.64/(0.64+0.94)), (1- 0.94/(0.64+0.94))))
[1] 3.041139
-
Use the Delta-method ? – Stéphane Laurent Jun 29 '12 at 13:48 | 216 | 699 | {"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} | 2.96875 | 3 | CC-MAIN-2013-48 | latest | en | 0.8656 |
http://slideplayer.com/slide/3824246/ | 1,576,468,007,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00091.warc.gz | 131,664,864 | 19,901 | Why present value leads to better investment decisions than other criteria Yes, there are two paths you can go by, but in the long run there’s still time.
Presentation on theme: "Why present value leads to better investment decisions than other criteria Yes, there are two paths you can go by, but in the long run there’s still time."— Presentation transcript:
Why present value leads to better investment decisions than other criteria Yes, there are two paths you can go by, but in the long run there’s still time to change the road you're on. And it makes me wonder. – Page, Plant
CFO Decision Tools Survey Data on CFO Use of Investment Evaluation Techniques SOURCE: Graham and Harvey, “The Theory and Practice of Finance: Evidence from the Field,” Journal of Financial Economics 61 (2001), pp. 187-243.
3 competitors of NPV Payback & Discounted Payback Average Return on Book Internal Rate of Return
Payback NPV Year: 0 1 2 3 Payback At 10% A -2 +2 1 -0.2 B -2 +1 +1 +5 2 +3.5 Project A has shorter payback but lower NPV
Discounted Payback Discounted NPV Year: 0 1 2 3 Payback At 10% A -2 +2 -0.2 B -2 +1 +1 +5 2.07 +3.5 Project A does not get paid back
Problems with Payback Payback does not recognize time value of money (discounted payback does). Payback & Dicounted Payback both ignore any cash flows after payback period.
Book Rate of Return Book Rate of Return - Average income divided by average book value over project life. Also called accounting rate of return.
Average return on book Year: 0 1 2 3 Book 9 6 3 0 Gross Profit 6 5 4 (= cash flow) Depreciation 3 3 3 Net Profit 3 2 1 Average Return on Book = = 44% 2 4.5
Average return on book (continued) This project also has a 44% average return on book, but cash flows come later and NPV is less. Year: 0 1 2 3 Book 9 6 3 0 Gross Profit 4 5 6 (= cash flow) Depreciation 3 3 3 Net Profit 1 2 3 Average Return on Book = = 44% 2 4.5
Problems with Average Return on Book Value It ignores time value of money. It is not based on project cash flows.
Rate of return rule Suppose you invest 350 in office building, which you expect to sell next year for 400. Cost of capital = 7%. Rate profit C1 + C0 400 - 350 of = = = = 14.3% return investment -C0 350 The project return is higher than equivalent investments. Note: 1. Project return is also the discount rate which gives zero NPV. 2. If the cost of capital is less than the return, NPV is positive. NPV 60 0 - 60 0 14.340Discount rate % Return = 14.3%
Internal rate of return rule (IRR) C 1 C 2 NPV = C 0 + + +... = 0 1 + IRR (1 + IRR) 2 i.e., Find cost of capital at which project would be fair value (NPV=0). Accept project if actual cost of capital is less than this.
Internal Rate of Return Example You can purchase a turbo powered machine tool gadget for \$4,000. The investment will generate \$2,000 and \$4,000 in cash flows for two years, respectively. What is the IRR on this investment?
Internal Rate of Return Example You can purchase a turbo powered machine tool gadget for \$4,000. The investment will generate \$2,000 and \$4,000 in cash flows for two years, respectively. What is the IRR on this investment?
Internal Rate of Return IRR=28%
Internal Rate of Return Pitfall 3 - Mutually Exclusive Projects IRR ignores the magnitude of the project. The following two projects illustrate that problem.
But beware, there are potential problems with IRR---- Lending or Borrowing? Year: 0 1 IRR NPV @ 10% A -100 +150 +50% +36.4 B +100 -150 +50% -36.4 Multiple Rates of Return Year: 0 1 2 IRR NPV @ 10% C -4 +25 -25 25% & 400% -1.9 Mutually Exclusive Projects Year: 0 1 IRR NPV AT 10% D -10 +13 30% 1.8 E -20 +25 25 2.7
Decisions Under Capital Rationing If capital is rationed in only one period: Profitability Index = NPV / PV of outlays If capital is rationed in more than one period or when there are other constraints, we may use: -Linear Programing Problem with fractional projects -Interger (zero-one) Programming
Profitability Index When resources are limited, the profitability index (PI) provides a tool for selecting among various project combinations and alternatives A set of limited resources and projects can yield various combinations. The highest weighted average PI can indicate which projects to select.
Download ppt "Why present value leads to better investment decisions than other criteria Yes, there are two paths you can go by, but in the long run there’s still time."
Similar presentations | 1,194 | 4,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-51 | latest | en | 0.907387 |
https://www.thefreedictionary.com/Kinetic+friction | 1,579,305,989,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591431.4/warc/CC-MAIN-20200117234621-20200118022621-00492.warc.gz | 1,131,005,297 | 11,973 | # kinetic friction
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Related to kinetic friction: Static friction
## kinetic friction
n.
The friction arising between bodies in motion with respect to each other, as a body sliding along a floor.
References in periodicals archive ?
In the case of the kinetic friction, the slip force opposes the relative movement between the contact points (which may differ from the relative movement between the bodies).
[4] carried out a research in the framework of the theory of plasticity, considered internal friction and cohesion as the variables of the plastic strain function, and proposed a hyperbolic hardening function for kinetic friction and a mixed parabolic exponential equation for kinetic cohesion.
The static friction coefficient and kinetic friction coefficient of each fiber were recorded at the standard testing conditions (i.e., 20[degrees]C and 60% RH).
As these four equations show the coefficient of kinetic friction [[micro].sub.k] and the wall-normal restitution coefficient [e.sub.n] have to be known to calculate the particle velocity after the wall collision.
However, most FE models above did not realize the modelling of real asphalt pavement and failed in considering the fact that the kinetic friction coefficient varied with different vehicle speed.
Static and kinetic friction coefficients are fundamental parameters for friction force simulation of drilling string.
Static friction coefficients for a given pair of surfaces usually are higher than those for kinetic friction. Most dry, clean surface combinations, including coatings, have SCOF values between 0.2 and 0.6.
In order to have a model that considers both rolling and sliding interaction between the gear teeth, a modification of Coulomb's law is applied to obtain the equivalent kinetic friction coefficient.
That said, the authors did an excellent job of compiling an in-depth appendix of coefficients of static and kinetic friction. It is by far the most complete collection I have ever seen in a theatrical textbook.
The kinetic friction coefficient is a useful indicator for describing tribological performance of a solid lubricant coating.
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Open / Close | 415 | 2,224 | {"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-2020-05 | latest | en | 0.923031 |
https://www.lmfdb.org/EllipticCurve/Q/75150a1/ | 1,575,906,435,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00430.warc.gz | 778,317,297 | 29,663 | # Properties
Label 75150a1 Conductor 75150 Discriminant -923443200 j-invariant $$-\frac{1543893517395}{1368064}$$ CM no Rank 1 Torsion Structure $$\mathrm{Trivial}$$
# Related objects
Show commands for: Magma / SageMath / Pari/GP
## Minimal Weierstrass equation
magma: E := EllipticCurve([1, -1, 0, -2112, -36864]); // or
magma: E := EllipticCurve("75150a1");
sage: E = EllipticCurve([1, -1, 0, -2112, -36864]) # or
sage: E = EllipticCurve("75150a1")
gp: E = ellinit([1, -1, 0, -2112, -36864]) \\ or
gp: E = ellinit("75150a1")
$$y^2 + x y = x^{3} - x^{2} - 2112 x - 36864$$
## Mordell-Weil group structure
$$\Z$$
### Infinite order Mordell-Weil generator and height
magma: Generators(E);
sage: E.gens()
$$P$$ = $$\left(\frac{2571}{25}, \frac{107154}{125}\right)$$ $$\hat{h}(P)$$ ≈ 7.53037581983
## Integral points
magma: IntegralPoints(E);
sage: E.integral_points()
None
## Invariants
magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E)[1] Conductor: $$75150$$ = $$2 \cdot 3^{2} \cdot 5^{2} \cdot 167$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$-923443200$$ = $$-1 \cdot 2^{13} \cdot 3^{3} \cdot 5^{2} \cdot 167$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$-\frac{1543893517395}{1368064}$$ = $$-1 \cdot 2^{-13} \cdot 3^{6} \cdot 5 \cdot 167^{-1} \cdot 751^{3}$$ Endomorphism ring: $$\Z$$ (no Complex Multiplication) Sato-Tate Group: $\mathrm{SU}(2)$
## BSD invariants
magma: Rank(E); sage: E.rank() Rank: $$1$$ magma: Regulator(E); sage: E.regulator() Regulator: $$7.53037581983$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega[1] Real period: $$0.352097843289$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] Tamagawa product: $$2$$ = $$1\cdot2\cdot1\cdot1$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E)[1] Torsion order: $$1$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$1$$ (exact)
## Modular invariants
#### Modular form 75150.2.a.p
magma: ModularForm(E);
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
$$q - q^{2} + q^{4} - q^{8} + 2q^{11} - q^{13} + q^{16} + 2q^{17} + 5q^{19} + O(q^{20})$$
magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 49920 $$\Gamma_0(N)$$-optimal: yes Manin constant: 1
#### Special L-value
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
$$L'(E,1)$$ ≈ $$5.30285817063$$
## Local data
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
sage: E.local_data()
gp: ellglobalred(E)[5]
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$2$$ $$1$$ $$I_{13}$$ Non-split multiplicative 1 1 13 13
$$3$$ $$2$$ $$III$$ Additive 1 2 3 0
$$5$$ $$1$$ $$II$$ Additive 1 2 2 0
$$167$$ $$1$$ $$I_{1}$$ Non-split multiplicative 1 1 1 1
## Galois representations
The 2-adic representation attached to this elliptic curve is surjective.
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ .
## $p$-adic data
### $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]
Note: $$p$$-adic regulator data only exists for primes $$p\ge5$$ of good ordinary reduction.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 167 nonsplit add add ss ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary nonsplit 2 - - 1,1 1 1 1 1 1 1 1 1 1 1 1 1 0 - - 0,0 0 0 0 0 0 0 0 0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has no rational isogenies. Its isogeny class 75150a consists of this curve only.
## Growth of torsion in number fields
The number fields $K$ of degree up to 7 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ (which is trivial) are as follows:
$[K:\Q]$ $K$ $E(K)_{\rm tors}$ Base-change curve
3 3.1.100200.1 $$\Z/2\Z$$ Not in database
6 6.0.40240480320000.1 $$\Z/2\Z \times \Z/2\Z$$ Not in database
We only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database. | 1,772 | 4,749 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-51 | latest | en | 0.195663 |
https://numbersworksheet.com/6th-grade-mixed-numbers-to-improper-fractions-worksheet/ | 1,721,725,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00726.warc.gz | 362,674,362 | 14,324 | # 6th Grade Mixed Numbers To Improper Fractions Worksheet
Fraction Amounts Worksheets are an effective way to apply the very idea of fractions. These worksheets are meant to train individuals in regards to the inverse of fractions, and can enable them to understand the relationship in between fractions and decimals. Many students have trouble converting fractions to decimals, but they can benefit from these worksheets. These computer worksheets will help your student to get far more familiar with fractions, and they’ll be sure to have fun doing them! 6th Grade Mixed Numbers To Improper Fractions Worksheet.
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Regardless of whether you could require a free pre-algebra fraction numbers worksheet or you want a computer model to your students, the Internet can provide you with a variety of alternatives. Many sites offer free pre-algebra worksheets, by incorporating notable conditions. When several of these worksheets could be personalized, a few free of charge pre-algebra fraction numbers worksheets can be acquired and imprinted for added training.
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Free mathematics worksheets for switching decimal to small fraction
You have come to the right place if you are looking for some free math worksheets for converting decimal to a fraction. These decimal to fraction worksheets can be obtained from numerous formats. You can download them inhtml and PDF, or random format. A lot of them come with a response key and can also be coloured by children! You can use them for summertime discovering, math locations, or as part of your normal math concepts programs.
To convert a decimal to your fraction, you should simplify it initial. If the denominator is ten, Decimals are written as equivalent fractions. Moreover, there are also worksheets on how to change merged figures to your small percentage. Totally free arithmetic worksheets for transforming decimal to fraction include mixed numbers and examples of these two conversion operations. However, the process of converting a decimal to a fraction is easier than you might think. Follow these steps to get going. | 869 | 4,864 | {"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-2024-30 | latest | en | 0.929734 |
http://certiguide.com/netplus/cg_np_Base10.htm | 1,542,279,011,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121814-00241.warc.gz | 63,832,194 | 4,902 | NEW! Network+ N11-003 2005 Beta Exam Study Guide - Just \$9! Get It Here!
Base 10
Our everyday lives typically use some math that we call "checkbook" math. The scientific name for this is known as Base 10 math. Let's look at why it is called that.
Begin by looking at the value of zero, or nothing. It is displayed as a 0. While zero is empty of a positive value, it still is a value. It isn't positive, and it isn't negative. It is used to either indicate the value of neither positive nor negative, or can be a placeholder that contains nothing. To make the point, let's look at the following example, and use US dollars as currency. \$1,000,000.00
That is a base 10 expression of one million United States dollars. Not one penny more or less. The six zeros before the decimal point indicates no amounts of hundred(s)-thousands, ten(s) thousand(s), thousand(s), hundred(s), ten(s), or single dollar(s). The zeros after the period indicate no fractions of a US dollar. Now visualize yourself having \$1,000,000.00
OK, now mentally give it to your authors. Now tell us the zero has NO value. Great, will just remove the items with no value. Now you have \$1.00. "Oops". No value with the two zeros after the decimal point. Now you have \$.01, or a US copper penny. Still want to say zero has no value? As you can see, it is more accurate to see the zero as neither a positive or negative value. It does indeed have a value, even if it is as a place holder.
OK. Let's do some counting with checkbook math. Starting with zero.
0
1
2
3
4
5
6
7
8
9
Now count the number of digits used. See? Ten, including the zero. Viola. Base 10 math.
We're willing to bet that you don't have to be as skilled as an NSA cryptographer to guess the next highest value is 10. Let's pick this apart a bit more. The zero is a placeholder meaning no positive values are contained in the units’ column. And the value in the second column is indicated with a 1 for one unit of tens. If you continue counting you get 11. This creates one value in the unit column, and one in the tens column. This continues until you get to twenty, or 20. This is because you have exhausted all the possibilities in the units’ column using base ten, and have to notch up the ten column. Of course this continues until you exhaust both columns with ninety-nine or 99. Once again, the two columns zero out and a third column gets a one or 1, and becomes one-hundred or 100.
In math theory, this can go on forever.
Now as clever as computers seem to be, they really are only made up of a form of an earth element known as silicon. In other words, sand. Now, how smart is sand? It’s about as smart as a rock. In fact, computers are nothing more than "sand with an attitude". Let’s take a look at how dumb computers really are in the next section.
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# Polynomial multiplication
Multiply. (m - 3n)(m + 3n)
#### Solution Summary
This provides an example of multiplying polynomials.
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https://boredofstudies.org/search/280048/ | 1,568,877,110,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573444.87/warc/CC-MAIN-20190919060532-20190919082532-00299.warc.gz | 391,994,145 | 12,614 | # Search results
Anyone?
Find the limit as x and y approach 0 of (7x^2y^2) / ( x^2+2y^4) Using sandwich theorem I haven't done this shit in ages and just can't brain how to do it right now I feel like something to do with proving that fraction is less than 1 and clearly above 0 so the limit must be zero, but idk...
3. ### Good tv shows
WATCH THE WALKING DEAD SERIOUSLY Ive never had to skip a few minutes ahead at a time to resolve the tension and there is so much goddamn dramatic tension Thoroughly engaging drama, 9/10 at the very least for intensity. Recommend wholeheartedly.
4. ### Please give me a list of NBN supported modems
https://imgflip.com/i/1y05wr
5. ### Please give me a list of NBN supported modems
^Also an absolutely useless, irrelevant comment.
7. ### What is your iq?
Are you insecure about your intelligence or something? :/ Dont worry dude we all have our skills in life :)
8. ### What is your iq?
hahah you plagiarise i plagiarise come up with your own IQ problem plagiarising boy
9. ### What is your iq?
Nice copypasta 10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die. Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is...
I cant believe all you kids were born in 2000 holy shit i feel so old from 1995 jesus
11. ### Relationships ..?
Relationship between the four fundamental forces of physics. Provocative discovery: Reconciling quantum mechanics and gravity is much, much harder than anticipated if you can pull that off I bet you'll get prime minister's medal
12. ### What is your iq?
Ive actually done all of the commercially available tests available with supervision of my psychiatrist so not "free online test" Lol 157 is the highest ive gotten for any component of the tests i did and hence I quote it and say "the rest are secret" hahah
o fuck
14. ### What is your iq?
This right here is proof of your horrific and mind blowing ignorance to reality and the real world. Grow up some day :)
15. ### What is your iq?
Lmao no youre shit Trump is literally greatest president to have ever lived, thank fucking christ allah and buddha for Trump's election man. World would have been gone to shit already if Clinton won. WW3, civil wars no doubt. WE're heading in that direction anyway but Trump is lessening the...
16. ### What is your iq?
My nonverbal IQ score is 157 My other ones are a bit lower and im embarrassed to tell. Heh. Yay Donald Trump! <3 Build the wall! make it tall!
17. ### Lower North Shore Tutoring 99.95, State #1 Chem 2011, Med/UMAT
Yes it is. You pay insurance so next time you dont have to pay for the losses.
18. ### Lower North Shore Tutoring 99.95, State #1 Chem 2011, Med/UMAT
Who are you bumping? you know bumping damages property? if you damage property you have to pay insurance.
19. ### q’s for those who got a 95+ atar
tfw 94.5 ATAR *cries*
20. ### 99.7 ATAR Oxbridge Student HSC Tutoring and Online Editing
What does one do with a History degree? .... | 779 | 3,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-39 | latest | en | 0.914842 |
https://www.coursehero.com/file/6091616/HW1-10/ | 1,487,662,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170696.61/warc/CC-MAIN-20170219104610-00288-ip-10-171-10-108.ec2.internal.warc.gz | 818,472,960 | 58,815 | # HW1_10 - described in class) and then use Matlab to obtain...
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1 ECE2280 Homework #1 Spring 2010 1. Given V g =10mV, find V o . Find the Thevenin equivalent between terminals a-b. (Note: v 1 Vg) 2. Sketch the following waveforms. Identify the dc component of the waveform and the ac component of the waveform. a. Vs=5cos( 20 t) V b. Vs=7V +7cos(2 π t) V c. Vs=8V ± 1V 3. Explain in your own words the procedural steps for plotting Bode Plots. (Note: I would prepare this question for use during an exam) Use the following equation for problem 4 and 5: ) 50 )( 10 ( ) 5 . ( 2000 ) ( + + + = s s s s s H 4. (a) Plug in values of ω from 0.1 to 10 5 rad/sec to obtain the magnitude and phase plots. (b) Sketch the Bode plots using a straight-line approximation (procedures described in class) 5. Using the equation from problem 4 above. (a) Use Matlab to obtain the Bode Plots. (b) Compare the 4(a), 4(b) and 5(a). What differences do you see? 6. Sketch the Bode plots for the equation below using a straight-line approximation (procedures
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Unformatted text preview: described in class) and then use Matlab to obtain the Bode Plot. Compare the two. ( ) ( )( ) k s k s s s s H 1 100 10 000 , 20 ) ( 2 + + + = + _ 10 g V 20 5v 1 2 i + _ o V a b 1 i 3i 2 40 + v 1 - 3 40 20 5 20 _ + 2 7. Use PSPICE to draw the circuit below. Derive the transfer function ) ( 1 ) ( ) ( s V s Vo s H = 8. Use PSPICE to obtain the Bode Plots for the circuit below 9. Sketch the straight-line approximation over the bode plot of problem 7 and 8. 10. Analyze the following circuit to find the transfer function Vo/V1. Solve the circuit symbolically first (with R 1 , R 2 , and C 1 ) and then plug in their values. Create a rough sketch of the transfer function using a straight-line approximation procedure. C1 125m V V1 FREQ = 1 VAMPL = 1 VOFF = 0 R1 40 R2 10 V1 AC = 1 TRAN = 0 DC = 1 R2 10k R1 1k C1 1n 20 + Vo - + Vo -...
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## This note was uploaded on 01/24/2011 for the course ECE 2280 taught by Professor Kal during the Spring '10 term at Utah State University.
### Page1 / 2
HW1_10 - described in class) and then use Matlab to obtain...
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Ask a homework question - tutors are online | 772 | 2,524 | {"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} | 3.546875 | 4 | CC-MAIN-2017-09 | longest | en | 0.818694 |
https://abh21.wordpress.com/2012/10/27/math-gently/?like=1&source=post_flair&_wpnonce=01a7653db7 | 1,397,674,921,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00015-ip-10-147-4-33.ec2.internal.warc.gz | 794,084,940 | 20,310 | ## Math, Gently
After three years of school, my eight-year-old believed he hated math. I didn’t send him to school thinking that, so yes, I consider this a failure on School’s part. I myself am quite comfortable with math; I got all the way up to Calculus, which was required for my BS in Natural Resource Science. I can remember doing my Calc homework as a break. I liked how orderly math was, the problem-solving required, the way the Right and Wrong are clearly defined. My point is that my children aren’t picking up on math anxiety from me.
When I decided to homeschool him, I decided to start with Life of Fred, gently. We began over the summer with the first book, Apples. We began with me reading the chapter aloud and going over the questions with him while he wrote the answers. We did it all together for the first two books, a chapter per day. Some days he fought it, but it got better. By the time we hit fall, we continued with one chapter per day, but I began having him read the chapter on his own, do the questions, and then we’d go over the answers together. Again, he fought this sometimes, but it gradually got better. Now he reads the chapter, does the questions, checks his answers, and moves the paperclip to the next chapter, all on his own. If he doesn’t understand a question or needs help, he tells me. Sometimes he still whines about it. But he does it. He’ll be starting the fifth book soon.
I’ve toyed with the idea of introducing something more rigorous at the half-year point, but I want to tread lightly. I don’t want him hating math. I hope he comes to see it as I do, a useful tool to get to where you need to be. I use math all the time–in recipes, in knitting, in sewing–and I try to talk about it when I do. One day not too long ago we were talking about I don’t even know what when I mentioned an Ancient Greek had used only math to figure out the circumference of the earth, and he came super close to the actual measurement. How? my kids wanted to know.
So I looked up the details. My eight-year-old and I looked at our globe pillow and at the map. He said he pictured the equator as a rope around the earth. Perfect, I said, and we moved to our circle rug. Pretend the edge of the rug is the rope. Pretend we’ve cut the earth in half. He picked two points on the edge of the circle, we tried to find the middle of the rug, we measured the angle, we did the math. Mostly I did the math–it’s not easy math. But he was so excited by it.
This is not like school math! He exclaimed. This is math that really does something!
We borrowed The Librarian Who Measured the Earth from the library and read it aloud. I bought him his own protractor and showed him how to use it. I printed out a page of angles (from this awesome site) for him to practice measuring. Next up is drawing circles with the compass so he knows exactly where the center is, and trying to find the circumference using Eratosthenes’ method, a bit more precisely than we did with the circle rug. I predict we will be outside measuring the angles of shadows before too long.
I can just imagine if I’d decided it was time for him to measure a sheet’s worth of angles just because. Fight, fight, fight. (What’s the point? What do I need this for?) But because he is excited about the way this Ancient Greek used angles to satisfy his own desire to know how big around the earth is, he wants to know how to measure angles. And in this way, I remind myself (deep breath), he will come to learn what he needs to know.
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### 4 Responses to Math, Gently
1. Lori says:
“This is not like school math! He exclaimed. This is math that really does something!” — love!
i had an interesting conversation with a friend who hated math in school but learned to love it when he became a machinist. he said he didn’t understand why school doesn’t let kids learn math in the context of what you can *do* with it.
2. Michelle says:
It’s amazing what you can learn in context and in following interests, even with math. Today, the 4yr old was telling us how one person in our family of four threw up so far, and the 9 yr old told us the percentage. And mama got to smile.
Wee’ve been doing a mix of things, so we’re just starting dogs now, but ‘ll definitely be doing like you are and doing less supplementation as we go.
3. Amy, I’m always so impressed with how in tune you are with your children’s needs and interests. We should all aspire to raise our children to love learning more than accumulate knowledge. Great post!
• amy says:
Rachelle, thank you for your very kind comment. Trust me, some days I do far better than others. Some days I’m just all WILL YOU PLEASE DO YOUR MATH CHAPTER! :) | 1,100 | 4,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2014-15 | longest | en | 0.982834 |
https://www.shaalaa.com/question-bank-solutions/the-number-diagonals-that-can-be-drawn-joining-vertices-octagon-a-20-b-28-c-8-d-16-concept-combinations_54429 | 1,618,241,077,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00572.warc.gz | 974,729,154 | 9,466 | # The Number of Diagonals that Can Be Drawn by Joining the Vertices of an Octagon is (A) 20 (B) 28 (C) 8 (D) 16 - Mathematics
MCQ
The number of diagonals that can be drawn by joining the vertices of an octagon is
• 20
• 28
• 8
• 16
#### Solution
20
An octagon has 8 vertices.
The number of diagonals of a polygon is given by
$\frac{n \left( n - 3 \right)}{2}$ .
∴ Number of diagonals of an octagon = $\frac{8 \left( 8 - 3 \right)}{2} = 20$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Q 22 | Page 26 | 201 | 596 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-17 | latest | en | 0.815972 |
http://physics401.one-school.net/2008/06/understanding-base-and-derived.html | 1,508,191,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00431.warc.gz | 327,840,187 | 12,199 | ## 1.2 Physical Quantities
### Physical Quantity
1. A physical quantity is a quantity that can be measured.
2. Physical quantities are usually expressed as the product of a numerical value and a physical unit
Example
2.34 x 102kJ
where
· E represents the physical quantity of energy
· is the numerical value
· k is the SI prefix kilo representing 103
· J is the symbol for the unit of energy, the joule
A physical quantity can be divided into base quantity and derived quantity. | 113 | 489 | {"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-2017-43 | longest | en | 0.861095 |
http://elsenaju.eu/Calculator/system-of-equations-Gauss-NxN.htm | 1,553,280,720,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202688.89/warc/CC-MAIN-20190322180106-20190322202106-00451.warc.gz | 67,239,622 | 3,096 | # Calculator for NxN equation systems
## Calculator NxN
### Linear Equation System NxN
$\left(A|b\right)=\left(\begin{array}{c}{a}_{11}\phantom{\rule{1em}{0ex}}{a}_{12}\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}}{a}_{1n}\\ {a}_{21}\phantom{\rule{1em}{0ex}}{a}_{22}\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}}{a}_{2n}\\ ⋮\\ {a}_{m1}\phantom{\rule{1em}{0ex}}{a}_{m2}\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}}{a}_{mn}\end{array}\right|\begin{array}{c}{b}_{1}\\ {b}_{2}\\ ⋮\\ {b}_{n}\end{array})$
#### Note
If leading coefficients zero then should be columns or rows are swapped accordingly so that a divison by the leading coefficient is possible. The value of the determinant is correct if, after the transformations the lower triangular matrix is zero, and the elements of the main diagonal are all equal to 1.
Dimension of the equation system N =
Number of digits =
Enter the coefficients: a11, a12, ... und b1, ...
Solution with the Gauss algorithm.
The entered matrix is: | 332 | 1,021 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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} | 3.234375 | 3 | CC-MAIN-2019-13 | latest | en | 0.677411 |
https://gist.github.com/Kadrian | 1,601,264,588,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401583556.73/warc/CC-MAIN-20200928010415-20200928040415-00596.warc.gz | 401,235,534 | 16,773 | {{ message }}
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// Trying to find the best solution to the following problem: // // Display a large number with a small space in between steps of thousand. // Thus => Convert a number to an array of max. three digits // // E.g. // Input: 10 Output: [10] // Input: 1234 Output: [1, 234] // Input: 24521280 Output: [23, 521, 280]
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You can’t perform that action at this time. | 462 | 1,639 | {"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} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.500773 |
https://curriculum.illustrativemathematics.org/k5/teachers/grade-5/unit-2/lesson-9/lesson.html | 1,720,990,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00717.warc.gz | 164,834,512 | 29,876 | # Lesson 9
Relate Area to Multiplication
## Warm-up: Which One Doesn’t Belong: Area (10 minutes)
### Narrative
This warm-up prompts students to compare four images. It gives students a reason to use language precisely (MP6). It gives the teacher an opportunity to hear how students use terminology to describe the characteristics of area.
### Launch
• Groups of 2
• Display the image.
• “Pick one that doesn’t belong. Be ready to share why it doesn’t belong.”
• 1 minute: quiet think time
### Activity
• 2–3 minutes: partner discussion
• Share and record responses.
### Student Facing
Which one doesn’t belong?
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “Let’s find at least one reason why each one doesn’t belong.”
## Activity 1: Find the Area (15 minutes)
### Narrative
The purpose of this activity is for students to find the area of rectangles with one fractional side length and one whole number side length. Students begin by considering a rectangle with whole number side lengths and then look at a series of rectangles with unit fraction side length. All of the rectangles have the same whole number width to help students see how the area changes when the fractional width changes. Students should use a strategy that makes sense to them. These strategies might include counting the individual shaded parts in the diagram or thinking about moving them to fill in unit squares. Some students may use multiplication or division. These ideas will be brought out in future lessons. During discussion, connect the different strategies students use to calculate the areas. As they choose a strategy, they have an opportunity to use appropriate tools, whether it be expressions that represent the shaded area or physical manipulations of the diagrams, strategically (MP5).
MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations and take turns sharing their understanding with their partner. Listen for and clarify any questions about the context.
Action and Expression: Internalize Executive Functions. Invite students to plan a strategy, including the tools they will use, for finding the area of a rectangle that has one side length that is a fraction. If time allows, invite students to share their plan with a partner before they begin.
Supports accessibility for: Conceptual Processing, Memory, Language
### Launch
• Groups of 2
• Display the images of the shaded rectangles.
• “What is the same about all of the rectangles? What is different?” (They are all shaded. They have different amounts shaded. They have different widths.)
• “We are going to figure out how much of each rectangle is shaded. We call this finding the area of the shaded region. What are some strategies we could use to find the area of each of the shaded regions?” (Move the pieces around to make full squares, count the number of blue pieces and multiply the number of pieces by their size.)
### Activity
• 5–7 minutes: partner work time
• Encourage students to find the area in a way that makes sense to them.
• Monitor for students who:
• count the number of shaded parts and multiply the total number of parts by their fractional area.
• visualize moving the shaded parts to fill whole unit squares.
### Student Facing
Find the area of the shaded region. Explain or show your reasoning.
1.
2.
3.
4.
### Student Response
For access, consult one of our IM Certified Partners.
If students do not find the area of the shaded region, ask “How can you use the rectangle that has 6 unit squares shaded in to help you find the area of the other shaded regions?”
### Activity Synthesis
• Ask previously selected students to share their reasoning.
• “What is the same about the strategies? What is different?” (They all counted the number of shaded parts, but they counted them in different ways. Some people multiplied and some people moved the parts to make whole unit squares.)
• Display image from final problem.
• “How does the expression $$6 \times \frac{1}{4}$$ represent the shaded area in square units?” (There are 6 shaded parts and each one has an area of $$\frac{1}{4}$$ square unit.)
• “How does the expression $$\frac{1}{4} \times 6$$ represent the shaded area in square units?” (There is a rectangle whose area is 6 square units and $$\frac{1}{4}$$ of the rectangle is shaded.)
## Activity 2: Draw Rectangles (20 minutes)
### Narrative
The purpose of this activity is for students draw and shade rectangles with a unit fraction side length and a whole number side length. Then they find the areas of the shaded regions. The tactile experience of drawing and shading encourages students to count the number of shaded parts and then either reason about their size or think about moving them to make full unit squares. They also consider a diagram where not all of the unit squares are shown. Students estimate how many of the unit squares are hidden. This helps to highlight that finding the total area can be done with multiplication where one factor is the area of each shaded part and the other factor is the total number of shaded parts (MP7).
### Required Materials
Materials to Copy
• Grid Paper 5
### Launch
• Groups of 2
• Give students grid paper.
### Activity
• 5–7 minutes: independent work time
• 5 minutes: partner work time
• Monitor for students who:
• use the grid structure on the paper to draw their rectangles
• count the number of unit squares that might be hidden under the yellow rectangle
### Student Facing
1. Represent each rectangle on grid paper:
• $$\frac{1}{2}$$ unit by 1 unit
• $$\frac{1}{2}$$ unit by 2 units
• $$\frac{1}{2}$$ unit by 3 units
• $$\frac{1}{2}$$ unit by 4 units
2. Find the area of each rectangle that you drew.
3. What information do you need to find the area of the shaded region?
4. What might the area of the shaded region be? Explain or show your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
If students do not draw the rectangles correctly, show them the shaded region from the previous activity with side lengths 6 units and $$\frac{1}{2}$$ unit and ask “How could you adapt this diagram to show a rectangle that is 4 units by $$\frac{1}{2}$$ unit?”
### Activity Synthesis
• Display a student generated image of the $$\frac{1}{2}$$ unit by 4 unit rectangle.
• Ask previously selected students to describe how they drew they 4 by $$\frac{1}{2}$$ rectangle.
• Display image from student workbook of the rectangle that is partly covered.
• “What do you need to know to determine the area of the shaded region?” (We need to know how many unit squares are under the yellow rectangle.)
• “How many unit squares do you think make up the rectangle?” (6 or 7)
• “How did you use the number of these unit squares to make an estimate for the shaded region?” (I know that each shaded blue rectangle is $$\frac{1}{2}$$ of a square so if there are 6 or 7 of those, that would be $$\frac{6}{2}$$ or $$\frac{7}{2}$$ square units.)
## Lesson Synthesis
### Lesson Synthesis
Display the shaded rectangle that has an area of 6 whole units:
“What strategies do we use to find the area of rectangles with 2 whole number side lengths?” (We can count the number of squares. We can multiply the side lengths.)
“What strategies did we use today to find the area of rectangles with a whole number side length and a unit fraction side length?” (We counted the number of unit squares and multiplied by the size of the shaded region in each unit square.)
“How are the strategies we used to find the area of rectangles with whole number side lengths the same as and different from the strategies we use to find the area of rectangles with a whole number side length and a fractional side length?” (We can use the same strategies, but we count area that is less than one unit square. We are still multiplying, but one of the numbers is a fraction.)
## Cool-down: Fractional Pieces (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners. | 1,794 | 8,077 | {"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} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.854735 |
https://gre-test-prep.com/iowa-algebra-test/hyperbolas/geometry-solver.html | 1,527,306,314,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00227.warc.gz | 575,586,764 | 12,060 | Algebra Tutorials!
Saturday 26th of May
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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• pre algebra everything book | 1,068 | 4,610 | {"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-2018-22 | latest | en | 0.893081 |
http://list.seqfan.eu/pipermail/seqfan/2009-December/055040.html | 1,696,153,411,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00302.warc.gz | 22,303,927 | 4,857 | # [seqfan] Re: "integers >4 which cannot be written as the sum of a nontrivial triangular number (>1) and a prime"
Jim Nastos nastos at gmail.com
Mon Dec 21 01:17:09 CET 2009
```On Sun, Dec 20, 2009 at 3:33 PM, Jonathan Post <jvospost3 at gmail.com> wrote:
> Playing on paper with pencil I think that
> 7, 16, 36, 42, 61, ...
> is "integers >4 which cannot be written as the sum of a nontrivial
> triangular number (>1) and a prime"
Hi, here is my output:
Numbers of form prime+triangular in exactly 0 ways:
2 3 4 7 36 42 54 61 90 105 171 210 211 216 325 351 406 528 561 630 741
780 990 1081 1176 1275 1596 1711 1830 1953 2016 2145 2346 2628 2775
3003 3081 3240 3321 3655 3741 3916 4278 4371 4465 4560 4851
(Note that 16 is not in this list... 16 = 13(prime) + 3(nontrivial triangular)
Numbers of form prime+triangular in exactly 1 ways:
5 6 9 10 11 12 14 15 16 18 19 21 24 25 27 30 31 37 45 48 55 60 63 66
70 72 75 78 84 87 91 120 121 126 129 132 135 136 140 150 153 168 180
186 198 231 240 246 252 253 276 300 315 333 345 378 390 396 405 420
435 462 465 468 540 556 588 595 660 666 690 717 798 801 820 858 861
903 924 946 960 1035 1080 1128 1225 1290 1326 1378 1410 1431 1485 1540
1620 1653 1710 1770 1863 1891 2080 2211 2278 2376 2415 2485 2556 2701
2826 2850 2926 3403 3486 3570 3828 3900 4005 4095 4186 4656 4950
Numbers of form prime+triangular in exactly 2 ways:
8 13 20 22 28 29 33 35 40 43 46 49 51 67 80 93 96 99 100 101 102 106
111 114 117 123 127 130 144 147 151 156 162 165 169 174 181 183 189
190 204 213 220 222 225 228 234 241 255 258 264 265 270 285 288 294
295 297 301 306 309 312 318 330 331 339 340 357 360 375 393 399 414
421 423 426 432 441 450 456 480 483 486 496 510 511 525 546 552 570
576 582 591 594 615 621 624 651 681 703 720 735 738 750 765 771 795
804 813 819 840 870 916 921 945 981 1008 1020 1032 1041 1050 1056 1095
1155 1158 1161 1170 1185 1191 1200 1218 1224 1260 1281 1330 1341 1365
1368 1371 1386 1396 1401 1440 1446 1470 1494 1533 1575 1632 1638 1701
1785 1836 1848 1860 1875 2010 2073 2085 2106 2157 2220 2250 2286 2430
2460 2463 2580 2616 2625 2646 2655 2670 2760 2880 2886 2922 3024 3066
3108 3123 3156 3160 3216 3270 3276 3330 3360 3501 3690 3876 3993 4140
4206 4455 4530 4605 4753 4875 4938
Numbers of form prime+triangular in exactly 3 ways:
17 23 26 32 34 38 39 41 44 50 52 53 56 57 59 64 65 69 71 73 76 79 81
85 88 94 97 103 108 113 115 131 138 141 146 148 154 157 159 160 161
175 177 192 195 199 201 205 219 223 224 230 235 237 243 249 261 267
271 273 280 281 282 291 308 321 322 324 336 342 346 348 355 363 366
367 369 372 381 384 387 400 402 411 444 463 471 474 477 481 490 492
495 498 501 504 513 516 520 531 534 543 549 555 558 560 565 567 571
573 600 601 606 609 618 631 636 639 648 661 663 670 672 675 678 682
696 699 700 702 705 711 715 723 726 747 759 762 768 777 781 786 810
822 825 826 828 831 843 846 876 879 882 885 886 888 900 906 915 918
927 930 933 936 939 948 955 966 976 978 1002 1011 1014 1023 1029 1044
1065 1068 1071 1092 1098 1101 1107 1110 1116 1125 1140 1143 1146 1149
1156 1221 1230 1233 1245 1251 1257 1263 1302 1305 1311 1320 1323 1332
1347 1359 1380 1398 1428 1443 1452 1458 1471 1473 1491 1500 1518 1521
1527 1539 1554 1560 1566 1590 1602 1605 1650 1662 1665 1668 1671 1680
1716 1740 1746 1755 1791 1794 1806 1815 1821 1827 1845 1851 1872 1881
1905 1911 1914 1935 1947 1950 1956 1965 1971 1974 1995 1998 2031 2040
2046 2052 2061 2068 2076 2100 2124 2151 2175 2199 2226 2238 2241 2268
2295 2325 2355 2380 2418 2424 2436 2478 2481 2511 2520 2562 2565 2577
2610 2619 2640 2661 2673 2685 2691 2715 2751 2800 2811 2820 2841 2856
2883 2895 2898 2976 3069 3090 3135 3141 3171 3186 3198 3246 3291 3300
3315 3396 3408 3426 3450 3468 3480 3483 3531 3555 3606 3615 3654 3675
3693 3696 3711 3720 3726 3840 3885 3915 3930 3948 3981 3996 4026 4032
4059 4080 4116 4125 4131 4161 4236 4311 4350 4368 4386 4446 4458 4551
4641 4695 4740 4791 4872 4920 4980
Numbers of form prime+triangular in exactly 4 ways:
58 77 82 83 86 92 98 109 110 112 118 124 133 142 143 145 163 166 170
178 185 187 191 193 196 197 206 207 208 226 238 245 250 259 262 268
274 275 279 283 286 292 303 304 310 316 327 350 354 358 361 364 365
371 376 379 391 397 408 416 417 418 430 436 438 447 448 453 455 459
466 469 489 491 505 507 521 522 523 526 535 547 550 553 564 585 586
603 610 612 625 627 633 640 642 645 654 657 680 684 687 693 714 721
729 736 745 756 766 783 791 792 796 807 815 816 834 836 850 855 856
864 871 873 883 891 894 897 913 925 931 942 951 954 967 969 972 984
991 994 996 1005 1021 1026 1030 1038 1046 1062 1074 1120 1122 1123
1131 1134 1150 1167 1188 1197 1206 1209 1210 1212 1215 1242 1248 1266
1269 1284 1296 1308 1335 1345 1350 1353 1366 1375 1390 1392 1395 1400
1404 1408 1413 1420 1455 1461 1483 1488 1506 1512 1524 1530 1536 1543
1548 1551 1561 1578 1581 1606 1608 1611 1635 1645 1659 1683 1695 1698
1704 1722 1725 1728 1731 1737 1743 1749 1761 1771 1776 1779 1782 1788
1800 1809 1818 1839 1842 1854 1866 1869 1885 1890 1893 1896 1915 1920
1926 1932 1941 1975 1980 1986 1989 2001 2025 2055 2067 2070 2088 2091
2115 2118 2131 2136 2148 2160 2166 2169 2181 2185 2190 2205 2208 2223
2235 2244 2256 2262 2265 2271 2280 2289 2292 2310 2316 2334 2335 2340
2341 2358 2370 2382 2400 2421 2442 2451 2466 2472 2493 2496 2499 2505
2523 2535 2536 2544 2551 2586 2595 2604 2605 2613 2688 2694 2709 2736
2745 2766 2781 2787 2790 2793 2796 2805 2814 2835 2859 2871 2910 2913
2925 2961 2964 3000 3006 3021 3030 3036 3045 3048 3057 3060 3096 3111
3195 3234 3255 3256 3303 3306 3318 3333 3345 3354 3366 3370 3372 3384
3402 3411 3423 3435 3465 3471 3510 3516 3525 3528 3546 3552 3564 3591
3594 3618 3633 3705 3708 3732 3735 3750 3756 3765 3771 3780 3798 3801
3816 3825 3927 3936 3942 3951 3963 3975 3990 4014 4038 4047 4075 4110
4146 4158 4221 4230 4245 4251 4290 4305 4320 4326 4329 4341 4356 4395
4410 4420 4431 4440 4515 4518 4525 4557 4581 4644 4650 4665 4668 4680
4686 4719 4761 4785 4788 4815 4821 4840 4860 4866 4881 4917 4941 4971
``` | 3,203 | 5,992 | {"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-2023-40 | latest | en | 0.258378 |
https://1-texasholdempoker.com/gamble/quick-answer-when-two-dice-are-rolled-determine-the-probability-that-the-sum-is-at-least-4.html | 1,696,462,704,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00263.warc.gz | 93,372,575 | 18,951 | Quick Answer: When two dice are rolled determine the probability that the sum is at least 4?
Contents
Answer: The probability of rolling two dice and getting a sum of 4 is 1/12.
What is the probability of rolling two dice and obtaining at least one 4?
Probability of getting 4 in both is 1/36.
What is the probability that a sum of 4 is rolled on any given roll of the pair of dice?
There are six faces for each of two dice, giving 36 possible outcomes. If the two dice are fair, each of 36 outcomes is equally likely. Three outcomes sum to 4: (1+3), (2+2) and (3+1). Probability of getting a sum of 4 on one toss of two dice is 3/36, or 1/12.
IT IS SURPRISING: Does the New Orleans Casino have bingo?
When two dice are rolled find the probability of getting a sum of 4 or 6?
Thus P(two dice sum to either 4 or 6)=836=29.
When two dice are rolled determine the probability that the sum is at least 5?
The probability of rolling a pair of dice whose numbers add to 5 is 4/36 = 1/9.
What is the probability of rolling 2 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 2/36 (5.556%)
4 3/36 (8.333%)
5 4/36 (11.111%)
What is probability of rolling two dice?
If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.
What is the probability that the sum of the numbers on two dice is 4 or 8?
There are 36 outcomes in total. Five of them (2,6), (3,5), (4,4), (5,3) and (6,2) result in sum 8. So, assuming all outcomes are equiprobable, the answer is 5/36.
When two dice are rolled find the probability of getting a sum of 5 or 6 a sum greater than 9?
Probability of getting a sum of 5 or 6 = 9/36 = 1/4.
When 2 dice are rolled find the probability of getting a sum of 8?
Therefore the probability that we get the sum as 8 when two dice are thrown is 5/36.
When two dice are rolled find the probability of getting a sum of 9?
The probability of getting 9 as the sum when 2 dice are thrown is 1/9.
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When two dice are rolled what is the probability of getting a sum of 6 or 7?
Hence probability of getting a sum of 6 or 7 is 1136 .
What is the probability that when two dice are rolled the sum of the numbers on the two dice is 7?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
When two dice are rolled find the probability of getting a sum less than 13?
If we’re talking about standard number cubes, then the maximum answer possible is 12 (=6+6) . Therefore, all possible rolls are less than 13, giving a probability of 100%.
When two dice are rolled find the probability of getting a sum divisible by 3?
There are 3×3=9 total possibilities for the two dice, which makes the probability of getting a multiple of three 3/9=1/3. Since there are 6×6=36 total dice rolls and 1/3 of those are a multiple of three, the number which are divisible by three is (1/3)(36)=12. | 932 | 3,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-40 | latest | en | 0.931498 |
http://www.okmij.org/ftp/Prolog/minimizer.prl | 1,501,030,048,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00404.warc.gz | 491,095,696 | 2,919 | /* ************************************************************************ * * A Hopcroft-Ullman Algorithm * for determining equivalence classes of a finite automaton * * Algorithm 2.6 of Hopcroft & Ullman * * Representation of a finite automaton: * A data base with the following facts * states([state1,...stateN]). gives state labels * initial(state). Specifies an initial state * final(state1). Specifies a final state * final(stateK). another final state (if any). * alphabet([symbol1,...symbolM]). Alphabet symbols * eats(state_from,symbol,state_to). Means Delta(state_from,symbol) * is a state_to. * .... an 'eats' fact for every edge of the FSM. * * Output of the algorithm * A list of lists, each list standing for a one equivalence class. * ************************************************************************ */ /* This is a representation of the automaton to */ /* minimize from Aho & Ullman, I, p.127 */ states([q0,q1,q2,q3,q4,q5]). initial(q0). final(q0). final(q5). alphabet([a,b]). eats(q0,a,q5). eats(q0,b,q2). eats(q1,a,q1). eats(q1,b,q0). eats(q2,a,q3). eats(q2,b,q1). eats(q3,a,q2). eats(q3,b,q4). eats(q4,a,q4). eats(q4,b,q5). eats(q5,a,q5). eats(q5,b,q3). /* *------------------------------------------------------------------------ * Initial classification - build lists of the final and non-final states */ final_states(States) :- setof(State,final(State),States),!. not_final_states(States) :- states(All_states), not_final_states(All_states,States),!. not_final_states([Trial|Others],Not_finals) :- final(Trial), /* Trial is final, don't include */ not_final_states(Others,Not_finals). not_final_states([Trial|Others],[Trial|Other_non_final]) :- not_final_states(Others,Other_non_final). not_final_states([],[]). /* *------------------------------------------------------------------------ * Estimate the size of the "quotient" set * * Given Set_of_states and the set of symbols (alphabet) * count_quotient_set(Set_of_states,Alphabet,Frequencies) * computes the set of frequencies, * Frequency[i] = ||{q | q in Set_of_states and there exists * p such that eats(p,Alphabet[i],q) is true}|| * */ /* Loop over all the symbols in the alphabet */ count_quotient_set(Set_of_states,[A|Other_symbols],[F|Other_freqs]) :- count_quotient_set1(Set_of_states,A,0,F), count_quotient_set(Set_of_states,Other_symbols,Other_freqs). count_quotient_set(_,[],[]). /* The quotient set with respect to a particular*/ /* symbol A */ count_quotient_set1([],_,F,F). count_quotient_set1([State|Other_states],A,F,Fnew) :- eats(_,A,State),!, /* There is somebody who eats A and gives State*/ F1 is F+1, count_quotient_set1(Other_states,A,F1,Fnew). count_quotient_set1([State|Other_states],A,F,Fnew) :- count_quotient_set1(Other_states,A,F,Fnew). /* *------------------------------------------------------------------------ * Body of the classification algorithm * * While working, the program appends/deletes the facts reflecting the * current status in the classification. The facts are of the form * class(ListOfStates,ListOfSymbols) * where the ListOfSymbols contains those symbols of the alphabet * that should be applied to the class in attempts to split it. */ classification(List_of_classes) :- initialize, write('Initial classification'),nl,listing(class),nl, classify, setof(X,class(X,_),List_of_classes). /* Build two initial classes, Final and */ /* Not_final states */ initialize :- final_states(Final), not_final_states(Not_final), set_indices(Final,Not_final), !. /* Build the index list for the two */ /* classes of states and write all */ /* the info into the data base */ set_indices(SetStates1,SetStates2) :- alphabet(Alphabet), count_quotient_set(SetStates1,Alphabet,ListFreq1), count_quotient_set(SetStates2,Alphabet,ListFreq2), partition_alph(ListFreq1,ListFreq2,Alphabet,Symbols1,Symbols2), assertz(class(SetStates1,Symbols1)), assertz(class(SetStates2,Symbols2)), !. /* partition_alph(ListFreq1,ListFreq2,Alphabet,Symbols1,Symbols2) given two lists of frequencies and the alphabet, returns two list of symbols, Alphabet[i] is put to Symbols1 if ListFreq1[i] <= ListFreq2[i], Alphabet[i] is put to Symbols2 otherwise. */ partition_alph([Fr1|OthersFr1],[Fr2|OthersFr2],[A|OthersAl],[A|Oss1],SS2) :- Fr1 =< Fr2, partition_alph(OthersFr1,OthersFr2,OthersAl,Oss1,SS2). partition_alph([Fr1|OthersFr1],[Fr2|OthersFr2],[A|OthersAl],SS1,[A|Oss2]) :- Fr1 > Fr2, partition_alph(OthersFr1,OthersFr2,OthersAl,SS1,Oss2). partition_alph([],[],[],[],[]). partition_alph(_,_,_,_,_) :- write('partition_alph: illegal call'),nl, break. classify :- class(States,Indices), Indices = [A|Other_symbs], /* If the index set is not empty */ retract(class(States,_)), asserta(class(States,Other_symbs)), /* with one index removed*/ write('Iteration with the basis class '),write(States),nl, split(States,A), nl, write('End of iteration, classes are'),nl, listing(class), classify. classify. /* End of classification - All the index sets are empty */ /* split(Basis_set,A) tries to partition each class of states with respect to the symbol A and the Basis_set. In the other words, for each class(States,_) the predicate finds SubSet1 = { q | q in States and eats(q,a,p), p in Basis_set } SubSet2 = States - SubSet1 If SubSet1 and SubSet2 are both non-empty, actually split the States into two more classes SubSet1 and SubSet2, and construct indices in the same way like we did for the initial classes */ split(Basis_set,A) :- class(States,_), /* Loop over all the states */ split1(Basis_set,States,A,SubSet1), SubSet1 = [_|_], /* Proceed if SubSet1 isn't empty*/ subtract(States,SubSet1,SubSet2), SubSet2 = [_|_], /* Proceed if SubSet2 isn't empty*/ retract(class(States,Indices)), /* Remove the old class */ write('Splitting the class '),write(States),nl, write('Into the '),write(SubSet1),write(' and '),write(SubSet2),nl, set_indices(SubSet1,SubSet2), fail. /* Repeat for other classes */ split(_,_) :- !. /* We've checked all the states */ /* split1(Basis_set,States,A,SubSet) finds SubSet = { q | q in States and eats(q,a,p), p in Basis_set } */ split1(Basis_set,[S|Other_states],A,[S|OtherSubSet]) :- eats(S,A,P), /* Two conditions to include */ member(P,Basis_set), /* S in the Subset */ split1(Basis_set,Other_states,A,OtherSubSet), !. /* Otherwise */ split1(Basis_set,[S|Other_states],A,OtherSubSet) :- split1(Basis_set,Other_states,A,OtherSubSet), !. split1(_,[],_,[]) :- !. /* A deterministic (!) predicate to check if an element is in the set*/ member(P,[P|_]) :- !. member(P,[_|Other_els]) :- member(P,Other_els),!. /* subtract(Set,SubSet,Diff) finds the difference Diff = Set - Subset */ subtract([S|Set],SubSet,Diff) :- member(S,SubSet), subtract(Set,SubSet,Diff),!. subtract([S|Set],SubSet,[S|Other_diff]) :- subtract(Set,SubSet,Other_diff),!. subtract([],_,[]) :- !. | 1,813 | 6,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-30 | latest | en | 0.4777 |
https://www.numwords.com/words-to-number/en/1362 | 1,563,607,746,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00490.warc.gz | 790,425,484 | 1,831 | NumWords.com
# How to write One thousand three hundred sixty-two in numbers in English?
We can write One thousand three hundred sixty-two equal to 1362 in numbers in English
< One thousand three hundred sixty-one :||: One thousand three hundred sixty-three >
Two thousand seven hundred twenty-four = 2724 = 1362 × 2
Four thousand eighty-six = 4086 = 1362 × 3
Five thousand four hundred forty-eight = 5448 = 1362 × 4
Six thousand eight hundred ten = 6810 = 1362 × 5
Eight thousand one hundred seventy-two = 8172 = 1362 × 6
Nine thousand five hundred thirty-four = 9534 = 1362 × 7
Ten thousand eight hundred ninety-six = 10896 = 1362 × 8
Twelve thousand two hundred fifty-eight = 12258 = 1362 × 9
Thirteen thousand six hundred twenty = 13620 = 1362 × 10
Fourteen thousand nine hundred eighty-two = 14982 = 1362 × 11
Sixteen thousand three hundred forty-four = 16344 = 1362 × 12
Seventeen thousand seven hundred six = 17706 = 1362 × 13
Nineteen thousand sixty-eight = 19068 = 1362 × 14
Twenty thousand four hundred thirty = 20430 = 1362 × 15
Twenty-one thousand seven hundred ninety-two = 21792 = 1362 × 16
Twenty-three thousand one hundred fifty-four = 23154 = 1362 × 17
Twenty-four thousand five hundred sixteen = 24516 = 1362 × 18
Twenty-five thousand eight hundred seventy-eight = 25878 = 1362 × 19
Twenty-seven thousand two hundred forty = 27240 = 1362 × 20
Sitemap | 404 | 1,372 | {"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} | 2.96875 | 3 | CC-MAIN-2019-30 | latest | en | 0.812446 |
https://mathstodon.xyz/@eseprimo | 1,656,539,241,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00635.warc.gz | 441,217,014 | 20,342 | If there is anyone reading this who knows a little about the Riemann Hypothesis, could you let me know?
I have a thing and I need someone with a little pre-existing knowledge to have a glance and tell me why it's rubbish.
It passes "The Crank Test".
Hi, and welcome to Mathstodon, a Mastodon instance largely focused on maths and related topics.
It's still pretty quiet here, but growing slowly, and we look forward to seeing what you choose to share. The more you say about what you're doing, the more likely you are to get people replying, following, and generally interacting.
So, tell us about yourself, using the hashtag
Cheers!
If you're an academic, you're probably unhappy with Chegg. I know I am. Not so much for promoting intellectual property theft, but because they profit from miseducating students by luring them into copying rather than getting the benefit of their own work, and because they greatly boost my workload by forcing me to make up new problems instead of reusing them. But did you know that your university's retirement fund investors may well be financial supporters of Chegg? See chronicle.com/article/work-in-
Does anyone have any idea what this character is for: ⋳
"element of with vertical bar at end of horizontal stroke (U+22F3)"
I chuckled when I saw my 3rd-grader's math homework today.
I just released #Tusky 18.0, but do you know what is even more exciting?
The feature drop coming to Tusky nightly today!
Unified push 👀
ugly parser hack
From an old Donald Knuth paper, discovered a scheme for implementing operator precedence that was used in the first Fortran compiler:
Just do a search-and-replace of + with )))+(((, of * with ))*((, etc., then put ((())) around the whole thing. More parens = lower precedence operator. Since this is purely textual preprocessing, the compiler doesn't need to know about operator precedence after that. Surprisingly, it actually works!
kmjn.org/notes/operator_preced
Anyone know a Masto instance for gardeners? It might be the gateway for my wife to get off FB. #ask #garden
Just saw a post that compared website cookie pop-ups to cigarette warnings and..
Yes! Let's make "having your data stolen" be the new "smoking".
Minimizing harm from corporate data theft is WAY easier than quitting nicotine, too.
While taking shots of this Ichneumonid Wasp (family Ichneumonidae), I noticed something on its back leg. Zooming in, I saw it had a hitch-hiker: a Chernetid Pseudoscorpion (family Chernetidae). Taken at Sungei Buloh Wetland Reserve, Singapore, on 1 May 2022.
Terry Tao tries to make mathematical sense of notations like ± or O(...) that specify something partially rather than exactly: terrytao.wordpress.com/2022/05
It's a long post, but I think much less technical than most of Tao's posts.
Dear , , and experts, I want to find a sparse distance matrix of a subgraph, using matrices. I appreciate any ideas, here are the details: stackoverflow.com/questions/72.
Hey coding peeps! 👋 What project practice makes you smile when you join a community:
* concise commit messages?
* tons of inline code documentation?
* well maintained changelogs?
* code of conduct?
And what makes you go nuts?
#TryingToDoThingsRightFromTheStart
Here's something I made a couple of weeks ago, before The Stressful Week: Farmyard Befunge.
checkmyworking.com/posts/2022/
Digital marbling: amandaghassaei.com/projects/di
A recent physics-based simulation project for paper marbling, by Amanda Ghassaei, who was also responsible for an origami simulator (origamisimulator.org/) that I linked with a different url a few years ago.
I think these two grids are all you need to memorise. Suppose you're multiplying by $$10a+b$$, [^0]. Rotate the grid so that $$b$$ is in the top left. If $$b \lt 5$$, write down the $$a$$ times table in front, but bump it up by one every time you cross a vertical line. If $$b\gt5$$, write down the $$a+1$$ times table, but drop it by one each time you cross a vertical line.
[^0] Sigh, because mathematicians: $$a$$ and $$b$$ integers, $$0\le a\lt10$$ and $$1\le b\le 9$$, $$b\ne 5$$. Honestly.
For example, for the 47 times table, write down the grid with 7 in the top right:
| 7 4 1
| 8 5 2
| 9 6 3
Then prepend the five times table, knocked down by one for each line you've crossed:
$$\begin{pmatrix} 47 & 94 & 141 \\ 188 & 235 & 282 \\ 329 & 376 & 423 \end{pmatrix}$$
Boom!
Optimist: The glass is ½ full.
Pessimist: The glass is ½ empty.
Excel: The glass is January 2nd.
Show older
The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon! | 1,166 | 4,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-27 | longest | en | 0.957532 |
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