url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://mathsux.org/page/6/ | 1,606,614,504,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195967.34/warc/CC-MAIN-20201129004335-20201129034335-00606.warc.gz | 379,637,742 | 24,591 | ## Algebra 2: Factor by Grouping
Factor By Grouping: Hard to solve? No. Hard to remember? Yes.
If you have any questions please don’t hesitate to post in the comments or to message me on Twitter. Happy mathing! 🙂
## Brain Teaser
School’s out!!!:) Since the summer season is in full swing, I thought I’d celebrate by looking at a brain teaser. I found this online, it involves some common sense, some logic, and some colorful boats. Well here goes……
At the local model boat club, four friends were talking about their boats.
There were a total of eight boats, two in each color, red, green, blue and yellow. Each friend owned two boats. No friend had two boats of the same color.
Alan didn’t have a yellow boat. Brian didn’t have a red boat, but did have a green one. One of the friends had a yellow boat and a blue boat and another friend had a green boat and a blue boat. Charles had a yellow boat. Darren had a blue boat, but didn’t have a green one.
Can you work out which friend had which colored boats? | 249 | 1,019 | {"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} | 3.4375 | 3 | CC-MAIN-2020-50 | latest | en | 0.979393 |
https://oeis.org/A074993 | 1,624,128,611,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00414.warc.gz | 390,912,684 | 3,816 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A074993 a(n) = floor((concatenation of n, n+1)/2). 9
0, 6, 11, 17, 22, 28, 33, 39, 44, 455, 505, 556, 606, 657, 707, 758, 808, 859, 909, 960, 1010, 1061, 1111, 1162, 1212, 1263, 1313, 1364, 1414, 1465, 1515, 1566, 1616, 1667, 1717, 1768, 1818, 1869, 1919, 1970, 2020, 2071, 2121, 2172, 2222, 2273, 2323, 2374 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The first differences follow a pattern. Odd-indexed terms and even-indexed terms form separate A.P.s with the same common difference for all n except n = 10^k -1. The corresponding common differences are the repunits = (10^(d+1)-1)/9 where d = the number of digits in n. LINKS MATHEMATICA cc[n_]:=Floor[FromDigits[Join[IntegerDigits[n], IntegerDigits[n+1]]]/2]; Array[cc, 40, 0] (* Harvey P. Dale, Nov 11 2011 *) CROSSREFS Cf. A001704, A074991, A074992, A074994, A074995, A074996, A074997, A073086, A074999, A127421. Sequence in context: A315522 A315523 A315524 * A315525 A231001 A315526 Adjacent sequences: A074990 A074991 A074992 * A074994 A074995 A074996 KEYWORD base,easy,nonn AUTHOR Amarnath Murthy, Aug 31 2002 STATUS approved
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Last modified June 19 14:40 EDT 2021. Contains 345140 sequences. (Running on oeis4.) | 582 | 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} | 3.15625 | 3 | CC-MAIN-2021-25 | latest | en | 0.575235 |
http://www.math.brown.edu/tbanchof/art/PAC-9603/tour/horizon/horizon-intro.html | 1,686,420,740,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00651.warc.gz | 83,650,367 | 2,641 | SB3D (Tour/Horizon)
# The Artist's Comments on Math Horizons
"Math Horizon" is so named because it appeared as the cover image for an article in the journal "Math Horizons", published by the Mathematical Association of America. The object under investigation is an immersion of the two-dimensional sphere into four-space in such a way that there is exactly one point where the surface intersects itself. In this sense, it represents an analogue of the figure-eight, an immersion of the one-sphere into two-space self-intersecting at a single point. In neither case can the immersion be deformed into an embedding, with no self-intersections, without introducing local singularities such as cusps. The particular image is obtained by projecting centrally from a point on the three-sphere so that the observer appears to be inside the object. Color depicts different circles of latitude on the original two-sphere. The north and south poles are both mapped to the origin in four-space and no other pair of points is mapped to a common point in four-space. Projecting into three-space does introduce a curve of double points, including the image of the origin. The example was originally introduced in a research/expository article on the geometry of characteristic classes for surfaces in four-space written together with Frank Farris.
When this image was used as the cover for the journal, it was rotated ninety degrees from how it appears in this show. The placement on the exhibit wall was voted on by a group of artists at the Art Club. This orientation was preferred since it more closely suggested a sunset over the water. One artist strongly wanted the image to be hung upside down from the position finally chosen, precisely because it presented the same geometrical picture but with an inversion of the expected sunset color values. There is, of course, no right way to hang such an abstract image. One exhibitor at the Providence Art Club expressed this ambiguity in his show by mounting some pieces on rotating discs but this does not work for a basically rectangular piece.
Surfaces Beyond the Third Dimension Last modified: 08 Oct 2000 08:14:23 Comments to: Thomas F. Banchoff (`thomas_banchoff@brown.edu`) | 449 | 2,222 | {"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-2023-23 | latest | en | 0.960522 |
https://percentagecalculator.pro/What-is_5_percent-of_six%20tenths_Dollars | 1,537,340,044,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155942.15/warc/CC-MAIN-20180919063526-20180919083526-00081.warc.gz | 591,959,715 | 14,058 | # 5 percent of six tenths Dollars
Using this tool you can find any percentage in three ways. So, we think you reached us looking for answers like: What is 5 (5%) percent (%) of six tenths (6/10)? Or may be: 5 percent of six tenths Dollars. See the solutions to these problems just after the percentage calculator below.
See the solutions to these problems just after the percentage calculator below.
What is % of ?
Answer: 6
### Percentage Calculator 2
is what percent of ?
Answer: %
### Percentage Calculator 3
is % of what?
Answer:
If you are looking for a Discount Calculator, please click here.
## How to work out percentages - Step by Step
Here are the solutions to the questions stated above:
### 1) What is 5% of 6/10?
Always use this formula to find a percentage:
% / 100 = Part / Whole replace the given values:
5 / 100 = Part / 6/10
Cross multiply:
5 x 6/10 = 100 x Part, or
3 = 100 x Part
Now, divide by 100 and get the answer:
Part = 3 / 100 = 0.3
### 2) 5 is what percent of 6/10?
Use again the same percentage formula:
% / 100 = Part / Whole replace the given values:
% / 100 = 5 / 6/10
Cross multiply:
% x 6/10 = 5 x 100
Divide by 6/10 and get the percentage:
% = 5 x 100 / 6/10 = 833.33333333333% | 357 | 1,242 | {"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} | 4.34375 | 4 | CC-MAIN-2018-39 | latest | en | 0.848266 |
https://nrich.maths.org/2199/index | 1,679,754,716,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945333.53/warc/CC-MAIN-20230325130029-20230325160029-00484.warc.gz | 477,371,380 | 6,141 | # The Knapsack Problem and Public Key Cryptography
##### Age 16 to 18
Published 2004 Revised 2011
Public-Key cryptography was invented in the 1970s by Whitfield Diffie, Martin Hellman and Ralph Merkle.
Public-key cryptography needs two keys. One key tells you how to encrypt (or code) a message and this is "public" so anyone can use it. The other key allows you to decode (or decrypt) the message. This decryption code is kept secret (or private) so only the person who knows the key can decrypt the message. It is also possible for the person with the private key to encrypt a message with the private key, then anyone holding the public key can decrypt the message, although this seems to be of little use if you are trying to keep something secret!
The First General Public-Key Algorithm used what we call the Knapsack Algorithm. Although we now know that this algorithm is not secure we can use it to look at how these types of encryption mechanisms work.
The knapsack algorithm works like this:
Imagine you have a set of different weights which you can use to make any total weight that you need by adding combinations of any of these weights together.
Let us look at an example:
Imagine you had a set of weights 1, 6, 8, 15 and 24. To pack a knapsack weighing 30, you could use weights 1, 6, 8 and 15. It would not be possible to pack a knapsack that weighs 17 but this might not matter.
You might represent the weight 30 by the binary code 11110 (one 1, one 6, one 8, one 15 and no 24).
Example:
Plain text 10011 11010 01011 00000 Knapsack 1 6 8 15 24 1 6 8 15 24 1 6 8 15 24 1 6 8 15 24 Cipher text 1 + 15 + 24 = 40 1 + 6 + 15 = 22 6 + 15 + 24 = 45 0 = 0
What total weights is it possible to make?
So, if someone sends you the code 38 this can only have come from the plain text 01101.
When the Knapsack Algorithm is used in public key cryptography, the idea is to create two different knapsack problems. One is easy to solve, the other not. Using the easy knapsack, the hard knapsack is derived from it. The hard knapsack becomes the public key. The easy knapsack is the private key. The public key can be used to encrypt messages, but cannot be used to decrypt messages. The private key decrypts the messages.
##### The Superincreasing Knapsack Problem
An easy knapsack problem is one in which the weights are in a superincreasing sequence. A superincreasing sequence is one in which the next term of the sequence is greater than the sum of all preceding terms. For example, the set {1, 2, 4, 9, 20, 38} is superincreasing, but the set {1, 2, 3, 9, 10, 24} is not because 10 < 1+2+3+9.
It is easy to solve a superincreasing knapsack. Simply take the total weight of the knapsack and compare it with the largest weight in the sequence. If the total weight is less than the number, then it is not in the knapsack. If the total weight is greater then the number, it is in the knapsack. Subtract the number from the total, and compare with the next highest number. Keep working this way until the total reaches zero. If the total doesn't reach zero, then there is no solution.
So, for example, if you have a knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, 38} then it does not contain the weight 38 (as 38 > 23)
but it does contain the weight 20; leaving 3;
which does not contain the weight 9 still leaving 3;
which does not contain the weight 4 still leaving 3;
which contains the weight 2, leaving 1; which contains the weight 1.
The binary code is therefore 110010.
It is much harder to decrypt a non-superincreasing knapsack problem. Give a friend a non-superincreasing knapsack and a total and see why this is the case.
One algorithm that uses a superincreasing knapsack for the private (easy) key and a non-superincreasing knapsack for the public key was created by Merkle and Hellman They did this by taking a superincreasing knapsack problem and converting it into a non-superincreasing one that could be made public, using modulus arithmetic.
Making the Public Key
To produce a normal knapsack sequence, take a superincreasing sequence; e.g. {1, 2, 4, 10, 20, 40}. Multiply all the values by a number, n, modulo m. The modulus should be a number greater than the sum of all the numbers in the sequence, for example, 110. The multiplier should have no factors in common with the modulus. So let's choose 31. The normal knapsack sequence would be:
1$\times$31 mod(110) = 31
2$\times$31 mod(110) = 62
4$\times$31 mod(110) = 14
10$\times$31 mod(110) = 90
20$\times$31 mod(110) = 70
40$\times$31 mod(110) = 30
So the public key is: {31, 62, 14, 90, 70, 30} and
the private key is {1, 2, 4, 10, 20.40}.
Let's try to send a message that is in binary code:
100100111100101110
The knapsack contains six weights so we need to split the message into groups of six:
100100
111100
101110
This corresponds to three sets of weights with totals as follows
100100 = 31 + 90 = 121
111100 = 31+62+14+90 = 197
101110 = 31+14+90+70 =205
So the coded message is 121 197 205.
Now the receiver has to decode the message...
The person decoding must know the two numbers 110 and 31 (the modulus and the multiplier). Let's call the modulus "m" and the number you multiply by "n".
We need $n^{-1}$, which is a multiplicative inverse of n mod m, i.e. $n(n^{-1})$ = 1 mod m
In this case I have calculated $n^{-1}$ to be 71.
All you then have to do is multiply each of the codes 71 mod 110 to find the total in the knapsack which contains {1, 2, 4, 10, 20, 40} and hence to decode the message.
The coded message is 121 197 205:
121$\times$71 mod(110) = 11 = 100100
197$\times$71 mod(110) = 17 = 111100
205$\times$71 mod(110) = 35 = 101110
The decoded message is:
100100111100101110.
Just as I thought!
Simple and short knapsack codes are far too easy to break to be of any real use. For a knapsack code to be reasonably secure it would need well over 200 terms each of length 200 bits. | 1,687 | 5,956 | {"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} | 4.1875 | 4 | CC-MAIN-2023-14 | latest | en | 0.916958 |
https://www.6miu.com/read-1450271.html | 1,713,583,201,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817474.31/warc/CC-MAIN-20240420025340-20240420055340-00500.warc.gz | 572,069,845 | 5,696 | 101. 删除排序数组中的重复数字 II
xiaoxiao2021-02-28 22
public: /** * @param A: a list of integers * @return : return an integer */ int removeDuplicates(vector<int> &nums) { // write your code here int num = nums.size(); if(num == 0) return 0; int flag = 0; int j = 0; for(int i = 1; i < num; i++) { if(nums[i] != nums[j]) { nums[++j] = nums[i]; flag = 0; } else if(nums[i] == nums[j] && flag == 0) { nums[++j] = nums[i]; flag++; } } nums.resize(j + 1); return nums.size(); } };
2018/1/25 | 172 | 479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.062685 |
https://slidetodoc.com/sections-of-solids-me-111-engineering-drawing-sectional/ | 1,721,645,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00059.warc.gz | 449,697,271 | 12,633 | # Sections of Solids ME 111 Engineering Drawing Sectional
• Slides: 16
Sections of Solids ME 111 Engineering Drawing
Sectional Views • The internal hidden details of the object are shown in orthographic views by dashed lines. • The intensity of dashed lines in orthographic views depends on the complexity of internal structure of the object. • If there are many hidden lines, it is difficult to visualize the shape of the object – unnecessarily complicated and confusing. • Therefore, the general practice is to draw sectional views for complex objects in addition to or instead of simple orthographic views. • A sectional view, as the name suggests, is obtained by taking the section of the object along a particular plane. An imaginary cutting plane is used to obtain the section of the object.
Theory of Sectioning • Whenever a section plane cuts a solid, it intersects (and or coincides with) the edges of the solids. • The point at which the section plane intersects an edge of the solid is called the point of intersection (POI). • In case of the solids having a curved surface, viz. , cylinder, cone and sphere, POIs are located between the cutting plane and the lateral lines.
True Shape of Sections • A section will show its true shape when viewed in normal direction. • To find the true shape of a section, it must be projected on a plane parallel to the section plane. • For polyhedra, the true shape of the section depends on the number of POIs. The shape of the section will be a polygon of the sides equal to the number of POIs. • The true shape of the section of a sphere is always a circle. • The sections of prisms and pyramids are straight line segmented curves. • The sections of cylinders and cones will mostly have smooth curves.
Types of Cutting Planes and Their Representation • A cutting plane is represented by a cutting plane line • The cutting plane line indicates the line view of the cutting plane. • The two ends of the cutting plane line are made slightly thicker and provided with arrows. • The direction of the arrow indicates the direction of viewing of the object. • In the first-angle method of projection, the direction of the arrows is toward the POP, i. e. , toward XY (or X 1 Y 1). • Types of section planes • Vertical Section plane • Horizontal Section Plane • Profile Section plane • Auxiliary Section plane • Oblique Section plane
Hatching of the Sections • The surface created by cutting the object by a section plane is called as section. • The section is indicated by drawing the hatching lines (section lines) within the sectioned area. • The hatching lines are drawn at 45° to the principal outlines or the lines of symmetry of the section • The spacing between hatching lines should be uniform and in proportion to the size of the section. 2 H H or HB
Section plane parallel to one of the Reference Planes
Remember: - • After launching a section plane either in FV or TV, the part towards observer is assumed to be removed. • As far as possible the smaller part is assumed to be removed.
1. Pentagonal pyramid resting on HP with on of its edge perpendicular to VP 2. A vertical sectional place cuts the solid 20 mm away from the farthest corner of solid o’ from XY reference line. 3’ 2’ 1’ a’(e’) b’(d’) 4’ c’ d e o c a b
A cone of diameter 60 mm and height 60 mm is resting on HP on one of its generators. A section plane whose VT is parallel to HP and 15 mm above HP, cuts the solid removing the top portion. Draw the front view and sectional top view of the solid. O’ 4 2 3 Assume cone is resting on HP 61 f 71 81 d 11 51 41 d’ b’ e’ c’ a’ O’ Φ 60 5 7’ O 1 5’ 1’ 2’, 8’ 3’, 7’ 4’, 6’ 5’ 7 8 6 4’, 6’ X f’ g’ 1 ’ 2 ’, 8 ’ 3’, 60 g 21 c a e 15 O 1 b 31 Tilt cone about its corner Y
Section plane inclined to one of the Reference Planes
1. Square prism resting on HP with one of its edges making 35 o to VP 2. Section plane perpendicular to HP cuts the solid 3 mm away from the axis of solid and is inclined at 50 o to VP. a’ 1’ d’ b’ 2’ c’ p’ s’ 35 o 4’ 50 o d(s) q’ 3’ r’ 1(4) c(r) a(p) 2(3) b(q)
• Square prism resting on HP with one of its edges inclining 30 o to VP • A section plane is inclined 60 o to HP and perpendicular to VP passing through the axis at 20 mm below the top face cuts it. d’ b’ c’ 20 a’ 5 d(s) 1 c(r) a(p) 4 2 3 b(q)
A Cone base 75 mm diameter and axis 80 mm long is resting on its base on H. P. It is cut by a section plane perpendicular to the V. P. , inclined at 45º to the H. P. and cutting the axis at a point 35 mm from the apex. Draw the front view, sectional top view, sectional side view and true shape of the section. (LAB QUESTION : COMMON) c 1 h 1 d 1 g 1 f 1 e 1 X 1 o” l 1 a 1 d’ j’ c’ k’ 4 10 c d b 6 8 7 6 e f g 7 h 11 i k 8 j 10 4” 5” 3” 5 o l h” i” e” 5 9 4 a 12 g” d” j” c” k” l” a” 3 1 f” b” 2 3 1 12 11 2 g’ f’ e’ h’ i’ b’ l’ a’ X 35 k 1 j 1 b 1 i 1 o’ 9 Y 1 6” 2” 7” 1” 8” 9” 10” 12” 11” Y
Comparison between section views when the section plane is parallel and inclined to one of the Reference Planes
Horizontal sectional plane Inclined sectional plane | 1,413 | 5,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.922502 |
https://www.physicsforums.com/threads/earths-spin.103399/ | 1,553,118,256,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202471.4/warc/CC-MAIN-20190320210433-20190320232433-00424.warc.gz | 883,235,303 | 13,420 | # Earth's spin
1. Dec 9, 2005
### neo143
Why the earth keeps on spinning?
2. Dec 9, 2005
### Staff: Mentor
Because it has rotational inertia. It's just Newton's 1st law applied to the rotational motion of a solid object. (In other words: It doesn't need a reason to keep spinning--that's what things do. But it would need a reason (a torque) to change its spinning.)
3. Dec 9, 2005
### neo143
What about friction due to atmosphere??
4. Dec 9, 2005
### Danger
Atmospheres are part of the planets that they're associated with, and as a basic rule rotate with them. There's nothing in space for the atmosphere to drag against. Winds are 'internal' phenomena of the system, caused by convection, and aren't reflective of the planet's motion.
5. Dec 9, 2005
### vaishakh
Better the question-What initiated the earth to be in a state of spinning? Why did earth start rotating around its axis when it was formed? Can anyone here clear my doubt?
6. Dec 9, 2005
### DaveC426913
The coalescence of matter into a body will conserve the angular momentum of its individual parts. The Earth did not "start rotating", the Earth was a cloud of particles, rocks and gas, in a loose orbit around a cloud of particles rocks and gas that was to become the Sun and the rest of the Solar System. It was already rotating before it formed.
Where does the rotation come from in the first place?
1] Take two particles in empty space.
2] Give them each a random velocity in a random direction.
3] Place them close enough so that they gravitationally interact.
4] When they eventually collide, they will not exactly cancel each others' motion out (There is only a single, astronomically small chance of this happening - it requires both particles to collide with each other in a direct line, with no tangential motion relative to each other i.e. they were on a collision course to begin with). So, when they collide, they will stick together. They will retain any residual tangential motion that doesn't completely cancel out. This effect is cumulative. | 483 | 2,045 | {"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-2019-13 | latest | en | 0.955336 |
https://www.jiskha.com/display.cgi?id=1336961234 | 1,503,209,532,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00427.warc.gz | 932,173,895 | 3,553 | # Algebra
posted by .
how do i find the fractional notation of 11.443 and equal it to ?/1000
• Algebra -
Treat 11.443 as a fraction over 1:
11.443
--------
1
Multiply both top and bottom by 1000:
11443
--------
1000
which is then the required fraction.
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A330856 Total sum of divisors of all the parts in the partitions of n into 2 parts. 2
0, 2, 4, 11, 15, 25, 33, 48, 56, 75, 87, 111, 127, 149, 165, 204, 220, 251, 277, 315, 339, 383, 407, 459, 491, 536, 564, 628, 660, 714, 762, 825, 857, 923, 959, 1046, 1098, 1156, 1196, 1294, 1342, 1416, 1480, 1560, 1608, 1710, 1758, 1866, 1930, 2018, 2080, 2194, 2250 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 Index entries for sequences related to partitions FORMULA a(n) = Sum_{i=1..floor(n/2)} sigma(i) + sigma(n-i), where sigma(n) is the sum of divisors of n (A000203). a(n) = ((n+1) mod 2) * sigma(floor(n/2)) + Sum_{i=1..n-1} sigma(i), where sigma(n) is the sum of divisors of n (A000203). EXAMPLE a(4) = 11; 4 has two partitions into 2 parts, (3,1) and (2,2). The total sum of all divisors of the parts is sigma(3) + sigma(1) + sigma(2) + sigma(2) = 4 + 1 + 3 + 3 = 11. MAPLE N:= 100: # for a(1) ... a(N) S:= map(numtheory:-sigma, [\$1..N]): T:= ListTools:-PartialSums(S): [0, seq(T[i-1]+`if`(i::even, S[i/2], 0), i=2..N)]; # Robert Israel, Apr 29 2020 MATHEMATICA Table[Sum[DivisorSigma[1, i] + DivisorSigma[1, n - i], {i, Floor[n/2]}], {n, 80}] CROSSREFS Cf. A000203, A330857 (distinct parts). Sequence in context: A214429 A002382 A356478 * A180384 A023168 A134419 Adjacent sequences: A330853 A330854 A330855 * A330857 A330858 A330859 KEYWORD nonn,easy AUTHOR Wesley Ivan Hurt, Apr 27 2020 STATUS approved
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Last modified September 24 19:02 EDT 2023. Contains 365581 sequences. (Running on oeis4.) | 744 | 1,969 | {"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} | 3.5625 | 4 | CC-MAIN-2023-40 | latest | en | 0.627652 |
https://kingdomship.net/new-south-wales/direct-and-indirect-proportion-word-problems-pdf.php | 1,620,729,557,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991982.8/warc/CC-MAIN-20210511092245-20210511122245-00461.warc.gz | 372,517,421 | 6,973 | Direct indirect and partitive proportion [PPT Powerpoint]. Algebra: proportion word problems - explanation and examples [direct and inverse proportion explained] cat and mouse: direct and inverse proportion - the problem is actually a great opportunity to teach direct variation and inverse variation., indirect proportion. showing top 8 worksheets in the category - indirect proportion. some of the worksheets displayed are direct and indirect proportions, mathematics linear 1ma0 direct inverse proportionality, proportion and rates, solving proportions date period, georgia performance 7e indirect measurement, direct and inverse variation work 4.
## Direct Variation Word Problems Moomoomath
Direct Proportion Worksheets Teacher Worksheets. Remember that these problems might use the word 'proportion' instead of 'variation,' but it means the same thing. sometimes you'll have to work out proportional relationships that get a little bit, the amount of green and white paint we need increase in direct proportion to each other. understanding direct proportions can help you to work out the value or amount of quantities either bigger or smaller than the one about which you have information..
In direct proportion, same in the case of inverse proportion. вђў when two quantities x and y are in direct proportion (or vary directly), they are written as x в€ќ y. symbol вђњв€ќвђќ stands for вђis proportional toвђ™. 12/04/18 вђў when two quantities x and y are in inverse proportion (or vary inversely) they are written as x в€ќ 1 y. in examples 1 to 3, th direct proportion. as one values increases, so does the other. unitary method : find the value of 1 unit, multiply to find the required units.
Algebra: proportion word problems - explanation and examples [direct and inverse proportion explained] cat and mouse: direct and inverse proportion - the problem is actually a great opportunity to teach direct variation and inverse variation. direct indirect and partitive proportion. ratio worksheets ratio worksheets for teachers. proportion word problem worksheet . free printable ratio and proportion worksheets mreichert kids . 1000 images about inb ratios rates proportions on pinterest . proportion sorting 2. math ratio problems. 1000 ideas about ratios and proportions on pinterest middle . ratio and proportions worksheet with
Remember that these problems might use the word 'proportion' instead of 'variation,' but it means the same thing. sometimes you'll have to work out proportional relationships that get a little bit direct vs indirect proportion keyword after analyzing the system lists the list of keywords related and the list of websites with related content, direct and indirect proportion pdf. direct and indirect proportions in maths . direct and indirect proportion word problems. compare search ( please select at least 2 keywords ) most searched keywords. weather live in lahore 1 . fashion land
Quick tips for tutors. factsheet: understanding direct proportion. n1/l1.7 work out simple ratio and direct proportion. introduces the idea of direct proportion. 28/05/2016в в· in this video we look at how to solving direct proportion problems using two methods: 1) cross multiplication and 2) solving for the constant of proportionality (k).
Quick tips for tutors. factsheet: understanding direct proportion. n1/l1.7 work out simple ratio and direct proportion. introduces the idea of direct proportion. direct proportion. as one values increases, so does the other. unitary method : find the value of 1 unit, multiply to find the required units.
В©d f2t0l1 u3s ok xu ttga e js vo 6fgtxwfa sr dec hlxl2cn.c 7 ya hl2la mrpi fg 3hptfsp lr re5sreir cvse kdj. w j smwaqd he1 kw ji 8t3h2 4iqn1fsi 6nbi etuex ga vl 8gse jb lrba1 q1 3.s worksheet by kuta software llc the amount of green and white paint we need increase in direct proportion to each other. understanding direct proportions can help you to work out the value or amount of quantities either bigger or smaller than the one about which you have information.
Direct variation word problems example: a local fast food restaurant takes in \$9000 in a 4 hour period. write a direct variation equation for the relationship between income and number of hours. estimate how many hours it would take the restaurant to earn \$20,250. a. write a direct variation equation for the income in any number of hours. step 1: assign variables: let i = income and h = вђ¦ direct, indirect and partitive proportion 1. word problem #1 chris earns p 5000.00 in 20 days. how much will he earn in 30 days? 5000 : 20 = n : 30 n = p 7500.00
Use this "quiz" to practice solving problems involving direct, inverse, partitive, and indirect proportions. show all work on looseleaf. when solving proportion word problems, make sure it is set up correctly. once you set up your proportion correctly, all you have to do if to replace values that you know and use an x or any other variable for the value you don't know
Use this "quiz" to practice solving problems involving direct, inverse, partitive, and indirect proportions. show all work on looseleaf. direct proportion direct proportion graphs direct proportion formula and calculations inverse direct proportion direct proportion other direct proportion. solving problems direct or indirect proportion. 1-read the problems 2- decide if it is direct or indirect proportion 3- solve them steps.
Direct vs indirect proportion" Keyword Found Websites. Worksheets are answer each question and round your answer to the nearest, mathematics linear 1ma0 direct inverse proportionality, direct and inverse variation work 4, partitive proportion word problems, nat 03, proportion and rates, direct and indirect proportions, proportions date period., вђў two quantities x and y are said to be in direct proportion if they increase or decrease together in such a manner that the ratio of their corresponding values remains constant. that is, x y = k where k is a positive number, if x and y are in direct proportion or vary directly. in case of direct proportion, if y 1, y 2 are the values of y corresponding to the values x 1, x 2 of x.
## Direct Proportion HelpingWithMath.com
Direct vs indirect proportion" Keyword Found Websites. A worksheet with word problems on inverse proportion. questions start with integer multiples and progress through to non-integer multiples. extension task to create their own question., a worksheet with word problems on inverse proportion. questions start with integer multiples and progress through to non-integer multiples. extension task to create their own question..
Direct Proportion Worksheets Teacher Worksheets. Rate problems in primary school. they will also have met probability in primary school they will also have met probability in primary school which involves ratio., quick tips for tutors. factsheet: understanding direct proportion. n1/l1.7 work out simple ratio and direct proportion. introduces the idea of direct proportion..
## Solving Direct Proportion Problems YouTube
Indirect Proportion Worksheets Printable Worksheets. Word problems involving partitive proportion.pptx - download as powerpoint presentation (.ppt / .pptx), pdf file (.pdf), text file (.txt) or view presentation slides online. scribd is the world's largest social reading and publishing site. Direct vs indirect proportion keyword after analyzing the system lists the list of keywords related and the list of websites with related content, direct and indirect proportion pdf. direct and indirect proportions in maths . direct and indirect proportion word problems. compare search ( please select at least 2 keywords ) most searched keywords. weather live in lahore 1 . fashion land.
• Direct and Inverse Variation Problems Definition & Examples
• Direct vs indirect proportion" Keyword Found Websites
• Direct indirect and partitive proportion [PPT Powerpoint]
• Direct and indirect proportion word problems worksheets,direct and inverse proportion word problems worksheet pdf,direct proportion word problems worksheets, this is the direct variation formula y equals k times x and i always think it is the same as y equals mx plus b but b is gone, it is zero m and k are very similar they are kind of like slope, then we will solve for k by plugging in our x and y values and with direct variation we are going to end up dividing y by x and then you take the k and plug it back into the formula .you will have your
Ratio/proportion word problems name_____ set up a proportion to solve each problem, show all work, and label all answers. the ratio of boys to girls is 3 to 2. remember that these problems might use the word 'proportion' instead of 'variation,' but it means the same thing. sometimes you'll have to work out proportional relationships that get a little bit
Indirect proportion. showing top 8 worksheets in the category - indirect proportion. some of the worksheets displayed are direct and indirect proportions, mathematics linear 1ma0 direct inverse proportionality, proportion and rates, solving proportions date period, georgia performance 7e indirect measurement, direct and inverse variation work 4 proportions worksheet author: maria miller subject: proportions worksheet keywords: proportions, word problems, worksheet created date: 12/29/2018 2:51:38 am
When solving proportion word problems, make sure it is set up correctly. once you set up your proportion correctly, all you have to do if to replace values that you know and use an x or any other variable for the value you don't know direct and inverse variation how to identify a direct variation problem: (a) when the problem states a direct variation exists or states that a variable is directly proportional to another variable. (b) by observing that if one quantity increases the other quantity increases or if one quantity decreases the other quantity decreases. how to solve a direct variation problem: (1) вђ¦
3. the time, t seconds, it takes a water heater to boil some water is directly proportional to the mass of water, m kg, in the water heater. when m = 250, t = 600 suppose that 20 men build a house in 6 days. if the number of men is increased to 30 then they take 4 days to build the same house. if the number of men becomes вђ¦
Proportions worksheet author: maria miller subject: proportions worksheet keywords: proportions, word problems, worksheet created date: 12/29/2018 2:51:38 am direct indirect and partitive proportion. ratio worksheets ratio worksheets for teachers. proportion word problem worksheet . free printable ratio and proportion worksheets mreichert kids . 1000 images about inb ratios rates proportions on pinterest . proportion sorting 2. math ratio problems. 1000 ideas about ratios and proportions on pinterest middle . ratio and proportions worksheet with
Direct indirect and partitive proportion. ratio worksheets ratio worksheets for teachers. proportion word problem worksheet . free printable ratio and proportion worksheets mreichert kids . 1000 images about inb ratios rates proportions on pinterest . proportion sorting 2. math ratio problems. 1000 ideas about ratios and proportions on pinterest middle . ratio and proportions worksheet with direct variation word problems example: a local fast food restaurant takes in \$9000 in a 4 hour period. write a direct variation equation for the relationship between income and number of hours. estimate how many hours it would take the restaurant to earn \$20,250. a. write a direct variation equation for the income in any number of hours. step 1: assign variables: let i = income and h = вђ¦
Worksheets are answer each question and round your answer to the nearest, mathematics linear 1ma0 direct inverse proportionality, direct and inverse variation work 4, partitive proportion word problems, nat 03, proportion and rates, direct and indirect proportions, proportions date period. remember that these problems might use the word 'proportion' instead of 'variation,' but it means the same thing. sometimes you'll have to work out proportional relationships that get a little bit
This is the direct variation formula y equals k times x and i always think it is the same as y equals mx plus b but b is gone, it is zero m and k are very similar they are kind of like slope, then we will solve for k by plugging in our x and y values and with direct variation we are going to end up dividing y by x and then you take the k and plug it back into the formula .you will have your direct proportion direct proportion graphs direct proportion formula and calculations inverse direct proportion direct proportion other direct proportion. solving problems direct or indirect proportion. 1-read the problems 2- decide if it is direct or indirect proportion 3- solve them steps. | 2,662 | 12,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-21 | latest | en | 0.897333 |
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### Grade 5 - Mathematics6.5 Units of Mass/Weight (Metric System)
Example: Change 3.5 kg to grams. 1 kg = 1000 3.5 kg = 3.5 x 1000 Answer: 3500 gms Directions: Answer the following questions. Also write at least ten examples of your own.
Q 1: An orange is weighed ingml Q 2: A postage stamp is weighed inkglmg Q 3: 5,000 mg = __ g550050 Q 4: You are weighed inmkgml Q 5: A computer is weighed inkgmcm Q 6: A book is weighed ingtm Q 7: Change 962 grams (g) to kilograms (kg).9.62 kg9,620 kg0.962 kg962,000 kg Q 8: A television is weighed inmkgkm Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 259 | 834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-30 | latest | en | 0.831506 |
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# The Limewood Furniture Company produces tables and chairs from two resources: labor and wood. The company has 80 hours of labor and 34 pounds of wood...
The Limewood Furniture Company produces tables and chairs from two resources: labor and wood. The company has 80 hours of labor and 34 pounds of wood available each day. Demand for chairs is limited to 6 per day. Each chair requires 8 hours of labor and 2 pounds of wood to produce, whereas a table requires 10 hours of labor and 7 pounds of wood. The profit from each chair is \$350 and from each table is \$150. The company wants to determine the number of chairs and tables to produce to maximize profit.
a. Formulate a linear programming model for this problem.
b. Solve this model using graphical analysis.
A restaurant sells ice cream and frozen yogurt. The restaurant makes one order for ice cream and frozen yogurt per week, and the store has enough freezer space for 110 gallons of both products. A gallon of frozen yogurt costs \$ 0.85 and a gallon of ice cream costs \$0.95, and the restaurant budgets \$90 each week for these products. The manager estimates that each week the restaurant sells at least twice as much ice cream as frozen yogurt. Profit per gallon of ice cream is \$4.30 and profit per gallon of frozen yogurt is \$3.90.
a. Formulate a linear programming model for this problem.
b. Solve this model using graphical analysis.
The Collins family owns 405 acres of farmland in NC on which they grow corn and tobacco. Each acre of corn costs \$110 to plan, cultivate, and harvest; each acre of tobacco costs \$215. The Collins budget \$55,000 for next year. The government wants to limit the number of acres of tobacco that can be planted to 100 acres. The profit for each acre of corn is \$325; the profit from each acre of tobacco is \$525. How many acres of each crop should be planted in order to maximize profit?
a) Formulate a linear programming model and solve.
b) How many acres of farmland will not be cultivated at the optimal solution? Do the Collins use the entire 100 acre tobacco allotment?
c) What would the profit for corn have to be for the Collins to plant only corn?
d) If the Collins can obtain an additional 100 acres of land, will the number of acres of corn and tobacco they plan to grow change?
e) If they decide not to cultivate a 50 acre section as part of a crop recovery-program, how will it affect their crop plans?
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# Remainders and Functions | AIME I, 1994 | Question 7
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.
## Remainders and Functions – AIME I, 1994
The function f has the property that, for each real number x, $$f(x)+f(x-1)=x^{2}$$ if f(19)=94, find the remainder when f(94) is divided by 1000.
• is 107
• is 561
• is 840
• cannot be determined from the given information
### Key Concepts
Integers
Remainder
Functions
But try the problem first…
Source
AIME I, 1994, Question 7
Elementary Number Theory by David Burton
## Try with Hints
First hint
f(94)=$$94^{2}-f(93)=94^{2}-93^{2}+f(92)$$
=$$94^{2}-93^{2}+92^{2}-f(91)$$
Second Hint
=$$(94^{2}-93^{2})+(92^{2}-91^{2})$$
$$+….+(22^{2}-21^{2})+20^{2}-f(19)$$
Final Step
=94+93+…..+21+400-94
=4561
$$\Rightarrow$$ remainder =561.
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https://www.metroleague.org/what-law-of-motion-is-volleyball/ | 1,708,639,164,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00485.warc.gz | 906,242,036 | 21,841 | # What Law Of Motion Is Volleyball?
Victor Holman
Gravity is a powerful force that can affect objects at a distance. Netting helps to keep things in place and prevents them from moving horizontally or vertically.
Upward acceleration creates an intense feeling of weightlessness, which can be exhilarating for some people. Moving objects through the air has its own set of risks, so it’s important to be aware of them before doing anything.
With practice, you’ll learn how to use the power of gravity to your advantage.
## What Law Of Motion Is Volleyball?
The force of gravity is what keeps objects in place on the Earth’s surface. Objects with mass are attracted to other masses, and this includes our planet’s gravitational pull.
In order to counteract or resist the force of gravity, netting can be used to create a barrier between two masses that are trying to attract each other. When an object moves vertically or horizontally, it will experience upward acceleration as well (due to the greater gravitational forces).
Understanding how these different types of movement interact with one another is key for creating safe environments.
## How is Newton’s 1st law used in volleyball?
The law is used in volleyball to calculate the trajectory of a ball. It helps players and coaches judge when a serve should be delivered, based on its arc and distance traveled.
Newton’s 1st Law can also be used to predict how an opposing player will react after receiving a service toss. Knowing this information allows teams to strategize effectively during matches by selecting which opponent they want to attack first.
In order for players to follow Newton’s First Law accurately, practice tossing balls with different arcs and speeds so that you become familiar with their effects on the court.
## What is the law of motion in basketball?
The law of motion in basketball is based on Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. This law governs the movement of objects both in space and time.
By understanding this principle, we can predict the motions of various basketball players with accuracy. For example, when a player dribbles the ball, it will hit the ground with a force due to its mass and momentum.
Likewise, if someone shoots a basket or passes the ball downcourt, these actions will result in corresponding movements.
## Which law of motion defines sports?
Newton’s Laws of Motion govern how forces create motion in sports. These laws are usually referred to as the Laws of Inertia, Acceleration, and Reaction.
These laws help athletes control their movements on the field or court by understanding how different forces work together to produce movement. Understanding these principles can make a big difference when playing any physical activity, including sports.
Keep these laws in mind when watching an athletic event and you’ll be able to understand why some athletes seem to move with more grace than others.
## What law of motion is badminton?
The law of motion is badminton because when the racket hits the birdie both sides create movement. Newton’s third law applies to badminton because when the shuttlecock goes flying across the court.
Vibrations created by a good smasher can often go unnoticed, making this sport one of stealthy action. Badminton players must be quick and agile in order to win–this makes it a great example of Newton’s 3rd Law of Motion.
Be sure to check out our selection of rackets and shuttlecocks to get started playing this exciting sport today.
## How does Newton’s 2nd Law relate to volleyball?
Newton’s 2nd Law of Motion is important when it comes to playing volleyball because it explains the relationship between force, mass, and acceleration. A spiked ball creates a net external force that stings your hands when you stop it- just like in real life.
Understanding this equation can help you better control the ball while playing Volleyball. The more practice you have with Newton’s 2nd Law, the better player you’ll be. Keep learning and practicing so that you can dominate on the court.
## Is volleyball a law of acceleration?
Volleyball is a sport that requires quick reflexes and agility, which is why the law of acceleration comes into play every time someone on the court moves.
Smaller athletes are more agile on the court because their lower mass accelerates and decelerates more quickly, which is particularly critical on defense. The law of acceleration can be used to your advantage when playing volleyball by taking advantage of its effects in game situations.
If you want to improve your skills as a volleyball player, it’s important to learn about this fundamental law of motion. Understanding how the law of acceleration works will give you an edge in the competition.
## How are laws of motion applied in sports?
In sports, the laws of motion are used to determine how objects move and interact with each other. The force that a player applies when dribbling a ball causes it to accelerate down towards the ground.
Newton’s First Law states that an object will continue in its state of rest or uniform motion unless acted upon by a resultant force. As the ball accelerates, air resistance becomes significant and affects its movement trajectory substantially.
Every action has an equal and opposite reaction- this is embodied in Newton’s Second Law of Motion which states that forces always act on pairs of particles simultaneously.
How does Newton’s 3rd law apply to shot put?
Newton’s Third Law of Motion applies to shot put. When the thrower exerts a force on the shotput, this equals the opposite of what was applied when the shot was fired.
How does Newton’s 2nd law apply to sports?
Tennis players apply Newton’s 2nd law to sports by applying a force on the ball and equal and opposite forces to the racket.
How is Newton’s third law used in sport?
Newton’s third law states that the more force you use to hit a tennis ball, the more reaction force your arm receives from the racket. Sports injuries can happen when someone uses too much force when hitting a tennis ball.
Which game is example of law of inertia?
There is no law of inertia in the world.
How is Newton’s second law used in badminton?
Badminton is played with a shuttlecock. When the shuttlecock is in the air and it’s smashed, its force is equal to the acceleration times its mass. The harder you can hit it, the more force it will have.
What is projectile motion in badminton?
Badminton is a sport where an object is thrown and it makes contact with the racket. The object moves along a curved path under the force of gravity, so it needs to be controlled in order to make sure that you hit your opponent’s shuttlecock as close to their court as possible.
What role does friction play in badminton?
There are three types of friction in badminton: static, sliding and rolling. Static friction occurs when two objects are stuck together. Sliding or rolling friction happens when something is moving quickly across a surface. Fluid friction happens when water moves through the air
How is momentum used in volleyball?
Technique in Volleyball
The change in momentum a player feels when they spike the ball has the same value as the impulse exerted on the ball. When spiking a ball, players go from having momentum when moving their hand to smack the ball, to having less momentum when they follow through after hitting.
What is the action and reaction force of a volleyball player hitting volleyball?
When a volleyball player hit the ball, she exerts an upward action force. The ball also experiences an equal but opposite downward reaction force. These forces act on different objects, which can be seen in the video below.
## To Recap
The Law of motion is a physical law that describes the behavior of objects in motion. It states that an object will continue moving at a constant speed and direction unless acted upon by an external force.
In sports, laws of motion are used to describe how people move around the court or field.
Victor Holman
I am a sports analytics expert with an extensive background in math, statistics and computer science. I have been working in the field for over 10 years, and have published several academic articles. I am a sports analytics expert with an extensive background in math, statistics and computer science. I have been working in the field for over 10 years, and have published several academic articles. I also run a blog on sports analytics where I share my thoughts on the latest developments in this field. But I specially love Volleyball. LinkedIn | 1,727 | 8,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-10 | longest | en | 0.961845 |
https://www.askiitians.com/forums/Trigonometry/i-need-the-full-solution-of-the-questions-no-24-p_194390.htm | 1,709,182,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00624.warc.gz | 663,711,817 | 42,659 | I need the full solution of the questions no. 24 plese help me finding this solution
Shashank Singh
6 years ago
(sinx+cosx)^2= sin^2x+cos^2x + 2sinxcosx
a^2= 1+2sinx cosx
2sinxcosx= a^2-1............. (1)
now (sin^2x+cos^2x)^3= sin^6x + 3sin^4xcos^2x + 3cos^4xsin^2x + cos^6x
1= sin^6x + cos^6x +3sin^2xcos^2x(sin^2x+cos^2x)
1=sin^6x+cos^6x+3sin^2xcos^2x............(2)
putting the value from (1) in (2), we get
1= sin^6x+cos^6x+ 3/4(a^2-1)^2
therefore sin^6x+cos^6x= 1-3/4(a^2-1)^2............. [hence proved] | 258 | 511 | {"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} | 3.8125 | 4 | CC-MAIN-2024-10 | latest | en | 0.530937 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/10%3A_Approximate_Quantum__Mechanical_Methods/10.37%3A_Hybrid_Variational_Calculation_for_the_1D_Hydrogen_Atom_with_Delta_Function_Potential | 1,709,613,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00113.warc.gz | 180,984,211 | 32,227 | # 10.37: Hybrid Variational Calculation for the 1D Hydrogen Atom with Delta Function Potential
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
The following normalized trial wave function is used in a variational calculation for the energy of a one‐dimensional model of the hydrogen atom that postulates a delta‐function potential energy interaction between the electron and the proton. The variational parameter, $$\alpha$$, is a decay constant which controls the spatial extent of the wave function.
$\Psi (x, \alpha ) = \sqrt{ \alpha} exp( - \alpha |x|) \nonumber$
$\int_{- \infty}^{ \infty} \Psi (x, \alpha)^{2} dx~assume,~ \alpha > 0 \rightarrow 1 \nonumber$
The Hamiltonian energy operator in coordinate space in atomic units (h/2π = me = e = 1) is:
$H = \frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \Delta (x) \blacksquare \nonumber$
The problem this tutorial seeks to solve is that it is obvious that the coordinate wave function is unsuitable for the calculation of the expectation value for kinetic energy because it is not well‐behaved at x = 0. The wave function and its first and second derivatives are discontinuous at x = 0. In spite of this defect, calculation of the expectation value for potential energy presents no problem in coordinate space.
Therefore the plan is to calculate the potential energy in coordinate space and Fourier transform the coordinate wave function into momentum space, where the wave function is well behaved, for the calculation of kinetic energy. Then the energy will be minimized with respect to $$\alpha$$, yielding the following results: $$\alpha$$ = 1, E = ‐0.5.
Using several values for x and the optimum value of $$\alpha$$ it is shown that solving Schrödingerʹs equation in the coordinate representation for the energy gives the correct value except at the discontinuity, x = 0.
$x = 0~~~ \frac{-1}{2} \frac{d^2}{dx^2} \Psi (x, 1) - \Delta (x) \Psi (x, 1) = E \Psi (x, 1)~solve,~E \rightarrow 0 \nonumber$
$x = 2~~~ \frac{-1}{2} \frac{d^2}{dx^2} \Psi (x, 1) - \Delta (x) \Psi (x, 1) = E \Psi (x, 1)~solve,~E \rightarrow \frac{-1}{2} \nonumber$
$x = -1~~~ \frac{-1}{2} \frac{d^2}{dx^2} \Psi (x, 1) - \Delta (x) \Psi (x, 1) = E \Psi (x, 1)~solve,~E \rightarrow \frac{-1}{2} \nonumber$
## Calculation of Potential Energy in Coordinate Space
$V( \alpha ) = \int_{- \infty}^{ \infty} \Psi (x, \alpha ) - \Delta (x) \Psi (x, ]alpha ) dx~ \rightarrow - \alpha \nonumber$
## Fourier Transform of the Coordinate Space Wave Function into Momentum Space
$\Phi (p, \alpha ) = \int_{- \infty}^{ \infty} \frac{exp(-pix)}{ \sqrt{2 \pi}} \Psi (x, \alpha )dx ~\big|_{simplify}^{assume,~ \alpha > 0} \rightarrow \frac{ \sqrt{2} \alpha ^{ \frac{3}{2}}}{ \sqrt{ \pi} ( \alpha^2 + p^2)} \nonumber$
Before proceeding, we demonstrate that the momentum space wavefunction is normalized and well behaved.
$\int_{ \infty}^{- \infty} \Phi (p, \alpha )^2 dp~assume,~ \alpha > 0 \rightarrow 1 \nonumber$
The kinetic energy operator in momentum space for an electron is: $$\frac{p^2}{2} \blacksquare$$
Therefore, the kinetic energy is:
$T ( \alpha ) = \int_{- \infty}^{ \infty} \Phi (p, \alpha ) \frac{p^2}{2} \Phi (p, \alpha ) dp~ asssume,~ \alpha > 0 \rightarrow \frac{ \alpha ^2}{2} \nonumber$
Now the coordinate and momentum space calculations are combined and the total energy is minimized with respect to the variational parameter, $$\alpha$$.
$E ( \alpha ) = T ( \alpha ) + V ( \alpha ) \nonumber$
$\alpha = \frac{d}{d \alpha} E( \alpha ) = 0~solve,~ \alpha \rightarrow 1 \nonumber$
$E ( \alpha ) \rightarrow \frac{-1}{2} \nonumber$
The optimum coordinate and momentum wavefunctions are compared below: x = -6, -5.99 .. 6
The uncertainty principle is illustrated by displaying the coordinate and momentum wave functions for different values of the decay constant, $$\alpha$$. For $$\alpha$$ = 1.2 the spatial distribution contracts and the momentum distribution expands relative to the optimum value $$\alpha$$ =1. In other words, less uncertainty in position requires greater uncertainty in momentum.
For $$\alpha$$ = 0.7 the spatial distribution expands and the momentum distribution contracts. More uncertainty in position leads to less uncertainty in momentum.
Numerically the Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum must be greater than or equal to 0.5 in atomic units.
$\Delta x = \sqrt{ \langle x^2 \rangle - \langle x \rangle^2} \nonumber$
$\Delta p = \sqrt{ \langle x^2 \rangle - \langle x \rangle^2} \nonumber$
$\Delta x \Delta p \geq \frac{h}{2} \nonumber$
The momentum wave function is now used to verify that the uncertainty principle is satisfied.
$\begin{pmatrix} Operator & Coordinate~Space & Momentum~Space\\ position & x \blacksquare & i \frac{d}{dp} \blacksquare\\ momentum & \frac{1}{i} \frac{d}{dx} \blacksquare & p \blacksquare \end{pmatrix} \nonumber$
$x_{ave} = \int_{- \infty}^{ \infty} \Phi (p, 1) i \frac{d}{dp} \Phi (p, 1) dp \rightarrow 0 \nonumber$
$x2_{ave} = \int_{- \infty}^{ \infty} \Phi (p, 1) - \frac{d}{dp} \Phi (p, 1) dp \rightarrow \frac{1}{2} \nonumber$
$p_{ave} = \int_{- \infty}^{ \infty} \Phi (p, 1)^2 dp \rightarrow 0 \nonumber$
$p2_{ave} = \int_{- \infty}^{ \infty} p^2 \Phi (p, 1)^2 dp \rightarrow 1 \nonumber$
$\sqrt{x2_{ave} - x_{ave}^2} \sqrt{p2_{ave} - p_{ave}^2} = 0.707 \nonumber$
## Concluding Remarks
This hybrid variational calculation on a one‐dimensional model of the hydrogen atom using a delta function potential interaction between the proton and electron yields the correct ground state energy. The model is composed of two peculiar elements: a trial wave function that has discontinuity at x = 0 and a potential energy interaction that is zero everywhere except at x = 0.
In Molecular Quantum Mechanics, 3rd ed., p. 43 Atkins and Friedman list the following criteria for an acceptable wave function.
1. It must be single valued (strictly Ψ*Ψ must be single valued).
2. It must not be infinite over a finite range.
3. It must be continuous everywhere.
4. It must have a continuous first derivative, except at ill‐behaved regions of the potential.
Ψ(x,α) does not satisfy criterion 3, making it impossible to calculate kinetic energy in the coordinate representation. However, it does satisfy criterion 4, so it is possible to calculate potential energy in the coordinate representation. Φ(p,α) is well‐behaved for all values of p. It is used to calculate kinetic energy, but it is not suitable for the calculation of potential energy. Thatʹs why a hybrid variational calculation was used.
This page titled 10.37: Hybrid Variational Calculation for the 1D Hydrogen Atom with Delta Function Potential is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 2,363 | 7,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.489216 |
https://www.numbersaplenty.com/9978599 | 1,611,387,804,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00672.warc.gz | 926,282,230 | 3,245 | Search a number
9978599 is a prime number
BaseRepresentation
bin100110000100…
…001011100111
3200202222001202
4212010023213
510023303344
6553513115
7150550061
oct46041347
920688052
109978599
1156a6084
12341279b
1320b4bc7
14147a731
15d2194e
hex9842e7
9978599 has 2 divisors, whose sum is σ = 9978600. Its totient is φ = 9978598.
The previous prime is 9978587. The next prime is 9978601. The reversal of 9978599 is 9958799.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 9978599 - 24 = 9978583 is a prime.
It is a Sophie Germain prime.
Together with 9978601, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (9978799) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4989299 + 4989300.
It is an arithmetic number, because the mean of its divisors is an integer number (4989300).
Almost surely, 29978599 is an apocalyptic number.
9978599 is a deficient number, since it is larger than the sum of its proper divisors (1).
9978599 is an equidigital number, since it uses as much as digits as its factorization.
9978599 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 1837080, while the sum is 56.
The square root of 9978599 is about 3158.8920526033. The cubic root of 9978599 is about 215.2896690459.
The spelling of 9978599 in words is "nine million, nine hundred seventy-eight thousand, five hundred ninety-nine". | 498 | 1,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-04 | latest | en | 0.874583 |
https://onlinejudge.org/board/viewtopic.php?f=5&t=8221&p=47340 | 1,611,103,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00316.warc.gz | 493,456,049 | 12,284 | ## 412 - Pi
Moderator: Board moderators
Jan
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Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Contact:
I think the problem number is wrong. It should be 412 - PI. However, if you are trying to solve 412, you can use gcd method.
Ami ekhono shopno dekhi...
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thomas1016
New poster
Posts: 19
Joined: Mon May 29, 2006 4:12 pm
### 412 WA
It seems alright but I got WA.
Code: Select all
``````#include <iostream>
#include <cmath>
//#include <ctype>
#include <iomanip>
using namespace std;
int gcd(int,int);
int main(void){
int a[50];
int k,u,j,t,m;
while(cin>>k){
if(k==0){
break;
}
m=0;
t=0;
for(u=0;u<k;cin>>a[u],u++);
// for(u=0;u<k;cout<<a[u]<<" ",u++);
for(u=0;u<k-1;u++){
for(j=u+1;j<k;j++){
m++;
// cout<<" u="<<u<<" j="<<j<<endl;
// cout<<" a[]="<<a[u]<<" a[]="<<a[j]<<endl;
// cout<<" gcd"<<gcd(a[u],a[j])<<endl;
if(gcd(a[u],a[j])==1){
t++;
}
}
}
// cout<<endl<<m<<" "<<t<<endl;
// cout<<sqrt(6*m/t)<<endl;
if(t==0){
cout<<"No estimate for this data set."<<endl;
continue;
}
cout.setf(ios::fixed);
cout.precision(6);
cout<<sqrt(6*m/t)<<endl;
}
// system("pause");
return 0;
}
int gcd(int a,int b){
int tmp;
if(a>=b){tmp=a;
a=b;
b=tmp;
}
while(b%a!=0){
// cout<<" a="<<a<<" b="<<b<<endl;
b=b%a;
tmp=a;
a=b;
b=tmp;
if(a>=b){tmp=a;
a=b;
b=tmp;
}
// cout<<"after a="<<a<<" b="<<b<<endl;
}
return a;
}``````
New poster
Posts: 45
Joined: Sun Jun 26, 2005 6:21 am
Contact:
### pi 412(WA)
Code: Select all
``````#include <stdio.h>
#include <cmath>
//using namespace std;
int gcd(int m,int n)
{
int t;
if (m<n)
{
t=m;
m=n;
n=t;
}
while (n!=0)
{
t=m;
m=n;
n=t%n;
}
return m;
}
int main()
{
int n,i,j,angka[100]={0},counter,counter2;
scanf("%d",&n);
while (n!=0)
{
counter = counter2 = 0;
for (i=1;i<=n;i++)
scanf("%d",&angka[i]);
for (i=1;i<=n-1;i++)
{
for (j=i+1;j<=n;j++)
{
counter2++;
if (gcd(angka[i],angka[j])==1)
{
counter++;
// printf("%d %d\n",angka[i],angka[j]);
}
}
}
// printf("%d %d\n",counter,counter2);
if (counter==0) puts("No estimate for this data set.");
else
printf("%.6Lf\n",(sqrt(6.0*counter2/counter)));
scanf("%d",&n);
}
return 0;
}``````
little joey
Guru
Posts: 1080
Joined: Thu Dec 19, 2002 7:37 pm
You say you've read the other threads for this problem? So why don't you post in on of them, but create yet another one?
To keep the forum useable for everyone, the number of threads per problem should be as low as possible. If there are tens of threads per problem, nobody can find anything.
New poster
Posts: 45
Joined: Sun Jun 26, 2005 6:21 am
Contact:
Code: Select all
``````#include <stdio.h>
#include <cmath>
//using namespace std;
int gcd(int m,int n)
{
int t;
if (m<n)
{
t=m;
m=n;
n=t;
}
while (n!=0)
{
t=m;
m=n;
n=t%n;
}
return m;
}
int main()
{
int n,i,j,angka[100]={0},counter,counter2;
scanf("%d",&n);
while (n!=0)
{
counter = counter2 = 0;
for (i=1;i<=n;i++)
scanf("%d",&angka[i]);
for (i=1;i<=n-1;i++)
{
for (j=i+1;j<=n;j++)
{
counter2++;
if (gcd(angka[i],angka[j])==1)
{
counter++;
// printf("%d %d\n",angka[i],angka[j]);
}
}
}
// printf("%d %d\n",counter,counter2);
if (counter==0) puts("No estimate for this data set.");
else
printf("%.6Lf\n",(sqrt(6.0*counter2/counter)));
scanf("%d",&n);
}
return 0;
}``````
payton1
New poster
Posts: 5
Joined: Mon Oct 02, 2006 10:31 pm
### Using c++ to write 412
Anybody can tell me in c++ how to show the '0' in 3.000000?
pankaj trivedi
New poster
Posts: 11
Joined: Wed Jun 21, 2006 5:11 pm
Contact:
Though i am not very sure about your question. Are you looking for this
Code: Select all
``````#include<iostream>
#include<iomanip>
using namespace std;
main()
{
float x=3;
cout.setf(ios::fixed, ios::floatfield);
cout.precision(7);
cout<<x;
}
``````
Anything other than Accepted is irritating,even Presentation Error
http://acm.uva.es/problemset/usersjudge.php?user=40301
payton1
New poster
Posts: 5
Joined: Mon Oct 02, 2006 10:31 pm
thank you,it solve my problem
srivatsan r
New poster
Posts: 2
Joined: Sat Mar 17, 2007 8:38 am
### WA 412
im getting the correct output for all the test cases given in the question..but still im getting a WA..pls help me with my code..
#include<iostream.h>
#include<stdlib.h>
#include<math.h>
int cf(int a,int b)
{
while(a!=0&&b!=0)
{
if(a>b)
a=a%b;
else
b=b%a;
if(a==1||b==1)
return 0;
}
return 1;
}
int main()
{
int n;
int i,j;
double count=0,c=0;
int a[50];
double pi,y,z;
while(cin>>n)
{
if(n==0)
exit(0);
if(n>1&&n<=50)
{
for(i=0;i<n;i++)
cin>>a;
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
c++;
if(!cf(a,a[j]))
count++;
}
}
if(count>0)
{
z=(c/count);
y=6*z;
pi=sqrt(y);
cout<<pi<<"\n";
}
else
cout<<"No estimate for this data set.\n";
}
else
exit(0);
n=0;
i=0;
j=0;
count=0.0;
c=0.0;
pi=0.0;
}
return 0;
}
Jan
Guru
Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Contact:
Your code doesnot even pass the samples! And you are printing using 'cout' (without any precision setting). So, if the output is '3.000000', your code will prodce '3'.
Hope these help.
Ami ekhono shopno dekhi...
HomePage
srivatsan r
New poster
Posts: 2
Joined: Sat Mar 17, 2007 8:38 am
Thank u..now accepted.
stalf
New poster
Posts: 7
Joined: Fri Apr 06, 2007 7:57 pm
i'm getting wa for this code...
also tried what they say on other posts like not counting repeated numbers and using double. Still don't work..
Code: Select all
``````#include <iostream>
#include <math.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
int mdc(int a,int b){
if(b>a)
return mdc(b,a);
if(a%b == 0)
return b;
return mdc(b,a%b);
}
int main(){
int n,m,count;
double pi;
bool first = true;
while(cin>>n){
if(n==0)
break;
if(first)
first = false;
else
cout<<endl;
int spaces[n];
count = 0;
for(int i=0;i<n;i++){
cin>>spaces[i];
}
int size = sizeof(spaces)/sizeof(int);
sort(spaces, spaces + size);
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(mdc(spaces[i],spaces[j]) == 1){
count ++;
if(i>0 && spaces[i]==spaces[i-1])
count--;
else if(j>0 && spaces[j]==spaces[j-1])
count--;
}
}
}
pi = 0;
if(count != 0){
double pair = n*(n-1)/2;
pi = sqrt(6*pair/count);
}
if(pi==0)
printf("No estimate for this data set.");
else
printf("%.6lf",pi);
}
}
``````
any help is appreciated
Jan
Guru
Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Contact:
Try the set below
Input:
Code: Select all
``````14
11701
31316
20671
5786
12263
4313
24355
31185
20053
912
10808
1832
20945
4313
0``````
Output:
Code: Select all
``2.833622``
Hope it helps.
Ami ekhono shopno dekhi...
HomePage
stalf
New poster
Posts: 7
Joined: Fri Apr 06, 2007 7:57 pm
hi there,
i got the output you said if i comment the following lines:
Code: Select all
``````if(i>0 && spaces[i]==spaces[i-1])
count--;
else if(j>0 && spaces[j]==spaces[j-1])
count--;``````
but this way, if the set is 2 2 3 4 5, i'll consider combinations like 2,5 twice... Is this right?
Anyways, I tried to submit it without these lines and also got wa.
Thanks for the help
stalf
New poster
Posts: 7
Joined: Fri Apr 06, 2007 7:57 pm
hey
I thought I had already submitted the code without these two lines, but I hadn't. Now got ac.
Thanks a lot.
Last edited by stalf on Sat Apr 21, 2007 8:33 pm, edited 1 time in total. | 2,520 | 7,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-04 | latest | en | 0.384952 |
https://thejediwife.com/makeup/how-many-grams-are-in-1-mole-of-kmno4.html | 1,657,005,234,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00113.warc.gz | 598,991,353 | 18,000 | # How many grams are in 1 mole of KMnO4?
Contents
1 mole is equal to 1 moles KMnO4, or 158.033949 grams.
## How many moles are in KMnO4?
Explanation: And moles of potassium permanganate = 23.7⋅g158.03⋅g⋅mol−1 ≅16⋅mol .
## How many grams are in a molecule of KMnO4?
The molar mass of KMnO4 is 158.034 g mol1. One mole of any substance contains 6.022×1023 atoms/molecules.
## How do you convert 1 mole to grams?
In order to convert the moles of a substance to grams, you will need to multiply the mole value of the substance by its molar mass.
## How many grams are in 1.70 moles of KMnO4?
1.70 moles KMnO4 = 269g KMnO4 (molar mass KMnO4 = 158.0g)
## How many moles are in 17.7 g of KMnO4?
Mole practice
12)Find the formula mass of KMnO4 158.00 g/mol
13)Find the formula mass of Mg3(PO4)2 262.79 g/mol
14)Find the molecular mass of C12H22O11 342.23 g/mol
15) How many moles are in 17.7 g of KMnO4? 0.112 mol KMnO4
## How do you convert grams to moles calculator?
To correctly estimate the number of moles, n , of a substance of a specific mass, m , (in grams), you need to follow the grams to moles formula: n = m / M , where, M is the molar mass of this material.
THIS IS EXCITING: Best answer: Does vitamin C cream exfoliate?
## What is the molar mass of Mn?
So, 10 mole of water will weigh (18×10) = 180g.
## How do I calculate moles?
How to find moles?
1. Measure the weight of your substance.
2. Use a periodic table to find its atomic or molecular mass.
3. Divide the weight by the atomic or molecular mass.
4. Check your results with Omni Calculator.
## How many moles of feso4 react with one mole of KMnO4?
Thus 5 moles of iron sulphate reats with 1 mole of potassium permanganate.
## How many grams are in 1 mole of HCl?
Thus, for hydrochloric acid, 1 mole is equal to 36.5 grams of HCl.
## How many grams are in 5 moles?
Therefore, 5 moles × 32.0 g/mol = 160 grams is the mass (m) when there are 5 moles of O2 .
## How much is a mole?
A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance. | 696 | 2,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-27 | latest | en | 0.862641 |
https://mathoverflow.net/questions/334186/does-int-02-pi-ei-thetat-phitn-dt-0-forall-n-in-mathbbn | 1,638,925,195,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00375.warc.gz | 448,691,928 | 29,787 | # Does $\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt=0$ $\forall \; n\in\mathbb{N}_0$ imply $\phi$ periodic?
PROBLEM. Let $$\theta(t)$$ and $$\phi(t)$$ be two real analytic non-constant functions $$[0,2\pi]\rightarrow \mathbb{R}$$. I am trying to prove the following claim
If the integral $$\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt=0$$ for all $$n\in\mathbb{N}_0$$ than the first derivative $$\theta'$$ and $$\phi$$ are periodic of common period $$2\pi/l$$ with $$1\neq l\in\mathbb{N}$$.
Note that this is equivalent to $$F(\lambda):=\int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt=0$$ for all $$\lambda \in \mathbb{R}$$. In fact, $$F(\lambda)$$ is analytic in $$\lambda$$ and its being constantly equal to 0 is equivalent to the vanishing of all its derivatives $$F^{(n)}(0)=\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt$$. Geometrically this means that the curve obtained by integrating the (tangent) vector function $$(\cos(\theta+\lambda\phi),\sin(\theta+\lambda\phi))$$ over $$[0,2\pi]$$ is closed $$\forall \lambda$$.
Just in case, a back-up less general claim for which I would like to see a clean solution is
If, in the hypotesis above, $$\phi$$ is a polynomial, then $$\phi$$ is constantly $$0$$.
OBSERVATION. If $$\theta'$$ and $$\phi$$ are periodic of common period $$\frac{2\pi}{l}$$ with $$1\neq l \in \mathbb{N}$$ and $$\int_0^{\frac{2\pi}{l}} e^{i\theta}\neq 0$$ then the converse implication is true. In fact, in this setting $$\theta=c\cdot t+\theta_p(t)$$ with $$c=\frac{2\pi}{l}(\theta(\frac{2\pi}{l})-\theta(0))$$ and $$\theta_p$$ periodic of period $$\frac{2\pi}{l}$$. Then \begin{align} \int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt &=& \sum_{j=0}^{l-1} \int_{j \frac{2\pi}{l}}^{(j+1) \frac{2\pi}{l}} e^{i(c\cdot t+\theta_p(t)+\lambda\phi(t))} dt \\ &=& \sum_{j=0}^{l-1} e^{i\cdot j \cdot \frac{2\pi}{l}} \int_{0}^{\frac{2\pi}{l}} e^{i(c\cdot t+\theta_p(t)+\lambda\phi(t))} dt, \end{align} where the last equality is obtained by repetedly applying the substitution $$t'=t-\frac{2\pi}{l}$$. Since we know $$\sum_{j=0}^{l-1} e^{i\cdot j \cdot \frac{2\pi}{l}} \int_{0}^{\frac{2\pi}{l}} e^{i\theta(t)}dt=\int_0^{2\pi} e^{i\theta(t)} dt=0$$ then also the integral above must be $$0$$. In the following picture the curve associated to $$\theta(t)=t + \cos( 12 t)$$ deformed in the direction $$\cos(3 t)$$. In this case $$l=3$$ and the curve is closed $$\forall \lambda$$.
$\theta(t)=t + \cos( 12 t)$ deformed in the direction $$\cos( 3 t)$$. In this case $$l=3$$ and the curve is closed $$\forall \lambda$$." />
IDEA. If $$\theta$$ monotone one can substitute $$s=\theta(t)$$ in the integral and get $$\int_{\theta(0)}^{\theta(2\pi)} e^{i s} \frac{(\phi(\theta^{-1}(s)))^n}{\theta'(\theta^{-1}(s))} ds=0.$$ In this case the idea behind the hypotesis becomes apperent: $$\phi(\theta^{-1}(s))$$ is periodic of non-trivial period iff $$\phi$$ and $$\theta'$$ have the common period property. It seems here that looking at the Fourier expansion of our functions on $$[\theta(0),\theta(2\pi)]$$ could be a good idea: the condition we have means indeed that, $$\forall n$$, the first harmonic of the function $$\frac{(\phi(\theta^{-1}(s)))^n}{\theta'(\theta^{-1}(s))}$$ is $$0$$. Fourier coefficients of a product are obtained by convolutions and therefore the condition above becomes, $$\forall n$$: $$\sum_{k_n=-\infty}^{+\infty} \sum_{k_{n-1}} ... \sum_{k_{2}}\sum_{k_{1}} \widehat{\frac{1}{\theta'}}(1-\sum_{i=1}^{n} k_i) \prod_{i=1}^{n} \widehat{\phi}(k_i)=0.$$ Is this approach viable? Can one from here exploit the fact that a function is periodic of non-trivial period iff there exists $$k$$ such that only harmonics multiple of $$k$$ are different from 0? Other way round, do non-zero harmonics of coprime orders imply a contradiction with our constraints? As for a toy example, if $$\theta(t)=t$$,$$\theta'(s)=1$$ and $$\phi(s)=\cos(2s)+\cos(3s)$$ already $$\widehat{f^2}(1)= 2 \widehat{f}(3)\widehat{f}(-2) \neq 0$$; in the general setting interaction of coefficients is not straightforward.
NOTE: This question originated from Orthogonality relation in $L^2$ implying periodicity. As suggested in the comments to the previous post, since the target of the question changed over time and edits were major, here I hope I gave a clearer and more consistent presentation of my problem.
• You do not need to move 6 points in a highly nontrivial way. Just rotate 3 opposite pairs in some independent ways for a while recovering the original star after 30 degree rotation and then move the 2 equilateral triples in some independent ways arriving at the 60 degree rotation. From pairs $\ell$ can be only $2$, but from triples it can be only $3$, so no $\ell$ exists in the end. Jun 17 '19 at 22:03
I missed the real analyticity condition (my comment makes perfect sense for $$C^\infty$$ though), so let's move points in a fancy way to satisfy it.
First, observe that if $$a_0,a_1,a_2$$ are positive reals close to $$1$$, then there exist unique $$\theta_1\approx \frac {2\pi} 3$$ and $$\theta_2\approx \frac {4\pi}3$$ such that $$a_0+a_1e^{i\theta_1}+a_2e^{i\theta_2}=0$$. Moreover, $$\theta_{1,2}$$ are real analytic functions of $$a_{0,1,2}$$ in some neighborhood of $$1$$. This is just the implicit function theorem.
Now choose your favorite $$2\pi$$-periodic real analytic function $$F(\tau)$$ with uniformly small derivative that is not periodic with any smaller period (say, $$\varepsilon\cos\tau$$) and put $$t(\tau)=\tau+F(\tau)$$. Then $$\tau$$ is uniquely determined by $$t$$ and the dependence is real analytic as well.
Next define $$\theta_{1,2}(\tau)$$ by $$\theta_j(\tau)\approx \frac {2\pi j}3$$ such that $$t'(\tau)+t'(\tau+\tfrac{2\pi}3)e^{i\theta_1(\tau)}+t'(\tau+\tfrac{4\pi}3)e^{i\theta_2(\tau)}=0$$ Everything is real analytic so far.
By uniqueness, we must have the relations $$\theta_1(\tau+\frac{2\pi}{3})=\theta_2(\tau)-\theta_1(\tau)$$ and $$\theta_1(\tau+\frac{4\pi}{3})=2\pi -\theta_2(\tau)$$. Thus $$\theta_1(\tau)+\theta_1(\tau+\frac{2\pi}{3})+\theta_1(\tau+\frac{4\pi}{3})=2\pi$$. This implies that there exists a real analytic $$\Theta(\tau)$$ such that $$\Theta(\tau+\frac{2\pi}3)=\Theta(\tau)+\theta_1(\tau)$$ (just divide the Fourier coefficients by appropriate numbers to get the periodic part and add $$\tau$$; note that the identity for $$\theta_1$$ implies that $$\widehat\theta_1(3k)=0$$ for $$k\ne 0$$, so no division by $$0$$ will be encountered). Then, automatically, $$\Theta(\tau+2\pi)=\Theta(\tau)+2\pi$$ and $$\Theta(\tau+\frac{4\pi}3)=\Theta(\tau+\frac{2\pi}3)+\theta_1(\tau+\frac{2\pi}3) \\ =\Theta(\tau)+\theta_1(\tau)+\theta_1(\tau+\frac{2\pi}3)=\Theta(\tau)+\theta_2(\tau)$$
Hence, we have the identity $$\sum_{j=0}^2 t'(\tau+\tfrac{2\pi j}3)e^{i\Theta(\tau+\frac{2\pi j}3)}=0$$ We can now pick up any $$\frac{2\pi}3$$-periodic real analytic function $$\Psi(\tau)$$, multiply the terms by the corresponding values of $$\Psi^n$$ (they are equal), integrate in $$\tau$$ from $$0$$ to $$\frac{2\pi}3$$, and use the standard change of variable formula to get $$\int_0^{2\pi} e^{i\Theta(\tau(t))}\Psi(\tau(t))^n\,dt=0$$ but $$\psi(t)=\Psi(\tau(t))$$ is no longer $$\frac{2\pi}3$$ periodic in $$t$$ because the composition kills periodicity.
As I said from the beginning, "there are many fancy ways to move six (well, even three) points around the circle and keep the sum balanced".
• Cool and nicely explained. Thank you very much! I am gonna update the post with some drawings that present your solution :) Jun 19 '19 at 16:41
• @Leonardo You are cordially welcome. :-) Jun 19 '19 at 16:50
Another way of constructing a counterexample and a picture.
Consider a curve allowing a parametrization as trigonometric polynomial, that is $$\gamma(t)=\Bigl(a_0+\sum_{j=1}^N a_j \cos(j\cdot t + \alpha_j),b_0+\sum_{j=1}^N b_j \cos(j\cdot t + \beta_j)\Bigl),$$ with $$N\in \mathbb{N}$$, $$(a_j),(b_j)\in \mathbb{R}^{N+1}$$ and $$(\alpha_j),(\beta_j) \in \mathbb{R}^N$$, and such that $$\|\gamma'\|>0$$. Such curves are closed and satisfy for any natural number $$l>N$$ $$\int_0^{2\pi} \gamma'(t) \cos(h\cdot l\cdot t) dt =(0,0), \; \forall h \in \mathbb{N}.$$ Using the complex notation we can write $$\gamma'(t)$$ as $$v(t)\mathrm{e}^{i\theta(t)}$$ where $$v(t)=\|\gamma'(t)\|$$ is the speed of $$\gamma$$ and $$\theta$$ is the turning angle associated to the parametrization. After reparametrizing via $$t(s)$$ to the arc-length (always possible as long as the curve is regular) we obtain, possibly scaling our curve to a length of $$2\pi$$, $$\int_0^{2\pi} \mathrm{e}^{i\theta(t(s))} \cos(h\cdot l\cdot t(s)) ds =0, \; \forall h \in \mathbb{N}.$$ Since $$\cos(l\cdot t(s))^n$$ can be rewritten as a linear combination of terms of the form $$\cos(h\cdot l\cdot t(s))$$ we also get $$\int_0^{2\pi} \mathrm{e}^{i\theta(t(s))} \cos(l\cdot t(s))^n ds =0, \; \forall n \in \mathbb{N},$$ and in general the function $$\cos(l\cdot t(s))$$ won't be periodic of any period smaller than $$2\pi$$.
In the next figure the turning angle $$\theta$$ of a trigonometric curve of degree $$3$$ is linearly changed to $$\theta+\lambda\phi$$ with $$\phi(t)=\mathrm{e}^{\cos(4t)}+2\cos(4t)$$ while the curve remains closed (it is possible to prove that any composition with $$L^1$$ functions mantains our property). From top left to bottom right $$\lambda$$ goes from $$0$$ to $$0.8$$ by $$0.1$$ increments. | 3,180 | 9,364 | {"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": 128, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.750696 |
https://cstheory.stackexchange.com/questions/17016/measurable-language-which-is-not-omega-regular?noredirect=1 | 1,653,771,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00392.warc.gz | 230,640,460 | 67,481 | Measurable language which is not $\omega$-regular
Let $\Sigma$ be a finite alphabet and let $\Sigma^\omega$ be the set of all infinite words over $\Sigma$. Consider $$d(x,y):=2^{-\min(n \in \Bbb N_0:x_n\neq y_n)}$$ to be the metric on $\Sigma^\omega$ which makes the latter being the Cantor space. Denote by $\mathscr B(\Sigma^\omega)$ the Borel $\sigma$-algebra of this space, and let's call its elements measurable languages. It follows from Proposition 6 here that any $\omega$-regular language is measurable. I am looking for an example of a measurable language which is not $\omega$-regular, so any hints are appreciated.
A simple class of examples can be found by considering singleton languages $\{w\}$. These are measurable (Let $C_n(w)$ be the set of words agreeing with $w$ up to the $n$-th letter, then $\{w\}$ is the intersection of all $C_n(w)$). However, unless $w$ is eventually periodic, the language is not $\omega$-regular.
For a concrete example, consider the word $w=ababbabbba\dots$. If $w$ is accepted by a nondeterministic Büchi automaton $A$ with $k$ states, then while parsing the $n$-th block of $b$s, with $n>k$, $A$ needs to visit some state twice, so we can construct another word accepted by $A$ by pumping this block suitably. Therefore no automaton recognizes $\{w\}$.
• One can also obtain non-regular languages at arbitrary Borel levels, by adding a nonperiodic $w$ to a regular language $L$ (with $w\notin L$). The resulting language $L'=L\cup\{w\}$ will always be non-regular, because otherwie $L'\setminus L=\{w\}$ would also be regular. Mar 25, 2013 at 13:09
• That's a very insightful answer, thank you very much. Can you suggest many any "canonical" references dealing with Borel structures over $\Sigma^\omega$? I've only seen the paper I cited in OP, but it only touches upon a few number of concepts. Clearly, $\Sigma^\omega$ is just a Cantor space, but I'm interested how the topological/measure-theoretical properties of it are related to those of regularity etc. For example, if I'm not mistaken, regular languages (embedded in $\Sigma^\omega$) are precisely open sets?
– Ilya
Mar 25, 2013 at 15:19
• Not sure about "canonical", but two starting points would be (1) Halpern and Schneider, Defining Liveness (contains the characterization safety = closed and liveness = dense), and (2) numerous papers by Alain Finkel, in particular "Borel Hierarchy and Omega Context Free Languages". Mar 25, 2013 at 15:56
• Cool, I'll take a look into those sources - perhaps references there will also steer me. Btw, could you please use this @username stuff? Otherwise it does not appear in the inbox, I passed by your comment by coincidence
– Ilya
Mar 25, 2013 at 16:00
• Nitpick: it's Bowen Alpern and Fred Schneider. Joseph Halpern did a lot of great things but didn't contribute to this paper.
– Kai
Dec 13, 2019 at 9:29 | 778 | 2,860 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.899296 |
http://sci4um.com/post-55311---Mon-Jul-04--2005-4-39-pm.html | 1,537,879,025,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00024.warc.gz | 216,650,042 | 6,880 | Search Memberlist Usergroups
Page 1 of 1 [4 Posts]
Author Message
amcwill417
Joined: 04 May 2005
Posts: 65
Posted: Sun Jul 03, 2005 2:31 am Post subject: Representation of Integers
Eah positive integer can be uniquely represented by three ordered smaller
integers per the following examples: 11 = (2,1,2), 18 = (2,3,1), 31 =
(3,2,0). What is the rule?
Alex
Timothy Little
science forum Guru Wannabe
Joined: 30 May 2005
Posts: 295
Posted: Sun Jul 03, 2005 4:58 am Post subject: Re: Representation of Integers
amcwill417 wrote:
Quote: Eah positive integer can be uniquely represented by three ordered smaller integers per the following examples: 11 = (2,1,2), 18 = (2,3,1), 31 = (3,2,0). What is the rule?
Although there are infinitely many such rules, the first one that
sprang to mind worked:
N is represented by (a, b, c), where
a = floor(N ^ (1/3)),
b = floor((N - a^3) ^ (1/2)),
c = N - a^3 - b^2.
Or more colloquially, a is the cube root, b is the square root of
what's left, and c is what's left after that.
The reason why it sprang to mind is that I thought to myself that the
most compact such representation would asymptotically have
max(a,b,c) ~= N^(1/3).
If you take out the largest possible a^3, you'd have a remainder on
the order of 3a^2, so taking out a largest possible square would be
useful. Then you'd be left with a linear remainder.
- Tim
amcwill417
Joined: 04 May 2005
Posts: 65
Posted: Mon Jul 04, 2005 1:25 am Post subject: Re: Representation of Integers
"Timothy Little" <tim-usenet@little-possums.net> wrote in message
news:slrndces37.soe.tim-usenet@soprano.little-possums.net...
Quote: amcwill417 wrote: Eah positive integer can be uniquely represented by three ordered smaller integers per the following examples: 11 = (2,1,2), 18 = (2,3,1), 31 = (3,2,0). What is the rule? Although there are infinitely many such rules, the first one that sprang to mind worked: N is represented by (a, b, c), where a = floor(N ^ (1/3)), b = floor((N - a^3) ^ (1/2)), c = N - a^3 - b^2. Or more colloquially, a is the cube root, b is the square root of what's left, and c is what's left after that. The reason why it sprang to mind is that I thought to myself that the most compact such representation would asymptotically have max(a,b,c) ~= N^(1/3). If you take out the largest possible a^3, you'd have a remainder on the order of 3a^2, so taking out a largest possible square would be useful. Then you'd be left with a linear remainder. - Tim
Yes. How do primes behave in this representation?
Alex
amcwill417
Joined: 04 May 2005
Posts: 65
Posted: Mon Jul 04, 2005 4:39 pm Post subject: Re: Representation of Integers
"amcwill417" <amcwill417@email.msn.com> wrote in message
news:b22ye.128\$4H2.1394@eagle.america.net...
Quote: "Timothy Little" wrote in message news:slrndces37.soe.tim-usenet@soprano.little-possums.net... amcwill417 wrote: Eah positive integer can be uniquely represented by three ordered smaller integers per the following examples: 11 = (2,1,2), 18 = (2,3,1), 31 = (3,2,0). What is the rule? Although there are infinitely many such rules, the first one that sprang to mind worked: N is represented by (a, b, c), where a = floor(N ^ (1/3)), b = floor((N - a^3) ^ (1/2)), c = N - a^3 - b^2. Or more colloquially, a is the cube root, b is the square root of what's left, and c is what's left after that. The reason why it sprang to mind is that I thought to myself that the most compact such representation would asymptotically have max(a,b,c) ~= N^(1/3). If you take out the largest possible a^3, you'd have a remainder on the order of 3a^2, so taking out a largest possible square would be useful. Then you'd be left with a linear remainder. - Tim Yes. How do primes behave in this representation? Alex What I have in mind here is this: 7 = (1,2,2), 11 = (2,1,2) and 13 =
(2,2,1) so that all permutations here yield primes. This is by no means
general but one can ask the question thus: Are there any other primes p =
(a,b,c) such that all permutations are also primes?
Alex
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# New Question Complex Numbers
0
334
1
+816
In the complex plane, what is the length of the diagonal of the square with vertices 4, 3+5i, -2+4i, and -1-i?
Dec 26, 2018
#1
+111328
+1
The two vertices we can look at are 3 + 5i and -1 - i
Taking the difference between these complex numbers produces a complex vector =
(3 + 5i) - ( -1 - i) = 4 + 6i
The length of the diagonal is the length of this vector =
√[ 4^2 + 6^2 ] = √ [ 16 + 36 ] = √ 52 = 2√13 units
Checking that this is correct the other two vertices are 4 + 0i and - 2 + 4i
Take the difference of these two complex numbers and we get
6 - 4i
And the length of the diagonal is √ [ 6^2 + (-4)^2 ] = √52 = 2√13 units
So....true!!!
Dec 26, 2018
edited by CPhill Dec 26, 2018 | 293 | 768 | {"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} | 3.984375 | 4 | CC-MAIN-2020-24 | latest | en | 0.703768 |
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1 – 50 of 174 | 1,061 | 4,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-22 | longest | en | 0.81286 |
https://percent.info/minus/70/how-to-calculate-910-minus-70-percent.html | 1,716,148,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00517.warc.gz | 406,444,620 | 2,692 | 910 minus 70 percent
This is where you will learn how to calculate nine hundred ten minus seventy percent (910 minus 70 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 910 minus 70 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 910 of something.
910
(100%)
70 percent means 70 per hundred, so for each hundred in 910, you want to subtract 70. Thus, you divide 910 by 100 and then multiply the quotient by 70 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(910 ÷ 100) × 70
= 637
We made a pink square that we put on top of the image shown above to illustrate how much 70 percent is of the total 910:
The dark blue not covered up by the pink is 910 minus 70 percent. Thus, we simply subtract the 637 from 910 to get the answer:
910 - 637
= 273
The explanation and illustrations above are the educational way of calculating 910 minus 70 percent. You can also, of course, use formulas to calculate 910 minus 70%.
Below we show you two formulas that you can use to calculate 910 minus 70 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
910 - ((910 × 70/100))
910 - 637
= 273
Formula 2
Number × (1 - (Percent/100))
910 × (1 - (70/100))
910 × 0.3
= 273
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
911 minus 70 percent
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Topic: [ap-stat] RE: Probability Help
Replies: 0
Search Thread: Advanced Search
Pat ballew Posts: 455 Registered: 12/3/04
[ap-stat] RE: Probability Help
Posted: Nov 15, 2004 10:45 AM
Plain Text Reply
Amber,
I would find the probability that none of the first three worked, and with
that information, you know the answer to the question you seek.
One of the rhythms of proability students need to pick up is that NOT NONE
is the same as AT LEAST ONE.
Pat Ballew
Lakenheath, UK
MathWords <a href="http://www.pballew.net/etyindex.html">http://www.pballew.net/etyindex.html</a>
<<a href="http://www.pballew.net/etyindex.html">http://www.pballew.net/etyindex.html</a>>
_____
From: Kandravy, Amber [<a href="mailto://Amber.Kandravy@fortbend.k12.tx.us]">mailto://Amber.Kandravy@fortbend.k12.tx.us]</a>
Sent: Monday, November 15, 2004 3:28 PM
To: for Teachers of AP Statistics
Subject: [ap-stat] Probability Help
Hello everyone,
I am working on a problem and I am stumped to find the answer.
The problem is this: There are a dozen old batteries in a junk box. 5 of
which are totally dead, but you don't know which ones specifically. You
start picking them one at at time and testing them. We are asked to find the
probabilty that at least one of the first three works.
Thank you for your time,
Amber Kandravy
Statistics/Algebra Teacher
George Bush High School
281-234-6060
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© The Math Forum at NCTM 1994-2018. All Rights Reserved. | 916 | 3,149 | {"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} | 3.15625 | 3 | CC-MAIN-2018-17 | latest | en | 0.779682 |
http://share.sis.org.cn/cdens1/author/cdens1/page/2/ | 1,576,246,977,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00140.warc.gz | 128,504,023 | 15,774 | Elephant Awareness Fair
3D is hosting an Elephant Awareness Fair, and any class that donates 150 RMB or more to support their chosen Elephant sanctuary is invited to attend their fair on April 30th.
Learning Update
This week we will take a look at balanced and unbalanced forces. Students will continue to explore using hands on activities where they are using the qualities of a scientist and engineer. Students will also dig a bit deeper using text to learn more about forces and motion.
Mathematicians will learn that different shapes (e.g., squares, rectangles, and rhombuses) have shared attributes that can fall within a larger category (parallelograms, quadrilaterals, and trapezoids). The will also look for shared attributes and learn to recognize polygons with sides that are equal. Finally, mathematicians will draw quadrilaterals that do not fit any subcategories.
This week, writers will continue writing their own adapted fairy tales. Students will revise as they draft. We will focus on the usage of dialogue, descriptive language (verbs and adjectives), and language structures in compound and complex sentences.
This week, readers will be continuing with reading fairy tales, focusing on the traits of characters. They will learn that characters have internal traits and external features/characteristics. Readers will learn to use these traits by applying them to the heroes and villains in their own fairy tales. If you are supporting your child with this work at home, then this anchor chart contains the type of vocabulary that we will be using:
Learning Update
Mathematicians will wrap up their work with generating and analyzing different types of data by drawing scaled picture graphs and scaled bar graphs. Then, students will get intensive practice with word problems, as well as hands-on investigation experiences with geometry and perimeter. Students will solve word problems, classify shapes based on their attributes, learn what a tessellation is, study perimeter and area, and end with a review of Grade 3 fundamental skills.
Readers will begin this week by categorizing the fairy tale stories that they are reading by identifying elements of a fairy tale (use of magic, types of characters, etc.). They will continue by studying the sequence that fairy tales often follow (opening > introduce the setting > introduce the characters > the problem > the solution > the ending > the lesson) and by then ‘mapping’ out the most important parts of the story.
As writers last week, we immersed ourselves in Fairy Tales, becoming more familiar with their commonalities. As we move forward this week, we will choose a Fairy Tale to adapt, by putting our own creative spin on a story already familiar to us. We will story tell and act out as we draft!
This week, students will discover qualities of a scientist by doing two different simple investigations. Students will make observations and or measurements of an object’s motion to provide evidence that a pattern can be used to predict future motion. The investigations will include creating a paper/recyclable airplane and measuring how far it flies, and also having a balloon travel across the classroom using a straw and string. It should be a fun week. Ask your kids what the qualities of a scientist are!
Learning Update
This week mathematicians will measure different items and create a line plot. They will also have to analyze the data given and answer questions about the information provided.
We are launching a new literacy unit this week! Readers will begin immersing themselves in fairy tales as well as “fractured” fairy tales, which are twists on our classical tales. Readers will read many familiar and new fairy tales and discuss the story elements specific to fairy tales. We will have guest readers, including Mrs. Kummerlin and Mr. Sheppard read us their favorite fairy tales.
Writers will use their knowledge and understanding about Fairy Tales to write adapted versions. This week they will explore many Fairy Tales and inquire into the specific writing elements of Fairy Tales. We encourage you to share with your child your knowledge about Fairy Tales in your own culture.
This week is an exciting one! Students will design a model to support an egg being dropped off the building. We will focus on the design process and thinking like an engineer. Students will use qualities such as observing, collecting data, making predictions and testing prototypes. It’s great fun and the kids learn a lot through the process.
Writing Celebration
This Wednesday all four third grade classrooms will celebrate our writing. 3A will showcase their Narrative Nonfiction work & their individual Persuasive Writing. Please join us to celebrate your child’s hard work and to celebrate the work of all SIS third grade students.
Writing Celebration – Wednesday, April 10th at 8:15am
Greetings Parents,
This Wednesday all four third grade classrooms will celebrate our writing. 3A will showcase their Narrative Nonfiction work & their individual Persuasive Writing. Please join us to celebrate your child’s hard work and to celebrate the work of all SIS third grade students.
Learning Update
We are starting a new unit this week! The central idea is “Scientists plan investigations to understand the forces and interactions in the world.” Our tentative lines of inquiry are; 1) qualities of a scientist, 2) balanced and unbalanced forces, 3) observable and measurable patterns. Our first provocations will be on Wednesday, so please ask your child questions about their experience.
In Module 6, we build on Grade 2 concepts about data, graphing, and line plots. We focus on generating and analyzing different types of data. By the end of the module, students are working with a mixture of scaled picture graphs, bar graphs, and line plots to problem solve using categorical and measurement data.
This week Mathematicians begin with collecting data and creating a bar graph. They will learn how to understand a table and create a bar graph from that information.
Readers this week will wrap up their current study with their research book clubs. They will practice reading like a non-fiction narrator, review non-fiction note taking and text structures as well as formulate questions with their partners and teams.
Learning Update
Last Minute Mystery Reader Opportunity for anytime this week!
In our classroom this week, we have the opportunity to participate in a student-favorite: Mystery Readers!
All week, we will host parents, grandparents, siblings, neighbors, etc. to bring in and read aloud a favorite picture book to the class. Students love when their family members are the Mystery Reader and all students benefit from an adult sharing their love of reading.
Because this is a mystery, all information must be kept TOP SECRET from your child, including the date you are coming and the picture book you will be reading. To make it fun I will ask for five clues about yourself. These clues will be read to the class before your arrival to build suspense and keep the students guessing about who our Mystery Reader will be!
Please send me an email (cdenson@sis.org.cn) if you would like to schedule a time to come in and read to 3A this week.
Let me know if you have any questions and I look forward to seeing you in class this year!
Sincerely,
Mr. Denson
Book Week!
Monday: SIS will Drop Everything and Read!
Tuesday: Meet with Author Eric Rohmann
Wednesday: Mo Willems Celebration and Door Decoration
Thursday: Read with our KC Reading buddies and meet Eric Rohmann again!
Friday: Book Character Dress up and Books Come Alive activity sponsored by PSA.
Learning Update
Mathematicians will begin to understand the concept of greater than and less than when comparing fractions. As they study and compare different fractions, students continue to reason about their size. They develop the understanding that the numerator or number of copies of the fractional unit does not necessarily determine the size of the fraction.
As writers this week we will be continuing to move through the writing process within our Narrative Nonfiction books. Our writing community is using so much creativity within this unit to both teach and entertain their reader! We can’t wait to share these with you once back from Spring break!
Readers will be celebrating book week by focusing on the love of reading, all that reading teaches us and the connections reading allows us to make. Reading will be mixed into every aspect of the day with special author visits, mystery readers, activities at recess and read alouds in class.
As inquires we will continue to explore ways to take action by sharing our learning with each other. We will have an in-house gallery walk where students can show off their hard work and enjoy the work of their friends in all of third grade.
Learning Update
Reminder: Field trip on Wednesday. Please help your child pack a lunch, water bottle and snack in a light backpack. They can also bring bug spray (or you can help put it on in the morning) and a hat if it’s sunny.
Readers will continue to work Literary Non Fiction. After reading, readers will share facts about their topic and ask questions in order to have a deeper understanding of the topic. Finally, they will present what they’ve learned about the topic with the class.
Writers are drafting and drawing within the writing process of their Literary Nonfiction books. We have used mentor texts to guide our decisions in making a plan and choosing a topic we are passionate about. There has been high interest around this genre of combining fiction and nonfiction elements…ask your child about the decisions they have made as authors so far within their books.
Mathematicians will continue to deepen their understanding that a fraction must be the same size but may not always have the same shape. Students will also demonstrated their knowledge of fractions including equivalent fractions by using number bonds, tape diagrams, a number line and physical models.
We continue to explore ways to take action for our current unit. Ask your child ways you can help at home and in your daily lives to support endangered animals. | 2,015 | 10,245 | {"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-2019-51 | latest | en | 0.948249 |
http://forum.philosophynow.org/viewtopic.php?f=26&t=22404&sid=eb2c7a1f1681997fd5577d7af8c0dd3f | 1,516,545,731,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00554.warc.gz | 131,809,141 | 9,237 | Patterns upon patterns
What is the basis for reason? And mathematics?
Moderators: AMod, iMod
Philosophy Explorer
Posts: 4533
Joined: Sun Aug 31, 2014 7:39 am
Patterns upon patterns
How do you get from these four numbers: 1 2 3 4 to 24?
How do you get from these four numbers: 2 3 4 5 to 48?
How do you get from these four numbers: 3 4 5 6 to 80?
Study them well.
PhilX
Arising_uk
Posts: 10665
Joined: Wed Oct 17, 2007 2:31 am
Re: Patterns upon patterns
Philosophy Explorer wrote:How do you get from these four numbers: 1 2 3 4 to 24?
Add one to the last number to get a successor and add one to each successive successor nineteen times.
How do you get from these four numbers: 2 3 4 5 to 48?
Add one to the last number to get a successor and add one to each successive successor forty two times.
How do you get from these four numbers: 3 4 5 6 to 80?
Add one to the last number to get a successor and add one to each successive successor seventy three times.
vegetariantaxidermy
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Re: Patterns upon patterns
Arising_uk wrote:
Thu Jul 13, 2017 10:46 pm
How do you get from these four numbers: 1 2 3 4 to 24?
Add one to the last number to get a successor and add one to each successive successor nineteen times.
I was thinking 1+2+3x4
Philosophy Explorer
Posts: 4533
Joined: Sun Aug 31, 2014 7:39 am
Re: Patterns upon patterns
Philosophy Explorer wrote:
Thu Jul 13, 2017 3:23 am
How do you get from these four numbers: 1 2 3 4 to 24?
How do you get from these four numbers: 2 3 4 5 to 48?
How do you get from these four numbers: 3 4 5 6 to 80?
Study them well.
PhilX
I didn't phrase the question "How do you get from the last number to the target number?" All four starting numbers in each set participate equally toward the target number.
Try again.
PhilX
vegetariantaxidermy
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Re: Patterns upon patterns
Towards idiot. And unless you have made these up yourself there's not a lot of point in posting them. You could just google them or get them out of your number puzzle book and post thread after thread of them.
Philosophy Explorer
Posts: 4533
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Re: Patterns upon patterns
vegetariantaxidermy wrote:
Thu Jul 13, 2017 11:55 pm
Towards idiot. And unless you have made these up yourself there's not a lot of point in posting them. You could just google them or get them out of your number puzzle book and post thread after thread of them.
Your frustration is showing. You won't find them on the internet as I did make them up. The only number puzzle book I would have is the one I write myself. Btw I don't figure you to be a mathematician so I wonder what attracts you here?
PhilX
vegetariantaxidermy
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Re: Patterns upon patterns
Philosophy Explorer wrote:
Fri Jul 14, 2017 12:13 am
vegetariantaxidermy wrote:
Thu Jul 13, 2017 11:55 pm
Towards idiot. And unless you have made these up yourself there's not a lot of point in posting them. You could just google them or get them out of your number puzzle book and post thread after thread of them.
Your frustration is showing. You won't find them on the internet as I did make them up. The only number puzzle book I would have is the one I write myself. Btw I don't figure you to be a mathematician so I wonder what attracts you here?
PhilX
Right. Just like you 'wrote' that thing you copy-pasted. Plagiarist. You are the biggest idiot I have ever encountered, and on the internet that's really saying something.
Philosophy Explorer
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Re: Patterns upon patterns
vegetariantaxidermy wrote:
Fri Jul 14, 2017 12:30 am
Philosophy Explorer wrote:
Fri Jul 14, 2017 12:13 am
vegetariantaxidermy wrote:
Thu Jul 13, 2017 11:55 pm
Towards idiot. And unless you have made these up yourself there's not a lot of point in posting them. You could just google them or get them out of your number puzzle book and post thread after thread of them.
Your frustration is showing. You won't find them on the internet as I did make them up. The only number puzzle book I would have is the one I write myself. Btw I don't figure you to be a mathematician so I wonder what attracts you here?
PhilX
Right. Just like you 'wrote' that thing you copy-pasted. Plagiarist. You are the biggest idiot I have ever encountered, and on the internet that's really saying something.
I defy you to find where I copy-pasted from and you can get Arising to help you out because I know it's not on the internet.
"You are the biggest idiot I have ever encountered." Can I put that on my resume.
PhilX
vegetariantaxidermy
Posts: 6155
Joined: Thu Aug 09, 2012 6:45 am
Location: Narniabiznus
Re: Patterns upon patterns
Not everything is on the internet. I mean, please. A rabidly nationalistic, barely literate yank moron, who has never before strung together more than one sentence on here, two at the most, suddenly writes an ENTIRE paragraph in English rather than bastardised-by-Americans English?? Give me a break. It's not hard to tell when someone hasn't written something him/herself.
Philosophy Explorer
Posts: 4533
Joined: Sun Aug 31, 2014 7:39 am
Re: Patterns upon patterns
vegetariantaxidermy wrote:
Fri Jul 14, 2017 12:51 am
Not everything is on the internet. I mean, please. A rabidly nationalistic, barely literate yank moron, who has never before strung together more than one sentence on here, two at the most, suddenly writes an ENTIRE paragraph in English rather than bastardised-by-Americans English?? Give me a break. It's not hard to tell when someone hasn't written something him/herself.
Prove it, visitor from outer space.
"...barely literate yank moron..." Can I add this one to my resume too?
PhilX
vegetariantaxidermy
Posts: 6155
Joined: Thu Aug 09, 2012 6:45 am
Location: Narniabiznus
Re: Patterns upon patterns
Philosophy Explorer wrote:
Fri Jul 14, 2017 1:02 am
vegetariantaxidermy wrote:
Fri Jul 14, 2017 12:51 am
Not everything is on the internet. I mean, please. A rabidly nationalistic, barely literate yank moron, who has never before strung together more than one sentence on here, two at the most, suddenly writes an ENTIRE paragraph in English rather than bastardised-by-Americans English?? Give me a break. It's not hard to tell when someone hasn't written something him/herself.
Prove it, visitor from outer space.
"...barely literate yank moron..." Can I add this one to my resume too?
PhilX
I don't need to prove it. Just the fact that you used the correct 'towards' not just once, but twice in the paragraph was proof enough, not to mention the fact that you didn't have a clue what most of it was about. This is how examiners can weed out plagiarists. Apart from obvious clues it's an almost instinctive thing. Not that I care, but you are such a little weasel who likes to be as annoying as possible that you deserve to be exposed as a plagiarist.
Arising_uk
Posts: 10665
Joined: Wed Oct 17, 2007 2:31 am
Re: Patterns upon patterns
Philosophy Explorer wrote:I didn't phrase the question "How do you get from the last number to the target number?" All four starting numbers in each set participate equally toward the target number.
Try again.
I don't see why as my answer is correct but if you wish to be picky.
Philosophy Explorer wrote:How do you get from these four numbers: 1 2 3 4 to 24?
Start with the first number and add one to get a successor(which will be the same number as in your pattern) and add one to each successive successor twenty two times.
How do you get from these four numbers: 2 3 4 5 to 48?
Start with the first number and add one to get a successor(which will be the same number as in your pattern) and add one to each successive successor forty five times.
How do you get from these four numbers: 3 4 5 6 to 80?
Start with the first number and add one to get a successor(which will be the same number as in your pattern) and add one to each successive successor seventy six times.
All the numbers are participating equally towards the result.
Philosophy Explorer
Posts: 4533
Joined: Sun Aug 31, 2014 7:39 am
Re: Patterns upon patterns
vegetariantaxidermy wrote:
Fri Jul 14, 2017 1:11 am
Philosophy Explorer wrote:
Fri Jul 14, 2017 1:02 am
vegetariantaxidermy wrote:
Fri Jul 14, 2017 12:51 am
Not everything is on the internet. I mean, please. A rabidly nationalistic, barely literate yank moron, who has never before strung together more than one sentence on here, two at the most, suddenly writes an ENTIRE paragraph in English rather than bastardised-by-Americans English?? Give me a break. It's not hard to tell when someone hasn't written something him/herself.
Prove it, visitor from outer space.
"...barely literate yank moron..." Can I add this one to my resume too?
PhilX
I don't need to prove it. Just the fact that you used the correct 'towards' not just once, but twice in the paragraph was proof enough, not to mention the fact that you didn't have a clue what most of it was about. This is how examiners can weed out plagiarists. Apart from obvious clues it's an almost instinctive thing. Not that I care, but you are such a little weasel who likes to be as annoying as possible that you deserve to be exposed as a plagiarist.
Is that in your secret files on me? My oh my. (btw your secret files on me are woefully inadequate).
"...you are such a little weasel who likes to be as annoying as possible..." I like to add this to my resume too. (if I'm so annoying, then why don't you add me to your foes list?)
PhilX
Philosophy Explorer
Posts: 4533
Joined: Sun Aug 31, 2014 7:39 am
Re: Patterns upon patterns
Arising_uk wrote:
Fri Jul 14, 2017 1:39 am
Philosophy Explorer wrote:I didn't phrase the question "How do you get from the last number to the target number?" All four starting numbers in each set participate equally toward the target number.
Try again.
I don't see why as my answer is correct but if you wish to be picky.
Philosophy Explorer wrote:How do you get from these four numbers: 1 2 3 4 to 24?
Start with the first number and add one to get a successor(which will be the same number as in your pattern) and add one to each successive successor twenty two times.
How do you get from these four numbers: 2 3 4 5 to 48?
Start with the first number and add one to get a successor(which will be the same number as in your pattern) and add one to each successive successor forty five times.
How do you get from these four numbers: 3 4 5 6 to 80?
Start with the first number and add one to get a successor(which will be the same number as in your pattern) and add one to each successive successor seventy six times.
All the numbers are participating equally towards the result.
Not really because you're adding 22 with the first set of numbers, 45 with the second set and 76 with the last set.
PhilX
Arising_uk
Posts: 10665
Joined: Wed Oct 17, 2007 2:31 am
Re: Patterns upon patterns
Philosophy Explorer wrote: Not really because you're adding 22 with the first set of numbers, 45 with the second set and 76 with the last set.
PhilX
Not really, I'm adding one to the first and then producing the rest by adding one to the next. It fits the pattern you gave, i.e. it produces the pattern you gave and it reaches the correct result.
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Users browsing this forum: No registered users and 2 guests | 3,064 | 11,451 | {"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} | 3.546875 | 4 | CC-MAIN-2018-05 | latest | en | 0.89263 |
https://stats.stackexchange.com/questions/466363/how-to-conclude-experiment-is-statistically-significant?noredirect=1 | 1,660,073,966,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00483.warc.gz | 491,184,300 | 61,460 | # How to conclude experiment is statistically significant? [duplicate]
Say I have 100 control group and 100 experimental group(drug received). I've gathered result of test and want to conclude that there is difference btwn two groups => drug is effective.
We must start by creating null hypothesis
Ho = two groups are the same.
Ha = two groups are NOT the same.
our goal is to reject the null hypothesis to conclude that Ha is true.
To reject null hypothesis p < alpha = 0.05. To calculate p value we calculate z-score of Xi (each person in exp group) and then find their p-values.
Here is what I'm confused about --> so we get p-value for each individual and if its p-value < 0.05 then we say that one person's result is statistically significant? and do the same for rest 99 people and if more than 95 people's results are statistically significant we say from our experiment we can conclude that drug is effective with 95% confidence?
Another Q: Why not simply try to prove that Ha is true? instead of rejecting Ho?
To reject null hypothesis p < alpha = 0.05. To calculate p value we calculate z-score of Xi (each person in exp group) and then find their p-values.
This is incorrect. We don't find the p-value for each subject, rather the p-value is computed from the test statistic, which is a function of the data observed from each patient.
• Thanks, by test statistic you are talking about mean, median or others right? so if we use mean than we would find p-value for the mean then if it is less than alpha we conclude our drug is effective with 95% confidence? May 14, 2020 at 1:47
• Test statistics may involve things like the mean or the median, but they are not themselves test statistics. Also, the 95% in 95% confidence interval is not what you think. Before clarifying that, I think it would be wise to revisit some introductory material on statistics to re-familiarize yourself with hypothesis testing. May 14, 2020 at 2:06
• So you are saying we can use one of the test statistics to calculate its p-value to see statistical significance right? May 14, 2020 at 2:12
• Yes, that is correct. May 14, 2020 at 3:23
This is not a full answer but, nevertheless :
IF your "null" hypothesis is true, i.e. if the treatment has no effect, i.e. if the two samples are really from the same population, THEN what is the probability of getting an absolute DIFFERENCE, between the two groups' means, as high as the one you observe in your CURRENT experiment, imagining you could repeat the entire experiment an infinite number of time to estimate that probability ?
IF that probability, calculated over an infinite number of replicated experiment, is INFERIOR to a threshold that YOU determined to be the THRESHOLD OF SIGNIFICANT EFFECT or THRESHOLD OF "HEY, THAT IS STRANGE" or THRESHOLD OF "HEY, THAT DIFFERENCE IS TOO MUCH TO BE POSSIBLE", and that threshold is YOUR p-value. THEN you will tell the story of a possible effect. Else you will tell the story of how you did your experiment, and you did not observe a signify effect at p-value 0.05 (natural sciences for example) or maybe 0.01 or 0.00001 (physics). | 725 | 3,128 | {"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} | 3.78125 | 4 | CC-MAIN-2022-33 | latest | en | 0.955644 |
https://jp.mathworks.com/matlabcentral/cody/problems/2063-a-matrix-of-extroverts/solutions/1124766 | 1,582,130,752,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00095.warc.gz | 440,700,044 | 16,406 | Cody
# Problem 2063. A matrix of extroverts
Solution 1124766
Submitted on 16 Feb 2017 by Jihye Sofia Seo
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = magic(3); y = extroverts(x); y_c = [4.2500 4.7500 ; 5.2500 5.7500]; assert(max(max(abs(y-y_c)))<1e-9);
r = 3 c = 3 fullMat(:,:,1) = 8 1 3 5 fullMat(:,:,2) = 3 5 4 9 fullMat(:,:,3) = 1 6 5 7 fullMat(:,:,4) = 5 7 9 2 y = 4.2500 4.7500 5.2500 5.7500
2 Pass
x = [1 2 3 ; 4 5 6]; y = extroverts(x); y_c = [3 4]; assert(max(max(abs(y-y_c)))<1e-9);
r = 2 c = 3 fullMat(:,:,1) = 1 2 fullMat(:,:,2) = 4 5 fullMat(:,:,3) = 2 3 fullMat(:,:,4) = 5 6 y = 3 4
3 Pass
x=[magic(4) -magic(4)]; y = extroverts(x); y_c=[8.5 6.5 8.5 0 -8.5 -6.5 -8.5 8 8.5 9 1.5 -8 -8.5 -9 8.5 10.5 8.5 0 -8.5 -10.5 -8.5]; assert(max(max(abs(y-y_c)))<1e-9);
r = 4 c = 8 fullMat(:,:,1) = 16 2 3 13 -16 -2 -3 5 11 10 8 -5 -11 -10 9 7 6 12 -9 -7 -6 fullMat(:,:,2) = 5 11 10 8 -5 -11 -10 9 7 6 12 -9 -7 -6 4 14 15 1 -4 -14 -15 fullMat(:,:,3) = 2 3 13 -16 -2 -3 -13 11 10 8 -5 -11 -10 -8 7 6 12 -9 -7 -6 -12 fullMat(:,:,4) = 11 10 8 -5 -11 -10 -8 7 6 12 -9 -7 -6 -12 14 15 1 -4 -14 -15 -1 y = 8.5000 6.5000 8.5000 0 -8.5000 -6.5000 -8.5000 8.0000 8.5000 9.0000 1.5000 -8.0000 -8.5000 -9.0000 8.5000 10.5000 8.5000 0 -8.5000 -10.5000 -8.5000
4 Pass
x = ones(20); y = extroverts(x); y_c = ones(19); assert(max(max(abs(y-y_c)))<1e-9);
r = 20 c = 20 fullMat(:,:,1) = Columns 1 through 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 17 through 19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 fullMat(:,:,2) = Columns 1 through 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 17 through 19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 fullMat(:,:,3) = Columns 1 through 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 17 through 19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 fullMat(:,:,4) = Columns 1 through 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 17 through 19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y = Columns 1 through 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ... | 4,026 | 4,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-10 | latest | en | 0.365762 |
https://www.ngpf.org/blog/math/math-monday-piecewise-functions-introducing-marginal-and-effective-tax-rates/ | 1,656,687,244,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00251.warc.gz | 1,006,148,138 | 21,491 | Feb 13, 2022
Math
Math Monday: Piecewise Functions - Introducing Marginal and Effective Tax Rates
This Math Monday, fall in love with Desmos as we explore the activity Piecewise Functions: Introducing Marginal and Effective Income Tax Rates
This Desmos activity corresponds to Level 1 of Application FA-1.7; it uses self-checking questions to practice writing equations for piecewise functions and using function notation. The interactive graph reinforces how income taxes are calculated using marginal tax rates and introduces students to effective tax rates.
Part I: Write the Piecewise Equation for Marginal Tax Rate
First, students complete the equation for the piecewise function representing marginal tax rate. They can self-check their answers and get feedback on which part of the equation they need to fix. As the teacher, you’ll see a warning if students check their work more than three times.
Part II: Marginal Tax Rate Graph
Next, students see the stepped graph for the marginal income tax rate. They can interact with the graph by dragging a point along the x-axis. This interaction conceptually reinforces the way income taxes are calculated using tax brackets; the shaded area represents the total taxes paid.
Students find Sam’s total taxes paid and calculate their effective tax rate. Then, students reflect on why Sam’s effective tax rate differs from their marginal tax rate.
Part III: Effective Tax Rate
Finally, students reveal the graph for the effective income tax rate as a function of taxable income. Again, the blue shading represents total taxes paid.
One the following slide, students will make a prediction about the end behavior of this function: what is the highest possible effective income tax rate? Then, students can reveal the same graph, scaled to show higher x-values, and reflect on their prediction.
Where to Find NGPF’s Desmos Activities
If you’re using our Financial Algebra curriculum, you’ll find Desmos activities built directly into the Lesson Guides and Application Answer Keys.
To look through NGPF’s library of Desmos activities, you can check out our Financial Algebra Collection.
Want More? | 422 | 2,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-27 | latest | en | 0.920694 |
https://iwarelogic.com/methodologies/how-much-force-is-in-a-scrum.html | 1,656,576,880,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00467.warc.gz | 382,248,245 | 18,674 | # How much force is in a scrum?
Contents
Reported group mean values for average sustained forces against a machine generally ranged from 1000 to 2000 N in individual scrums and 4000–8000 N for full packs of male rugby players older than high school age.
## How hard is a rugby tackle?
In the short time the devices have been used at the club, most of the recorded impacts have been around 10 to 20 G’s, with some up to 40 G’s. The average impact is 22 G’s.
## How far can a scrum be pushed?
Maximum 1.5m travel in scrum.
## Is rugby more brutal than NFL?
Conclusion. The data would suggest that rugby is indeed a more dangerous sport in the sense that a player is more likely to get hurt while playing. However, the severity of injury is likely higher in football, considering the nature of the collisions to be at a greater speed and with less control.
## Why don’t they wear helmets in rugby?
Rugby players are taught to never use their head in making the tackle, and without a helmet to protect them, the logic is pretty clear. Any player leading with the head is almost certain to get hurt as badly as the person he’s trying to hit.
IT IS IMPORTANT: What is discussed in scrum of scrum meeting?
## What does the ref say before a scrum?
Since 2013, rugby referees say three words to initiate a scrum. “Crouch, Bind, Set” is the current sequence: “Crouch” tells the forwards to drop into a low position. “Bind” tells the props to grip their opponent’s jersey.
## Does the ball have to go in straight in a scrum?
The scrum-half must put the ball in straight to the scrum, but they are allowed to align their shoulder to the middle line of the scrum. … So the ball has to be put in straight, but rather than being put in down the middle of the tunnel it is put in slightly towards the scrum-half’s own team.
## Why is scrum called scrum?
Why is it called Scrum? When Jeff Sutherland co-created the Scrum process in 1993, he borrowed the term “scrum” from an analogy put forth in a 1986 paper by Takeuchi and Nonaka, published in the Harvard Business Review.
## What happens when a scrum collapses?
Twisting, dipping or collapsing a scrum will result in a penalty against the offending team. Rather than engaging square on with their opponent, tight-head props can bore their heads into the hooker. … Sometimes you may see a tight-head prop’s body pop out of a scrum while it is still taking place.
## What does a scrum master do?
Summary: The scrum master helps to facilitate scrum to the larger team by ensuring the scrum framework is followed. He/she is committed to the scrum values and practices, but should also remain flexible and open to opportunities for the team to improve their workflow.
IT IS IMPORTANT: What is Agile and its benefits? | 643 | 2,771 | {"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-2022-27 | latest | en | 0.974609 |
https://optimization.cbe.cornell.edu/index.php?title=Set_covering_problem&diff=1725&oldid=1689 | 1,628,065,167,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00009.warc.gz | 446,790,799 | 10,886 | Difference between revisions of "Set covering problem"
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Authors: Sherry Liang, Khalid Alanazi, Kumail Al Hamoud
Introduction
The set covering problem is a significant NP-hard problem in combinatorial optimization. In the set covering problem, two sets are given: a set U of elements and a set S of subsets of the set U. Each subset in S is associated with a predetermined cost, and the union of all the subsets covers the set U. This combinatorial problem then concerns finding the optimal number of subsets whose union covers the universal set while minimizing the total cost.1 The problem has many applications in the airline industry, and it was explored on an industrial scale as early as the 1970s.2
Methodology
The mathematical formulation of the set covering problem is define as follows. We define U = {u1, ….. um} as the universe of elements and S = {s1, ….sn} as a collection of subsets such that si ${\displaystyle \in }$U and the union of si cover U (i.e. ${\displaystyle \cup }$si = U ). Addionally, each set si must cover at least one element of U and has associated cost c that is larger than zero (i.e. ci > 0). The objective is to find the minimum cost sub-collection of sets X ${\displaystyle \in }$ S that covers all the elements in the universe U.
An integer linear program (ILP) model can be formulated for the minimum set covering problem as follows:
Decision variables
${\displaystyle y_{i}={\begin{cases}1,&{\text{if subset }}i{\text{ is selected}}\\0,&{\text{otherwise }}\end{cases}}}$
Objective function
minimize ${\displaystyle \sum _{i=1}^{n}c_{i}y_{i}}$
Constraints
${\displaystyle \sum _{i=1}^{n}a_{i}jy_{i}>=1,\forall j=1,....,m}$
${\displaystyle y_{i}\in \{0,1\},\forall i=1,....,n}$
The objective function ${\displaystyle \sum _{i=1}^{n}c_{i}y_{i}}$ is defined to minimize the number of subset si that cover all elements in the universe by minimizing its total cost. The first constraint implies that every element j in the universe U must be be covered where aij =1 if j ${\displaystyle \in }$ si and 0 otherwise. The second constraints
Applications
The applications of the set covering problem span a wide range of applications, but its usefulness is evident in industrial and governmental planning. Variations of the set covering problem that are of practical significance include the following.
The optimal location problem
This set covering problems is concerned with maximizing the coverage of some public facility placed at different locations.3 Consider the problem of placing fire stations to serve the towns of some city.4 If each fire station can serve its town and all adjacent towns, we can formulate a set covering problem where each subset consists of a set of adjacent towns. The problem is then solved to minimize the required number of fire stations to serve the whole city.
The optimal route selection problem
Consider the problem of selecting the optimal bus routes to place pothole detectors. Due to scarcity of the physical sensors, the problem does not allow for placing a detector at every road. The task of finding the maximum coverage using a limited number of detectors could be formulated as a set covering problem.5 Specifically, giving a collection of routes U, where each route itself is divided into segments. The segments of two routes can overlap. The goal is then to select the routes that maximize number of covered segments.
The airline crew scheduling problem
An important application of large-scale set covering is the airline crew scheduling problem, which pertains to assigning airline staff to work shifts.2,6 Thinking of the collection of flights as a universal set to be covered, we can formulate a set covering problem to search for the optimal assignment of employees to flights.
Conclusion
The set covering problem, which aims to find the least number of subsets that cover some universal set, is a widely known NP-hard combinatorial problem. Due to its applicability to route planning and airline crew scheduling, several methods have been proposed to solve it. Its straightforward formulation allows for the use of off-the-shelf optimizers to solve it. Moreover, heuristic techniques and greedy algorithms can be used to solve large-scale set covering problems for industrial applications.
References
1. Grossman, T., & Wool, A. (1997). Computational experience with approximation algorithms for the set covering problem. European Journal of Operational Research, 101(1), 81-92. doi:10.1016/s0377-2217(96)00161-0
2. RUBIN, J. (1973). A Technique for the Solution of Massive Set Covering Problems, with Application to Airline Crew Scheduling. Transportation Science, 7(1), 34-48. Retrieved November 23, 2020, from http://www.jstor.org/stable/25767684
3. Church, R., ReVelle, C. The maximal covering location problem. Papers of the Regional Science Association 32, 101–118 (1974). https://doi.org/10.1007/BF01942293
4. Aktaş, E., Özaydın, Ö, Bozkaya, B., Ülengin, F., & Önsel, Ş. (2013). Optimizing Fire Station Locations for the Istanbul Metropolitan Municipality. Interfaces, 43(3), 240-255. doi:10.1287/inte.1120.0671
5. Ali, J., & Dyo, V. (2017). Coverage and Mobile Sensor Placement for Vehicles on Predetermined Routes: A Greedy Heuristic Approach. Proceedings of the 14th International Joint Conference on E-Business and Telecommunications. doi:10.5220/0006469800830088
6. Marchiori, E., & Steenbeek, A. (2000). An Evolutionary Algorithm for Large Scale Set Covering Problems with Application to Airline Crew Scheduling. EvoWorkshops. | 1,327 | 5,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "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} | 3.796875 | 4 | CC-MAIN-2021-31 | latest | en | 0.871568 |
https://jp.mathworks.com/matlabcentral/answers/1667019-extracting-information-from-multiple-tables-within-a-cell-to-plot | 1,660,471,501,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00735.warc.gz | 323,177,409 | 31,159 | # Extracting information from multiple tables within a cell to plot
13 ビュー (過去 30 日間)
Nicholas Scott 2022 年 3 月 8 日
コメント済み: Peter Perkins 2022 年 3 月 10 日
I have a cell with multiple tables and I am trying to extract information from to plot the median line of all the separate lines.
Currently, I plot all lines in grey.
for i= 1:length(ArrayPR) %i=Nucleus number
if height(ArrayPR{i}) >= 0.75*max(table_heightsPR)
plot(ArrayPR{i,1}.ImageNumber,ArrayPR{i,1}.Intensity_MeanIntensity_Alexa488,'color',[0,0,0]+0.75)
else
end
end
I also would like to plot the median line out of all of those as a thicker line like so:
plot(ArrayPR{:,1}.ImageNumber,median(ArrayPR{i,1}.Intensity_MedianIntensity_Alexa488),'k', 'LineWidth', 2)
However, I am not telling it to use the correct value to median, and I can't wrap my head around how to do so.
What i need it to do, is go into every table and look at the same image number (column one in ArrayPR{i,1}). I then need it to find the median of that one image number across all 'i', and then proceed to the next image number. Then plot that final line. I'm basically making an average line from all the different lines plotted, but I want the line to be less susceptible to outliers, which these data sets frequently see.
It is also possible that some of the elements within the ArrayPR don't have the maximum table height (some are even ignored with the if statement I have created), but when looking through all of the different rows of the ArrayPR, if it doesn't find a specific image number in a given 'i' it needs to ignore that specific 'i' for that given image number.
How does one accomplish this? Thanks,
Nick
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### 採用された回答
Voss 2022 年 3 月 9 日
figure();
hold on
table_heightsPR = cellfun(@height,ArrayPR);
height_threshold = 0.75*max(table_heightsPR);
for i = 1:numel(ArrayPR)
if table_heightsPR(i) >= height_threshold
plot(ArrayPR{i,1}.ImageNumber,ArrayPR{i,1}.Intensity_MeanIntensity_Alexa488,'color',[0,0,0]+0.75);
else
end
end
% median intensity for each image number calculation:
% first, concatenate all the tables into one big table:
% t = vertcat(ArrayPR{:});
% or concatenate just the tables whose height is >= height_threshold:
t = vertcat(ArrayPR{table_heightsPR >= height_threshold});
% get all the image numbers:
all_image_number = t.ImageNumber;
% get all the intensities:
all_intensity = t.Intensity_MeanIntensity_Alexa488;
% get the unique image numbers:
u_image_number = unique(all_image_number);
% calculate the median intensity for each image number:
% number of unique image numbers:
n_images = numel(u_image_number);
% pre-allocating the median intensity array:
median_intensity = zeros(1,n_images);
% for each unique image number:
for i = 1:n_images
% get a logical index saying whether each element of all_image_number
% is equal to this image number (idx is true where
% all_image_number == u_image_number(i) and false elsewhere):
idx = all_image_number == u_image_number(i);
% use that logical index to calculate the median intensity for this
% image number:
median_intensity(i) = median(all_intensity(idx));
end
% plot
plot(u_image_number,median_intensity,'k','LineWidth',2);
##### 2 件のコメント表示非表示 1 件の古いコメント
Voss 2022 年 3 月 9 日
You're welcome!
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### その他の回答 (1 件)
Peter Perkins 2022 年 3 月 9 日
Here's a version that uses grouped rowfun and varfun to do the heavy lifting:
table_heightsPR = cellfun(@height,ArrayPR);
height_threshold = 0.75*max(table_heightsPR);
keep = cellfun(@(t) height(t)>=height_threshold, ArrayPR);
t = vertcat(ArrayPR{keep});
t = t(:,["ImageNumber" "ObjectNumber" "Intensity_MeanIntensity_Alexa488"]);
figure; hold on
rowfun(@(inum,int)plot(inum,int,'color',[.75 .75 .75]), t, "GroupingVariables","ObjectNumber","NumOutputs",0);
med = varfun(@median, t,"InputVariables","Intensity_MeanIntensity_Alexa488","GroupingVariables","ImageNumber");
plot(med.ImageNumber,med.median_Intensity_MeanIntensity_Alexa488,'k-','LineWidth',2);
hold off
##### 2 件のコメント表示非表示 1 件の古いコメント
Peter Perkins 2022 年 3 月 10 日
A grouped varfun applies a function to groups of rows in each variable in the table, separately. In this case, the function is median.
A grouped rowfun applies a function to groups of rows in a table, where that function accepts multiple variables. In this case, the function is plot, with some extra arguments (thus the anonymous function).
There are ungrouped versions of both.
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Translated by | 1,214 | 4,558 | {"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-2022-33 | latest | en | 0.770023 |
http://www.instructables.com/id/TheSUN-Arduino-powered-design-wall-clock/ | 1,496,072,139,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612399.20/warc/CC-MAIN-20170529145935-20170529165935-00486.warc.gz | 658,235,294 | 25,994 | Hi again Instructables-folks! :-P
Because of shipping-problems I wasn´t able to continue my ABTW Project so I decided to show you another, my newest creation.
I think many of us, like me, like those nice addressable LED Stripes (also called NEOPIXEL LED). You can get them from ADAFRUIT. Other vendors will also provide similar products. There is a library available on ADAFRUITS - GitHub (click me) including some sample code.
So coding should be straight-forward...
I saw those NEOPIXELS an thought, what the h... can I do with those shiny little things.
• A LED-matrix Display? -> To complicated and I don´t use it (at the moment)
• X-Mas Lighting? -> It fits the season but it would be cheaper to buy one :-P
• a clock? -> Why not! But it should be stylish and unconventional
So, let´s make a wall clock.
If we have a close look at our wrist watch (if you have an analog like me) we will notice that we have 12 hours and 60 minutes (hopefully). That will mean, that we need 60 addressable LED´s, phu!
If we take a stripe with 60 LED´s / meter we will get a diameter of ~318mm (radius = scope /(2*Π)) that´s definitely too big.
The truth is, if you ask someone for the time, no one will say it is 2 minutes past 3! You will get "It´s 5 past 3" as an answer.
So why shouldn´t we scale everything down to 5min steps? For that we will only need 12 LEDs which means we get a diameter of 63.6mm. We are also able to differentiate hours and minutes by giving them a separate colour.
We will also be able to provide the "missing" single minute steps with an additional Strip of 4 LEDs(or single addressable LED.
THAT`S THE PLAN!
Let´s have a look how I did everything.
As always I will provide a list/bill of materials and instructions on how to build it.
If you think, only Swiss people can make cool clocks, let´s proof you´re wrong (sorry Switzerland :-P)
## Step 1: Design & Choice of Materials
Design:
If we have a close look at our analog watch/clock again we see that the circle is divided into 12 * 30° steps we know, that we need 63.6mm for the LED-Strip. So it should be possible to align the strip around a tube somehow. I decided to use acrylic glass, because it looks nice and it is possible to encapsulate the LED light into it and at every flaw in the glass some light scattering will happen. So, let´s say: more impurities will lead to more light scattering! That´s exactly what we want. So feel free to grab your engraving tools and be creative :-)
If you refer to my BoM-list and the name i gave to the clock, I have chosen a sun like design.
I got all the acrylic parts from a German seller on E-Bay (link provided in the BoM).
For my design you will need:
• acrylic ground plate, transparent thickness = 6mm, diameter = 300mm
• acrylic middle plate, transparent thickness = 3mm, diameter = 150mm
• acrylic front plate, satin, thickness = 3mm, diameter = 90mm
• acrylic tube, transparent, outer diameter = 64mm (will mean we have to tweak a little with the LED strip)
• acrylic rod, transparent, diameter = 5mm (this will be our beams); There are also acrylic rods around with bubbles inside, I recommend them but i don´t have them around.
• acrylic glue
Electronics (refer to the Fritzing-files):
• Arduino mini (or similar)
• 1 addressable LED Strip (12 LEDs for hour and 5min steps)
• 4 addressable LEDs (single minutes)
• 2 330Ohm Resistors
• 1 1000µF Capacitor
• 1 powersupply (5V/500mA)
• an RTC DS-1307 (optional!)
• Bluetooth module (optional! yes you can set the time via BT and an Android Smartphone)
If you ask yourself why i have MAX485 chips on my BoM. The answer is, that I want to syncronize the clock with the home automation system I´m about to make (never ever have to set a clock for daylight-saving again :-P).
I will describe that in my blog in the next couple of weeks/month.
As you noticed, I will also try to get the clock off-grid with some solar panels and a LiPo, but I don´t cover that in this Instructable feel free to try it yourself.
## Step 2: Prepare the Acrylic Parts
The tools:
First of all it is really helpful if you print the DWG plan i added in the scale 1:1.
This will help you to align all the parts and will serve you as a drilling plan.
further you will need:
• hobbyknife
• miter gauge
• hacksaw
• clamps
• hand drill
• can drill, diameter 65mm
• a set of metal drills
• a small metal file
• acrylic glue
Let´s start:
Take the ground plate and align it on the plan, so you can get the center of the circle. Now take your hand drill with the can drill mounted onto it and drill (very slow! not to much pressure!) a hole in the center of the ground plate, the outer circle should be ~2-3mm deep. This is to sink the LED strip into the ground plate (LED strip ~10mm wide, beams only 5mm in diameter) and to align them with the beams (refer to picture 1).
Now we need the hacksaw, the miter gauge and the acrylic tube. Just cut it into pieces i decided to make the housing (tube) 40mm long (picture 2). Now grab the hacksaw again and make a little rabbet on one side of the tube, make it smooth with the metal file. That´s where the wires will come out ;-) (refer to picture 3)
Time for some glue...
Take the middle plate (d=150mm) and the front plate (the satinated one).
align them on the plan again, put some glue on the center of the middle plate, align the front plate and wait till the glue is slightly hardened. The glue i used is light hardening and it can take up to 2-3h, so mybe you want to use a clamp... (picture 3 and 4)
Do the same for gluing the tube onto the ground plate, make sure the rabbet is facing to the plate and is alligned somewhere where you want the first LED (12 o´clock) to be.
Wait till it is hardened!
We can now align the 2 parts (bookmatched) onto the plan an drill our 4 single minute holes (5mm in diameter or the diameter of the LED you have chosen; drill it slow with not too much pressure). Drill about 8-9mm deep. Be careful, the satinated plate is very brittle and can brake if you drill to deep.
You can now glue them together or you decide, like me to cut a threat into the ground plate and attach it with a screw.
Again, wait till glue has hardened.
Now align and glue the beams onto the ground plate. (picture 6)
Guess what... wait till the glue has hardened :-)
Let´s proceed to the electronics...
## Step 3: Electronics
The tools:
• soldering iron
• solderwire
• hobby knife
• a small piece of prototyping PCB
• enamelled wire or any other wire you prefer
• hot glue
I strated with the single LEDs.
If you use enamelled wire don´t forget to scrape of the lacquer before soldering. You can use a hobby knife for that.
Wire them up, you can refer to the picture with the pinout on flikto.de. Note that DOUT goes to DIN on the next LED! (see picture 2)
After that you can cut the LED strip into 4 elements each with 3 LEDs. Remember, we have 63.6mm LED Strip and 64mm outer diameter of the tube so we need some "extra length to align it precisely to the beams. Wire it up with enamelled wire like in picture 4.
I made a little proto PCB which will serve as a "power harness" and will contain the components for the LED Strips (the two 330Ohm Resistors and the 1000µF Capacitor, picture 7). Refer to the Fritzing Image for that.
Now mount the Strip around the tube, allign the LEDs to the beams. The first Pixel matches 12 o´clock. If you have turned around your housing, don´t forget that everything is mirrored. Proceed counter clockwise! Use some hot glue to attach it to the tube. A small drop for every segment will do it!
You can do the same for the single LEDs (eventually mirrored), just add some hot glue and press them into the predrilled holes.
Don´t wire the Arduino yet, we will use the hardware-serial for the BT connection, so first check the next steps where i describe the software.
## Step 4: Code
You can now load the sketch to the Arduino. You´re also able to wire the LED strips now. Don´t connect the BT Module!!! We first want to have a look at the code, you should know where you can tweak several things...
Install the IDE and put the librarys into the library-folder.
Open the attached INO file and upload it to your arduino. The code described here is the same but with additional comments! If you have done everything right, you can now see the "bootanimation". It is possible to set the time over the serialmonitor. Just type @"hour"/"min"/"sec" eg. @10/33/00 (10:33).
Feel free to play with the code...
Here ill give you a brief description of the Code (Setup without RTC!)
DEFINITIONS:
#define PIN 6 //Hour LED Strip
#define MINPIN 5 //Singelminute LED
#define NUMPIXELS 12 //Number of Pixels for hour
#define MINNUMPIXELS 4 //Number of Pixels for single minute
#define BAUDRATE 115200 //Baudrate, should match the baudrate of BT Module
#define utch '@' //start BYTE of TimeSync
int timeset = 0; //flag to store if the time was set after boot
int delayval = 20; //delay for fading animation
int clocktimer = 10000; //time refresh
int timebright = 250; //brightness of hour Strip
int mtimebright = 50; //brightness of singelmin
int initialize = 0; //flag to call the clearpixels function after boot
int ahour;
int oldahour = 0; // store prev. hour
int aminute;
int oldamin = 0; //stores the previous minute for refresh
int asecond;
int amonth;
int ayear;
int mmin;
tmElements_t tm;
//Setup for the 2 NeoPixel LED arrays ( NAME = TYPE(NUMBER OF PIXELS, WHICH PIN, FORMAT RGB OR GRB, FREQ); Refer to the Adafruit guide for more information.
SETUP:
void setup() {
Serial.begin(BAUDRATE);
Wire.begin();
// Initialice the strips, all OFF
pixels.begin();
minpixels.begin();
pixels.show();
minpixels.show();
//Make a little animation
Serial.println("SUNRISE");
sunrise();
delay(1000);
Serial.println("SUNSET");
sunset();
pixels.show();
}
LOOP:
void loop() {
// check for timesync
while (Serial.available() >0){
if(c == utch) //if there is a @ on the line, read the coming bytes / ints
{
}
}
// initialice the LEDs, clear boot animation
if(initialize == 0){
clearpixels();
initialize = 1;
}
ahour = hour();
aminute = minute();
if(timeset == 1 || timeset == 0) // here you can check if the Time was set, you can stop the program here if Timeset = FALSE, just remove "|| timeset == 0" !
{
if(oldamin < aminute || oldahour < ahour) //check if the time has changed. TRUE = refresh -> set all to OFF, display new time
{
clearpixels();
ClockDisplay();
}
}
}
Display the Clock:
void ClockDisplay()
{
oldahour = ahour;
oldamin = aminute;
int xhour, xmin;
if (ahour >= 12){
xhour = ahour-12; //we only have 12 LEDs for 24h display
}
else {
xhour = ahour;
}
//scale it into 5min steps
xmin = (aminute / 5);
if(oldamin < aminute)
{
oldamin = aminute;
clearpixels();
}
//take the rest of the division dor the singelmin LED
mmin = (aminute % 5); // modulo operator eg. 24 % 5 = 4! very useful :-P
pixels.setBrightness(timebright);
pixels.setPixelColor(xmin, pixels.Color(5,125,255)); // you can change the colours here! play around!
pixels.setPixelColor(xhour, pixels.Color(255,50,0));
pixels.show();
//display the singel mins
for (int m=0; m
minpixels.setBrightness(mtimebright);
minpixels.setPixelColor(m, pixels.Color(255,255,0));
minpixels.show();
}
}
Read and process TIMEinformation from Serial
void readtime() // if we already got the leading "@" process the coming data and store the time for the TIME Lib
{
ahour = Serial.parseInt();
aminute = Serial.parseInt();
asecond = Serial.parseInt();
amonth = Serial.parseInt();
ayear = Serial.parseInt();
Serial.println("TIMESET");
Serial.print(ahour);
Serial.print(" : ");
Serial.println(aminute);
}
Clear all!
void clearpixels() // set every single PIXEL to off to refresh the display
{
pixels.begin();
minpixels.begin();
for(int i=0;ipixels.setPixelColor(i, pixels.Color(0,0,0));
minpixels.setPixelColor(i, pixels.Color(0,0,0));
pixels.show();
minpixels.show();
}
}
## Step 5: The Android APP and BT Connection
If you were successful with the previous steps, you can now wire up your BT Module. (i hope you made sure, that the baudrates match). don´t forget to cross TX & RX lines :-)
Download and install the app, pair with your BT dongle, start the app, connect to the dongle and Sync the time with your mobile. The APP basically does the same like we did before. It just sends @hh/mm/ss/dd/mm/YYYY generated from its systemtime.
I also provided the APPInventor AIA File and an explanation on the next step (for those who are interested).
## Step 6: APPInventor
APP Inventor is pretty easy to use and worth the effort for such a simple program.
If you make a new project you will find your self at the DESIGNER screen. (picture 1)
This is where we add tables, buttons, sensors and other elements for further use.
In our case we need:
• a table (to allign all elements)
• a listpicker (for selection of the BT device we connect to)
• a button (to fire the TIME over BT)
• some labels (display the actual time and date)
• the clock sensor (refresh the time)
• the bluetooth client sensor (connectivity)
Adding them is just as easy as drag & drop!
On Picture 2 you can see an overview of the "APP" in the BLOCKS screen. Well, that´s basically where all the "magic" happens.
On the top I created some variables to store the Time and Date.
The first block on the upper left will initialize the listpicker element with the list of paired BT devices.
With the second block we decide what to do with the previously picked element. Well, we want to connect to it.
If you have a close look at the next Block, you can see, that we generate, if the BT status "is connected", the BT message. It´s the same we typed into the SerialMonitor before.
The last block on the left will provide us the leading zeroes to display the time (eg. 01:08).
On the right side you can find our last block, that´s where we use the clock element.
Here we update the variables and merge them with the digits procedure, this will happen every 1000ms (default setting, change it in the designer mode) and display the updated values with the label.
That´s just a brief description, but APPInventor is really as easy as that :-)
Maybe there is someone in the community who wants to write a software for iOS or WindowsPhone. (would be great)
I hope you liked my Instructable!
Have fun with your new wall clock!
Maybe you want to gift it to someone you love (Its X-Mas season) :-)
And if there are any questions, feel free to ask me!
Best regards and Merry X-Mas.
Ok
<p>I´m afraid, there was a small bug in the code where the code gets stuck during daychange. So please use the V0.3 of the code! </p>
Hi
definitely on my to-make list.
thank you, hope you are going to show us some pictures of your approach...
<p>Very nicely done! I love the way this turned out. So cool!</p>
<p>Thanks mate</p> | 3,775 | 14,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-22 | longest | en | 0.927677 |
https://mirjamglessmer.com/2017/09/28/who-is-faster-the-currents-or-the-waves-the-froude-number/ | 1,726,502,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00386.warc.gz | 357,601,758 | 70,951 | # Who is faster, the currents or the waves? The Froude number
A very convenient way to describe a flow system is by looking at its Froude number. The Froude number gives the ratio between the speed a fluid is moving at, and the phase velocity of waves travelling on that fluid. And if we want to represent some real world situation at a smaller scale in a tank, we need to have the same Froude numbers in the same regions of the flow.
For a very strong example of where a Froude number helps you to describe a flow, look at the picture below: We use a hose to fill a tank. The water shoots away from the point of impact, flowing so much faster than waves can travel that the surface there is flat. This means that the Froude number, defined as flow velocity devided by phase velocity, is larger than 1 close to the point of impact.
At some point away from the point of impact, you see the flow changing quite drastically: the water level is a lot higher all of a sudden, and you see waves and other disturbances on it. This is where the phase velocity of waves becomes faster than the flow velocity, so disturbances don’t just get flushed away with the flow, but can actually exist and propagate whichever way they want. That’s where the Froude number changes from larger than 1 to smaller than 1, in what is called a hydraulic jump. This line is marked in red below, where waves are trapped and you see a marked jump in surface height. Do you see how useful the Froude number is to describe the two regimes on either side of the hydraulic jump?
Obviously, this is a very extreme example. But you also see them out in nature everywhere. Can you spot some in the picture below?
But still, all those examples are a little more drastic than what we would imagine is happening in the ocean. But there is one little detail that we didn’t talk about yet: Until now we have looked at Froude numbers and waves at the surface of whatever water we looked at. But the same thing can also happen inside the water, if there is a density stratification and we look at waves on the interface between water of different densities. Waves running on a density interface, however, move much more slowly than those on a free surface. If you are interested, you can have a look at that phenomenon here. But with waves running a lot slower, it’s easy to imagine that there are places in the ocean where the currents are actually moving faster than the waves on a density interface, isn’t it?
For an example of the explanatory power of the Froude number, you see a tank experiment we did a couple of years ago with Rolf Käse and Martin Vogt (link). There is actually a little too much going on in that tank for our purposes right now, but the ridge on the right can be interpreted as, for example, the Greenland-Scotland-Ridge, making the blue reservoir the deep waters of the Nordic Seas, and the blue water spilling over the ridge into the clear water the Denmark Strait Overflow. And in the tank you see that there is a laminar flow directly on top of the ridge and a little way down. And then, all of a sudden, the overflow plume starts mixing with the surrounding water in a turbulent flow. And the point in between those is the hydraulic jump, where the Froude number changes from below 1 to above 1.
Nifty thing, this Froude number, isn’t it? And I hope you’ll start spotting hydraulic jumps every time you do the dishes or wash your hands now! :-)
## 14 thoughts on “Who is faster, the currents or the waves? The Froude number”
1. Joe Buchanan
I am really enjoying this blog. I am trying to understand hydraulic jumps. Can you explain why faster flowing water slows down? Friction? ( I thought water flows had negligible friction). Collision with slow water?
But how is this maintained? And how does fast flowing water entrain slower water? Is turbulence required? And finally, are hydraulic jumps really any different from standing waves? Do the former just have enough energy to break upstream?
1. Mirjam
Hi Joe, Thanks for your comment! Always makes me happy to read people are enjoying my blog :-)
There are a lot of reasons why the water might be slowing down. You mention two: Friction, but probably with the boundary or the bottom, and collision with slower water are definitely important! Other reasons might include for example a sudden opening of the cross section of the flow (if the same amount of water suddenly has to fill a larger cross section, it has to slow down), a change in the slope of the bottom (if water isn’t accelerated any more by a sloping bottom, or even has to go up hill, it will slow down).
I don’t quite understand what you mean about how it is maintained?
When fast water is flowing through slow water, there is going to be some friction between the two, which will lead to disturbances of the boundary between the two, i.e. turbulence. And that will then entrain water.
In my mind, hydraulic jumps and standing waves are very much linked: In a standing wave, the wave’s phase velocity is exactly as fast as the current underneath, which is why the wave stays in place. The hydraulic jump is the spot where the phase velocity changes goes through exactly the same velocity as the current, so if that happens over a longer stretch of water, you will have a standing wave there: A wave that is caught because it is too slow to move upstream against the current, but also to fast to be washed away downstream with the current. In case of the water hose in the post above however, there is the area around where the water hits the floor where the water is moving a lot faster than a wave’s phase velocity, and then over a really short distance the water’s speed changes from faster than phase velocity to slower than phase velocity, hence a distinct hydraulic jump. Does this make sense?
1. Joe Buchanan
Yes, thanks that makes sense. Is there a simple explanation for why hy a river wave never moves upstream? ( At least I have never seen this except with a tidal bore). Waves with more speed seem to change from a smooth standing wave to a steeper wave then a breaking wave, but never move upstream against the current.
My question about how a collision with slower water is maintained is to do with this quandry: why does a body of water moving at speed not push the slower water out of the way and create a path of faster moving water through the slower water? I would expect this to happen with other moving objects. If friction is low in water, why does water moving down slope slow so quickly? Newton’s laws suggest it would keep moving, even if the slope flattened out, until it was acted on by a force like friction. But water does not seem to make a path for itself. Is turbulence between the two bodies of water (like the fast moving water entering a slow pool) the main source energy loss?
I am really enjoying thinking about this – I used to kayak obsessively, these days I body surf, and I completed a PhD in marine biology a few years ago, so water has been a big thing in my life, but I have never studied the physics of water. I live in New Zealand which has some great standing waves and hydraulic jumps.
Thanks again, Joe.
1. Mirjam
Hi Joe!
A wave can only move upstream if its phase velocity is larger than the velocity of the current that it is moving against. So if you have a slowly moving river, then it is definitely possible for waves to go upstream. Think of the mouth of the river where it meets the ocean — technically, you are still in the river and the water is flowing out into the ocean, but swell from the ocean can propagate into the mouth for a bit. And tidal bores do basically just that: propagating against the current with a phase velocity that is higher than the velocity of the current.
To try and answer your question: To a certain degree the faster water DOES push the slower water out of the way. But think about billiard for a second: two equal balls on a flat surface, one at rest and one moving at a given velocity. When they collide (assuming that you hit the ball right in the center), what happens is that the ball that was at rest now moves at (pretty much) the velocity the other ball was moving at, and that the ball that was moving is now still in the spot where the two balls collided. The momentum got transferred from one ball to the other. And almost the same thing happens with the water, too. Of course, two bodies of water don’t just collide and then one bounces away and the other one stays at rest; there is mixing going on etc.. But the physics are the same in so far that one volume of water moving at speed can’t just hit a second volume of water and make it move at the same speed. Either all of it starts moving, but then more slowly, or some of it continues to move and the rest stays put (and the proportions of that depend on the topography in which this is happening). So friction does cause some energy loss, but also, more importantly, it isn’t even about energy lost: the energy of the system is only enough for one of the bodies of water moving at the initial speed, or both of them moving a lot more slowly. Does this make sense? I think I might have a different picture in my mind than you do, maybe if you have an image it’s easier to talk about what’s really going on there?
I love kayaking! Although I suck in white water, I find it super scary. But I used to play canoe polo and that is so much fun! And I’d love to visit New Zealand at some point, I have heard the most amazing things! Best, Mirjam
2. Joe Buchanan
Thanks Mirjam, that makes sense and the billiard ball analogy is a good one. I think I was thinking that if I have a large number of billiard balls, and I am shooting at a group of stationery balls at the other end of the table, then eventually I would be able to knock all the stationery balls out of the way and create a path that my fast moving balls could move through without collisions. So in a river I would expect to have a fast current down the middle of the river and slow or still pools on either side. But instead rivers at normal flow tend to speed up (where the gradient is steeper, forming rapids) then slow down (at a hydraulic jump or the pools between rapids).
Swollen or flooded rivers often do move at fairly constant speed – the speed is similar bank to bank and for long stretches of river. I have also seen this in a kind of “river without banks”. I was kayaking across Cook Strait (between the North and South Islands of New Zealand). It was a still day and the sea was flat and glassy. But Cook Strait is funnel shaped in three dimensions – there is a wide deep area to the south-east, and a narrow shallow area to the north-west. This makes for some impressive tidal currents. What was strange was that the tidal currents formed a distinct “river” in the sea. In a few kayak lengths I went from flat, glassy, apparently still water into a river of fast moving (at least 5 knots) water with large (1-2 metre) standing waves). The “river” was maybe 100 metres wide, and after several minutes of hard paddling I crossed the “river” to the other side – more still glassy water. I don’t know how far this current extended, probably 10s of kilometres.
So can I try and answer my own question: If there is sufficient energy – in a flood state river or a tidal current (the one I am describing must have been several thousand cumecs) there is enough energy for most of the water to gain speed and maintain it. In a normal river turbulent interactions with slow moving water in eddies and pools transfers energy from fast to slow water decreasing the speed of the main current. Does this make sense?
But now I am wondering about the interactions between the fast and slow water in a normal river. Even if the velocity of the water at the steep part of a rapid is more or less linear – directed straight downstream, the water will slow down and push sideways into an eddy or become turbulent at the eddy-line. It seems almost non-newtonian for the linear water to start pushing or swirling towards still water at the sides of the river – as if the billiard balls not only hit the stationary balls directly in their way, but curve towards balls lying to the side and entrap them. This would make sense if there was friction between the fast moving and stationery balls, but I am used to thinking of water as having very little friction. So why does waterin a current apparently curve to the sides?
1. Mirjam
I guess the billiard ball analogy only carries so far. One thing that it doesn’t include is continuity, which we do have in the river. In your example, yes, you can kick all the stationary billiard balls away and get a path for the fast moving ones. In the river, however, every water parcel you kicked away will be replaced by another one (since you don’t just create empty space in the river the way you do on a pool table). Therefore, you can’t just get rid of a parcel of water one after the other. For every one you got rid of to clear your path, there will be a new one coming in from the sides or from underneath or wherever (depending on the topography and situation).
And since you really dislike thinking about friction in water: This effect — water coming in from the sides — also creates a link between the fast flowing and stationary parts of the river, since water coming in from the sides will have to speed up to join the moving water. And even if you were moving water almost as if through a pipe, so in a confined flow, what will happen is that there will be sometimes instabilities in the flow; meanders, parts slowing down a little more, that kind of thing. And every bend in the current will be filled with slower water from the side, and as more fast water pushes from upstream, the slower water will speed up a little, creating more disturbances behind it where now more slow water will be sucked in to replace it. And this is exactly what creates turbulence, eddies, and behaves a lot like friction, except on a larger scale, not on a molecular one. So in oceanography, we’d still think of it as friction and parameterise it as friction in numerical models. Does this make sense?
The tidal current you describe sounds fascinating. If I had to guess, I’d say that the river without banks was coming through the narrow end of the tunnel, moving out the wider part? And if I had to guess further, I’d say that the fast-flowing river was slowing down and widening the further you got away from the narrow end of the funnel — albeit probably very little. For a situation where the water moves through the funnel “the right way”, I’d suspect that there would be a faster current right upstream the narrowest part of the funnel, but that overall the flow would be a lot more sluggish and over a much wider part of the funnel. I’d love to go there now to check :-)
1. Joe Buchanan
Mmm, Thanks, yes that all makes sense, but after a long break I have hit on the crux of my problem – I have to ask you this otherwise it will keep me up at night!
The conventional explanation I have heard for the jump is that the water slows down, losing kinetic energy and this is changed into potential energy through a rise in height. That is fine, but if the water is slowing down because it is losing energy through collisions that transfer of energy to slower moving water etc., then why is energy left over to be changed to potential energy? This implies that the water is slowed down by some process other than loss of energy, with the surplus kinetic energy having to change into potential energy – i.e. height.
The explanation seems circular – the jump is created because the water slows down, the slowing down is caused kinetic energy transferring to other water. So why does some energy get changed to potential energy and not just act to move more stationery water out of the way?
It is as if a moving billiard ball hits a stationery billiard ball, and not only accelerates the stationery ball – as expected – but also jumps into the air. I can understand this happening if the moving water is hitting stationery water at an angle and is deflected upwards, but this is a simple collision which doesn’t need the “change in energy from one form to another” part of the explanation. Obviously anything moving upwards in a gravitational field loses kinetic energy and gains potential energy, but the conventional explanation seems to say that moving upwards is a consequence of energy loss, not a cause of energy loss.
Is this a semantic problem or confused phrasing, or does the water really slow down and therefore rise in height? Help Mirjam!
BTW I really enjoyed the dead water experiment posts. I haven’t experienced dead water but I know how frustrating it can be when the chop has just the wrong period for a vessel’s length. However hard you paddle your kayak you are always pushing against a wave ahead of you and can’t gain any more speed. Nansen must have experienced similar frustration.
1. Mirjam
Hi Joe,
I am loving this conversation! And thanks — we really enjoyed playing with the dead water experiments, too. So cool to see the internal wave slowing down and stopping a ship even though it’s being pulled with the same force that was sufficient to move it before!
I have heard the kinetic and potential energy explanation, too, but I am not too happy with it, either. Or, rather, in my head, I use a different explanation:
While you are in a super critical flow, the water is flowing faster than waves on it can travel, so all waves/disturbances are being flushed downstream and away. In a subcritical flow, however, waves can travel both up- and downstream without any problems. So I imagine the hydraulic jump (where the flow changes from supercritical to subcritical) as the place where all the waves that travel upstream in the subcritical part get “bunched up”: They are running as far upstream as they can, but then at some point the current velocity is so fast that they can’t compete against it and are locked in place running exactly as fast upstream as the current is carrying them downstream. And if all those waves are stuck in this place, in my mind it makes sense that the water level must go up (since there are all those waves accumulating there). Does that make any sense to you?
Another point about the water slowing down: That can happen due to friction as you describe. But it could also happen for other reasons: If, for example, you turn on the water in your sink when doing the dishes, the water hits the sink and then spreads to all sides, forming some kind of circle of supercritical flow, and then eventually an hydraulic jump and then subcritical flow, right? So the area that each segment of flow covers as it moves radially outwards gets bigger and bigger. Since the amount of water per segment stays the same, it has to move slower and slower as it moves outwards, since there is less and less water available to cover the sink’s floor. So eventually the water will move at the same speed as the waves (i.e. you’ll have the hydraulic jump) and then after that it’s subcritical. So in this scenario we don’t actually need to “loose” any energy at all, we just spread the same amount over larger and larger areas until something happens. Does this make sense?
Looking forward to hearing from you again :-)
1. Joe Buchanan
Thanks Mirjam, I am so relieved that you are unhappy with the potential/kinetic energy explanation – It doesn’t work for me and your explanation is so much better. You are really good at this!
But – and I always have a but – now I need to know what is the source of the waves? and why do they usually stay in the same place? Here I am thinking of what I call a typical river rapid. The water accelerates as the gradient of the river bed steepens. In this part of the rapid the water flow is smooth, apparently linear, but at the bottom of the rapid there is a series of standing waves. This pattern it seems similar whether the river bottom is gravel, a rock shelf or a concrete weir.
What bothers me is that waves in rivers rarely move upstream, at least not visibly (I accept that a tidal bore is an exception). In turbulent, flooded rivers I can pick out undulations, maybe waves, being washed downstream, and I can accept that this is happening in any supercritical flow – but the waves don’t form clearly visible peaks until the velocity of the water changes to subcritical, where the wave energy would reinforce itself until a visible peak forms.
OK, but the typical series of standing waves do not move relative to the river bed. You can surf them in a kayak and you stay in the same place relative to the river bed. For this to happen the velocity of the wave must be upstream and equal to the velocity of the water downstream. I have always assumed that the speed of the wave is in some way determined by the speed of the current, just because standing waves in rivers are so common, whereas downstream or upstream moving waves are rare, not visible and not easily felt ( in a kayak you can sometimes feel a wave that’s not clearly visible).
So what is causing these waves, or the waves that form a hydraulic jump? Obviously irregularities in flow, or obstacles or whatever might propagate waves, but why do they so often propagate upstream at the same speed as the current?
On a more minor note I am a bit unhappy with the explanation with your suggestion that the water in a sink has to slow down as it spreads out. This seems anti-newtonian – surely the water molecules will move outward at a constant speed until acted on by another force (friction or something, according to Newton’s first law). If the water spreads out, as it must, surely it just gets shallower. At this scale I imagine friction with the sink bottom is quite important, but I also wonder about drag caused by polar forces. I can imagine in a thin layer these forces might act to pull the water together as the geometry of the flow causes it to spread out. Is this a source of drag? I always get a little uneasy when explanations suggest that something has to happen to a moving particle or something (I have similar problems with the usual explanations for lift produced by air flowing over an aircraft wing). It suggests that the particle somehow “knows” what is happening around it. In the sink example each particle has to “know” that the current is spreading out over a larger area. Surely each particle moves independent of the particles around it unless there are intermolecular forces or something?
It is the beginning of summer here in New Zealand and we have had a couple of days of very clean medium sized swell at the beach – great bodysurfing conditions so I have been spending a bit of time among waves. I’m very grateful for your help in intellectualising the experience!
2. Mirjam
Hi Joe,
Great to hear from you again! :-)
Let’s see if I can answer those questions. Phew! :-)
What’s the source of the waves? There can be several sources. A wave produced by any of the “normal” sources (wind, animals, boats, …) will get locked in place at the hydraulic jump because it can’t travel further upstream. But more relevant at the rapids is probably the shape of the ground underneath: Usually there is a rock or something that the water “jumps over” (similar to those mountain bikers jumping over the little hills — they drive up the slope and then somewhat overshoot at the peak before they land a bit down on the opposite slope). Or if the river suddenly becomes a lot more narrow, the water level might also have risen up before the narrowest bit, and then water “falls down” afterwards when the river widens again. The water overshooting the normal resting water level in either direction is a wave in itself, and then those two or three smaller wave crests downstream are descendants of that first one. Does that make sense?
“I have always assumed that the speed of the wave is in some way determined by the speed of the current, just because standing waves in rivers are so common, whereas downstream or upstream moving waves are rare, not visible and not easily felt ( in a kayak you can sometimes feel a wave that’s not clearly visible).” – Well, this depends on how you measure the speed of the wave. Relative to the current, the wave speed does not depend on the current, it is just moved with it. But then relative to the river bed, of course faster flowing rivers seem to have faster waves on them. And rapids are more easily spotted than “normal” waves, because they are locked in place and persist over longer times. Also typically rapids have a larger wave height than the other “normal” waves caused by wind or whatever (Wind-forced waves in rivers have to be quite small, because there isn’t a lot of fetch for the wind to put energy into generating waves. Animals or paddlers don’t put that much energy into making waves, either)
“… but why do they so often propagate upstream at the same speed as the current?” — Waves propagate in all directions simultaneously, but the part that is propagating downstream is a lot more difficult to spot than the one locked into place, because as it gets swept away, its amplitude is decreasing fairly quickly (since no more energy is supplied to keep it up)
“On a more minor note I am a bit unhappy with the explanation with your suggestion that the water in a sink has to slow down as it spreads out.” — You are right, that’s really not a good way of putting it. The volume flow per circle segment decreases as the radius of the circle increases. So as the water is spread out over larger and larger areas, it becomes shallower and friction with the bottom becomes more and more important. So that’s probably the mechanism there. I don’t think that molecular forces are important for slowing down the flow, but that’s just a gut feeling.
Is this making things any clearer? I need to think about this more, I am getting confused now (Which is awesome, I love thinking about this kind of stuff and don’t do it enough :-D) | 5,548 | 26,031 | {"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-2024-38 | latest | en | 0.951003 |
https://www.physicsforums.com/threads/momentum-planetary-motion-problem.676064/ | 1,521,839,775,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648594.80/warc/CC-MAIN-20180323200519-20180323220519-00658.warc.gz | 869,890,614 | 18,711 | # Momentum Planetary Motion Problem
1. Mar 4, 2013
### Plutonium88
1. The problem statement, all variables and given/known data
Two spheres of mass m and radius r, are released from rest in empty space. The centers of the spheres are separated by a distance R. They end up colliding due to gravitation attraction. Find the magnitude of the impulse just before they collide.
2. Relevant equations
Eg= -Gmm/r
ek= 1/2mv^2
J=F*T
J=ΔP
3. The attempt at a solution
I first drew my problem out with two frames.
The first frame of the two masses separated by a distance R. The second frame, where they are just about to touch separated by a distance of 2r.
I calculated Change in momentum for both objects.
ΔP1=ΔP2
ΔP1=mV
i found that the change in momentum was reliant upon the final velocity..
Initially when i observe the energy of the two spheres, They are seperated by a distance of R, and have no kinetic energy. So if i calculate the Initial energy total for 1 mass, (Which is the same as the other) i get.
Et1= Eg1
Et1= -G*(m^2)/R
In my second frame when i calculate total energy the two spheres are now separated by a distance of 2r. The spheres now have kinetic energy after having moved toward each other..
Et2= Eg2 + Ek2
Et2= -G*((m^2)/2r
Now when i look at this question, i was thinking that it is similar to an orbital question where you have to move an object on the surface to an orbital level and it will have a certain speed... The kinetic energy in a question like that is based on the difference of the First Gravitational Potential Energy Subtract The Second.
I came up with this relation using conservation of energy (i'm assuming no energy is lost cause they are in space and no external forces have acted upon the objects only the conservative force of gravity.)
Et1=Et2
Eg1 = Eg2 + Ek2
Ek2= Eg1-Eg2
Ek2 = -G(m^2)/R - (-G(m^2)/2r)
Ek2= G(m^2)/2r - Gm^2/R
so 1/2mv^2 = G(m^2)/2r - Gm^2/R
this leaves me with
V = √(Gm(1/r - 2/R))
so J = ΔP
so J = mV
so J = m*√(Gm(1/r - 2/R))
so j = √(Gm^3(1/r - 2/R))
This is my dilemma... I found a textbook online which has this same question stated, with the answer to the question i'm solving.... There answer is different than mine so i'm wondering if some one could point out to me what i'm doing wrong. There answer seems to have 1/2r where i have 1/r and 1/R where i have 2/R...
J = √(Gm^3(1/2r - 1/R))
http://books.google.ca/books?id=Kuh...r centers separated by the distance R&f=false
Question #48 P382 (SHOULD BE A DIRECT LINK)
Anyways i would really appreciate some help cause i can't seem to find where i'm going wrong, and i'm not exactly positive if the textbook answer is even correct, because it is just stated in the question itself... I don't even really have a solid "Answer" to check my answer with other than this one... ^.^
Last edited: Mar 4, 2013
2. Mar 4, 2013
### Staff: Mentor
The gravitational energy you calculated is the energy of the whole system, not the energy of a single object.
You have so many equations for something which can be done in a few lines...
Initial energy of the system: Ei=-Gm^2/R
Final energy of the system: Ef=-Gm^2/(2r) + 2*(1/2)mv^2
Energy conservation gives Ei = Ef or
$-\frac{Gm^2}{R}=-\frac{Gm^2}{2r}+mv^2$
=> $m^2v^2=Gm^3\left(\frac{1}{2r}-\frac{1}{R}\right)$
=> $mv=\sqrt{Gm^3\left(\frac{1}{2r}-\frac{1}{R}\right)}$
3. Mar 4, 2013
### Plutonium88
So taking Eg1 = -Gm^2/R is not taking the the energy of a single object? Cause i thought you did the same thing in the next line...? I don't understand can you explain what you mean please?
And here for the EFinal
You are taking -Gm^2/2r + (1/2mv^2 + 1/2mv^2) <=== this is where you get 2MV^2 i'm assuming(please explain)?
but again isn't this taking the energy of the whole system? So which energy am i taking, of the entire system or of a single object?
Thank you for your time and i would really appreciate a response <3.
4. Mar 4, 2013
### Staff: Mentor
It is the energy you need to separate those objects to "infinite" distance. Once you moved one object to "infinite" distance, there is nothing to do for the other object. In that way, the gravitational potential energy of a single object is the same as the gravitational potential energy of the whole system.
Due to symmetry, both masses have the same velocity, and 2*(1/2)=1. I don't get 2mv^2 anywhere.
Correct.
It is easier to consider the whole system, otherwise you have to split the gravitational energy in two parts without a real physical meaning.
5. Mar 4, 2013
### Plutonium88
Thank you sir for the explanations they are clear. I appreciate all the feedback and thank you for brightening the start of my day. | 1,342 | 4,681 | {"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} | 3.6875 | 4 | CC-MAIN-2018-13 | longest | en | 0.900135 |
https://www.thedonutwhole.com/should-i-count-my-net-calories-or-total-calories/ | 1,721,376,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514900.59/warc/CC-MAIN-20240719074314-20240719104314-00579.warc.gz | 866,986,699 | 27,474 | # Should I count my net calories or total calories?
When it comes to weight loss, the most important number to pay attention to is calories. Specifically, you need to create a calorie deficit, meaning you consume fewer calories than you burn each day. This causes your body to dip into its fat stores for energy, resulting in weight loss. But should you count total calories or net calories to most accurately track your deficit? Here’s a quick overview of each approach:
## Counting Total Calories
Counting your total calorie intake for the day is the more straightforward method. You simply add up all the calories from the foods, drinks and snacks you consume. The advantage is you don’t need to do any complicated math to tally your calorie intake. However, this approach doesn’t take into account the calories you burn from exercise and other daily movement. So it may overestimate your actual calorie intake if you are very active.
## Counting Net Calories
Net calories refer to your total calories consumed minus the calories burned through exercise and normal daily activity. For example, if you consumed 2,000 calories but burned 500 calories through a workout, your net calorie intake would be 1,500 calories. The advantage of this method is it provides a more accurate picture of your true calorie balance for the day. However, it requires tracking your exercise calories burned, which can be complicated.
## How to Decide Which to Use
So which calorie counting approach is better for weight loss? Here are some tips:
• If you are sedentary, counting total calories will likely be sufficient since your activity calories will be minimal.
• If you exercise moderately (30-60 minutes a day), you may want to count net calories for more accuracy.
• If you are very active or training for an event, net calories will be important to ensure you eat enough to fuel your workouts.
• If counting net calories, use a fitness tracker or online calculator to estimate your burned calories rather than eating back all the calories your machine reports.
• Consider your lifestyle – if tracking net calories feels like too much work, opt for total calories instead.
## Total Calorie Tracking Tips
Here are some tips for accurately tracking your total calorie intake:
• Use a calorie counting app to look up the calories for all foods and drinks.
• Weigh foods on a kitchen scale for portion accuracy.
• Log calories as you eat throughout the day.
• Include every bite of food, small tastes, and liquid calories.
• Read nutrition labels carefully and track serving sizes.
• For packaged foods, go with the higher calorie listing when a range is given.
Apps like MyFitnessPal or Lose It make tracking total calories easy by having huge databases of food and recipes already logged. Going by package labels and weighing portions will ensure your counts are as precise as possible.
### calorie counting tips
• Use measuring cups and spoons for foods without a label.
• Check calorie counts online if you eat out and don’t know the calories.
• Log a few go-to meals in advance to easily insert them into your log.
• Input home-cooked recipes by adding up the calories for each ingredient.
• When in doubt, overestimate calories to be on the safe side.
## Net Calorie Tracking Tips
To track net calories, you’ll need to follow these steps:
1. Calculate your basal metabolic rate (BMR) – the calories your body burns at rest – using an online calculator.
2. Track your total calories consumed as described in the previous section.
3. Calculate your exercise calories burned using a tracking device or online calculator.
4. Subtract your exercise calories from your total calories consumed to get your net calories for the day.
Some tips for determining exercise calories:
• Use a heart rate monitor along with an activity calculator to estimate calorie burn.
• Wearing a fitness tracker will provide calorie estimates based on your movement and heart rate.
• Stick to calculators from reputable sources and use the more conservative estimates.
• Track BMR calories burned in addition to exercise calories.
## Should You Eat Back Exercise Calories?
A common question when counting net calories is whether or not to eat back the calories burned during exercise. Here are some guidelines:
• Eat back about 30-50% of hard workout calories like running or lifting weights.
• Don’t eat back low intensity workout calories like yoga or walking.
• On heavy training days, eat back more calories if needed to fuel your workouts and recovery.
• If trying to lose weight, err on the side of eating slightly fewer exercise calories.
The reason you may not want to eat back all exercise calories is that activity trackers tend to overestimate calorie burn. So if you eat it all back, you could erase your deficit.
## Example Daily Calorie Tracking
Here is an example of how to track net versus total calories for a day:
### Total Calories
Meal Food Calories
Breakfast Oatmeal 300
Banana 105
Coffee 50
Lunch Tuna sandwich 400
Yogurt 150
Dinner Chicken stir fry 550
Rice 200
Snacks Protein bar 200
Popcorn 120
Total 2075
Without accounting for exercise, this person consumed 2,075 calories for the day.
### Net Calories
Description Calories
Total calories consumed 2075
Morning walk (30 min) -150
Afternoon jog (45 min) -350
Weightlifting workout (60 min) -250
Net Calories 1325
By subtracting 750 exercise calories from the total 2075 consumed, this person has a net intake of only 1325 calories for the day.
## Pros and Cons of Each Approach
Let’s summarize the key advantages and disadvantages of tracking total vs net calories:
### Total Calories:
Pros:
• Simple to calculate.
• No guesswork estimating calories burned.
• Accounts for all food intake.
Cons:
• Doesn’t account for high activity levels.
• Calorie deficit may be overestimated.
### Net Calories:
Pros:
• Accounts for exercise calories burned.
• Provides a more accurate deficit picture.
• Better for active lifestyles.
Cons:
• Harder to accurately track burned calories.
• More time consuming to calculate.
• Higher chance of user error.
## Conclusion
In summary, while tracking total calories is simpler, going with net calories is likely to provide a more accurate view of your true daily calorie balance if you are very active. It ultimately comes down to choosing the method that you will be able to track consistently and that fits best with your lifestyle and goals.
The most important thing, regardless of which method you choose, is accuracy. Underestimating calorie intake or overestimating burn can sabotage your weight loss efforts. Invest in a food scale, use a tracking app diligently, and be conservative in your calorie burn estimates. Consistency is also key – stick with your tracking every single day.
As long as you are able to maintain a consistent daily calorie deficit through your preferred tracking approach, you should see successful weight loss over time. Don’t get hung up on total vs net calories – pick the method that works best for your needs and commit to accurate tracking. | 1,443 | 7,096 | {"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 | 0.920601 |
https://www.physicsforums.com/threads/geometry-of-sphere.66100/ | 1,508,198,569,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00564.warc.gz | 1,181,939,936 | 15,733 | # Geometry of sphere
1. Mar 6, 2005
### kevinalm
Subject wise, this seems the most appropriate forum, so I'll post here. Feel free to move.
Given a circle of radius r, whose center is distance R from point P and which always lies in a plane perpendicular to line defined by P and center of circle (r) at distance R. All points of the circle will always lie within the surface of a sphere centered on P of radius= sqr( r^2 +R^2). I'm pretty confident on my result but intuitively I didn't expect this.
2. Mar 6, 2005
### Andrew Mason
I am a little confused by your terms, but I think this is equivalent to slicing a sphere at a distance R from the centre. This generates a circular cross section and the points on the circumference of that circular cross section are all on the surface of the sphere.
AM
3. Mar 6, 2005
### kevinalm
Yes, that is what I'm doing. I just came at it from a strange viewpoint. I was setting up some kinematic equations for a gyro rim, the axel of which is constrained to pass through point P, and the center of which is constrained to distance R from P. What surprised me was that for all r,R >0 all points of the gyro rim (idealized to a circle of course) lie within a sphere. On reflection, I see this is to be expected. Thanks.
4. Mar 6, 2005
### HallsofIvy
Staff Emeritus
Take any of the points, X, on the circle and draw the line from the center of the circle, O, to X, the line from O to P, and the line from X to P. Since the circle is in a plane perpendicular to OP, those three line segments form a right triangle. The distance from X to O is r and the distance from O to P is R. Use the Pythagorean to find the distance from X to P.
5. Mar 6, 2005
### kevinalm
Yes, that was my original solution, in essence. I was looking at the "axel" rotating in the xy plane and was curious as to the nature of the surface of revolution. What I wasn't expecting was that for all non-negative R,r all points of circle (r) lie in a single sphere of radius= sqr(r^2 +R^2) . It even fails sensibly in the cases R=0, r=0 and R,r=0,0 collapsing to sphere radius r, sphere radius R and point P respectively. For some silly reason I was thinking of varying oblateness, depending on R,r. Didn't want to believe myself. Leads to some interesting implications I may explore, like the instantaneous v always in the tangent plane.(to the sphere at that point.)
Again Thanks. | 612 | 2,403 | {"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-2017-43 | longest | en | 0.957832 |
https://www.statsmedic.com/ced-apstats-unit9-chapter12-day4 | 1,720,975,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00145.warc.gz | 903,664,297 | 224,811 | top of page
## Chapter 12 - Day 4
##### Students Struggled With…
• Remembering what needed to be checked for each of the five conditions.
• Understanding the difference between a residual plot and a dotplot of the residuals.
• Mixing up the slope and y-intercept from computer output.
Remedies…
• (Condition) - how to check it --> SO WHAT?
• (Linear) - scatterplot linear or residual plot with no leftover pattern --> so there is a fairly linear relationship between x and the mean of y.
• (Independent) - 10% condition --> so sampling without replacement is OK
• (Normal) - dotplot of residuals is approximately normal --> so for each x, the distribution of y is approximately normal.
• (Equal Variance) - residual plot shows similar size residuals for each x-value --> so for each x, the standard deviation of y is the same.
• (Random) - random sample or random assignment --> so we can generalize to the population or we can show causation.
• Remind students that statisticians use y = a + bx rather than y = mx + b. It makes sense that the computer output would give us the y-intercept first. ​hey should calculate a z-score and find the area above it.
bottom of page | 278 | 1,187 | {"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-2024-30 | latest | en | 0.861759 |
https://chemistry.stackexchange.com/questions/53625/nernst-equation-half-cell-vs-overall-reaction/53681 | 1,563,439,867,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525587.2/warc/CC-MAIN-20190718083839-20190718105839-00001.warc.gz | 332,942,855 | 35,477 | # Nernst equation: half cell vs overall reaction
Can the Nernst equation always be applied to the overall redox reaction as opposed to only half cell reactions?
For example, in the following silver-silver chloride electrode: $$\ce{Ag (s)|AgCl (s)|Cl- (aq)}$$ rather than applying the Nernst eqution to the half cell reaction $$\ce{AgCl + e- -> Ag + Cl-}$$ can I apply it to $$\ce{AgCl -> Ag+ + Cl-}$$
The answer is the same for the question I was trying to solve (whether I use the overall reaction or the half-cell reaction), but I was wondering if this was just a coincidence or if it was the same.
If they are the same, are there any other factors I should consider such as whether or not the electrolyte fully dissociates in the given conditions?
• $$\ce{AgCl -> Ag+ + Cl-}$$ is not a redox equation! – Yomen Atassi Jun 13 '16 at 14:33
It is coincidence, since $$\ce{AgCl -> Ag+ + Cl-}$$ is not a half cell reaction.
The silver chloride electrode functions as a redox electrode and the reaction is between the silver metal (Ag) and its salt — silver chloride (AgCl). There are two ways to represent the processes (half reactions) happening on the electrode: $$\ce{AgCl + e- -> Ag + Cl-}$$ or $$\ce{Ag+ + e- -> Ag}$$ $$\ce{AgCl -> Ag+ + Cl-}$$
$$E = E^0 - \frac{\mathrm RT}{\mathrm F}\ln(a(\ce{Cl^-}))$$ | 374 | 1,312 | {"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} | 2.703125 | 3 | CC-MAIN-2019-30 | longest | en | 0.92465 |
https://tickeron.com/trading-investing-101/what-dividend-discount-model/ | 1,726,458,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00369.warc.gz | 522,686,962 | 19,258 | What is the Dividend Discount Model?
The Dividend Discount Model (DDM) is a widely used method for estimating the intrinsic value of a stock, particularly in the world of finance and investment analysis. The model's primary focus is on the expected future dividend payouts of a company, which are then adjusted to their present value. By comparing the calculated value to the current trading price, investors can gain insight into whether a security may be overvalued or undervalued. Although the DDM is a helpful tool, it is not without its flaws and should not be relied upon as the sole basis for valuing a company. In this article, we will delve into the mechanics of the DDM, its advantages, and its limitations.
The Dividend Discount Model Explained
At its core, the Dividend Discount Model is a valuation method that seeks to estimate the intrinsic value of a stock by considering the present value of its expected future dividends. The basic formula for DDM can be expressed as:
Stock Value = D1 / (r - g)
Where:
• Stock Value is the intrinsic value of the stock
• D1 is the expected annual dividend payment in the next year
• r is the required rate of return (also referred to as the discount rate)
• g is the expected constant growth rate of the dividends
The model assumes that dividends will grow at a constant rate indefinitely, and the present value of those dividends is calculated using the discount rate (r). If the resulting value is lower than the current trading price, the security may be considered undervalued, and vice versa.
Advantages of the Dividend Discount Model
1. Simplicity: The DDM's straightforward nature makes it easy for investors to understand and apply in their valuation process. The formula itself is relatively simple, requiring only basic financial data and assumptions.
2. Focus on dividends: The model emphasizes the importance of dividends in a company's overall value, as they represent a significant source of return for investors, especially for those seeking income-generating investments.
3. Basis for comparison: By calculating the intrinsic value of a stock, the DDM provides a benchmark for comparing the current trading price, allowing investors to identify potential opportunities for investment.
Limitations of the Dividend Discount Model
Despite its advantages, the DDM has its fair share of limitations, which investors must be aware of before relying solely on this valuation method.
1. Uncertainty in dividend projections: One of the most significant limitations of the DDM is the reliance on future dividend projections, which can be highly uncertain. Companies may change or eliminate dividends at any time, and unforeseen factors such as economic downturns, changes in industry dynamics, or shifts in company strategy can significantly impact dividend payouts.
2. Applicability to non-dividend paying companies: The DDM is not suitable for valuing companies that do not pay dividends or have an irregular dividend history. In such cases, alternative valuation methods, such as the Discounted Cash Flow (DCF) model, may be more appropriate.
3. Constant growth assumption: The DDM assumes that dividends will grow at a constant rate indefinitely, which may not be a realistic assumption for many companies. Dividend growth can be influenced by factors such as competition, regulatory changes, and technological advancements, making it challenging to accurately predict a constant growth rate.
4. Sensitivity to input assumptions: The DDM's calculated intrinsic value is highly sensitive to changes in the discount rate and growth rate assumptions. Small changes in these inputs can lead to significant differences in the estimated stock value, making the model susceptible to errors in estimation.
The Dividend Discount Model is a valuable tool for estimating the intrinsic value of a stock, particularly for dividend-paying companies. Its simplicity and focus on dividends make it an attractive option for investors seeking income-generating investments. However, it is crucial to understand its limitations, including the uncertainty of dividend projections, the inapplicability to non-dividend paying companies, the constant growth assumption, and the sensitivity to input assumptions.
To overcome these limitations, investors should not rely solely on the DDM for valuation purposes. Instead, it should be used in conjunction with other valuation methods, such as the Discounted Cash Flow (DCF) model, Price-to-Earnings (P/E) ratio, and Enterprise Value-to-EBITDA (EV/EBITDA) ratio, among others. By incorporating multiple valuation methods into their analysis, investors can develop a more comprehensive understanding of a company's intrinsic value, mitigating the risks associated with the limitations of the DDM.
The Dividend Discount Model is a useful tool for investors, but it should be used with caution and as one of many inputs when valuing a company. By understanding its strengths and limitations, investors can make more informed decisions, ultimately leading to more successful investment outcomes.
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Why Does the Price of a Stock Change?
Disclaimers and Limitations | 1,132 | 5,965 | {"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-2024-38 | latest | en | 0.921466 |
https://ncertmcq.com/rs-aggarwal-class-8-solutions-chapter-15-quadrilaterals-ex-15/ | 1,723,605,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00228.warc.gz | 323,861,473 | 10,560 | RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15
These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15.
Question 1.
Solution:
(i) Four
(ii) Four
(iii) 4, collinear
(iv) two
(v) opposite
(vi) 360°
Question 2.
Solution:
(i) There are four pairs of adjacent sides which are (AB, BC), (BC, CD), (CD, DA) and (DA, AB)
(ii) There are two pairs of opposite sides which are (AB, CD) and (BC, AD)
(iii) There are four pairs of adjacent angles which are (∠ A, ∠ B), (∠ B, ∠ C), (∠ C, ∠ D) and (∠ D, ∠ A)
(iv) There are two pairs of opposite angles which are (∠A, ∠C) and (∠B, ∠D)
(v) There are two diagonals which are AC and BD.
Question 3.
Solution:
Given : ABCD is a quadrilateral
To prove : ∠A + ∠B + ∠C + ∠D = 360°
Construction : Join BD
Proof : In ∆ ABD,
∠ A + ∠1 + ∠ 4 = 180° (sum of angles of a triangle)
Similarly ∠2 + ∠C + ∠ 3 = 180° Adding we get,
∠ A + ∠1 + ∠4 + ∠2 + ∠C + ∠ 3
= 180° + 180°
=> ∠A + ∠1 + ∠2 + ∠C + ∠3 + ∠4 = 360°
=> ∠A + ∠B + ∠C + ∠D = 360° Hence proved.
Question 4.
Solution:
We know that
Sum of 4 angles of a quadrilateral = 360°
But sum of 3 angles = 76° + 54° + 108°
= 238°
4th angle = 360 – 238°
= 122°
Hence, measure of fourth angle = 122° Ans
Question 5.
Solution:
Ratio of four angles of a quadrilateral = 3 : 5 : 7 : 9
Let these angles be 3x, 5x, 7x and 9x
then 3x + 5x + 7x + 9x = 360° (sum of angles)
=> 24x = 360°
First angle = 3x = 3 x 15° = 45°
Second angle = 5x = 5 x 15° = 75°
Third angle = 7x = 7 x 15° = 105°
Fourth angle = 9x = 9 x 15° = 135° Ans.
Question 6.
Solution:
Three acute angles of a quadrilateral are 75° each
Sum of three angles = 3 x 75° = 225°
But sum of 4 angles = 360°
Fourth angle = 360° – 225°
= 135° Ans.
Question 7.
Solution:
Sum of 4 angles of a quadrilateral 360°
One angles = 120°
Sum of other three angles = 360° – 120° = 240°
But each of these 3 angles are equal
Each of equal angles = $$\frac { 240^{ o } }{ 3 }$$
= 80°
Question 8.
Solution:
Sum of 4 angles of a quadrilateral = 360°
Sum of two angles = 85° + 75° = 160°
Sum of other two angles = 360° – 160° = 200°
But each of these two angles are equal
Measure of each equal angle = $$\frac { 200^{ o } }{ 2 }$$
= 100° Ans.
Question 9.
Solution:
∠C = 100°, ∠D = 60°
and ∠A + ∠B + ∠C + ∠D = 360°
(sum of angles of a quadrilateral)
∴ ∠ A + ∠ B = 360° – (100° + 60°)
= 360° – 160° = 200°
But AP and BP are the bisectors of ∠ A and ∠ B
∴ $$\\ \frac { 1 }{ 2 }$$ – (∠ A + ∠B) = 200° x $$\\ \frac { 1 }{ 2 }$$ = 100°
i.e. ∠ 1 + ∠2 = 100°
But in ∆ APB,
∠1 + ∠2 + ∠P = 180°
=> 100° + ∠P = 180°
=> ∠P = 180° – 100° = 80°
or ∠APB = 80° Ans.
Hope given RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for yo | 1,163 | 2,874 | {"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} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.816949 |
https://www.convertunits.com/molarmass/Calcium+Peroxide | 1,566,098,983,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313589.19/warc/CC-MAIN-20190818022816-20190818044816-00188.warc.gz | 788,241,956 | 9,370 | # ››Calcium Peroxide molecular weight
Molar mass of CaO2 = 72.0768 g/mol
Molecular weight calculation:
40.078 + 15.9994*2
# ››Percent composition by element
Element Symbol Atomic Mass # of Atoms Mass Percent Calcium Ca 40.078 1 55.605% Oxygen O 15.9994 2 44.395%
# ››Calculate the molecular weight of a chemical compound
Enter a chemical formula:
Browse the list of common chemical compounds.
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Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.
Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.
Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. | 510 | 2,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-35 | longest | en | 0.919723 |
https://www.sscadda.com/latitude-and-longitude-of-india/ | 1,675,872,581,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500837.65/warc/CC-MAIN-20230208155417-20230208185417-00169.warc.gz | 996,156,151 | 121,482 | Latest SSC jobs » GK Questions » Latitudes and Longitudes
# Latitude and Longitude of India – Definition and Difference
## Latitude and Longitude
Latitude and Longitude: The Latitude and Longitude are commonly referred to as geographical coordinates as they provide a systematic network of lines upon which the position of various surface features of the earth can be represented. With the help of these coordinates, the location, distance, and direction of various points can be easily determined. Latitude and Longitude are measured in degrees (°) because they represent angular distances. Each degree is further divided into 60 minutes ( ‘ ) and each minute into 60 seconds ( “ ).
Prime Minister of India
## What are Latitude and Longitude?
### Latitude:
• Latitude is defined as the angular distance of any point, north or south of the equator, i.e. it is a coordinate system, that is used as a reference point to locate places on earth.
• The Equator is an imaginary circular line drawn on the earth, which divides it into two equal parts, wherein the upper half is called Northern Hemisphere, and the lower half is known as Southern Hemisphere. The circular lines, parallel to the equator, and up to the North and South poles are the parallels of latitude.
• Latitude ranges from 0 to 90 degrees, with the equator indicating 0 degrees and the poles indicating 90 degrees. The Equator is the longest line of latitude. Parallels in the northern hemisphere are referred to as north latitudes, while those in the southern hemisphere are referred to as south latitudes. The following are some major latitudes:
1. Tropic of Cancer (23.5° N)
2. Tropic of Capricorn (23.5° S)
3. Arctic Circle (66.5° N)
4. Antarctic Circle (66.5° S)
Latitude and Longitude of India in Hindi
### Longitude:
• Longitude is the angular distance of any point east or west of the Prime Meridian or west of the Standard Meridian. It determines how far a specific location is from the reference line. Meridians of longitude are reference lines that run from the north pole to the south pole. These are semicircles, and the distance between them decreases sharply as they approach the poles.
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• All meridians are the same length and the Greenwich Meridian is used to count meridians. The prime meridian has a value of 0° longitude and divides the earth into two equal parts, the Eastern and Western Hemispheres.
States and Capitals
Union Territories in India
## Difference between Latitude and Longitude
The difference between Latitude and Longitude is given in the table below:
BASIS FOR COMPARISON LATITUDE LONGITUDE
Meaning Latitude implies the geographic coordinate which determines the distance of a point, north-south of the equator. Longitude alludes to the geographic coordinate, which identifies the distance of a point, east-west of the Prime Meridian.
Direction East to west North to south
Stretches from 0 to 90° 0 to 180°
Lines of reference Known as parallels Known as meridians
Number of lines 180 360
Length of lines Different Same
Parallel Yes, the lines are parallel. No, the lines are not parallel.
Classifies Heat Zones Time Zones
## Latitude and Longitude of India
India is a vast country. Lying entirely in the Northern hemisphere the mainland extends between latitudes 8°4’N and 37°6’N and longitudes 68°7’E and 97°25’E. The Indian Government has accepted the meridian of 82.5° east for the standard time which is 5 hours 30 mins, ahead of Greenwich Mean Time.
Check related links Largest State in India Important Days and Dates National Symbols Of India Preamble of Indian constitution
## FAQs
### What do you mean by Latitudes and Longitudes?
Latitudes and Longitudes are imaginary lines used to determine the location of a place on earth.
### What do you mean by Parallels of Latitude?
The parallels of latitude refer to the angular distance, in degrees, minutes and seconds of a point north or south of the Equator. Lines of latitude are often referred to as parallels.
### How are latitudes and longitudes measured?
Latitudes and longitudes are measured in degrees (°) because they represent angular distances. Each degree is further divided into 60 minutes ( ‘ ) and each minute into 60 seconds ( “ ).
#### Congratulations!
General Awareness & Science Capsule PDF | 943 | 4,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | longest | en | 0.892194 |
https://internal.ncl.ac.uk/ask/numeracy-maths-statistics/core-mathematics/pure-maths/algebra/factorials.html | 1,601,135,970,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244231.61/warc/CC-MAIN-20200926134026-20200926164026-00630.warc.gz | 456,718,317 | 5,824 | Factorials
Definition
A factorial of a positive number $n$ is defined to be the product of all positive whole numbers less than and equal to $n$. E.g. $n=5$ $5! = 5 \times 4 \times 3 \times 2 \times 1.$ Note: We define the factorial $0!=1$.
Worked Examples
Example 1
Calculate $7!$.
Solution
From the definition we have $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1.$ Hence $7! =5040.$
Example 2
Simplify $\dfrac{9!}{6!}$.
Solution
Firstly we can expand the factorials using the definition $\dfrac{9!}{6!} = \dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}.$ This can then be simplified by canceling terms on the top and bottom to give $\dfrac{9!}{6!} = 9 \times 8 \times 7.$ Therefore, $\dfrac{9!}{6!} = 504.$
Example 3
Simplify $\dfrac{16!}{13!\cdot 8!}$
Solution
As above start by expanding the factorials $\dfrac{16!}{13!\cdot 8!} = \dfrac{16 \times 15 \times \dots \times 3 \times 2 \times 1}{(13 \times 12 \times 11 \times \dots \times 3 \times 2 \times 1 )( 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}.$ Then cancel out any terms that appear on the top and the bottom of the fraction $\dfrac{16!}{13!\cdot 8!} = \dfrac{16 \times 15 \times 14}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}.$ This can be canceled further if we factorize the numerator $\dfrac{16!}{13!\cdot 8!} = \dfrac{8 \times 2 \times 5 \times 3 \times 2 \times 7}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}.$ Then finally we get \begin{align} \dfrac{16!}{13! \cdot 8!} &= \dfrac{1}{4 \times 3},\\ &=\dfrac{1}{12}. \end{align}
Example 4
Simplify $\dfrac{(n-1)!}{(n+1)!}.$
Solution
\begin{align} \dfrac{(n-1)!}{(n+1)!} &= \dfrac{(n-1) \times (n-2) \times \dots \times 2 \times 1}{ (n+1) \times n \times (n-1) \times \dots \times 2 \times 1},\\ &= \dfrac{1}{(n+1) \times n},\\ &= \dfrac{1}{n^2 + n}. \end{align} | 767 | 1,948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2020-40 | latest | en | 0.472352 |
http://www.thestudentroom.co.uk/showthread.php?t=4012785 | 1,477,533,462,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721027.15/warc/CC-MAIN-20161020183841-00109-ip-10-171-6-4.ec2.internal.warc.gz | 727,208,901 | 36,779 | You are Here: Home >< Maths
# Integrating parametric equations
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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. I'm stuck on a question from the C4 book (chapter 6, Mixed Exercises 15 (d)).
I understand finding the area of the whole triangle, but not how to find the area of A1. Why is the curve 4sinθ(-5sinθ)? I know how to go from there, but not how they got that equation.
Thanks guys!
2. (Original post by кяя)
I understand finding the area of the whole triangle, but not how to find the area of A1. Why is the curve 4sinθ(-5sinθ)? I know how to go from there, but not how they got that equation.
Thanks guys!
3. (Original post by кяя)
...
As an alternative to the above (I prefer this once since you get used to the idea of substitutions being scalings and this will help once you do multivariate calculus and need to deal with Jacobians, but I digress):
You know the area under a curve is given in cartesian co-ordinates as , yes? But at the same time you can think of since they 's kind of cancel.
Let's go ahead and re-write this now:
Now you know that and and but then the negative sign means you flip them, so you're left with:
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Top tips from students who have already aced their exams | 535 | 2,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2016-44 | latest | en | 0.941913 |
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posted by .
Solve using distributive property
(Show all Work)
15-3y=y+13+y
• algebra -
15-3y=y+13+y
=13+2y
15-3y=13+2y
+3y +3y
15=13+5y
-13 -13
12=5y
y=2.4
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You exercised 24 hours each month for a year. How many hours did you exercise by the end of the year?
More Similar Questions | 588 | 1,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-34 | latest | en | 0.861401 |
https://datascience.stackexchange.com/questions/13359/identifying-waveform-segments-using-training-waveforms | 1,702,330,681,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00545.warc.gz | 229,521,805 | 43,750 | # Identifying Waveform Segments Using Training Waveforms
Problem : There are several events (eventA, eventB,....) represented by waveforms. For each event there are several csv files (eventA1.csv, eventA2.csv,...eventAn.csv) having the points(x,y) from which the corresponding waveform can be generated.
Using these waveforms as training data, waveform segments on (test data) larger data should be identified i.e. the interval during which an event occurs is needed to be labelled on the test data (See image for reference).
1. What would be the best strategy, tools and algorithms to address this problem?
2. If classification model is to be generated then,
1. The points are just to describe the nature of the waveform i.e. the test waveform will not be on same points but with similar nature e.g. the concavity, convexity, slope etc.
2. Different events in training graphs are represented using different number of points i.e. event A can be determined by 20 points, while event B has 200 points.
So,
1. How do I create feature vectors to include all these things for classification?
2. How would the test data be analysed, as it has large number of points on the large waveform i.e. how to segment it to generate features and give it to classification model?
• This seems to be a signal processing problem. Could you either resample your signal to be relative to your signal features, or vary the size of your signal features? Then you could convolve the signal features with the test signal and retrieve an impulse where there are similarities. Aug 10, 2016 at 22:14
## Statistical Data Analysis
1. Statistically analyse your training events and see the mean and the variance of length and width and other properties to have an overall image of what they are.
2. For each event take the mean and the standard deviation.
3. Walk through the test data and compute the error between test signal and learned events and plot it.
4. At the end for each event type, you will have an error curve whit many local minimums.
5. Choose a threshold based on standard deviation of training events and detect minimums behind that threshold as events.
## Pattern Recognition
More advanced option is to learn a model to classify events. At the beginning I should say the last question is theoretically wrong because you do not generate features from test data but only compute them. Features have been extracted (generated) from training data. I assume you mean how to feed the model with test data for classification according to its size. Well, I'd say you can segment it to chunk which may lead to losing some events because of cutting points but if test data is large enough that would not be a problem.
The point is that time-series are a special kind of data and there are methods specifically for them so the classic feature extraction and learning may lead you to an ocean of confusing methods. I suggest to search the term Time-series Mining instead to see more direct solutions.
## Time-Series Mining
The very first approach I proposed was kind of this! There are many more algorithms for detecting similarity based on two time-series mostly based on Euclidean distance. For example look at LCSS or DTW. These methods are what you are looking for and both are advanced version of my first suggested algorithm. Please note that for all of them normalizing your training and test data plays an important role.
See this answer as well however that is about unsupervised segmentation of time-series.
• Thanks for providing a deeper insight into possible solutions! Can you provide some links to statistical segmentation examples using time series, as I couldn't find any relevant source? Moreover, on discussions, I also came across that "Hidden Markov Models" can also be used here, whats your say? Any links? Aug 22, 2016 at 8:34
• Actually I provided the link to a paper in the answer I hyeperlinked for you. It's here lancs.ac.uk/~khaleghi/Publications_files/khaleghi16a.pdf . But 2 points: 1) it's an unsupervised method and you need to modify it to your question as I stated in my answer. 2) The paper is deeply mathematical and if you are not that into theoretical background of time-series analysis then that might be pretty difficult to follow. I would suggest you to have a look at web.science.mq.edu.au/~cassidy/comp449/html/ch11s02.html for some insight into DTW plus explanations of HMM usage as u asked Aug 22, 2016 at 11:47
• But my favourite tutorial is one published by Springer in which DTW is explained theoretically rigorous but totally understandable. The great point about that tutorial is that it contains subsequence matching which is exactly your problem. I could not copy the link here but if you google "dynamic time warping" one of top 5 results would be a PDF file from Springer. I'd say go for that directly. Aug 22, 2016 at 11:50
• About HMMs: As you will see in the second link, HMMs are pretty fine for your application but again; it depends on your background. If you are from Machine Learning community go for it otherwise it might be a bit overdeep mathematically. But after all, HMMs are great and do the job like perfect. The best HMM tutorial is by Andrew Moore from Carnegie Mellon. His tutorials are widely read in ML community because they are just great! Here you can find them: autonlab.org/tutorials/hmm14.pdf Aug 22, 2016 at 11:54
• @KasraManshaei great summary for anybody starting with solving similar problems. Do you see any problems in using this approach to analyse data from Googles project Soli radar? Aug 21, 2019 at 17:56 | 1,227 | 5,588 | {"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-2023-50 | latest | en | 0.932177 |
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The "Hardware" of Life Argument for a...
# The "Hardware" of Life Argument for an Intelligent Designer, part 1
There are 10 comments on the Examiner.com story from Jul 24, 2013, titled The "Hardware" of Life Argument for an Intelligent Designer, part 1. In it, Examiner.com reports that:
The common scientific view of the "hardware of life" is, as Biologist Richard Dawkins puts it, "the study of complicated things that give the appearance of having been designed for a purpose."
Join the discussion below, or Read more at Examiner.com.
“you must not give faith”
Since: Jul 12
Leicester, UK
#1 Jul 24, 2013
I like another way to refute the "odds of a single bacterium reassembling by chance is one in 10 to the 100,000,000,000th power" argument.
First state the obvious, this is not how evolution works, their would have been less complex things before the bacterium.
Second convert the 10 to the 100,000,000,000th power to the number 10,000,000,000,000 and then show the probability of a single leaf falling (19,750,348,800,000 to 1) and DarkAntics video "Beating Astronomical Odds"
It doesn't directly relate to the odds of a single... argument but it does show that probability doesn't matter, it is the rate of attempts that matters. Role 30 dice for long enough and you will get them to be all six's at some point just like you would if you got many sets of dice rolling at the same time. Time and mass numbers can make the near impossible a daily occurrence.
Judged:
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Thinking
Royston, UK
#2 Jul 24, 2013
You don't even need to roll all 6s.
Benjamin Frankly wrote:
I like another way to refute the "odds of a single bacterium reassembling by chance is one in 10 to the 100,000,000,000th power" argument.
First state the obvious, this is not how evolution works, their would have been less complex things before the bacterium.
Second convert the 10 to the 100,000,000,000th power to the number 10,000,000,000,000 and then show the probability of a single leaf falling (19,750,348,800,000 to 1) and DarkAntics video "Beating Astronomical Odds"
It doesn't directly relate to the odds of a single... argument but it does show that probability doesn't matter, it is the rate of attempts that matters. Role 30 dice for long enough and you will get them to be all six's at some point just like you would if you got many sets of dice rolling at the same time. Time and mass numbers can make the near impossible a daily occurrence.
“you must not give faith”
Since: Jul 12
Leicester, UK
#3 Jul 24, 2013
Thinking wrote:
You don't even need to roll all 6s.
<quoted text>
It's just more impressive if say all six's, I know the chances are the same as any other combination but (for better or worse) the human heart doesn't take that into consideration.
EdSed
Hamilton, UK
#4 Jul 24, 2013
Thinking
Royston, UK
#5 Jul 24, 2013
I recently posted on how unlikely it was that you dealt your last 52 card deck in that order - about 1.2*10^-66 of a percent - but it 100% just happened.
Benjamin Frankly wrote:
<quoted text>
It's just more impressive if say all six's, I know the chances are the same as any other combination but (for better or worse) the human heart doesn't take that into consideration.
“you must not give faith”
Since: Jul 12
Leicester, UK
#6 Jul 24, 2013
Thinking wrote:
I recently posted on how unlikely it was that you dealt your last 52 card deck in that order - about 1.2*10^-66 of a percent - but it 100% just happened.
<quoted text>
I assume that's bigger because my brain can't grasp that number.
Thinking
Royston, UK
#7 Jul 24, 2013
Looking at it the other way up, there are 80,658,175,170,943,878,571,660 ,636,856,403,766,975,289,505,4 40,883,277,824,000,000,000,000 possible combinations of a 52 card deck.
Yet any dealt previously is 100% probable.
The next time anyone says "life is too unlikely to have happened without [insert deity here]", maybe this will help them see how flawed their argument is.
Benjamin Frankly wrote:
<quoted text>
I assume that's bigger because my brain can't grasp that number.
Judged:
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Amused
#8 Jul 29, 2013
The puddle fits exactly into the depression in the ground. Therefore, the whole thing has to have been designed, otherwise they would not fit so perfectly together.
Thinking
Royston, UK
#9 Jul 29, 2013
I can see why you link depression with religion.
Amused wrote:
The puddle fits exactly into the depression in the ground. Therefore, the whole thing has to have been designed, otherwise they would not fit so perfectly together.
Amused
#10 Jul 30, 2013
Thinking wrote:
I can see why you link depression with religion.
<quoted text>
It is depressing that so many people with the 'hardware' needed to think choose instead to put their brains in neutral and coast with 'goddidit'.
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# How will you divide 8kg rice into 2 equal halves only with the help of a 5kg and a 3kg pot?
Updated: 4/28/2022
Wiki User
12y ago
you want to end upwith 2 pots each with 4 kg
Here is one way:
1. Fill the 3kg pot from the 8 kg bag; yhe 3kg pot has 3 kg in it and the bag is left with 5 kg
2. Pour the 3 kg pot into the 5kg pot and it now has zero, the 5kg pot has 3 kg
3.Pour the 5kg from the bag into the 3 kg pot. Thereis now 2 kg in the bag , 3 kg in the 3 kg pot, and still 3kg in the 5 kg pot.
4. Pour the 3 kg from the 3 kg pot into the 5 kg pot until full. Now there is 1 kg in the 3 kg pot, 5 kg in the 5 kg pot, still 2 kg in the bag of rice
5. Pour the 5 kg fromthe 5 kg pot into the bag and then the 1 kg from the 3 kg pot into the 5 kg pot. Now the bag has 7 kg, the 5 kg pot has 1 kg and the 3 kg pot has zero
6. Fillthe 3 kg pot from the 7 kg bag; the bag has 4 kg now, the 5 kg pot has 1 kg and the 3 kg pot has 3 kg
7. Pour the 3 kg from the 3 kg pot into the 5 kg pot; the 5 kg pot now has 4 kg in it and the bag also has 4 kg in it
That's it - there has to be an easier way?
Wiki User
12y ago
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Q: How will you divide 8kg rice into 2 equal halves only with the help of a 5kg and a 3kg pot?
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### How many syllables are in the word halves?
There is only 1 syllable in "halves." | 821 | 3,046 | {"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} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.92851 |
https://www.memrise.com/course/700001/learn-mathematics/176/ | 1,576,186,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540547165.98/warc/CC-MAIN-20191212205036-20191212233036-00407.warc.gz | 784,256,612 | 60,994 | Level 175 Level 177
Level 176
## Ignore words
Check the boxes below to ignore/unignore words, then click save at the bottom. Ignored words will never appear in any learning session.
Ignore?
point slope form
where m is the slope and (x1, y1) is a point on the line
y - intercept
the y-coordinate of the point where the line crosses the y-axis
1 - Write ƒ(0) = 5 as (0, 5) and ƒ(4) = 17 as (4, 17)
Ex: Write an equation for the linear function ƒ with the values ƒ(0) = 5 and ƒ(4) = 17.
How to Write an Equation of a Line in Slope-Intercept Form
1 - Identify the slope m. You can use the slope formula if you know two points on the line.
standard form
can be used to find the x- and y-intercepts.
the slope and the y-intercept
What do you need to graph a line?
plug the slope and a point into y = mx + b
Write the equation of a line that passes through the point with the given slope
use the formula to find the slope
Write the equation of a line in slope intercept form that passes through two points
Converse
(noun) A statement that switches the hypothesis and conclusion
parallel lines
2 lines that never cross/intersect and stay the same distance apart
1 - Identify the slope (in this case = 3)
Write an equation of the line that passes through (-3, -5) and is parallel to the line y = 3x - 1
Perpendicular lines
lines in a plane that intersect to form 4 right angles
To determine whether lines are parallel or perpendicular
write all equations in slope-intercept form ( y = mx + b)
Scatter plot
is a graph used to determine whether there is a relationship between paired data. Scatter plots can show trends in data.
positive correlation
y tends to increase, as x increases
negative correlation
y tends to decrease, as x increases
Relatively No Correlation
x and y have no apparent relationship
line of best fit
The line that most closely follows a trend in data is called the best-fitting line.
Linear Regression
the process of finding the best-fitting line to model a set of data; you can use technology to do this
Linear Interpolation
using a line or its equation to approximate a value between two known values
Interpolate Using an Equation
1 - Make a scatter plot of the data
Linear Extrapolation
using a line or its equation to approximate a value outside the range of known values
Zeros of a Function
for the function ƒ, any number x such that ƒ(x) = 0 | 590 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-51 | longest | en | 0.90079 |
http://innoludic.com/puzzle/sudoku/even/507-sudoku-even-1.html?tmpl=component&print=1&layout=default | 1,566,789,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00331.warc.gz | 101,497,670 | 7,958 | ## Sudoku Even # 1
This variant of Sudoku is easy to do. The usual rules apply: in each row, column and region the numbers from 1 to 9 must appear only once. But as an additional constraint the grey cells must have even numbers : 2,4,6 and 8. This means that the white cells have odd numbers: 1,3,5 and 7. So I have created a new item in the menu where all the futur Sudoku Even will appear. Also this variant could be applied to Super Sudoku of size 16x16 (see the last one). Sometimes we do not want to rackle our brain then a gentle puzzle is welcome. The Sudoku even is perfect for that.
To begin, I offer you three puzzles of size 9x9.
1)
The printable files PDF.
Puzzle: sudo_even_00001.pdf
Solution : sudo_even_00001_sol.pdf
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Puzzle: sudo_even_00002.pdf
Solution : sudo_even_00002_sol.pdf
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The printable files PDF.
Puzzle: sudo_even_00003.pdf
Solution : sudo_even_00003_sol.pdf
And the last one is a Super Sudoku of Size 16x16
The printable files PDF.
Puzzle: super_even_00222.pdf
Solution : super_even_00222_sol.pdf
I would like to remind you that you can find Super Sudoku even in my latest book : Super Sudoku 16x16 Variations. This book is full of Super Sudoku in many forms and shapes. It is a must for all fans of Super Sudoku | 348 | 1,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | latest | en | 0.79724 |
https://stackoverflow.com/questions/16266809/convert-from-latitude-longitude-to-x-y | 1,721,132,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00499.warc.gz | 481,477,953 | 51,123 | # Convert from latitude, longitude to x, y
I want to convert GPS location (latitude, longitude) into x,y coordinates. I found many links about this topic and applied it, but it doesn't give me the correct answer!
I am following these steps to test the answer: (1) firstly, i take two positions and calculate the distance between them using maps. (2) then convert the two positions into x,y coordinates. (3) then again calculate distance between the two points in the x,y coordinates and see if it give me the same result in point(1) or not.
one of the solution i found the following, but it doesn't give me correct answer!
``````latitude = Math.PI * latitude / 180;
longitude = Math.PI * longitude / 180;
latitude -= 1.570795765134; // subtract 90 degrees (in radians)
// and switch z and y
xPos = (app.radius) * Math.sin(latitude) * Math.cos(longitude);
zPos = (app.radius) * Math.sin(latitude) * Math.sin(longitude);
``````
also i tried this link but still not work with me well!
any help how to convert from(latitude, longitude) to (x,y) ?
Thanks,
• Why you // subtract 90 degrees ? Commented May 28, 2020 at 14:53
# No exact solution exists
There is no isometric map from the sphere to the plane. When you convert lat/lon coordinates from the sphere to x/y coordinates in the plane, you cannot hope that all lengths will be preserved by this operation. You have to accept some kind of deformation. Many different map projections do exist, which can achieve different compromises between preservations of lengths, angles and areas. For smallish parts of earth's surface, transverse Mercator is quite common. You might have heard about UTM. But there are many more.
The formulas you quote compute x/y/z, i.e. a point in 3D space. But even there you'd not get correct distances automatically. The shortest distance between two points on the surface of the sphere would go through that sphere, whereas distances on the earth are mostly geodesic lengths following the surface. So they will be longer.
# Approximation for small areas
If the part of the surface of the earth which you want to draw is relatively small, then you can use a very simple approximation. You can simply use the horizontal axis x to denote longitude λ, the vertical axis y to denote latitude φ. The ratio between these should not be 1:1, though. Instead you should use cos(φ0) as the aspect ratio, where φ0 denotes a latitude close to the center of your map. Furthermore, to convert from angles (measured in radians) to lengths, you multiply by the radius of the earth (which in this model is assumed to be a sphere).
• x = r λ cos(φ0)
• y = r φ
This is simple equirectangular projection. In most cases, you'll be able to compute cos(φ0) only once, which makes subsequent computations of large numbers of points really cheap.
• Thanks @MvG, I am not involved with these topics, i need just to draw points on a map, these points the distances between them don't exceed 50 meter, is there an easy way for that? Commented Apr 29, 2013 at 13:03
– MvG
Commented Apr 29, 2013 at 14:27
• Thanks, but i have these questions: here λ and φ in radian? also about φ0, what do you suggest to me put it, if my area nearly 50mx50m Commented Apr 29, 2013 at 20:07
• @alsadi90: Whether you measure your angles in degrees or radians only affects the size but not the shape of the map. You have to make sure to compute the cosine correctly, which in most programming languages means using radians. For φ₀ I'd suggest you simply take the average between the lower bound and the upper bound of your latitudes. But for such small maps, taking any latitude from within that area will likely work well enough.
– MvG
Commented Apr 30, 2013 at 4:48
• Thanks for your help, but i applied the steps on this example:: point1(32.49194144785378,35.99031005054712) ,, point2(32.492285325127966,35.99031541496515) ,, the actual distance between them = 38.284 meter _ using this sight daftlogic.com/projects-google-maps-distance-calculator.htm _ then i applied the steps assuming φ₀ = 32.492113386490873 which is the avg of lat of the two points ,, then in x,y coordination i got point1(0.3559,0.5668) and ppoint2(0.3559,0.5668), then distance between them = 5.9990e-06 !! which is wrong value! Commented Apr 30, 2013 at 7:16
I want to share with you how I managed the problem. I've used the equirectangular projection just like @MvG said, but this method gives you X and Y positions related to the globe (or the entire map), this means that you get global positions. In my case, I wanted to convert coordinates in a small area (about 500m square), so I related the projection point to another 2 points, getting the global positions and relating to local (on screen) positions, just like this:
First, I choose 2 points (top-left and bottom-right) around the area where I want to project, just like this picture:
Once I have the global reference area in lat and lng, I do the same for screen positions. The objects containing this data are shown below.
``````//top-left reference point
var p0 = {
scrX: 23.69, // Minimum X position on screen
scrY: -0.5, // Minimum Y position on screen
lat: -22.814895, // Latitude
lng: -47.072892 // Longitude
}
//bottom-right reference point
var p1 = {
scrX: 276, // Maximum X position on screen
scrY: 178.9, // Maximum Y position on screen
lat: -22.816419, // Latitude
lng: -47.070563 // Longitude
}
//## Now I can calculate the global X and Y for each reference point ##\\
// This function converts lat and lng coordinates to GLOBAL X and Y positions
function latlngToGlobalXY(lat, lng){
//Calculates x based on cos of average of the latitudes
let x = radius*lng*Math.cos((p0.lat + p1.lat)/2);
//Calculates y based on latitude
return {x: x, y: y}
}
// Calculate global X and Y for top-left reference point
p0.pos = latlngToGlobalXY(p0.lat, p0.lng);
// Calculate global X and Y for bottom-right reference point
p1.pos = latlngToGlobalXY(p1.lat, p1.lng);
/*
* This gives me the X and Y in relation to map for the 2 reference points.
* Now we have the global AND screen areas and then we can relate both for the projection point.
*/
// This function converts lat and lng coordinates to SCREEN X and Y positions
function latlngToScreenXY(lat, lng){
//Calculate global X and Y for projection point
let pos = latlngToGlobalXY(lat, lng);
//Calculate the percentage of Global X position in relation to total global width
pos.perX = ((pos.x-p0.pos.x)/(p1.pos.x - p0.pos.x));
//Calculate the percentage of Global Y position in relation to total global height
pos.perY = ((pos.y-p0.pos.y)/(p1.pos.y - p0.pos.y));
//Returns the screen position based on reference points
return {
x: p0.scrX + (p1.scrX - p0.scrX)*pos.perX,
y: p0.scrY + (p1.scrY - p0.scrY)*pos.perY
}
}
//# The usage is like this #\\
var pos = latlngToScreenXY(-22.815319, -47.071718);
\$point = \$("#point-to-project");
\$point.css("left", pos.x+"em");
\$point.css("top", pos.y+"em");
``````
As you can see, I made this in javascript, but the calculations can be translated to any language.
P.S. I'm applying the converted positions to an HTML element whose id is "point-to-project". To use this piece of code on your project, you shall create this element (styled as position absolute) or change the "usage" block.
• There's an error: the radius of the Earth in km is `6371`, not `6.371` (which in JS is a little more than 6). Of course over such a small area it doesn't change much, but future readers should take note of it. Commented Sep 8, 2019 at 10:32
• I am trying to use the above JS to do exactly the same thing you did: I have top-left and bottom-right lat\log coord. and related x\y values for a map of WA, OR and CA. I am testing the code by plotting lat 48,46,44,42 and 40 with long of -122. I was expecting my plotting points to be on top of the cross-hair of these coordinates but they just go on a straight line down (i.imgur.com/GuCP0iz.png). They do not follow the "curve" of the long line. Your map (as you mentioned) is small and mine is not. Is this a limitation of the code? Or this is not the code to solve my issue? Thank you Commented Oct 13, 2023 at 1:55
Since this page shows up on top of google while i searched for this same problem, I would like to provide a more practical answers. The answer by MVG is correct but rather theoratical.
I have made a track plotting app for the fitbit ionic in javascript. The code below is how I tackled the problem.
``````//LOCATION PROVIDER
index.js
var gpsFix = false;
var circumferenceAtLat = 0;
function locationSuccess(pos){
if(!gpsFix){
gpsFix = true;
circumferenceAtLat = Math.cos(pos.coords.latitude*0.01745329251)*111305;
}
pos.x:Math.round(pos.coords.longitude*circumferenceAtLat),
pos.y:Math.round(pos.coords.latitude*110919),
plotTrack(pos);
}
``````
plotting.js
``````plotTrack(position){
let x = Math.round((this.segments[i].start.x - this.bounds.minX)*this.scale);
let y = Math.round(this.bounds.maxY - this.segments[i].start.y)*this.scale; //heights needs to be inverted
//redraw?
let redraw = false;
//x or y bounds?
if(position.x>this.bounds.maxX){
this.bounds.maxX = (position.x-this.bounds.minX)*1.1+this.bounds.minX; //increase by 10%
redraw = true;
}
if(position.x<this.bounds.minX){
this.bounds.minX = this.bounds.maxX-(this.bounds.maxX-position.x)*1.1;
redraw = true;
};
if(position.y>this.bounds.maxY){
this.bounds.maxY = (position.y-this.bounds.minY)*1.1+this.bounds.minY; //increase by 10%
redraw = true;
}
if(position.y<this.bounds.minY){
this.bounds.minY = this.bounds.maxY-(this.bounds.maxY-position.y)*1.1;
redraw = true;
}
if(redraw){
reDraw();
}
}
function reDraw(){
let xScale = device.screen.width / (this.bounds.maxX-this.bounds.minX);
let yScale = device.screen.height / (this.bounds.maxY-this.bounds.minY);
if(xScale<yScale) this.scale = xScale;
else this.scale = yScale;
//Loop trough your object to redraw all of them
}
``````
• Are you converting to SI (km) or Imperial (mi) unit? Commented Jan 11, 2020 at 9:22
• Hi @CloudCho. I use metric, as any programming language does. I do a conversion only for visual reasons in the very end of my code.111305 is the difference in meters between 1 degree of latitude. Commented Jan 15, 2022 at 22:15
For completeness I like to add my python adaption of @allexrm code which worked really well. Thanks again!
``````radius = 6371 #Earth Radius in KM
class referencePoint:
def __init__(self, scrX, scrY, lat, lng):
self.scrX = scrX
self.scrY = scrY
self.lat = lat
self.lng = lng
# Calculate global X and Y for top-left reference point
p0 = referencePoint(0, 0, 52.526470, 13.403215)
# Calculate global X and Y for bottom-right reference point
p1 = referencePoint(2244, 2060, 52.525035, 13.405809)
# This function converts lat and lng coordinates to GLOBAL X and Y positions
def latlngToGlobalXY(lat, lng):
# Calculates x based on cos of average of the latitudes
# Calculates y based on latitude
return {'x': x, 'y': y}
# This function converts lat and lng coordinates to SCREEN X and Y positions
def latlngToScreenXY(lat, lng):
# Calculate global X and Y for projection point
pos = latlngToGlobalXY(lat, lng)
# Calculate the percentage of Global X position in relation to total global width
perX = ((pos['x']-p0.pos['x'])/(p1.pos['x'] - p0.pos['x']))
# Calculate the percentage of Global Y position in relation to total global height
perY = ((pos['y']-p0.pos['y'])/(p1.pos['y'] - p0.pos['y']))
# Returns the screen position based on reference points
return {
'x': p0.scrX + (p1.scrX - p0.scrX)*perX,
'y': p0.scrY + (p1.scrY - p0.scrY)*perY
}
pos = latlngToScreenXY(52.525607, 13.404572);
``````
pos['x] and pos['y] contain the translated x & y coordinates of the lat & lng (52.525607, 13.404572)
I hope this is helpful for anyone looking like me for the proper solution to the problem of translating lat lng into a local reference coordinate system.
Best
• Latitude and longitude are in degrees but `math.cos` takes arguments in radian. Is `latlngToGlobalXY` correct? Commented Feb 15, 2021 at 17:51
Its better to convert to utm coordinates, and treat that as x and y.
``````import utm
u = utm.from_latlon(12.917091, 77.573586)
``````
The result will be (779260.623156606, 1429369.8665238516, 43, 'P') The first two can be treated as x,y coordinates, the 43P is the UTM Zone, which can be ignored for small areas (width upto 668 km).
• I would assume that just converting spherical coordinates to UTM and then ignoring the zone might cause problems for regions that cross zone boundaries. You'd get results from different zones that don't fit together. There should be a way to force all points to use the same zone, then they should fit better.
– MvG
Commented May 1, 2022 at 8:24
The easiest way to do it is to use Mercator Projection. Be careful that there might be some stretching around the poles.
And the reference code (Kotlin):
``````fun latLngToXY(point: LatLng): Pair<Double, Double> {
Note that it returns values in `[0;pi/2]` so you might want to scale it. | 3,508 | 13,054 | {"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} | 3.515625 | 4 | CC-MAIN-2024-30 | latest | en | 0.863643 |
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# ch7 - CHAPTER 7 Techniques of Integration Integration...
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CHAPTER 7 Techniques of Integration 7.1. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given by complicated formulae, and practi- tioners consult a Table of Integrals in order to complete the integration. There are certain methods of integration which are essential to be able to use the Tables effectively. These are: substitution, integration by parts and partial fractions. In this chapter we will survey these methods as well as some of the ideas which lead to the tables. After the examination on this material, students will be free to use the Tables to integrate. The idea of substitution was introduced in section 4.1 (recall Proposition 4.4). To integrate a differen- tial f x dx which is not in the table, we first seek a function u u x so that the given differential can be rewritten as a differential g u du which does appear in the table. Then, if g u du G u C , we know that f x dx G u x C . Finding and employing the function u often requires some experience and ingenuity as the following examples show. Example 7.1 x 2 x 1 dx ? Let u 2 x 1, so that du 2 dx and x u 1 2. Then x 2 x 1 dx u 1 2 u 1 2 du 2 1 4 u 3 2 u 1 2 du 1 4 2 5 u 5 2 2 3 u 3 2 C (7.1) 1 30 u 3 2 3 u 5 C 1 30 2 x 1 3 2 6 x 2 C 1 15 2 x 1 3 2 3 x 1 C (7.2) where at the end we have replaced u by 2 x 1. Example 7.2 tan xdx ? 107
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Chapter 7 Techniques of Integration 108 Since this isn’t on our tables, we revert to the definition of the tangent: tan x sin x cos x . Then, letting u cos x du sin xdx we obtain (7.3) tan xdx sin x cos x dx du u ln u C lncos x C lnsec x C Example 7.3 sec xdx ?. This is tricky, and there are several ways to find the integral. However, if we are guided by the principle of rewriting in terms of sines and cosines, we are led to the following: (7.4) sec x 1 cos x cos x cos 2 x cos x 1 sin 2 x Now we can try the substitution u sin x du cos xdx . Then (7.5) sec xdx du 1 u 2 This looks like a dead end, but a little algebra pulls us through. The identity (7.6) 1 1 u 2 1 2 1 1 u 1 1 u leads to (7.7) du 1 u 2 dx 1 2 1 1 u 1 1 u du 1 2 ln 1 u ln 1 u C Using u sin x , we finally end up with (7.8) sec xdx 1 2 ln 1 sin x ln 1 sin x C 1 2 ln 1 sin x 1 sin x C Example 7.4 As a circle rolls along a horizontal line, a point on the circle traverses a curve called the cycloid . A loop of the cycloid is the trajectory of a point as the circle goes through one full rotation. Let us find the length of one loop of the cycloid traversed by a circle of radius 1. Let the variable t represent the angle of rotation of the circle, in radians, and start (at t 0) with the point of intersection P of the circle and the line on which it is rolling. After the circle has rotated through t radians, the position of the point is as given as in figure 7.1. The point of contact of the circle with the line is now t units to the right of the original point of contact (assuming no slippage), so (7.9) x t t sin t y t 1 cos t To find arc length, we use ds 2 dx 2 dy 2 , where dx 1 cos t dt dy sin
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Ask a homework question - tutors are online | 1,082 | 3,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-13 | latest | en | 0.84821 |
https://www.doorsteptutor.com/Exams/NSO/Class-3/Questions/Part-4.html | 1,575,982,556,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00149.warc.gz | 687,735,362 | 11,282 | # NSO Level 1- Science Olympiad (SOF) Class 3: Questions 22 - 29 of 504
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## Question number: 22
Edit
MCQ▾
### Question
A movie started at pm. It ended 1 hour 30 minutes later. At what time did it end?
### Choices
Choice (4) Response
a.
10: 45 pm
b.
11: 15 pm
c.
11: 15 am
d.
10: 45 am
## Question number: 23
Edit
MCQ▾
### Question
The sum of two numbers is 183. The smaller number is 11. What is the bigger number?
### Choices
Choice (4) Response
a.
194
b.
172
c.
164
d.
188
## Question number: 24
Edit
MCQ▾
### Question
432 written in words in ________
### Choices
Choice (4) Response
a.
Four hundred thirty and two
b.
Four hundred and thirty two
c.
Four thirty two
d.
Forty three hundred and two
## Question number: 25
Edit
MCQ▾
### Question
What is the weight of the books. If weight of
### Choices
Choice (4) Response
a.
1050 g
b.
1055 g
c.
1100 g
d.
782 g
## Question number: 26
Edit
MCQ▾
### Question
________ The missing number is ________.
### Choices
Choice (4) Response
a.
25
b.
35
c.
53
d.
20
## Question number: 27
Edit
MCQ▾
### Question
________ The missing number is ________.
### Choices
Choice (4) Response
a.
40
b.
45
c.
39
d.
64
## Question number: 28
Edit
MCQ▾
### Question
If Ruksana sells 356 eggs each day, how many eggs will he sell in a week?
### Choices
Choice (4) Response
a.
2012 eggs
b.
2145 eggs
c.
2492 eggs
d.
363 eggs
## Question number: 29
Edit
MCQ▾
### Question
Find the value of P and Q.
### Choices
Choice (4) Response
a.
P , Q
b.
P , Q
c.
P , Q
d.
P , Q
Developed by: | 589 | 1,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-51 | longest | en | 0.678927 |
https://en.wikipedia.org/wiki/W/m2 | 1,653,199,051,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00005.warc.gz | 287,946,847 | 21,723 | (Redirected from W/m2)
Spectral irradiance is the irradiance of a surface per unit frequency or wavelength, depending on whether the spectrum is taken as a function of frequency or of wavelength. The two forms have different dimensions: spectral irradiance of a frequency spectrum is measured in watts per square metre per hertz (W⋅m−2⋅Hz−1), while spectral irradiance of a wavelength spectrum is measured in watts per square metre per metre (W⋅m−3), or more commonly watts per square metre per nanometre (W⋅m−2⋅nm−1).
## Mathematical definitions
Irradiance of a surface, denoted Ee ("e" for "energetic", to avoid confusion with photometric quantities), is defined as[2]
${\displaystyle E_{\mathrm {e} }={\frac {\partial \Phi _{\mathrm {e} }}{\partial A}},}$
where
If we want to talk about the radiant flux emitted by a surface, we speak of radiant exitance.
Spectral irradiance in frequency of a surface, denoted Ee,ν, is defined as[2]
${\displaystyle E_{\mathrm {e} ,\nu }={\frac {\partial E_{\mathrm {e} }}{\partial \nu }},}$
where ν is the frequency.
Spectral irradiance in wavelength of a surface, denoted Ee,λ, is defined as[2]
${\displaystyle E_{\mathrm {e} ,\lambda }={\frac {\partial E_{\mathrm {e} }}{\partial \lambda }},}$
where λ is the wavelength.
## Property
Irradiance of a surface is also, according to the definition of radiant flux, equal to the time-average of the component of the Poynting vector perpendicular to the surface:
${\displaystyle E_{\mathrm {e} }=\langle |\mathbf {S} |\rangle \cos \alpha ,}$
where
• ⟨ • ⟩ is the time-average;
• S is the Poynting vector;
• α is the angle between a unit vector normal to the surface and S.
For a propagating sinusoidal linearly polarized electromagnetic plane wave, the Poynting vector always points to the direction of propagation while oscillating in magnitude. The irradiance of a surface is then given by[3]
${\displaystyle E_{\mathrm {e} }={\frac {n}{2\mu _{0}\mathrm {c} }}E_{\mathrm {m} }^{2}\cos \alpha ={\frac {n\varepsilon _{0}\mathrm {c} }{2}}E_{\mathrm {m} }^{2}\cos \alpha ,}$
where
This formula assumes that the magnetic susceptibility is negligible, i.e. that μr ≈ 1 where μr is the magnetic permeability of the propagation medium. This assumption is typically valid in transparent media in the optical frequency range.
## Point source
A point source of light produces spherical wavefronts. The irradiance in this case varies inversely with the square of the distance from the source.
${\displaystyle E={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}.\,}$
where
• r is the distance;
• P is the radiant power;
• A is the surface area of a sphere of radius r.
For quick approximations, this equation indicates that doubling the distance reduces irradiation to one quarter; or similarly, to double irradiation, reduce the distance to 0.7. When it is not a point source, for real light sources, the irradiance profile may be obtained by the image convolution of a picture of the light source.[4]
## Solar energy
The global irradiance on a horizontal surface on Earth consists of the direct irradiance Ee,dir and diffuse irradiance Ee,diff. On a tilted plane, there is another irradiance component, Ee,refl, which is the component that is reflected from the ground. The average ground reflection is about 20% of the global irradiance. Hence, the irradiance Ee on a tilted plane consists of three components:[5]
${\displaystyle E_{\mathrm {e} }=E_{\mathrm {e} ,\mathrm {dir} }+E_{\mathrm {e} ,\mathrm {diff} }+E_{\mathrm {e} ,\mathrm {refl} }.}$
The integral of solar irradiance over a time period is called "solar exposure" or "insolation".[5][6]
Quantity Unit Dimension Notes
Name Symbol[nb 1] Name Symbol Symbol
Radiant energy density we joule per cubic metre J/m3 ML−1T−2 Radiant energy per unit volume.
Radiant flux Φe[nb 2] watt W = J/s ML2T−3 Radiant energy emitted, reflected, transmitted or received, per unit time. This is sometimes also called "radiant power", and called luminosity in Astronomy.
Spectral flux Φe,ν[nb 3] watt per hertz W/Hz ML2T−2 Radiant flux per unit frequency or wavelength. The latter is commonly measured in W⋅nm−1.
Φe,λ[nb 4] watt per metre W/m MLT−3
Radiant intensity Ie,Ω[nb 5] watt per steradian W/sr ML2T−3 Radiant flux emitted, reflected, transmitted or received, per unit solid angle. This is a directional quantity.
Spectral intensity Ie,Ω,ν[nb 3] watt per steradian per hertz W⋅sr−1⋅Hz−1 ML2T−2 Radiant intensity per unit frequency or wavelength. The latter is commonly measured in W⋅sr−1⋅nm−1. This is a directional quantity.
Ie,Ω,λ[nb 4] watt per steradian per metre W⋅sr−1⋅m−1 MLT−3
Radiance Le,Ω[nb 5] watt per steradian per square metre W⋅sr−1⋅m−2 MT−3 Radiant flux emitted, reflected, transmitted or received by a surface, per unit solid angle per unit projected area. This is a directional quantity. This is sometimes also confusingly called "intensity".
Spectral radiance Le,Ω,ν[nb 3] watt per steradian per square metre per hertz W⋅sr−1⋅m−2⋅Hz−1 MT−2 Radiance of a surface per unit frequency or wavelength. The latter is commonly measured in W⋅sr−1⋅m−2⋅nm−1. This is a directional quantity. This is sometimes also confusingly called "spectral intensity".
Le,Ω,λ[nb 4] watt per steradian per square metre, per metre W⋅sr−1⋅m−3 ML−1T−3
Flux density
Ee[nb 2] watt per square metre W/m2 MT−3 Radiant flux received by a surface per unit area. This is sometimes also confusingly called "intensity".
Spectral flux density
Ee,ν[nb 3] watt per square metre per hertz W⋅m−2⋅Hz−1 MT−2 Irradiance of a surface per unit frequency or wavelength. This is sometimes also confusingly called "spectral intensity". Non-SI units of spectral flux density include jansky (1 Jy = 10−26 W⋅m−2⋅Hz−1) and solar flux unit (1 sfu = 10−22 W⋅m−2⋅Hz−1 = 104 Jy).
Ee,λ[nb 4] watt per square metre, per metre W/m3 ML−1T−3
Radiosity Je[nb 2] watt per square metre W/m2 MT−3 Radiant flux leaving (emitted, reflected and transmitted by) a surface per unit area. This is sometimes also confusingly called "intensity".
Spectral radiosity Je,ν[nb 3] watt per square metre per hertz W⋅m−2⋅Hz−1 MT−2 Radiosity of a surface per unit frequency or wavelength. The latter is commonly measured in W⋅m−2⋅nm−1. This is sometimes also confusingly called "spectral intensity".
Je,λ[nb 4] watt per square metre, per metre W/m3 ML−1T−3
Radiant exitance Me[nb 2] watt per square metre W/m2 MT−3 Radiant flux emitted by a surface per unit area. This is the emitted component of radiosity. "Radiant emittance" is an old term for this quantity. This is sometimes also confusingly called "intensity".
Spectral exitance Me,ν[nb 3] watt per square metre per hertz W⋅m−2⋅Hz−1 MT−2 Radiant exitance of a surface per unit frequency or wavelength. The latter is commonly measured in W⋅m−2⋅nm−1. "Spectral emittance" is an old term for this quantity. This is sometimes also confusingly called "spectral intensity".
Me,λ[nb 4] watt per square metre, per metre W/m3 ML−1T−3
Radiant exposure He joule per square metre J/m2 MT−2 Radiant energy received by a surface per unit area, or equivalently irradiance of a surface integrated over time of irradiation. This is sometimes also called "radiant fluence".
Spectral exposure He,ν[nb 3] joule per square metre per hertz J⋅m−2⋅Hz−1 MT−1 Radiant exposure of a surface per unit frequency or wavelength. The latter is commonly measured in J⋅m−2⋅nm−1. This is sometimes also called "spectral fluence".
He,λ[nb 4] joule per square metre, per metre J/m3 ML−1T−2
Hemispherical emissivity ε N/A 1 Radiant exitance of a surface, divided by that of a black body at the same temperature as that surface.
Spectral hemispherical emissivity εν
or
ελ
N/A 1 Spectral exitance of a surface, divided by that of a black body at the same temperature as that surface.
Directional emissivity εΩ N/A 1 Radiance emitted by a surface, divided by that emitted by a black body at the same temperature as that surface.
Spectral directional emissivity εΩ,ν
or
εΩ,λ
N/A 1 Spectral radiance emitted by a surface, divided by that of a black body at the same temperature as that surface.
Hemispherical absorptance A N/A 1 Radiant flux absorbed by a surface, divided by that received by that surface. This should not be confused with "absorbance".
Spectral hemispherical absorptance Aν
or
Aλ
N/A 1 Spectral flux absorbed by a surface, divided by that received by that surface. This should not be confused with "spectral absorbance".
Directional absorptance AΩ N/A 1 Radiance absorbed by a surface, divided by the radiance incident onto that surface. This should not be confused with "absorbance".
Spectral directional absorptance AΩ,ν
or
AΩ,λ
N/A 1 Spectral radiance absorbed by a surface, divided by the spectral radiance incident onto that surface. This should not be confused with "spectral absorbance".
Hemispherical reflectance R N/A 1 Radiant flux reflected by a surface, divided by that received by that surface.
Spectral hemispherical reflectance Rν
or
Rλ
N/A 1 Spectral flux reflected by a surface, divided by that received by that surface.
Directional reflectance RΩ N/A 1 Radiance reflected by a surface, divided by that received by that surface.
Spectral directional reflectance RΩ,ν
or
RΩ,λ
N/A 1 Spectral radiance reflected by a surface, divided by that received by that surface.
Hemispherical transmittance T N/A 1 Radiant flux transmitted by a surface, divided by that received by that surface.
Spectral hemispherical transmittance Tν
or
Tλ
N/A 1 Spectral flux transmitted by a surface, divided by that received by that surface.
Directional transmittance TΩ N/A 1 Radiance transmitted by a surface, divided by that received by that surface.
Spectral directional transmittance TΩ,ν
or
TΩ,λ
N/A 1 Spectral radiance transmitted by a surface, divided by that received by that surface.
Hemispherical attenuation coefficient μ reciprocal metre m−1 L−1 Radiant flux absorbed and scattered by a volume per unit length, divided by that received by that volume.
Spectral hemispherical attenuation coefficient μν
or
μλ
reciprocal metre m−1 L−1 Spectral radiant flux absorbed and scattered by a volume per unit length, divided by that received by that volume.
Directional attenuation coefficient μΩ reciprocal metre m−1 L−1 Radiance absorbed and scattered by a volume per unit length, divided by that received by that volume.
Spectral directional attenuation coefficient μΩ,ν
or
μΩ,λ
reciprocal metre m−1 L−1 Spectral radiance absorbed and scattered by a volume per unit length, divided by that received by that volume.
1. ^ Standards organizations recommend that radiometric quantities should be denoted with suffix "e" (for "energetic") to avoid confusion with photometric or photon quantities.
2. Alternative symbols sometimes seen: W or E for radiant energy, P or F for radiant flux, I for irradiance, W for radiant exitance.
3. Spectral quantities given per unit frequency are denoted with suffix "ν" (Greek)—not to be confused with suffix "v" (for "visual") indicating a photometric quantity.
4. Spectral quantities given per unit wavelength are denoted with suffix "λ" (Greek).
5. ^ a b Directional quantities are denoted with suffix "Ω" (Greek). | 3,102 | 11,154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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-2022-21 | latest | en | 0.816848 |
https://www.univerkov.com/find-the-sharp-corners-of-a-right-triangle-if-one-of-them-is-20-degrees-larger-than-the-other/ | 1,708,654,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00757.warc.gz | 1,061,513,795 | 6,276 | # Find the sharp corners of a right triangle if one of them is 20 degrees larger than the other.
Find the sharp corners of a right triangle if one is 20 degrees larger than the other.
Given: a right-angled triangle, the angles will be 1, 2, 3; let angle 3 be a straight line (90 °), and angle 1 is greater than angle 2 by 20 °.
Find: the magnitude of the angle 1 and the magnitude of the angle 2.
Solution: The angles in the triangle add up to 180 degrees. 1 + 2 + 3 = 180o; 1 + 2 = 180o – 90o = 90o; suppose the angles are 1 = 2, then 1 + 2 = 90 ° + 20 ° = 110 °, which means angle 1 = 110 °: 2 = 55 °, and angle 2 = 90 ° – 55 = 35 °.
Answer: one angle is 55 °, and the second angle is 35 °.
You can calculate the magnitude of the angles in different ways.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 300 | 1,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-10 | latest | en | 0.879325 |
https://worldsworstsportsblog.com/tag/willie-mays/ | 1,685,949,483,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00360.warc.gz | 683,682,678 | 28,031 | ## The 600 Home Run AlmanacJuly 28, 2010
Posted by tomflesher in Baseball, Economics.
Tags: , , , , , , , , , , , , , ,
People are interested in players who hit 600 home runs, at least judging by the Google searches that point people here. With that in mind, let’s take a look at some quick facts about the 600th home run and the people who have hit it.
Age: There are six players to have hit #600. Sammy Sosa was the oldest at 39 years old in 2007. Ken Griffey, Jr. was 38 in 2007, as were Willie Mays in 1969 and Barry Bonds in 2002. Hank Aaron was 37. Babe Ruth was the youngest at 36 in 1931. Alex Rodriguez, who is 35 as of July 27, will almost certainly be the youngest player to reach 600 home runs. If both Manny Ramirez and Jim Thome hang on to hit #600 over the next two to three seasons, Thome (who was born in August of 1970) will probably be 42 in 2012; Ramirez (born in May of 1972) will be 41 in 2013. (In an earlier post that’s when I estimated each player would hit #600.) If Thome holds on, then, he’ll be the oldest player to hit his 600th home run.
Productivity: Since 2000 (which encompasses Rodriguez, Ramirez, and Thome in their primes), the average league rate of home runs per plate appearances has been about .028. That is, a home run was hit in about 2.8% of plate appearances. Over the same time period, Rodriguez’ rate was .064 – more than double the league average. Ramirez hit .059 – again, over double the league rate. Thome, for his part, hit at a rate of .065 home runs per plate appearance. From 2000 to 2009, Thome was more productive than Rodriguez.
Standing Out: Obviously it’s unusual for them to be that far above the curve. There were 1,877,363 plate appearances (trials) from 2000 to 2009. The margin of error for a proportion like the rate of home runs per plate appearance is
$\sqrt{\frac{p(1-p)}{n-1}} = \sqrt{\frac{.028(.972)}{1,877,362}} = \sqrt{\frac{.027}{1,877,362}} \approx \sqrt{\frac{14}{1,000,000,000}} = .00012$
Ordinarily, we expect a random individual chosen from the population to land within the space of $p \pm 1.96 \times MoE$ 95% of the time. That means our interval is
$.027 \pm .00024$
That means that all three of the players are well without that confidence interval. (However, it’s likely that home run hitting is highly correlated with other factors that make this test less useful than it is in other situations.)
Alex’s Drought: Finally, just how likely is it that Alex Rodriguez will go this long without a home run? He hit his last home run in his fourth plate appearance on July 22. He had a fifth plate appearance in which he doubled. Since then, he’s played in five games totalling 22 plate appearances, so he’s gone 23 plate appearances without a home run. Assuming his rate of .064 home runs per plate appearance, how likely is that? We’d expect (.064*23) = about 1.5 home runs in that time, but how unlikely is this drought?
The binomial distribution is used to model strings of successes and failures in tests where we can say clearly whether each trial ended in a “yes” or “no.” We don’t need to break out that tool here, though – if the probability of a home run is .064, the probability of anything else is .936. The likelihood of a string of 23 non-home runs is
$.936^{23} = .218$
It’s only about 22% likely that this drought happened only by chance. The better guess is that, as Rodriguez has said, he’s distracted by the switching to marked baseballs and media pressure to finally hit #600.
## Back when it was hard to hit 55…July 8, 2010
Posted by tomflesher in Baseball, Economics.
Tags: , , , , , , , , ,
Last night was one of those classic Keith Hernandez moments where he started talking and then stopped abruptly, which I always like to assume is because the guys in the truck are telling him to shut the hell up. He was talking about Willie Mays for some reason, and said that Mays hit 55 home runs “back when it was hard to hit 55.” Keith coyly said that, while it was easy for a while, it was “getting hard again,” at which point he abruptly stopped talking.
Keith’s unusual candor about drug use and Mays’ career best of 52 home runs aside, this pinged my “Stuff Keith Hernandez Says” meter. After accounting for any time trend and other factors that might explain home run hitting, is there an upward trend? If so, is there a pattern to the remaining home runs?
The first step is to examine the data to see if there appears to be any trend. Just looking at it, there appears to be a messy U shape with a minimum around t=20, which indicates a quadratic trend. That means I want to include a term for time and a term for time squared.
Using the per-game averages for home runs from 1955 to 2009, I detrended the data using t=1 in 1955. I also had to correct for the effect of the designated hitter. That gives us an equation of the form
$\hat{HR} = \hat{\beta_{0}} + \hat{\beta_{1}}t + \hat{\beta_{2}} t^{2} + \hat{\beta_{3}} DH$
The results:
Estimate Std. Error t-value p-value Signif B0 0.957 0.0328 29.189 0.0001 0.9999 t -0.0188 0.0028 -6.738 0.0001 0.9999 tsq 0.0004 0.00005 8.599 0.0001 0.9999 DH 0.0911 0.0246 3.706 0.0003 0.9997
We can see that there’s an upward quadratic trend in predicted home runs that together with the DH rule account for about 56% of the variation in the number of home runs per game in a season ($R^2 = .5618$). The Breusch-Pagan test has a p-value of .1610, indicating a possibility of mild homoskedasticity but nothing we should get concerned about.
Then, I needed to look at the difference between the predicted number of home runs per game and the actual number of home runs per game, which is accessible by subtracting
$Residual = HR - \hat{HR}$
This represents the “abnormal” number of home runs per year. The question then becomes, “Is there a pattern to the number of abnormal home runs?” There are two ways to answer this. The first way is to look at the abnormal home runs. Up until about t=40 (the mid-1990s), the abnormal home runs are pretty much scattershot above and below 0. However, at t=40, the residual jumps up for both leagues and then begins a downward trend. It’s not clear what the cause of this is, but the knee-jerk reaction is that there might be a drug use effect. On the other hand, there are a couple of other explanations.
The most obvious is a boring old expansion effect. In 1993, the National League added two teams (the Marlins and the Rockies), and in 1998 each league added a team (the AL’s Rays and the NL’s Diamondbacks). Talent pool dilution has shown up in our discussion of hit batsmen, and I believe that it can be a real effect. It would be mitigated over time, however, by the establishment and development of farm systems, in particular strong systems like the one that’s producing good, cheap talent for the Rays. | 1,790 | 6,817 | {"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": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.973616 |
https://mathematica.stackexchange.com/questions/284607/assigning-a-value-to-a-functions-argument?noredirect=1 | 1,716,769,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00264.warc.gz | 333,764,891 | 40,277 | # Assigning a value to a functions argument
I´m struggeling now a few hours on a very basic problem. I want to assign the value 0 to variable i, if i has another value, then it should be changed to 0 in the function.
Function
ClearAll["Global*"]
ExpandSum[summand_, range_List] := Total[(Function @@ {range[[1]],HoldForm[summand]}) /@ Array[range[[2]] - 1 + # &, (range[[3]] - range[[2]] + 1)]]
Input:(It works perfectly with that:)
ExpandSum[i^4, {i, 3, 20}]
It works not with that:
i = 10;
ExpandSum[i^4, {i, 3, 20}]
I tried clear[i], but it doesnt work. i=0 leeds to "Cannot unset raw object 10). tried to set i_Object or i_Integer - no help.
Sorry for that very basic question but I´m stuck...
Many thanks!
• It is not really clear what you want to achieve. First, Table is a built-in symbol, so you "cannot" change its definitions – you should use some other symbol. Second, I suppose you are not showing us your actual code with the actual problem. Please edit the question and include your code that will clearly show what is the input and what is the desired output. May 3, 2023 at 12:08
• @Domen I added the function. May 3, 2023 at 12:34
• Please look up docs for Module and With. Set i inside the module (locally) to zero and it doesn't need to be an argument bringing in a global value to the function.
– Syed
May 3, 2023 at 15:58
Roberta, this is essentially the same problem as Passing argument to a function
You seem to be overthinking these scenarios. You're trying to layer abstractions on top of the abstractions that already exist and could be used directly.
ExpandSum[i^4, {i, 3, 20}]
Now, just from a design perspective, there is an immediate problem: you must pass in i in both arguments. That kind of repetition should be a big red flag. But other than that, we're right back to the problem from your other question, which is that you need to figure out how to control i when it's inside your function so that it doesn't take on its expected value from outside your function. As soon as the global i gets a value, your function will break if you try to pass in expressions with i. There are ways you could do that, but they are just way more complicated than just letting the built in functions do this for you:
theRange = Range[3, 20];
theFunction = HoldForm[#^4] &;
theResult = Total[theFunction /@ theRange]
Or:
Total@Table[theFunction[index], {index, theRange}]
(*which is equivalent to ...*)
Total@Table[theFunction[index], {index, 3, 20}]
(*which is equivalent to ...*)
Total@Table[HoldForm[#^4] &[index], {index, 3, 20}]
I'm not sure if you really wanted HoldForm in there or if you were just debugging. These could be simplified without the HoldForm.
If you do this:
ExpandSum[summand_, range_List] := Block[{},
Echo[range];
Total[(Function @@ {range[[1]], HoldForm[summand]}) /@
Array[range[[2]] - 1 + # &, (range[[3]] - range[[2]] + 1)]]
]
Then you will see that your function is operating on {10,3,20}
So the function that ends up being mapped is function ends up being mapped
Function[10,10000]/@yourList
Perhaps you are looking for something like:
Total[HoldForm[#^4] & /@ Range[3, 20]]
`
• PS. Clear[i] works for me. It doesn't give the Unset warning. May 3, 2023 at 13:46
• Many thanks, but it doenst work. Range[1] is 10 but should be i May 3, 2023 at 14:18 | 922 | 3,335 | {"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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.918903 |
https://www.moorcroftwood.net/21st-january-2021/ | 1,679,764,868,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00531.warc.gz | 1,002,752,455 | 23,347 | A founder member of the Atlas Federation
# 21st January 2021
21st January 2021
Today’s Time Table
8:40-9:10 9:10-10:10 10:10-10:30 10:30-10:50 10:50-11:50 11:50-12:30 12:30-1:15 1:15 – 2:35 2:35-2:50 Watch Newsround Times Tables Practice Maths BREAK Reading English History LUNCH Art Miss Wall Reads
08:40 – 9:10
Watch Newsround for the day’s news and happenings in the world.
Follow the link below for the up-to-date news from around the world. If you wish to discuss anything you see on Newsround please remember it for our first live lesson of the day.
https://www.bbc.co.uk/newsround/news/watch_newsround
Once you have watch Newsround it’s time to practice your times tables. You may want to go to the Hit the Button website or head over to Times Tables Rock stars
https://www.topmarks.co.uk/maths-games/hit-the-button
Let me know your high score and remember to challenge yourself with the level you choose.
https://ttrockstars.com/
If you need your login for this please send me a message and I can let you know what it is.
9:10-10:10
Maths
Key Stage Two Maths Year 5 Targeted Question Book – page 32 and 33
Multiplying Fractions
When multiplying with fractions it’s very similar to working out fractions of amounts.
When you see ¼ x 20 the x sign means of, so your question is ¼ of 20. Remember to do this you need to divide by the bottom and times by the top.
20 ÷ 4 = 5
5 x 1 = 5
So ¼ x 20 = 5
When you are working with larger numbers you need to work it out slightly differently.
If you had 2/6 of 54 you would need to convert 54 into a fraction – at the moment you have 54 whole ones so the fraction becomes 54/1.
Now you have 2/6 x 54/1. This is now a simple multiply the numerators and multiply the denominators. So 2 x 54 = 108 and 6x1 = 6
To be able to convert this into a mixed number you need to carry out a division
108 ÷ 6.
You all know how to do short division using bus stop so you can do this and write your remainders as a fraction.
In this example you don’t have any remainders so your answer would just be 18.
If you had to find 3/5 x 136
3/5 x 136/1
So 3 x 136 = 408
And 5 x 1 = 5
This would give you a fraction of 408/5
Remember to convert mixed numbers to improper fractions then work out your multiplication as above.
When you are asked to find missing digits from the question you need to work it backwards.
So if you had 6/8 x ___ = 12
So 12 ÷ 6 = 2 then 2 x 8 = 16
16 would be your missing number.
10:30-10:50
Quiet Reading Time – Bug Club
Find a comfortable spot for your reading or you may want to find an unusual place for a change. Ask your adult to send me a picture of you reading in your spot or tell me where it is on Class Dojo.
Make sure you have your reading record signed by an adult at home to show that you have kept up your reading and send over a picture on Class Dojo, this will earn you more dojos.
10:50 – 11:50
English- Comprehension
Comprehension Book One (the one with the train on) – Pages 28 and 29 – Wolves in the UK
Comprehension Book One (the one with Stonehenge on- if you have this one) – Pages 10 and 11 – Let’s get growing.
The first part of the lesson will be online to share a guided text about the Mayan writing and maths systems they used, this document is attached below. If you want to print it to be able to follow it you can but if you don’t have this option it will be shared on the screen so you can see it.
Miss Watton
11:50 – 12:30
History
Mayan Number System
Today we are going to be looking at the Maya number system. Their number system is quite different to the one we use, the Maya used symbols instead of numbers. We will be looking at what the symbols represent, what their values are and how the system works.
Read through the presentation where you will discover what symbols were used, you will see a Maya number list, shown how the system works and how to work out larger numbers.
I have attached the presentation and the worksheet, have fun.
Miss Watton
1:15 – 2:35
Art
The Maya made many different masks. Some were delicate mosaics made out of jade, some were created to represent gods, some showed faces of snakes and other animals. I have attached some examples of face masks you might want to use for your art work today. You need to either use one of these templates or draw one of them and create a mosaic using the design.
Think about the colours that the Maya would have used for these face masks.
2:35 – 2:45 | 1,183 | 4,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-14 | latest | en | 0.893462 |
https://socratic.org/questions/how-many-molecules-are-in-7-8-moles-of-nh-3 | 1,566,443,720,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00149.warc.gz | 631,759,960 | 5,811 | # How many molecules are in 7.8 moles of NH_3?
In one mole there are $6.022 \times {10}^{23}$ ammonia particles. So in $7.8$ $m o l$ there are approx. $48 \times {10}^{23}$ such particles.
So $7.8$ $\cancel{m o l}$ $\times$ $6.022$ $\times$ ${10}^{23}$ particles $\cancel{m o {l}^{-} 1}$ $=$ ?? particles? How many atoms in such a quantity. | 124 | 341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-35 | longest | en | 0.770421 |
http://us.metamath.org/ileuni/com23.html | 1,653,213,779,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00779.warc.gz | 49,550,023 | 4,150 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > com23 GIF version
Theorem com23 77
Description: Commutation of antecedents. Swap 2nd and 3rd. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Wolf Lammen, 4-Aug-2012.)
Hypothesis
Ref Expression
com3.1 (𝜑 → (𝜓 → (𝜒𝜃)))
Assertion
Ref Expression
com23 (𝜑 → (𝜒 → (𝜓𝜃)))
Proof of Theorem com23
StepHypRef Expression
1 com3.1 . 2 (𝜑 → (𝜓 → (𝜒𝜃)))
2 pm2.27 39 . 2 (𝜒 → ((𝜒𝜃) → 𝜃))
31, 2syl9 71 1 (𝜑 → (𝜒 → (𝜓𝜃)))
Colors of variables: wff set class Syntax hints: → wi 4 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 This theorem is referenced by: com3r 78 com13 79 pm2.04 81 pm2.86d 98 impancom 256 con2d 587 impidc 789 pm2.61dc 796 3com23 1145 expcomd 1371 spimth 1664 sbiedh 1711 eqrdav 2081 necon4bbiddc 2320 ralrimdva 2442 ralrimdvva 2447 ceqsalt 2626 vtoclgft 2650 reu6 2782 sbciegft 2845 reuss2 3251 reupick 3255 reusv3 4218 ssrel 4454 ssrel2 4456 ssrelrel 4466 funssres 4972 funcnvuni 4999 fv3 5229 fvmptt 5294 funfvima2 5423 isoini 5488 isopolem 5492 f1ocnv2d 5735 f1o2ndf1 5880 nnmordi 6155 nnmord 6156 xpdom2 6375 findcard2 6423 findcard2s 6424 findcard2d 6425 findcard2sd 6426 ordiso2 6505 genpcdl 6771 genpcuu 6772 distrlem5prl 6838 distrlem5pru 6839 lemul12a 8007 divgt0 8017 divge0 8018 lbreu 8090 bndndx 8354 elnnz 8442 nzadd 8484 fzind 8543 fnn0ind 8544 eqreznegel 8780 lbzbi 8782 irradd 8812 irrmul 8813 ledivge1le 8884 iccid 9024 uzsubsubfz 9142 fzrevral 9198 elfz0fzfz0 9214 fz0fzelfz0 9215 elfzmlbp 9220 elfzodifsumelfzo 9287 ssfzo12bi 9311 elfzonelfzo 9316 flqeqceilz 9400 le2sq2 9648 facdiv 9762 facwordi 9764 faclbnd 9765 cau3lem 10138 mulcn2 10289 climcau 10322 climcaucn 10326 dvdsdivcl 10395 ltoddhalfle 10437 halfleoddlt 10438 ndvdssub 10474 dfgcd2 10547 coprmdvds1 10617 coprmdvds 10618 coprmdvds2 10619 divgcdcoprm0 10627 cncongr1 10629 cncongr2 10630 prmfac1 10675 bj-rspgt 10747
Copyright terms: Public domain W3C validator | 1,199 | 2,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.125746 |
https://nostalgia.wikipedia.org/wiki/Disjunctive_syllogism | 1,653,393,595,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00391.warc.gz | 475,871,260 | 6,351 | # Disjunctive syllogism
Disjunctive syllogism is a valid, simple argument form:
Either P or Q.
Not P.
Therefore, Q.
Roughly, we are told that it has to be one or the other that is true; then we are told that it is not the one that is true; so we infer that it has to be the other that is true. The reason this is called "disjunctive syllogism" is that, first, it is a syllogism--a three-step argument--and second, it contains a disjunction, which means simply an "or" statement. "Either P or Q" is a disjunction; P and Q are called the statement's disjuncts.
Here is an example:
Either I will choose soup or I will choose salad.
I will not choose soup. | 184 | 657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.959745 |
https://www.jiskha.com/display.cgi?id=1322787946 | 1,500,800,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424296.90/warc/CC-MAIN-20170723082652-20170723102652-00558.warc.gz | 807,442,672 | 4,058 | posted by .
Is 1,725+346 more or less than 2,000? Explain how you can tell without actually adding.
You can tell because 725 is almost already to 1000. To get to 1000 from 725 all you have to do to it is to add 275
You can tell because just by looking at the number 1,725 you don't need to add 346 to get to 2000. All you need to do is add 275 to 725 to see you can get 1000...
How to see if 275 is the correct answer to add it with do 1000-725. Youll get 275 and to check if it's correct add 725 and 275
725 is almost 1000. All u need is 275 | 168 | 547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-30 | longest | en | 0.958767 |
https://qiita.com/koyopro/items/1f55153b117f7c561909 | 1,642,920,391,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00110.warc.gz | 527,172,373 | 15,046 | 40
Help us understand the problem. What are the problem?
More than 5 years have passed since last update.
posted at
updated at
背景
ちょっと調べないと分からなかったので。
変換方法
Array
``````// Array -> NSArray
var nsarr: NSArray = NSArray(array: ["A", "B", "C"])
println(nsarr)
// NSArray -> Array
var arr: [String] = nsarr as [String]
println(arr)
``````
``````(
A,
B,
C
)
[A, B, C]
``````
Dictionary
``````// Dictionary -> NSDictionary
var nsdic: NSDictionary = NSDictionary(dictionary: ["key1": "value1", "key2": "value2"])
println(nsdic)
// NSDictionary -> Dictionary
var dic: Dictionary = nsdic as Dictionary
println(dic)
``````
``````{
key1 = value1;
key2 = value2;
}
[key2: value2, key1: value1]
``````
おまけ: Array of Dictionary
``````var sample: [Dictionary<String, AnyObject>] = [
["key1": "value1", "key2": "value2"],
["key3": "value3", "key4": "value4"]]
// Array of Dictionary -> NSArray of NSDictionary
var nsdicarr: NSArray = NSArray(array: sample)
println(nsdicarr)
// NSArray of NSDictionary -> Array of Dictionary
var dicarr: [Dictionary<String, AnyObject>] = nsdicarr as [Dictionary<String, AnyObject>]
println(dicarr)
``````
output
``````(
{
key1 = value1;
key2 = value2;
},
{
key3 = value3;
key4 = value4;
}
)
[[key2: value2, key1: value1], [key4: value4, key3: value3]]
``````
参考
Mac Developers blog — Convert NSMutableArray to Swift Array...
http://dev.eltima.com/post/96538497489/convert-nsmutablearray-to-swift-array
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Help us understand the problem. What are the problem? | 521 | 1,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-05 | latest | en | 0.635696 |
https://brainacademy.pro/question/333293 | 1,679,539,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00152.warc.gz | 188,478,700 | 14,957 | The rectangle is 20 by 16 and the scale factor is 0. 25. What are the dimensions of the reduced rectangle?.
0 (0 stars)
0
ashishdwivedilVT 11 months ago
175 response - 0 helps
The dimensions of the reduced rectangle are 5(length) and 4(width).
The original length of the rectangle= 20
The original width of the rectangle=16
### What is a scale factor?
It is the factor by which the size or dimension of an object is dilated.
Reduced length/Original length = 0.25
Reduced length = 0.25 * original length
Reduced length = 0.25 * 20
Reduced length = 5
Reduced width/Original width = 0.25
Reduced width = 0.25 * original width
Reduced width = 0.25 * 16
Reduced width = 4
Therefore, the dimensions of the reduced rectangle are 5(length) and 4(width).
To get more about dilation visit: | 225 | 794 | {"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} | 3.765625 | 4 | CC-MAIN-2023-14 | longest | en | 0.890003 |
http://xine.org/~jwest/documentation/alg/node40.html | 1,553,359,057,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202889.30/warc/CC-MAIN-20190323161556-20190323183556-00383.warc.gz | 395,678,037 | 2,847 | Search
Next: 0.2.10.1 Octree Source Code Up: 0.2 Data Searching and Previous: 0.2.9.1 Source Code
A quadtree is a hierarchical data structure often used for image representation. Quadtrees encode two-dimensional images by placing subsections of the image into a tree structure much like a binary tree. Unlike a binary tree where every tree node has up to two children, in a quadtree every node can have up to four children. Each node stores a particular piece of the overall image.
Quadtrees operate by means of recursively decomposing space. While several specialized types of quadtrees exist, the most common type are known as region quadtrees. In such trees the image to be encoded and stored is partitioned into four quadrants. Each quadrent is then represented by a node in the tree. The root of the quadtree represents the entire stored image while each of it's four children represent a different fourth of the image. This process continues; each of the quadrants represented by the children of the root is again partitioned into four sub-quadrants and represented by the root's grandchildren nodes. This process continues until a region being stored is sufficiently simple that it can be wholly encoded in one node. Sometimes this means that we continue breaking the image down until every individual pixel is stored at some leaf node. Other times a region larger than a pixel is sufficiently homogeneous that it can be represented at a leaf node and requires no further decomposition. For example, if an entire quadrent is one color it does not make sense to continue to decompose it.
The regions represented by nodes in a quadtree are sometimes named after map directions (northwest, northeast, southwest, southeast).
An octree is essentially the same thing as a quadtree however each node in an octree has eight children instead of four. This makes octrees good candidates for representing three-dimensional objects in memory because instead of breaking the image into four quadrants the octree breaks the image into eight blocks. These blocks are then recursively decomposed into eight sub-blocks. This process, as in the quadtree, continues until a blocks is sufficiently homogenous that it can be represented by one node.
An interesting property of both quadtrees and octrees is that they support a form of data compression. By never traversing below a certain level of the tree structure it is possible to filter out the "details" of the image opting, instead, for a less-precise but smaller version of the data. Such a compression could be accomplished by means of a breadth-first traversal of the tree data structure.
Scott Gasch
1999-07-09 | 549 | 2,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-13 | latest | en | 0.955893 |
http://www.math-drills.com/money.shtml | 1,398,018,216,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00481-ip-10-147-4-33.ec2.internal.warc.gz | 560,148,596 | 10,360 | # U.S. and Canadian Money Worksheets
Welcome to the U.S./Canadian version of our money worksheets page at Math-Drills.com where you will not be short-changed! This page includes U.S. and Canadian money worksheets including money worksheets with dollars and counting coins.
Students encounter money early on, and they must be able to manage it themselves in their everyday lives and into adulthood. There are many activities that you can do related to counting, adding, and subtracting money, so save some coins or purchase some play money to complete some of the activities on the money math worksheets below.
Thanks to the policies of the United States Mint, we are able to use images of their coins on our worksheets. The use of the Canadian coin images was permitted by the Royal Canadian Mint, so we could also have Canadian coin counting worksheets for our Canadian visitors.
## Counting Coins Worksheets
This is a great place to start with younger students as they are likely to encounter coins before they encounter too many bills. Including children in money transactions helps them to develop important money management skills and has benefits in other math topics such as fractions (think a quarter of a dollar, for example). Some of the coin worksheets are quite large due to the images contained, so be patient while they load or right-click the direct link on the next page and save.
This is a nice way to gently move students into thinking about decimals and what a great opportunity to use some manipulatives. Students generally perform better with math that has meaning. Decimal numbers to hundredths have most likely been in students' lives since very early on as stores display prices, parents comment about the prices to help develop critical thinking in their children, so if there is one decimal that students understand well, it is money. They sometimes have trouble relating it to paper and pencil which is why manipulatives come in handy. Who doesn't have a few thousand dollars in small bills and coins kicking around to bring into a classroom.... no? Well, good thing there is play money! It's a little cheaper than real money and doesn't disappear quite as quickly.
## Subtracting Money Worksheets
The strategy of choice for subtracting money is the counting up strategy. It works especially well for giving change from whole dollar amounts. Following is an example of how it works. Let's say the bill at the grocery store came out to \$13.46 and the cashier was presented with a \$20.00 bill and for some odd reason, the monitor on his cash machine had a crack in it that obscured the amount of change to give. What would he do?! The first thing to do is to take four pennies out of the drawer because he needs to add the \$13.46 to an unknown amount to make \$20, and four pennies will bring the \$13.46 up to \$13.50 which makes things rounder and, for most, a little easier. Next, he needs to pull out two quarters to bring the amount up to \$14 even. He can then remove a dollar bill to make the amount \$15, and finally pull out a \$5 bill to count up to \$20. Now, if he wanted to know how much change he gave, he just needs to think back to what he pulled out of the drawer: \$5 + \$1 + \$0.50 + \$0.04 = \$6.54.
## Adding Money Worksheets With Varying Increments
These worksheets are useful for scaffolding student learning as money can be easier to think of in increments for students who aren't quite ready for the hundredths yet.
## Subtracting Money Worksheets With Varying Increments
Like the adding money worksheets with varying increments, these worksheets are useful for students who aren't quite ready for cents yet. They are also useful if you happen to live somewhere that has scrapped the penny.
## Next Dollar Up Strategy
The next dollar up strategy is used for students who are not able to make change, but can determine how many dollars it will take to cover an amount. For example, if something is \$5.45, then they would need to give \$6 to cover that amount since the next dollar up from \$5(.45) is \$6. | 856 | 4,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2014-15 | longest | en | 0.969985 |
https://cs.stackexchange.com/questions/159695/optimal-worse-case-sorting-algorithm-when-some-comparisons-are-cheap | 1,702,146,713,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00486.warc.gz | 221,282,591 | 42,276 | # Optimal worse-case sorting algorithm when some comparisons are cheap
I want to sort a set of items, where some comparisons are expensive and others are cheap. What is an algorithm that will minimize the worse-case number of expensive comparisons?
For instance, I want to rank sports players (assuming performance is completely deterministic, i.e. a game between two players will always have the same result), where some comparisons are already known, but others will require the players to play a game.
More formally, my machine has access to two oracles $$f(a, b) \in \{<,=,>,\mathrm{unknown}\}$$ and $$g(a, b) \in \{<,=,>\}$$, and I want to sort my list of items with the smallest worse-case number of calls to $$g$$ (which of course depends on the specific cases for which $$f$$ returns "unknown").
One approach I thought of is to extend $$f$$ into $$f'$$, which removes the unknown results that can be determined by transitivity, then use a standard sorting algorithm, calling $$f'$$ first, then $$g$$ if $$f'$$ returns "unknown". I'm not sure whether this is optimal (it may depend on the specific sorting algorithm).
I'd be satisfied with an algorithm that is near-optimal. I have about 100 items.
• We don't know an optimal procedure (in the sense of minimum comparisons) in the case of all unknown results. Your problem is even harder.
– user16034
Apr 18 at 15:34
• For a tiny number of elements ($n\le10$ ?), brute-forcing all possible sequences of non-redundant comparisons, and every comparison outcomes can be an option.
– user16034
Apr 18 at 15:43
• Would you be satisfied with a solution that is close to optimal?
– D.W.
Apr 19 at 0:29
• The following paper gives a polynomial time algorithm which performs an asymptoticaly optimal number of comparisons: arxiv.org/abs/0911.0086 (of course the hidden constant factor might be prohibitive for your practical case, I don't know) Apr 19 at 18:32
Your problem amounts to finding a linear extension of a given partial order, using the minimum number of comparisons between two elements. @Tassle highlights the following paper, which I believe is the state-of-the-art:
Sorting under Partial Information (without the Ellipsoid Algorithm). Jean Cardinal, Samuel Fiorini, Gwenaël Joret, Raphaël Jungers, J. Ian Munro. Combinatorica 33, 6 (2013), 655–697.
I have seen some papers call this the "SUPI" problem (Sorting Under Partial Information).
However, this algorithm is fairly complicated. If you want to a practical heuristic, that comes with no guarantees, you could try using a greedy heuristic.
Define some metric that represents how much uncertainty their remains about the final order, and that can be computed efficiently. For example, perhaps you can count the number of pairs whose order remains unclear (i.e., they are incomparable in the transitive closure). Then apply the following algorithm:
1. Compute the value of the metric on the current partial order.
2. For each pair $$(x,y)$$ you could possibly query next:
• Compute the value of the metric if this query returns $$x, and the value if it returns $$x>y$$, and average those two metrics (or better yet, compute the max of these two metrics).
3. Pick the pair $$(x,y)$$ that reduces this metric the most. Query this pair $$(x,y)$$.
4. Update the current partial order with the result of this query (and apply the transitive closure). Go back to step 1 and repeat, until you have obtained a total order.
I don't know how well this will perform in practice, but it would be simple to implement and you could try it out.
The ideal metric would be the entropy, i.e., the log of the number of linear extensions compatible with the current partial order, but unfortunately it is #P-hard to compute that metric, and approximating it takes you deep into complicated algorithms, which I suspect you might want to avoid. | 899 | 3,855 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-50 | longest | en | 0.919757 |
https://www.thestudentroom.co.uk/showthread.php?t=4549104 | 1,529,344,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00305.warc.gz | 935,281,161 | 41,897 | You are Here: Home >< Maths
# inverse function question help watch
1. why is that when i want to solve for value of x so a function equals its inverse function
f(x) = f-1(x) (the -1 is suppose to be small)
how come the inverse function can be substituted with x?
for example , f(x) = x^2 - 3
√x+3 = x^2 - 3 can be worked out with x^2 - 3 = x
why is f(x) = f-1(x) equals to f(x) = x ?
could that be because the x and y are reversed in inverse function so by inputting x value i'm actually imputing its y value and outputting the x value?
2. (Original post by alvan15)
why is that when i want to solve for value of x so a function equals its inverse function
Look at the graph of and . You'll see that they intersect only along the line . Hence any solution to will be equivalent to or .
This phenomenon occurs precisely due to the definition of what it means for a function to be an inverse to another. Specifically, their composition is the identity.
3. [QUOTE=Zacken;69983322]
(Original post by alvan15)
why is that when i want to solve for value of x so a function equals its inverse function
/QUOTE]
Look at the graph of and . You'll see that they intersect only along the line . Hence any solution to will be equivalent to or .
This phenomenon occurs precisely due to the definition of what it means for a function to be an inverse to another. Specifically, their composition is the identity.
thank you so much , i totally get it now , appreciate the help, have a nice day
4. (Original post by alvan15)
thank you so much , i totally get it now , appreciate the help, have a nice day
No problem
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Top tips from students who have already aced their exams | 557 | 2,287 | {"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} | 3.890625 | 4 | CC-MAIN-2018-26 | latest | en | 0.927835 |
https://www.infobloom.com/what-is-strain-energy.htm | 1,627,460,765,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00411.warc.gz | 852,843,118 | 15,657 | # What Is Strain Energy?
M. McGee
M. McGee
Strain energy is a form of potential energy that is produced by deforming a solid. As a solid item is deformed, the energy used to deform the item is stored. This energy is released when the deformed item returns to its original shape. This stored energy is called strain energy as a way of describing the effect the energy has on the solid. This energy is a part of a larger class of energy called potential energy; this group is made up of stored energy that is waiting to go into use.
When an item is strained, it is being forced to do something that it is not supposed to do. With strain energy, the strain is caused by forcing the item to hold a shape or position that is different from the one it wants to have. This can be as simple as bending a tree branch or as complex as the deformation that occurs in the systems of a landing fighter jet.
The strain energy within an object is generally equal to the amount of energy applied to the object that cased the strain. This basically means that a solid object will absorb and store the energies applied to it. When the energy is released, it is generally equal to the amount held. As a result, an object can have energy applied to it, then hold it until it is needed and finally release the total amount of the energy back into the system with no loss.
The only time that a loss will occur is when the strain energy approaches or passes the solid’s point of plastic deformation. At this point, the solid will undergo a permanent change into another shape. The energy stored in the item is used to catalyze the change. This deformation will slowly bleed off the strain energy until it has all gone into the transformation or into heat. If the object is released before the deformation is complete, a smaller amount of the energy will release as normal.
The best simple example of strain energy is a simple spring. When a person pushes the spring down, he is exerting force on the solid spring. This energy goes from the person and into the object. As long as the spring is held down all that energy is just sitting within the object. When the spring is released, the energy transforms from potential strain energy to kinetic energy. If the spring were held down so long that its shape changed, then some or all of the energy would be lost to deformation. | 492 | 2,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-31 | latest | en | 0.959318 |
https://newurbanhabitat.com/find-out-how-to-convert-12-fluid-oz-to-ml/ | 1,675,900,190,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500983.76/warc/CC-MAIN-20230208222635-20230209012635-00404.warc.gz | 434,411,988 | 14,433 | # Learn how to transform 12 fluid ozto ml
Changing 12 fluid ozto ml needs to be straightforward, however it’s actually not that easy. That’s as a result of “oz” refers back to the ounce of measurement in English, and that’s a time period that may truly imply various things in numerous contexts. It’s a part of the rarity and unwieldy nature of English measurements that has induced many of the world to transform to the metric system. A liter is 1,000 ml.
That’s fairly easy to grasp. However you want 437.5 grains to make an oz. and 16 ounces to make a pound. So if you will commerce or need to convert 12 ozto ml, most individuals world wide could not have an correct thought of what you might be speaking about. So, with this internet web page, you’ll know the milliliter (ml) equal of 12 fl oz.
## 12 fluid ozto ml – 12 OZ in ML
How a lot is 12 ozin ml? Right here you’ll find the converter program to alter your models from ozto ml (ounces to milliliters). It’s fairly straightforward to make use of. Enter 12 (which is how a lot you’ve gotten in fluid ounces) within the discipline marked OZ.
Subsequent, select the suitable quantity unit. You may have 3 choices right here:
• the common US fluid ounce,
• the US fluid ounce for meals and diet labeling, and
• the UK (imperial) model.
The converter will routinely course of the 12 ozto ml conversion and get the specified milliliter equal displayed within the ML discipline. It’s that straightforward.
Click on the “Reset” button if you wish to carry out one other conversion. You could have a special quantity of liquid in fluid ounces that you just need to convert to milliliters (fl ozto ml conversion).
For those who’re in a rush and wish the reply proper now, right here they’re:
• For those who’re utilizing common US fluid ounces, your 12 fluid ounces equals 354.88 mL when rounded.
• For those who’re placing this on a meals label, it ought to say that your 12 fluid ounces equals 360 mL.
• In case you are utilizing British or Imperial ounces, then 12 ounces equals 340.96 ml rounded.
Abbreviations for fluid ounces: “fl oz”, “fl. oz.”, “oz. Florida.”.
## 12 OZ to ML – Unit Definition
What’s an oz.? For this internet web page, we’re speaking about fluid ounces (fl oz). We’re not speaking about ounces which can be used to measure solids like metals. One ounce Fluid is only for measuring the quantity of a liquid.So
what are these totally different variations of fluid ounces about?The issue is that you just first have to know what sort of fluid ounce you might be utilizing.In case you are a US citizen, you might be most likely utilizing the usual US fluid ounce (US fl oz).
In case you are within the US or enter the milliliter equal on a meals label or diet details, there’s a totally different ml equal. For those who use British or Imperial fluid ounces (UK fl oz), you’ll once more have a barely totally different ml equal.
What’s a milliliter? This one is less complicated. As soon as you know the way a lot liquid is in a liter, you simply have to divide it by the issue of conversion that equals 1,000.That leads to 1 m milliliter. That’s what “milli-” means: 1/1000. When you’ve gotten 1,000 ml, it’s truly 1 liter.
## 12 OZ to ML Conversion Desk
For those who don’t need to use the converter, here’s a fast and helpful desk you should utilize in case you’ll want to memorize conversion components. In case you need an excellent simpler answer, be happy to bookmark this web page.
Quantity US Normal Fl oz UK (Imperial) Fl oz US Vitamin labeling 12 fl oz 354.88 ml 340.96 ml 360 ml
## 12 OZ to ML
So how did we get our outcomes? How do you get the identical outcomes utilizing your personal calculator? Too straightforward. All you’ll want to do is first decide the kind of fluid ounce (fl oz) measurement you’ve gotten. Every sort of fluid ounce (fl oz) has its personal ml equal. So that you simply have to multiply by the equal in ml.
So, you begin by coming into the quantity of fluid ounce you’ve gotten. Right here, we’re assuming it’s 12 ounces, so we will use that for example. However the identical formulation apply no matter what number of fluid ounces you’ve gotten.
### 12 US FL OZ to ML
More often than not, if you’re American, you might be utilizing the usual or typical US fluid ounce (fl oz). Which means 1 fluid ounce (fl oz) is precisely 29.5735295625 milliliters (ml). So multiply 12 by 29.5735295625 and also you get rounded 354.88 mL.
### 12 US FL OZ to ML for Meals Labeling (Vitamin)
Nevertheless, the rule for US Fluid Ounces (US fl oz) adjustments once you put the ml equal on a meals label. This time, 1 fluid ounce (fl oz) equals 30 ml. You merely multiply 12 by 30 and also you get 360 ml. Positive, there’s a distinction, however for many functions, that’s inconsequential.
### 12 UK (Imperial) FL OZ to ML
Now UK residents have their very own thought of what a fluid ounce is. For them, it’s the equal of 28.4130625 ml. So you must multiply 12 by 28.4130625 and you’re going to get 340.96 ml.
## Convert 12 OZ to ML
So why do you’ll want to convert 12 ozto ml? There are a number of good causes for this. One is that it might probably current the amount extra precisely for individuals who use the metric system as a substitute of the English system. Not everybody understands the thought of ounces, pints, and gallons.
The second motive is that you would be able to be sure that everyone seems to be on the best web page. If you use fluid ounces, there might be some confusion as to what you imply. Are you utilizing the common US liquid as soon as (US fl oz), the model for meals labels, or the British/Imperial ounce? Upon getting transformed the determine to ml, it is going to be clear. A milliliter is a milliliter no matter the place your reader is positioned.
## What number of ML is 12 OZ?
It’s principally between 340.96 and 360 ml. For many conditions, the distinction could not imply that a lot. However it may be a special story for a chemistry experiment or a really actual meals recipe.
In any case, 12 fl ozin ml is:
• 354.88 ml if we’re speaking concerning the typical US fluid ounce (US fl oz).
• 360 ml once you need to put this equal on a meals label.
• 340.96 ml if the info is from a UK publication (Imperial UK fl oz).
You’ll be able to all the time around the numbers if you wish to make it extra sensible. So, within the US, a 12-ounce quantity of gasoline is definitely 354.88 ml. Whether it is milk, put it to 360 ml. For those who’re within the UK, it’s actually 340.96 ml. In case you’ll want to do one other ozto ml conversion, don’t neglect to bookmark this web page.
### Tips on how to convert 12 fluid ounces to milliliters?
To calculate 12 fluid ounces to the corresponding worth in milliliters, multiply the quantity in fluid ounces by 29.5735296875 (conversion issue). On this case we should multiply 12 Fluid Ounces by 29.5735296875 to acquire the equal end in Milliliters:
12 Fluid Ounces x 29.5735296875 = 354.88235625 Milliliters
12 Fluid Ounces equals 354.88235625 Milliliters.
### Tips on how to convert from Fluid Ounces to Milliliters?
The conversion issue from Fluid Ounces to Milliliters is 29.5735296875. To search out what number of fluid ounces are in milliliters, multiply by the conversion issue or use the quantity converter above. Twelve Fluid Ounces equals to a few hundred fifty-four level eight eight two Milliliters.
### What’s the fluid ounce?
A fluid ounce (abbreviated fl oz, fl. oz., or fl. oz.) is a unit of quantity. It is the same as about 28.41 ml within the imperial system or about 29.57 ml within the US system. The fluid ounce is usually referred to easily as an “ounce” in functions the place its use is implied.
### What’s Milliliter?
A milliliter (additionally spelled “milliliter”, SI image ml) is a non-SI metric system unit of quantity generally used as a unit of liquid. It is the same as 1/1000 liter, or one cubic centimeter, due to this fact 1ml = 1/1000 L = 1 cm3.
Thanks for studying and please comply with us on Auto Oil And Fluid! | 2,018 | 8,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-06 | latest | en | 0.865396 |
https://math.stackexchange.com/questions/3147961/find-the-cdf-of-z-x-1x-2-x-1x-3 | 1,569,198,324,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575844.94/warc/CC-MAIN-20190923002147-20190923024147-00280.warc.gz | 562,932,386 | 29,990 | # Find the CDF of $Z=\{X_1+X_2, X_1<X_3\}$
Let $$X_1, X_2$$ and $$X_3$$ be three independent exponential random variables. The PDF and CDF of $$X_i$$ with parameter $$\beta_i$$ are $$f_{X_i}(x_i)=\beta_i e^{-\beta_i x_i}$$
$$F_{X_i}(x_i)=1-e^{-\beta_i x_i}$$
What is the CDF of random variable $$Z$$ define the sum of $$X_1$$ and $$X_2$$ where $$X_1$$ is less then $$X_3$$. $$Z=\{X_1+X_2, X_1.
Thanks.
• What are your thoughts? What have you tried? – Easymode44 Mar 14 at 13:19
• I want to find the probability that $X_1+X_2\leq z$ where $X_1$ is upper bound or less then another random variable $X_3$. – Monir Mar 14 at 13:21
• Yes, I can see that. Where exactly are you stuck? – Easymode44 Mar 14 at 13:22
• I tride $$\int_{x_3=0}^{\infty}\int_{x_2=0}^{x_3}\left(\int_{x_2=0}^{z-x_1}f_{X_2}(x_2) dx_2\right)f_{X_1}(x_1) dx_1f_{X_3}(x_3) dx_2.$$ – Monir Mar 14 at 13:24
• I stuck in the last integral. And i think we need to take all possible cases – Monir Mar 14 at 13:25 | 412 | 978 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-39 | latest | en | 0.680152 |
https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_110A%3A_Physical_Chemistry__I/UCD_Chem_110A%3A_Physical_Chemistry_I_(Larsen)/Text/MathChapters/D%3A_Spherical_Coordinates | 1,580,069,830,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00517.warc.gz | 379,383,924 | 25,046 | D: Spherical Coordinates
Objectives
• Understand the concept of area and volume elements in cartesian, polar and spherical coordinates.
• Be able to integrate functions expressed in polar or spherical coordinates.
• Understand how to normalize orbitals expressed in spherical coordinates, and perform calculations involving triple integrals.
Coordinate Systems
The simplest coordinate system consists of coordinate axes oriented perpendicularly to each other. These coordinates are known as cartesian coordinates or rectangular coordinates, and you are already familiar with their two-dimensional and three-dimensional representation. In the plane, any point $$P$$ can be represented by two signed numbers, usually written as $$(x,y)$$, where the coordinate $$x$$ is the distance perpendicular to the $$x$$ axis, and the coordinate $$y$$ is the distance perpendicular to the $$y$$ axis (Figure $$\PageIndex{1}$$, left). In space, a point is represented by three signed numbers, usually written as $$(x,y,z)$$ (Figure $$\PageIndex{1}$$ , right).
Often, positions are represented by a vector, $$\vec{r}$$, shown in red in Figure $$\PageIndex{1}$$. In three dimensions, this vector can be expressed in terms of the coordinate values as $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$, where $$\hat{i}=(1,0,0)$$, $$\hat{j}=(0,1,0)$$ and $$\hat{z}=(0,0,1)$$ are the so-called unit vectors.
We already know that often the symmetry of a problem makes it natural (and easier!) to use other coordinate systems. In two dimensions, the polar coordinate system defines a point in the plane by two numbers: the distance $$r$$ to the origin, and the angle $$\theta$$ that the position vector forms with the $$x$$-axis. Notice the difference between $$\vec{r}$$, a vector, and $$r$$, the distance to the origin (and therefore the modulus of the vector). Vectors are often denoted in bold face (e.g. r) without the arrow on top, so be careful not to confuse it with $$r$$, which is a scalar.
While in cartesian coordinates $$x$$, $$y$$ (and $$z$$ in three-dimensions) can take values from $$-\infty$$ to $$\infty$$, in polar coordinates $$r$$ is a positive value (consistent with a distance), and $$\theta$$ can take values in the range $$[0,2\pi]$$.
The relationship between the cartesian and polar coordinates in two dimensions can be summarized as:
$\label{eq:coordinates_1} x=r\cos\theta$
$\label{eq:coordinates_2} y=r\sin\theta$
$\label{eq:coordinates_3} r^2=x^2+y^2$
$\label{eq:coordinates_4} \tan \theta=y/x$
In three dimensions, the spherical coordinate system defines a point in space by three numbers: the distance $$r$$ to the origin, a polar angle $$\phi$$ that measures the angle between the positive $$x$$-axis and the line from the origin to the point $$P$$ projected onto the $$xy$$-plane, and the angle $$\theta$$ defined as the is the angle between the $$z$$-axis and the line from the origin to the point $$P$$:
Before we move on, it is important to mention that depending on the field, you may see the Greek letter $$\theta$$ (instead of $$\phi$$) used for the angle between the positive $$x$$-axis and the line from the origin to the point $$P$$ projected onto the $$xy$$-plane. That is, $$\theta$$ and $$\phi$$ may appear interchanged. This can be very confusing, so you will have to be careful. When using spherical coordinates, it is important that you see how these two angles are defined so you can identify which is which.
Spherical coordinates are useful in analyzing systems that are symmetrical about a point. For example a sphere that has the cartesian equation $$x^2+y^2+z^2=R^2$$ has the very simple equation $$r = R$$ in spherical coordinates. Spherical coordinates are the natural coordinates for physical situations where there is spherical symmetry (e.g. atoms). The relationship between the cartesian coordinates and the spherical coordinates can be summarized as:
$\label{eq:coordinates_5} x=r\sin\theta\cos\phi$
$\label{eq:coordinates_6} y=r\sin\theta\sin\phi$
$\label{eq:coordinates_7} z=r\cos\theta$
These relationships are not hard to derive if one considers the triangles shown in tFigure $$\PageIndex{4}$$:
Area and Volume Elements
In any coordinate system it is useful to define a differential area and a differential volume element. In cartesian coordinates the differential area element is simply $$dA=dx\;dy$$ (Figure $$\PageIndex{1}$$), and the volume element is simply $$dV=dx\;dy\;dz$$.
We already performed double and triple integrals in cartesian coordinates, and used the area and volume elements without paying any special attention. For example, in example [c2v:c2vex1], we were required to integrate the function $${\left | \psi (x,y,z) \right |}^2$$ over all space, and without thinking too much we used the volume element $$dx\;dy\;dz$$ (see page ). We also knew that “all space” meant $$-\infty\leq x\leq \infty$$, $$-\infty\leq y\leq \infty$$ and $$-\infty\leq z\leq \infty$$, and therefore we wrote:
$\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{\left | \psi (x,y,z) \right |}^2\; dx \;dy \;dz=1 \nonumber$
But what if we had to integrate a function that is expressed in spherical coordinates? Would we just replace $$dx\;dy\;dz$$ by $$dr\; d\theta\; d\phi$$? The answer is no, because the volume element in spherical coordinates depends also on the actual position of the point. This will make more sense in a minute. Coming back to coordinates in two dimensions, it is intuitive to understand why the area element in cartesian coordinates is $$dA=dx\;dy$$ independently of the values of $$x$$ and $$y$$. This is shown in the left side of Figure $$\PageIndex{2}$$. However, in polar coordinates, we see that the areas of the gray sections, which are both constructed by increasing $$r$$ by $$dr$$, and by increasing $$\theta$$ by $$d\theta$$, depend on the actual value of $$r$$. Notice that the area highlighted in gray increases as we move away from the origin.
The area shown in gray can be calculated from geometrical arguments as
$dA=\left[\pi (r+dr)^2- \pi r^2\right]\dfrac{d\theta}{2\pi}.$
Because $$dr<<0$$, we can neglect the term $$(dr)^2$$, and $$dA= r\; dr\;d\theta$$ (see Figure $$10.2.3$$).
Let’s see how this affects a double integral with an example from quantum mechanics. The wave function of the ground state of a two dimensional harmonic oscillator is: $$\psi(x,y)=A e^{-a(x^2+y^2)}$$. We know that the quantity $$|\psi|^2$$ represents a probability density, and as such, needs to be normalized:
$\int\limits_{all\;space} |\psi|^2\;dA=1 \nonumber$
This statement is true regardless of whether the function is expressed in polar or cartesian coordinates. However, the limits of integration, and the expression used for $$dA$$, will depend on the coordinate system used in the integration.
In cartesian coordinates, “all space” means $$-\infty<x<\infty$$ and $$-\infty<y<\infty$$. The differential of area is $$dA=dxdy$$:
$\int\limits_{all\;space} |\psi|^2\;dA=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} A^2e^{-2a(x^2+y^2)}\;dxdy=1 \nonumber$
In polar coordinates, “all space” means $$0<r<\infty$$ and $$0<\theta<2\pi$$. The differential of area is $$dA=r\;drd\theta$$. The function $$\psi(x,y)=A e^{-a(x^2+y^2)}$$ can be expressed in polar coordinates as: $$\psi(r,\theta)=A e^{-ar^2}$$
$\int\limits_{all\;space} |\psi|^2\;dA=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi} A^2 e^{-2ar^2}r\;d\theta dr=1 \nonumber$
Both versions of the double integral are equivalent, and both can be solved to find the value of the normalization constant ($$A$$) that makes the double integral equal to 1. In polar coordinates:
$\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi} A^2 e^{-2ar^2}r\;d\theta dr=A^2\int\limits_{0}^{\infty}e^{-2ar^2}r\;dr\int\limits_{0}^{2\pi}\;d\theta =A^2\times\dfrac{1}{4a}\times2\pi=1 \nonumber$
Therefore1, $$A=\sqrt{2a/\pi}$$. The same value is of course obtained by integrating in cartesian coordinates.
It is now time to turn our attention to triple integrals in spherical coordinates. In cartesian coordinates, the differential volume element is simply $$dV= dx\,dy\,dz$$, regardless of the values of $$x, y$$ and $$z$$. Using the same arguments we used for polar coordinates in the plane, we will see that the differential of volume in spherical coordinates is not $$dV=dr\,d\theta\,d\phi$$. The geometrical derivation of the volume is a little bit more complicated, but from Figure $$\PageIndex{4}$$ you should be able to see that $$dV$$ depends on $$r$$ and $$\theta$$, but not on $$\phi$$. The volume of the shaded region is
$\label{eq:dv} dV=r^2\sin\theta\,d\theta\,d\phi\,dr$
We will exemplify the use of triple integrals in spherical coordinates with some problems from quantum mechanics. We already introduced the Schrödinger equation, and even solved it for a simple system in Section 5.4. We also mentioned that spherical coordinates are the obvious choice when writing this and other equations for systems such as atoms, which are symmetric around a point.
As we saw in the case of the particle in the box (Section 5.4), the solution of the Schrödinger equation has an arbitrary multiplicative constant. Because of the probabilistic interpretation of wave functions, we determine this constant by normalization. The same situation arises in three dimensions when we solve the Schrödinger equation to obtain the expressions that describe the possible states of the electron in the hydrogen atom (i.e. the orbitals of the atom). The Schrödinger equation is a partial differential equation in three dimensions, and the solutions will be wave functions that are functions of $$r, \theta$$ and $$\phi$$. The lowest energy state, which in chemistry we call the 1s orbital, turns out to be:
$\psi_{1s}=Ae^{-r/a_0} \nonumber$
This particular orbital depends on $$r$$ only, which should not surprise a chemist given that the electron density in all $$s$$-orbitals is spherically symmetric. We will see that $$p$$ and $$d$$ orbitals depend on the angles as well. Regardless of the orbital, and the coordinate system, the normalization condition states that:
$\int\limits_{all\;space} |\psi|^2\;dV=1 \nonumber$
For a wave function expressed in cartesian coordinates,
$\int\limits_{all\;space} |\psi|^2\;dV=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x,y,z)\psi(x,y,z)\,dxdydz \nonumber$
where we used the fact that $$|\psi|^2=\psi^* \psi$$.
In spherical coordinates, “all space” means $$0\leq r\leq \infty$$, $$0\leq \phi\leq 2\pi$$ and $$0\leq \theta\leq \pi$$. The differential $$dV$$ is $$dV=r^2\sin\theta\,d\theta\,d\phi\,dr$$, so
$\int\limits_{all\;space} |\psi|^2\;dV=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}\psi^*(r,\theta,\phi)\psi(r,\theta,\phi)\,r^2\sin\theta\,dr d\theta d\phi=1 \nonumber$
Let’s see how we can normalize orbitals using triple integrals in spherical coordinates.
Example $$\PageIndex{1}$$
When solving the Schrödinger equation for the hydrogen atom, we obtain $$\psi_{1s}=Ae^{-r/a_0}$$, where $$A$$ is an arbitrary constant that needs to be determined by normalization. Find $$A$$.
Solution
In spherical coordinates,
$\int\limits_{all\; space} |\psi|^2\;dV=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}\psi^*(r,\theta,\phi)\psi(r,\theta,\phi)\,r^2\sin\theta\,dr d\theta d\phi=1 \nonumber$
because this orbital is a real function, $$\psi^*(r,\theta,\phi)\psi(r,\theta,\phi)=\psi^2(r,\theta,\phi)$$. In this case, $$\psi^2(r,\theta,\phi)=A^2e^{-2r/a_0}$$.
Therefore,
$\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}\psi^*(r,\theta,\phi)\psi(r,\theta,\phi) \, r^2 \sin\theta \, dr d\theta d\phi=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}A^2e^{-2r/a_0}\,r^2\sin\theta\,dr d\theta d\phi=1 \nonumber$
$\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}A^2e^{-2r/a_0}\,r^2\sin\theta\,dr d\theta d\phi=A^2\int\limits_{0}^{2\pi}d\phi\int\limits_{0}^{\pi}\sin\theta \;d\theta\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr \nonumber$
The result is a product of three integrals in one variable:
$\int\limits_{0}^{2\pi}d\phi=2\pi \nonumber$
$\int\limits_{0}^{\pi}\sin\theta \;d\theta=-\cos\theta|_{0}^{\pi}=2 \nonumber$
$\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr=? \nonumber$
From the formula sheet:
$\int_{0}^{\infty}x^ne^{-ax}dx=\dfrac{n!}{a^{n+1}}, \nonumber$
where $$a>0$$ and $$n$$ is a positive integer.
In this case, $$n=2$$ and $$a=2/a_0$$, so:
$\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr=\dfrac{2!}{(2/a_0)^3}=\dfrac{2}{8/a_0^3}=\dfrac{a_0^3}{4} \nonumber$
Putting the three pieces together:
$A^2\int\limits_{0}^{2\pi}d\phi\int\limits_{0}^{\pi}\sin\theta \;d\theta\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr=A^2\times2\pi\times2\times \dfrac{a_0^3}{4}=1 \nonumber$
$A^2\times \pi \times a_0^3=1\rightarrow A=\dfrac{1}{\sqrt{\pi a_0^3}} \nonumber$
The normalized 1s orbital is, therefore:
$\displaystyle{\color{Maroon}\dfrac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}} \nonumber$ | 3,956 | 13,053 | {"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} | 4.71875 | 5 | CC-MAIN-2020-05 | latest | en | 0.894033 |
http://www.sparknotes.com/physics/rotationalmotion/rotationalkinetics/section2.rhtml | 1,369,286,988,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702849682/warc/CC-MAIN-20130516111409-00000-ip-10-60-113-184.ec2.internal.warc.gz | 715,818,622 | 20,286 | # Rotational Kinetics
### Rotational Kinematics
In this section we will use our new definitions for rotational variables to generate kinematic equations for rotational motion. In addition, we will examine the vector nature of rotational variables and, finally, relate linear and angular variables.
#### Kinematic Equations
Because our equations defining rotational and translational variables are mathematically equivalent, we can simply substitute our rotational variables into the kinematic equations we have already derived for translational variables. We could go through the formal derivation of these equations, but they would be the same as those derived in One-Dimensional Kinematics. Thus we can simply state the equations, alongside their translational analogues:
v f = v o + at σ f = σ o + αt x f = x o + v o t + at 2 μ f = μ o + σ o t + αt 2 v f 2 = v o 2 + 2ax σ f 2 = σ o 2 +2αμ x = (v o + v f)t μ = (σ o + σ f)t
These equations for rotational motion are used identically as the corollary equations for translational motion. In addition, like translational motion, these equations are only valid when the acceleration, α , is constant. These equations are frequently used and form the basis for the study of rotational motion.
#### Relationships Between Rotational and Translational Variables
Now that we have established both equations for our variables and kinematic equations relating them, we can also relate our rotational variables to translational variables. This can sometimes be confusing. It is easy to think that because a particle is engaged in rotational motion, it is not also defined by translational variables. Simply remind yourself that no matter what path a given particle is traveling in, it always has a position, velocity and acceleration. The rotational variables we generated do not substitute for these traditional variables; instead, they simplify calculations involving rotational motion. Thus we can relate our rotational and translational variables.
#### Translational and Angular Displacement
Recall from our definition of angular displacement that:
μ = s/r
Implying that
s = μr
Thus the displacement, s , of a particle in rotational motion is given by the angular displacement multiplied by the radius of the particle from the axis of rotation. We can differentiate both sides of the equation with respect to time:
=
Thus
v = σr
#### Translational and Angular Velocity
Just as linear displacement is equal to angular displacement times the radius, linear velocity is equal to angular velocity times the radius. We can relate α and a , by the same method we used before: differentiating with respect to time.
= r
#### Translational and Angular Acceleration
We must be careful in relating translationa and angular acceleration because only gives us the change in velocity with respect to time in the tangential direction. We know from Dynamics that any particle traveling in a circle experiences a radial force equal to . We must therefore generate two different expressions for the linear acceleration of a particle in rotational motion:
a T = αr a R = = σ 2 r
These two equations may seem a bit confusing, so we shall examine them closely. Consider a particle moving around a circle with a constant speed. The rate at which the particle makes a revolution about the axis is constant, so α = 0 and a T = 0 . However, the particle is being constantly accelerated towards the center of the circle, so a R is nonzero, and varies with the square of the angular velocity of the particle.
#### The Power of Rotational Equations
With these equations we can describe the motion of any given particle through rotational and translational variables. So why even bother with rotational variables if everything can be expressed in terms of the more familiar linear variables? The answer lies in the fact that every particle in a rigid body has the same value for rotational variables. This characteristic makes rotational variables a far more powerful means of predicting the motion of rotating bodies, and not just particles.
#### Vector Notation of Rotational Variables
Every equation we have derived so far has been in terms of the magnitude of our rotational variables. But what about their direction? Can we give our variables both magnitude and direction? It would seem as though the direction of our rotational variables would be the same as our linear ones. For instance, it would make sense to make the direction of angular velocity always tangent to the circle through which the particle travels. However, with this definition, the direction of σ is always changing, even if the particle is moving with constant angular velocity. Clearly, such inconsistency is a problem; we must define the direction for our variables in a new way.
For reasons too complicated to discuss here, angular displacement μ cannot be represented as a vector. However, σ and α can, and we shall describe how to find their direction through the right hand rule.
#### Right Hand Rule
Take your right hand, curl your fingers, and stick your thumb straight up. If you let the curl of your fingers follow the path of the rotating particle or body, your thumb will point in the direction of the angular velocity of the body. This way, the direction is constant throughout the rotation. Below are shown a few examples of rotation, and of the resultant direction of σ :
Figure %: Three different directions of rotation, shown with direction of σ
Angular acceleration is defined in a similar way. If the magnitude of the angular velocity increases, then the angular acceleration is in the same direction as the angular velocity. Conversely, if the magnitude of the velocity decreases, the angular acceleration points in the direction opposite the angular velocity.
Though the direction of these vectors may seem trivial for now, they become quite important when studying concepts such as torque and angular momentum. Now, equipped with kinematic equations for rotational motion, relations between angular and linear variables, and a sense of the vector notation of rotational variables, we are able to develop and explore the dynamics and energetics of rotational motion.
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Get Our FREE NOOK Reading Apps | 1,385 | 6,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2013-20 | longest | en | 0.886225 |
https://www.geeksforgeeks.org/count-how-many-times-the-given-digital-clock-shows-identical-digits/ | 1,591,002,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347415315.43/warc/CC-MAIN-20200601071242-20200601101242-00379.warc.gz | 739,214,673 | 30,060 | # Count how many times the given digital clock shows identical digits
Given a generic digital clock, having h number of hours and m number of minutes, the task is to find how many times the clock shows identical time. A specific time is said to be identical if every digit in the hours and minutes is same i.e. the time is of type D:D, D:DD, DD:D or DD:DD.
Note that the time is written on the digital clock without any leading zeros and the clock shows time between 0 to h – 1 hours and 0 to m – 1 minutes. Few examples of identical times are:
• 1:1
• 22:22
• 3:33
• 11:1
Examples:
Input: hours = 24, minutes = 60
Output: 19
The clock has 24 hours and 60 minutes.
So the identical times will be:
Single digit hours and single digit minutes -> 0:0, 1:1, 2:2, …., 9:9
Single digit hours and double digit minutes -> 1:11, 2:22, 3:33, 4:44 and 5:55
Double digit hours and single digit minutes -> 11:1 and 22:2
Double digit hours and double digit minutes -> 11:11, 22:22
Total = 10 + 5 + 2 + 2 = 19
Input: hours = 34, minutes = 50
Output: 20
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: As we can see in the explained example, we have to first count the single digit (of hours) identical times and then double-digit hours. During each of these counts, we need to consider single digit minutes as well as double-digit minutes.
There will be two loops. First loop deals with single digit hours. And the second deals with double digit hours. In each of the loops, there should be two conditions. First, if iterator variable is less than total minutes, then increment the counter. Second, if (iterator variable + iterator variable * 10) is less than total minutes, increment the counter. At the end, we will have the total identical times that clock shows.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` ` `// Function to return the count of ` `// identical times the clock shows ` `int` `countIdentical(``int` `hours, ``int` `minutes) ` `{ ` ` ` ` ``// To store the count of identical times ` ` ``// Initialized to 1 because of 0:0 ` ` ``int` `i, count = 1; ` ` ` ` ``// For single digit hour ` ` ``for` `(i = 1; i <= 9 && i < hours; i++) { ` ` ` ` ``// Single digit minute ` ` ``if` `(i < minutes) ` ` ``count++; ` ` ` ` ``// Double digit minutes ` ` ``if` `((i * 10 + i) < minutes) ` ` ``count++; ` ` ``} ` ` ` ` ``// For double digit hours ` ` ``for` `(i = 11; i <= 99 && i < hours; i = i + 11) { ` ` ` ` ``// Single digit minute ` ` ``if` `(i < minutes) ` ` ``count++; ` ` ` ` ``// Double digit minutes ` ` ``if` `((i % 10) < minutes) ` ` ``count++; ` ` ``} ` ` ` ` ``// Return the required count ` ` ``return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `hours = 24; ` ` ``int` `minutes = 60; ` ` ``cout << countIdentical(hours, minutes); ` ` ` ` ``return` `0; ` `} `
## Java
`// Java implementation of the above approach ` `class` `GFG ` `{ ` ` ` ` ``// Function to return the count of ` ` ``// identical times the clock shows ` ` ``static` `int` `countIdentical(``int` `hours, ``int` `minutes) ` ` ``{ ` ` ` ` ``// To store the count of identical times ` ` ``// Initialized to 1 because of 0:0 ` ` ``int` `i, count = ``1``; ` ` ` ` ``// For single digit hour ` ` ``for` `(i = ``1``; i <= ``9` `&& i < hours; i++) ` ` ``{ ` ` ` ` ``// Single digit minute ` ` ``if` `(i < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// Double digit minutes ` ` ``if` `((i * ``10` `+ i) < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// For double digit hours ` ` ``for` `(i = ``11``; i <= ``99` `&& i < hours; i = i + ``11``) ` ` ``{ ` ` ` ` ``// Single digit minute ` ` ``if` `(i < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// Double digit minutes ` ` ``if` `((i % ``10``) < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// Return the required count ` ` ``return` `count; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `hours = ``24``; ` ` ``int` `minutes = ``60``; ` ` ``System.out.println(countIdentical(hours, minutes)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
## Python3
`# Python 3 implementation of the approach ` ` ` `# Function to return the count of ` `# identical times the clock shows ` `def` `countIdentical(hours, minutes): ` ` ` ` ``# To store the count of identical times ` ` ``# Initialized to 1 because of 0:0 ` ` ``count ``=` `1` ` ``i ``=` `1` ` ` ` ``# For single digit hour ` ` ``while``(i <``=` `9` `and` `i < hours): ` ` ` ` ``# Single digit minute ` ` ``if` `(i < minutes): ` ` ``count ``+``=` `1` ` ` ` ``# Double digit minutes ` ` ``if` `((i ``*` `10` `+` `i) < minutes): ` ` ``count ``+``=` `1` ` ` ` ``i ``+``=` `1` ` ` ` ``# For double digit hours ` ` ``i ``=` `11` ` ``while``(i <``=` `99` `and` `i < hours): ` ` ` ` ``# Single digit minute ` ` ``if` `(i < minutes): ` ` ``count ``+``=` `1` ` ` ` ``# Double digit minutes ` ` ``if` `((i ``%` `10``) < minutes): ` ` ``count ``+``=` `1` ` ` ` ``i ``+``=` `11` ` ` ` ``# Return the required count ` ` ``return` `count ` ` ` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ``hours ``=` `24` ` ``minutes ``=` `60` ` ``print``(countIdentical(hours, minutes)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar `
## C#
`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ``// Function to return the count of ` ` ``// identical times the clock shows ` ` ``static` `int` `countIdentical(``int` `hours, ``int` `minutes) ` ` ``{ ` ` ` ` ``// To store the count of identical times ` ` ``// Initialized to 1 because of 0:0 ` ` ``int` `i, count = 1; ` ` ` ` ``// For single digit hour ` ` ``for` `(i = 1; i <= 9 && i < hours; i++) ` ` ``{ ` ` ` ` ``// Single digit minute ` ` ``if` `(i < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// Double digit minutes ` ` ``if` `((i * 10 + i) < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// For double digit hours ` ` ``for` `(i = 11; i <= 99 && i < hours; i = i + 11) ` ` ``{ ` ` ` ` ``// Single digit minute ` ` ``if` `(i < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// Double digit minutes ` ` ``if` `((i % 10) < minutes) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// Return the required count ` ` ``return` `count; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main(String[] args) ` ` ``{ ` ` ``int` `hours = 24; ` ` ``int` `minutes = 60; ` ` ``Console.WriteLine(countIdentical(hours, minutes)); ` ` ``} ` `} ` ` ` `// This code has been contributed by 29AjayKumar `
## PHP
` `
Output:
```19
```
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Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. | 2,716 | 8,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-24 | latest | en | 0.858541 |
https://hackage.haskell.org/package/Yampa-0.13.1/src/tests/AFRPTestsLoopLaws.hs | 1,590,747,300,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00174.warc.gz | 387,069,280 | 1,907 | {-# OPTIONS_GHC -fno-warn-tabs #-} {- \$Id: AFRPTestsLoopLaws.hs,v 1.2 2003/11/10 21:28:58 antony Exp \$ ****************************************************************************** * A F R P * * * * Module: AFRPTestsLoopLaws * * Purpose: Test cases based on the arrow laws for loop * * Authors: Antony Courtney and Henrik Nilsson * * * * Copyright (c) Yale University, 2003 * * * ****************************************************************************** -} module AFRPTestsLoopLaws (looplaws_trs, looplaws_tr) where import Data.Tuple(swap) import FRP.Yampa import AFRPTestsCommon ------------------------------------------------------------------------------ -- Test cases based on the arrow laws for loop ------------------------------------------------------------------------------ -- For a description of the laws, see Ross Paterson: Embedding a Class of -- Domain-Specific Languages in a Functional Language. -- Only a very rudimentary sanity check. Obviously not intended to "prove" -- this implementation indeed do respect the laws. simple_loop :: ((a,c) -> (b,c)) -> (a -> b) simple_loop f a = b where (b, c) = f (a, c) -- Left tightening looplaws_t0_f = second integral >>> arr swap looplaws_t0_h :: Fractional a => SF a a looplaws_t0_h = arr (+10.0) looplaws_t0_lhs :: [Double] looplaws_t0_lhs = testSF1 (loop (first looplaws_t0_h >>> looplaws_t0_f)) looplaws_t0_rhs :: [Double] looplaws_t0_rhs = testSF1 (looplaws_t0_h >>> loop looplaws_t0_f) -- Right tightening looplaws_t1_f = second integral >>> arr swap looplaws_t1_h :: Fractional a => SF a a looplaws_t1_h = arr (+10.0) looplaws_t1_lhs :: [Double] looplaws_t1_lhs = testSF1 (loop (looplaws_t1_f >>> first looplaws_t1_h)) looplaws_t1_rhs :: [Double] looplaws_t1_rhs = testSF1 (loop looplaws_t1_f >>> looplaws_t1_h) -- Sliding -- Used to work with only signature t2_f :: Fractional a -> SF a a looplaws_t2_f :: SF (Double, Double) (Double, Double) looplaws_t2_f = integral looplaws_t2_k = id *** (+42.0) looplaws_t2_lhs :: [Double] looplaws_t2_lhs = testSF1 (loop (looplaws_t2_f >>> arr looplaws_t2_k)) looplaws_t2_rhs :: [Double] looplaws_t2_rhs = testSF1 (loop (arr looplaws_t2_k >>> looplaws_t2_f)) -- Vanishing -- The lazy pattern matching (~) is necessary to avoid a black hole in the -- RHS due to premature forcing of tuples. As far as I can tell, loop is -- as lazy as it can be, and this problem could not have been solved by -- "fixing" the loop definition. looplaws_t3_f = second integral >>> first (arr swap) >>> arr (\ ~((a,b),c) -> ((a,c),b)) looplaws_t3_lhs :: [Double] looplaws_t3_lhs = testSF1 (loop (loop looplaws_t3_f)) looplaws_t3_rhs :: [Double] looplaws_t3_rhs = testSF1 (loop (arr assocInv >>> looplaws_t3_f >>> arr assoc)) -- Superposing looplaws_t4_f = second integral >>> arr swap looplaws_t4_lhs :: [(Double,Double)] looplaws_t4_lhs = testSF1 (arr dup >>> (second (loop looplaws_t4_f))) looplaws_t4_rhs :: [(Double, Double)] looplaws_t4_rhs = testSF1 (arr dup >>> (loop (arr assoc >>> second looplaws_t4_f >>> arr assocInv))) -- Extension looplaws_t5_f = \(a,c) -> (take 5 c, a : c) looplaws_t5_lhs :: [[Double]] looplaws_t5_lhs = testSF1 (loop (arr looplaws_t5_f)) looplaws_t5_rhs :: [[Double]] looplaws_t5_rhs = testSF1 (arr (simple_loop looplaws_t5_f)) looplaws_trs = [ looplaws_t0_lhs ~= looplaws_t0_rhs, looplaws_t1_lhs ~= looplaws_t1_rhs, looplaws_t2_lhs ~= looplaws_t2_rhs, looplaws_t3_lhs ~= looplaws_t3_rhs, looplaws_t4_lhs ~= looplaws_t4_rhs, looplaws_t5_lhs ~= looplaws_t5_rhs ] looplaws_tr = and looplaws_trs | 1,000 | 3,520 | {"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} | 2.8125 | 3 | CC-MAIN-2020-24 | latest | en | 0.587013 |
https://m.lotterypost.com/thread/187208 | 1,656,836,436,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00587.warc.gz | 422,386,096 | 8,113 | # How Do I Rundown Numbers?? Part 2
On yesteday I introduced you to how I do my rundown. When any number is drawn basically all that is done is that they go up and down on a number that was prevously drawn. Now going back to my example from yesterday: which was 972...if I was to go up on 972 it would be...072...and if I went down on it it would be 872...now those particuliar numbers might not be drawn...remmember I said that their nine other possibilities in each each family that could be drawn. Let's look at them...in the 072 rundown you would have 072-183-294-305-416-527-638-749-850-961-072...in the 872 rundown you have 872-983-094-105-216-327-438-549-650-761-872. Keep in mind that these numbers do not have to be drawn the very next draw...it might be three draws or more before a number from one or both of these family of numbers is drawn. Cash 4 would work the same way but it a little more complicate because you still have to determine what the front number will be...I will discuss this in a later posting... Good Bye and Good Luck!!!
Suggestion: Place all your rundown information in one thread instead of multiple threads: It'll make it easier to follow the Topic and make it easier to locate the information in the future if its all in one thread.
<Moved to Lottery Systems forum>
Please post in the appropriate forum ... thank you. | 333 | 1,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-27 | latest | en | 0.96187 |
https://cs.stackexchange.com/questions/1296/solve-a-recurrence-using-the-master-theorem/1297 | 1,620,309,364,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00229.warc.gz | 225,932,320 | 37,869 | # Solve a recurrence using the master theorem
This is the recursive formula for which I'm trying to find an asymptotic closed form by the master theorem: $$T(n)=9T(n/27)+(n \cdot \lg(n))^{1/2}$$
I started with $a=9,b=27$ and $f(n)=(n\cdot \lg n)^{1/2}$ for using the master theorem by $n^{\log_b(a)}$, and if so $n^{\log_{27}(9)}=n^{2/3}$ but I don't understand how to play with the $(n\cdot \lg n)^{1/2}$.
I think that the $(n\cdot \lg n)^{1/2}$ is bigger than $n^{2/3}$ but I'm sure I skip here on something.
I think it fits to the third case of the master theorem.
• No. $(n\lg n)^{1/2} = o(n^{2/3})$. – JeffE Apr 16 '12 at 8:38
• There is no question here. – Raphael Apr 16 '12 at 19:00
$f(n) = (n\cdot \lg n)^{1/2}$ and $n^{\log_b a}=n^{2/3}$, thus $f(n) = O(n^{\log_b a})$ and even $f(n) = O(n^{\log_b a - \epsilon})$ for $\epsilon < 1/6$.
Why? because $$\lim_{n\to\infty} \frac{f(n)}{n^{\log_b a - \epsilon}} = \lim_{n\to\infty}\frac{n^{1/2}\lg^{1/2}n}{n^{2/3-\epsilon}} = \lim_{n\to\infty} \frac{\lg^{1/2}n}{n^{1/6-\epsilon}} =0 \quad\text{for }\epsilon< 1/6$$
Thus case 1 of the Master theorem should apply, and $T(n) = \Theta(n^{2/3})$. | 467 | 1,154 | {"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} | 4.1875 | 4 | CC-MAIN-2021-21 | latest | en | 0.831445 |
https://robotics.stackexchange.com/questions/12838/how-to-move-a-robot-to-a-point-while-avoiding-obstacles/12843 | 1,632,093,270,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00551.warc.gz | 549,490,922 | 39,791 | # How to move a robot to a point while avoiding obstacles?
I am trying to move a robot in a straight line from point A, to point B. The robot's primary sensor is a Hokuyo URG-04LX-UG01 LIDAR that gives me the magnitude and direction of each point it detects in the form of two arrays.
I also have the wheel encoders on each motor of the robot, so I can obtain some odometry of the robot even though its accuracy will diminish over time.
The problem that I have is that there are obstacles in the path of point A and point B that the robot must go around. I am not sure how to take the readings from the LIDAR, and convert them into movements that go around the obstacle, while also heading to B. Sorry if that doesn't make sense, but I am really confused on how to solve this problem.
Is this an easy problem to solve and I am just over complicating it? Does anyone know of a way to do this? Any help or guidance would be greatly appreciated.
• I would recommend looking up RRT (rapidly exploring random trees). There is likely an open source implementation for your software stack (ROS has many). Jul 19 '17 at 19:23
First off, I would highly recommend checking ROS (Robot Operating System) out, the ROS community includes researchers and enthusiasts as well as professionals from all around the world and thus most of what you need to solve the problem is already available there. You would have to just implement what others have done, on your system.
I, rather me and my team, have done the exact same thing as you want and the problem becomes much easier if you are using ROS.
There are pre-built SLAM packages available open source which you can use and for making maps and localizing your robot. Once you have the robot's position in a map that also contains all of the obstacles, you just have to do the path planning so that you do not hit any of the obstacles and reach your goal as fast as possible. Then break your path into neared goal points which are simpler to reach so that just by getting to these points your robot can get to the final goal.
As the problem you have put forth is a rather large problem, I feel it is hard to explain it completely here, but I will try to do as much as I can (When I get time I will add more links to help).
Let us break this problem into several parts,
(Don't worry about some occasional ROS jargon, you'll catch them during the tutorials).
1. If you are unaware of ROS, I suggest taking a look at it and here is a place where you can learn it hands-on.
2. Next you will have to get your sensors working with ROS, this should be easy as for most of them you will have packages already made and open source.
3. Once all your sensors are working with ROS, you will have topics which contain have current sensor data, now you can do anything with them, i would use filters on the data and fuse to find your robots estimated pose. (You should also look into Kalman and Extended Kalman Filters at this point.)
4. Now is the time when you can use some SLAM (Simultaneous Localization and Mapping) algorithms to have a map with obstacles and your robot at all times.
5. Do the Motion Planning, break the path into smaller pieces and feed them to your robot in the format it wants (You will need some knowledge about Control Systems at this point).
From my experience you can use these packages : gmapping, robot-poe-ekf, moveit. (Google these by adding ros)
I hope this helps, and all the best man.
What you are talking about is Robotic Path Planning.
Regarding your specific questions
1. "Is this an easy problem to solve" - For the specific problem you show, no. It is 2D Configuration Space with defined goal and obstacles. If you expand that to an loosely defined goal and undefined obstacles in a 3D space then NO!!
2. "Does anyone know of a way to do this?" - There are LOTS of people who have been doing this and researching it for a while. Google "Robotics Path Planning". There are products that do it from the Roomba to the Mars Exploration Rover. They can both path plan; but, I'm sure both have situations where they will get stuck or confused.
3. "Any help or guidance"
• Read and learn from others
• Try something simple
• Once you have simple working, add some complexity and the go back to the first step (read) and continue
start small and grow from there | 964 | 4,338 | {"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} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.971962 |
https://kupdf.net/download/8698462-short-circuit-calculations_59c5678808bbc56b49686f44_pdf | 1,656,926,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00782.warc.gz | 394,656,882 | 14,673 | # 8698462 Short Circuit Calculations
September 23, 2017 | Author: Mohamed Wahid | Category: Alternating Current, Force, Electrical Engineering, Electricity, Electromagnetism
#### Description
Short Circuit Calculation Sector Energy D SE PTI NC Steffen Schmidt
Standards and Terms
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Steffen Schmidt
Purpose of Short-Circuit Calculations Dimensioning of switching devices Dynamic dimensioning of switchgear Thermal rating of electrical devices (e.g. cables) Protection coordination Fault diagnostic Input data for Earthing studies Interference calculations EMC planning …..
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Steffen Schmidt
Short-Circuit Calculation Standards IEC 60909: Short-Circuit Current Calculation in Three-Phase A.C. Systems European Standard EN 60909 German National Standard DIN VDE 0102 further National Standards Engineering Recommendation G74 (UK) Procedure to Meet the Requirements of IEC 60909 for the Calculation of Short-Circuit Currents in Three-Phase AC Power Systems ANSI IIEEE Std. C37.5 (US) IEEE Guide for Calculation of Fault Currents for Application of a.c. High Voltage Circuit Breakers Rated on a Total Current Basis. Page 4
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Short-Circuit Calculations Standard IEC 60909 IEC 60909 : Short-circuit currents in threephase a.c. systems Part 0: Part 1: Part 2: Part 3:
Part 4:
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Calculation of currents Factors for the calculation of short-circuit currents Electrical equipment; data for short-circuit current calculations Currents during two separate simultaneous line-to-earth short circuits and partial short-circuit currents flowing through earth Examples for the calculation of short-circuit currents Copyright © Siemens AG 2008. All rights reserved. E D SE PTI NC
Short-Circuit Calculations Scope of IEC 60909 three-phase a.c. systems low voltage and high voltage systems up to 500 kV nominal frequency of 50 Hz and 60 Hz balanced and unbalanced short circuits three phase short circuits two phase short circuits (with and without earth connection) single phase line-to-earth short circuits in systems with solidly earthed or impedance earthed neutral two separate simultaneous single-phase line-to-earth short circuits in a systems with isolated neutral or a resonance earthed neutral (IEC 60909-3) maximum short circuit currents minimum short circuit currents
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Short-Circuit Calculations Types of Short Circuits
3-phase
2-phase
1-phase
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Steffen Schmidt
Variation of short circuit current shapes
fault at voltage peak
fault at voltage zero crossing
fault located in the network
fault located near generator
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Short-Circuit Calculations Far-from-generator short circuit Ik” Initial symmetrical short-circuit current ip Peak short-circuit current Ik Steady-state short-circuit current A
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Steffen Schmidt
Initial value of the d.c component
Short-Circuit Calculations Definitions according IEC 60909 (I) initial symmetrical short-circuit current Ik” r.m.s. value of the a.c. symmetrical component of a prospective (available) short-circuit current, applicable at the instant of short circuit if the impedance remains at zero-time value initial symmetrical short-circuit power Sk” fictitious value determined as a product of the initial symmetrical shortcircuit current Ik”, the nominal system voltage Un and the factor √3:
Sk" = 3 ⋅ Un ⋅ Ik" NOTE: Sk” is often used to calculate the internal impedance of a network feeder at the connection point. In this case the definition given should be used in the following form:
c ⋅ Un2 Z= Sk" Page 10
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Steffen Schmidt
Short-Circuit Calculations Definitions according IEC 60909 (II) decaying (aperiodic) component id.c. of short-circuit current mean value between the top and bottom envelope of a short-circuit current decaying from an initial value to zero peak short-circuit current ip maximum possible instantaneous value of the prospective (available) short-circuit current NOTE: The magnitude of the peak short-circuit current varies in accordance with the moment at which the short circuit occurs.
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Short-Circuit Calculations Near-to-generator short circuit Ik” ip Ik A IB
Initial symmetrical short-circuit current Peak short-circuit current Steady-state short-circuit current Initial value of the d.c component Symmetrical short-circuit breaking current
2 ⋅ 2 ⋅ IB
tB
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Short-Circuit Calculations Definitions according IEC 60909 (III) steady-state short-circuit current Ik r.m.s. value of the short-circuit current which remains after the decay of the transient phenomena symmetrical short-circuit breaking current Ib r.m.s. value of an integral cycle of the symmetrical a.c. component of the prospective short-circuit current at the instant of contact separation of the first pole to open of a switching device
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Short-Circuit Calculations Purpose of Short-Circuit Values Design Criterion
Physical Effect
Relevant short-circuit current
Breaking capacity of circuit breakers
Thermal stress to arcing chamber; arc extinction
Symmetrical short-circuit breaking current Ib
Mechanical stress to equipment
Forces to electrical devices (e.g. bus bars, cables…)
Peak short-circuit current ip
Thermal stress to equipment
Temperature rise of electrical Initial symmetrical shortcircuit current Ik” devices (e.g. cables) Fault duration Selective detection of partial Minimum symmetrical shortcircuit current Ik short-circuit currents
Protection setting Earthing, Interference, EMC
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Potential rise; Magnetic fields
Steffen Schmidt
Maximum initial symmetrical short-circuit current Ik”
Standard IEC 60909 Simplifications and Assumption Assumptions quasi-static state instead of dynamic calculation no change in the type of short circuit during fault duration no change in the network during fault duration arc resistances are not taken into account impedance of transformers is referred to tap changer in main position neglecting of all shunt impedances except for C0
-> safe assumptions
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Equivalent Voltage Source
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Short-circuit Equivalent voltage source at the short-circuit location real network Q
A
F
equivalent circuit ZN
Q
ZT
A
ZL ~
I"K
c.U n 3
Operational data and the passive load of consumers are neglected Tap-changer position of transformers is dispensable Excitation of generators is dispensable Load flow (local and time) is dispensable Page 17
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Short circuit in meshed grid Equivalent voltage source at the short-circuit location real network
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equivalent circuit
Steffen Schmidt
Voltage Factor c c is a safety factor to consider the following effects: voltage variations depending on time and place, changing of transformer taps, neglecting loads and capacitances by calculations, the subtransient behaviour of generators and motors. Voltage factor c for calculation of Nominal voltage
maximum short circuit currents
minimum short circuit currents
-systems with a tolerance of 6%
1.05
0.95
-systems with a tolerance of 10%
1.10
0.95
Medium voltage >1 kV – 35 kV
1.10
1.00
High voltage >35 kV
1.10
1.00
Low voltage 100 V – 1000 V
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Maximum and minimum Short-Circuit Currents maximum
minimum
short circuit currents
short circuit currents
Voltage factor
Cmax
Cmin
Power plants
Maximum contribution
Minimum contribution
Network feeders
Minimum impedance
Maximum impedance
Motors
shall be considered
shall be neglected
at 20°C
at maximum temperature
Resistance of lines and cables
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Short Circuit Impedances and Correction Factors
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Short Circuit Impedances For network feeders, transformer, overhead lines, cable etc. impedance of positive sequence system = impedance of negative sequence system impedance of zero sequence system usually different topology can be different for zero sequence system Correction factors for generators, generator blocks, network transformer factors are valid in zero, positive, negative sequence system
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Network feeders At a feeder connection point usually one of the following values is given: the initial symmetrical short circuit current Ik” the initial short-circuit power Sk” c ⋅ Un c ⋅ Un2 ZQ = = " Sk" 3 ⋅ Ik XQ =
ZQ 1 + (R / X)2
If R/X of the network feeder is unknown, one of the following values can be used: R/X = 0.1 R/X = 0.0 for high voltage systems >35 kV fed by overhead lines Page 23
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Network transformer Correction of Impedance
ZTK = ZT KT general
c max K T = 0,95 ⋅ 1 + 0,6 ⋅ x T
at knownUconditions c max of operation K T = nb ⋅ U 1 + x T (IbT IrT ) sin ϕbT
no correction for impedances between star point and ground Page 24
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Network transformer Impact of Correction Factor 1.05 1.00
KT
0.95 0.90 cmax = 1.10 cmax = 1.05
0.85 0.80 0
5
10
15
20
xT [%]
The Correction factor is KT7.5 %. Reduction of transformer impedance Increase of short-circuit currents Page 25
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Steffen Schmidt
Generator with direct Connection to Network Correction of Impedance
ZGK = ZG KG general KG =
Un c max ⋅ UrG 1 + x′d′ ⋅ sin ϕrG
for continuous operation above rated voltage: UrG (1+pG) instead of UrG turbine generator:
X(2) = X(1)
salient pole generator:
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E D SE PTI NC
Generator Block (Power Station) Correction of Impedance Q
ZS(O) = (tr2 ZG +ZTHV) KS(O)
G
power station with on-load tap changer: 2 2 UnQ UrTLV c max KS = 2 ⋅ 2 ⋅ UrG UrTHV 1 + x′d′ − x T ⋅ sin ϕrG
power station without on-load tap changers: UnQ U c max K SO = ⋅ rTLV ⋅ (1 ± p t ) ⋅ UrG (1 + pG ) UrTHV 1 + x′d′ ⋅ sin ϕrG
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Asynchronous Motors Motors contribute to the short circuit currents and have to be considered for calculation of maximum short circuit currents
2 1 UrM ZM = ⋅ ILR / IrM SrM
XM =
ZM 1 + (RM / XM )2
If R/X is unknown, the following values can be used: R/X = 0.1 medium voltage motors power per pole pair > 1 MW R/X = 0.15 medium voltage motors power per pole pair ≤ 1 MW R/X = 0.42 low voltage motors (including connection cables) Page 28
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Special Regulations for low Voltage Motors
low voltage motors can be neglected if ∑IrM ≤ Ik” groups of motors can be combined to a equivalent motor ILR/IrM = 5 can be used
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Calculation of initial short circuit current
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Calculation of initial short circuit current Procedure Set up equivalent circuit in symmetrical components Consider fault conditions in 3-phase system transformation into symmetrical components Calculation of fault currents in symmetrical components transformation into 3-phase system
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Calculation of initial short circuit current Equivalent circuit in symmetrical components
(1)
(0)
(1)
(1)
(1)
(2)
(2)
(2)
(2)
(1)
(1)
(1)
positive sequence system
(2)
negative sequence system
(1)
(2)
(2)
(2)
(0)
(0)
(0) (0)
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(0)
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(0)
(0)
Steffen Schmidt
zero sequence system
Calculation of initial short circuit current 3-phase short circuit L1-L2-L3-system
Z(1)l
L1 L2 L3 ~ ~ ~ -Uf
012-system
~
~
~
′′ = I sc3
c ⋅ Ur 3 ⋅ Z (1)
Z(1)r
c Un √3
(1)
Z(2)l
Z(2)r
~
~ (2)
Z(0)l
Z(0)r
~
~ (0)
UL1 =
network left of fault location
– Uf
U(1) = – Uf
UL2 = a2 (– Uf) UL3 = a Page 33
network right of fault location
U(2) = 0
(– Uf)
28.06.2008
fault location
U(0) = 0 Steffen Schmidt
Calculation of 2-phase initial short circuit current L1-L2-L3-system
Z(1)l
012-system
~
L1 L3
~
c ⋅U r ′′ = I sc2 Z ( 1) + Z ( 2 )
-Uf
′′ = I sc2
c ⋅U r 2 Z ( 1)
′′ I sc2 3 = ′′ I sc3 2 U = −c n 3
IL1 = 0
U (1) − U ( 2 )
IL2 = – IL3
I(0) = 0
UL3 – UL2 = – Uf
I(1) = – I(2)
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28.06.2008
~
~
L2
Steffen Schmidt
Z(1)r
c Un
(1)
√3
Z(2)l
Z(2)r
~
~ (2)
Z(0)l
Z(0)r
~
~ (0)
network left of fault location
fault location
network right of fault location
Calculation of 2-phase initial short circuit current with ground connection L1-L2-L3-system
Z(1)l
012-system
~
L1
~
~
L2 L3
~
′′ I scE2E
-Uf
Z(1)r
3⋅ c ⋅ U r = Z ( 1) + 2 Z ( 0 )
c Un
(1)
√3
Z(2)l
Z(2)r
~
~ (2)
Z(0)l
Z(0)r
~
~ (0)
I L1 = 0 2
U L2 = − a c U L3 = − a c Page 35
network left of fault location
Un 3
U (1) − U ( 2) = − c
Un 3
28.06.2008
fault location
Un 3
network right of fault location
= U (1) − U ( 0)
I(0) = I(1) = I(2) Steffen Schmidt
Calculation of 1-phase initial short circuit current L1-L2-L3-System
Z(1)l
012-System
Z(1)r
~
~ (1)
L1 L2 L3
~
" I sc1 =
-Uf
3⋅ c ⋅ U r Z (1) + Z ( 2 ) + Z ( 0 )
Z(2)l
Z(2)r
~
~ c Un √3
(2)
~
Z(0)l
Z(0)r
~
~ (0)
U L1 = − c
network left of fault location
Un 3
U ( 0) + U (1) + U ( 2) = − c
IL2 = 0
network right of fault location
Un 3
I(0) = I(1) = I(2)
IL3 = 0 Page 36
fault location
28.06.2008
Steffen Schmidt
Largest initial short circuit current Because of Z1 ≅ Z2 the largest short circuit current can be observed for Z1 / Z0 < 1 3-phase short circuit for Z1 / Z0 > 1 2-phase short circuit with earth connection (current in earth connection)
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Steffen Schmidt
Feeding of short circuits single fed short circuit
S"kQ
I sc"
k3
ür:1
UnQ
multiple fed short circuit G 3~ M 3~ I“scG
I“scN
I sc" =
I“scM
" ∑ I sc_part ≅ ∑ I sc_part "
Fault Page 38
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Steffen Schmidt
Calculation of short circuit currents by programs (1/3) Basic equation i=Yu Y: matrix of admittances (for short circuit) 0 Y 11 0 Y 21 . . . . . . '' = I sci Y i1 . . . . . . 0 Y n1
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. .
. .
. .
.
.
.
.
.
.
. Y 1n . Y 2n . . . . Y in . . . . Y nn
U1 U 2 . . . Ur − c ⋅ 3 . . . U n
Steffen Schmidt
Calculation of short circuit currents by programs (2/3) Inversion of matrix of admittances u = Y-1 i U1 U 2 . . . Ur = − c ⋅ 3 . . . U n
Page 40
Z 11 Z 21 . . . Z i1 . . . Z n1
28.06.2008
. .
. .
. .
. .
.
Z ii
.
.
.
.
.
.
Z 1n Z 2n . . . Z in . . . Z nn
Steffen Schmidt
0 0 . . . '' I sci . . . 0 Copyright © Siemens AG 2008. All rights reserved. E D SE PTI NC
Calculation of short circuit currents by programs (3/3)
from line i: − c Ur " = Z ii ⋅ I sci 3
⇒I " = − c U r sci 3 ⋅ Z ii
from the remaining lines: "
U sc = Z sci ⋅ I sci
calculation of all node voltages from there -> calculation of all short circuit currents
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Steffen Schmidt
Short Circuit Calculation Results Faults at all Buses
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Steffen Schmidt
Short Circuit Calculation Results Contribution for one Fault Location
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Steffen Schmidt
Example
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Steffen Schmidt
Data of sample calculation
Network feeder:
Transformer:
110 kV 3 GVA R/X = 0.1
110 / 20 kV 40 MVA uk = 15 % PkrT = 100 kVA
20 kV 10 km R1’ = 0.3 Ω / km X1’ = 0.4 Ω / km
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Impedance of Network feeder
c ⋅ Un2 ZI = Sk" 1.1⋅ ( 20 kV ) ZI = 3 GVA
2
ZI = 0.1467 Ω
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RI = 0.0146 Ω
Steffen Schmidt
XI = 0.1460 Ω
Impedance of Transformer
Un2 Z T = uk ⋅ Sn Z T = 0.15 ⋅
( 20 kV ) 2 40 MVA
Z T = 1.5000 Ω
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Un2 R T = PkrT ⋅ 2 Sn
( 20 kV ) 2 R T = 100 kVA ⋅ ( 40 MVA ) 2 R T = 0.0250 Ω
Steffen Schmidt
X T = 1.4998 Ω
Impedance of Transformer Correction Factor
K T = 0.95 ⋅
c max 1 + 0.6 ⋅ x T
K T = 0.95 ⋅
1 .1 1 + 0.6 ⋅ 0.14998
K T = 0.95873 Z TK = 1.4381 Ω
Page 48
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R TK = 0.0240 Ω
Steffen Schmidt
X TK = 1.4379 Ω
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RL = R'⋅
XL = X'⋅
RL = 0.3 Ω / km ⋅ 10 km
XL = 0.4 Ω / km ⋅ 10 km
RL = 3.0000 Ω
XI = 4.0000 Ω
Steffen Schmidt
Initial Short-Circuit Current – Fault location 1
R = RI + R TK
X = XI + X TK
R = 0.0146 Ω + 0.0240 Ω
X = 0.1460 Ω + 1.4379 Ω
R = 0.0386 Ω
X = 1.5839 Ω Ik" = Ik" =
c ⋅ Un 3 ⋅ ( R1 + j ⋅ X1 ) 1.1⋅ 20 kV 3⋅
( 0.0386 Ω ) 2 + (1.5839 Ω ) 2
Ik" = 8.0 kA Page 50
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Steffen Schmidt
Initial Short-Circuit Current – Fault location 2
R = RI + R TK + RL
X = XI + X TK + XL
R = 0.0146 Ω + 0.0240 Ω + 3.0000 Ω
X = 0.1460 Ω + 1.4379 Ω + 4.0000 Ω
R = 3.0386 Ω
X = 5.5839 Ω Ik" = Ik" =
c ⋅ Un 3 ⋅ ( R1 + j ⋅ X1 ) 1.1⋅ 20 kV 3⋅
( 3.0386 Ω ) 2 + ( 5.5839 Ω) 2
Ik" = 2.0 kA Page 51
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Steffen Schmidt
Calculation of Peak Current
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Steffen Schmidt
Peak Short-Circuit Current Calculation acc. IEC 60909 maximum possible instantaneous value of expected short circuit current equation for calculation: ip = κ ⋅ 2 ⋅ Ik" κ = 1.02 + 0.98 ⋅ e −3R / X
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Steffen Schmidt
Peak Short-Circuit Current Calculation in non-meshed Networks The peak short-circuit current ip at a short-circuit location, fed from sources which are not meshed with one another is the sum of the partial short-circuit currents: M G
M
ip1
ip2
ip3
ip4
i p = ip1 + ip2 + ip3 + ip4 Page 54
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Steffen Schmidt
Peak Short-Circuit Current Calculation in meshed Networks Method A: uniform ratio R/X smallest value of all network branches quite inexact Method B: ratio R/X at the fault location factor κb from relation R/X at the fault location (equation or diagram) κ =1,15 κb Method C: procedure with substitute frequency factor κ from relation Rc/Xc with substitute frequency fc = 20 Hz R R c fc = ⋅ X Xc f best results for meshed networks Page 55
28.06.2008
Steffen Schmidt
Peak Short-Circuit Current Fictitious Resistance of Generator
RGf = 0,05 Xd" for generators with UrG > 1 kV and SrG ≥ 100 MVA RGf = 0,07 Xd" for generators with UrG > 1 kV and SrG < 100 MVA RGf = 0,15 Xd" for generators with UrG ≤ 1000 V
NOTE: Only for calculation of peak short circuit current
Page 56
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Steffen Schmidt
Peak Short-Circuit Current – Fault location 1
Ik" = 8.0 kA R = 0.0386 Ω
X = 1.5839 Ω
R / X = 0.0244
κ = 1.02 + 0.98 ⋅ e −3R / X κ = 1.93 ip = κ ⋅ 2 ⋅ Ik" ip = 21.8 kA
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Steffen Schmidt
Peak Short-Circuit Current – Fault location 2
Ik" = 2.0 kA R = 3.0386 Ω
X = 5.5839 Ω
R / X = 0.5442
κ = 1.02 + 0.98 ⋅ e −3R / X κ = 1.21 ip = κ ⋅ 2 ⋅ Ik" ip = 3.4 kA
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Steffen Schmidt
Calculation of Breaking Current
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Steffen Schmidt
Breaking Current Differentiation Differentiation between short circuits ”near“ or “far“ from generator Definition short circuit ”near“ to generator for at least one synchronous machine is: Ik” > 2 ∙ Ir,Generator or Ik”with motor > 1.05 ∙ Ik”without motor Breaking current Ib for short circuit “far“ from generator Ib = Ik”
Page 60
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Steffen Schmidt
Breaking Current Calculation in non-meshed Networks The breaking current IB at a short-circuit location, fed from sources which are not meshed is the sum of the partial short-circuit currents:
M G
M
IB1 = μ∙I“ k
IB2 = I“k
IB3 = μ∙q∙I“ k
I B4 = μ∙q∙I“ k
I B = IB1 + I B2 + IB3 + IB4 Page 61
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Steffen Schmidt
Breaking current Decay of Current fed from Generators IB = μ ∙ I“k Factor μ to consider the decay of short circuit current fed from generators.
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Steffen Schmidt
Breaking current Decay of Current fed from Asynchronous Motors IB = μ ∙ q ∙ I“k Factor q to consider the decay of short circuit current fed from asynchronous motors.
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Steffen Schmidt
Breaking Current Calculation in meshed Networks Simplified calculation: Ib = Ik” For increased accuracy can be used: ∆U"Mj ∆U"Gi " " Ib = I − ∑ ⋅ (1 − µi ) ⋅ IkGi − ∑ ⋅ (1 − µ jq j ) ⋅ IkMj i c ⋅ Un / 3 j c ⋅ Un / 3 " k
"
"
∆UGi = jX "diK ⋅ IkGi
X“diK X“Mj I“kGi , I“kMj
"
"
" ∆UMj = jXMj ⋅ IkMj
subtransient reactance of the synchronous machine (i) reactance of the asynchronous motors (j) contribution to initial symmetrical short-circuit current from the synchronous machines (i) and the asynchronous motors (j) as measured at the machine terminals
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Steffen Schmidt
Continuous short circuit current
Continuous short circuit current Ik r.m.s. value of short circuit current after decay of all transient effects depending on type and excitation of generators statement in standard only for single fed short circuit calculation by factors (similar to breaking current) Continuous short circuit current is normally not calculated by network calculation programs. For short circuits far from generator and as worst case estimation Ik = I”k
Page 65
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Steffen Schmidt
Page 66
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Steffen Schmidt
A
Load flow calculation that considers all network parameters, such as loads, tap positions, etc.
B
Place voltage source with the voltage that was determined by the load flow calculation at the fault location.
C
Superposition of A and B
Page 67
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Steffen Schmidt
B Short circuit calculation
Page 68
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Steffen Schmidt
40A 3 Ω
40A
Superposition: Load flow + feed back 2Ω
10A 2Ω 50A
10A
1000V
50. A 153.95A 203.95A
720V 900V ~
90 Ω
780V
700V
14 Ω
1000V -0V 1000V
~
40. A 157.37A 197.37A
40A 208A 168A
900. V -307.89V 592.11V
~
10A 182A 192A 720V -0V 720V
700V -364V 336V
~
365.37A
Short-circuit: feed back 153.95A 0V
3.42A
203.95A
182A
365.3A 157.37A
307.89V
208.0A
780V
26A
0V
197.37A 6.58A
1000V
592.11V
364V
168A
192.0A 24A 720V
336V ~
~ 365.37A
Page 69
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Steffen Schmidt
Break time!
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Steffen Schmidt
Contact Steffen Schmidt Senior Consultant Siemens AG, Energy Sector E D SE PTI NC Freyeslebenstr. 1 91058 Erlangen Phone: +49 9131 - 7 32764 Fax: +49 9131 - 7 32525 E-mail: [email protected]
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Steffen Schmidt | 7,866 | 22,516 | {"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-2022-27 | latest | en | 0.713839 |
https://users.math.yale.edu/public_html/People/frame/Fractals/FlT/PerCount/PerCount.html | 1,659,988,922,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00186.warc.gz | 542,657,060 | 1,510 | # Something dynamics can tell us about number theory
## Count the periodic points
Recall the fixed points of a function occur at the intersection of the graph of the function and the diagonal.
Consequently, each funcion fn has n fixed points. For example,
f2 f3 f4
It is not hard to see that the graph of fnm consists of line segments having slope nm. Here is a simple, sample calculation, illuminated by this graph.
f22(x) = f2(f2(x)) =
2⋅f2(x) for 0 ≤ f2(x) < 1/2 2⋅f2(x) - 1 for 1/2 ≤ f2(x) ≤ 1
=
2⋅f2(x) for 0 ≤ x < 1/4 and 1/2 ≤ x < 3/4 2⋅f2(x) - 1 for 1/4 ≤ x < 1/2 and 3/4 ≤ x ≤ 1
=
2⋅2x = 4x for 0 ≤ x < 1/4 2⋅2x - 1 = 4x - 1 for 1/4 ≤ x < 1/2 2⋅(2x - 1) 4x - 2 for 1/2 ≤ x < 3/4 2⋅(2x - 1)f - 1 = 4x - 3 for 3/4 ≤ x ≤ 1
Because the segments of the graph of fnm consists of segments of slope nm, and the left-most starts at (0, 0), the graph of fnm consists of nm straight line segments, each going from x = 0 to x = 1.
Consequently, fnm has nm fixed points. | 397 | 972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-33 | latest | en | 0.91254 |
http://stackoverflow.com/questions/4915046/math-question-regarding-a-fantasy-sports-snake-draft | 1,395,027,357,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678704624/warc/CC-MAIN-20140313024504-00064-ip-10-183-142-35.ec2.internal.warc.gz | 133,607,891 | 15,369 | # Math question regarding a Fantasy Sports (snake) draft
If you're familiar with any fantasy sports draft, the draft-order grid looks something like this:
``````EXAMPLE 1 (3-teams):
Round Team 1 Team 2 Team 3
1 1 (1.1) 2 (1.2) 3 (1.3)
2 6 (2.3) 5 (2.2) 4 (2.1)
3 7 (3.1) 8 (3.2) 9 (3.3)
``````
The numbers 1-9 represent the `overall pick number` of the draft.
The items in parentheses represent the `round_number` and `pick_number_of_that_round`.
I cannot figure out a formula which converts my `overall_pick_number` into it's proper `pick_number_of_that_round`.
In the above example, the number 8 equals 2 (the 2nd pick of the 3rd round). But in a 4-team league, the number 8 equals 4 (the 4th pick of the 2nd round).
``````EXAMPLE 2 (4-teams):
Round Team 1 Team 2 Team 3 Team 4
1 1 (1.1) 2 (1.2) 3 (1.3) 4 (1.4)
2 8 (2.4) 7 (2.3) 6 (2.2) 5 (2.1)
3 9 (3.1) 10 (3.2) 11 (3.3) 12 (3.4)
``````
I thought about trying to dynamically build an associative array based on the number of teams in the league containing every pick and which pick it belonged to, but it's just beyond me.
-
``````round_number = ((overall-1) / number_of_teams) + 1
pick_number_of_round = ((overall-1) % number_of_teams) + 1
``````
-
I had to cast the `round_number` to `(int)` to achieve my desired result. Many thanks for taking the time to help. – Jeff Feb 6 '11 at 18:58
``````round_number = ((overall_pick_number - 1) / number_of_teams) + 1 | 525 | 1,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2014-10 | latest | en | 0.914807 |
https://math.stackexchange.com/questions/1956433/why-is-the-exponential-function-not-in-the-subspace-of-all-polynomials | 1,722,761,160,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00722.warc.gz | 302,994,083 | 40,927 | # Why is the exponential function not in the subspace of all polynomials?
The exponential function can be written as
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots.$$
The subspace of all polynomials is $$\text{span}\{1, x,x^2, \dots \}$$
Sure $e^x$ is in this set?
• Any function in the subspace must be a linear combination of a finite number of basis elements. Commented Oct 6, 2016 at 11:53
• What Parcly wrote, and adding: what is true is that $\;e^x\;$ is in the closure (with respect to uniform convergence) of the span of the polynomials (in fact, any continuous function in a closed bounded interval) is. This is the famous Weierstrass approximation theorem. Commented Oct 6, 2016 at 11:56
• @DonAntonio The restriction of $e^x$ to a closed bounded interval is in the closure of $\mathbb{R}[x]$, but not $e^x$ itself. An experienced reader can deduce that from your comment in parentheses, but I'm not sure a beginner would get the correct idea from your comment. Commented Oct 6, 2016 at 13:43
• With your definition, for any $n$, i'ts not hard to deduce $\lim_{x\rightarrow\infty }{e^x\over x^n}=\infty$. This is not true for any given polynomial. Commented Oct 7, 2016 at 9:26
• Similarly, $\sqrt2 = 1 + 1/(2 + 1/(2 + 1/(2 +\cdots$, but we don't expect to find $\sqrt2$ in the rationals. Commented Oct 7, 2016 at 13:13
If $p$ is a polynomial of degree $n$, then the $n$th derivative of $p$ is constant. Note that the $n$th derivative of $e^x$ is $e^x$. Now all you have to do is prove that $e^x$ is not constant.
• I find it more natural to consider the $n+1$st, not only but also for avoiding to consider "the $0$th derivative" which might confuse somebody.
– quid
Commented Oct 6, 2016 at 14:30
• I think this is what the OP wanted, whereas the accepted answer, albeit true, doesn't really address the reason why $e^x$ isn't a polynomial.
– Vim
Commented Oct 6, 2016 at 14:52
• @Vim if you consider what a finite linear combination of the basis vectors looks like, then this answer is equivalent to the accepted one. Commented Oct 6, 2016 at 20:25
• (+1) In the same vein: the only polynomial equal to its own derivative is 0. Honestly, I don't think "this answer is equivalent to the accepted one". Realizing this doesn't require much "reading competence". (+ linear combinations are " finite" by definition) Commented Feb 15, 2023 at 5:23
The function $e^x$ is not in $\text{span}\{1, x,x^2, \dots \}$ because it is no finite linear combination of basis elements (but a countable one). What is true is that $$e^x \in \overline{\text{span}\{1, x,x^2, \dots \}}$$ is in the closure because you can find a sequence in $\text{span}\{1, x,x^2, \dots \}$ which converges to $e^x$. I hope it helps you :)
• What you mention about the closure is a bit unclear. You need to put a topology on the space of continuous functions to even speak about "closure" and "convergence". For example the restriction of $\exp$ to $[0,1]$ is in the closure of $\mathbb{R}[x]$ for the uniform convergence topology, as also mentioned by DonAntonio, but you really have to restrict. If you want you can also say that the sequence of polynomials converges pointwise, but that's not quite satisfactory either. Commented Oct 6, 2016 at 13:45
• @NajibIdrissi You are right. Thanks for pointing that out, I just tried to keep it brief. The closure I meant is respecting the space $(C(K), \Vert \cdot \Vert_\infty)$ for some compact $K \subset \mathbb R$ and that works just fine! Commented Oct 6, 2016 at 14:09
• or closure in formal power series ... :) Commented Oct 6, 2016 at 22:20
• Anyway, " it is no finite linear combination of basis elements" is a rewording of the statement, not a proof of it. Commented Feb 14, 2023 at 8:23
• @AnneBauval Awkward to answer a comment of a 7 year old question. Anywas, this question was about why $\mathrm e^x$ is not contained in the span of the monomials and that was just an explanation. I never claimed that my explanation was a proof, to begin with. Reading competence is also a valuable skill. Commented Feb 15, 2023 at 2:01
$\mathrm{span}(A)$ is the set of finite linear combinations of terms from $A$. Infinite sums require notions of limits and bring up issues of convergence radii (there are plenty of infinite polynomial that converge only at a single point). | 1,249 | 4,313 | {"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 | 4 | CC-MAIN-2024-33 | latest | en | 0.923489 |
http://tenminutetutor.com/generative-art/generativepy/tick/ | 1,611,491,982,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00159.warc.gz | 93,010,458 | 3,607 | # Tick
Martin McBride, 2020-08-23
Tags geometry tick
Categories generativepy generative art
The tick function adds a small cross line marker to a line, of the type often used in geometric diagrams to show that two lines are the same length. They look like this:
You can add 1, 2 or 3 ticks.
## tick function
tick(ctx, a, b, count=1, length=4, gap=1)
Parameter Type Description
ctx Context The Pycairo Context to draw to
a (number, number) Tuple (x, y) for point a
b (number, number) Tuple (x, y) for point b
count int Number of ticks on the line, 1, 2 or 3.
length number length of the tick in user units.
gap number Gap between ticks in user units.
Draws an tick across the line ab. The tick is half way between points a and b.
## Example
See the example in the angle_marker article. | 215 | 793 | {"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} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.777992 |
https://www.asvabtestbank.com/study-guide/mos/0511/401 | 1,627,101,571,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00224.warc.gz | 654,626,029 | 8,780 | # Marine Corps 0511 MAGTF Planning Specialist ASVAB Study Guide
Branch Marine Corps MOS 0511 Title MAGTF Planning Specialist Description The enlisted MAGTF planning specialist is responsible for functional support in the areas of FDP&E. Typical duties include updating plan and unit information to unit level detail for force deployment planning; operating and managing force deployment planning automated data processing tools; producing force reports; and properly formatting and forwarding electronic mail, files, and newsgroup message traffic.Access networked information; Utilize newsgroups; Utilize formal message application; Operate the Joint Forces Requirements Generator; Operate the JOPES Editing Tool (JET) ; Operate Web Scheduling and Movement; Generate reports. Subtests Arithmetic Reasoning, Paragraph Comprehension, Word Knowledge
# Arithmetic Reasoning
• 13 Questions
• 54 Problems
• 36 Flash Cards
### Fundamentals
##### Number Properties
###### Integers
An integer is any whole number, including zero. An integer can be either positive or negative. Examples include -77, -1, 0, 55, 119.
###### Rational Numbers
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form $${a \over b}$$ where a is the numerator and b is the denominator. An improper fraction ($${5 \over 3}$$) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number ($$1 {2 \over 3}$$) which has a whole number part and a fractional part.
###### Absolute Value
The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: $$\left|-5\right| = 5$$. In the case of a variable absolute value ($$\left|a\right| = 5$$) the value of a can be either positive or negative (a = -5 or a = 5).
###### Factors & Multiples
A factor is a positive integer that divides evenly into a given number. The factors of 8 are 1, 2, 4, and 8. A multiple is a number that is the product of that number and an integer. The multiples of 8 are 0, 8, 16, 24, ...
###### Greatest Common Factor
The greatest common factor (GCF) is the greatest factor that divides two integers.
###### Least Common Multiple
The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.
###### Prime Number
A prime number is an integer greater than 1 that has no factors other than 1 and itself. Examples of prime numbers include 2, 3, 5, 7, and 11.
##### Operations on Fractions
###### Simplifying Fractions
Fractions are generally presented with the numerator and denominator as small as is possible meaning there is no number, except one, that can be divided evenly into both the numerator and the denominator. To reduce a fraction to lowest terms, divide the numerator and denominator by their greatest common factor (GCF).
Fractions must share a common denominator in order to be added or subtracted. The common denominator is the least common multiple of all the denominators.
###### Multiplying & Dividing Fractions
To multiply fractions, multiply the numerators together and then multiply the denominators together. To divide fractions, invert the second fraction (get the reciprocal) and multiply it by the first.
##### Operations on Exponents
###### Defining Exponents
An exponent (cbe) consists of coefficient (c) and a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1).
To add or subtract terms with exponents, both the base and the exponent must be the same. If the base and the exponent are the same, add or subtract the coefficients and retain the base and exponent. For example, 3x2 + 2x2 = 5x2 and 3x2 - 2x2 = x2 but x2 + x4 and x4 - x2 cannot be combined.
###### Multiplying & Dividing Exponents
To multiply terms with the same base, multiply the coefficients and add the exponents. To divide terms with the same base, divide the coefficients and subtract the exponents. For example, 3x2 x 2x2 = 6x4 and $${8x^5 \over 4x^2}$$ = 2x(5-2) = 2x3.
###### Exponent to a Power
To raise a term with an exponent to another exponent, retain the base and multiply the exponents: (x2)3 = x(2x3) = x6
###### Negative Exponent
A negative exponent indicates the number of times that the base is divided by itself. To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal: $$b^{-e} = { 1 \over b^e }$$. For example, $$3^{-2} = {1 \over 3^2} = {1 \over 9}$$
Radicals (or roots) are the opposite operation of applying exponents. With exponents, you're multiplying a base by itself some number of times while with roots you're dividing the base by itself some number of times. A radical term looks like $$\sqrt[d]{r}$$ and consists of a radicand (r) and a degree (d). The degree is the number of times the radicand is divided by itself. If no degree is specified, the degree defaults to 2 (a square root).
The radicand of a simplified radical has no perfect square factors. A perfect square is the product of a number multiplied by itself (squared). To simplify a radical, factor out the perfect squares by recognizing that $$\sqrt{a^2} = a$$. For example, $$\sqrt{64} = \sqrt{16 \times 4} = \sqrt{4^2 \times 2^2} = 4 \times 2 = 8$$.
To add or subtract radicals, the degree and radicand must be the same. For example, $$2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$$ but $$2\sqrt{2} + 2\sqrt{3}$$ cannot be added because they have different radicands.
To multiply or divide radicals, multiply or divide the coefficients and radicands separately: $$x\sqrt{a} \times y\sqrt{b} = xy\sqrt{ab}$$ and $${x\sqrt{a} \over y\sqrt{b}} = {x \over y}\sqrt{a \over b}$$
###### Square Root of a Fraction
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately. For example, $$\sqrt{9 \over 16}$$ = $${\sqrt{9}} \over {\sqrt{16}}$$ = $${3 \over 4}$$
##### Miscellaneous
###### Scientific Notation
Scientific notation is a method of writing very small or very large numbers. The first part will be a number between one and ten (typically a decimal) and the second part will be a power of 10. For example, 98,760 in scientific notation is 9.876 x 104 with the 4 indicating the number of places the decimal point was moved to the left. A power of 10 with a negative exponent indicates that the decimal point was moved to the right. For example, 0.0123 in scientific notation is 1.23 x 10-2.
###### Factorials
A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.
### Applications
##### Order of Operations
###### PEMDAS
Arithmetic operations must be performed in the following specific order:
1. Parentheses
2. Exponents
3. Multiplication and Division (from L to R)
4. Addition and Subtraction (from L to R)
The acronym PEMDAS can help remind you of the order.
###### Distributive Property - Multiplication
The distributive property for multiplication helps in solving expressions like a(b + c). It specifies that the result of multiplying one number by the sum or difference of two numbers can be obtained by multiplying each number individually and then totaling the results: a(b + c) = ab + ac. For example, 4(10-5) = (4 x 10) - (4 x 5) = 40 - 20 = 20.
###### Distributive Property - Division
The distributive property for division helps in solving expressions like $${b + c \over a}$$. It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: $${b + c \over a} = {b \over a} + {c \over a}$$. For example, $${a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6$$.
###### Commutative Property
The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.
##### Ratios
###### Ratios
Ratios relate one quantity to another and are presented using a colon or as a fraction. For example, 2:3 or $${2 \over 3}$$ would be the ratio of red to green marbles if a jar contained two red marbles for every three green marbles.
###### Proportions
A proportion is a statement that two ratios are equal: a:b = c:d, $${a \over b} = {c \over d}$$. To solve proportions with a variable term, cross-multiply: $${a \over 8} = {3 \over 6}$$, 6a = 24, a = 4.
###### Rates
A rate is a ratio that compares two related quantities. Common rates are speed = $${distance \over time}$$, flow = $${amount \over time}$$, and defect = $${errors \over units}$$.
###### Percentages
Percentages are ratios of an amount compared to 100. The percent change of an old to new value is equal to 100% x $${ new - old \over old }$$.
##### Statistics
###### Averages
The average (or mean) of a group of terms is the sum of the terms divided by the number of terms. Average = $${a_1 + a_2 + ... + a_n \over n}$$
###### Sequence
A sequence is a group of ordered numbers. An arithmetic sequence is a sequence in which each successive number is equal to the number before it plus some constant number.
###### Probability
Probability is the numerical likelihood that a specific outcome will occur. Probability = $${ \text{outcomes of interest} \over \text{possible outcomes}}$$. To find the probability that two events will occur, find the probability of each and multiply them together.
##### Word Problems
###### Practice
Many of the arithmetic reasoning problems on the ASVAB will be in the form of word problems that will test not only the concepts in this study guide but those in Math Knowledge as well. Practice these word problems to get comfortable with translating the text into math equations and then solving those equations. | 2,561 | 10,104 | {"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} | 4.34375 | 4 | CC-MAIN-2021-31 | longest | en | 0.80791 |
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Prepare for the LSAT or discuss it with others in this forum.
confusedlawyer
Posts: 136
Joined: Thu May 27, 2010 11:21 am
### Prep Test 40 LG #2
How do you guys tackle In/out logic games? I always find myself wasting time doing hypotheticals
mz253
Posts: 319
Joined: Thu Mar 25, 2010 11:18 pm
### Re: Prep Test 40 LG #2
i usually draw a long chain, linking everything together. guaranteed to work. can solve everything less than 8 minutes.
Shrimps
Posts: 269
Joined: Sun Feb 14, 2010 10:04 pm
### Re: Prep Test 40 LG #2
Are we on the same page? PT40 Game 2? Cold medications? There are two hypos only, really: one starts with FL, the other with GL. The second hypo must have I, so it's GL..I.. It cannot have H (H before G if both), and it cannot have M, which would require the presence of both F and H. Thus, the only two medications that are left are K and F. Thus, if G is the top ranked medication, we know all five: G, L, (I, K..F).
The first hypo (FL..I..) can have M, H, and G two fill the other two spots.
Notor
Posts: 391
Joined: Sat May 23, 2009 11:32 am
### Re: Prep Test 40 LG #2
So glad I don't have to do any of this anymore
confusedlawyer
Posts: 136
Joined: Thu May 27, 2010 11:21 am
### Re: Prep Test 40 LG #2
Shrimps wrote:Are we on the same page? PT40 Game 2? Cold medications? There are two hypos only, really: one starts with FL, the other with GL. The second hypo must have I, so it's GL..I.. It cannot have H (H before G if both), and it cannot have M, which would require the presence of both F and H. Thus, the only two medications that are left are K and F. Thus, if G is the top ranked medication, we know all five: G, L, (I, K..F).
The first hypo (FL..I..) can have M, H, and G two fill the other two spots.
I understand where I went wrong I think, I neglected to write out the contrapositive (if no F no M. If no H no M). That could have saved me a lot of time even though I only got one wrong (Which one must be included-->F). Thanks for the reply
confusedlawyer
Posts: 136
Joined: Thu May 27, 2010 11:21 am
### Re: Prep Test 40 LG #2
Notor wrote:So glad I don't have to do any of this anymore
I envy you Eazy E
confusedlawyer
Posts: 136
Joined: Thu May 27, 2010 11:21 am
### Re: Prep Test 40 LG #2
mz253 wrote:i usually draw a long chain, linking everything together. guaranteed to work. can solve everything less than 8 minutes.
Can you elaborate please?
mz253
Posts: 319
Joined: Thu Mar 25, 2010 11:18 pm
### Re: Prep Test 40 LG #2
well, i create a diagram as expansive as possible... try to put every condition in one or two chains (if possible), something like a network. i wouldn't write contrapositive. i feel it's more confusing trying to find some elements in numerous if-then conditions. so when i do the question, i just refer to the "network", trying to trace the chain. sometimes go to the right direction, sometimes go to the left (contrapositive), is it clear?
well, i guess all i wanted to say is try to create a simple set-up in case you will confuse yourself. because i find under time pressure, it's easy to confuse myself if i have a very complicated set up.
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# Math Background and Competitiveness
Author Message
Intern
Joined: 27 Dec 2011
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27 Dec 2011, 15:51
Currently, I have taken through Calculus 1. I am considering Calculus 2, but I don't know how important Math is to top MBA programs. I'm more concerned that it could be potentially hurtful to my GPA since Calculus 2 seems to be the hardest in the Calculus sequence. However, if Math can increase how desirable I am to MBA programs, I am willing to take the risk. Does anyone have any insight into the Math background of those at top MBA schools and how favorably they look to people with backgrounds in Math?
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Re: Math Background and Competitiveness [#permalink]
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28 Dec 2011, 03:30
Taking Calculus 2, will make about 0 difference. It's all the other stuff that counts a whole lot more. So preserve your GPA if you have any doubts. Unless you want to challenge yourself!
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Jon Frank
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Re: Math Background and Competitiveness [#permalink]
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07 Jan 2012, 10:56
Plus, taking Calculus 2 will hardly be considered "having a background in math." The people who truly have backgrounds in math are those who studied astrophysics or engineering, won national math competitions, taught themselves calculus at the age of 14, etc. (for each of these things, I know someone applying to business school who has accomplished it).
As precision essay said, focus on other aspects of your profile, making sure to demonstrate that your quantitative background is "good enough", i.e. good GMAT quant score, some reasonable quant coursework like stats, etc.
Also, you sound like you're still an undergrad. Why are you thinking about business school now? Have you ever worked in a corporate environment full time?
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Re: Math Background and Competitiveness [#permalink]
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10 Jan 2012, 11:31
I'm still an undergraduate, but I want to apply to the programs at Harvard and Yale, so I'm planning my application.
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Re: Math Background and Competitiveness [#permalink]
### Show Tags
11 Jan 2012, 23:41
Ya calculus 2 is okay, Its good enough in MBA,,
Re: Math Background and Competitiveness [#permalink] 11 Jan 2012, 23:41
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spanish 2
to dance = bailar Then write out the verb tense chart with the '-ar' endings. After that chop off the 'ar' from bailar and add the verb endings.
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Lol, I'm taking the same class ;)
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There are two situations in which it possible for a charged particle to be in a magnetic field but not experiencing a magnetic force. What are they? Given that Fm=qVBsintheta, I am going to say 1. If the particle is NOT in motion. 2. The direction of currect and the direction ...
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FEATURES | 376 | 1,369 | {"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-2013-20 | longest | en | 0.901418 |
https://locke-movie.com/2022/08/20/how-do-you-bet-on-college-basketball/ | 1,670,356,982,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00041.warc.gz | 409,184,606 | 10,358 | # How do you bet on college basketball?
## How do you bet on college basketball?
How do you bet on college basketball? A: You can bet on college basketball by picking the winner of the game (moneyline), which team will cover the spread (point spread), or combined points scored in the match (Over/Under).
## How do odds work in NCAA basketball?
The NCAAB betting sites set a line of how many points they think a team will win by (which automatically is how many points they expect the other team to lose by). If the team expected to win does so by more points than they were expected to, that side of the bet wins.
The minus sign (e.g.-130) always indicates the favorite and the amount you must bet to win \$100. The plus sign (e.g.+120) always indicates the underdog and the amount you win for every \$100 bet.
Do college basketball teams score more in the first or second half?
College basketball totals are usually made on the assumption that there will be 10 more points scored in the second half. A game with a total of 140 will see a first half total closer to 65 than it will closer to 70. There are several reasons that bettors consider first and second half bets.
### Why are NCAA scores so low?
Points per game statistics via NCAA.com. Since there is no way to mandate coaches sending more players to the glass, the obvious solution to the pace-of-play problem is to reduce the shot clock. When the NCAA reduced the shot clock from 45 seconds to 35 seconds in 1994, scoring went from 73.6 points per game to 75.0.
### What is the average NCAA score?
The average total score was 145.4 We rounded to 145. The average individual score, then, was 72.5, which we rounded up to 73. The average margin was 9.1 which we rounded down to 9. So, accounting for those numbers, the best prediction for the score of the NCAA Tournament men’s championship game is 77-68.
What’s the most common college basketball score?
The average total score was 143.6, meaning the average individual score was 71.8 for both of the final teams. The average winning margin was 8.9 points.
What is the highest score in college basketball?
What was the highest hockey score ever?
• What is the highest score in a college basketball game?
• What was the highest team score in college basketball history?
• ## What is the average score for college basketball?
What is the average total score of a college basketball game? On average, NCAA Division I men’s teams manage 67.875 points in a game. NCAA Division I women’s teams score an average of 60.937 points each game. These figures are calculated by adding 16 teams’ point totals together on a particular night and dividing that number by 16.
## Who won Basketball Last Night?
Seeded second in the Division III tournament, they ultimately rolled to a 53-17 first-round victory over overmatched Belmont High at The Den Wednesday night.
Who will win NCAA Tonight?
© Robert Deutsch, USA TODAY Sports Final Four: Gonzaga guard Jalen Suggs celebrates with teammates after hitting a buzzer-beating 3-pointer in overtime that lifted the Bulldogs past No. 11 UCLA and into the men’s national championship game. Start the day smarter. Get all the news you need in your inbox each morning. | 719 | 3,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-49 | latest | en | 0.96734 |
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# help pls?
0
392
1
When the expression 3x^2-24x+55 written in the form a(x+d)^2+e, where a, d, and e are constants, then what is the sum a+d+e?
Dec 22, 2017
#1
+101760
+2
3x^2 -24x + 55 complete the square
3 (x^2 - 8x + 55/3)
3(x^2 - 8x + 16 + 55/3 - 16)
3 [ (x -4)^2 + 55/3 - 48/3 ]
3[ (x - 4 )^2 + 7/3)
3(x - 4)^2 + 7
So
a = 3
d = -4
e = 7
And their sum = 6
Dec 22, 2017 | 232 | 421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-30 | latest | en | 0.162472 |
https://thesororitysecrets.com/bake/frequent-question-can-you-cook-an-egg-on-the-sidewalk-in-arizona.html | 1,642,677,839,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301737.47/warc/CC-MAIN-20220120100127-20220120130127-00609.warc.gz | 606,222,908 | 18,924 | # Frequent question: Can you cook an egg on the sidewalk in Arizona?
Contents
Spoiler alert: The sidewalk cannot conduct enough heat to cook an egg, as concrete’s maximum temperature — according to the laws of thermodynamics — is 145°F, while the minimum heat required to really firm up the proteins in an egg is 158°F.
## Can you cook eggs on a hot sidewalk?
According to the Library of Congress, it’s possible, but not probable, that you could fry an egg on a sidewalk during a hot day. Eggs need to reach a temperature of 158*F to cook through. Sidewalks can usually get up to 145*F. … A concrete sidewalk isn’t the best conductor of heat.
## Can you cook an egg on a rock in Arizona?
Metal conducts heat better and gets hotter, so people actually have been able to cook an egg on a car hood’s surface. Still, the idea of cooking an egg on a sidewalk won’t die. … It turns out that eggs also have a bit of an advantage in Arizona, the land of low humidity and high heat.
IT IS IMPORTANT: Why do we need heat to boil water?
## How do you cook an egg on the sidewalk?
According to the American Egg Board, eggs need to be heated to a temperature between 144° F and 158° F (62.2° C to 70° C) to be cooked. That’s why sidewalks don’t make such great frying pans. Although sidewalks get plenty hot during a heat wave, they don’t usually get above 140° F (60° C).
## How long does it take to cook an egg on the sidewalk?
In the hottest part of the day, it takes the egg about 20 minutes to cook through. If you don’t live somewhere that hot, you could probably do it in a solar oven that concentrates the rays of the sun under glass.
## Could fry an egg on the sidewalk meaning?
That means they soak up the heat, but they don’t distribute it efficiently to other things, such as eggs. When you crack an egg on a hot sidewalk, the egg will cool the sidewalk slightly. Without another heat source present, the sidewalk is unlikely to cook the egg much, if at all.
## How hot does Pavement get in Phoenix?
“Because if you look at hot pavement or asphalt at two o’clock in the afternoon in direct sunlight, the temperature is usually somewhere around 170 to 180 degrees Fahrenheit.” Arizona Burn Center Valleywise Health entrance shown Monday in Phoenix.
## Can you cook an egg at 113 degrees?
“One hundred-ten degrees is not an acceptable temperature,” the spokesman said. “The only safe methods as used in our kitchen, whether frying on a skillet, boiling eggs or baking, is at temperatures of 325 degrees minimum.” The USDA recommends cooking eggs until firm and to a temperature of at least 160 degrees.
## Can you cook an egg at 90 degrees?
Therefore, the best temperature of letting the egg to set but not being overcooked is at the temperature just above 80°C/180°F and below 90°C/194°F.
## How hot does it have to be to cook an egg on a car?
In vehicles left in sunlight, the average cabin air temperature reached 116 degrees. The surface temperatures of the dashboard, seats, and steering wheel reached 157, 123, and 127 degrees, respectively. A temperature of 150 degrees is thought to be hot enough to cook an egg.
## What’s the best way to fry an egg?
butter in nonstick skillet over medium-high heat until hot. BREAK eggs and SLIP into pan, 1 at a time. IMMEDIATELY reduce heat to low. COOK SLOWLY until whites are completely set and yolks begin to thicken but are not hard. | 788 | 3,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.922445 |
http://www.openbookproject.net/pybiblio/gasp/course/L-loops.html | 1,558,459,861,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256494.24/warc/CC-MAIN-20190521162634-20190521184634-00148.warc.gz | 333,591,571 | 5,278 | # Loops¶
## Introduction¶
Computers are good at repetitive things, so we often want them to do something over and over again (perhaps with slight changes from one time around to the next):
• Add up all the numbers in some list
• Move all the Evil Alien Invaders one step closer to Earth
• Print out all the numbers from 1 to 100
and so on. This is called looping, for some reason.
Python has two different kinds of loop. This sheet tells you about them.
## Two kinds of loop¶
There’s an important difference between Python’s two kinds of loop. One (called a `for` loop) is used when you know in advance how many times you want to do whatever-it-is that the loop does. The other (called a `while` loop) is used when you don’t know in advance.
## `for` loops: When you know how many times¶
The `for` loop is called a `for` loop because the first thing you have to type when setting one up is the word `for`. The simplest form looks like this:
```for x in 1, 2, 3, 4, 5, 1000:
print('Here is a number: ')
print(x)
```
So, you need to give a variable name (`x`), a list of things (`1, 2, 3, 4, 5, 1000`), and some stuff to do once for each item in the list. The stuff will get done once with `x` naming the value 1, once with it naming the value 2, and so on.
The list doesn’t have to be written out like that. You can, for instance, say:
```my_list = [1, 2, 3, 4, 5, 1000]
for x in my_list:
print("Here's a number:")
print(x)
```
And, of course, `my_list` might actually get its value in some more complicated way: it might be the result of a lengthy calculation instead of being typed in directly.
### Ranges¶
Annoyingly, the commonest sort of loop is rather fiddly to do in Python. Often, you just want to do something 10 times (or 93 times, or whatever). You could say `for x in 1, 2, 3, 4, 5, 6, 7, 8, 9, 10:` but you’d quickly get bored of typing all that – and what if you wanted 1000 repetitions? Or if the number of repetitions might vary?
Fortunately, you can say this instead:
```for x in range(10):
blah blah blah
```
This will do `blah blah blah` 10 times. It may not do it in quite the way you’d expect, though. The sequence of numbers named by `x` isn’t 1, 2, 3, …, 10; it’s 0, 1, 2, …, 9. That’s still 10 numbers in all.
Incidentally, `range` isn’t only for using in `for` loops. You can use it elsewhere too:
```>>> print(list(range(5)))
[0, 1, 2, 3, 4]
```
But all `for` loops have the feature that, when the loop begins, the computer has to know what list it’s going through. What if you don’t know when to stop until after you’ve started?
## `while` loops: When you don’t know how many times¶
For this, Python has another kind of loop. It’s called `while` because that’s the first word you type when setting up this kind of loop:
```number = 1
while number < 1000:
print(number)
number = 2 * number
```
When the computer sees `while number < 1000:`, what it does is:
• See whether `number < 1000` is true or not.
• If it isn’t, abandon the loop: carry on with whatever comes after the end of the loop.
• If it is, do the stuff inside the loop…
• …and then go back to the first step, seeing whether ```number < 1000``` is true or false.
In other words, it does the stuff inside the loop over and over again, but only while the condition `number < 1000` is true.
## Leaving a loop early¶
Sometimes you want to leave a loop early . For instance, you might have a `for` loop adding up 100 numbers, but want to stop at once if any of the numbers is 0. (Why? We don’t know. It’s just an example.)
For this, you need the `break` statement. It means abandon whatever loop you’re in the middle of . So, for instance, to add up all the numbers in a list but stop if you ever hit 0:
```total = 0
for x in the_list:
if x == 0:
break
total = total + x
```
The `break` statement is particularly useful when you have a loop that’s like a `while` loop, but where the condition to be tested doesn’t actually come up at the start of the loop. For instance, suppose you want to add up lots of random numbers, and stop if any of the numbers is ever equal to 3. (Yes, this is a pointless example. There are plenty of less pointless examples, but they’re all longer and more complicated.) Here’s how you could do that.
```total = 0
while True:
r = random_between(1, 10)
if r == 3:
break
total = total + r
```
The only really weird thing here is the `True` in `while True:`, which means Do the following stuff for ever, until you hit a `break`.
Sometimes you want to abandon, not a whole loop, but just a single iteration of it: in other words, one trip around the loop. The `continue` statement does that. It’s a bit like `break` except that instead of leaping out of the loop it effectively goes back to the start of the loop and begins the next trip around it. If you were already on the last iteration of the loop, `continue` thus does the same as `break`.
The following strange-looking construction is sometimes useful.
```for n in range(10):
if a[n] == 'aardvark': break
else:
print 'No aardvark found!'
```
At first sight, it looks like the `else` here is at the wrong level of indentation. But actually the `else` doesn’t go with the `if`; it goes with the `for`. What it means is: Do the following stuff if the loop finished normally, and not by `break` being done.
If you’re confused by this, don’t worry about it - try creating two lists, one that contains `'aardvark'`, and one that doesn’t. Then try using them in that loop and see what happens. | 1,499 | 5,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-22 | longest | en | 0.907833 |
https://en.unionpedia.org/c/Three-twist_knot/vs/Twist_knot | 1,679,520,683,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00778.warc.gz | 253,482,624 | 10,070 | Faster access than browser!
# Three-twist knot and Twist knot
## Difference between Three-twist knot and Twist knot
### Three-twist knot vs. Twist knot
In knot theory, the three-twist knot is the twist knot with three-half twists. In knot theory, a branch of mathematics, a twist knot is a knot obtained by repeatedly twisting a closed loop and then linking the ends together.
## Similarities between Three-twist knot and Twist knot
Three-twist knot and Twist knot have 6 things in common (in Unionpedia): Alexander polynomial, Chiral knot, Crossing number (knot theory), Invertible knot, Jones polynomial, Knot theory.
### Alexander polynomial
In mathematics, the Alexander polynomial is a knot invariant which assigns a polynomial with integer coefficients to each knot type.
### Chiral knot
In the mathematical field of knot theory, a chiral knot is a knot that is not equivalent to its mirror image.
### Crossing number (knot theory)
In the mathematical area of knot theory, the crossing number of a knot is the smallest number of crossings of any diagram of the knot.
### Invertible knot
In mathematics, especially in the area of topology known as knot theory, an invertible knot is a knot that can be continuously deformed to itself, but with its orientation reversed.
### Jones polynomial
In the mathematical field of knot theory, the Jones polynomial is a knot polynomial discovered by Vaughan Jones in 1984.
### Knot theory
In topology, knot theory is the study of mathematical knots.
### The list above answers the following questions
• What Three-twist knot and Twist knot have in common
• What are the similarities between Three-twist knot and Twist knot
## Three-twist knot and Twist knot Comparison
Three-twist knot has 13 relations, while Twist knot has 18. As they have in common 6, the Jaccard index is 19.35% = 6 / (13 + 18).
## References
This article shows the relationship between Three-twist knot and Twist knot. To access each article from which the information was extracted, please visit:
Hey! We are on Facebook now! » | 446 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-14 | latest | en | 0.923886 |
https://minuteshours.com/74-32-hours-in-hours-and-minutes | 1,656,730,438,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00571.warc.gz | 436,109,041 | 4,745 | # 74.32 hours in hours and minutes
## Result
74.32 hours equals 74 hours and 19.2 minutes
You can also convert 74.32 hours to minutes.
## Converter
Seventy-four point three two hours is equal to seventy-four hours and nineteen point two minutes. | 65 | 250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.86876 |
https://origin.geeksforgeeks.org/construction-of-longest-increasing-subsequence-using-dynamic-programming/?ref=lbp | 1,685,441,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645595.10/warc/CC-MAIN-20230530095645-20230530125645-00351.warc.gz | 475,822,163 | 42,224 | GFG App
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# Construction of Longest Increasing Subsequence(LIS) and printing LIS sequence
The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.
Examples:
Input: [10, 22, 9, 33, 21, 50, 41, 60, 80]
Output: [10, 22, 33, 50, 60, 80] OR [10 22 33 41 60 80] or any other LIS of same length.
In the previous post, we have discussed The Longest Increasing Subsequence problem. However, the post only covered code related to the querying size of LIS, but not the construction of LIS. In this post, we will discuss how to print LIS using a similar DP solution discussed earlier.
Let arr[0..n-1] be the input array. We define vector L such that L[i] is itself is a vector that stores LIS of arr that ends with arr[i]. For example, for array [3, 2, 6, 4, 5, 1],
```L[0]: 3
L[1]: 2
L[2]: 2 6
L[3]: 2 4
L[4]: 2 4 5
L[5]: 1```
Therefore, for index i, L[i] can be recursively written as –
```L[0] = {arr[O]}
L[i] = {Max(L[j])} + arr[i]
where j < i and arr[j] < arr[i] and if there is no such j then L[i] = arr[i]```
Below is the implementation of the above idea –
## C++
`/* Dynamic Programming solution to construct Longest` ` ``Increasing Subsequence */` `#include ` `#include ` `using` `namespace` `std;` `// Utility function to print LIS` `void` `printLIS(vector<``int``>& arr)` `{` ` ``for` `(``int` `x : arr)` ` ``cout << x << ``" "``;` ` ``cout << endl;` `}` `// Function to construct and print Longest Increasing` `// Subsequence` `void` `constructPrintLIS(``int` `arr[], ``int` `n)` `{` ` ``// L[i] - The longest increasing sub-sequence ` ` ``// ends with arr[i]` ` ``vector > L(n);` ` ``// L[0] is equal to arr[0]` ` ``L[0].push_back(arr[0]);` ` ``// start from index 1` ` ``for` `(``int` `i = 1; i < n; i++)` ` ``{` ` ``// do for every j less than i` ` ``for` `(``int` `j = 0; j < i; j++)` ` ``{` ` ``/* L[i] = {Max(L[j])} + arr[i]` ` ``where j < i and arr[j] < arr[i] */` ` ``if` `((arr[i] > arr[j]) &&` ` ``(L[i].size() < L[j].size() + 1))` ` ``L[i] = L[j];` ` ``}` ` ``// L[i] ends with arr[i]` ` ``L[i].push_back(arr[i]);` ` ``}` ` ``// L[i] now stores increasing sub-sequence of` ` ``// arr[0..i] that ends with arr[i]` ` ``vector<``int``> max = L[0];` ` ``// LIS will be max of all increasing sub-` ` ``// sequences of arr` ` ``for` `(vector<``int``> x : L)` ` ``if` `(x.size() > max.size())` ` ``max = x;` ` ``// max will contain LIS` ` ``printLIS(max);` `}` `// Driver function` `int` `main()` `{` ` ``int` `arr[] = { 3, 2, 6, 4, 5, 1 };` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// construct and print LIS of arr` ` ``constructPrintLIS(arr, n);` ` ``return` `0;` `}`
## Java
`// Java program for ` `// the above approach` `// Dynamic Programming ` `// solution to construct Longest` `// Increasing Subsequence` `import` `java.util.*;` `class` `GFG{` `// Utility function to print LIS` `static` `void` `printLIS(Vector arr)` `{` ` ``for` `(``int` `x : arr)` ` ``System.out.print(x + ``" "``);` ` ``System.out.println();` `}` `// Function to construct and print ` `// Longest Increasing Subsequence` `static` `void` `constructPrintLIS(``int` `arr[], ` ` ``int` `n)` `{` ` ``// L[i] - The longest increasing ` ` ``// sub-sequence ends with arr[i]` ` ``Vector L[] = ``new` `Vector[n];` ` ``for` `(``int` `i = ``0``; i < L.length; i++)` ` ``L[i] = ``new` `Vector();` ` ` ` ``// L[0] is equal to arr[0]` ` ``L[``0``].add(arr[``0``]);` ` ``// Start from index 1` ` ``for` `(``int` `i = ``1``; i < n; i++)` ` ``{` ` ``// Do for every j less than i` ` ``for` `(``int` `j = ``0``; j < i; j++)` ` ``{` ` ``//L[i] = {Max(L[j])} + arr[i]` ` ``// where j < i and arr[j] < arr[i]` ` ``if` `((arr[i] > arr[j]) &&` ` ``(L[i].size() < L[j].size() + ``1``))` ` ``L[i] = (Vector) L[j].clone(); ``//deep copy` ` ``}` ` ``// L[i] ends with arr[i]` ` ``L[i].add(arr[i]);` ` ``}` ` ``// L[i] now stores increasing sub-sequence of` ` ``// arr[0..i] that ends with arr[i]` ` ``Vector max = L[``0``];` ` ` ` ``// LIS will be max of all increasing sub-` ` ``// sequences of arr` ` ``for` `(Vector x : L)` ` ``if` `(x.size() > max.size())` ` ``max = x;` ` ``// max will contain LIS` ` ``printLIS(max);` `}` `// Driver function` `public` `static` `void` `main(String[] args)` `{` ` ``int` `arr[] = {``3``, ``2``, ``4``, ``5``, ``1``};` ` ``int` `n = arr.length;` ` ``// print LIS of arr` ` ``constructPrintLIS(arr, n);` `}` `}` `// This code is contributed by gauravrajput1`
## Python3
`# Dynamic Programming solution to construct Longest` `# Increasing Subsequence` `# Utility function to print LIS` `def` `printLIS(arr: ``list``):` ` ``for` `x ``in` `arr:` ` ``print``(x, end``=``" "``)` ` ``print``()` `# Function to construct and print Longest Increasing` `# Subsequence` `def` `constructPrintLIS(arr: ``list``, n: ``int``):` ` ``# L[i] - The longest increasing sub-sequence` ` ``# ends with arr[i]` ` ``l ``=` `[[] ``for` `i ``in` `range``(n)]` ` ``# L[0] is equal to arr[0]` ` ``l[``0``].append(arr[``0``])` ` ``# start from index 1` ` ``for` `i ``in` `range``(``1``, n):` ` ``# do for every j less than i` ` ``for` `j ``in` `range``(i):` ` ``# L[i] = {Max(L[j])} + arr[i]` ` ``# where j < i and arr[j] < arr[i]` ` ``if` `arr[i] > arr[j] ``and` `(``len``(l[i]) < ``len``(l[j]) ``+` `1``):` ` ``l[i] ``=` `l[j].copy()` ` ``# L[i] ends with arr[i]` ` ``l[i].append(arr[i])` ` ``# L[i] now stores increasing sub-sequence of` ` ``# arr[0..i] that ends with arr[i]` ` ``maxx ``=` `l[``0``]` ` ``# LIS will be max of all increasing sub-` ` ``# sequences of arr` ` ``for` `x ``in` `l:` ` ``if` `len``(x) > ``len``(maxx):` ` ``maxx ``=` `x` ` ``# max will contain LIS` ` ``printLIS(maxx)` `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` ` ``arr ``=` `[``3``, ``2``, ``6``, ``4``, ``5``, ``1``]` ` ``n ``=` `len``(arr)` ` ``# construct and print LIS of arr` ` ``constructPrintLIS(arr, n)` `# This code is contributed by` `# sanjeev2552`
## C#
`// Dynamic Programming solution to construct Longest` `// Increasing Subsequence` `using` `System;` `using` `System.Collections.Generic; ` `class` `GFG ` `{` ` ` ` ``// Utility function to print LIS` ` ``static` `void` `printLIS(List<``int``> arr)` ` ``{` ` ``foreach``(``int` `x ``in` `arr)` ` ``{` ` ``Console.Write(x + ``" "``);` ` ``}` ` ``Console.WriteLine();` ` ``}` ` ` ` ``// Function to construct and print Longest Increasing` ` ``// Subsequence` ` ``static` `void` `constructPrintLIS(``int``[] arr, ``int` `n)` ` ``{` ` ` ` ``// L[i] - The longest increasing sub-sequence ` ` ``// ends with arr[i]` ` ``List> L = ``new` `List>();` ` ``for``(``int` `i = 0; i < n; i++)` ` ``{` ` ``L.Add(``new` `List<``int``>());` ` ``}` ` ` ` ``// L[0] is equal to arr[0]` ` ``L[0].Add(arr[0]);` ` ` ` ``// start from index 1` ` ``for` `(``int` `i = 1; i < n; i++)` ` ``{` ` ``// do for every j less than i` ` ``for` `(``int` `j = 0; j < i; j++)` ` ``{` ` ``/* L[i] = {Max(L[j])} + arr[i]` ` ``where j < i and arr[j] < arr[i] */` ` ``if` `((arr[i] > arr[j]) && (L[i].Count < L[j].Count + 1))` ` ``L[i] = L[j];` ` ``}` ` ` ` ``// L[i] ends with arr[i]` ` ``L[i].Add(arr[i]);` ` ``}` ` ` ` ``// L[i] now stores increasing sub-sequence of` ` ``// arr[0..i] that ends with arr[i]` ` ``List<``int``> max = L[0];` ` ` ` ``// LIS will be max of all increasing sub-` ` ``// sequences of arr` ` ``foreach``(List<``int``> x ``in` `L)` ` ``{` ` ``if` `(x.Count > max.Count)` ` ``{` ` ``max = x;` ` ``}` ` ``}` ` ` ` ``// max will contain LIS` ` ``printLIS(max);` ` ``}` ` ``// Driver code` ` ``static` `void` `Main() ` ` ``{` ` ``int``[] arr = { 3, 2, 4, 5, 1 };` ` ``int` `n = arr.Length;` ` ` ` ``// construct and print LIS of arr` ` ``constructPrintLIS(arr, n);` ` ``}` `}` `// This code is contributed by divyesh072019`
## Javascript
``
Output
`2 4 5 `
Note that the time complexity of the above Dynamic Programming (DP) solution is O(n^3) (n^2 for two nested loops and n for copying another vector in a vector eg: L[i] = L[j] contributes O(n) also) and space complexity is O(n^2) as we are using 2d vector to store our LIS and there is a O(n Log n) non-DP solution for the LIS problem. See below post for O(n Log n) solution.
Construction of Longest Monotonically Increasing Subsequence (N log N) | 3,457 | 9,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-23 | latest | en | 0.746875 |
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Approaches to the measurement of excess risk
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Approaches to the measurement of excess risk
Approaches to the measurement of excess risk
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Presentation Transcript
1. Approaches to the measurement of excess risk • 1. Ratio of RISKS • 2. Difference in RISKS: • (risk in Exposed)-(risk in Non-Exposed) Risk in Exposed Risk in Non-Exposed
2. Measures of RISK in Epidemiologic Studies • Without an explicit Comparison • Absolute risk • With an explicit comparison • Relative risk • Odds ratio • Attributable risk
3. An example comparing two ways of calculating excess risk
4. Cohort Study …then follow to see whether: a a+c c c+d = Incidence in exposed a a+b c c+d Relative Risk = = Incidence in Non-exposed
5. Hypothetical cohort study of the 1-year incidence of Acute Myocardial Infarction in indivduals with Severe Hpertension (180 mm Hg) and Normal Systolic blood Pressure (<120 mm Hg) 180 9820 30 9970 0.018 0.003 0.01833 0.00301 = = 6.09 Adapted from M. Szkly & J Nieto; Epidemiology: beyond the basics.
6. When is the Odds Ratio a Good Estimate of the Relative Risk? • When the “cases” studied are representative of all people with the disease in the population from which the cases were drawn, with regard to history of the exposure; • When the “controls” studied are representative of all people without the disease in the population from which the cases were drawn, with regard to history of exposure; • When the disease being studied is not a frequent one.
7. Incidence of Local Reactions in the Vaccinated and Placebo Groups, Influenza Vaccination trial Note: Based on data for individuals 40 years old or older in Seltser et al. To avoid rounding ambiguities in these and subsequent examples based on these data, the original sample sizes in Seltser et al.’s study (257 vaccinees and 241 placebo recipients) were multiplied by 10. Source: Data from R Seltser, PE Sartwell, and JA Bell, A Controlled test of Asian Influenza Vaccine in Population of Families, Am J. of Hygiene, 1962 (75):112-135. Adapted from M. Szkly & J Nieto; Epidemiology: beyond the basics.
8. Cross-tabulation of exposure and disease in a cohort study a b c d c c+d c c+d q- c d = = 1-q- 1 -
9. An expression of of the mathematical relationship between the OR on the one hand and the relative risk on the other, can be derived as follows. Assume that q+ is the incidence (probability) in exposed (e.g. vaccinated) and q- the incidence in unexposed individuals. The odds ratio is then: q+ 1-q+ q- 1-q- q+ 1-q+ 1-q- q- q+ q- 1-q- 1-q+ = x OR = = x 1-q- 1-q+ Notice that the term q+/q- in the equation is the relative risk. Thus the term defines the bias responsible for the discrepancy between the relative risk and odds ratio estimates (built-in bias). If the association between the exposure and the outcome is positive, q- < q+, thus (1-q-) > (1-q+). The bias term will therefor be greater than 1.0, leading to an overestimation of the relative risk by the odds ratio. By analogy, if the factor is protective, the opposite occurs - that is, (1-q-) < (1-q+) - and the odds ratio will again overestimate the strength of the association. In general, the odds ratio tends to yield an estimate further away from 1.0 than the relative risk on both sides of the scale (above or below 1.0).
10. Do the math: 1. Using the hypertension/myocardial infarct example RR= OR= OR=RR x “built in bias = 2. Using the example of local reactions to the influenza vaccine RR= OR= OR=RR x “built in bias =
11. Cohort studyA sample of exposed and non-exposed Incidence among smokers = 84/3000=28.0 Incidence among non-smokers = 87/5000=17.4
12. Attributable Risk The incidence in smokers which is attributable to their smoking - Incidence in smokers Incidence in Non-Smokers The ARexp of CHD attributable to smoking is:
13. Percent ARexp • A percent ARexp (%ARexp) is simply the ARexp expressed as a percentage of the risk in the exposed (q+). The excess risk associated with the exposure as a percentage of the total q+. • For a binary exposure, it is • %ARexp = q+ - q- x 100 • q+ • The %ARexp in the CHD/Smoking example is:
14. Population Attributable Risk • PAR is dependent on the population prevalence of exposure. • As the population is composed of exposed and unexposed individuals, the incidence in the population is similar to the incidence in the unexposed when the exposure is rare (A). • Incidence in the population is closer to that in the exposed, when the exposure is common (B).
15. Population Attributable Risk (PAR) and its dependence on the population prevalence of the exposure A. Pop ARexp ARexp Pop ARexp B. ARexp
16. Incidence in the total population, which is due to the exposure, can be calculated by subtracting: [Incidence in the total population] – [Incidence in the non-exposed group] In order to calculate this, one must know: EITHER: the incidence in the total population, OR: the incidence among smokers, the incidence among non-smokers, AND: the proportion of the total population with the exposure, i.e., the proportion of the population that smokes in the population under study or from which incidence can be calculated
17. We KNOW: incidence among smokers = 28.0/1,000/year, And, the incidence among non-smokers = 17.4/1,000/year. If we assume that from some other source of information, we know that the proportion of smokers in the population is 44% (and therefor the proportion of non-smokers is 56%), then the incidence in the total population can be calculated as: [28.0/1,000] [.44] + [17.4/1,000] [.56] = 22.0/1,000 SO THAT: [Incidence in the total population] – [Incidence in the non-exposed group] = [22.0/1,000/year] – [17.4/1,000/year] – 4.6/1,000/year
18. And the proportion of the incidence in the total population, which is attributable to the exposure, can be calculated by: • [Incidence in the total population] – [Incidence in the non-exposed group]-- • Incidence in the total population • = 22.0 – 17.4 = 20.9% • 22.0 • i.e., there should be a total reduction of 20.9% in the incidence of CHD in this population if smoking were eliminated.
19. Levin’s formula for the attributable risk for the Total Population After simple arithmetic manipulation, the previous formula can be expressed as a function of the prevalence of the exposure in the population and the relative risk : p (RR-1) x 100 p (RR-1) + 1 where p = proportion of the population with the characteristic or exposure and RR= relative risk or Odds Ratio (if applicable)
20. PAR: dependence on prevalence of exposure and relative risk RR=10 RR=5 RR=3 RR=2 For all values of the relative risk, the population AR increased markedly as the exposure prevalence increases. | 1,788 | 6,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-10 | latest | en | 0.882727 |
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Simulate Early, Simulate Often... In Rhino
# Simulation and Analysis of the FIU Bridge Collapse
NOTE: This is an engineering exercise looking at the FIU Bridge as an interesting application of Scan&Solve ONLY. We are not bridge designers. Theories presented here must be tempered by the limitations inherent in the geometric detail, material assumptions, and mathematical models being used.
Introduction:
On March 15, 2018, a bridge in Florida collapsed during construction killing 6 people with 9 people also injured. The bridge was designed to be 274 feet in length with a tower splitting the bridge into 175 and 99-foot sections. The bridge, being constructed for Florida International University, was built using a new technique in bridge design called Accelerated Bridge Construction (ABC). In this case, the main span and supports were first constructed before the main span was rotated from the staging zone to the supports over the roadway. This was meant to decrease the construction time and increase safety because the construction would not take place over the top of the roadway. Power lines also played a role in using the technique as they were present at one end of the bridge making it difficult to use cranes for the process [1]. Shortly after the 175-foot section was positioned, it collapsed onto the roadway crushing both cars and people.
Figure 0: Rendered image of the planned FIU bridge
A key aspect in the design was the aesthetic features of the truss members and cables. The members were aligned to be colinear with their respective cables that were to be attached to the tower. This caused an asymmetrical arrangement of the trusses along the length of the bridge. The cables and tower were not a main support structure for the bridge since the main span was designed to meet the loading of its own weight and pedestrians without needing the cables. Their main purpose was to handle vibrations from the loads the bridge would experience from pedestrians and to provide a view in the skyline for residents around the bridge.
Materials and Design:
The 175-foot section of the bridge collapsed after it was positioned over the roadway and standing alone on the supports at both ends. We wish to test why the bridge collapsed, considering it was claimed to have been able to support its own weight once it was positioned. With a simulation, we will be able to look at where the structure is weak to find some critical areas. To do this, the bridge needs to be modeled to size and our material needs to be chosen. The bridge was made of mostly concrete with some steel reinforcement. Since plain concrete is much stronger under compression than tension, reinforcement steel is added to increase the overall strength and hold the concrete in compression. Although the density of the steel is greater than concrete, which increases the weight of the structure when it is added, the elastic modulus of the steel makes a much larger impact. This makes the structure stiffer overall and the bridge will experience a smaller deformation under loading despite being heavier.
Fairly high strength concrete without reinforcement was simulated along with 2.5% and 5% steel reinforcement by volume on the main span. For this simulation, it is estimated that the actual amount of steel is somewhere between those percentages. Custom materials were made in Scan&Solve Pro to replicate the reinforced materials <link to customizing materials>. Material properties for the reinforced concrete were calculated based on the volume fraction of steel to concrete along with their respective material properties.
Table 0: Material Properties Used
Since the company chose to make a bridge with uneven trusses, we also wish to test if creating a uniform structure would have been more advantageous. Considering that the main purpose of the bridge is not for it to be a landmark, but to allow people to walk over the road, it makes sense to design a stronger bridge at the expense of aesthetics if the structure requires it. These designs were modeled with the walkway and overhead of the bridge remaining the same. The truss members are at 45-degree angles and it is expected that the new arrangements will strengthen the bridge by decreasing the stresses in some of the members.
Figure 1: Designs of alternate bridges tested
Another setup to test is how the bridge reacted during the transportation process. The self-propelled modular transporters that moved the bridge into place were located under the bridge just like the permanent supports, except they were closer to the center of the structure. The images below show the permanent structure and the transporters where the bridge was supported during movement.
Figure 2: Entire design proposed not including the pylon and cables
Figure 3: Photograph of the FIU bridge being positioned with self-propelled modular transporters
Scan&Solve Pro was used to run all the simulations using the bridge models created in Rhino 5. First, under the components section, the model that was desired to be simulated was selected. Then fairly high strength concrete or one of the custom materials added was chosen as the material of the structure. The restraints varied between the final position simulations and the transported case. When simulating the final case, each side of the walkway was selected and the restraints were edited to only be in the z direction. At only one end of the bridge, small edges were used to restrain it in the x and y directions so that the restraints act like a pin and a roller. The only load applied in the simulations was gravity because the bridge collapsed under its own weight. The same set up was used for the transportation simulations. Except this time restraints were moved from the outside surfaces to edges that were created in the position of the transporters.
Figure 4: Final bridge design with restraints and restraint editor showing the settings used
Figure 5: Restraints on the underside of the transported case design
Results:
The data in Table 1 follows a trend that we would expect. As the material contains more steel, we expect it to get stiffer while the maximum deflection should decrease. This occurred in all the designs tested even though a slight increase in the maximum principal tension and compression stress was generally observed as the percentage of steel increases. All the tensile stress values are of concern in that column because they are all larger than the tensile strength of fairly high strength concrete. This means that concrete alone will not be strong enough for the structure. It gets a little harder to directly judge the reliability of the reinforced concrete because the tensile strength is unknown. This will lead to us looking at the stress in each individual member along with the pre-tensions added to them.
Table 1: Overall Bridge Simulation with Designated Materials
The member numbers and pre-tensioning data in Table 3 correlate to the numbers given in the diagram shown below it. Although the concrete materials containing steel have pre-tension, the simulation gives us the results as if there was no pre-tension. The maximum principal tension and compression stress columns in Table 2 show these values. The positive numbers highlighted in red denote tension and the negative numbers in green show a member in compression. The pre-tension stresses can later be analyzed by comparing them with the stresses from the simulation.
Figure 6: Principal tensile/compression stress comparison among models with the given slider settings in lb/ft^2. From top to bottom: FIU Design, FIU Design Transported, Trusses at 45˚, Trusses at opposite 45˚
Figure 7: Deflection diagrams of the designs tested. From top to bottom: FIU Design, FIU Design Transported, Trusses at 45˚, Trusses at opposite 45˚
Table 2: Original FIU Bridge Design: Data of Each Member Without P.T. Considered
Member numbers are given in Figure 10 and the stress values for each member were found using the point marking tool in Scan&Solve Pro. This can be used after the simulation by going to the “View” tab and clicking on “Point” in the section labeled “Mark”. Points of interest on the model can now be clicked on to show the stress. This can also be verified by placing a small range on the legend to isolate a member from the structure and view the stress distribution in the truss. In this case, any stress above the maximum value is red and any value below the minimum is blue. Since tension is of more concern than compression, the value of -333.20KIPS/ft^2 was used as the critical value because it was the most positive value in the member.
Figure 8: Value displayed using the point tool and stress range colored on the individual member in lb/ft^2
Analysis:
A large part of the analysis is finding out where and why the bridge collapsed. Looking at the bridge from the view below, you can see the highlighted member, number 10, consistently contains the highest tensile stress through all the simulations conducted when the bridge is in its final position.
Figure 9: Full FIU bridge design
This is the same end of the bridge that appears to break first during the collapse. Even though this is the case, it might not actually contain as much tensile stress as the simulation indicates because the steel pre-tensioning helps hold the concrete in compression.
In the real bridge design, the designers defined the number of bars in each member and the pre-tension in each bar. As you can see in Table 3, members 3, 8, and 10, were all specified to have much larger total pre-tension forces on them than the other members in the truss. Designing it that way makes sense because those three also need to hold the largest tensile forces out of all the members. These along with the other pre-tensions all seem to do their job and keep the concrete in compression since these forces are larger than any of the forces felt from the tension loads in the simulation. This is noticed in Tables 4 and 5 where the maximum stresses with the pre-tension accounted for are all negative. This means that any tensile forces acting on the members are still less than the compressive force from the steel and that the material at these locations should still be safe.
Table 3: Pre-Tension in each truss member
Table 4: Analysis of each member in FIU 2.5 and 5% Steel: Transporting Bridge
Table 5: Analysis of each member in FIU 2.5 and 5% Steel: Final Position
In the simulations where the bridge is being moved into position, the bridge is supported toward the center of the structure more than when it is in its final position. Since this is the case, there is less deflection and smaller maximum principal stresses from tension and compression. This can also be seen in Table 1 where it is confirmed that smaller deflections and stresses take place relative to the final design. If the bridge is designed to stand in its final position, there should be no issues during transportation if the positioning process goes smoothly. Since there is a different support system here, it is possible that some structures are now in tension that are usually in compression when the bridge is in place. If this happens to a member with little reinforcement it could potentially cause a problem since the concrete is so much weaker in tension than compression. This happens in both members 2 and 11 during the simulation of the process where the bridge is moved. The only difference between the two is that number 2 was given enough pre-tension to effectively keep it in compression where member 11 was not given any. Even though this happens, the tensile strength of the non-reinforced concrete is still larger than the stress seen in the member. This means that the bridge appears strong enough when looking at the individual members for both the transported and final reinforcement cases.
We also wanted to analyze the more traditional truss layout structures with the members located at 45-degree angles. They are similar except for the fact that the first member on the left design goes upward whereas the one on the right starts downward. In Table 1, you can see that both designs behaved better than the final design because they both had smaller maximum stresses and deflections. It seems the final design could have been based off these initially and altered to satisfy the visual goal they had.
Conclusion:
Since we previously saw that the pre-tensions were large enough to keep all the members in compression or under the tensile strength of concrete, it seems like it was reasonable for them to use the design they ultimately went with even though some strength was given up. All the simulations and analysis done point to the possibility of a mishap during the construction process. It was reported that at the same location where the bridge collapsed, work was being done to strengthen it. It was also stated that cracks had been found near the same edge of the structure which could explain some testing and strengthening that was being conducted. The cracks were specifically located at the bottom of truss element number 11. Ongoing investigations are focusing on the cracks even though the design engineer stated that they caused no safety concerns [1]. The Miami Herald reported that support rods in numbers 2 and 11 were kept tight during the transportation process to support the ends of the bridge and were to be loosened once it was resting on the final supports where this tension was no longer needed. This directly correlates with the analysis of those members because the simulations showed they were in tension when the bridge was moving but only compression afterwards. They also reported that a possibility of the collapse is because number 11 shattered while the support rods were being tightened [2]. In the analysis conducted, I used the pre-tension numbers given in the technical proposal of the design as seen in Figure 10. One surprising thing is that the rods were specified in member 2 but none were given for number 11 in the proposal despite the report claiming they were tightening rods in member 11. Although the simulations pointed to member 11 being safe without any rods, it seems like either reinforcement was not documented in the proposal or the plans changed somewhere down the line and it was added.
Since the bridge appears to fail near members 10 and 11, it is reasonable to conclude that the investigations are correct to analyze number 11 extensively considering the high number of concerns that arose. Having the reports claim that the bridge failed during the process of tightening the rods even though no rods were planned in the report, and adding the cracks into the equation, makes it believable that a mistake could have been made with either the design or the adjusting process. As investigations continue, everybody will have to wait to see what details emerge and if they can pin down the actual cause of the collapse.
References:
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