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https://math.stackexchange.com/questions/1403260/lim-n-to-infty-fracna-n-lim-x-to-infty-frac1x-sharp-n-leq | 1,653,205,940,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00549.warc.gz | 444,272,323 | 65,161 | # $\lim_{n\to\infty} \frac{n}{a_n} = \lim_{x\to\infty} \frac{1}{x}\sharp \{n \leq x: n \in A\}$ when the limit exists.
This question is about natural density $d(A) = \lim_{x\to\infty}\frac{1}{x}\sharp\{n \leq x: n \in A\}$. I'm trying to prove that when either that limit or this limit:
$\lim_{n\to \infty} \frac{n}{a_n}$ exists, then the other exists and they're equal.
I've got, working backwards, that I'm trying to prove $\forall \epsilon \gt 0, \exists n_{\epsilon} : \forall n \geq n_{\epsilon}, |\frac{n}{a_n} - d(A)| \lt \epsilon$. But I'm not seeing what to do next.
($a_n$ is the ordered sequence that is $A$: $a_1 \lt a_2 \lt \dots$. In other words take the set $A$ and retrieve its elements in order to generate the sequence.)
• Take care of $x$ in $a_n < x < a_{n+1}$. Aug 19, 2015 at 22:02
For $A$ finite, $d(A) = 0$ and $\lim_{n\to \infty} \frac{n}{a_n}$, let a finite sequence be defined such that $a_k = 0$ for all remaining indexes $k$. Done.
For $A$ infinite, thanks to i707107's comment, I considered $$a_{n-1} \lt x \leq a_{n}$$ ie. for every $x$ there exists an $n$ such that the above holds since $A$ is infinite in number of elements. So let $x = m \in \Bbb{N}$, then write:
$$m \leq a_{n}$$
Since $x \gt 0$ (we're concerned only with $x \to +\infty$) we have $\frac{m}{a_n} \leq 1$.
We need an example. Say $A = {1, 3, 7, 11, 17, \dots}$
Then as $x$ takes on values from $\Bbb{N}$, the sequence is: $$1/1, 2/3, 4/7, 5/7, 6/7, 7/7, 8/11, 9/11, \dots$$
In other words this approach wont work. What we did learn though is that $n \leq a_n \implies \frac{n}{a_n} \leq 1$. Which is more importantly, a decreasing bounded sequence (bounded below by $0$) so has a limit.
I can't seem to prove what this limit equals though. | 644 | 1,754 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.827405 |
https://www.tes.com/en-au/teaching-resources/hub/primary/mathematics/advanced-mechanics/projectiles | 1,527,254,358,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00049.warc.gz | 843,109,175 | 16,749 | #### Bundle - One and Two Dimensional Motion - (Freely Falling Objects & Projectile Motion)
This source contains; 1. Motion in One Dimension - Freely Falling Objects - PPT This resource contains 24 slides; - Freely Falling Objects; (Release, fall, drop, thrown or jumping directly downwards & upwards questions) (Answer key) - Free Fall Problems solving strategies, - Physics around us… - Review, - HW. 2. Motion in Two Dimensions - Projectile Motion - PPT This resource contains 16 slides; - PROJECTILE; (Questions AND Answer key) - Physics around us… - Review, 3. Motion - One and Two Dimensional Motion - Freely Falling Objects & Projectile Motion - Teachers Note This document has 4 pages New Teachers can use this document as a lesson plan. 4. Motion - One and Two Dimensional Motion - Freely Falling Objects & Projectile Motion - Question Bank This Source contains 2 pages, (30 questions) Questions are about; - Motion in One Dimension - Free Falling Objects - Motion in Two Dimensions - Projectile Motion Everything is ready to be used or easily edited to suit your learners. To attract attention of students to the subject, daily life animations were used. You should install these Fonts for this document; (Marker Felt & Tempus Sans ITC)
#### Motion in Two Dimensions; Projectile Motion – Lesson Presentation (PPT)
This resource includes 16 slides; - PROJECTILE MOTION; (Questions & Answer key) - Physics around us… - Review, Everything is ready to be used or easily edited to suit your learners. To attract the attention of students to the subject, daily life animations were used. You should install these Fonts for this document; (Marker Felt & Tempus Sans ITC)
#### Motion - One and Two Dimensional Motion - Freely Falling Objects & Projectile Motion - Question Bank
This Worksheet has pages, (26 questions) Questions are about; - Motion in One Dimension - Free Falling Objects - Motion in Two Dimensions - Projectile Motion You should install these Fonts for this document; (Marker Felt & Tempus Sans ITC)
#### National 5 / KS3 Problem Solving #3 - Rocket Science
A problem solving scenario to combine skills, useful for revision and group work. The skills tested here are: • Parabolas (completing square, turning points, sketching, factorising) • Rounding to significant figures • Bearings • Area of sector and arc length • Area of triangle, sine rule, cosine rule • SOH CAH TOA • Pythagoras • Vectors • Lines • Percentages Pupils will also need good communication skills and teamwork (if working in groups). Provided with teacher notes and full worked solutions. | 573 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.775998 |
https://www.jiskha.com/questions/16301/a-scuba-diver-breathing-normal-air-descend-to-100-m-of-depth-where-the-toal-pressure-is | 1,620,440,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00469.warc.gz | 874,597,320 | 4,850 | # chemistry... confusing!
a scuba diver breathing normal air descend to 100 m of depth, where the toal pressure is 11 atm. What is the partial pressure of oxygen that the diver experiences at this depth?
I know that at 100m that a diver would have to breathe Heliox, meaning that the air in the tank is a combo of O2 and He.
But I don't know how to figure out the problem.
Assume it is not heliox, but atmospheric air. What percent of atmospheric pressure is the partial pressure of O2.
would that be 21%?
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A diver is currently located at a depth of 50m in the ocean. (if the density of sea water is 1025ππ/π3 and π=10/π 2), what is the absolute Pressure of the diver? | 898 | 3,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.914017 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-11-infinite-series-11-3-convergence-of-series-with-positive-terms-exercises-page-557/54 | 1,579,443,143,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00002.warc.gz | 901,254,438 | 13,036 | ## Calculus (3rd Edition)
The series $\Sigma_{n=1}^{\infty} \frac{1}{n^2+\sin n}$ converges.
Using the limit comparison test by $b_n=\frac{1}{n^2}$, now we have $$L=\lim_{n\to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{1/(n^2+\sin n)}{1/n^2}\\ =\lim_{n\to \infty} \frac{n^2}{n^2+\sin n}=\lim_{n\to \infty} \frac{1}{1+(\sin n/n^2)}=1$$ Since $\Sigma_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent p-series and $L=1\gt 0$ then the series $\Sigma_{n=1}^{\infty} \frac{1}{n^2+\sin n}$ converges. | 226 | 497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-05 | longest | en | 0.37579 |
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# Mission SSC CGL 2017 Tier-2-Day34-Mathematics Comprehensive revision series
Mission SSC CGL 2017 Tier-2-Day34-Mathematics Comprehensive revision series
Dear Students,
Welcome to Pinnacle Online SSC Coaching Portal. SSC CGL Tier-1 has concluded. Now there is almost 18 days time for Tier-2 preparation. We all know the importance of this section. While entire Tier-1 was of total 200 marks, this paper alone carry 200 marks and mind it. it is the most scoring section of SSC CGL examination. In the past we have many examples where students had scored 200/200. So this is the section which will either make or break your dream of a reputed government job. A very diligent effort can put you in a commanding position. We at pinnacle brings revision series for mathematics where test yourself while socializing with friends. We use it as a time off from preparation. Now, think if the time spent online becomes more productive than the normal preparation itself. Sounds crazy, isn’t it? We at Pinnacle want your time spent online to get as meaningful as it can be. Here comes Tier-2-Day33-Mathematics of MISSION SSC CGL 2017 day wise Comprehensive revision series.
## Tier-2-Day34-Mathematics
Q1. Ravindra walks at 6 km/hr and Riya cycles at 11 km/hr towards each other. What was the distance between them when they started if they meet after 42 minutes?
रविंद्र 6 कि..मी./घंटा की गति से चल कर और रिया 11 कि.मी./घंटा की गति से साइकिल चलाते हुए एक दूसरे की तरफ आते है | यदि वे 42 मिनट के बाद मिलते है, तो शुरुआत में उनके बीच कितनी दूरी थी?
(a) 17.9 kms / 17.9 कि..मी.
(b) 14.9 kms / 14.9 कि..मी.
(c) 8.9 kms / 8.9 कि..मी.
(d) 11.9 kms / 11.9 कि..मी.
Q2. In a Kite _____.
एक पतंग में_______ .
(a) One diagonal bisects the other / एक विकर्ण अन्य को दो भागो में बांटता है
(b) Both diagonals form two congruent triangles / दोनों विकर्ण दो संगत त्रिभुज बनाते है
(c) Adjacent angles are supplementary / निकटस्थ कोण अनुपूरक होते हैं
(d) Opposite sides are parallel / विपरीत बाजुएँ समांतर होती है
Q3. Akhilesh is two times as good a workman as Bansi and therefore is able to finish a job in 39 days less than Bansi. Working together, they can do it in
अखिलेश बंसी से दो गुना अधिक अच्छा कर्मचारी है और इसी कारण वह एक काम उससे 39 दिन कम में पूरा कर सकता है | दोनों मिलकर वह काम कितने दिन में करेंगे?
(a) 13 days / 13 दिन
(b) 24 days / 24 दिन
(c) 26 days / 26 दिन
(d) 12 days / 12 दिन
Q4. When a discount of 20% is given on a pizza, the profit is 32%. If the discount is 18%, then the profit is
जब एक पिज़्ज़ा पर 20% की छूट दी जाती है, तो 32% का लाभ होता है| यदि छूट 18% हो,तो लाभ कितना होगा?
(a) 50 percent / 50 प्रतिशत
(b) 64.7 percent / 64.7 प्रतिशत
(c) 35.3 percent / 35.3 प्रतिशत
(d) 20.6 percent / 20.6 प्रतिशत
Q5. Two students appeared for an examination. One of them secured 16 marks more than the other and his marks were 75% of the sum of their marks. The marks obtained by them are
दो छात्र एक परीक्षा में बैठे | उनमें से एक ने दूसरे की तुलना में 16 अंक अधिक हासिल किए और उसके अंक उन दोनों के अंक के योग का 75% थे | उन दोनों ने कितने अंक प्राप्त किए?
(a) 98 and 82 / 98 और 82
(b) 49 and 33 / 49 और 33
(c) 96 and 80 / 96 और 80
(d) 24 and 8 / 24 और 8
Q6 .Rehman goes on a trip on his motor-cycle and rides for 661 kms. If he rides for 8 hours at a speed of 31 km/hr, find at what speed he travels for the remaining 7 hours of the journey?
रेहमान उसकी मोटर साइकिल पर की गई एक यात्रा में 661 कि.मी. की दुरी तय करता है| यदि वह 31 कि.मी./घंटा की गति से 8 घंटे तक मोटर साइकिल चलता है,तो वह शेष 7 घंटे की यात्रा किस गति से तय करता है?
(a) 50 km/hr / 50 3 कि.मी./घंटा
(b) 42 km/hr / 42 3 कि.मी./घंटा
(c) 53 km/hr / 53 3 कि.मी./घंटा
(d) 59 km/hr / 59 3कि.मी./घंटा
Q7. The price of an article is cut by 31%, to restore to its original value, the new price must be increased by
एक वस्तु की कीमत में 31% की छूट दी जाती है| पुराना मूल्य वापस लाने के लिए कितने प्रतिशत से वृद्धि करनी चाहिए?
(a) 66 percent/ 66 प्रतिशत
(b) 44.93 percent/ 44.93 प्रतिशत
(c) 39.76 percent/ 39.76 प्रतिशत
(d) 82.5 percent/ 82.5 प्रतिशत
Q8. Akka can bake 100 cakes in 20 hours, Akka and Tai together can bake 75 cakes in 10 hours. How many cakes Tai can bake in 20 hours?
अक्का 20 घंटे में 100 केक बना सकती है, अक्का और ताई साथ मिलकर 10 घंटे में 75 केक बना सकती हैं| ताई 20 घंटे में कितने केक बना सकती हैं?
(a) 25
(b) 50
(c) 30
(d) 75
Q9. Among three numbers, the first is twice the second and thrice the third. If the average of three numbers is 396, then what is the difference between the first and the third number?
तीन संख्याओं के बीच में, पहली संख्या दूसरी संख्या से दोगुनी और तीसरी संख्या से तीन गुनी है| यदि उन तीन संख्याओं का औसत 396 है, तो पहली और तीसरी संख्याओं के बीच कितना अंतर है?
(a) 594
(b) 448
(c) 432
(d) 453
Q10. Raghavendra sells a machine for Rs 43 lakhs at a loss. Had he sold it for Rs 52 lakh, his gain would have been 5 times the former loss. Find the cost price of the machine.
राघवेंद्र एक मशीन 43 लाख में घाटे में बेचता है| यदि उसने उसे 52 लाख में बेचा होता, तो उसका लाभ उसके पूर्व घाटे का 5 गुणा होता| मशीन का क्रय मूल्य प्राप्त करें|
(a) Rs 50.5 lakhs/ 50.5 लाख रु
(b) Rs 60.6 lakhs/ 60.6 लाख रु
(c) Rs 44.5 lakhs/ 44.5 लाख रु
(d) Rs 38.5 lakhs/ 38.5 लाख रु
For Remaining Questions and other Tier 2 Maths Paper,
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Now Get All Notifications And Updates In Your E-mail Account Just Enter Your E-mail Address Below And Verify Your Account To Get More Updates : | 2,377 | 6,547 | {"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} | 3.671875 | 4 | CC-MAIN-2019-22 | latest | en | 0.72759 |
http://strathprints.strath.ac.uk/33798/ | 1,503,284,142,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00029.warc.gz | 392,082,994 | 8,419 | # Permutations sortable by n-4 passes through a stack
Steingrimsson, Einar and Dukes, Mark and Claesson, Anders (2010) Permutations sortable by n-4 passes through a stack. Annals of Combinatorics, 14. pp. 45-51.
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## Abstract
The subject of pattern avoiding permutations has its roots in computer science, namely in the problem of sorting a permutation through a stack. A formula for the number of permutations of length n that can be sorted by passing it twice through a stack (where the letters on the stack have to be in increasing order) was conjectured by West, and later proved by Zeilberger. Goulden and West found a bijection from such permutations to nonseparable planar maps, and later, Jacquard and Schaeffer presented a bijection from these planar maps to certain labeled plane trees, called beta(1,0)-trees. Using generating trees, Dulucq, Gire and West showed that nonseparable planar maps are equinumerous with permutations avoiding the (classical) pattern 2413 and the barred pattern 41\bar{3}52; they called these permutations nonseparable. We give a new bijection between beta(1,0)-trees and permutations avoiding the dashed patterns 3-1-4-2 and 2-41-3. These permutations can be seen to be exactly the reverse of nonseparable permutations. Our bijection is built using decompositions of the permutations and the trees, and it translates seven statistics on the trees into statistics on the permutations. Among the statistics involved are ascents, left-to-right minima and right-to-left maxima for the permutations, and leaves and the rightmost and leftmost paths for the trees. In connection with this we give a nontrivial involution on the beta(1,0)-trees, which specializes to an involution on unlabeled rooted plane trees, where it yields interesting results. Lastly, we conjecture the existence of a bijection between nonseparable permutations and two-stack sortable permutations preserving at least four permutation statistics. | 471 | 2,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.904241 |
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# Ex.6.2 Q1 Squares and Square Roots Solutions - NCERT Maths Class 8
Go back to 'Ex.6.2'
## Question
Find the squares of the following numbers.
(i) $$32$$
(ii) $$35$$
(iii) $$86$$
(iv) $$93$$
(v) $$71$$
(vi) $$46$$
Video Solution
Squares And Square Roots
Ex 6.2 | Question 1
## Text Solution
What is known?
Numbers
What is unknown?
Squares of the numbers.
Reasoning 1:
There is a way to find this by without multiplication.
Use identity:
$$(a+b)^2=a^2+2ab+b^2$$
$$(a+b)^2=a^2-2ab+b^2$$
Steps (i):
\begin{align}32 &= 30 + 2\\{32^2} &= {(30 + 2)^2}\\ &= 30(30 + 2) + 2(30 + 2)\\ &= {30^2} + 30 \times 2 + 2 \times 30 + {2^2}\\ &= 900 + 60 + 60 + 4\\ &= 1024\end{align}
Reasoning 2:
If a number have its unit digit $$5$$ i.e. $$a5$$, its square number will be [$$a (a+1)\; \rm hundred+25$$].
Steps (ii):
Hence $$a = 3$$
Square of the number $$35$$
\begin{align}&= [3(3+1) \text{hundreds}+25]\\&= [ (3\times 4) \;\text{hundreds}+25]\\&= 1200+25\\&= 1225 \end{align}
Reasoning:
There is a way to find this by without multiplication
Steps (iii):
\begin{align}86 &= 80 + {86^2}\\ &= {(80 + 6)^2}{\rm{ }}\\ &= 80(80 + 6) + 6(80 + 6)\\ &= {80^2} + 80 \times 6 + 6 \times 80 + {6^2}\\ &= 6400 + 480 + 480 + 36\\&= 7396 \end{align}
Reasoning:
There is a way to find this by without multiplication.
Steps (iv):
\begin{align}93 &= 90 + 3\\{93^2} &= {(90 + 3)^2}\\ &= 90(90 + 3) + 3(90 + 3)\\ &= {90^2} + 90 \times 3 + 3 \times 90 + {3^2}\\ &= 8100 + 270 + 270 + 9\\ &= 8649 \end{align}
Reasoning
There is a way to find this by without multiplication.
Steps (v):
\begin{align}71 &= 70 + 1\\{71^2} &= {(70 + 1)^2}\\ &= 70(70 + 1) + 1(70 + 1)\\ &= {70^2} + 70 \times 1 + 1 \times 70 + {1^2}\\ &= 4900 + 70 + 70 + 1\\ &= 5041 \end{align}
Reasoning
There is a way to find this by without multiplication.
Steps (vi):
\begin{align} 46 &= 40 + 6\\{46^2} &= {(40 + 6)^2}\\ &= 40(40 + 6) + 6(40 + 6)\\ &= {40^2} + 40 \times 6 + 6 \times 40 + {6^2}\\ &= 1600 + 240 + 240 + 36\\ &= 2116 \end{align}
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harangzsolt33 has asked for the wisdom of the Perl Monks concerning the following question:
Dear Monks,
This question is about an algorithm, which I am writing in Perl.
Recently I tried to write a simple script for an accountant friend of mine as a hobby, and when I sat down to put this thing together, then I realized that it's not as simple as I thought!
The problem is we have a long list of numbers. The accountant types in one big number which we call TARGET NUMBER. And my program immediately lists all possible combination of numbers from the list whose sum equals the TARGET NUMBER. So, if that number is 100 and our list is made up of the following numbers: 1 99 2 40 50 60 90 3 5 95 100, then the result should look like this:
100
1+99
2+3+95
5+95
2+3+5+90
2+3+5+40+50
40+60
Unfortunately, I don't know how to write the algorithm that finds all possible combinations. My program sorts the list of numbers first. Then it picks the smallest number from the list and adds it to the largest to see if it equals the TARGET NUMBER. If it's bigger, then it tries to add the smallest number to the second from the last and so forth until it finds a combination of two numbers that is equal or smaller than the TARGET NUMBER.
If the sum of two numbers is LESS than the TARGET NUMBER, then we try to add a third number to see if it equals and so forth... The problem is that this requires the numbers to occur in a certain order. If they occur in the wrong order, then we will miss some combinations! For example, we're looking to find combinations that equal 100. This is our list: 5 5 5 5 10 15 80 99
As you can see, in this scenario, we will find 5+15+80, because they occur in a specific order. But we will completely miss 5+5+5+5+80, because the algorithm has a bug. I don't know how to make this work. Can anyone suggest a fix or a different algorithm?
```#!/usr/bin/perl -w
use strict;
use warnings;
my @RESULT;
my \$TARGET = 100;
my @LIST = qw(0 5 10 5 5 5 15 80 99);
#34 111.38 55 3.93 100 100 100 100 88 6.3 99 400 1020 -2.43
#73 39 3 12 -0.999 228 104 12377.31 390 399 212 315 5.8 405 4402 16252
#10 3600 18209 288.62 3384 12 450 902 151 396.07 44 88 52 107 244 1 52
+0);
print "\n This program finds a combination of numbers from a long li
+st";
print "\n whose sum equals the \"TARGET\" total number. Ideally, we
+want";
print "\n to find ALL possible combinations! For example:";
print "\n When Target = 5 and our list is 1, 3, 6, 3.38, -9.8, 4, 72
+, 2";
print "\n then the solution would be : 5 = 1 + 4 and 5 = 2 + 3.";
print "\n 1 + 2 + 2 will not appear in the list, because we only hav
+e one number 2.";
print "\n\n\n";
FindCombinations();
exit;
##################################################
#
# This is the main algorithm. First of all, it sorts
# all the numbers in ascending order. Then we get rid
# of all the numbers that are larger than the TARGET
# or if they're zero or smaller. Then it looks at
# the first number in the list and tries to add another
# number to it until it equals the TARGET. Then it takes
# the second number, and so forth, looking for pairs.
#
sub FindCombinations
{
# First, we get rid of all the numbers that are larger
# than the TARGET or ZERO or less than 0. We also remove
# numbers if number == TARGET record a number if it equals the targe
+t.
print "\nSTAGE 1\n>>>LIST=", join(' ', @LIST), "<<<\nRESULTS=|", joi
+n(' ', @RESULT), "|\n";
my \$NUMBER;
for (my \$i = 0; \$i < @LIST; \$i++)
{
\$NUMBER = \$LIST[\$i];
if (\$NUMBER <= 0 || \$NUMBER >= \$TARGET)
{
if (\$NUMBER == \$TARGET) { \$RESULT[0] = (\$TARGET); }
\$LIST[\$i] = '';
}
}
print "\nSTAGE 2\n>>>LIST=", join(' ', @LIST), "<<<\nRESULTS=|", joi
+n(' ', @RESULT), "|\n";
@LIST = RemoveBlankLines(@LIST);
@LIST = SortNumbers(@LIST);
print "\nSTAGE 3\n>>>LIST=", join(' ', @LIST), "<<<\nRESULTS=|", joi
+n(' ', @RESULT), "|\n";
############# SEARCH ALGORITHM BEGINS HERE #############
my @SKIP_LIST;
for (my \$j = 0; \$j < @LIST; \$j++)
{
my \$Total = 0;
\$Total = \$LIST[\$j] * 1;
+s number.
for (my \$i = @LIST - 1; \$i >= 0; \$i--)
{
if (\$i == \$j) { next; } # Skip this number, because it's alread
### Try to add numbers and see if the sum is exactly what we are
+ looking for.
\$Total += \$LIST[\$i];
if (\$Total > \$TARGET)
{
\$Total = \$LIST[\$j];
next;
}
if (\$Total == \$TARGET)
{
last;
}
}
}
@RESULT = ExtractDuplicates(@RESULT);
print "\nSTAGE 4\n>>>LIST=", join(' ', @LIST), "<<<\nRESULTS=|", joi
+n(' ', @RESULT), "|\n";
}
##################################################
# v2020.11.19
# This function removes duplicate lines from an array
# by sorting it and comparing each line with case-sensitive
# comparison. Returns a new array.
#
# Usage: NEW_ARRAY = ExtractDuplicates(ARRAY)
#
sub ExtractDuplicates
{
my @A = @_;
@A > 1 or return @A;
@A = sort(@A);
my \$i = 0;
my \$j = 1;
while (\$j < @A)
{
if (\$A[\$i] eq \$A[\$j])
{
splice(@A, \$j, 1);
}
else
{
\$i++; \$j++;
}
}
return @A;
}
##################################################
# v2020.11.19
# This function trims each element of the input array
# and removes empty strings elements. This function
# shortens the original array.
#
# Usage: RemoveBlankLines(ARRAY)
#
sub RemoveBlankLines
{
@_ or return;
my @A = @_;
my (\$j, \$i, \$LINE) = 0;
for (\$i = 0; \$i < @A; \$i++)
{
\$LINE = Trim(\$A[\$i]);
if (length(\$LINE))
{
if (\$j < \$i) { \$A[\$j] = \$LINE; }
\$j++;
}
}
\$#A = \$j - 1;
return @A;
}
##################################################
# v2019.8.25
# Removes whitespace from before and after STRING.
# Whitespace is here defined as any byte whose
# ASCII value is less than 33. That includes
# tabs, esc, null, vertical tab, new lines, etc.
# Usage: STRING = Trim(STRING)
#
sub Trim
{
defined \$_[0] or return '';
(my \$L = length(\$_[0])) or return '';
my \$P = 0;
while (\$P <= \$L && vec(\$_[0], \$P++, 8) < 33) {}
for (\$P--; \$P <= \$L && vec(\$_[0], \$L--, 8) < 33;) {}
substr(\$_[0], \$P, \$L - \$P + 2)
}
##################################################
sub SortNumbers { return sort {\$a <=> \$b} @_; }
##################################################
Replies are listed 'Best First'.
Re: Find combination of numbers whose sum equals X
by choroba (Archbishop) on Nov 20, 2020 at 09:47 UTC
```#!/usr/bin/perl
use warnings;
use strict;
sub _summands {
my (\$target, @numbers) = @_;
return [[]] if 0 == \$target;
my @results;
for my \$index (0 .. \$#numbers) {
my \$number = \$numbers[\$index];
my @remaining = @numbers[ grep \$_ != \$index, 0 .. \$#numbers ];
next if \$target - \$number < 0;
my \$result = _summands(\$target - \$number, @remaining);
push @results,
map [\$number, @\$_],
grep ! @\$_ || \$number <= \$_->[0],
@\$result;
}
return \@results
}
sub summands {
my \$results = _summands(@_);
my %unique;
for my \$result (@\$results) {
undef \$unique{"@\$result"};
}
return [ map [split ' '], keys %unique ]
}
use Test::More tests => 2;
use Test::Deep;
cmp_deeply summands(100, 1, 99, 2, 40, 50, 60, 90, 3, 5, 95, 100),
bag([100], [1, 99], [2, 3, 95], [5, 95], [2, 3, 5, 90], [2, 3, 5,
+40, 50], [40, 60]);
cmp_deeply summands(100, 5, 5, 5, 5, 10, 15, 80, 99),
bag([5, 15, 80], [5, 5, 5, 5, 80], [5, 5, 10, 80]);
Update: Sorry, I'm kind of busy, so I don't have much time to explain it. It's a classical example of Dynamic Programming - the only complication is the numbers can be repeated, which I solved using the \$unique hash. It's probably possible to build the solutions in a unique way right away in the recursive function, so there won't be any postprocessing needed.
Update 2: For a speed-up, add
```use Memoize;
memoize('_summands');
and, as found by haukex in Fastest way to "pick without replacement", replace
``` my \$number = \$numbers[\$index];
my @remaining = @numbers[ grep \$_ != \$index, 0 .. \$#numbers ];
by
``` my @remaining = @numbers;
my (\$number) = splice @remaining, \$index, 1;
map{substr\$_->[0],\$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]
Wow, thank you very much for all your answers! These replies have been very enlightening! :-)
Re: Find combination of numbers whose sum equals X
by QM (Parson) on Nov 20, 2020 at 10:20 UTC
I think this is known as the Subset Sum Problem.
-QM
--
Quantum Mechanics: The dreams stuff is made of
> I think this is known as the Subset Sum Problem.
Well ... almost.
The Subset Sum Problem asks if there is one solution.
But the OP asks to "list all possible combination of numbers"
Algorithm wise that's a huge difference, because one can often optimize searching for a single solution, while happily ignoring the rest. °
And that's also why I'm hesitant solving this, you can easily show that the solution space of all possible combinations will explode quickly, in a way that already the time needed to print them out will take an eternity.
I.O.W. such problems don't make much sense, unless you are singling out a single (or a few) solution which are optimal regarding a second value-function.
Cheers Rolf
(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery
°)The Knapsack Problem will also demand to optimize a second "value" function and only require the "weight" to be less or equal the "target". It's a generalization of Subset Sum b/c if you choose the weight as value, the equal case - if it exists - will be maximal.
Re: Find combination of numbers whose sum equals X
by tybalt89 (Prior) on Nov 20, 2020 at 12:44 UTC
```#!/usr/bin/perl
use strict; # https://perlmonks.org/?node_id=11123870
use warnings;
use List::Util qw( uniq );
print "\$_\n" for find( 100, '1 99 2 40 50 60 90 3 5 95 100' );
print "\n";
print "\$_\n" for find( 100, '5 5 5 5 10 15 80 99' );
sub find
{
my (\$target, \$from, \$have) = @_;
\$have //= '';
\$target == 0 and return \$have =~ s/.//r;
\$target > 0 && \$from =~ s/\d+// or return ();
uniq find( \$target - \$&, \$from, "\$have+\$&"), find( \$target, \$from,
+\$have );
}
Outputs:
```1+99
2+40+50+3+5
2+90+3+5
2+3+95
40+60
5+95
100
5+5+5+5+80
5+5+10+80
5+15+80
Re: Find combination of numbers whose sum equals X
by johngg (Canon) on Nov 20, 2020 at 17:38 UTC
The word "combination" in the title brought Algorithm::Combinatorics to mind. I'm not sure if duplicate sums, e.g. the several 5+5+10+80 combinations in the third example, should all be shown but I have eliminated them. This code
```use strict;
use warnings;
use feature qw{ say };
use Algorithm::Combinatorics qw{ combinations };
use List::Util qw{ sum };
my @tests = (
{
target => 100,
values => [ 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 ],
},
{
target => 10,
values => [ 1, 3, 2, 4 ],
},
{
target => 100,
values => [ 5, 5, 5, 5, 10, 15, 80, 99 ],
},
);
foreach my \$rhTest ( @tests )
{
say
qq{\nFind sums from },
join( q{, }, @{ \$rhTest->{ values } } ),
qq{ making \$rhTest->{ target }};
say for do {
my %seen;
grep { ! \$seen{ \$_ } ++ }
grep { \$_ == \$rhTest->{ target } }
@{ \$rhTest->{ values } };
};
for my \$sumsOf ( 2 .. scalar @{ \$rhTest->{ values } } )
{
my \$combIter = combinations( \$rhTest->{ values }, \$sumsOf );
my %seen;
while ( my \$raComb = \$combIter->next() )
{
next if \$seen{ join q{+}, sort { \$a <=> \$b } @{ \$raComb }
+} ++;
say join q{+}, @{ \$raComb }
if \$rhTest->{ target } == sum @{ \$raComb };
}
}
}
produces
```
Find sums from 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 making 100
100
1+99
40+60
5+95
2+3+95
2+90+3+5
2+40+50+3+5
Find sums from 1, 3, 2, 4 making 10
1+3+2+4
Find sums from 5, 5, 5, 5, 10, 15, 80, 99 making 100
5+15+80
5+5+10+80
5+5+5+5+80
Cheers,
JohnGG
Re: Find combination of numbers whose sum equals X
by LanX (Cardinal) on Nov 22, 2020 at 00:35 UTC
here another approach, it calculates the possible partial sums in %step with increasing possible \$delta.
the solutions are than printed by walking back from \$target to zero.
NB: it creates two kind of outputs, a tree with partial solutions and a result hash.
unfortunately this only works efficiently for unique deltas, I'll probably try to fix it tomorrow.
(or better leave it open for the interested reader ;)
```use strict;
use warnings;
use Data::Dump qw/pp dd/;
use Data::Dumper;
# --- input
my @input = (1,99,2,40,50,60,90,3,5,95,100);
my \$target = 100;
my %steps = ( 0 => []);
# --- processing
my @deltas = sort { \$a <=> \$b } @input ;
for my \$delta ( @deltas ) {
for my \$last (keys %steps) {
my \$next = \$last + \$delta;
unshift @{\$steps{\$next}},\$last
if \$next <= \$target-\$delta # \$delta grows!
or \$next == \$target; # goal
}
# pp \$delta, \%steps;
}
pp \%steps;
# --- output
my %free;
\$free{\$_}++ for @deltas;
our \$level = -1;
sub walk_back {
my (\$target,\$h_path)=@_;
local \$level = \$level +1;
for my \$last (@{\$steps{\$target}}) {
my \$delta = \$target-\$last;
next unless \$free{\$delta};
local \$free{\$delta} = \$free{\$delta}-1;
print "\t" x \$level , "+\$delta\n";
my \$sub_path = \$h_path->{\$delta} = {};
if ( \$last>0 ) {
walk_back(\$last,\$sub_path)
} else {
print "\n\n";
}
}
}
my %path;
walk_back(\$target,\%path);
pp \@deltas;
print Dumper \%path;
```
{
"0" => [],
"1" => [0],
"2" => [0],
"3" => [0, 1],
"4" => [1],
"5" => [0, 2],
"6" => [1, 3],
"7" => [2],
"8" => [3],
"9" => [4],
"10" => [5],
"11" => [6],
"40" => [0],
"41" => [1],
"42" => [2],
"43" => [3],
"44" => [4],
"45" => [5],
"46" => [6],
"47" => [7],
"48" => [8],
"49" => [9],
"50" => [0, 10],
"51" => [11],
"100" => [0, 1, 5, 10, 40, 50],
}
[1, 2, 3, 5, 40, 50, 60, 90, 95, 99, 100]
+100
+99
+1
+95
+5
+3
+2
+90
+5
+3
+2
+60
+40
+50
+40
+5
+3
+2
\$VAR1 = {
'99' => {
'1' => {}
},
'50' => {
'40' => {
'5' => {
'3' => {
'2' => {}
}
}
}
},
'95' => {
'5' => {},
'3' => {
'2' => {}
}
},
'100' => {},
'60' => {
'40' => {}
},
'90' => {
'5' => {
'3' => {
'2' => {}
}
}
}
}; | 4,639 | 13,965 | {"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} | 3.328125 | 3 | CC-MAIN-2021-04 | latest | en | 0.897021 |
http://medialab.di.unipi.it/web/IUM/Waterloo/node51.html | 1,558,949,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00452.warc.gz | 129,290,352 | 2,090 | Next: Transformation Applications and Up: Projections and Projective Previous: 3D Clipping
# Homogeneous Clipping
Projection: linear transformations then normalize
• Linear transformation
• Normalization
Region mapping:
\
\
Clipping not good after normalization:
• Ambiguity after normalization
• Numerator can be positive or negative
• Denominator can be positive or negative
• Normalization expended on points that are subsequently clipped
Clip in homogeneous coordinates:
• Compare unnormalized coordinate against
Clipping Homogeneous Coordinates
• Assume NDC window of
• To clip to X=-1 (left):
• Projected coordinates: Clip to X=-1
• Homogeneous coordinate: Clip to
• Homogeneous plane:
\
• Point is visible if
• For line segment want a such that
lies on the plane w+x=0
• Solving for a
gives us
• Repeat for remaining boundaries:
• Near and far clipping planes
Readings: Jim Blinn's Corner, Chapters 13, 18
CS488/688: Introduction to Interactive Computer Graphics
University of Waterloo
Computer Graphics Lab
cs488@cgl.uwaterloo.ca | 243 | 1,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-22 | latest | en | 0.771616 |
https://ccssmathanswers.com/8th-grade-math/ | 1,716,898,460,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00013.warc.gz | 129,974,907 | 63,056 | 8th Grade Math Practice, Topics, Test, Problems, and Worksheets
Do you have a test coming up and scared about what to prepare for in 8th Grade Math? Don’t Fret as we have included the bunch of topics that you might come across in Grade 8 Maths all in one place.
Check out Chapterwise Go Math 8th Grade Answer Key available here during your practice sessions. Make the most out of them and score better grades in your exams. You can access whichever chapter you feel like preparing by tapping on the quick links listed below. Once you click on them you will be redirected to the concerned chapter in no time.
Access our Big Ideas Math 8th Grade Answers listed below to resolve all your queries on the Chapters involved. Don’t worry about the accuracy of the Big Ideas Math Grade 8 Solutions as they are given after extensive research. Start Practicing the Chapterwise BIM 8th Grade Answers and no longer feel the concepts of Big Ideas Math Grade 8th difficult.
Teaching Grade 8 Math Topics effectively will help your kids to advance their math reasoning and logical ability. Confidence and ability to learn will be improved by referring to the 8th Standard Math Topics available. Thus, they will be prepared for high school studies. If you want to learn more about the 8th Grade Math Concepts have an insight into the below topics and get an idea of what is included in the 8th Grade Curriculum.
All you have to do is simply tap on the quick links available to avail the respective topics and get a grip on them. We included both the theoretical part as well as worksheets for your practice. Our 8th Grade Math Worksheets make it easy for you to test your preparation standard on the corresponding topics. Identify the knowledge gap and improvise on the topics you are facing difficulty with.
Ratio and Proportion – Worksheets
• Worksheet on Ratio and Proportion
Significant Figures – Worksheets
• Worksheet on Significant Figures
Square – Worksheets
• Worksheet on Squares
Square Root- Worksheets
• Worksheet on Square Root using Prime Factorization Method
• Worksheet on Square Root using Long Division Method
• Worksheet on Square Root of Numbers in Decimal and Fraction Form
Algebraic Expression
• Algebraic Expression
• Subtraction of Algebraic Expressions
• Multiplication of Algebraic Expression
• Division of Algebraic Expressions
Formula – Worksheets
• Worksheet on Framing the Formula
• Worksheet on Changing the Subject of a Formula
• Worksheet on Changing the Subject in an Equation or Formula
Factorization – Worksheet
• Worksheet on Factoring Algebraic Expression
• Worksheet on Factoring Binomials
• Worksheet on Factoring by Grouping
• Worksheet on Factorization by Regrouping
• Worksheet on Factoring out a Common Binomial Factor
• Worksheet on Factoring Identities
• Worksheet on Factoring the Differences of Two Squares
• Worksheet on Factorization using Formula
• Worksheet on Factoring Trinomials
• Worksheet on Factoring Simple Quadratics
• Worksheet on Factoring Quadratic Trinomials
• Worksheet on Factoring Trinomials by Substitution
• Worksheet on Factorization
Equations – Worksheets
• Worksheet on Linear Equations
• Worksheet on Word Problems on Linear Equation
Simultaneous Linear Equations – Worksheets
• Worksheet on Simultaneous Linear Equations
• Worksheet on Problems on Simultaneous Linear Equations
Linear Inequations – Worksheets
• Worksheet on Linear Inequations
Percentage
• Fraction into Percentage
• Percentage into Fraction
• Percentage into Ratio
• Ratio into Percentage
• Percentage into Decimal
• Decimal into Percentage
• Percentage of the given Quantity
• How much Percentage One Quantity is of Another?
• Percentage of a Number
• Increase Percentage
• Decrease Percentage
• Basic Problems on Percentage
• Solved Examples on Percentage
• Problems on Percentage
• Real-Life Problems on Percentage
• Word Problems on Percentage
• Application of Percentage
Percentage – Worksheets
• Worksheet on Fraction into Percentage
• Worksheet on Percentage into Fraction
• Worksheet on Percentage into Ratio
• Worksheet on Ratio into Percentage
• Worksheet on Percentage into Decimal
• Worksheet on Percentage of a Number
• Worksheet on Finding Percent
• Worksheet on Finding Value of a Percentage
• Worksheet on Percentage of a Given Quantity
• Worksheet on Word Problems on Percentage
• Worksheet on Increase Percentage
• Worksheet on Decrease Percentage
• Worksheet on increase and Decrease Percentage
• Worksheet on Expressing Percent
• Worksheet on Percent Problems
• Worksheet on Finding Percentage
Profit, Loss, and Discount – Worksheets
• Worksheet to Find Profit and Loss
• Worksheets on Profit and Loss Percentage
• Worksheet on Gain and Loss Percentage
• Worksheet on Discounts
Time and Work
• Time and Work
• Pipes and Cistern
• Practice Test on Time and Work
Time and Work – Worksheets
• Worksheet on Time and Work
Time and Distance – Worksheets
• Worksheet on Conversion of Units of Speed
• Worksheet on Calculating Time
• Worksheet on Calculating Speed
• Worksheet on Calculating Distance
• Worksheet on Train Passes through a Pole
• Worksheet on Train Passes through a Bridge
• Worksheet on Relative Speed
• Worksheet on Decimal into Percentage
Simple Interest
• What is Simple Interest?
• Calculate Simple Interest
• Practice Test on Simple Interest
Simple Interest – Worksheets
• Simple Interest Worksheet
Compound Interest
• Compound Interest
• Compound Interest with Growing Principal
• Compound Interest with Periodic Deductions
• Compound Interest by Using Formula
• Compound Interest when Interest is Compounded Yearly
• Compound Interest when Interest is Compounded Half-Yearly
• Compound Interest when Interest is Compounded Quarterly
• Problems on Compound Interest
• Variable Rate of Compound Interest
• Difference of Compound Interest and Simple Interest
• Practice Test on Compound Interest
• Uniform Rate of Growth
• Uniform Rate of Depreciation
• Uniform Rate of Growth and Depreciation
Compound Interest – Worksheet
• Worksheet on Compound Interest
• Worksheet on Compound Interest when Interest is Compounded Half-Yearly
• Worksheet on Compound Interest with Growing Principal
• Worksheet on Compound Interest with Periodic Deductions
• Worksheet on Variable Rate of Compound Interest
• Worksheet on Difference of Compound Interest and Simple Interest
• Worksheet on Uniform Rate of Growth
• Worksheet on Uniform Rate of Depreciation
• Worksheet on Uniform Rate of Growth and Depreciation
Ratio and Proportion (Direct & Inverse Variation)
• Direct Variation
• Inverse Variation
• Practice Test on Direct Variation and Inverse Variation
Ratio and Proportion – Worksheets
• Worksheet on Direct Variation
• Worksheet on Inverse Variation
• Probability
Probability – Worksheets
• Worksheet on Probability
Geometry
Parallelogram – Worksheet
• Worksheet on Parallelogram
• Worksheet on Construction on Quadrilateral
• Worksheet on Different Types of Quadrilaterals
Three-Dimensional Figures – Worksheets
• Worksheet on Three Dimensional Figures
Mensuration – Worksheets
• Worksheet on Area and Perimeter of Rectangles
• Worksheet on Area and Perimeter of Squares
• Worksheet on Area of the Path
• Worksheet on Circumference and Area of Circle
• Worksheet on Area and Perimeter of Triangle
Area of a Trapezium – Worksheet
• Worksheet on Trapezium
• Worksheet on Area of a Polygon
Area Proposition – Worksheets
• Worksheet on Same Base and Same Parallels
Data Handling
• Data Handling
• Frequency Distribution
• Grouping of Data
Data Handling – Worksheet
• Worksheet on Data Handling
Constructing and Interpreting Bar Graphs or Column Graphs
• Bar Graph or Column Graph
Bar Graphs or Column Graphs – Worksheets
• Worksheet on Bar Graphs or Column Graphs
• Pie Chart
Pie Charts or Pie Graphs – Worksheets
• Worksheet on Pie Chart
Coordinate Geometry
• Coordinate Graph
• Ordered pair of a Coordinate System
• Plot Ordered Pairs
• Coordinates of a Point
• Signs of Coordinates
• Find the Coordinates of a Point
• Coordinates of a Point in a Plane
• Plot Points on Co-ordinate Graph
• Graph of Linear Equation
• Simultaneous Equations Graphically
• Graphs of Simple Function
• Graph of Perimeter vs. Length of the Side of a Square
• Graph of Area vs. Side of a Square
• Graph of Simple Interest vs. Number of Years
• Graph of Distance vs. Time
Coordinate Geometry – Worksheet
• Worksheet on Coordinate Graph
Set Theory
• Sets Theory
• Representation of a Set
• Types of Sets
• Finite Sets and Infinite Sets
• Power Set
• Problems on Union of Sets
• Problems on Intersection of Sets
• Difference of two Sets
• Complement of a Set
• Problems on Complement of a Set
• Problems on Operation on Sets
• Word Problems on Sets
• Venn Diagrams in Different Situations
• Relationship in Sets using Venn Diagram
• Union of Sets using Venn Diagram
• Intersection of Sets using Venn Diagram
• Disjoint of Sets using Venn Diagram
• Difference of Sets using Venn Diagram
• Examples on Venn Diagram
Relations and Mapping
• Ordered Pair
• Cartesian Product of Two Sets
• Relation
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Domain Co-domain and Range of Function
• Math Practice Test on Function
Cube and Cube Roots
• Cube
• To Find if the Given Number is a Perfect Cube
• Cube Root
• Method for Finding the Cube of a Two-Digit Number
• Table of Cube Roots
Cube and Cube Roots – Worksheets
• Worksheet on Cube
• Worksheet on Cube and Cube Root
• Worksheet on Cube Root
Rational Numbers
• Introduction of Rational Numbers
• What are Rational Numbers?
• Is Every Rational Number a Natural Number?
• Is Zero a Rational Number?
• Is Every Rational Number an Integer?
• Is Every Rational Number a Fraction?
• Positive Rational Number
• Negative Rational Number
• Equivalent Rational Numbers
• Equivalent form of Rational Numbers
• Rational Number in Different Forms
• Properties of Rational Numbers
• Lowest form of a Rational Number
• Standard form of a Rational Number
• Equality of Rational Numbers using Standard Form
• Equality of Rational Numbers with Common Denominator
• Equality of Rational Numbers using Cross Multiplication
• Comparison of Rational Numbers
• Rational Numbers in Ascending Order
• Rational Numbers in Descending Order
• Representation of Rational Numbers on the Number Line
• Rational Numbers on the Number Line
• Addition of Rational Number with Same Denominator
• Addition of Rational Number with Different Denominator
• Properties of Addition of Rational Numbers
• Subtraction of Rational Number with Same Denominator
• Subtraction of Rational Number with Different Denominator
• Subtraction of Rational Numbers
• Properties of Subtraction of Rational Numbers
• Rational Expressions Involving Addition and Subtraction
• Simplify Rational Expressions Involving the Sum or Difference
• Multiplication of Rational Numbers
• Product of Rational Numbers
• Properties of Multiplication of Rational Numbers
• Rational Expressions Involving Addition, Subtraction, and Multiplication
• Reciprocal of a Rational Number
• Division of Rational Numbers
• Rational Expressions Involving Division
• Properties of Division of Rational Numbers
• Rational Numbers between Two Rational Numbers
• To Find Rational Numbers
Rational Numbers – Worksheets
• Worksheet on Rational Numbers
• Worksheet on Equivalent Rational Numbers
• Worksheet on Lowest form of a Rational Number
• Worksheet on Standard form of a Rational Number
• Worksheet on Equality of Rational Numbers
• Worksheet on Comparison of Rational Numbers
• Worksheet on Representation of Rational Number on a Number Line
• Worksheet on Adding Rational Numbers
• Worksheet on Properties of Addition of Rational Numbers
• Worksheet on Subtracting Rational Numbers
• Worksheet on Addition and Subtraction of Rational Number
• Worksheet on Rational Expressions Involving Sum and Difference
• Worksheet on Multiplication of Rational Number
• Worksheet on Properties of Multiplication of Rational Numbers
• Worksheet on Division of Rational Numbers
• Worksheet on Properties of Division of Rational Numbers
• Worksheet on Finding Rational Numbers between Two Rational Numbers
• Worksheet on Word Problems on Rational Numbers
• Worksheet on Operations on Rational Expressions
• Objective Questions on Rational Numbers
Fun with Numbers
• Playing with Numbers
• Test of Divisibility
• Number Puzzles and Games
Fun with Numbers – Worksheets
• Worksheet on Number Puzzles and Games
Exponents
• Exponents
• Laws of Exponents
• Rational Exponent
• Integral Exponents of a Rational Numbers
• Solved Examples on Exponents
• Practice Test on Exponents
Exponents – Worksheets
• Worksheet on Exponents
Simplification of Algebraic Fractions
• Algebraic Fractions
• Arithmetic Fraction and Algebraic Fraction
• Highest Common Factor of Monomials
• Highest Common Factor of Monomials by Factorization
• Lowest Common Multiple of Monomials
• Lowest Common Multiple of Monomials by Factorization
• Highest Common Factor of Polynomials
• Lowest Common Multiple of Polynomials
• H.C.F. of Polynomials by Factorization
• Highest Common Factor of Polynomials by Factorization
• L.C.M. of Polynomials by Factorization
• Lowest common multiple of Polynomials by Factorization
• H.C.F. of Polynomials by Division Method
• H.C.F. of Polynomials by Long Division Method
• Relation Between H.C.F. and L.C.M. of Two Polynomials
• Reduce Algebraic Fractions to its Lowest Term
• Simplification of Algebraic Fractions
• Rule of Separation of Division
• Sum and Difference of Algebraic Fractions
• Problems on Algebraic Fractions
• Solving Algebraic Fractions
• Multiplication of Algebraic Fractions
• Division of Algebraic Fractions
Simplification of Algebraic Fractions – Worksheets
• Worksheet on H.C.F. of Monomials
• Worksheet on L.C.M. of Monomials
• Worksheet on H.C.F. and L.C.M. of Monomials
• Worksheet on H.C.F. of Polynomials
• Worksheet on L.C.M. of Polynomials
• Worksheet on H.C.F. and L.C.M. of Polynomials
• Worksheet on Reducing Algebraic Fractions
• Worksheet on Algebraic Expressions to the Lowest Terms
• Worksheet on Algebraic Fractions
• Worksheet on Simplifying Algebraic Fractions
We know the Concepts learned in Grade 8 go a long way in higher grades concepts. Grade 8 Children are advised to practice these concepts regularly and get a good hold of them. Higher-Order Concepts such as Rational Numbers, Irrational Numbers, Exponents, etc. are all introduced in the 8th Standard.
8th Grade Math Topics covered here help you to tackle any kind of Math Problem easily. Grasp the Formulas and Techniques needed for learning 8th Standard Math Concepts in no time. Improve your mathematical abilities by practicing from our 8th Grade Math Problems provided.
Below is the general list of math objectives that 8th Grade Students should attain
• Identify Rational and Irrational Numbers
• Calculate and Approximate Principal Square Roots
• Solve Linear Inequalities in Two Variables.
• Differentiate between Types of Sampling Techniques.
• Determine Mean, Median, Mode, and Range of a Set of Real World Data.
• Why Choose our 8th Grade Math Curriculum? | 3,230 | 15,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.942282 |
https://developer.apple.com/documentation/accelerate/1513106-cblas_ztbsv | 1,579,628,560,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00472.warc.gz | 416,329,118 | 11,004 | Function
# cblas_ztbsv(_:_:_:_:_:_:_:_:_:_:)
Solves a triangular banded system of equations.
## Parameters
`Order`
Specifies row-major (C) or column-major (Fortran) data ordering.
`Uplo`
Specifies whether to use the upper or lower triangle from the matrix. Valid values are `'U'` or `'L'`.
`TransA`
Specifies whether to use matrix A (`'N'` or `'n'`) or the transpose of A (`'T'`, `'t'`, `'C'`, or `'c'`).
`Diag`
Specifies whether the matrix is unit triangular. Possible values are `'U'` (unit triangular) or `'N'` (not unit triangular).
`N`
Order of matrix `A`.
`K`
Number of superdiagonals or subdiagonals of matrix `A` (depending on the value of `Uplo`).
`A`
Matrix `A`.
`lda`
The leading dimension of matrix `A`.
`X`
Contains vector `B` on entry. Overwritten with vector `X` on return.
`incX`
Stride within `X`. For example, if `incX` is 7, every 7th element is used.
## Discussion
Solves the system of equations `A*X=B` or `A'*X=B`, depending on the value of `TransA`.
### Double-Precision Complex Matrix Functions
`func cblas_dzasum(Int32, UnsafeRawPointer!, Int32) -> Double`
Computes the sum of the absolute values of real and imaginary parts of elements in a vector (single-precision complex).
`func cblas_dznrm2(Int32, UnsafeRawPointer!, Int32) -> Double`
Computes the unitary norm of a vector (double-precision complex).
`func cblas_zdscal(Int32, Double, UnsafeMutableRawPointer!, Int32)`
Multiplies each element of a vector by a constant (double-precision complex).
`func cblas_zher(CBLAS_ORDER, CBLAS_UPLO, Int32, Double, UnsafeRawPointer!, Int32, UnsafeMutableRawPointer!, Int32)`
Adds the product of a scaling factor, vector `X`, and the conjugate transpose of `X` to matrix `A`.
`func cblas_zher2(CBLAS_ORDER, CBLAS_UPLO, Int32, UnsafeRawPointer!, UnsafeRawPointer!, Int32, UnsafeRawPointer!, Int32, UnsafeMutableRawPointer!, Int32)`
Hermitian rank 2 update: adds the product of a scaling factor, vector `X`, and the conjugate transpose of vector `Y` to the product of the conjugate of the scaling factor, vector `Y`, and the conjugate transpose of vector `X`, and adds the result to matrix `A`.
`func cblas_zhpr(CBLAS_ORDER, CBLAS_UPLO, Int32, Double, UnsafeRawPointer!, Int32, UnsafeMutableRawPointer!)`
Scales and multiplies a vector times its conjugate transpose, then adds a matrix.
`func cblas_zhpr2(CBLAS_ORDER, CBLAS_UPLO, Int32, UnsafeRawPointer!, UnsafeRawPointer!, Int32, UnsafeRawPointer!, Int32, UnsafeMutableRawPointer!)`
Multiplies a vector times the conjugate transpose of a second vector and vice-versa, sums the results, and adds a matrix.
`func cblas_zscal(Int32, UnsafeRawPointer!, UnsafeMutableRawPointer!, Int32)`
Multiplies each element of a vector by a constant (double-precision complex). | 764 | 2,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-05 | longest | en | 0.663936 |
http://mathhelpforum.com/differential-geometry/172544-prove-lemma-norms-function.html | 1,513,630,809,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948623785.97/warc/CC-MAIN-20171218200208-20171218222208-00305.warc.gz | 172,844,662 | 12,253 | # Thread: Prove Lemma on the norms of a function.
1. ## Prove Lemma on the norms of a function.
Hi.
Problem:
(i) Suppose that the real-valued weight function $w$ is defined, continuous, positive and integrable on the interval $(a,b)$. Then, for any function $f\in C[a,b]$,
$||f||_2 \leq W||f||_{\infty}$,
where
$W=\bigg[\int^b_aw(x)dx\bigg]^{1/2}$.
(ii) Given any two positive numbers $\epsilon$ (however small) and $M$ (however large), there exists a function $f\in C[a,b]$ such that
$||f||_2 < \epsilon$,
$||f||_{\infty}>M$.
Attempt:
(i) By using integration by parts I get,
\begin{aligned}
\int^b_a|f(x)|^2w(x)dx=&\bigg[|f(x)|^2\int w(x)dx\bigg]^b_a-2\int^b_a|f(x)|f'(x)\bigg(\int w(x)dx\bigg)dx \\
\leq& \bigg[|f(x)|^2\int w(x)dx\bigg]^b_a \\
\leq& \bigg(\max_{x\in [a,b]}|f(x)|\bigg)^2\int^b_a w(x)dx \\
\Rightarrow \bigg(\int^b_a|f(x)|^2w(x)dx\bigg)^{1/2}\leq& \max_{x\in [a,b]}|f(x)| \bigg(\int^b_a w(x)dx\bigg)^{1/2} \\
\Rightarrow ||f||_2 \leq& W||f||_{\infty}
\end{aligned}
To me that looks good, but I've been mistaken before
(ii) I do not see how I can prove existence of the function. This somehow reminds me of the dirac delta function, as I can make my function as pointy as I want...
The book continues by saying that the Lemma indicates that the norms are not equivalent as they would be if this was a finite-dimensional linear space and we were dealing with vectors. That is if $||*||_{\infty}$ and $||*||_2$ are vector norms on $\mathbb{R}^n$ then
$n^{-1/2}||v||_{\infty}\leq ||v||_2 \leq n^{1/2}||v||_{\infty}, \forall v\in\mathbb{R}^n.$
Does this mean that if we had $W^{-1}||f||_{\infty} \leq ||f||_2$, then we could use the two norms interchangeably? I do not immediately see how the Lemma indicates that this is not so.
Thanks!
2. No, I don't think your approach to (i) works. You have not assumed that f is differentiable. However, f is a continuous function on a closed interval. What do you know about such functions? Incidentally, I think there's an error in the problem statement. It should have
$\displaystyle W=\left(\int_{a}^{b}w^{2}(x)\,dx\right)^{1/2}.$
As for (ii), for the first inequality, I would experiment with (constant) functions like
$f(x)=\dfrac{\epsilon}{2(b-a)}.$
You may have to take square roots, or squares, or something to get that to work with the 2-norm.
For (ii) on the second inequality, again, you're dealing with continuous functions on closed intervals. Just think about a constant function. Is there a constant function that might satisfy that inequality?
3. Originally Posted by Mollier
(i) Suppose that the real-valued weight function $w$ is defined, continuous, positive and integrable on the interval $(a,b)$. Then, for any function $f\in C[a,b]$,
$||f||_2 \leq W||f||_{\infty}$,
where
$W=\bigg[\int^b_aw(x)dx\bigg]^{1/2}$.
Attempt:
(i) By using integration by parts I get,
\begin{aligned}
\int^b_a|f(x)|^2w(x)dx=&\bigg[|f(x)|^2\int w(x)dx\bigg]^b_a-2\int^b_a|f(x)|f'(x)\bigg(\int w(x)dx\bigg)dx \\
\leq& \bigg[|f(x)|^2\int w(x)dx\bigg]^b_a \\
\leq& \bigg(\max_{x\in [a,b]}|f(x)|\bigg)^2\int^b_a w(x)dx \\
\Rightarrow \bigg(\int^b_a|f(x)|^2w(x)dx\bigg)^{1/2}\leq& \max_{x\in [a,b]}|f(x)| \bigg(\int^b_a w(x)dx\bigg)^{1/2} \\
\Rightarrow ||f||_2 \leq& W||f||_{\infty}
\end{aligned}
To me that looks good, but I've been mistaken before
You are making this too complicated. All you need is to use the fact that $|f(x)|\leqslant\|f\|_\infty$ for all x in [a,b]. The weighted L_2-norm is then given by
$\displaystyle \|f\|_2^2 = \int^b_a|f(x)|^2w(x)\,dx \leqslant \int^b_a\|f(x)\|_\infty^2w(x)\,dx=\|f(x)\|_\infty^ 2W^2.$
Now take square roots.
Originally Posted by Mollier
(ii) Given any two positive numbers $\epsilon$ (however small) and $M$ (however large), there exists a function $f\in C[a,b]$ such that
$||f||_2 < \epsilon$,
$||f||_{\infty}>M$.
...
(ii) I do not see how I can prove existence of the function. This somehow reminds me of the dirac delta function, as I can make my function as pointy as I want...
That is correct. It is easiest to construct such a function on the unit interval, for example $f(x) = n^kx^n$, where n is large, and $n^k$ is some suitable power of n, probably $n^{1/2}$ or something like that. Once you have found a function that works on the unit interval, you can translate it to the interval [a,b], to get $f(x) = n^k\bigl(\frac{x-a}{b-a}\bigr)^n$
4. I do tend to overcomplicate, which I guess is fine if the result is correct...
Thanks. | 1,564 | 4,474 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 35, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-51 | longest | en | 0.841639 |
https://dietoad.org/poj-solution-3158/ | 1,726,682,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00615.warc.gz | 186,513,279 | 10,010 | # Poj Solution 3158
```http://poj.org/problem?id=3158
//* @author mekarlos@gmail.com
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
String s1,s2,s;
int[] m;
int k,v1,v2;
boolean band;
while(scan.hasNext()){
s1=scan.next();
s2=scan.next();
s=s1;
for(int i=0;i< s2.length();i++)s+="1";
k=-1;
for(int i=0;i<=s.length();i++){
band=true;
for(int j=0;j< s2.length();j++){
if((Integer.parseInt(s2.charAt(j)+"")+Integer.parseInt(s.charAt(i+j)+""))>3)
{
band=false;break;
}
}
if(band){k=i;break;}
}
v1=Math.max(s1.length(),k+s2.length());
s=s2;
for(int i=0;i< s1.length();i++)s+="1";
k=-1;
for(int i=0;i<=s.length();i++){
band=true;
for(int j=0;j< s1.length();j++){
if((Integer.parseInt(s1.charAt(j)+"")+Integer.parseInt(s.charAt(i+j)+""))>3)
{band=false;break;}
}
if(band){k=i;break;}
}
v2=Math.max(s2.length(),k+s1.length());
System.out.println(Math.min(v1, v2));
}
}
}
```
This entry was posted in poj. Bookmark the permalink. | 325 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.180316 |
https://gauravtiwari.org/best-statistical-mechanics-books/ | 1,721,269,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00395.warc.gz | 218,190,710 | 36,465 | # Best Statistical Mechanics Books for Physics Majors
Looking for recommendations on the best statistical mechanics textbooks? In this article, I have listed some of the best statistical mechanics books for Statistical Physics majors. In this list, I have collected only those books that are loved by both teachers and students. These books are considered the top textbooks in all the colleges and universities around the world.
## What is Statistical Mechanics/Statistical Physics?
If you are new to statistical physics or statistical mechanics, let me help you understand what exactly it deals with.
As the name suggests, statistical mechanics is more or less the statistical study of physics.
If you are looking for a technical definition, Statistical mechanics can be defined as the study that deals mostly with collecting and analysis of data in Physics.
It is mainly used to calculate probabilities, keep records, and provide conclusions in Physics. It helps us to understand the world a little better through numbers and other quantitative information.
Statistical methods are involved in carrying out a study that includes planning, analyzing, collecting data, designing, drawing meaningful interpretation, and reporting the research findings.
Statistical analysis is a subject that gives meaning to meaningless numbers, hence giving life to lifeless data. It is a valuable tool. It relates microscopic parameters to a system's macroscopic properties, which include a large number of colloidal particles and molecules.
Statistical mechanics rely heavily on the law of probability. It doesn't focus on the behavior of every single particle in a microscopic substance but the standard behavior of a more significant number of particles of the same kind.
## 10 Best Statistical Mechanics Books
If you are genuinely interested in learning statistical mechanics and all the essential theories behind them, then you are in the right place. I have listed some of the best books on statistical Mechanics based on readers' reviews and recommendations.
### Statistical mechanics by Kerson Huang
Unlike most textbooks on the subject, this textbook gives a lot of information about the theory of microscopic bodies that treat the modern theory of critical phenomena. This textbook also provides updated recent significant advances, including the classical kinetic theory of gases and the complete description of thermodynamics.
It includes topics covering fascinating applications such as the quantum hall effect, superfluids, and many other research applications. The last three chapters of this textbook are all about Landau-Wilson's approach to critical phenomena. Several new question answers and pictures have been added to this edition.
### A modern course in statistical physics by Linda E. Reichl
Exceeding regular textbook materials, A modern course in statistical physics focuses on the research in an introductory course on statistical Mechanics.
From the latest outcome in the spectral properties of the decay process to the universal nature of matter, this textbook highlights the probability theory and the theoretical framework obtained from thermodynamics underlying all the other theories in statistical physics.
This is a completely revised and updated Third Edition. It continues comprehensive coverage of special applications and many core topics, allowing the professors flexibility in designing particular individual courses.
This textbook includes extensive references and advanced topics, making this book an invaluable resource for students and researchers.
### Statistical physics by L D Landau
Statistical Physics is an excellent book of theoretical statistical physics. This book is brief and clearly to the point. It is written in a way that's easily understandable to readers. The book explains the fundamental concepts of statistical physics very clearly in a self-contained manner.
In this book, thermodynamics and statistical physics are presented in an easy and understandable way. This textbook includes all important topics like the fundamental principles of theoretical physics, ensembles, Ideal gases, Chemical reactions, surfaces, Gibbs/Boltzmann distribution, Fluctuations, Solids, and Non-ideal gases. It's an essential book to get knowledge of both the worlds, experimental and theoretical physics.
### Statistical mechanics by R.K. Pathria and Paul D. Beale
Statistical mechanics by R.K. Pathria mainly discusses the physical properties of matter based on the dynamic behavior of its Microscopic component.
This book comprises chapters on statistical mechanics of interacting systems, simple gases theory, thermodynamics, ideal bose Fermi systems, computer simulation, ensemble theory, and phase transitions. This volume contains new topics such as chemical equilibrium, degenerate Fermi gas behavior in ultracold atomic gases, and Bose-Einstein condensation.
This book also describes the fluctuation-dissipation theorem, the Clausius-Clapeyron equation, correlation functions and scattering, phase equilibrium, exact solutions of one-dimensional fluid models, and two-dimensional Ising model on a finite lattice.
Rest is all about the two new topics, 'The Monte Carlo and molecular dynamics Simulation' and 'The thermodynamics of the early universe.' This book is extremely useful to practitioners and students who are interested in statistical mechanics and physics.
### Statistical Physics of Particles by Mehran Kardar
Statistical Physics of particles appears to be a book replacing Pathria as the recommended basic Graduate textbook.
Based on Professor Kardar's works at MIT, this textbook introduces the central concepts and tools of statistical physics. It has a chapter on probability and issues related to it, such as information theory, the central limit theorem, and covers interacting particles, with a substantial explanation of Van Der Waals equation and its derivation, by mean-field approximation. This textbook also contains an integrated set of problems with detailed solutions.
### Thermal Physics by Kittel and Kroemer
Courses for Freshman or sophomore in thermodynamics or statistical mechanics written by Kroemer and Kittel offers a contemporary approach to thermal physics based on the aim that all physical systems can be characterized in terms of their distinct quantum states.
Rather than drawing on nineteenth-century classical mechanics concepts. This book immediately starts with statistical mechanics that give an extensive introduction to each of the three crucial heart-related topics.
This textbook includes chapters on, Temperature, Heat, Enthalpy, Entropy, Limitations to the First Law of Thermodynamics, the Third Law of Thermodynamics, the Second Law of Thermodynamics, Limitations to the First Law of Thermodynamics.
### Statistical Physics of Fields by Mehran Kardar
Based on the lectures taught by Professor Kardarat at MIT, this textbook explains how statistical field theories are formulated and studied. Exact solutions, Perturbation theory, renormalization groups.
Other tools are applied to explain the emergency of universality, the non-equilibrium Dynamics of interfaces, scale invariance, and directed paths in random media. The chapters at the beginning of this book connect the particulate perspective elaborated in the companion volume of the farinaceous statistical fields are deliberated here.
This textbook is perfect for advanced graduate courses in statistical physics. At the end of the book, there is also a combined set of selected problems with their solutions.
### Thermodynamics and Statistical Mechanics by Walter Greiner, Ludwig Neise, Horststöcker, D. Rischke
This textbook is one of the best among all other books written by Walter Greiner. It is full of insights about statistical mechanics and thermodynamics, easy to understand, and mathematically rigorous.
This textbook is an important source, particularly for those who teach introductory statistical physics at universities. The essential point of this textbook is that all results in statistical mechanics and equilibrium thermodynamics follow from one single non-provable axiom, namely elementary classical mechanics, elementary quantum mechanics, the principle of equal theoretical probabilities combined with elementary probability theory.
### Introduction to Modern Statistical Mechanics by David Chandler
This textbook briefly talks about the essential contemporary statistical ideas. David Candler has done a fantastic job introducing some of the most fundamental concepts of statistical mechanics understandably.
The author provides concise information on the importance of this branch of physics and discussion on many of its standard and simple applications.
This textbook is also interspersed with a microcomputer program and also over a hundred fifty exercises.
In this textbook, the top physical chemist, David Candler, took a new perspective to provide the initial-level work on the contemporary concept of Time correlation function, Monte Carlo simulation, liquid structure, and renormalization group theory.
### Statistical mechanics: Theory and Molecular Simulation by Mark E. Tuckerman
If you compare every other book on this subject, you will find Statistical Mechanics by Tuckerman as unique, interesting, and extremely valuable. One of the unique characteristics of this textbook is the way the computational and theoretical topics are intertwined, as equally treated fundamentally.
The ultimate goal of this textbook is to assemble growing users to become active participants in this exciting and fast-advancing research area for the first time. This textbook includes fundamental concepts of time-dependent statistical mechanics and equilibrium, with the contemporary methods used to resolve the complicated problems in real-world applications.
This textbook also contains an in-depth discussion of the most commonly used at the same time with contemporary computational techniques, such as Monte Carlo and dynamics, and complete analysis of classical and quantum mechanics.
There are important topics, including linear-response theory, critical phenomena and free-energy calculations, advanced conformational sampling methods and generalized Langevin equation, and the harmonic baths. This textbook will be a handy reference tool for experienced practitioners.
Get this book on Amazon.in / Get this book on Amazon.com
## Conclusion
To conclude, here are the top 10 books on statistical physics and mechanics:
Among the statistical mechanics books suggested above, most, if not all, but many are best for those who enrolled in graduate programs in physics. These books will help you to understand statistics in a more straightforward and practical way.
Last update on 2024-07-18 using Amazon Product Advertising API. | 1,942 | 10,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.932203 |
http://irodovsolutionsmechanics.blogspot.com/2008/07/irodov-problem-1200.html | 1,532,229,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00150.warc.gz | 182,605,797 | 11,293 | ## Wednesday, July 2, 2008
### Irodov Problem 1.200
Let the mass of sun be Ms and the radius of orbit of the planet be r. The force of gravity acting on the planet is given by . This gravitational force provide for the centripetal acceleration of the planet and so we have,
The total distance traveled by the planet in one rotation is the circumference of the circle and so the time taken for one rotation is given by,
. Note that in Irodov the answer seems like the period of rotation depends on the mass of the planet M, but that false, it will depend on the mass of the central mass (sun in this case) and not on the mass of the planet. | 151 | 642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-30 | latest | en | 0.873186 |
https://www.physicsforums.com/threads/partial-fractions.278860/ | 1,527,496,272,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872114.89/warc/CC-MAIN-20180528072218-20180528092218-00439.warc.gz | 817,852,480 | 15,028 | # Homework Help: Partial fractions
1. Dec 11, 2008
### leopard
$$\frac{s-1}{s(s-2)^2}$$
How can I expand this fraction?
$$\frac{A}{s} + \frac{B}{(s-2)} + \frac{C}{(s-2)^2}$$
right?
This gives me the equation
$$As^3 - 6As^2 + 12As - 8A Bs^3 - 4Bs^2 + 4Bs + Cs^2 - 2Cs = s-1$$
so that
(1) A + B =0
(2)- 6A - 4B + C = 0
(3) 12A + 4B - 2C = 1
(4) -8A = -1
(4) gives A = 1/8
(1) gives B = -1/8
(2) gives C = 1/4
A = -1/4
B = 1/4
C = 2/4
What's wrong?
2. Dec 11, 2008
### gabbagabbahey
So far so good ...
3. Dec 11, 2008
### leopard
I figured it out, thanks!
4. Dec 11, 2008
### HallsofIvy
leopard, for future use, you may find it easier (and less error prone) to do this:
to find A, B, C so that
$$\frac{A}{s}+ \frac{B}{s-2}+ \frac{C}{(x-2)^2}= \frac{s-1}{s(s-2)^2}$$
Multiply both sides by the denominator, s(s-2)2, leaving it as
$$A(s-2)^2+ Bs(s-2)+ Cs= s- 1$$
Now, since this must be true for all s, select simple values of s:
if s= 0, 4A= -1
if s= 2, 2C= 1
if s= 1, A- B+ C= 0 | 479 | 998 | {"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} | 4.1875 | 4 | CC-MAIN-2018-22 | longest | en | 0.662635 |
http://www.slideserve.com/brad/classical-decomposition | 1,506,347,356,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691830.33/warc/CC-MAIN-20170925130433-20170925150433-00643.warc.gz | 577,292,627 | 17,151 | 1 / 37
# Classical Decomposition - PowerPoint PPT Presentation
Classical Decomposition. Boise State University By: Kurt Folke Spring 2003. Overview:. Time series models & classical decomposition Brainstorming exercise Classical decomposition explained Classical decomposition illustration Exercise Summary Bibliography & readings list
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### Classical Decomposition
Boise State University
By: Kurt Folke
Spring 2003
• Time series models & classical decomposition
• Brainstorming exercise
• Classical decomposition explained
• Classical decomposition illustration
• Exercise
• Summary
• Appendix A: exercise templates
• Time series models are sequences of data that follow non-random orders
• Examples of time series data:
• Sales
• Costs
• Time series models are composed of trend, seasonal, cyclical, and random influences
• Decomposition time series models:
• Multiplicative: Y = T x C x S x e
• Additive: Y = T + C + S + e
• T = Trend component
• C = Cyclical component
• S = Seasonal component
• e = Error or random component
• Classical decomposition is used to isolate trend, seasonal, and other variability components from a time series model
• Benefits:
• Shows fluctuations in trend
• Provides insight to underlying factors affecting the time series
• Identify how this tool can be used in your organization…
Basic Steps:
• Determine seasonal indexes using the ratio to moving average method
• Deseasonalize the data
• Develop the trend-cyclical regression equation using deseasonalized data
• Multiply the forecasted trend values by their seasonal indexes to create a more accurate forecast
• Determine seasonal indexes
Y = TCSe
• Equate…
Se = (Y/TC)
• To find seasonal indexes, first estimate trend-cyclical components
Se = (Y/TC)
• Use centered moving average
• Called ratio to moving average method
• For quarterly data, use four-quarter moving average
• Averages seasonal influences
Example
• Four-quarter moving average will position average at…
• end of second period and
• beginning of third period
• Use centered moving average to position data in middle of the period
• Example
• Find seasonal-error components by dividing original data by trend-cyclical components
Se = (Y/TC)
• Se = Seasonal-error components
• Y = Original data value
• TC = Trend-cyclical components
(centered moving average value)
Example
• Unadjusted seasonal indexes (USI) are found by averagingseasonal-error components by period
• Develop adjusting factor (AF) so USIs are adjusted so their sum equals the number of quarters (4)
• Reduces error
Example
Example
• Adjusted seasonal indexes (ASI) are derived by multiplying the unadjusted seasonal index by the adjusting factor
ASI = USI x AF
• ASI = Adjusted seasonal index
• USI = Unadjusted seasonal index
Example
• Deseasonalized data is produced by dividing the original data values by their seasonal indexes
(Y/S) = TCe
• Y/S = Deseasonalized data
• TCe = Trend-cyclical-error component
Example
• Develop the trend-cyclical regression equation using deseasonalized data
Tt = a + bt
• Tt = Trend value at period t
• a = Intercept value
• b = Slope of trend line
Example
• Use trend-cyclical regression equation to develop trend data
• Create forecasted data by multiplying the trend data values by their seasonal indexes
• More accurate forecast
Example
Example
Summarized Steps:
• Determine seasonal indexes
• Deseasonalize the data
• Develop the trend-cyclical regression equation
• Create forecast using trend data and seasonal indexes
Classical Decomposition:Illustration
• Gem Company’s operations department has been asked to deseasonalize and forecast sales for the next four quarters of the coming year
• The Company has compiled its past sales data in Table 1
• An illustration using classical decomposition will follow
• (a) Compute the four-quarter simple moving average
Ex: simple MA at end of Qtr 2 and beginning of Qtr 3
(55+47+65+70)/4 = 59.25
Explain
Classical DecompositionIllustration: Step 1
• (b) Compute the two-quarter centered moving average
Ex: centered MA at middle of Qtr 3
(59.25+61.25)/2
= 60.500
Explain
Classical Decomposition Illustration: Step 1
• (c) Compute the seasonal-error component (percent MA)
Ex: percent MA at Qtr 3
(65/60.500)
= 1.074
Explain
Classical DecompositionIllustration: Step 1
• (d) Compute the unadjusted seasonal index using the seasonal-error components from Table 2
Ex (Qtr 1): [(Yr 2, Qtr 1) + (Yr 3, Qtr 1) + (Yr 4, Qtr 1)]/3
= [0.989+0.914+0.926]/3 = 0.943
Explain
Classical DecompositionIllustration: Step 1
• (e) Compute the adjusting factor by dividing the number of quarters (4) by the sum of all calculated unadjusted seasonal indexes
= 4.000/(0.943+0.851+1.080+1.130) = (4.000/4.004)
Explain
Classical DecompositionIllustration: Step 1
• (f) Compute the adjusted seasonal index by multiplying the unadjusted seasonal index by the adjusting factor
Ex (Qtr 1): 0.943 x (4.000/4.004) = 0.942
Explain
Classical DecompositionIllustration: Step 2
• Compute the deseasonalized sales by dividing original sales by the adjusted seasonal index
Ex (Yr 1, Qtr 1):
(55 / 0.942)
= 58.386
Explain
Classical DecompositionIllustration: Step 3
• Compute the trend-cyclical regression equation using simple linear regression
Tt = a + bt
t-bar = 8.5
T-bar = 69.6
b = 1.465
a = 57.180
Tt = 57.180 + 1.465t
Explain
Classical DecompositionIllustration: Step 4
• (a) Develop trend sales
Tt = 57.180 + 1.465t
Ex (Yr 1, Qtr 1):
T1 = 57.180 + 1.465(1) = 58.645
Explain
Classical DecompositionIllustration: Step 4
• (b) Forecast sales for each of the four quarters of the coming year
Ex (Yr 5, Qtr 1):
0.942 x 82.085
= 77.324
Explain
Classical DecompositionIllustration: Graphical Look
Classical Decomposition:Exercise
• Assume you have been asked by your boss to deseasonalize and forecast for the next four quarters of the coming year (Yr 5) this data pertaining to your company’s sales
• Use the steps and examples shown in the explanation and illustration as a reference
Basic Steps
Explanation
Illustration
Templates
• Time series models are sequences of data that follow non-arbitrary orders
• Classical decomposition isolates the components of a time series model
• Benefits:
• Insight to fluctuations in trend
• Decomposes the underlying factors affecting the time series
DeLurgio, Stephen, and Bhame, Carl. Forecasting Systems for Operations Management. Homewood: Business One Irwin, 1991.
Shim, Jae K. Strategic Business Forecasting. New York: St Lucie, 2000.
StatSoft Inc. (2003). Time Series Analysis. Retrieved April 21, 2003, from http://www.statsoft.com/textbook/sttimser.html
Appendix A:Exercise Templates
Appendix A:Exercise Templates
Appendix A:Exercise Templates
Appendix A:Exercise Templates
Appendix A:Exercise Templates | 1,900 | 7,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.814232 |
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[caption id="attachment_28677" align="alignright" width="480"] Fig. 1. Plot of cook time vs. temperature[/caption]A familiar technique used when collecting data is to graph the results. For instance, suppose you want to see how quickly the internal temperature of a roast of pork rises in a 250° F oven. Nine internal meat temperature measurements are taken over a period of an hour-and-a-half, or 180 minutes. Collecting the Data After 20 minutes, the internal temperature of our pork roast is 60° F. Twenty more minutes yields 95° F. At 60 minutes, the temperature is 118° F, whereas the temperature is 139° F after 80 minutes. At 100 minutes, we read 148° F. When two hours have passed, we obtain 156° F. 140 minutes of cooking puts us at 163° F, while 160 minutes gives…
## Cyclopentadecane the Simplest Catenane Interlocking Ring Structure?
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## Essay title: Circuit Analysis of Linear Network and one Nonlinear Element
Circuit Analysis of Linear Network and one Nonlinear Element
ETEE3153
Experiment #4
Submitted: September 5, 2005
By, David Scott
Lab partner: Blake Griffin
Prof. Jack Carter
Submitted: 10/27/2005
Microsoft Word
Main Body
The purpose of this laboratory experiment is to learn simple techniques for analyzing a circuit with a nonlinear element. It gives details as to what methods to uses when using nonlinear elements. This experiment will give us a way to get the characteristics of the circuit through simulation and actual lab measurements.
The lab begins with describing that most our time in network analysis deal with linear circuit. In order to analyze the circuit, we have to know the characteristic of the nonlinear element and the characteristic of the circuit at the points where nonlinear elements are connected. The characteristics of a device are described as voltage vs. current. The devise must be experimentally determined by either the manufacturer or determined experimentally by a user.
The characteristic of a resistor is a straight line, because voltage and current are proportional, through Ohms Law (V=IR or R=V/I). This equation shows that the slope is the value of the resistance which intercept at zero (line that passes through the origin). This information, along with a visual representation of it, is shown in the lab book with the visual representation shown in Graph 1 (Ambrose 4-1):
The DC characteristic of a network at two terminals is a DC load line which is a straight line in linear networks. The equation for the DC load line can be determined from two experimentally or calculated points. The two points are the open circuit voltage point (0, Voc) and the shout circuit current (Isc, 0) point. This information, along with a visual representation of it, is shown in the lab book with the visual representation shown in Graph 2 (Ambrose 4-2):
Once the two characteristics are know then the two curves provide the operating point (voltage and current).
I. PSPICE Results
PSPICE analysis was preformed before class to get an understanding of what was to be done in the lab. Figure 1 will be used throughout this laboratory experiment to show what happens when different loads are applied to a circuit with linear elements.
The equipment that was used in this part of the lab, are as follows as described in the lab book (Ambrose 4-2):
Power Supply: Variable voltage, variable current
Ammeter: 0-10 milliampere range
Volt Meter: 0-2.5 volt range
Resistors: 10, 50, 100, 500, 1000 ohm
Diode: 1N4148 PSPICE and 1N914 in lab
LED: Red LED in lab
Before class a PSPICE simulation was preformed on Figure 1. Below in the PSPICE program from the first simulations with 0ohms as the load (represented by 2Pica ohms):
R1 1 2 150
R2 2 0 300
R3 2 3 300
RL 3 0 2pohms
V1 1 0 DC 3.0V
.op
.DC V1 3 3 3
.print DC I(R1) I(R2) V(RL) V(V1) V(R2)
.probe
.end
PSPICE for 0Ohms
The next PSPICE program uses a 1000 ohm resistor as the load. This program can be seen below:
R1 1 2 150
R2 2 0 300
R3 2 3 300
RL 3 0 1000ohms
V1 1 0 DC 3.0V
.op
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https://chem.libretexts.org/Sandboxes/evlisitsyna_at_ualr.edu/Elena's_Book/Inorganic_Chemistry/6%3A_Ligands/6.02%3A_The_Chelate_Effect_(and_Macrocycle_Effect) | 1,701,520,088,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100399.81/warc/CC-MAIN-20231202105028-20231202135028-00528.warc.gz | 199,276,524 | 32,953 | # 6.02: The Chelate Effect (and Macrocycle Effect)
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## Introduction
Monodentate ligands bind through only one donor atom. Monodentate means "one-toothed". The halides, phosphines, ammonia and amines seen previously are monodentate ligands.
Bidentate ligands bind through two donor sites. Bidentate means "two-toothed". An example of a bidentate ligand is bis(dimethylphosphino)propane. It can bind to a metal via two donor atoms at once: it uses one lone pair on each phosphorus atom.
More examples of bidentate ligands are shown below. They all have at least two different atoms with lone pairs. In some cases, there are additional atoms with lone pairs, but only two of them are able to face the metal at one time. Oxalate and glycinate would act as bidentate donors, donating up to two sets of lone pairs at the same time.
Table CC $$\PageIndex{1}$$ Some common bidentate ligands
## The Chelate Effect
Chelating ligands have higher affinity for a metal ion than analogous monodentate ligands. The chelate effect is the enhanced affinity of a chelating ligand for a metal ion compared to its monodentate ligand counterpart(s). This term comes from the Greek chelos, meaning "crab". A crab does not have any teeth at all, but it does have two claws for tightly holding onto something. A very simple analogy is that, if you are holding something with two hands rather than one, you are not as likely to drop it. For example, ethylenediamine (en, H2NCH2CH2NH2) is a bidentate ligand that binds metal ions more strongly than monodentate amine ligands like ammonia (NH3) and methylamine (CH3NH2). Tridentate ligands, which bind through three donors, can bind even more tightly than bidentate, and so on.
Multidentate ligands bind more tightly because of the chelate effect
### Chemical reasoning for the Chelate Effect
The chelate effect can be explained using principles of thermodynamics. Recall that reactions are spontaneous when the Gibbs Free Energy change is negative $$-\Delta G$$; this is true when change in enthalpy is negative ($$-\Delta H$$) and the change in entropy is positive (disorder increases, $$+\Delta S$$. (From the equation $$\Delta G = \Delta H - T\Delta S$$.)
Consider the reaction shown below:
#### Enthalpy
In each the reactant Cu complex and product Cu-complex in Figure $$\PageIndex{4}$$, there are two N-Cu bonds. Electronically, the ammonia and en ligands are very similar, since both bind through N and since the Lewis base strengths of their nitrogen atoms are similar. The enthalpy change due to breaking two H3N-Cu binds and replacing them with two new N(en)-C bonds is almost zero. Thus, enthalpy is not a major driving factor in the chelate effect.
#### Entropy
In terms of entropy (disorder) there are two things to consider:
(1) The entropy from free rotation of the chelator. The chelator becomes somewhat constrained upon binding to the metal, and so this would result in small entropic penalty; loss in entropy. This is worth noting, but is a relatively small effect.
(2) The entropy from change in the number of molecules that can move freely. When a chelating ligand replaces several monodentate ligands, the result is an increase in the number of free molecules in the system, and a relatively large increase in entropy. This is the major energetic factor driving the chelate effect.
When a chelating ligand replaces monodentate ligands, there is a relatively large increase in entropy (+$$\Delta S$$). This is the primary driving factor for the Chelate Effect.
For example, when en replaces two ammonia ligands (Figure $$\PageIndex{4}$$), the number of total molecules increases from two to three. Increasing the number of molecules by just one is enough to drive the reaction forward.
The example above is gives a case when just one bidentate ligand is involved. When multiple bidentate ligands are involved, or when denticity increases, the chelate effect is enhanced further. Consider the two complexation equilibria in aqueous solution, between the cobalt (II) ion, Co2+(aq) and ethylenediamine (en) on the one hand and ammonia, NH3, on the other.
$[Co(H_2O)_6]^{2+} + 6 NH_3 \rightleftharpoons [Co(NH_3)_6]^{2+} + 6 H_2O \ (1)$
$[Co(H_2O)_6]^{2+} + 3 en \rightleftharpoons [Co(en)_3]^{2+} + 6 H_2O \ (2)$
This means that ΔH must be very similar for the two reactions, since six Co-N bonds are formed in each case. Interestingly however, we observe that the equilibrium constant is 100,000 times larger for the second reaction than it is for the first.
The big difference between these two reactions is that the second one involves "condensation" of fewer particles to make the complex. This means that the entropy changes for the two reactions are different. The first reaction has a ΔS value close to zero, because their is the same number of molecules on both sides of the equation. The second one has a positive ΔS° because four molecules come together but seven molecules are produced. The difference between them (ΔΔS) is about +100 J/mol-K. We can translate this into a ratio of equilibrium constants using:
$K_f(en)/K_f(NH_3) = e^{-\Delta \Delta G^\circ/RT} \approx e^{+\Delta \Delta S^\circ/R} \approx e^{12} \approx 10^5$
The bottom line is that the chelate effect is entropy-driven. It follows that the more binding groups a ligand contains, the more positive the ΔS° and the higher the Kf will be for complex formation. In this regard, the hexadentate ligand ethylenediamine tetraacetic acid (EDTA) is an optimal ligand for making octahedral complexes because it has six binding groups. In basic solutions where all four of the COOH groups are deprotonated, the chelate effect of the EDTA4- ligand is approximately 1015. This means, for a given metal ion, Kf is 1015 times larger for EDTA4- than it would be for the relevant monodentate ligands at the same concentration. EDTA4-tightly binds essentially any 2+, 3+, or 4+ ion in the periodic table, and is a very useful ligand for both analytical applications and separations.
Exercise $$\PageIndex{1}$$
Draw metal complexes using the ligands below, binding to Ni(2+) in a bidentate mode.
## The Macrocycle Effect
The macrocyclic effect follows the same principle as the chelate effect, but the effect is further enhanced by the cyclic conformation of the ligand. Macrocyclic ligands are not only multi-dentate, but because they are covalently constrained to their cyclic form, they allow less conformational freedom. The ligand is said to be "pre-organized" for binding, and there is little entropy penalty for wrapping it around the metal ion. For example heme b is a tetradentate cyclic ligand which is strongly complexes transition metal ions, including (in biological systems) Fe+2.
Figure 5.10.3: Heme b
Some other common cyclic ligands are shown below:
• Crown ethers such as 18-crown-6 (below center) are cyclic hard bases that can complex alkali metal cations. Crowns can selectively bind Li+, Na+, or K+ depending on the number of ethylene oxide units in the ring.
• The chelating properties of crown ethers are mimetic of the natural antibiotic valinomycin (below right), which selectively transports K+ ions across bacterial cell membranes, killing the bacterium by dissipating its membrane potential. Like crown ethers, valinomycin is a cyclic hard base. | 2,227 | 8,335 | {"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": 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} | 2.578125 | 3 | CC-MAIN-2023-50 | latest | en | 0.579991 |
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Magic Squares
Stage: 4 and 5
An account of some magic squares and their properties and and how to construct them for yourself.
Jam
Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
Mystic Rose
Stage: 4 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
Polycircles
Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon? | 2,149 | 9,245 | {"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} | 4.0625 | 4 | CC-MAIN-2017-43 | latest | en | 0.911847 |
http://azfoo.net/gdt/babs/numbers/n/number500000.html | 1,568,601,702,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572471.35/warc/CC-MAIN-20190916015552-20190916041552-00284.warc.gz | 23,099,967 | 2,035 | ### About the Number 500,000 (five hundred thousand )
According to TheMovieBit.com it took "500,000 polygons to create the Godzilla 3D model."
```MathBabbler Number Analyst (MBNA) output:
=========================================
500000 is a natural, whole, integer
500000 is even
500000 proper divisors are: 1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,
160,200,250,400,500,625,800,1000,1250,2000,
2500,3125,4000,5000,6250,10000,12500,15625,
20000,25000,31250,50000,62500,100000,125000,
250000,
500000 has 41 proper divisors
500000 is abundant (sum of divisors is 730453; ratio: 1.46091)
500000 is unhappy
500000 is Harshad (Niven) number
500000 is composite (not prime)
500000 has the prime factors: 2*2*2*2*2*5*5*5*5*5*5 (sum=40)
500000 is powerful (250^2 * 2^3)
500000 in octal is 01720440
500000 in hexadecimal is 0x7a120
500000 in binary is 1111010000100100000 (is odious)
500000 nearest square numbers: -151...1264 (499849...501264 [708])
sqrt(500000) = 707.107
ln(500000) = 13.1224
log(500000) = 5.69897
500000 reciprocal is .00000200000000000000000000000000
500000 is 159155 Pi years
500000 is 25000 scoreyears
500000 is 125001^2 - 124999^2 and 2 * (125001 + 124999)
```
TheMovieBit.com::Godzilla Ultimate Trivia
Creator: Gerald Thurman [gthurman@gmail.com]
Created: 03 May 2014
This work is licensed under a Creative Commons Attribution 3.0 United States License. | 487 | 1,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-39 | latest | en | 0.721593 |
http://board.flatassembler.net/topic.php?t=20946 | 1,568,881,365,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00332.warc.gz | 32,025,375 | 4,965 | flat assembler
Message board for the users of flat assembler.
flat assembler > Heap > Smallest Divide!
Author
rocketsoft
Joined: 26 Jan 2010
Posts: 189
Here U go...
01 Feb 2019, 21:28
vivik
Joined: 29 Oct 2016
Posts: 669
Code:
`use this: [code][/code] `
Code:
```#include <stdio.h>
using namespace std;
int q;
//standard 32 bit by 32 bit divide
//32 bit result!
//smallest divide?
void divide(int a,int b){
int n=1;q=0;loop:;a=a-b;
if (a<0) {a=a+b;n=n>>1;b=b>>1;
if (n==0) return;goto loop;}
q=q+n;n=n<<1;b=b<<1;goto loop;}
int main()
{
divide (1000000001,3);
printf ("The Quotient:%d\n",q);
return 0;
} ```
somebody unrape my eyes please (says the guy whose coding habbits are terrible)
Code:
```#include <stdio.h>
int q;
//standard 32 bit by 32 bit divide
//32 bit result!
//smallest divide?
void divide(int a, int b){
int n=1;
q=0;
while(true) {
a=a-b;
if (a<0) {
a=a+b;
n=n>>1;
b=b>>1;
if (n==0) return;
} else {
q=q+n;
n=n<<1;
b=b<<1;
}
}
}
int main()
{
divide (1000000001,3);
printf ("The Quotient: %d\n",q);
return 0;
} ```
02 Feb 2019, 09:41
bitRAKE
Joined: 21 Jul 2003
Posts: 2795
Location: dank orb
n is just a bit index and x86 supports bit strings. So,shift on b is needed, but nothing else. The general solution is to BSR to set initial shift on b, and value of n. Works for arbitrary size operands in O(Log2 a - Log2 b) time.
Assuming we couldn't use DIV/IDIV (locked out of using specific registers); small would be:
Code:
```; ECX <= EAX / EBX
or ecx,-1
.back: inc ecx
sub eax,ebx
jnc .back
retn ```
...not very useful.
02 Feb 2019, 19:26
rocketsoft
Joined: 26 Jan 2010
Posts: 189
I meant Small AND Fast... for implementing into a CPU!
12 Feb 2019, 03:31
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2012-04-02, 05:20 #1
princeps
Nov 2011
22×3 Posts
Proof of Primality Test for Fermat Numbers
Let $F_n$ be a Fermat number of the form :
$F_n=2^{2^n}+1$
Next , let's define sequence $S_i$ as :
$S_i=S^4_{i-1}-4\cdot S^2_{i-1}+2 ~ \text { with } ~ S_0=8$
Then :
$F_n ~; (n \geq 2) ~\text{ is a prime iff }~ F_n ~ \mid ~ S_{2^{n-1}-1$
Proof is attached . Any constructive comment is appreciated .
Attached Files
FermatProof.pdf (169.1 KB, 153 views)
2012-04-02, 06:01 #2 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·37·127 Posts Not too late for the April 1st! (at least in our TZ)
2012-04-02, 07:00 #3 rajula "Tapio Rajala" Feb 2010 Finland 13B16 Posts The proof was also posted on MO (and not on 1st of April).
2012-04-02, 07:24 #4 Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40
2012-04-02, 07:40 #5
princeps
Nov 2011
22·3 Posts
Quote:
Originally Posted by Dubslow Nice link. Going by the bottom comment on the only answer, the two tests are exactly the same, and the other test is from 1960, so... I guess you could see if your proof improves the original proof? (I don't have anywhere near the sort of knowledge to examine the proofs.)
The tests could be the same only in sense of computational complexity , although I think that this test should be faster than Inkeri's . Thank you anyway .
2012-04-02, 08:02 #6 Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40=0). Your test ends with i=2^(n-1)-1, while the other test ends with k=2^n-2=2i, so the sequences are identical. All your sequence does is the same as two iterationsof R, and then do half the iterations, but that's still exactly the same thing. Last fiddled with by Dubslow on 2012-04-02 at 08:05
2012-04-02, 08:59 #7 literka Mar 2010 26×3 Posts Sorry i did not notice the assumption n>=2 Last fiddled with by literka on 2012-04-02 at 09:05
2012-04-02, 09:03 #8
princeps
Nov 2011
148 Posts
Quote:
Originally Posted by literka I started to read this link and I am totally confused. On the page 2 there is equality 2(Fn-1)/2 = 1 (mod Fn) where Fn denotes Fermat number. Take n=1. Then Fn=F1=5, (Fn-1)/2 = 2, 2^2=4 and 4 is not equal 1 modulo 5. Am I missing something?
Yes , you have missed condition : $n \geq 2$
2012-04-02, 09:13 #9 rajula "Tapio Rajala" Feb 2010 Finland 4738 Posts As underlined by Dubslow, the most satisfactory answer one can give is already given in Emil Jeřábek's comment (behind the link I posted). The test is the same as Inkeri's. So, nothing new.
2012-04-02, 11:40 #10
literka
Mar 2010
26×3 Posts
Quote:
Originally Posted by princeps Yes , you have missed condition : $n \geq 2$
I cannot compare your work with work of Inkeri, which I do not know. I have few editorial remarks, which may be useful for you.
Everywhere you use sign of equivalence but in Lemma 2.1. you use sign of equality.
I would, in your place, specify the range of x, y, in Lemma 2.1., since you use Lemma 2.1. for non-integer elements.
2012-04-02, 12:31 #11
princeps
Nov 2011
148 Posts
Quote:
Originally Posted by literka I noticed this shortly after uploading post. I cannot compare your work with work of Inkeri, which I do not know. I have few editorial remarks, which may be useful for you. Everywhere you use sign of equivalence but in Lemma 2.1. you use sign of equality. I would, in your place, specify the range of x, y, in Lemma 2.1., since you use Lemma 2.1. for non-integer elements.
As far as I know Inkeri's proof isn't freely available...Thanks for your observations..
Similar Threads Thread Thread Starter Forum Replies Last Post Godzilla Miscellaneous Math 40 2018-10-17 00:11 Arkadiusz Math 6 2011-04-05 19:39 AntonVrba Math 96 2009-02-25 10:37 T.Rex Math 0 2004-10-26 21:37 T.Rex Math 2 2004-09-11 07:26
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Thu Apr 22 01:00:11 UTC 2021 up 13 days, 19:41, 0 users, load averages: 1.79, 2.42, 2.44 | 1,330 | 4,037 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 7, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-17 | latest | en | 0.869362 |
https://www.speedsolving.com/threads/orions-progression-thread-i-really-need-to-get-faster-at-square-1.89473/page-11 | 1,723,241,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00135.warc.gz | 775,123,910 | 20,627 | # Orion's progression thread (I really need to get faster at square-1)
## What should my 3x3 main be?
• Total voters
24
• Poll closed .
#### GeographicalCuber
##### Member
I use the mid-solve parity, you do it before PBL.
It's a shorter alg (7 slice I think)
Edit:
Here's the alg
/(3,3)/(1,0)/(4,-2)/(-4,2)/(-1,0)/(-3,-3)/
Last edited:
LBr
#### bldcuber314
##### Member
I use the mid-solve parity, you do it before PBL.
It's a shorter alg (7 slice I think)
Edit:
Here's the alg
/(3,3)/(1,0)/(4,-2)/(-4,2)/(-1,0)/(-3,-3)/
Yeah, that's what I use but I find it hard to figure out if a PLL is legal or not.
#### LBr
##### Member
One thing that Lin is better at is reducing damage that parity can cause to a solve. On vandenbergh there is the 7 slicer, which does nothing other than shift parity and preserve obl, but Lin combines it with EPLL. You just have to learn 5 cases, which can be reduced to 3 when considering that opp parity and o perms are very similar. Obviously with CSP this isn’t important but without could save slices
#### SpeakCuba
##### Member
Maybe I'm thinking of a different CSP lol. But what I meant is that this case (which I think is a parity case):
View attachment 31235
is a headache. I think I mean this parity case, not fixing parity during cubeshape. I'M NEW GIVE ME A BREAK
That’s quite easy once you know how to deal with it. I’ll try to explain, but it is very hard to do so in words only.
If you would like, I could try to make I video showing how I solve this case.
#### GeographicalCuber
##### Member
Yeah, that's what I use but I find it hard to figure out if a PLL is legal or not.
I just use my PLL recognition skills from 3x3
#### bldcuber314
##### Member
I just use my PLL recognition skills from 3x3
The problem is that I only use a small amount of information from the PLL case to actually recognise it, and sometimes the part that parity affected isn't what I use to recognise.
#### GeographicalCuber
##### Member
Maybe I'm thinking of a different CSP lol. But what I meant is that this case (which I think is a parity case):
View attachment 31235
is a headache. I think I mean this parity case, not fixing parity during cubeshape. I'M NEW GIVE ME A BREAK
I solve it like this:
/(0,2)/(3,6)/(-2,0)/
And then the case you get from that is pretty easy, being
(3,0)/(2,4)/(1,2)/(-3,-3)/
#### Bingus Ringus McDingus
##### Member
PB AO5 + PB3 SINGLE??????? HOW?!
Generated By csTimer on 2024-05-30
avg of 5: 18.46
Time List:
1. (14.38) U R2 D' B2 D' F2 L2 D' F2 U2 B2 R2 B' L' R2 D F' R2 B' D' F'
2. (24.99) U2 B R L2 D' F B U2 B2 U' R2 F2 R2 F2 U2 B2 R2 D' L'
3. 17.83 L D B' R F2 B' R' F' R2 U2 L2 U2 F2 R2 B' D2 F' U2 F2 L
4. 17.85 R2 F2 D2 R' D' F B R2 U' B' L2 D2 L2 D2 B' R2 F2 B L2 D2
5. 19.71 U2 B' L2 B L2 U2 B2 U2 F' U2 F R2 U F' U2 R' D L F' R D2
#### Bingus Ringus McDingus
##### Member
Behold:
Poor financial decisions. I bought the YuFeng 2x2 as a joke, the Lustrous 3x3 as a showpiece, the MGC as a collector's item, and the YJ Appari because I believe it deserves an even-handed opinion.
Oh yeah, I'm also getting lubed up. | 1,008 | 3,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-33 | latest | en | 0.942687 |
https://mathematica.stackexchange.com/questions/172028/sum-using-variables-then-evaluated-with-values-gives-different-result-than-sum-w | 1,716,336,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00777.warc.gz | 336,162,759 | 39,662 | # Sum using variables then evaluated with values gives different result than sum with values
I'm trying to do a sum symbolically. However, Mathematica is giving me a different result if I do the sum with numbers or symbols. What's causing this error?
$Assumptions = m \[Element] Integers && n \[Element] Integers && m >= 0 && n >= 0; f[i_, j_] := If[OddQ[Min[i, j]], (Min[i, j] + (-1)^Max[i, j])/2, Ceiling[Min[i, j]/2]]; Table[f[i, j], {i, 0, 1}, {j, 0, 4}] // TableForm F = Sum[f[i, j], {i, 0, m}, {j, 0, n}]; F /. {m -> 1, n -> 4} Sum[f[i, j], {i, 0, 1}, {j, 0, 4}] Result 0 0 0 0 0 0 0 1 0 1 4 2 • If you don't mind me asking, what do you intend to do with the symbolic result, if one can be obtained? I tried running your symbolic calculation, but it ran for minutes, after which I aborted it. How long did it take to run on your system? Apr 26, 2018 at 22:39 • About 1-2 minutes. And it's just for fun, I'm trying to come up with an analytical expression to count the number of triangles in this rug (youtube.com/watch?v=HViA6N3VeHw&t=12s) Apr 26, 2018 at 23:54 • OddQ on symbolic input will evaluate to False. Might be able to achieve the desired effect using Piecewise instead. Apr 27, 2018 at 14:24 ## 1 Answer I'm not entirely sure why MMA is ignoring the $Assumptions, but here is the correct expression:
Table[
Sum[f[i, j], {i, 0, n}, {j, 0, m}] ==
1/48 (-3 (-1 + (-1)^m) (-1 + (-1)^n) + 12 m n +
2 Min[m, n] (2 + 3 (-1)^m + 3 (-1)^n - 2 Min[m, n]^2 + 6 m n))
, {n, 0, 12}, {m, 0, 12}] // Flatten // Tally
(* {{True, 169}} *)
• How did you get that to evaluate? What version are you using? Apr 27, 2018 at 1:09
• @user1543042 I didn't get it to evaluate; I calculated the sum myself. Apr 27, 2018 at 1:25
• What method did you use to calculate this? May 11, 2018 at 19:22
• @user1543042 You should ask on math.SE. You will get much better answers than here. May 11, 2018 at 20:51 | 678 | 1,897 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-22 | latest | en | 0.869707 |
https://math.stackexchange.com/questions/791050/homogeneous-non-negative-least-squares | 1,575,674,870,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540491491.18/warc/CC-MAIN-20191206222837-20191207010837-00450.warc.gz | 461,364,457 | 31,741 | # Homogeneous non-negative least-squares
I would like to least-squares-"solve" a set of linear equations ($\underset{\mathbf{x}}{\mathrm{argmin}}\; \|\mathbf{Ax-b}\|_2$).
In my case, $\mathbf{b=0}$, e.g. the system is homogeneous. I also happen to know that all parameters must be positive, $\mathbf{x} \geq 0$. (The math works out because half of the coefficients are negative.)
• I am assuming that enforcing $\|\mathbf{x}\|_2=1$, e.g. constraining the solution to a hypersphere of constant radius, will take care of the homogeneneity (lest we simply get the trivial solution).
• I have found iterative methods (based on quadratic programming) to deal with the non-negativity constraint.
I have, however, not been able to find a way to combine the two constraints. When adding the fixed-norm constraint to the QP, the solver complains about non-convexity of the problem. Intuitively, I had assumed this was a convex problem. Am I wrong, and if yes, why? If not, how can I solve this problem in software?
• With the unit norm constraint, the problem is not convex because $(1,0,0,\ldots)$ and $(0,1,0,\ldots)$ are in the domain but the line segment joining them is not. – Rahul May 12 '14 at 2:36
• If you're willing to settle for enforcing $\|\mathbf x\|_1=1$ instead, then in conjunction with $\mathbf x\ge0$ this reduces to $\mathbf 1^T\mathbf x=1$ where $\mathbf 1$ is the vector of all ones, and you will obtain a quadratic programming problem that you can solve. – Rahul May 12 '14 at 2:43
• I see. Is there any downside to ditching the squares and going with a simple sum? – sebastian_k May 12 '14 at 2:46
• Yes: you will obtain a different answer with the one-norm than you would with the two-norm. But it is not clear to me that the answer you get from the two-norm is the only answer you want, anyway. – Rahul May 12 '14 at 2:49
• That isn't clear to me either. :) Is there any way to quantify the difference between the two? Are there other ways of "solving" homogeneous overconstrained systems? I appreciate your help! – sebastian_k May 12 '14 at 3:03
Following Rahuls comment, let us consider the convex optimization problem $$\min_{\mathbf{x}} \|\mathbf{Ax}\|_2 ~~~\text{subject to } \mathbf{x} \geq 0, ~~\mathbf 1^T\mathbf x =\alpha.$$ For some $\alpha^\star$, the solution $x^\star$ of the problem above will coincide with the solution of your original problem.
For any $\alpha_1 < \alpha^\star$, it follows $||x_1^\star|| < 1$. On the other hand, for $\alpha_2 > \alpha^\star$, it follows $||x_2^\star|| > 1$ (This is of course no rigorous proof). Therefore, one way to solve your optimization problem would be to search for the true $\alpha$ by solving the above stated problem combined with bi-section search:
Start with $\alpha_1=0$ and $\alpha_2$ great enough such that $||x_2||>1$.
Try $\alpha=(\alpha_1+\alpha_2)/2$. If the respective solution to the convex optimization problem has a norm greater than 1 set $\alpha_2:=\alpha$, otherwise $\alpha_1:=\alpha$. Then, repeat. | 849 | 3,003 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-51 | latest | en | 0.905568 |
http://bedtimemath.org/category/daily-math/entertainment/page/9/ | 1,534,746,835,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215858.81/warc/CC-MAIN-20180820062343-20180820082343-00409.warc.gz | 52,539,282 | 15,127 | # ‘Entertainment’
##### How to Get a Bigger Brownie
November 6, 2017 4:00 pm
Cake tastes delicious no matter what shape it’s in. But we’re loving this cake-cutter made by Matthias Wandel that cuts the cake into hexagons. Read on to find out why this hexagon knife is so cool – and bite into the cake math!
##### Can a Bird Take You for a Ride?
November 5, 2017 4:00 pm
Bedtime Math fan Kaien M. asked us a great question: how many birds would it take to pick you up and fly with you? Read on to fly away with the math and see what the answer to this question is!
##### B is for Birthday
November 2, 2017 4:00 pm
Today we celebrate a very important person’s birthday. Read on to find out who gets to blow out the candles and eat the (cookie) cake today, and get munching on the “monster” math!
##### Dinosaur Dogs
November 1, 2017 4:00 pm
Lots of kids dress up for Halloween. But people also spend a lot of time carefully planning out the costumes for their dogs, too! And pooches in New York City then get to strut their stuff in a Halloween Dog Parade. Read on to find out what some of the popular pup outfits were this year – and do the canine costume math.
##### Spotted on Halloween
October 27, 2017 4:00 pm
With Halloween just a few days away, lots of us have decorated our homes for the big day. And if you’re a little creeped out by all those giant spiders and webs on people’s bushes you can try glowsticks for something a little less yucky and a little more colorful. Read on to find out how to glow your way through Halloween – and get in the Halloween spirit with some math!
##### Pumpkin Face-Off
October 26, 2017 4:00 pm
What do you get when you divide the circumference of a jack-o-lantern by its diameter? Read on to find out the answer to this pumpkin math riddle, and learn more wacky number facts behind pumpkins!
##### Living in a Lego House
October 21, 2017 4:00 pm
If there was a life-size house made of Legos, would you live in it? Well, we don’t know about living in it, but people were lining up to help build a real house of Legos for a British TV show. And good thing, too, because the house used over 3 million Lego bricks! Read on to see how the math quickly stacked up when building this big house with tiny bricks.
##### What’s the Best Stuffed Animal?
October 20, 2017 4:00 pm
Ever wonder if your favorite stuffed animal is the same kind as other people’s? Even though the first stuffed animal was a bear, there are all types now – from dogs and dinosaurs to cows and crocodiles. Read on to cozy up with the math in your favorite plush pals!
##### Strut Like a Snack Machine
October 19, 2017 4:00 pm
It’s always tough to choose a Halloween costume from all the possible choices. And there are even more options if you think outside the box – like being a Transformer or piece of sushi! Read on to see where Halloween and numbers collide with some creative costume math.
##### Speedy Scooter Dog
October 14, 2017 4:00 pm
We’ve seen dogs ride skateboards and surfboards, which takes great balance. But for a dog to ride a scooter is a different story. Well, Norman the French sheepdog can do just that, and has set the doggie scooter-riding world record. When you think about all the hours of training and scooter rides it took for him to do this, it really adds up and makes his feat that much more impressive! | 829 | 3,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.94007 |
https://www.infogalactic.com/info/Logical_connective | 1,653,195,979,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00216.warc.gz | 931,278,462 | 23,234 | # Logical connective
In logic, a logical connective (also called a logical operator) is a symbol or word used to connect two or more sentences (of either a formal or a natural language) in a grammatically valid way, such that the sense of the compound sentence produced depends only on the original sentences.
The most common logical connectives are binary connectives (also called dyadic connectives) which join two sentences which can be thought of as the function's operands. Also commonly, negation is considered to be a unary connective.
Logical connectives along with quantifiers are the two main types of logical constants used in formal systems such as propositional logic and predicate logic. Semantics of a logical connective is often, but not always, presented as a truth function.
A logical connective is similar to but not equivalent to a conditional operator. [1]
## In language
### Natural language
In the grammar of natural languages two sentences may be joined by a grammatical conjunction to form a grammatically compound sentence. Some but not all such grammatical conjunctions are truth functions. For example, consider the following sentences:
A: Jack went up the hill.
B: Jill went up the hill.
C: Jack went up the hill and Jill went up the hill.
D: Jack went up the hill so Jill went up the hill.
The words and and so are grammatical conjunctions joining the sentences (A) and (B) to form the compound sentences (C) and (D). The and in (C) is a logical connective, since the truth of (C) is completely determined by (A) and (B): it would make no sense to affirm (A) and (B) but deny (C). However, so in (D) is not a logical connective, since it would be quite reasonable to affirm (A) and (B) but deny (D): perhaps, after all, Jill went up the hill to fetch a pail of water, not because Jack had gone up the hill at all.
Various English words and word pairs express logical connectives, and some of them are synonymous. Examples (with the name of the relationship in parentheses) are:
The word "not" (negation) and the phrases "it is false that" (negation) and "it is not the case that" (negation) also express a logical connective – even though they are applied to a single statement, and do not connect two statements.
### Formal languages
In formal languages, truth functions are represented by unambiguous symbols. These symbols are called "logical connectives", "logical operators", "propositional operators", or, in classical logic, "truth-functional connectives". See well-formed formula for the rules which allow new well-formed formulas to be constructed by joining other well-formed formulas using truth-functional connectives.
Logical connectives can be used to link more than two statements, so one can speak about "n-ary logical connective".
## Common logical connectives
Name / Symbol Truth table Venn
diagram
P = 0 1
Truth/Tautology 1 1
Proposition P 0 1
Negation ¬ 1 0
Binary connectives P = 0 0 1 1
Q = 0 1 0 1
Conjunction 0 0 0 1
Alternative denial 1 1 1 0
Disjunction 0 1 1 1
Joint denial 1 0 0 0
Material conditional 1 1 0 1
Exclusive or $\nleftrightarrow$ 0 1 1 0
Biconditional 1 0 0 1
Converse implication 1 0 1 1
Proposition P 0 0 1 1
Proposition Q 0 1 0 1
### List of common logical connectives
Commonly used logical connectives include
• Negation (not): ¬ , N (prefix), ~
• Conjunction (and): $\wedge$ , K (prefix), & , ∙
• Disjunction (or): $\or$, A (prefix)
• Material implication (if...then): $\rightarrow$ , C (prefix), $\Rightarrow$ , $\supset$
• Biconditional (if and only if): $\leftrightarrow$ , E (prefix), $\equiv$ , $=$
Alternative names for biconditional are "iff", "xnor" and "bi-implication".
For example, the meaning of the statements it is raining and I am indoors is transformed when the two are combined with logical connectives. For statement P = It is raining and Q = I am indoors:
• It is not raining (¬P)
• It is raining and I am indoors (P $\wedge$ Q)
• It is raining or I am indoors (P $\or$ Q)
• If it is raining, then I am indoors (P $\rightarrow$ Q)
• If I am indoors, then it is raining (Q $\rightarrow$ P)
• I am indoors if and only if it is raining (P $\leftrightarrow$ Q)
It is also common to consider the always true formula and the always false formula to be connective:
• True formula (⊤, 1, V [prefix], or T)
• False formula (⊥, 0, O [prefix], or F)
### History of notations
• Negation: the symbol ¬ appeared in Heyting in 1929.[2][3] (compare to Frege's symbol in his Begriffsschrift); the symbol ~ appeared in Russell in 1908;[4] an alternative notation is to add an horizontal line on top of the formula, as in $\overline{P}$; another alternative notation is to use a prime symbol as in P'.
• Conjunction: the symbol ∧ appeared in Heyting in 1929[2] (compare to Peano's use of the set-theoretic notation of intersection[5]); & appeared at least in Schönfinkel in 1924;[6] . comes from Boole's interpretation of logic as an elementary algebra.
• Disjunction: the symbol ∨ appeared in Russell in 1908[4] (compare to Peano's use of the set-theoretic notation of union ∪); the symbol + is also used, in spite of the ambiguity coming from the fact that the + of ordinary elementary algebra is an exclusive or when interpreted logically in a two-element ring; punctually in the history a + together with a dot in the lower right corner has been used by Peirce,[7]
• Implication: the symbol → can be seen in Hilbert in 1917;[8] ⊃ was used by Russell in 1908[4] (compare to Peano's inverted C notation); $\Rightarrow$ was used in Vax.[9]
• Biconditional: the symbol ≡ was used at least by Russell in 1908;[4] ↔ was used at least by Tarski in 1940;[10] ⇔ was used in Vax; other symbols appeared punctually in the history such as ⊃⊂ in Gentzen,[11] ~ in Schönfinkel[6] or ⊂⊃ in Chazal.[12]
• True: the symbol 1 comes from Boole's interpretation of logic as an elementary algebra over the two-element Boolean algebra; other notations include $\bigwedge$ to be found in Peano.
• False: the symbol 0 comes also from Boole's interpretation of logic as a ring; other notations include $\bigvee$ to be found in Peano.
Some authors used letters for connectives at some time of the history: u. for conjunction (German's "und" for "and") and o. for disjunction (German's "oder" for "or") in earlier works by Hilbert (1904); Np for negation, Kpq for conjunction, Dpq for alternative denial, Apq for disjunction, Xpq for joint denial, Cpq for implication, Epq for biconditional in Łukasiewicz (1929);[13] cf. Polish notation.
### Redundancy
Such a logical connective as converse implication "←" is actually the same as material conditional with swapped arguments; thus, the symbol for converse implication is redundant. In some logical calculi (notably, in classical logic) certain essentially different compound statements are logically equivalent. A less trivial example of a redundancy is the classical equivalence between ¬P ∨ Q and P → Q. Therefore, a classical-based logical system does not need the conditional operator "→" if "¬" (not) and "∨" (or) are already in use, or may use the "→" only as a syntactic sugar for a compound having one negation and one disjunction.
There are sixteen Boolean functions associating the input truth values P and Q with four-digit binary outputs.[14] These correspond to possible choices of binary logical connectives for classical logic. Different implementations of classical logic can choose different functionally complete subsets of connectives.
One approach is to choose a minimal set, and define other connectives by some logical form, as in the example with the material conditional above. The following are the minimal functionally complete sets of operators in classical logic whose arities do not exceed 2:
One element
{↑}, {↓}.
Two elements
{$\vee$, ¬}, {$\wedge$, ¬}, {→, ¬}, {←, ¬}, {→, $\bot$}, {←, $\bot$}, {→, $\nleftrightarrow$}, {←, $\nleftrightarrow$}, {→, $\nrightarrow$}, {→, $\nleftarrow$}, {←, $\nrightarrow$}, {←, $\nleftarrow$}, {$\nrightarrow$, ¬}, {$\nleftarrow$, ¬}, {$\nrightarrow$$\top$}, {$\nleftarrow$$\top$}, {$\nrightarrow$$\leftrightarrow$}, {$\nleftarrow$$\leftrightarrow$}.
Three elements
{$\lor$, $\leftrightarrow$, $\bot$}, {$\lor$, $\leftrightarrow$, $\nleftrightarrow$}, {$\lor$, $\nleftrightarrow$, $\top$}, {$\land$, $\leftrightarrow$, $\bot$}, {$\land$, $\leftrightarrow$, $\nleftrightarrow$}, {$\land$, $\nleftrightarrow$, $\top$}.
See more details about functional completeness in classical logic at Functional completeness in truth function.
Another approach is to use with equal rights connectives of a certain convenient and functionally complete, but not minimal set. This approach requires more propositional axioms, and each equivalence between logical forms must be either an axiom or provable as a theorem.
The situation, however, is more complicated in intuitionistic logic. Of its five connectives, {∧, ∨, →, ¬, ⊥}, only negation "¬" can be reduced to other connectives (see details). Neither conjunction, disjunction, nor material conditional has an equivalent form constructed of the other four logical connectives.
## Properties
Some logical connectives possess properties which may be expressed in the theorems containing the connective. Some of those properties that a logical connective may have are:
• Associativity: Within an expression containing two or more of the same associative connectives in a row, the order of the operations does not matter as long as the sequence of the operands is not changed.
• Commutativity: The operands of the connective may be swapped preserving logical equivalence to the original expression.
• Distributivity: A connective denoted by · distributes over another connective denoted by +, if a · (b + c) = (a · b) + (a · c) for all operands a, b, c.
• Idempotence: Whenever the operands of the operation are the same, the compound is logically equivalent to the operand.
• Absorption: A pair of connectives $\land$, $\lor$ satisfies the absorption law if $a\land(a\lor b)=a$ for all operands a, b.
• Monotonicity: If f(a1, ..., an) ≤ f(b1, ..., bn) for all a1, ..., an, b1, ..., bn ∈ {0,1} such that a1b1, a2b2, ..., anbn. E.g., $\vee$, $\wedge$, $\top$, $\bot$.
• Affinity: Each variable always makes a difference in the truth-value of the operation or it never makes a difference. E.g., $\neg$, $\leftrightarrow$, $\nleftrightarrow$, $\top$, $\bot$.
• Duality: To read the truth-value assignments for the operation from top to bottom on its truth table is the same as taking the complement of reading the table of the same or another connective from bottom to top. Without resorting to truth tables it may be formulated as a1, ..., ¬an) = ¬g(a1, ..., an). E.g., $\neg$.
• Truth-preserving: The compound all those argument are tautologies is a tautology itself. E.g., $\vee$, $\wedge$, $\top$, $\rightarrow$, $\leftrightarrow$, ⊂. (see validity)
• Falsehood-preserving: The compound all those argument are contradictions is a contradiction itself. E.g., $\vee$, $\wedge$, $\nleftrightarrow$, $\bot$, ⊄, ⊅. (see validity)
• Involutivity (for unary connectives): f(f(a)) = a. E.g. negation in classical logic.
For classical and intuitionistic logic, the "=" symbol means that corresponding implications "…→…" and "…←…" for logical compounds can be both proved as theorems, and the "≤" symbol means that "…→…" for logical compounds is a consequence of corresponding "…→…" connectives for propositional variables. Some many-valued logics may have incompatible definitions of equivalence and order (entailment).
Both conjunction and disjunction are associative, commutative and idempotent in classical logic, most varieties of many-valued logic and intuitionistic logic. The same is true about distributivity of conjunction over disjunction and disjunction over conjunction, as well as for the absorption law.
In classical logic and some varieties of many-valued logic, conjunction and disjunction are dual, and negation is self-dual, the latter is also self-dual in intuitionistic logic.
## Order of precedence
As a way of reducing the number of necessary parentheses, one may introduce precedence rules: $\neg$ has higher precedence than $\wedge$, $\wedge$ higher than $\vee$, and $\vee$ higher than $\rightarrow$. So for example, $P \vee Q \wedge{\neg R} \rightarrow S$ is short for $(P \vee (Q \wedge (\neg R))) \rightarrow S$.
Here is a table that shows a commonly used precedence of logical operators.[15]
Operator Precedence
$\neg$ 1
$\wedge$ 2
$\vee$ 3
$\rightarrow$ 4
$\leftrightarrow$ 5
However, not all authors use the same order; for instance, an ordering in which disjunction is lower precedence than implication or bi-implication has also been used.[16] Sometimes precedence between conjunction and disjunction is unspecified requiring to provide it explicitly in given formula with parentheses. The order of precedence determines which connective is the "main connective" when interpreting a non-atomic formula.
## Computer science
A truth-functional approach to logical operators is implemented as logic gates in digital circuits. Practically all digital circuits (the major exception is DRAM) are built up from NAND, NOR, NOT, and transmission gates; see more details in Truth function in computer science. Logical operators over bit vectors (corresponding to finite Boolean algebras) are bitwise operations.
But not every usage of a logical connective in computer programming has a Boolean semantic. For example, lazy evaluation is sometimes implemented for P ∧ Q and P ∨ Q, so these connectives are not commutative if some of expressions P, Q has side effects. Also, a conditional, which in some sense corresponds to the material conditional connective, is essentially non-Boolean because for if (P) then Q; the consequent Q is not executed if the antecedent P is false (although a compound as a whole is successful ≈ "true" in such case). This is closer to intuitionist and constructivist views on the material conditional, rather than to classical logic's ones.
## Notes
1. Cogwheel. "What is the difference between logical and conditional /operator/". Stack Overflow. Retrieved 9 April 2015.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
2. Heyting (1929) Die formalen Regeln der intuitionistischen Logik.
3. Denis Roegel (2002), Petit panorama des notations logiques du 20e siècle (see chart on page 2).
4. Russell (1908) Mathematical logic as based on the theory of types (American Journal of Mathematics 30, p222–262, also in From Frege to Gödel edited by van Heijenoort).
5. Schönfinkel (1924) Über die Bausteine der mathematischen Logik, translated as On the building blocks of mathematical logic in From Frege to Gödel edited by van Heijenoort.
6. Peirce (1867) On an improvement in Boole's calculus of logic.
7. Hilbert (1917/1918) Prinzipien der Mathematik (Bernays' course notes).
8. Vax (1982) Lexique logique, Presses Universitaires de France.
9. Tarski (1940) Introduction to logic and to the methodology of deductive sciences.
10. Gentzen (1934) Untersuchungen über das logische Schließen.
11. Chazal (1996) : Éléments de logique formelle.
12. See Roegel
13. Bocheński (1959), A Précis of Mathematical Logic, passim.
14. O'Donnell, John; Hall, Cordelia; Page, Rex (2007), Discrete Mathematics Using a Computer, Springer, p. 120, ISBN 9781846285981<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>.
15. Jackson, Daniel (2012), Software Abstractions: Logic, Language, and Analysis, MIT Press, p. 263, ISBN 9780262017152<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>. | 3,982 | 15,626 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 104, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-21 | latest | en | 0.966493 |
https://www.helpwithphysics.com/2013/05/speed-of-light-magnetic-field.html | 1,713,965,066,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00499.warc.gz | 742,772,132 | 14,744 | # Speed of Light. Magnetic Field
1) $c = omega/k = 1/sqrt{(epsilon_0*mu_0)} =1/sqrt{(8.856*10^{-12}*4pi*10^{-7})}=299761783.5 m/s$
2) only the component of the speed perpendicular to B (magnetic field ) contributres to the magnetic force
$v = vi +vj =32*i+40*j m/s$
$F = -e(v times B)$
x is the vector product between v and B
$F = e v j*B$ (in modulus)
this force is equal to the centripetal force
$F = m vj^2/R$
$e*vj*B = m*vj^2/R$
$R = mvj/(e*B) =9.1*10^{-31}*40000/1.6*10^{-19}*60*10^{-6} =3.792*10^{-3} m = 3.792 mm$
time T for one revolution is
$2*pi*R = vj*T$
$T = 2*pi*R/vj =2*pi*3.792*10^{-3}/40000=5.956*10^{-7} sec$
pitch of the path
$X = vi*T = 32000*5.956*10^{-7} =19.06*10^{-3} m =19.06 mm$
The electron is spiraling clockwise. See the picture below. | 317 | 781 | {"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} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.61635 |
http://at.metamath.org/ilegif/cbvexv.html | 1,716,415,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00106.warc.gz | 2,432,408 | 3,714 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > cbvexv Unicode version
Theorem cbvexv 1795
Description: Rule used to change bound variables, using implicit substitition. (Contributed by NM, 5-Aug-1993.)
Hypothesis
Ref Expression
cbvalv.1
Assertion
Ref Expression
cbvexv
Distinct variable groups: , ,
Allowed substitution hints: () ()
Proof of Theorem cbvexv
StepHypRef Expression
1 ax-17 1419 . 2
2 ax-17 1419 . 2
3 cbvalv.1 . 2
41, 2, 3cbvexh 1638 1
Colors of variables: wff set class Syntax hints: wi 4 wb 98 wex 1381 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-4 1400 ax-17 1419 ax-i9 1423 ax-ial 1427 This theorem depends on definitions: df-bi 110 This theorem is referenced by: eujust 1902 euind 2728 reuind 2744 r19.3rm 3310 r19.9rmv 3313 raaanlem 3326 raaan 3327 cbvopab2v 3834 bm1.3ii 3878 mss 3962 zfun 4171 xpiindim 4473 relop 4486 dmmrnm 4554 dmxpm 4555 dmcoss 4601 xpm 4745 cnviinm 4859 fv3 5197 fo1stresm 5788 fo2ndresm 5789 iinerm 6178 riinerm 6179 ac6sfi 6352 ltexprlemdisj 6704 ltexprlemloc 6705 recexprlemdisj 6728 climmo 9819 bdbm1.3ii 10011
Copyright terms: Public domain W3C validator | 652 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.169735 |
https://metanumbers.com/71854 | 1,643,068,457,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00262.warc.gz | 435,140,670 | 7,367 | # 71854 (number)
71,854 (seventy-one thousand eight hundred fifty-four) is an even five-digits composite number following 71853 and preceding 71855. In scientific notation, it is written as 7.1854 × 104. The sum of its digits is 25. It has a total of 3 prime factors and 8 positive divisors. There are 34,920 positive integers (up to 71854) that are relatively prime to 71854.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 25
• Digital Root 7
## Name
Short name 71 thousand 854 seventy-one thousand eight hundred fifty-four
## Notation
Scientific notation 7.1854 × 104 71.854 × 103
## Prime Factorization of 71854
Prime Factorization 2 × 37 × 971
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 71854 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 71,854 is 2 × 37 × 971. Since it has a total of 3 prime factors, 71,854 is a composite number.
## Divisors of 71854
1, 2, 37, 74, 971, 1942, 35927, 71854
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 110808 Sum of all the positive divisors of n s(n) 38954 Sum of the proper positive divisors of n A(n) 13851 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 268.056 Returns the nth root of the product of n divisors H(n) 5.18764 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 71,854 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 71,854) is 110,808, the average is 13,851.
## Other Arithmetic Functions (n = 71854)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 34920 Total number of positive integers not greater than n that are coprime to n λ(n) 17460 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7109 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 34,920 positive integers (less than 71,854) that are coprime with 71,854. And there are approximately 7,109 prime numbers less than or equal to 71,854.
## Divisibility of 71854
m n mod m 2 3 4 5 6 7 8 9 0 1 2 4 4 6 6 7
The number 71,854 is divisible by 2.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (71854)
Base System Value
2 Binary 10001100010101110
3 Ternary 10122120021
4 Quaternary 101202232
5 Quinary 4244404
6 Senary 1312354
8 Octal 214256
10 Decimal 71854
12 Duodecimal 356ba
20 Vigesimal 8jce
36 Base36 1jfy
## Basic calculations (n = 71854)
### Multiplication
n×y
n×2 143708 215562 287416 359270
### Division
n÷y
n÷2 35927 23951.3 17963.5 14370.8
### Exponentiation
ny
n2 5162997316 370982009143864 26656541285023203856 1915379117494057289869024
### Nth Root
y√n
2√n 268.056 41.5735 16.3724 9.36031
## 71854 as geometric shapes
### Circle
Diameter 143708 451472 1.622e+10
### Sphere
Volume 1.55397e+15 6.48801e+10 451472
### Square
Length = n
Perimeter 287416 5.163e+09 101617
### Cube
Length = n
Surface area 3.0978e+10 3.70982e+14 124455
### Equilateral Triangle
Length = n
Perimeter 215562 2.23564e+09 62227.4
### Triangular Pyramid
Length = n
Surface area 8.94257e+09 4.37206e+13 58668.5
## Cryptographic Hash Functions
md5 bc98021b16ce113cecd539e1555c6125 21b728c139373fb41cf2a9d499aed1c70fc469df a97ab24c9d57fe7b125b23625ffdfab27f679beffb9b78820bb68be2fa233834 57a56f1b8c2da66293b1ea4608fc9b92d178c6d01f54698cf5734bd0ca6c1484fdfe54af182a3f6a7dd97f6d73801cd863467da9bdb06d3c432458b586696004 832b6a0e334c6698f62d70df040e5b88e987a693 | 1,446 | 4,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.816374 |
https://scienceprog.com/tag/signal-power/ | 1,726,849,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00620.warc.gz | 471,841,348 | 14,273 | ## Electrical signal power and energy calculations by example
In electronics and signal processing, you have to deal with electrical signals. In many cases, you may need to calculate signal power and energy. Power and energy for DC In a standard situation, when DC power supply is applied to a known resistor or another device like an LED motor you can calculate its power very easy by applying Ohms law: If we run this device for time T then we can calculate total energy used: In some cases, you may not know the resistance of your circuit. In this case, you can measure the current flow. So your power formula can be transformed by using same Ohms law:
## Signal power and energy calculation
The most common signal characteristics are energy and power. In signal theory, these terms require additional comments because they are different from what we are using in AC or DC systems. What are power and energy? When we connect the R resistor to voltage U, the resistor will dissipate some power equal to P=U2/R. During time T, the energy loss on this resistor will be E= TU2/R. Now let us say that we add some signal s() instead of DC voltage. In this case, the power will depend on time as the signal is time-dependent. The term is called instantaneous power: p(s)=s(t)2/R | 272 | 1,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.936947 |
https://muddoo.com/tutorials/how-to-plot-scatter-plot-in-python-using-matplotlib/ | 1,606,290,064,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00289.warc.gz | 397,943,734 | 15,313 | Categories
# How To Plot Scatter Plot In Python Using Matplotlib
In this article, we will learn how to plot a Scatter plot in Python using Matplotlib. But before we do that, we will learn what a Scatter plot is and what all options are there for us to use. Sounds good? Great! Then let us start!
## What Is A Scatter Plot?
A scatter plot is a type of plot that we can use to display values from two sets of data. So what happens is, we will take two sets of data of same length & pair them together. We then use this pair to plot the scatter plot. It is very important for you to remember here that both the data sets have to be of the same length!
But there is another thing to note here. It is that the scatter plot will have only points drawn and no lines in them. So in other words the points in the plot will not be connected together. But just the points scattered across the chart. Hence the name scatter plot!
## What Is The Use Of A Scatter Plot?
We can use Scatter plot to see any correlation between two data sets. So what happens is, similar points get grouped together in the scatter plot. Now this can be a very valuable insight for us. Especially when looking at non linear relationships between the two datasets! Does that make sense?
So we can use a scatter plot to find any relationship between data points.
So now that we know the uses of a scatter plot, let us see how to plot it. But to do so, we need two sets of data. Right? So what do we do for that? Where can we find it?
## How To Plot A Scatter Plot In Python
Here is what we will do. We will use our good old random function randn( ) from the Numpy library for that. Alright? So check this code:
``````import matplotlib.pyplot as plt
import numpy as np
x = np.random.randn(100)
y = np.random.randn(100)
plt.scatter(x, y);
plt.show()``````
So this is the code that will generate us a scatter plot using two sets of random data. But what is going on here? Let us go through the code line by line:
So in line 1 and 2, we are importing our Matplotlib and Numpy libraries.
But what is going on in line 3 & 4? Well, as I said earlier, we need two sets of data for our scatter plot. So we are using Numpy’s randn( ) to generate these data sets. We will assign them to variables x & y.
Next in line 5, we call our good old plt module’s scatter( ) function and passing x & y to it. So this is the function that will generate our scatter plot using x & y data!
Finally we call the usual plt.show( ) function to display our resulting scatter plot. So here is how it looks like:
So as you can see, there are 100 points taken from x & y variables to be plotted along the X & Y axis. So if there are any points that are similar, they will converge together here!
## How To Change The Size Of A Marker
The above chart is all well and good, but is there a way to control the size, color & marker type in it? Well, lucky for us Matplotlib does give us an option for this. So to do so, we need to set the s, c & marker parameters in our plt.scatter( ) function!
So here is a simple example code that changes the size and color of our plot markers:
``````import matplotlib.pyplot as plt
import numpy as np
x = np.random.randn(100)
y = np.random.randn(100)
size = 20*np.random.randn(100)
colors = np.random.rand(100)
marker = "^"
plt.scatter(x, y, s=size, c=colors, marker=marker);
plt.show()``````
And here is how our final scatter plot would look like:
So there you have it! This is how we plot a Scatter Plot in Python. I hope this was easy enough for you to follow. But if you have any doubts, do let me know in the comments below. I will be more than happy to help!
Alright? So then see you until next time. take care! 🙂
What to learn how to plot a Pie chart using Matplotlib? Take a look at this!
## By Amar Nath
You can buy me a cup of coffee at: | 935 | 3,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-50 | longest | en | 0.917091 |
https://www.coursehero.com/file/5866682/CH15/ | 1,544,923,674,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827175.38/warc/CC-MAIN-20181216003916-20181216025916-00433.warc.gz | 839,788,004 | 43,793 | # CH15 - CHAPTER 15 Financial Statement Analysis ASSIGNMENT...
• Notes
• mobinil1
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• 67% (3) 2 out of 3 people found this document helpful
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CHAPTER 15 Financial Statement Analysis ASSIGNMENT CLASSIFICATION TABLE Study Objectives Questions Brief Exercises Exercises Problems Alternate Problems 1. Discuss the need for comparative analysis. 1, 2, 3, 5 2. Identify the tools of financial statement analysis. 2, 3, 5 3. Explain and apply horizontal analysis. 3, 4 1, 3, 4, 6 1, 3, 4 4. Describe and apply vertical analysis. 3, 4, 8 2, 5 2, 3, 4 1 1 5. Identify and compute ratios and describe their purpose and use in analyzing a firm's liquidity, profitability, and solvency. 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 7, 8, 9, 10, 11, 15 5, 6, 7, 8, 9, 10, 12 1, 2, 3, 4, 5, 6, 7 1, 2, 3, 4, 5 6. Understand the con- cept of earning power and indicate how ma- terial items not typical of regular operations are presented. 21, 22, 23, 24 12, 13, 14 11, 12 7. Recognize the limita- tions of financial statement analysis. 19, 20 12 15-1
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ASSIGNMENT CHARACTERISTICS TABLE Problem Number Description Difficulty Level Time Allotted (min.) 1 Prepare vertical analysis and comment on profitability. Simple 20-30 2 Compute ratios from balance sheet and income statement. Simple 20-30 3 Perform ratio analysis. Simple 20-30 4 Compute ratios, commenting on overall liquidity and profitability. In addition, ratios need to be computed before and after transactions have occurred. Moderate 30-40 5 Compute selected ratios and compare liquidity, profitability, and solvency for two companies. Moderate 50-60 6 Compute numerous ratios. Simple 30-40 7 Compute missing information, given a set of ratios. Complex 30-40 1A Prepare vertical analysis and comment on profitability. Simple 20-30 2A Compute ratios from balance sheet and income statement. Simple 20-30 3A Perform ratio analysis. Simple 20-30 4A Compute ratios, commenting on overall liquidity and profitability. In addition, ratios need to be computed before and after transactions have occurred. Moderate 30-40 5A Compute selected ratios and compare liquidity, profitability, and solvency for two companies. Moderate 50-60
Correlation Chart between Bloom's Taxonomy, Study Objectives and End-of-Chapter Exercises and Problems Study Objective Knowledg e Compr ehensi on Applicati on Analysis Synthesi s Evaluation 1. Discuss the need for comparative analysis. Q15-1 Q15-2 Q15-3 Q15-5 2. Identify the tools of financial statement analysis. Q15-2 Q15-3 Q15-5 3. Explain and apply horizontal analysis. Q15-3 Q15-4 BE15-1 BE15-3 BE15-4 BE15-6 E15-1 E15-3 E15-4 4. Describe and apply vertical analysis. Q15-8 Q15-3 Q15-5 Q15-4 BE15-2 BE15-5 E15-2 E15-3 E15-4 P15-1 P15-1A 5. Identify and compute ratios and describe their purpose and use in analyzing a firm's liquidity, profitability, and solvency. Q15-6 Q15-8 Q15-5 Q15-7 Q15-9 Q15-10 Q15-11 Q15-12 Q15-13 Q15-14 Q15-15 Q15-16 Q15-17 Q15-18 BE15-7 BE15-8 BE15-15 E15-6 E15-7 E15-8 E15-9 E15-10 P15-2 P15-6 P15-2A BE15-9 BE15-10 BE15-11 E15-5 E15-12 P15-1 P15-3 P15-4 P15-5 P15-7 P15-1A P15-3A P15-4A P15-5A 6. Understand the concept of earning power and indicate how material items not typi- cal of regular operations are presented.
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• Spring '06
• MR.SOMEONE
• Balance Sheet, Financial Ratio, Generally Accepted Accounting Principles
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Jill Tulane University ‘16, Course Hero Intern | 1,333 | 4,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.798649 |
https://stats.stackexchange.com/questions/442519/covariance-in-2-dimensional-field | 1,628,094,871,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00361.warc.gz | 539,625,281 | 37,235 | # covariance in 2 dimensional field
I want to know is the covariance equation which is shown below is how valid when we are in 2-dimensional fields (not 2-dimensional data)?
$$Cov(x,y)=\frac{\sum_{i=1}^{n}(x_i-\bar{x})(y_i-y)}{N-1}$$
for instance, suppose we want to find COV(x,y) given two elements of x and y:
x_samples={ [[1,2],[3,4]] , [[5,6],[7,8]] }
y_samples={ [[9,10],[11,12]] , [[13,14],[15,16]] }
can I use the above equation for this space?
This way unlike scaler space $$Cov(x,y)$$ is not equal to $$Cov(y,x)$$. is it true?
• Please tell us what you mean by "2-dimensional fields." Your bracket notation is not standard mathematical notation. – whuber Dec 28 '19 at 21:42
• suppose u have scaler numbers of x samples, {1,2}, and y samples, {3,4}, then you can compute COV(x,y) easily, but what if we have 2dimentional matrix instead of each sample ( 1, 2, 3, 4)? like the above samples I mentioned as an example. can we use same COV formula? – sakht Dec 30 '19 at 9:10
• Could you explain what this two-dimensional matrix represents? – whuber Dec 30 '19 at 14:32
• u can imagine that we have to class of images; x and y, and we have for each class 2 sample images, and each sample is a 2*2 matrix, like above, so I want to find covariance of these two classes. – sakht Jan 1 '20 at 8:54 | 403 | 1,306 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.871526 |
http://www.evi.com/q/convert_15_square_meters_to_sq_feet | 1,419,121,647,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770433.122/warc/CC-MAIN-20141217075250-00019-ip-10-231-17-201.ec2.internal.warc.gz | 496,191,263 | 5,297 | # convert 15 square meters to sq feet
• 15 square meters is 161.46 square feet.
• tk10npubl tk10ncanl
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• how many square feet in 15 square meters | 572 | 2,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2014-52 | latest | en | 0.902694 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=40717&p=147110 | 1,606,655,514,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00520.warc.gz | 362,770,892 | 12,446 | ## increasing pressure, what happens to concentration?
Josephine Lu 4L
Posts: 62
Joined: Fri Sep 28, 2018 12:18 am
### increasing pressure, what happens to concentration?
Why is it that when we increase pressure, the reaction favors the side with less moles of gas?
Julia Go 2L
Posts: 60
Joined: Sun Sep 30, 2018 12:17 am
Been upvoted: 1 time
### Re: increasing pressure, what happens to concentration?
According to Le Chatelier's Principle, a change in pressure will result in an attempt to restore equilibrium by creating more or less moles of gas. For example, if the pressure in a system increases, the equilibrium will shift to favor the side of the reaction that involves fewer moles of gas. Similarly, if the the pressure decreases, the production of additional moles of gas will be favored. A system will try to minimize the disturbance.
Selina Bellin 2B
Posts: 62
Joined: Fri Sep 28, 2018 12:25 am
### Re: increasing pressure, what happens to concentration?
increasing pressure will make the reaction go in the direction where there are LESS moles
Lopez_Melissa-Dis4E
Posts: 66
Joined: Fri Sep 28, 2018 12:20 am
### Re: increasing pressure, what happens to concentration?
According to Le Chatelier's Principle, if you increase the pressure the reaction will go towards the side with less moles.
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am
### Re: increasing pressure, what happens to concentration?
Increasing pressure will cause the reaction to proceed towards the side with less moles of gas.
Hannah Pham 1D
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am
### Re: increasing pressure, what happens to concentration?
Take volume into account. The reaction will favor the side with fewer moles.
Jeannine 1I
Posts: 73
Joined: Fri Sep 28, 2018 12:27 am
### Re: increasing pressure, what happens to concentration?
You can imagine that when pressure increases, the gas molecules on both sides of the reaction move faster, and are therefore more likely to bump into each other. The side with more moles of gas will have more molecules that bump into each other, which will drive the reaction to shift towards the side with less moles.
I hope that helps! I'm more of a visual learner myself, so imagining what things would look like helps me understand it better(:
Xingzheng Sun 2K
Posts: 62
Joined: Fri Sep 28, 2018 12:29 am
### Re: increasing pressure, what happens to concentration?
When pressure is increased, the reaction wants to do something to reduce the effects. Lowering the number of molecules can decrease the pressure which reduce the effect.
Tony Chung 2I
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am
### Re: increasing pressure, what happens to concentration?
Le Chatelier's Principle for pressure: if the pressure increases, the reaction will proceed towards the side with less moles of gas. if the pressure is decreased, the reaction will proceed towards the side with more moles of gas.
jlinwashington1B
Posts: 65
Joined: Fri Sep 28, 2018 12:22 am
### Re: increasing pressure, what happens to concentration?
when pressure increases, the reaction is attracted to the side with fewer moles.
Sarah_Kang_2K
Posts: 72
Joined: Fri Sep 28, 2018 12:23 am
### Re: increasing pressure, what happens to concentration?
When pressure increases, the gaseous molecules on both sides of the reaction increasingly collide with each other. In order to alleviate/compensate for these increased collisions, the reaction favors the side with the less number of moles of gaseous molecules.
Return to “Equilibrium Constants & Calculating Concentrations”
### Who is online
Users browsing this forum: No registered users and 3 guests | 901 | 3,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-50 | latest | en | 0.901549 |
https://www.jiskha.com/questions/755605/please-help-i-am-not-sure-if-this-is-correct-i-need-to-simplify-this-to-the-lowest-terms | 1,590,705,604,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00431.warc.gz | 784,977,028 | 5,998 | # Algebra
please help. I am not sure if this is correct. I need to simplify this to the lowest terms
a^2 - 4b ^2
_____ ______
a+2b a+2b
my answer is:
a^2-4b^2
_________
a+2b
is this really to the lowest term?? for some reason I think I am doing this wrong. Thank you!
1. 👍 0
2. 👎 0
3. 👁 148
1. you are correct, as far as you have gone. However, since
a^2 - 4b^2 = (a-2b)(a+2b), you wind up with just
a-2b
1. 👍 0
2. 👎 0
2. 12a^3b/-6ab^5
1. 👍 0
2. 👎 0
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Please check these: 1.When the fraction 2/3 is used as exponent, wht role does the denominator 3 play in evaluating expression? I think it indicates cube root of the base,correct? 2.sqrt of (2^4x/4^-2x) is equilvalent to 2^4x,
asked by Tabitha on December 12, 2011
4. ### Math
Simplify each expression (-216)^1/3 My answer is 1/6. Am I correct. (x^4y)^1/3 (xy^4)^2/3 I do not know how to simplify this.
asked by Hannah on September 6, 2010
5. ### Algebra 1
Simplify the difference (-7p^4+4p^3+6)-(3p^3-5p+3) I removed the parenthesis and changed the addition and subtraction signs to their opposites on the second set; -7p^4 + 4p^3 + 6 - 3p^3 + 5p - 3. When I combined the like terms, I
asked by Sami on October 26, 2016
6. ### Algebra
1.Simplify the radical expression. √5+6√√5 A.5√5 B.7√10 C.7√5 D.5√10 2.Simplify the radical expression. 2√6+3√96 A.14√6 B.14√96 C.5√96 D.50√6 3.Simplify the radical expression. (8+√11)(8-√11) A.53
asked by GummyBears16 on April 24, 2016
More Similar Questions | 1,295 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-24 | latest | en | 0.883799 |
http://www.riddlesandanswers.com/puzzles-brain-teasers/a-beer-with-a-suit-cover-and-other-suit-is-empty-riddles/ | 1,558,819,081,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258451.88/warc/CC-MAIN-20190525204936-20190525230936-00168.warc.gz | 332,079,800 | 27,380 | # A BEER WITH A SUIT COVER AND OTHER SUIT IS EMPTY RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
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## Solving A Beer With A Suit Cover And Other Suit Is Empty Riddles
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## A Suitor To Portia Riddle
Hint:
The Prince of Tangiers
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YES NO
## Portia's Favored Suitor
Hint:
Lead, as he is not deceived by 'outward shows'
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Solved: 17%
## Santas Suit Riddle
Hint:
Santa Clauset
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## A Suit In A Deck Of Cards
Hint:
Heart
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YES NO
Solved: 75%
## The Jug Full Of Beer
Hint:
First fill the 8 liters jug complete - 4, 8, 0
Fill the 5 liters jug with 8 liters jug - 4, 3, 5
Pour back the beer from 5 liters jug to 12 liters jug - 9, 3, 0
Pour the 3 liters from 8 liters jug to 5 liters jug - 9, 0, 3
Fill the 8 liters jug completely from 12 liters jug - 1, 8, 3
Fill the 5 liters jug from the 8 liters jug - 1, 6, 5
Pour the entire 5 liters jug back in 12 liters jug - 6, 6, 0
You have successfully split the beer into two equal parts.
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## I Cover The Planet
Hint:
The ocean
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Solved: 70%
## Covered In Lines Riddle
Hint:
I am paper!
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Solved: 80%
## An Empty Stomach Riddle
Hint:
One bite. After that, the stomach isnt empty.
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## Cover At The Club Riddle
Hint:
It's cheaper to take two friends at the same time. In this case, you would only be buying three covers, whereas if you take the same friend twice you are buying four covers.
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YES NO
Solved: 100%
## Gynecology Beer Riddle
Hint:
"Pabst Smir!"
Did you answer this riddle correctly?
YES NO
Solved: 17% | 722 | 2,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.837325 |
https://www.math10.com/forum/viewtopic.php?f=29&t=8489 | 1,596,897,097,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00583.warc.gz | 750,214,855 | 5,404 | # Calculate the length of the fourth side edge of pyramid.
### Calculate the length of the fourth side edge of pyramid.
The three lateral edges of a pyramid based on a square are 6016, 2370 and 4350 long. Calculate the length of the fourth side edge. We assume that edges 6016 and 2370 extend from the opposite tops of the base.
ghostfirefox
Posts: 10
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### Re: Calculate the length of the fourth side edge of pyramid.
ghostfirefox wrote:The three lateral edges of a pyramid based on a square are 6016, 2370 and 4350 long. Calculate the length of the fourth side edge. We assume that edges 6016 and 2370 extend from the opposite tops of the base.
Doesn't the edge of length 4350 do also? Are we to assume the base is a square?
HallsofIvy
Posts: 206
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 75 | 235 | 849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-34 | latest | en | 0.854269 |
http://www.enotes.com/homework-help/adult-tickets-cost-4-children-tickets-cost-1-285-147589 | 1,477,324,892,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719646.50/warc/CC-MAIN-20161020183839-00295-ip-10-171-6-4.ec2.internal.warc.gz | 434,365,553 | 10,848 | # Adult tickets cost \$4 and children tickets cost \$1. 285 tickets are sold. And \$765 is collected. How many adult tickets were sold.?Use substitution or elimination to solve the problem
bullgatortail | High School Teacher | (Level 1) Distinguished Educator
Posted on
If adult tickets cost \$4, and children's tickets cost \$1, and 285 tickets total were sold, and the total amount collected is \$765, there can only be one possible answer. I believe you will find that 160 adult tickets were sold (total of \$640) and 125 children's tickets were sold (total of \$125). Combining these two amounts will give you 285 tickets and a total of \$765. Any difference in tickets sold by either adults or children would result in a higher or lower total amount.
pohnpei397 | College Teacher | (Level 3) Distinguished Educator
Posted on
The answer to this is that 160 adult tickets were sold. This means that 125 children's tickets were sold. Here is how to solve this problem:
Let's call adult tickets A and children's tickets C.
We know that A + C = 285 because that's how many total tickets were sold.
We know that 4A + C = 765. That's because each adult ticket sold cost \$4 so 4 times the number of adult tickets, plus the number of kids tickets (\$1 each) make the total amount collected.
Let's use the first equation to find for C. C = 285 - A.
Now just substitute that into the other equation and you have
4A + 285 - A = 765.
3A = 480
A = 160
neela | High School Teacher | (Level 3) Valedictorian
Posted on
In the question the need is to determine the number of adult and child tickets tickets sold out. So one of the unknown is x.
We presume x number of tickets for sdults are sold. The child tickets is automatically must be 285-x.
The collection of revenue from x adult tickets = number of tickets* rate of tickets = 4x
The collection of revenue from 285-x child tickets = number*rate = (285*x)*1 =285-x.
The total collection = 4x+285-x algebraically.......(1)
The actual collection = \$765..............................(2)
Therefore the required equation of the problem:
Algebraic collection as at (1) = actual collection as at (2). So,
\$(4x+285-x) = %765. Or
4x-x +285 = 765. Or
3x- 765 - 285 = 480. Or
3x = 480. Or
3x/3 = 480/3. So
x = 160 is the number of adult tickets sold.
285 - x = 285 - 160 = 125 is the number of child tickets sold.
Check: 160+125 = 285 and revenue \$(160*4+125) =\$ (640+125) = \$765 | 667 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2016-44 | latest | en | 0.941607 |
https://www.tessshebaylo.com/solving-systems-of-equations-with-3-variables-practice-problems/ | 1,723,128,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00085.warc.gz | 803,355,757 | 13,905 | # Solving Systems Of Equations With 3 Variables Practice Problems
By | December 13, 2018
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This site uses Akismet to reduce spam. Learn how your comment data is processed. | 429 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.669557 |
https://customerthink.com/when_success_is_bad_the_math_behind_why_failure_is_essential/ | 1,723,059,633,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00248.warc.gz | 147,054,027 | 60,063 | # When Success is Bad – The Math Behind Why Failure Is Essential
0
45
We all know Murphy’s Law: “Whatever can go wrong, will.”
But, what happens when the right things happen for what we think are the right reasons?
Or, restating in a slightly different way:
In any system there are ways of achieving the correct result through a combination of known and unknown means.
As any product developer will tell you, there are times when you test a product (or code) and it works. You get excited. You decide to show your results to others. You do everything the exact same way you did last time, only this time, it doesn’t work.
Not. At. All.
How does this happen?
Let’s imagine a product as a mathematical expression: A+B+C=X (eq.1). A, B, and C are things that we know, things that we do to bring about X which is the result we want to have happen. Let’s call this the “Success Equation”.
The equation for some undesired outcome could be depicted as: A+B+C+W=Z (eq.2). W is some known wrong step or condition that causes Z, which is an undesired result. We can call this the “Devil We Know” equation.
Now, when working with a product prototypes we actually only know what we know. Sounds obvious right? Another way to say this is: We don’t know what we don’t know.
What this means is that the REAL equation for our product is very often: A+B+C+D+E+F=X. (eq. 3) This is one of many versions of what I’ll call the “Devil We Don’t Know” equations.
A, B and C are known and are BOLD in the equation. D, E and F, are grey because we don’t even know these variables exist. Nevertheless, they are a part of the equation and if they all come into play, X occurs, so we’re happy.
And that’s a problem.
Why?
What happens if D, or E, or F, or some combination of these disappear? We could get the formula A+B+C+E+F=Z, (eq.4) . What gives? We’re doing everything right, just like we were before, and getting the wrong result!!!
And it gets worse…
If n=”The number of variables we don’t know, but when all are present result in success”, then there are 2(n)-2 possible permutations of possible failure modes. In other words, if there are two unknown variables, then there are two possible failure mode combinations; 3 variables translates to 6; 4 to 14; and 5 unknown variables could lead to 30 possible failure modes!
Which brings us to the main point of this post.
Failure in the product development process is a necessity!
Why?
We ultimately want to get to the Success Equation (eq.1). We want to be able to know that every time we do A and B and C, we get X.
The best way to get there is to convert every “Devil We Don’t Know” equation (eq.3) to a “Devil We Know” equation (eq. 2). And that only happens through testing, experimenting, failure, and learning from those failures.
So the lesson from this Law is this: Next time you’re testing a product and it works as expected, don’t fall into the trap of thinking that your product is working perfectly. Test. Fail. Test again! Avail yourself of digital tools and novel testing techniques (and people that like to break things) to create failures and learn from them. Find out what you don’t know.
Fail early, fail often, fail to learn, fail to succeed.
Oh, and this Law needs a name. Any suggestions?
Republished with author's permission from original post.
Michael Plishka
Michael Plishka is the President and Founder of ZenStorming Solutions, LLC an innovation design consultancy. He believes in co-design methodologies, sharing design thinking essentials - empowering people and companies to make a difference with their products and services. | 848 | 3,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.946433 |
https://codedump.io/share/VuAkgem2cwlE/1/little-changes-in-the-given-algorithm | 1,477,260,825,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719437.30/warc/CC-MAIN-20161020183839-00337-ip-10-171-6-4.ec2.internal.warc.gz | 852,505,286 | 9,792 | mementomori - 4 months ago 6x
Java Question
# Little changes in the given algorithm
It's about changing a maze algorithm.
What I mean by that? We got a 2-dimensional array filled with 0 and 1 where 0 stands for "not possible to pass" and 1 for "possible to pass".
An that algorithm finds its way from x to y (also known example: cat to mouse).
And this exactly is what the following algorithm is doing.
As input we got:
``````{1, 0, 0,},
{1, 1, 0},
{0, 1, 1} };
``````
And the output:
``````(0,0) // ressembles the coordinates of the 1 in top left corner
(1,0) // ressembles the 1 under the first 1 I just explained
(1,1) // ...
(2,1)
(2,2)
``````
I want change some little things:
1. Change the starting and end position (this algorithm starts in top left and ends in bottom right) - I want mine to start in bottom left and end top right.
2. This algorithm can only move down and right - I want only move up and right.
What changes need to be done, I'm pretty sure but I don't know how to code that:
For 1.) the problem seems to be:
``````public List<Coordinate> solve() {
return getMazePath(0, 0, new Stack<Coordinate>());
}
``````
Somehow, I need to do 0-1 with the second zero but how if I haven't access to x and y declaration? I really believe 0-1 would make me start at bottom left instead of top left, is that right?
For 2.) changes for the column, also know as y need to be done.
Instead of +1 it requires -1, is that right?
Sorry for that wall of text, I really tried to keep it short but I seem to have failed :P
Anyway I hope someone will read this^^
Algorithm WITHOUT the changes:
``````import java.util.Arrays;
import java.util.*;
final class Coordinate {
private final int x;
private final int y;
public Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
public int getX() {
return x;
}
public int getY() {
return y;
}
}
public class Alg {
private final int[][] maze;
public Alg(int[][] maze) {
if (maze == null) {
throw new NullPointerException("The input maze cannot be null");
}
if (maze.length == 0) {
throw new IllegalArgumentException("The size of maze should be greater than 0");
}
this.maze = maze;
}
public List<Coordinate> solve() {
return getMazePath(0, 0, new Stack<Coordinate>());
}
private List<Coordinate> getMazePath(int row, int col, Stack<Coordinate> stack) {
assert stack != null;
if ((row == maze.length - 1) && (col == maze[0].length - 1)) {
Coordinate[] coordinateArray = stack.toArray(new Coordinate[stack.size()]);
return Arrays.asList(coordinateArray);
}
for (int j = col; j < maze[row].length; j++) {
if ((j + 1) < maze[row].length && maze[row][j + 1] == 1) {
return getMazePath(row, j + 1, stack);
}
if ((row + 1) < maze.length && maze[row + 1][col] == 1) {
return getMazePath(row + 1, col, stack);
}
}
return Collections.emptyList();
}
public static void main(String[] args) {
int[][] m = { {1, 0, 0,},
{1, 1, 0},
{0, 1, 1} };
Alg maze = new Alg(m);
for (Coordinate coord : maze.solve()) {
System.out.println("("+coord.getX() + "," + coord.getY()+")");
}
}
}
``````
Look at the method declaration for `getMazePath`. The current 0, 0 is passed as the `row` and `col` to that argument. So instead of sending 0, 0 to that method as currently coded, you'll send 2, 0 (for row 2, column 0, the bottom left).
The directional movements are in the `for` loop inside that `getMazePath()` method, the two if statements check if there is a 1 in column `j + 1` (the column to the right) or in row `row + 1` (the row below the current one). You'd modify those to use minus instead of plus to move to the left, or up, respectively. | 1,025 | 3,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-44 | longest | en | 0.87881 |
http://math.stackexchange.com/questions/799325/question-regarding-basic-probability | 1,467,007,547,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395621.98/warc/CC-MAIN-20160624154955-00085-ip-10-164-35-72.ec2.internal.warc.gz | 192,618,204 | 17,864 | # Question regarding basic probability.
I'm looking for confirmation that what I for did this question is correct:
"If two people are randomly chosen from a group of eight women and six men, what is the probability that (a) both are women (b) one is a man and the other is a woman?"
a) The probability that both are women: $\frac{8}{14}\times\frac{7}{13}\approx.31$
b) The probability that one is a man and the other is a woman: $\frac{6}{14}\times\frac{8}{13}\approx.26$
-
Why did you divide by $16$ in $b$ ? – Belgi May 17 '14 at 17:24
oops, that was a typo. – Oscar Flores May 17 '14 at 17:25
## 3 Answers
Unfortunately, no. Your method works in part (a), but fails in part (b).
For a more generally workable method, we could proceed as follows. There are $\binom{14}2=91$ ways to pick $2$ people from a group of $14$. For part (a), there are $\binom82=28$ ways to pick two people from a group of $8$, so the answer is $\frac{28}{91}=\frac4{13},$ as you calculated.
For part (b), though, there are $8$ ways to choose one woman out of $8$ and $6$ ways to choose one man out of $6$, so there are $48$ ways to choose one man and one woman. Thus, the probability is $\frac{48}{91},$ instead.
Basically, what you found is the probability that a man was picked first and then a woman. However, for our purposes, we could have picked a woman first, instead, so we need to multiply this probability by $2$ to get the correct answer.
-
Why do you call (a) coincidence? His method is what I'd have used, and it's correct. – JoeTaxpayer May 17 '14 at 19:42
True. It's perfectly fine. – Cameron Buie May 17 '14 at 22:09
The first one is correct.
However, to see the problem with part $b$ try to answer this question: What is the probability that the first person is men and the second one is a woman ?
-
There are two possibilities of getting one of each gender:
First "man" then "woman" or first "woman" then "man"
- | 542 | 1,925 | {"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} | 4.375 | 4 | CC-MAIN-2016-26 | latest | en | 0.959571 |
http://cn.metamath.org/mpeuni/txcn.html | 1,653,204,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00739.warc.gz | 17,553,433 | 11,659 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > txcn Structured version Visualization version GIF version
Theorem txcn 21477
Description: A map into the product of two topological spaces is continuous iff both of its projections are continuous. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 22-Aug-2015.)
Hypotheses
Ref Expression
txcn.1 𝑋 = 𝑅
txcn.2 𝑌 = 𝑆
txcn.3 𝑍 = (𝑋 × 𝑌)
txcn.4 𝑊 = 𝑈
txcn.5 𝑃 = (1st𝑍)
txcn.6 𝑄 = (2nd𝑍)
Assertion
Ref Expression
txcn ((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) → (𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ↔ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))))
Proof of Theorem txcn
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 txcn.1 . . . . 5 𝑋 = 𝑅
21toptopon 20770 . . . 4 (𝑅 ∈ Top ↔ 𝑅 ∈ (TopOn‘𝑋))
3 txcn.2 . . . . 5 𝑌 = 𝑆
43toptopon 20770 . . . 4 (𝑆 ∈ Top ↔ 𝑆 ∈ (TopOn‘𝑌))
5 txcn.5 . . . . . . 7 𝑃 = (1st𝑍)
6 txcn.3 . . . . . . . 8 𝑍 = (𝑋 × 𝑌)
76reseq2i 5425 . . . . . . 7 (1st𝑍) = (1st ↾ (𝑋 × 𝑌))
85, 7eqtri 2673 . . . . . 6 𝑃 = (1st ↾ (𝑋 × 𝑌))
9 tx1cn 21460 . . . . . 6 ((𝑅 ∈ (TopOn‘𝑋) ∧ 𝑆 ∈ (TopOn‘𝑌)) → (1st ↾ (𝑋 × 𝑌)) ∈ ((𝑅 ×t 𝑆) Cn 𝑅))
108, 9syl5eqel 2734 . . . . 5 ((𝑅 ∈ (TopOn‘𝑋) ∧ 𝑆 ∈ (TopOn‘𝑌)) → 𝑃 ∈ ((𝑅 ×t 𝑆) Cn 𝑅))
11 txcn.6 . . . . . . 7 𝑄 = (2nd𝑍)
126reseq2i 5425 . . . . . . 7 (2nd𝑍) = (2nd ↾ (𝑋 × 𝑌))
1311, 12eqtri 2673 . . . . . 6 𝑄 = (2nd ↾ (𝑋 × 𝑌))
14 tx2cn 21461 . . . . . 6 ((𝑅 ∈ (TopOn‘𝑋) ∧ 𝑆 ∈ (TopOn‘𝑌)) → (2nd ↾ (𝑋 × 𝑌)) ∈ ((𝑅 ×t 𝑆) Cn 𝑆))
1513, 14syl5eqel 2734 . . . . 5 ((𝑅 ∈ (TopOn‘𝑋) ∧ 𝑆 ∈ (TopOn‘𝑌)) → 𝑄 ∈ ((𝑅 ×t 𝑆) Cn 𝑆))
16 cnco 21118 . . . . . . 7 ((𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ 𝑃 ∈ ((𝑅 ×t 𝑆) Cn 𝑅)) → (𝑃𝐹) ∈ (𝑈 Cn 𝑅))
17 cnco 21118 . . . . . . 7 ((𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ 𝑄 ∈ ((𝑅 ×t 𝑆) Cn 𝑆)) → (𝑄𝐹) ∈ (𝑈 Cn 𝑆))
1816, 17anim12dan 900 . . . . . 6 ((𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ (𝑃 ∈ ((𝑅 ×t 𝑆) Cn 𝑅) ∧ 𝑄 ∈ ((𝑅 ×t 𝑆) Cn 𝑆))) → ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆)))
1918expcom 450 . . . . 5 ((𝑃 ∈ ((𝑅 ×t 𝑆) Cn 𝑅) ∧ 𝑄 ∈ ((𝑅 ×t 𝑆) Cn 𝑆)) → (𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) → ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))))
2010, 15, 19syl2anc 694 . . . 4 ((𝑅 ∈ (TopOn‘𝑋) ∧ 𝑆 ∈ (TopOn‘𝑌)) → (𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) → ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))))
212, 4, 20syl2anb 495 . . 3 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top) → (𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) → ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))))
22213adant3 1101 . 2 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) → (𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) → ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))))
23 cntop1 21092 . . . . . . . 8 ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) → 𝑈 ∈ Top)
2423ad2antrl 764 . . . . . . 7 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → 𝑈 ∈ Top)
25 txcn.4 . . . . . . . 8 𝑊 = 𝑈
2625topopn 20759 . . . . . . 7 (𝑈 ∈ Top → 𝑊𝑈)
2724, 26syl 17 . . . . . 6 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → 𝑊𝑈)
2825, 1cnf 21098 . . . . . . 7 ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) → (𝑃𝐹):𝑊𝑋)
2928ad2antrl 764 . . . . . 6 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (𝑃𝐹):𝑊𝑋)
3025, 3cnf 21098 . . . . . . 7 ((𝑄𝐹) ∈ (𝑈 Cn 𝑆) → (𝑄𝐹):𝑊𝑌)
3130ad2antll 765 . . . . . 6 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (𝑄𝐹):𝑊𝑌)
328, 13upxp 21474 . . . . . . 7 ((𝑊𝑈 ∧ (𝑃𝐹):𝑊𝑋 ∧ (𝑄𝐹):𝑊𝑌) → ∃!(:𝑊⟶(𝑋 × 𝑌) ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
33 feq3 6066 . . . . . . . . . 10 (𝑍 = (𝑋 × 𝑌) → (:𝑊𝑍:𝑊⟶(𝑋 × 𝑌)))
346, 33ax-mp 5 . . . . . . . . 9 (:𝑊𝑍:𝑊⟶(𝑋 × 𝑌))
35343anbi1i 1272 . . . . . . . 8 ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ↔ (:𝑊⟶(𝑋 × 𝑌) ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
3635eubii 2520 . . . . . . 7 (∃!(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ↔ ∃!(:𝑊⟶(𝑋 × 𝑌) ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
3732, 36sylibr 224 . . . . . 6 ((𝑊𝑈 ∧ (𝑃𝐹):𝑊𝑋 ∧ (𝑄𝐹):𝑊𝑌) → ∃!(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
3827, 29, 31, 37syl3anc 1366 . . . . 5 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ∃!(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
39 euex 2522 . . . . 5 (∃!(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) → ∃(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
4038, 39syl 17 . . . 4 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ∃(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
41 simpll3 1122 . . . . . . 7 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → 𝐹:𝑊𝑍)
4227adantr 480 . . . . . . 7 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → 𝑊𝑈)
431topopn 20759 . . . . . . . . . 10 (𝑅 ∈ Top → 𝑋𝑅)
443topopn 20759 . . . . . . . . . 10 (𝑆 ∈ Top → 𝑌𝑆)
45 xpexg 7002 . . . . . . . . . . 11 ((𝑋𝑅𝑌𝑆) → (𝑋 × 𝑌) ∈ V)
466, 45syl5eqel 2734 . . . . . . . . . 10 ((𝑋𝑅𝑌𝑆) → 𝑍 ∈ V)
4743, 44, 46syl2an 493 . . . . . . . . 9 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top) → 𝑍 ∈ V)
48473adant3 1101 . . . . . . . 8 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) → 𝑍 ∈ V)
4948ad2antrr 762 . . . . . . 7 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → 𝑍 ∈ V)
50 fex2 7163 . . . . . . 7 ((𝐹:𝑊𝑍𝑊𝑈𝑍 ∈ V) → 𝐹 ∈ V)
5141, 42, 49, 50syl3anc 1366 . . . . . 6 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → 𝐹 ∈ V)
52 eumo 2527 . . . . . . . 8 (∃!(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) → ∃*(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
5338, 52syl 17 . . . . . . 7 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ∃*(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
5453adantr 480 . . . . . 6 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ∃*(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
55 simpr 476 . . . . . 6 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
56 3anass 1059 . . . . . . . 8 ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ↔ (:𝑊𝑍 ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))))
57 coeq2 5313 . . . . . . . . . . . 12 (𝐹 = → (𝑃𝐹) = (𝑃))
58 coeq2 5313 . . . . . . . . . . . 12 (𝐹 = → (𝑄𝐹) = (𝑄))
5957, 58jca 553 . . . . . . . . . . 11 (𝐹 = → ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
6059eqcoms 2659 . . . . . . . . . 10 ( = 𝐹 → ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
6160biantrud 527 . . . . . . . . 9 ( = 𝐹 → (:𝑊𝑍 ↔ (:𝑊𝑍 ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))))
62 feq1 6064 . . . . . . . . 9 ( = 𝐹 → (:𝑊𝑍𝐹:𝑊𝑍))
6361, 62bitr3d 270 . . . . . . . 8 ( = 𝐹 → ((:𝑊𝑍 ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) ↔ 𝐹:𝑊𝑍))
6456, 63syl5bb 272 . . . . . . 7 ( = 𝐹 → ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ↔ 𝐹:𝑊𝑍))
6564moi2 3420 . . . . . 6 (((𝐹 ∈ V ∧ ∃*(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) ∧ ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ 𝐹:𝑊𝑍)) → = 𝐹)
6651, 54, 55, 41, 65syl22anc 1367 . . . . 5 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → = 𝐹)
67 eqid 2651 . . . . . . . . . 10 (𝑅 ×t 𝑆) = (𝑅 ×t 𝑆)
6867, 1, 3, 6, 5, 11uptx 21476 . . . . . . . . 9 (((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆)) → ∃! ∈ (𝑈 Cn (𝑅 ×t 𝑆))((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
6968adantl 481 . . . . . . . 8 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ∃! ∈ (𝑈 Cn (𝑅 ×t 𝑆))((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))
70 df-reu 2948 . . . . . . . . . 10 (∃! ∈ (𝑈 Cn (𝑅 ×t 𝑆))((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ↔ ∃!( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))))
71 euex 2522 . . . . . . . . . 10 (∃!( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ∃( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))))
7270, 71sylbi 207 . . . . . . . . 9 (∃! ∈ (𝑈 Cn (𝑅 ×t 𝑆))((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) → ∃( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))))
73 eqid 2651 . . . . . . . . . . . . . . 15 (𝑅 ×t 𝑆) = (𝑅 ×t 𝑆)
7425, 73cnf 21098 . . . . . . . . . . . . . 14 ( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) → :𝑊 (𝑅 ×t 𝑆))
751, 3txuni 21443 . . . . . . . . . . . . . . . . . 18 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top) → (𝑋 × 𝑌) = (𝑅 ×t 𝑆))
766, 75syl5eq 2697 . . . . . . . . . . . . . . . . 17 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top) → 𝑍 = (𝑅 ×t 𝑆))
77763adant3 1101 . . . . . . . . . . . . . . . 16 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) → 𝑍 = (𝑅 ×t 𝑆))
7877adantr 480 . . . . . . . . . . . . . . 15 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → 𝑍 = (𝑅 ×t 𝑆))
7978feq3d 6070 . . . . . . . . . . . . . 14 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (:𝑊𝑍:𝑊 (𝑅 ×t 𝑆)))
8074, 79syl5ibr 236 . . . . . . . . . . . . 13 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) → :𝑊𝑍))
8180anim1d 587 . . . . . . . . . . . 12 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → (:𝑊𝑍 ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)))))
8281, 56syl6ibr 242 . . . . . . . . . . 11 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))))
83 simpl 472 . . . . . . . . . . . 12 (( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ∈ (𝑈 Cn (𝑅 ×t 𝑆)))
8483a1i 11 . . . . . . . . . . 11 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ∈ (𝑈 Cn (𝑅 ×t 𝑆))))
8582, 84jcad 554 . . . . . . . . . 10 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ ∈ (𝑈 Cn (𝑅 ×t 𝑆)))))
8685eximdv 1886 . . . . . . . . 9 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (∃( ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ∧ ((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ∃((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ ∈ (𝑈 Cn (𝑅 ×t 𝑆)))))
8772, 86syl5 34 . . . . . . . 8 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → (∃! ∈ (𝑈 Cn (𝑅 ×t 𝑆))((𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) → ∃((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ ∈ (𝑈 Cn (𝑅 ×t 𝑆)))))
8869, 87mpd 15 . . . . . . 7 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ∃((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ ∈ (𝑈 Cn (𝑅 ×t 𝑆))))
89 eupick 2565 . . . . . . 7 ((∃!(:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ ∃((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) ∧ ∈ (𝑈 Cn (𝑅 ×t 𝑆)))) → ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) → ∈ (𝑈 Cn (𝑅 ×t 𝑆))))
9038, 88, 89syl2anc 694 . . . . . 6 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → ((:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄)) → ∈ (𝑈 Cn (𝑅 ×t 𝑆))))
9190imp 444 . . . . 5 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → ∈ (𝑈 Cn (𝑅 ×t 𝑆)))
9266, 91eqeltrrd 2731 . . . 4 ((((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) ∧ (:𝑊𝑍 ∧ (𝑃𝐹) = (𝑃) ∧ (𝑄𝐹) = (𝑄))) → 𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)))
9340, 92exlimddv 1903 . . 3 (((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) ∧ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))) → 𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)))
9493ex 449 . 2 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) → (((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆)) → 𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆))))
9522, 94impbid 202 1 ((𝑅 ∈ Top ∧ 𝑆 ∈ Top ∧ 𝐹:𝑊𝑍) → (𝐹 ∈ (𝑈 Cn (𝑅 ×t 𝑆)) ↔ ((𝑃𝐹) ∈ (𝑈 Cn 𝑅) ∧ (𝑄𝐹) ∈ (𝑈 Cn 𝑆))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 383 ∧ w3a 1054 = wceq 1523 ∃wex 1744 ∈ wcel 2030 ∃!weu 2498 ∃*wmo 2499 ∃!wreu 2943 Vcvv 3231 ∪ cuni 4468 × cxp 5141 ↾ cres 5145 ∘ ccom 5147 ⟶wf 5922 ‘cfv 5926 (class class class)co 6690 1st c1st 7208 2nd c2nd 7209 Topctop 20746 TopOnctopon 20763 Cn ccn 21076 ×t ctx 21411 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-ral 2946 df-rex 2947 df-reu 2948 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-op 4217 df-uni 4469 df-iun 4554 df-br 4686 df-opab 4746 df-mpt 4763 df-id 5053 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-1st 7210 df-2nd 7211 df-map 7901 df-topgen 16151 df-top 20747 df-topon 20764 df-bases 20798 df-cn 21079 df-tx 21413 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 9,499 | 12,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-21 | latest | en | 0.337932 |
https://kingscalculator.com/sn/math-macalculator/muzana-calculator | 1,701,884,585,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00171.warc.gz | 382,517,432 | 6,903 | # Percentage karukureta
### Chii chinonzi muzana
Percentage inowanzoreva kukosha kwehukama kubva pane hwese kukosha. Isu tinoshandisa muzana semuenzaniso seizvi:
1. Yedu yakazara kukosha pano imiriyoni imwe mota.
2. Uye isu tinoti: "yega yechipiri mota inodarika makore mashanu ekuberekwa"
3. Inoturikirwa kumapecents - "yega yega mota" zvinoreva makumi mashanu muzana (50%).
4. Mhinduro chaiyo ndeiyi: hafu yemamiriyoni emotokari yakura kupfuura makore mashanu.
Imwe muzana zvakare zvinoreva zana. Kubva pamuenzaniso wepamusoro - zana (1%) kubva kumamiriyoni angave zana rimwe chete rezviuru. $$x = \frac{1 000 000}{100} = 100 000 \\$$
#### Percentage fomula
$$Percentage = Kukosha / TotalValue \cdot 100 \\[1ex]$$
Semuenzaniso: Mangani muzana muzana mota shanu kubva pagumi mota
$$Percentage = (5 / 10) \cdot 100 \\ Percentage = 50\%$$
#### Kukosha fomura
$$Kukosha = Percentage \cdot (TotalValue / 100) \\[1ex]$$
Semuenzaniso: Mangani mota ari 10% emota makumi mashanu
$$Kukosha = 10 \cdot (50 / 100) \\ Kukosha = 5 \, mota$$
#### Yese kukosha kukosha
$$TotalValue = Kukosha \cdot (100 / Percentage) \\[1ex]$$
Semuenzaniso: Chii chinonzi TotalValue kana mota shanu dziri 50%
$$TotalValue = 5 \cdot (100 / 50) \\ TotalValue = 10\; mota$$ | 487 | 1,249 | {"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} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.218179 |
http://brainden.com/forum/topic/16128-at-book-store/ | 1,601,535,717,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00053.warc.gz | 22,478,791 | 18,398 | BrainDen.com - Brain Teasers
• 0
# At Book Store
Go to solution Solved by bonanova,
## Question
Two men, David and Clifton, and their wives, Kim and Allison, go out shopping for books. Each person paid for each book a number of dollars equal to their number of books. David bought 1 more book than Kim, while Allison bought only 1 book. If each couple spent the same two-digit sum, who is Kim's husband?
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• 0
• Solution
For some arrangement of M-F couples the sums of two sets
of squares lies in the interval [10, 99]. Two of the integers
being squared differ by 1, and one of the integers is 1.
Find Kim's husband.
There are two possibilities: David and Clifton.
Assume Kim's husband is David. Then,
9 < d2 + k2 = c2 + 1 < 100
There are no solutions. The closest ones are
(d k c sum) = (9 8 12 145 >99) and (2 1 2 5<10)
David is not Kim's husband; Clifton is
.
Verify OP claim that a solution exists for the Clifton-Kim marriage.
9 < (k+1)2 + 1 = c2 + k2 < 100
9 < 82 + 1 = 42 + 72 = 65 < 100
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OP says: "Each person paid for each book a number of dollars equal to their number of books."
Which is meant?
1. Each person paid for each book bought by that person a number of dollars equal to the number of books bought by that person.
2. Each person paid for each book bought by that person a number of dollars equal to the number of books they all bought together.
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• 0
Two men, David and Clifton, and their wives, Kim and Allison, go out shopping for books. Each person paid for each book a number of dollars equal to their number of books. David bought 1 more book than Kim, while Allison bought only 1 book. If each couple spent the same two-digit sum, who is Kim's husband?
Say Kim bought x books for x^2 dollars.
David bought x+1 books for (x+1)^2 dollars.
Allison bought only 1 book for 1 dollar.
Clifton bought y books for y^2 dollars.
If David and Kim were the couple, then they spent x^2 + (x + 1)^2 = two digit number = say z…....[1]
So x < 7 and x > 1
Then remaining couple Allison and Clifton would spend 1 + y^2 = z…………………………..…......[2]
So y can’t be greater than 9, i.e. z is not greater than 82.
Then from [1], x < 6 or z cannot be greater than 61.
Then from [2], y can’t be greater than 7, i.e. z can’t be greater than 50.
Then from [1], x < 5 or Z is not greater than 41.
And so on…
But we do not find suitable values of X and Y which satisfy both [1] & [2].
Note this can be prooved by equation y = SQRT 2x(x + 1), where we get only suitable values of x as 1 to get a whole number y. But from [1], X can't be lesser than 2, to get double digit number Z.
Therefore David and Kim are not the couple, so David and Allison are the husband & wife.
And Clifton is husband of Kim.
Then we have the two equations
(x + 1)^2 + 1 = z………………………………………………………………………………………..[3]
X^2 + y^2 = z…………………………………………………………………………………………….[4]
From [3] & [4], y^2 = 2x + 2 OR y = √(2x + 2)..…………………………………………………….[3]
[3] can be satisfied only when x= 7, or 1 to get y as a whole number.
But from [1], to get Z a double digit number X must be greater than 1.
Therefore X = 7, and y = 4.
Answer: David bought 8 books, his wife Allison bought 1 book. The spent total (64 + 1) = 65 dollars.
Clifton bought 4 books and his wife Kim bought 7 books, and they also spent 16 + 49 =65 dollars.
Edited by bhramarraj
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× You cannot paste images directly. Upload or insert images from URL. | 1,067 | 3,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | longest | en | 0.947519 |
https://smashville247.net/17-grams-equal-how-many-tablespoons/ | 1,653,030,721,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00016.warc.gz | 606,610,568 | 4,630 | ## 17 Grams to Tablespoons (17g to tbsp) General Conversion
Learn how to fast and easily convert 17 Grams to Tablespoons. So, how many tablespoons in 17 grams?17 Grams (g) is Equal to 1.133 Tablespoons (tbsp)or 17 g = 1.133 tbsp
## 17 Grams to Tablespoons Conversion Calculator
You are watching: 17 grams equal how many tablespoons
## How to Convert 17 g to tbsp
It’s easy to convert grams to tablespoons. For the general equation just divide the grams by 15 to convert them to tablespoons.17g to tbsp calculation:Conversion factortbsp = g ÷ 15 17 Grams to Tablespoons Conversion Equation17 g ÷ 15 = 1.133 tbsp
## Common Grams to Tablespoon Conversions
GramsTablespoonsGramsTablespoons1 g0.067 or 1/15 tbsp20 g1.333 tbsp2 g0.133 or 2/15 tbsp30 g2 tbsp3 g0.2 or 1/5 tbsp40 g2.667 tbsp4 g0.267 or 4/15 tbsp50 g3.333 tbsp5 g0.333 or 1/3 tbsp60 g4 tbsp6 g0.4 or 2/5 tbsp70 g4.667 tbsp7 g0.467 or 7/15 tbsp80 g5.333 tbsp8 g0.533 or 8 /15 tbsp90 g5 tbsp9 g0.6 or 3/15 tbsp100 g6.667 tbsp10 g0.667 or 2/3 tbsp1000 g66.667 tbsp
## Ingredient Gram to Tablespoon Conversions
Not all gram to tablespoon conversions are the same, and depend on the specific ingredients density. This conversion is not always clear, because a gram is a unit of weight, whereas a tablespoon is a unit of volume.For example let’s look at a tablespoon of sugar. 12.6 grams of sugar would be equal to 1 tablespoon, instead of the general 15 grams per tablespoon. The added sugar is less dense, so it takes less grams to equal a tablespoon.Here is a list of common ingredients with measurement in volume, and their gram to tablespoon conversion tables:Ingredient1 Tablespoon (tbsp) =Water14.79 gSugar12.6 gHoney21 gFlour7.83 gMilk15.3 gButter14.19 gBaking Powder13.32 g
Check out this link for conversions of cooking ingredients.
## Convert Tablespoons to Grams
It’s also easy to convert tablespoons to grams. For the general equation just multiply the tablespoons by 15 to convert them to grams.tbsp to g calculation:Conversion factor15 g = 1 tbsp * 15 Example Tablespoons to Grams Conversion Equation17tbsp * 15 = 255 g
## Convert 17 Grams to Other Units
Do you want to convert 17 grams into another unit? Here is a helpful table for converting 17 grams into other units:Unit17 Grams (g) =Micrograms (mcg)17,000,000 mcgMilligram (mg)17,000 mgKilogram (kg)0.017 kgOunce (oz)0.6 ozPound (lb)0.037 lbTeaspoon (tsp)3.4 Tsp
## What is a Gram?
The gram is a unit of mass in the commonly used metric system of measurement. The official definition is that a gram is one thousandth of the International Systems of Units (SI) base unit for mass, which is the kilogram.The abbreviated symbol for a gram is “g”. Example 56 grams is the same as 56 g.See the dictionary definition here.
## What is a Tablespoon?
The tablespoon is used as a measurement of volume, most commonly used as a measurement in cooking recipes.
See more: How To Substitute Butter Instead Of Vegetable Oil In Brownies
Most commonly it is 1/16 of a cup or 3 teaspoons.The abbreviated symbol for a tablespoon is “tbsp”. Examples 52 tablespoons is the same as 52 tbsp.See the dictionary definition here. | 893 | 3,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.739974 |
http://www.jiskha.com/display.cgi?id=1191722062 | 1,498,265,326,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320206.44/warc/CC-MAIN-20170623235306-20170624015306-00165.warc.gz | 567,761,745 | 3,889 | # physics
posted by .
A spelunker drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.50s after the stone is dropped. How deep is the hole?
how can i use the speed of sound to solve this problem?Please give me some hints!
Thanks a lot!
• physics -
The time required to hear the impact is the sum of the time for the stone to fall and the time for the sound wave to reach the top of the hole.
If the hole depth is X,
1.50 seconds = sqrt (2X/g) + X/v
where v is the speed of sound and g is the acceleration of gravity.
Solve for X
• physics -
10.6m | 167 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.904175 |
https://books.google.gr/books?id=NycAAAAAYAAJ&qtid=e07e7840&lr=&hl=el&source=gbs_quotes_r&cad=6 | 1,618,430,153,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00100.warc.gz | 258,789,700 | 6,151 | Αναζήτηση Εικόνες Χάρτες Play YouTube Ειδήσεις Gmail Drive Περισσότερα »
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The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - 1762 - 466 σελίδες
...triangle DEF. Wherefore if two triangles, &c. Q^ED PROP, Book VI. \^r«j PROP. VIII. THE OR. 8ee N. TN a right angled triangle, if a perpendicular be drawn from the right angle to the bafe; the triangles ->n each fide of it are firnilar to the whole triangle, ar.d to one another. Let...
A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the ...
Thomas Malton - 1774 - 440 σελίδες
...:EG : GI, and as EB : BD. Hence is deduced the 36th, a moft ufcful Problem. THEOREM VII. 8 Euclid. In a right angled Triangle, if a Perpendicular be drawn, from the Right Angle to the Hypothenufe, it will divide that Triangle into two Triangles, which are fimilar to each other; and...
The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ...
Robert Simson - 1775 - 520 σελίδες
...equiangular to the triangle DEF. Wherefore, if two triangles, &c. PROP Book vI. PROP. VIII. THEO R. See N. TN a right angled triangle, if a perpendicular be drawn from the right angle to the bafe ; the triangles on each fide of it are fimilar to the whole triangle, and to one another. Let...
Elements of Geometry;: Containing the First Six Books of Euclid, with Two ...
Euclid, John Playfair - 1795 - 400 σελίδες
...ABC may be proved to be equiangular to the triangle DEF, as in the firft cafe. PROP. VIII. THEO R. !N a right angled triangle, if a perpendicular be drawn from the right angle to the bafe; the triangles on each fide of it are fimilar to the whole triangle, and to one another. Let ABC...
The Elements of Euclid: Viz. the First Six Books, with the Eleventh and ...
Alexander Ingram - 1799 - 351 σελίδες
...Wherefore the triangle ABC is equiangular to the triangle DEF. Therefore, &c. Q^ED PROP. VIII. THEOR. IN a right angled triangle, if a perpendicular be drawn from the right angle to the bafe ; the triangles on each fide of it are fimilar to the whole triangle, and to one another. Let...
Elements of Geometry;: Containing the First Six Books of Euclid, with a ...
Euclid, John Playfair - 1804 - 440 σελίδες
...ABC may be proved, to be equiangular to the triangle DEF, as in the firft cafe. PROP. VIII. THEOR. IN a right angled triangle, if a perpendicular be drawn from the right angle to the bafe ; the triangles on each fide of it are limilar to the whole triangle, and to one another. Let...
The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The ...
Robert Simson - 1804
...triangle DEF, &c. QED . \Vhercfcre if two triangles, Book VI. V— v-"' PROP. VIII. THEOR. See N. TN a right angled triangle, if a perpendicular be drawn ,•. from the right angle to the bale ; the triangles on each fide of it are fimjlar to the whole triangle, and to one another. Let...
The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - 1806 - 518 σελίδες
...equiangular to the triangle DEF. Wherefore, if two triangles, &c. QED G D 5.1. F PROP. VIII. THEOR. See N IN a right angled triangle, if a perpendicular be...triangles on each side of it are similar to the whole triangle, and to one another. Let ABC be a right angled triangle, having the right angle BAC ; and...
Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1806 - 311 σελίδες
...to the triangle DEF, as in the first case. Therefore, if two triangles, &c. QED PROP. VIII. THEOR. IN a right angled triangle, if a perpendicular be...triangles on each side of it are similar to the whole triangle, and to each other* Let ABC be a right angled triangje, having the right angle BAC, and from...
Pantologia. A new (cabinet) cyclopædia, by J.M. Good, O. Gregory ..., Τόμος 5
John Mason Good - 1813
...equiangular, and have those angles equal about which the sities are proportionals. Prop. VllI.Thcor. In a right angled triangle, if a perpendicular be...triangles on each side of it are similar to the whole triangle, and to one another. Prop. IX. Prob. From a given strait line to cut off any part required.... | 1,263 | 4,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-17 | latest | en | 0.53129 |
http://www.jiskha.com/display.cgi?id=1296965840 | 1,498,326,404,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320270.12/warc/CC-MAIN-20170624170517-20170624190517-00322.warc.gz | 558,721,015 | 3,941 | # Calculus
posted by .
Consider the function f(x)=8.5x−cos(x)+2 on the interval 0¡Üx¡Ü1 . The Intermediate Value Theorem guarantees that there is a value c such that f(c)=k for which values of c and k? Fill in the following mathematical statements, giving an interval with non-zero length in each case.
For every k in ______¡Ük¡Ü______
there is a c in 0 ¡Üc¡Ü 1
• Calculus -
What is "c in 0 ¡Üc¡Ü 1" ?
You should always check your question after posting.
A tutor can't help if they cannot understand your 'symbols'.
• Calculus -
Sorry...
Consider the function f(x) = 8.5 x − cos(x) + 2 on the interval 0 ¡Ü x ¡Ü 1. The Intermediate Value Theorem guarantees that there is a value c such that f(c) = k for which values of c and k? Fill in the following mathematical statements, giving an interval with non-zero length in each case.
For every k in ____ ¡Ü k ¡Ü___ ,
there is a c in 0 ¡Ü c ¡Ü 1
such that f(c) = k. | 266 | 921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.780759 |
https://stats.stackexchange.com/questions/166520/basic-permutation-question/166524 | 1,639,012,155,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00462.warc.gz | 596,095,553 | 34,420 | # Basic Permutation Question
I've started studying permutations and I'm confused as to how to properly solve a problem. For example, select 4 digits from the set $\{1,2,3,4,5,6,7\}$ without replacement. What is the probability that the number formed is even?
Looking at this question it's pretty obvious that the answer is $\frac 37$ since there are 3 even numbers from the set so the last digit (ones digit) can be any of the even numbers. However, if I approach this problem step by step I cannot get the same answer:
To solve this problem step by step, let $A$ be the event that the number is even and S represent the sample space. Let $n(s)$ be the number of ways any number can be formed and $n(A)$ be the number of ways an even number can be formed.
$n(s) = 7^{(4)}$ since the first digit (the thousands digit) can be chosen from 7 the next from 6... Now I'm confused on how to calculate $n(A)$ since the first (thousands), second(hundreds), or third digit(tens) can be an even number. I calculated $n(A)$ as following: $n(A) = (7*6*5*3) + (7*6*5*2) + (7*6*5*1)$ the first group is when the first three digits are odd, the second group is for when one of the digits is odd and the last group is when two digits are odd. But my dilemma is that now $n(A) > n(s)$ which is impossible so I did something wrong.
## 1 Answer
Your notation is a bit strange. For instance, what is $7^{(4)}$? Also, why do you consider it necessary to count all the permutations of the other digits? Just consider the partition of the sample space where we only look at the first digit since the number is even if and only if the first digit is even.
In any case, the permutations that correspond to the first digit being even are $3 \cdot 6 \cdot 5 \cdot 4$ in number (the first must be even, the next can be any other value, etc.), and the total number of permutations is $7 \cdot 6 \cdot 5 \cdot 4$. If you divide the former by the latter you get $3/7$.
• This is a solution for the first sampled number being even. I thought the question was asking "what are the odds of the fourth number being even?" Aug 10 '15 at 18:02
• "What is the probability that the number formed is even?" Aug 10 '15 at 18:03
• If you pull {1,2,3,4} you can either form 1234 or 4321. That's why the questioner was concerned with "the first, second, or third digit can be even." Aug 10 '15 at 18:06
• I think it's understood that we're sampling the digits in order, otherwise the question makes no sense. Is $\{1, 2, 3, 4 \}$ even or odd? This has no answer if the selection can correspond to either $1234$ or $4321$. Also what would be the relevance of the first, second or third digit being even? Aug 10 '15 at 18:12
• dsaxton I am a bit confused as well with the statement "since the number is even if and only if the first digit is even." A number where the first digit is even does not mean the number itself will be even for example $4321$ is an odd number with an even first digit. When I wrote $7^{(4)}$ I meant $7 \cdot 6 \cdot 5 \cdot 4$. Aug 10 '15 at 18:13 | 827 | 3,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-49 | latest | en | 0.925813 |
http://at.metamath.org/ileuni/2p1e3.html | 1,723,461,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00083.warc.gz | 2,526,634 | 2,840 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > 2p1e3 GIF version
Theorem 2p1e3 8043
Description: 2 + 1 = 3. (Contributed by Mario Carneiro, 18-Apr-2015.)
Assertion
Ref Expression
2p1e3 (2 + 1) = 3
Proof of Theorem 2p1e3
StepHypRef Expression
1 df-3 7974 . 2 3 = (2 + 1)
21eqcomi 2044 1 (2 + 1) = 3
Colors of variables: wff set class Syntax hints: = wceq 1243 (class class class)co 5512 1c1 6890 + caddc 6892 2c2 7964 3c3 7965 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1336 ax-gen 1338 ax-ext 2022 This theorem depends on definitions: df-bi 110 df-cleq 2033 df-3 7974 This theorem is referenced by: 1p2e3 8044 cnm2m1cnm3 8176 6t5e30 8447 7t5e35 8452 8t4e32 8457 9t4e36 8464 decbin3 8472
Copyright terms: Public domain W3C validator | 422 | 889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.308066 |
https://electronics.stackexchange.com/questions/542959/measuring-battery-with-low-power-requirement | 1,660,791,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00376.warc.gz | 219,868,748 | 67,404 | # Measuring battery with low power requirement
I am trying to create a low power device, which would work off a LIR2450 or similar coin cell battery. athe device has long periods of sleep with short periods when it performs operation (e.g. reading sensor and/or sending LoRa data).
For this purpose, I am using an atmega328p. The idea is to have a stable 3.3V out of a voltage regulator and correct battery reading over the voltage divider R9/R10.
Here is the relevant part of the diagram:
I have selected high resistance values for the voltage divider in order to reduce current (this might have been a bad idea).
I am getting readings of Vmeas, but depending on the battery voltage they are somehow - not consistent. When I have approx. 3.5V I seem to get around expected measurement of ADC, however, with around 3V or 4V the readings are strange - either lower than expected or sometimes higher, as if the ADC is not linear in its readings.
How do I measure:
// calculate Vbat from adc value, assuming Vbat:Vmeas is over the resistor-devider 1M:2M
// first calculate adc measurement (10 bit adc - reference 3.3V)
double meas = ((double)3.3) * ((double)adcV) / 1023;
// calculate Vbat from voltage devider
double result = meas * (R1M + R2M) / R2M;
return result;
}
// read Vcc in miliVolts
}
Do you see anything wrong with my concept and should it, in general, work as I expect?
• Try putting a 100 nF capacitor across R10. Jan 16, 2021 at 10:06
• that voltage regulator stops working correctly at about 3.6V in. Jan 16, 2021 at 12:16
The problem is that the ADC input works best with impedances below 10k which reads in the datasheet.
The divider has an impedance much greater than that, above 500k.
It has no chance to charge the internal ADC sampling capacitance in the given sampling time.
If you don't plan to sample very often, put something like a 100nF capacitor at the ADC input pin.
If you sample too often, the ADC sampling discharges the 100nF capacitor faster than the resistors charge it so it results into lower voltage result.
• Thanks, I don't sample often, actually it is when the device "wakes up", and that is not expected to occur more than 2-3 times per hour. I will try to reduce resistance in divider circuit, at the expense of more battery consumption, it might still be acceptable. Jan 16, 2021 at 12:05
• The resistances would have to be around 15k and 30k to get source impedance for ADC down to 10k. If that is too much, then simply use the capacitor which allows for higher resistances. Note that the 100nF cap value was a random large enough value you already use. In reality, even a 10nF would hugely improve the reading. Jan 16, 2021 at 12:24
LIR2450 has a voltage output from 4.2 to 2.75V,
ATMEGA328P will operate from 2.7 to 5.5V so you don't need to waste energy running a voltage regulator.
so to measure voltage you need to compare the supply voltage with a known voltage. you can use the internal 1.1V reference.
So use a 4:1 divider from a GPIO pin to ground and set the GPIO high and measure the voltage. then set the GPIO low to save energy.
simulate this circuit – Schematic created using CircuitLab
• Thanks, my issue there is that I need to communicate over SPI with modules that are 3.3V only. My understanding is that atmega would output on pins the voltages that it is supplied with, therefore it would kill 3.3V modules. Jan 16, 2021 at 12:01
• well you could use transistors instead of the GPIO to turn on the top end of the voltage divider. Jan 16, 2021 at 12:18 | 910 | 3,538 | {"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} | 2.84375 | 3 | CC-MAIN-2022-33 | longest | en | 0.911143 |
http://www.statslab.cam.ac.uk/~rrw1/opt1998/solver.html | 1,511,137,041,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805881.65/warc/CC-MAIN-20171119234824-20171120014824-00470.warc.gz | 493,894,059 | 5,001 | Here is a Pascal program to solve small problems using the simplex algorithm.
From: Stephen Gale
Newsgroups: sci.op-research
Subject: Re: SIMPLEX code for PC
Date: Sun, 5 May 1996 07:21:50 -0600
Attached below is a fairly simple Simplex Solver (written for Turbo Pascal 3.0). Please let me know of how you use the program, as well as any problems and any successes. =====================================================================
Stephen F. Gale, B.Sc., CCP, I.S.P.
SFGale@FreeNet.Calgary.Ab.Ca
http://www.freenet.calgary.ab.ca/~sfgale/
=====================================================================
95/96 Program Director
Association for Systems Management - Calgary Chapter
http://www.freenet.calgary.ab.ca/populati/communit/asmcal/asmcal.html
=====================================================================
```************************
*** SIMPLEX SOFTWARE ***
************************
program linearoptimization;
const rowmx = 72;
colmx = 112;
mxval = 1.0E+35;
zero = 0.0;
eqzero = 0.00000001;
var matrix : array [0..rowmx, 0..colmx] of real;
basis : array [1..rowmx] of integer;
basisp : array [1..rowmx] of integer;
minmax : real;
error : integer;
name : string[70];
filename : string[14];
ncon : integer; {number of constraints}
nvar : integer; {number of variables}
nltcon : integer; {number of less than constraints}
neqcon : integer; {number of equal to constraints}
ngtcon : integer; {number of greater than contraints}
trows1 : integer;
trows2 : integer;
tcols1 : integer;
tcols2 : integer;
tcols3 : integer;
function getmatrix( row, col : integer ) : real;
begin
getmatrix := matrix[ row, col ];
end;
procedure putmatrix( row, col : integer; value : real );
begin
matrix[ row, col ] := value;
end;
procedure initmatrix;
var row, col : integer;
begin
for row := 0 to rowmx do
for col := 0 to colmx do
putmatrix( row, col, zero );
end;
procedure price( var xcol : integer; trow : integer; var error : integer );
var quant, val : real;
col : integer;
begin
quant := -eqzero;
for col := 1 to tcols3 do
begin
val := getmatrix( trow, col );
if ( val < quant ) then
begin
xcol := col;
quant := val;
end;
end;
error := 0;
if ( quant = -eqzero ) then
error := 1;
end;
procedure leave( var xrow : integer; xcol : integer; var error : integer );
var quant, val : real;
row : integer;
begin
quant := mxval;
for row := 1 to ncon do
begin
val := getmatrix( row, xcol );
if ( val > eqzero ) then
begin
val := getmatrix( row, tcols2 ) / val;
if ( val < quant ) then
begin
xrow := row;
quant := val;
end;
end;
end;
error := 0;
if ( quant = mxval ) then
error := 2;
end;
procedure pivot( xrow, xcol : integer );
var value, val, vl : real;
row, col : integer;
begin
value := getmatrix( xrow, xcol );
for row := 1 to trows2 do
if ( row <> xrow ) then
begin
vl := getmatrix( row, xcol );
for col := 1 to tcols2 do
if ( col <> xcol ) then
begin
val := getmatrix( row, col ) - vl * getmatrix( xrow, col ) / value;
if ( abs( val ) < eqzero ) then
val := zero;
putmatrix( row, col, val );
end;
end;
for col := 1 to tcols2 do
putmatrix( xrow, col, getmatrix( xrow, col ) / value );
for row := 1 to trows2 do
putmatrix( row, xcol, zero );
putmatrix( xrow, xcol, 1.0);
basis[ xrow ] := xcol;
end;
procedure optimize( trow : integer; var error : integer);
var xrow, xcol, iterate : integer;
begin
repeat
price( xcol, trow, error );
if ( error = 0 ) then
leave( xrow, xcol, error );
if ( error = 0 ) then
pivot( xrow, xcol );
until ( error <> 0 )
end;
procedure simplex( var error : integer );
var val : real;
row, col : integer;
flag : boolean;
label 1000;
begin
if ( ncon <> nltcon ) then
begin
optimize( trows1, error );
if ( error > 1 ) then exit;
error := 3;
for row := 1 to ncon do
if ( basis[ row ] > tcols3 ) then
begin
if ( getmatrix( row, tcols2 ) > eqzero ) then exit;
flag := false;
col := 1;
repeat
if ( abs( getmatrix( row, col ) ) >= eqzero ) then
begin
pivot( row, col );
flag := true;
end;
col := col + 1;
until ( (flag) or (col > tcols3) );
end;
end;
error := 0;
optimize( trows2, error );
end;
procedure reader( var error : integer );
var row, col, column : integer;
value, amt : real;
filevar : text;
begin
error := 0;
writeln(con,'Problem file should be in the following format:');
writeln(con,' Line 1 : Up to 70 character problem description');
writeln(con,' Line 2 : +1 (for max) or -1 (for min); # of constraints; # of variables');
writeln(con,' Line 3 : # of <= constraints; # of = constraints; # of >= constraints');
writeln(con,' Next : Constraints coefficients and RHS value for each constraint');
writeln(con,' Last : Objective function coefficients');
writeln(con);
write (con,'Enter the filename containing the problem: ');
assign(filevar,filename);
reset(filevar);
{ read the problem description }
readln( filevar, name );
{ read the minmax, number of constraints, number of variables }
readln( filevar, minmax, ncon, nvar );
minmax := -minmax;
{ read the number of less than, equal to, greater than contraints}
readln( filevar, nltcon, neqcon, ngtcon );
if ( ncon <> nltcon + neqcon + ngtcon ) then error := -1;
trows1 := ncon + 1;
trows2 := ncon + 2;
tcols1 := nvar + ncon + ngtcon;
tcols2 := tcols1 + 1;
tcols3 := nvar + nltcon + ngtcon;
{prepare matrix and basis}
for row := 1 to trows2 do
for col := 1 to tcols2 do
putmatrix( row, col, zero );
for row := 1 to ncon do
basis[ row ] := 0;
{prepare artificial and surplus variables}
for row := 1 to ncon do
if ( row <= nltcon ) then
begin
column := nvar + row;
basis[ row ] := column;
putmatrix( row, column, +1.0 );
end
else
begin
column := nvar + ngtcon + row;
basis[ row ] := column;
putmatrix( row, column, +1.0 );
if ( row > nltcon + neqcon ) then
begin
column := nvar - neqcon + row;
putmatrix( row, column, -1.0 );
putmatrix( trows1, column, +1.0 );
end
end;
{read matrix and right hand side}
for row := 1 to ncon do
begin
for col := 1 to nvar do
begin
read( filevar, value );
putmatrix( row, col, value );
end;
read( filevar, value );
putmatrix( row, 0, value );
putmatrix( row, tcols2, value );
end;
{ read the coefficients of the objective function }
for col := 1 to nvar do
begin
putmatrix( 0, col, value * minmax );
putmatrix( trows2, col, value * minmax );
end;
{ calculate artifical variables }
for col := 1 to nvar do
begin
value := zero;
for row := nltcon+1 to ncon do
value := value - getmatrix( row, col );
putmatrix( trows1, col, value );
end;
close(filevar);
end;
procedure stats;
begin
writeln;
writeln(' * Your Variables : 1 through ', nvar);
if ( nltcon > 0 ) then
writeln(' * Slack Variables : ',nvar+1,' through ',nvar+nltcon);
if ( ngtcon > 0 ) then
writeln(' * Surplus Variables : ',nvar+nltcon+1,' through ',tcols3);
if ( nltcon <> ncon ) then
writeln(' * Artificial Variables : ',tcols3+1,' through ',tcols1);
end;
procedure setbasis;
var row, col : integer;
flag : boolean;
begin
for col := 1 to nvar+ncon do
begin
flag := false;
row := 1;
repeat
if ( basis[ row ] = col ) then
flag := true
else
row := row + 1;
until ( (flag) or (row > ncon) );
if (flag) then
basisp[ col ] := row
else
basisp[ col ] := 0;
end;
end;
procedure problem;
var row, col : integer;
begin
{filename and problem description}
writeln('Filename: ',filename);
writeln(' Problem: ',name);
writeln;
{objective function}
if minmax < 0 then
writeln('Maximize:')
else
writeln('Minimize:');
for col := 1 to nvar do
begin
write(minmax*getmatrix( trows2, col ):18:8,' * Var#',col:3);
if col <> nvar then
write(' + ');
writeln;
end;
writeln;
{constraints}
writeln('Subject to:');
for row := 1 to ncon do
begin
for col := 1 to nvar do
begin
if (col = 1) then
write(' Constraint #',row:3,'...')
else
write(' ':20);
write(getmatrix( row, col):18:8,' * Var#',col:3);
if col <> nvar then
writeln(' + ');
end;
if (row <= nltcon) then
write(' <= ')
else if (row > nltcon+neqcon) then
write(' >= ')
else
write(' = ');
writeln(getmatrix( row, tcols2 ):18:8);
end;
end;
var smallpos : real;
smallptr : integer;
largeneg : real;
largeptr : integer;
value : real;
row, col, row1 : integer;
begin
setbasis;
stats;
{objective value}
writeln;
writeln('* Value of Objective Function: ',
-minmax*getmatrix( trows2, tcols2 ):18:8);
writeln;
writeln('* Variable Analysis *');
writeln('Var':3,'Value':18,'Reduced Cost':18);
writeln('---':3,'-----':18,'------------':18);
for col := 1 to nvar do
begin
write(col:3);
if (basisp[ col ] <> 0) then
write(getmatrix( basisp[ col ], tcols2 ):18:8)
else
write(0.0:18:8);
write(-minmax * getmatrix( trows2, col ):18:8);
writeln;
end;
writeln;
writeln('* Constraint Analysis *');
writeln('---':3,'---------':18,'-------------':18,'------------':18);
for row := 1 to ncon do
begin
if (row <= nltcon) then
col := nvar + row
else if (row > nltcon+neqcon) then
col := nvar + row - neqcon
else
col := nvar + ngtcon + row;
write(row:3);
write(getmatrix( row, 0 ):18:8);
if (basisp[ col ] <> 0) then
write(getmatrix( basisp[ col ], tcols2 ):18:8)
else
write(0.0:18:8);
write(-minmax * getmatrix( trows2, col ):18:8);
writeln;
end;
writeln(' ');
writeln('* Sensitivity Analysis - Right Hand Side Ranging *');
writeln('Row':3,'Lower':18,'Current':18,'Upper':18,'OutLo':6,'OutUp':6);
writeln('---':3,'-----':18,'-------':18,'-----':18,'-----':6,'-----':6);
for row1 := 1 to ncon do
begin
if (row1 <= nltcon) then
col := nvar + row1
else if (row1 > nltcon+neqcon) then
col := nvar + row1 - neqcon
else
col := nvar + ngtcon + row1;
smallpos := +mxval;
smallptr := 0;
largeneg := -mxval;
largeptr := 0;
for row := 1 to ncon do
begin
value := getmatrix( row, col );
if ( value <> zero ) then
begin
value := getmatrix( row, tcols2 ) / value;
if (value>zero) and (valuelargeneg) then
begin
largeneg := value;
largeptr := basis[ row ];
end;
end;
end;
if (row1<=nltcon+neqcon) then
begin
write(row1:3);
if (smallpos <> mxval) then
write((getmatrix( row1, 0 )-smallpos):18:8)
else
write('No Limit':18);
write((getmatrix( row1, 0 ) ):18:8);
if (largeneg <> -mxval) then
write((getmatrix( row1, 0 )-largeneg):18:8)
else
write('No Limit':18);
if (smallptr <> 0) then
write(smallptr:6)
else
write('None':6);
if (largeptr <> 0) then
write(largeptr:6)
else
write('None':6);
end
else
begin
write(row1:3);
if (largeneg <> -mxval) then
write((getmatrix( row1, 0 )+largeneg):18:8)
else
write('No Limit':18);
write((getmatrix( row1, 0 ) ):18:8);
if (smallpos <> mxval) then
write((getmatrix( row1, 0 )+smallpos):18:8)
else
write('No Limit':18);
if (largeptr <> 0) then
write(largeptr:6)
else
write('None':6);
if (smallptr <> 0) then
write(smallptr:6)
else
write('None':6);
end;
writeln;
end;
writeln(' ');
writeln('* Sensitivity Analysis - Objective Coefficient Ranging *');
writeln('Var':3,'Lower':18,'Current':18,'Upper':18,'InLo':6,'InUp':6);
writeln('---':3,'-----':18,'-------':18,'-----':18,'----':6,'----':6);
for col := 1 to nvar do
begin
smallpos := +mxval;
smallptr := 0;
largeneg := -mxval;
largeptr := 0;
if (basisp[ col ] = 0) then
if (minmax < 0) then
begin
smallpos := -minmax*getmatrix( trows2, col );
smallptr := col;
end
else
begin
largeneg := -minmax*getmatrix( trows2, col );
largeptr := col;
end
else
for row := 1 to tcols3 do
if (basisp[ row ] = 0) then
begin
value := getmatrix( basisp[ col ], row );
if ( value <> zero ) then
begin
value := minmax*getmatrix( trows2, row ) / value;
if (value>zero) and (valuelargeneg) then
begin
largeneg := value;
largeptr := row;
end;
end;
end;
write(col:3);
if (largeneg <> -mxval) then
write((minmax*getmatrix( 0, col )+largeneg):18:8)
else
write('No Limit':18);
write((minmax*getmatrix( 0, col ) ):18:8);
if (smallpos <> mxval) then
write((minmax*getmatrix( 0, col )+smallpos):18:8)
else
write('No Limit':18);
if (largeptr <> 0) then
write(largeptr:6)
else
write('None':6);
if (smallptr <> 0) then
write(smallptr:6)
else
write('None':6);
writeln;
end;
end;
procedure print;
var row, col : integer;
begin
writeln;
for row := 1 to trows2 do
begin
if (row > 0) and (row <= ncon) then
write(basis[ row ]:2,' ')
else
write(' ');
for col := 1 to tcols2 do
write(getmatrix( row, col ):9:3,' ');
writeln;
end;
writeln;
end;
begin
initmatrix;
writeln;
writeln('*** Linear Programming - Simplex Algorithm ***');
writeln;
problem;
if ( error = 0 ) then
simplex( error );
if ( error = 0 ) or ( error = 1 ) then
if ( error < 0 ) then
writeln('-- Inconsistent Data - Not Run --');
if ( error = 2 ) then
writeln('-- The Solution is Unbounded --');
if ( error = 3 ) then
writeln('-- The Problem is Infeasible --');
end.
********************
*** SAMPLE INPUT ***
********************
Photocopy - Sensitivity Analysis - Page 329
-1 5 4
2 0 3
1.00 0.00 0.00 0.00 2.00
0.00 1.00 0.00 0.00 1.50
300.00 1000.00 10.00 1500.00 2500.00
20.00 110.00 10.00 0.00 200.00
15.00 10.00 8.00 48.00 100.00
50.00 80.00 15.00 180.00
*********************
*** SAMPLE OUTPUT ***
*********************
*** Linear Programming - Simplex Algorithm ***
Filename: pc329.lp
Problem: Photocopy - Sensitivity Analysis - Page 329
Minimize:
50.00000000 * Var# 1 +
80.00000000 * Var# 2 +
15.00000000 * Var# 3 +
180.00000000 * Var# 4
Subject to:
Constraint # 1... 1.00000000 * Var# 1 +
0.00000000 * Var# 2 +
0.00000000 * Var# 3 +
0.00000000 * Var# 4 <= 2.00000000
Constraint # 2... 0.00000000 * Var# 1 +
1.00000000 * Var# 2 +
0.00000000 * Var# 3 +
0.00000000 * Var# 4 <= 1.50000000
Constraint # 3... 300.00000000 * Var# 1 +
1000.00000000 * Var# 2 +
10.00000000 * Var# 3 +
1500.00000000 * Var# 4 >= 2500.00000000
Constraint # 4... 20.00000000 * Var# 1 +
110.00000000 * Var# 2 +
10.00000000 * Var# 3 +
0.00000000 * Var# 4 >= 200.00000000
Constraint # 5... 15.00000000 * Var# 1 +
10.00000000 * Var# 2 +
8.00000000 * Var# 3 +
48.00000000 * Var# 4 >= 100.00000000
* Your Variables : 1 through 4
* Slack Variables : 5 through 6
* Surplus Variables : 7 through 9
* Artificial Variables : 10 through 12
* Value of Objective Function: 335.23437500
* Variable Analysis *
Var Value Reduced Cost
--- ----- ------------
1 0.00000000 -4.29687500
2 1.50000000 0.00000000
3 6.90104167 0.00000000
4 0.62065972 0.00000000
* Constraint Analysis *
Row RHS Value Slack/Surplus Shadow Price
--- --------- ------------- ------------
1 2.00000000 2.00000000 0.00000000
2 1.50000000 0.00000000 -0.46875000
3 2500.00000000 0.00000000 -0.06250000
4 200.00000000 34.01041667 0.00000000
5 100.00000000 0.00000000 -1.79687500
* Sensitivity Analysis - Right Hand Side Ranging *
Row Lower Current Upper OutLo OutUp
--- ----- ------- ----- ----- -----
1 0.00000000 2.00000000 No Limit 5 None
2 1.25469572 1.50000000 2.40506329 8 4
3 1606.25000000 2500.00000000 3316.25000000 4 8
4 No Limit 200.00000000 234.01041667 None 8
5 73.88000000 100.00000000 815.00000001 8 4
* Sensitivity Analysis - Objective Coefficient Ranging *
Var Lower Current Upper InLo InUp
--- ----- ------- ----- ---- ----
1 45.70312500 50.00000000 No Limit 1 None
2 No Limit 80.00000000 80.46875000 None 6
3 1.20000000 15.00000000 15.16363636 9 6
4 179.31645570 180.00000000 202.00000000 6 1
``` | 5,253 | 15,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-47 | latest | en | 0.464204 |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.19/share/doc/Macaulay2/Macaulay2Doc/html/_lead__Term_lp__Ring__Element_rp.html | 1,656,868,385,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-27/segments/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00450.warc.gz | 307,275,609 | 2,326 | leadTerm(RingElement) -- get the greatest term
Synopsis
• Usage:
i1 : R = QQ[a..d]; i2 : leadTerm (3*b*c^2-d^3-1) 2 o2 = 3b*c o2 : R i3 : S = QQ[a..d, MonomialOrder => Lex] o3 = S o3 : PolynomialRing i4 : leadTerm (3*b*c^2-d^3-1) 2 o4 = 3b*c o4 : S
i5 : R = ZZ[a..d][x,y,z]; i6 : leadTerm((a+b)*y^2 + (b+c)*x*z) 2 o6 = (a + b)y o6 : R | 162 | 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} | 2.8125 | 3 | CC-MAIN-2022-27 | latest | en | 0.231975 |
https://unitmaster.com/convert/length/convert-fractional-P-to-decimal-Picas.html | 1,685,836,133,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00047.warc.gz | 635,787,153 | 4,530 | # Free Online Converter to Convert from Fractional Picas to Decimal Picas and vice versa
Length Conversion Settings
Number of decimal places
Round fraction to the nearest
## Picas from Fractions to Decimals Conversion
1/2 P = 0.5 P
1/3 P = 0.33333333 P
2/3 P = 0.66666667 P
1/4 P = 0.25 P
2/4 P = 0.5 P
3/4 P = 0.75 P
1/5 P = 0.2 P
2/5 P = 0.4 P
3/5 P = 0.6 P
4/5 P = 0.8 P
1/6 P = 0.16666667 P
2/6 P = 0.33333333 P
3/6 P = 0.5 P
4/6 P = 0.66666667 P
5/6 P = 0.83333333 P
1/7 P = 0.14285714 P
2/7 P = 0.28571429 P
3/7 P = 0.42857143 P
4/7 P = 0.57142857 P
5/7 P = 0.71428571 P
6/7 P = 0.85714286 P
1/8 P = 0.125 P
2/8 P = 0.25 P
3/8 P = 0.375 P
4/8 P = 0.5 P
5/8 P = 0.625 P
6/8 P = 0.75 P
7/8 P = 0.875 P
1/9 P = 0.11111111 P
2/9 P = 0.22222222 P
3/9 P = 0.33333333 P
4/9 P = 0.44444444 P
5/9 P = 0.55555556 P
6/9 P = 0.66666667 P
7/9 P = 0.77777778 P
8/9 P = 0.88888889 P
1/10 P = 0.1 P
2/10 P = 0.2 P
3/10 P = 0.3 P
4/10 P = 0.4 P
5/10 P = 0.5 P
6/10 P = 0.6 P
7/10 P = 0.7 P
8/10 P = 0.8 P
9/10 P = 0.9 P
1/11 P = 0.09090909 P
1/12 P = 0.08333333 P
2/12 P = 0.16666667 P
3/12 P = 0.25 P
4/12 P = 0.33333333 P
5/12 P = 0.41666667 P
6/12 P = 0.5 P
7/12 P = 0.58333333 P
8/12 P = 0.66666667 P
9/12 P = 0.75 P
10/12 P = 0.83333333 P
11/12 P = 0.91666667 P
1/13 P = 0.07692308 P
1/14 P = 0.07142857 P
1/15 P = 0.06666667 P
1/16 P = 0.0625 P
3/16 P = 0.1875 P
5/16 P = 0.3125 P
7/16 P = 0.4375 P
9/16 P = 0.5625 P
11/16 P = 0.6875 P
13/16 P = 0.8125 P
15/16 P = 0.9375 P
1/17 P = 0.05882353 P
1/18 P = 0.05555556 P
1/19 P = 0.05263158 P
1/20 P = 0.05 P
1/25 P = 0.04 P
1/32 P = 0.03125 P
1/50 P = 0.02 P
1/64 P = 0.015625 P
1/100 P = 0.01 P
1/125 P = 0.008 P
1/128 P = 0.0078125 P
1/250 P = 0.004 P
1/256 P = 0.00390625 P
1/500 P = 0.002 P
1/1000 P = 0.001 P
-
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## Convert Picas from decimals to fractions
0.001 P = 1/1000 P
0.002 P = 1/500 P
0.003 P = 3/1000 P
0.004 P = 1/250 P
0.005 P = 1/200 P
0.006 P = 3/500 P
0.007 P = 7/1000 P
0.008 P = 1/125 P
0.009 P = 9/1000 P
0.01 P = 1/100 P
0.02 P = 1/50 P
0.03 P = 3/100 P
0.04 P = 1/25 P
0.05 P = 1/20 P
0.06 P = 3/50 P
0.07 P = 7/100 P
0.08 P = 2/25 P
0.09 P = 9/100 P
0.1 P = 1/10 P
0.11 P = 11/100 P
0.12 P = 3/25 P
0.13 P = 13/100 P
0.14 P = 7/50 P
0.15 P = 3/20 P
0.16 P = 4/25 P
0.17 P = 17/100 P
0.18 P = 9/50 P
0.19 P = 19/100 P
0.2 P = 1/5 P
0.21 P = 21/100 P
0.22 P = 11/50 P
0.23 P = 23/100 P
0.24 P = 6/25 P
0.25 P = 1/4 P
0.26 P = 13/50 P
0.27 P = 27/100 P
0.28 P = 7/25 P
0.29 P = 29/100 P
0.3 P = 3/10 P
0.31 P = 31/100 P
0.32 P = 8/25 P
0.33 P = 33/100 P
0.34 P = 17/50 P
0.35 P = 7/20 P
0.36 P = 9/25 P
0.37 P = 37/100 P
0.38 P = 19/50 P
0.39 P = 39/100 P
0.4 P = 2/5 P
0.41 P = 41/100 P
0.42 P = 21/50 P
0.43 P = 43/100 P
0.44 P = 11/25 P
0.45 P = 9/20 P
0.46 P = 23/50 P
0.47 P = 47/100 P
0.48 P = 12/25 P
0.49 P = 49/100 P
0.5 P = 1/2 P
0.51 P = 51/100 P
0.52 P = 13/25 P
0.53 P = 53/100 P
0.54 P = 27/50 P
0.55 P = 11/20 P
0.56 P = 14/25 P
0.57 P = 57/100 P
0.58 P = 29/50 P
0.59 P = 59/100 P
0.6 P = 3/5 P
0.61 P = 61/100 P
0.62 P = 31/50 P
0.63 P = 63/100 P
0.64 P = 16/25 P
0.65 P = 13/20 P
0.66 P = 33/50 P
0.67 P = 67/100 P
0.68 P = 17/25 P
0.69 P = 69/100 P
0.7 P = 7/10 P
0.71 P = 71/100 P
0.72 P = 18/25 P
0.73 P = 73/100 P
0.74 P = 37/50 P
0.75 P = 3/4 P
0.76 P = 19/25 P
0.77 P = 77/100 P
0.78 P = 39/50 P
0.79 P = 79/100 P
0.8 P = 4/5 P
0.81 P = 81/100 P
0.82 P = 41/50 P
0.83 P = 83/100 P
0.84 P = 21/25 P
0.85 P = 17/20 P
0.86 P = 43/50 P
0.87 P = 87/100 P
0.88 P = 22/25 P
0.89 P = 89/100 P
0.9 P = 9/10 P
0.91 P = 91/100 P
0.92 P = 23/25 P
0.93 P = 93/100 P
0.94 P = 47/50 P
0.95 P = 19/20 P
0.96 P = 24/25 P
0.97 P = 97/100 P
0.98 P = 49/50 P
0.99 P = 99/100 P
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Saturday, June 3, 2023
Privacy Policy
Terms & Conditions
Copyright © 2023 Intemodino Group s.r.o.
All rights reserved | 2,306 | 3,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.453655 |
https://www.reddit.com/user/filifjonk?sort=top&t=year | 1,477,379,147,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719960.60/warc/CC-MAIN-20161020183839-00054-ip-10-171-6-4.ec2.internal.warc.gz | 972,751,886 | 20,952 | [–] 41 points42 points (0 children)
And 53/17 is 3.14h a day. How spooky is that?
[–][S] 24 points25 points (0 children)
Ta det med din regering.
Helt sjukt att va så spydig mot någon som befinner sig i nöd.
[–] 15 points16 points (0 children)
There are lots of companion planting rules like this and once you start to consider them all, garden planning gets complicated. It can also be hard to know what rules actually have scientific backup.
[–][S] 9 points10 points (0 children)
I guess this part is the ball values:
`````` function c(e) {
if (e.rx == undefined || e.ry == undefined || e.lx == undefined || e.ly == undefined || e.a == undefined)
return;
e.sync = null ;
var r = new Box2D.Dynamics.b2FixtureDef
, i = new Box2D.Dynamics.b2BodyDef
, s = .19;
r.density = 1,
r.friction = .5,
r.restitution = .2,
r.shape = new Box2D.Collision.Shapes.b2CircleShape(s),
i.type = Box2D.Dynamics.b2Body.b2_dynamicBody,
i.linearDamping = .5,
i.angularDamping = .5,
tagpro.events.setPlayerPhysics && tagpro.events.setPlayerPhysics.forEach(function(e) {
e.setPlayerPhysics(Box2D, i, r)
});
var o = t.CreateBody(i)
, u = o.CreateFixture(r);
return o.SetPosition(new Box2D.Common.Math.b2Vec2(e.rx,e.ry)),
e.x = e.rx * 100,
e.y = e.ry * 100,
e.angle = e.ra,
o.player = e,
o.fixture = u,
n[e.id] = o,
o
}
``````
[–][S] 6 points7 points (0 children)
Compared how some of my co-workers write code, minified JS is easy to understand.
[–] 3 points4 points (0 children)
Programmet Medierna i P1 är ett exempel på att debatten är mer öppen på SR.
[–][S] 3 points4 points (0 children)
Låter lite dyrt med en miljard och är det inte lite väl många hållplatser?
[–][S] 4 points5 points (0 children)
I blame auto-correct on that. Better to have a human doing the job than a machine...
[–][S] 2 points3 points (0 children)
Yes, 4v4 up until last game, then only 2v2. Lots of high ranking players participated.
[–] 2 points3 points (0 children) | 621 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-44 | latest | en | 0.418751 |
https://i.reddit.com/r/theydidthemath/comments/ptjva0/request_how_accurate_is_this/hdzdk36/.compact | 1,638,791,944,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00488.warc.gz | 368,121,129 | 6,549 | # theydidthemath
1✓ 0 points1 point 2 months ago
thank you and this is in acoustic pressure, and it even neglect the volume of the cats. It could be even lower than that. a cat is about 11 litres (0.011 cubic meter. 2B cat is a ball of 36m of diameter. (4/3 pi r³)
we cannit take a measurement closer than 18m from the center of the serenade, thus the sound pressure is way lower.
0 points1 point 2 months ago
This would be sound energy, not pressure. dB measures energy!
A 36 meter diameter seems very small. I imagine the actual measurement would be very close to 163ish dB, given the outside cats, although far away, contribute so little to the total energy!
1✓ 2 points3 points 2 months ago
dB means it's on a log scale.
you can have sound power in dB, the base unit is watt.
you can have sound energy in dB, base unit is Joules
You can have sound pressure in dB, base unit is Pascal.
the subscript wil change, usually, we use "level" to let know the reader that the unit will be in dB. Sound pressure level (SPL) and sound powet level (SWL) are both in dB. whereas Sound power is in Watt and Sound pressure is in Pa.
I'm going back to my test but I'll use my software here to check the math and use sound power everywhere before midnigh EST.
0 points1 point 2 months ago
Goodness, I did not know that. Thank you for the info. I'm curious what you find out!
1✓ 0 points1 point 2 months ago
Let's assume a cat is 70 dB, this is 10^7 dB (neet hey?). two billin cat is 2 * 10^6*10^7 Pa = 2 * 10^13 Pa. Which means it's 133 dB at 2m (6foot) BUT a 2B cat ball is 18 m of diameter, then at 9 meter (radius), there is a r^2 reduction in acoustic pressure.
9.87 * 10^11 Pa, 10 * log(9.87*10^11) = 119 dB
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Apr 30, 2019 | Elisabeth Guillet | 688 | 2,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.9436 |
https://www.physicsforums.com/threads/lock-in-amplifier.587179/ | 1,542,751,871,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746800.89/warc/CC-MAIN-20181120211528-20181120233528-00253.warc.gz | 971,053,172 | 16,429 | # Lock-in amplifier
1. Mar 15, 2012
### Niles
Hi
Usually when a lock-in amplifier is introduced to students, it is done when the input signal is a harmonic function and the reference is one as well. This is a nice concept, and it is easy to see how a DC-term propotional to the (desired) incident amplitude can be obtained.
My question is, say my input signal is a square wave. Then I can decompose it into a Fourier series. I was wondering how to get a DC-term containing the incident amplitude out via phase sensitive detection in this case.
What I would say is that I can decompose the input signal and reference into their respective Fourier series. They will have the same harmonics. Then I would mix them, and this would give me a DC-component for each harmonic in the Fourier expansion. Is this the explanation?
Best,
Niles.
2. Mar 15, 2012
### f95toli
Is this a theoretical of practical question?
What will happen depends on what type of lock-in you are using.
E.g. RF lock-ins usually prefer a square-wave reference even if the input is sinusoidal.
I would suggest locking in the manuals for the Stanford Research SR830 and SR844 (I use both types), they both have pretty good explanation for how they work,
edit: the SR website www.thinksrs.com
3. Mar 15, 2012
### Niles
It is just a theoretical question. I haven't ever seen a lock-in amplifier in real life, I just come across them all the time in the litterature.
Best,
Niles.
4. Mar 15, 2012
Well I don't know how the lock in amplifier was introduced to you, but I find square waves much easier to understand.... Anyhow how the beast works:
The lock in reference output is any periodic function f(t) that fulfils:
$$\int_0^T f(t) dt = 0$$
Where T is the period. Let us assume that it is also normalized (otherwise we'd have to divide by some number)
$$\int_0^T f(t)^2 dt = 1$$
If we multiply the input with the reference and integrate we have a scalar product in the Hilbert space.
$$\left< f(t),g(t) \right> =\int_0^T f(t)g(t) dt$$
This is exactly what the lock in does. It gets the projection of the input signal onto the reference signal this way. The integration is usually done with a low pass filter, (this changes the measure of the integral a bit, but those are details...). You can get some phase sensitive detection if:
$$\left< f(t-T/4),g(t) \right> = 0$$
Which will give you two so called "quadrature components" of the reference signal or you manually adjust the phase shift (=time shift) until you get the maximum signal and see what the phase is.
For a square wave it is really easy. So the reference signal oscillates between say +1 and -1. If the input signal and the reference are in phase. The input signal changes sign at the same moment. So if you multiply the two numbers you always get the same positive value. If you integrate over it, then you get exactly the amplitude of the input signal. The phase detection is simply done by time shifting the reference. So there appear negative dips and the integral decreases until you are at T/4 when the negative dips and the positive ones cancel exactly and the integral is zero, if you shift further the signal becomes negative.
I do not know what the lock in manufacturers call phase, if the reference is not a sine wave. The "true phase" in my eyes would be the time shift of the input signal divided by the period times two pi, but I suppose it will usually be easier to use the arctan of the in-phase signal and the 90° shifted signal.
As you have probably noticed a phase shifted (=time shifted) signal will have different phase shifts on all the different frequency components so the Fourier transform dis not really useful for determining the phase.
5. Mar 18, 2012
### Niles
Thanks for taking the time to write that. The way I was introduced to it was by looking at a reference given by
$$V_{ref}\sin(\omega t+\Theta_{ref})$$
If the signal is varied harmonically with a frequency the same as the reference, then we get the amplitude of the signal out after mixing the terms and using a proper low-pass filter. So this explanation is pretty trivial. I am not sure where in this process the inner product is taken though.
I like yours a lot more, it is more detailed. However, I have to admit that I'm not 100% sure how the PSD works.
Do you have a reference for these things?
Best,
Niles.
Last edited: Mar 18, 2012
6. Mar 18, 2012
Sorry, but I don't have an English reference. Looking at the recommended manual is probably a pretty good idea.
The inner product is taken by the nonlinearity of the mixer if it is not done digitally.
If the input output relation is something like
$$V_\text{out} = \alpha_0 + \alpha_1 V_\text{in} +\alpha_2 V_\text{in}^2 + O(V_\text{in}^3)$$
Then the quadratic term will produce a multiplication.
$$V_\text{in} = V_\text{reference} + V_\text{signal}$$
$$V_\text{in}^2 = V_\text{reference}^2 + 2V_\text{signal}\cdot V_\text{reference} + V_\text{signal}^2$$
(You can see the mixed multiplication term in the middle)
Again. You can do the phase sensitive detection by changing the phase of your signal with a variable delay or an adjustable all pass filter, and adjust until you have the maximum signal.
The other way is to rewrite your signal
$$V_\text{ref} = \sin (\omega t + \Theta) = \cos(\Theta)\sin(\omega t) + \sin(\Theta)\cos(\omega t) =\beta_1\sin(\omega t) + \beta_2\cos(\omega t)$$
If you now find the two coefficients $$\beta_1,\beta_2$$ by projecting to the two quadrature components sin and cos. Then you can calculate the phase $$\Theta = \arctan \frac{\beta_2}{\beta_1}$$ With square waves you can do something similar, but in that case the result is not the phase, just something close to it.
7. Mar 18, 2012
### marcusl
Phase sensitive detection is often/usually done with devices that act like mixers, in the sense that they switch between states -1 and +1 rapidly when the reference voltage changes sign. Obviously a square wave reference r(t) commutates the switching elements; a sine wave reference is effectively converted to a square wave and also commutates the switch. The PSD output is then
$$y(t)=\int{x(t) \mathrm{\ signum}[r(t)] dt}$$
You can easily work out how both sine- and square-wave inputs x result, after low-pass filtration, in a DC output.
BTW, an RF mixer does the same. The sinusoidal local oscillator level is chosen to drive the mixer diodes rapidly into saturation, producing a "square wave" commutating action like the one described above.
8. Mar 19, 2012
### f95toli
There are -from a "practical" point of view- two ways to think about lock-ins; depending on the application.
*As a tunable band-pass filter, with the centre frequency given by the reference frequency and the BW by the time-constant/filter steepness.
*As a down-converting mixer with a tunable phase difference between the RF and the LO, and built in amplification both at the RF and IF stage (and the "distribution" of amplification set by the "dynamic reserve" button).
The latter way of thinking about is is quite useful if you are working at high frequencies (where a lock-in quite sometimes is used in place of an ordinary mixer), whereas the former is better for low frequencies/DC.
9. Mar 19, 2012
### marcusl
It is useful to point out that Robert Dicke's original lock-in, developed during World War II for microwave radiometry, operated at 30 Hz. The utility of locking in to a frequency, and of phase sensitive detection, extends to even lower frequencies as well. | 1,851 | 7,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-47 | latest | en | 0.935686 |
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Assignment Help Macroeconomics
##### Reference no: EM13181714
According to the American Metal Markets Magazine, the spot market price of U.S. hot rolled steel recently reached \$580 per ton. Less than a year ago this same ton of steel was only \$260. A number of factors are cited to explain the large price increase. The combination of China's increased demand for raw steel-due to expansion of its manufacturing base and infrastructure changes to prepare for the 2008 Beijing Olympics-and the weakening U.S. dollar against the euro and yuan partially explain the upward spiral in raw steel prices. Supply-side changes have also dramatically affected the price of raw steel. In the last 20 years there has been a rapid movement away from large integrated steel mills to minimills. The minimill production process replaces raw iron ore as its primary raw input with scrap steel. Today, minimills account for approximately 52 percent of all U.S. steel production. However, the worldwide movement to the minimill production model has bid up the price of scrap steel. In December, the per-ton price of scrap was around \$156 and soared to \$302 just two months later. Suppose that, as a result of this increase in the price of scrap, the supply of raw steel changed from Qs raw = 4,900 + 5P to Qs raw = 100 + 5P. Assuming the market for raw steel is competitive and that the current worldwide demand for steel is Qd raw = 8,800 - 10P, compute the equilibrium price and quantity when the per-ton price of scrap steel was \$156, and the equilibrium price-quantity combination when the price of scrap steel reached \$302 per ton. Suppose the cost function of a representative minimill producer is C(Q) = 1,000 + 10Q2. Compare the change in the quantity of raw steel exchanged at the market level with the change in raw steel produced by a representative firm. How do you explain this difference?
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#### dfidler
Hi All,
I'm looking at a PCB that has three relays marked TQ2-12V ATQ203 and I'm hoping that you guys can sanity check my questions/assumptions based on the datasheet and the readings I'm getting.
This image is from the Panasonic EW TQ2-12V datasheet (http://www.farnell.com/datasheets/22540.pdf):
Questions:
1. I'm not sure if each diagram represents a different part or different states of the same part based on how you hook it up?
From what i can tell from other datasheets, they do refer to different parts.
2. What is the significance of the color of the coil in the diagrams?
I have two theories on this.
The first is that the color of the signal pins (highlighted in yellow in the picture) represents the latch position for the latching relays. White means means that pins 2/3, 9/8 are closed and 4/3, 7/8 are open while black means that state is reversed.
The other theory is that they represent the polarity of the signal that should go through the pins and that there is a diode or resistor in place to prevent reverse voltage spikes.
3. Which pins are the power in and NC?
If I'm reading this right, pins 3/8 are ouptut, 2/9 are the NC input and 4/7 are the NO input. I'm not sure what the significance of the black dots for pins 2/9 are for though.
4. Which one do I have?
In my circuit, pins 5/6 are going to ground, which doesn't tell me much. Measuring the R between pins 1/10 gives me about 1K so I'm guessing that, if the diagrams represent different parts, I have a single side stable or 1-coil latching (this is further supported by the fact that pins 1/5 and 6/10 show no resistance/connectivity at all).
Other than powering up the board and/or desoldering the relay (which I don't want to do for fear of disturbing the IC that's adjacent to it) and powering the relay directly, is there a way of determining whether this is a stable or latching relay?
Or will I just have to guess based on the circuit that it's servicing?
Note: I haven't yet mapped out all of the pins between the various releays as this is a multi-layer PCB and it's taking some time to figure out what is connected to what. Therefore, I don't totally understand the purpose of this circuit yet, though it seems pretty likely that it's part of the amp's input circuit protection (which would make me want to say it's a latching relay).
Cheers,
Dave.
Last edited:
#### PedroDaGr8
OK it appears from what I can see of the datasheet the technical name would be
TQ2-Nil-Nil-12V
Nil of course means nothing (so they shorten the name to TQ2-12V)
So it would be appear that you have:
Single side stable AND Std. B.M.M Type (no clue whatthis means).
#### PedroDaGr8
Ahhh MBB = Make Before Break. Had not encountered that concept before (I am used to automotive relays which I guess are Break before make).
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button. | 771 | 3,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.945183 |
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# 6y-5=-3(2y+1)
6y-5=-3(2y+1)
## Research, Knowledge and Information :
### Equation Calculator & Solver | Wyzant Resources
Equation Calculator & Solver. Equation: Variable: Example Equation. Hint: Selecting "AUTO" in the variable box will make the ...
### Which expression is equivalent to 3 + 4y – y(9 – 2y) + 5? A ...
Which expression is equivalent to 3 + 4y – y(9 ... 2y^2-5y+8 D. - 6y^2+27y+8 1. Ask for details ; ... 3+4y-y(9-2y)+5 =>3+4y -9y+2y^2+5 =>2y^2-5y+8 so the answer is B
### 6y-5=-3(2y+1) - solution - Get Easy Solution
Simple and best practice solution for 6y-5=-3(2y+1) equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, ...
### 2[(2y-1)+y]=5(y+3) - solution - Get Easy Solution
... (2y-1)+y]=5(y+3) equation. Check how ... 2y + y = 3y 2[-1 + 3y] = 5(y + 3) [-1 * 2 + 3y * 2] = 5(y + 3) [-2 + 6y] = 5(y + 3) Reorder the terms: -2 + 6y = 5(3 ...
### SOLUTION: solve by subtitution: 1/5x + 1/2y = 8 x +y = 20 ...
6x + 5y = -3-x - 5/6y = -7 ... You can put this solution on YOUR website! 1/5x+1/2y=8 x+y=20 or x=20-y (20-y)/5+y/2=8 [2(20-y)+5y)]/10=8 (40-2y+5y)/10=8 40+3y=10*8
### The work of a student to solve the equation 2(6y − 3) = 20 ...
The work of a student to solve the equation 2(6y − 3) = 20 + 6y is shown below: Step 1: 2(6y − 3) = 20 + 6y Step 2: 8y − 5 = 20 + 6y Step 3: 2y − 5 = 20 Step ...
### 3x-2y=7, 9x+6y=14 - System of Equations Calculator - Symbolab
Free system of equations calculator ... x+2y=2x-5,\:x-y=3; ... xy=10,\:2x+y=1; system-of-equations-calculator. 3x-2y=7, 9x+6y=14. en.
### 6y-5=-3(2y+1) - OpenStudy
6y-5=-3(2y+1) surrely by now u ... 6y-5= -6y - 3, 12y = 2, y = 1/6. anonymous 6 years ago. yes i get that and thank you for trying but i really dont understand math ...
### How do you solve 6y-20=2y-4? | Socratic
How do you solve #6y-20=2y-4#? Algebra. 1 Answer ... How do you solve and graph #-2a+3>=6a-1>3a-10#? Answer 18 minutes ago. I would like to ask questions ...
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• Be clear and specific
• Use proper spelling and grammar | 951 | 2,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.707713 |
https://www.analystforum.com/t/positive-and-negative-convexity/102953 | 1,638,770,212,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00203.warc.gz | 698,064,380 | 5,230 | Positive and negative convexity
For a security that exhibits positive convexity, the duration changes in the desired direction; for a security that exhibits negative convexity, there is an adverse change in the duration. If interest rates rise, the duration of the MBS increases while the duration of the Treasury decreases. Why does duration of the MBS increase, but duration of Treasury decreases?
because of the negative convexity which is a result prepayment risk
• basically the slope of the tangent line to the MBS curve (the duration) decreases in absolute value (but increases not in absolute value)
Not to put too fine a point on it, but the slope of the price-yield curve is not the (negative of) the duration; it’s the (negative of) the _ dollar _ duration.
Thanks for the graph very useful. But where on the graph can we see duration increasing as interest rates increase? Aren’t the axis yield and price?
so draw a tangent line to two different yield points along the negative convex portion of the price-yield curve and compare their slopes…you will see
assume yield = interest rate for the sake of illustration
thanks S2000, you always got my back!
[/quote]
so draw a tangent line to two different yield points along the negative convex portion of the price-yield curve and compare their slopes…you will see
assume yield = interest rate for the sake of illustration
[/quote]
But where is duration on the price-yield curve? Duration is a function of payments and time, so not sure how that is on the graph?
Thats true, it is afunction of payments and time, BUT it can be used to approximate the % change in the bond price given a 100 bps change in the yield, hence, it can be graphed in the yield-price space
Thats true, it is afunction of payments and time, BUT it can be used to approximate the % change in the bond price given a 100 bps change in the yield, hence, it can be graphed in the yield-price space
I see. So a flattening slope as yield increases means duration decreases? | 438 | 2,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-49 | latest | en | 0.926086 |
https://docstore.mik.ua/orelly/perl4/prog/ch03_20.htm | 1,501,246,748,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550967030.98/warc/CC-MAIN-20170728123647-20170728143647-00662.warc.gz | 637,087,765 | 5,296 | home | O'Reilly's CD bookshelfs | FreeBSD | Linux | Cisco | Cisco Exam
## 3.20. Logical and, or, not, and xor
As lower precedence alternatives to &&, ||, and !, Perl provides the and, or, and not operators. The behavior of these operators is identical--in particular, and and or short-circuit like their counterparts, which makes them useful not only for logical expressions but also for control flow.
Since the precedence of these operators is much lower than the ones borrowed from C, you can safely use them after a list operator without the need for parentheses:
```unlink "alpha", "beta", "gamma"
or gripe(), next LINE;```
With the C-style operators you'd have to write it like this:
```unlink("alpha", "beta", "gamma")
|| (gripe(), next LINE);```
But you can't just up and replace all instances of || with or. Suppose you change this:
`\$xyz = \$x || \$y || \$z;`
to this:
`\$xyz = \$x or \$y or \$z; # WRONG`
That wouldn't do the same thing at all! The precedence of the assignment is higher than or but lower than ||, so it would always assign \$x to \$xyz, and then do the ors. To get the same effect as ||, you'd have to write:
`\$xyz = ( \$x or \$y or \$z );`
The moral of the story is that you still must learn precedence (or use parentheses) no matter which variety of logical operators you use.
There is also a logical xor operator that has no exact counterpart in C or Perl, since the only other exclusive-OR operator ((^)) works on bits. The xor operator can't short-circuit, since both sides must be evaluated. The best equivalent for \$a xor \$b is perhaps !\$a != !\$b. One could also write !\$a ^ !\$b or even \$a ? !\$b : !!\$b, of course. The point is that both \$a and \$b have to evaluate to true or false in a Boolean context, and the existing bitwise operator doesn't provide a Boolean context without help. | 464 | 1,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.919666 |
https://forum.godotengine.org/t/cant-get-rigidbody2d-to-rotate/9843 | 1,719,236,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00752.warc.gz | 219,599,770 | 5,513 | Can't get Rigidbody2d to rotate
Attention Topic was automatically imported from the old Question2Answer platform.
Hello,
I’m trying to create a character with completely physics-based movement. It’s supposed to be able to move along the X and Y axes, as well as rotate. The movement works fine, but I can’t get the thing to rotate. Can anyone help me?
I tried to use the code found at
However, even when copy-pasting the code, the movement works fine, but the thing still won’t rotate. No error, it just won’t do anything when the keys are pressed.
“Can sleep” is turned off.
This is my version of the code (slightly modified but the rotation part is the same):
``````extends RigidBody2D
var max_speed = 250.0
var thrust = 250
var torque = 20000
func _integrate_forces(state):
if state.linear_velocity.length()>max_speed:
state.linear_velocity = state.linear_velocity.normalized()*max_speed
if Input.is_action_pressed("move_forward"):
set_applied_force(Vector2(0, -thrust).rotated(rotation))
elif Input.is_action_pressed("move_backward"):
set_applied_force(Vector2(0, thrust).rotated(rotation))
elif Input.is_action_pressed("strafe_left"):
set_applied_force(Vector2(-thrust, 0).rotated(rotation))
elif Input.is_action_pressed("strafe_right"):
set_applied_force(Vector2(thrust, 0).rotated(rotation))
else:
set_applied_force(Vector2())
var rotation_dir = 0
if Input.is_action_pressed("rotate_right"):
rotation_dir += 1
if Input.is_action_pressed("rotate_left"):
rotation_dir -= 1
set_applied_torque(rotation_dir * torque)
``````
Not to be patronising, but did you make sure to bind the input actions in the project settings?
CardboardComputers | 2021-08-30 02:53
Not patronizing, I wondered about that, too. I double-checked the input, it works fine.
I found some articles that said a collision shape or collision polygon is required for set_applied_torque to function. I didn’t have that. Added one in now. Still doesn’t work.
Edit: I’m using set_angular_velocity now. That works. Multiple articles state one should use set_applied_torque, but if that doesn’t work, what can I do.
I’d still like to know why set_applied_torque doesn’t work, especially since that’s the one all the rigidbody tutorials are using, so if anyone knows, please, tell me.
Irolan | 2021-08-30 19:48 | 566 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.78404 |
https://www.analystforum.com/t/cfai-vol-5-page-78-a-7/8015 | 1,652,972,814,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00657.warc.gz | 762,483,418 | 4,884 | # CFAI Vol 5, page 78 (A-7)
Can some explain why increase in Expected return would give a lower VAR as the answer key on page A-7 sugguested? VAR = expected returns - 1.65*(standard deviation) If expected returns increase, shouldn’t VAR increase??? The answer key stated the opposite.
VAR measure the downside part of the normal curve. so if you increase the expected return then the bottom 5% of the normal curve rises, thus the VAR decrases. For example. e(x) = 10% with a deviation of 10%. a 5% var = -6.5%…(that is, 5% of the time you’re expected to return less than -6.5%) ex = 20% with a deviation of 10%. A 5% var = 3.5% (that is, 5% of the time you’ve expected to return less than 3.5%)… so your Value at Risk decreases as expected return increases… this help?
I still didn’t get this. Let’s take another example - Case 1 : e(x) = 10% with a deviation of 5%. a 5% var = 1.75%…(that is, 5% of the time you’re expected to return less than 1.75%) Case 2 : ex = 20% with a deviation of 5%. A 5% var = 11.75% (that is, 5% of the time you’ve expected to return less than 11.75%)… If we assume a portfolio of 1 million, then the \$VARs would be Case 1: \$17500 Case 2: \$117500 Did the Var not increase? Am I making some silly mistake?
CareerChange…the example you gave is not exactly a “good” VAR example. Using your figures, those portfolio will never (mathmatically) expect a loss at 95% of the time. Case (1), E® range is from +1.75 to +18.25 (10+/-1.65*5)…no negative return. Case (2), E® range is from +11.75 to +28.25 (20+/-1.65*5)…no negative return. Bottom line, given a 5% VAR (95% significant), the examples you used will not experience a loss 95% of the time. Therefore, in those case, VAR\$ is meaningless. Does this make sense? In general, if a portfoilo has loss in its range of returns, higher return will give lower \$VAR.
Just to add…in your example case (1) \$17500 is the 5% min. gain. case (2) \$117500 is the 5% min. gain. Gain is good, right? | 600 | 1,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-21 | latest | en | 0.882774 |
https://www.enotes.com/homework-help/log-10-xy-4-z-5-use-properties-logarithms-expand-521475 | 1,508,720,097,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00177.warc.gz | 962,850,602 | 8,465 | # `log_10 ((xy^4)/(z^5))` Use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.)
loves2learn | Student
Expand using addition (log multiplication) and subtraction (log division)
`(log_10x+log_10y^4)-log_10z^5 `
`(log_10x+4log_10y)-5log_10z ` | 107 | 346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-43 | latest | en | 0.771988 |
https://www.instasolv.com/question/calculate-the-que-dill-no-of-meles-in-following-5-56-lit-of-hz-1-ing-of-c-3j3cke | 1,611,185,963,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00680.warc.gz | 833,718,260 | 11,223 | Calculate the Que Dill no. of meles...
Question
Calculate the Que Dill no. of meles in following (5)56 Lit of Hz (1) ing of Cog v) ag of 40 + 11g of CO2 (1) 99 of the in 10xN. Od som
NEET/Medical Exams
Chemistry
Solution
116
4.0 (1 ratings)
( mathbb{A} ) (1) 5.6 lit ( 7 mathrm{Hr}= ) ; ( 22 cdot 4 / c t=1 mathrm{m} cdot mathrm{c} ) ( 5.6 mathrm{l} cdot mathrm{r}=frac{5.6}{22.4} mathrm{mol} . Rightarrow frac{1}{4}=0.25 mathrm{mol} ) M. ( operatorname{l丨g} m ) of ( cos ) ( =44 g m=1 m a u ) ( operatorname{lig} m=frac{1}{44} x |=0.25 mathrm{mol} ) ( v_{9 m}+H_{20}+1 operatorname{lgm}left(s_{2}right. ) ( Rightarrow frac{9}{18} mathrm{mol} mathrm{H}_{2} mathrm{s}+frac{11}{44} mathrm{mi}^{2} mathrm{Cur}_{2} ) ( 5+0.25=0.75 mathrm{m} d mathrm{s} )
Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free | 364 | 855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.398264 |
https://www.isnt.org.in/the-closest-in-between-with-code-examples.html | 1,722,657,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00395.warc.gz | 664,961,273 | 50,581 | # The Closest In Between With Code Examples
In this article, we will look at how to get the solution for the problem, The Closest In Between With Code Examples
## What is the distance between gold atoms?
∴ Nearest neighbour distance =a√22=4.070×√22 oA=2.878 oA.
``````int num;
int x;
printf("Enter the total number of our men: ");
scanf("%d", &num);
int men[num];
int n,n1;
for(x=1; x<=num; x++){
printf("Men #%d: ", x);
scanf("%d", &men[x]);
//if(x>=num+1)
}
printf("Enemy #1: ");
scanf("%d",&n);
printf("Enemy #2: ");
scanf("%d",&n1);
printf("\nMen Cornered:\n");
for(x=1; x<=num; x++){
if(men[x]<=n1 && men[x]>=n){
printf("%d ",men[x]);
}}```
```
## What is the packing fraction in FCC lattice?
The packing efficiency of FCC lattice is 74%. Let r be the radius of the sphere and a be the edge length of the cube and the number of atoms or spheres is n that is equal to 4.
## What is the coordination number in BCC?
Coordination number – the number of nearest neighbor atoms or ions surrounding an atom or ion. For FCC and HCP systems, the coordination number is 12. For BCC it's 8.
## What is packing efficiency of BCC?
4. What is the packing efficiency in BCC? The packing efficiency of body-centred cubic unit cell (BCC) is 68%.
4g/cm3.
## When gold crystallizes it forms face Centred cubic cells .the unit cell edge length is 408 pm calculate the density of gold molar mass of gold is 197g Mol?
Therefore, the density of the gold crystal is found to be 192.67 × 10⁵ gm/m³.
## How do you find the volume of a mole of gold?
Since one mole of gold contains Avogadro's number of atoms, we can therefore calculate the volume of 1 atom as 7.55 cm3 ÷ 6.02 × 1023 = 1.25 × 10–23 cm3.
## What is the distance between next nearest Neighbour in BCC unit cells?
The distance between two nearest neighbour in a bcc cell = `1/2xx` the length of body diagonal = `1/2xx4r=2r` <br> `=2xxsqrt3/4a=sqrt3/2a`.
## What is CCP unit cell?
The unit cell is composed of four layers of atoms arranged in a cubic close-packed (CCP) arrangement. Each of the top and bottom layers have six atoms at the hexagon corners and one atom at the center of each hexagon. The packing quality is the ratio of the atoms directly occupied by the crystal (or unit cell).
## What is the number and closest distance between octahedral voids and tetrahedral voids in fcc unit cell?
Distance between an octahedral and tetrahedral void in fcc lattice is 4body diagonal =4 a.
## Python How To Import Library Absoluth Path With Code Examples
In this article, we will look at how to get the solution for the problem, Python How To Import Library Absoluth Path With Code Examples How do I import absolute path? import * as file4 from './file4'; An absolute import path is a path that starts from a root, and you need to define a root first. In a typical JavaScript/TypeScript project, a common root is the src directory. For file1. import sys sys.path.append('/foo/bar/my_module') # Considering your module contains a funct
## Negate Regular Expression With Code Examples
In this article, we will look at how to get the solution for the problem, Negate Regular Expression With Code Examples How do you negate a regex in Python? An regex of '[^abdfgh]' will match any single character which is NOT one of 'a', 'b', 'd', 'f', 'g' or 'h'. This is a negated character class, and is indicated by the '^' character at the start of the character class. The character '^' has a special mean
## To Expend Hidden Columns And Rows With Code Examples
In this article, we will look at how to get the solution for the problem, To Expend Hidden Columns And Rows With Code Examples How do I unhide all columns quickly? Unhide All Columns At One Go Click on the small triangle at the top left of the worksheet area. This will select all the cells in the worksheet. Right-click anywhere in the worksheet area. Click on Unhide. # set up display area to show dataframe in jupyter qtconsole pd.set_option('display.height', 1000) pd.set_option( | 1,119 | 4,128 | {"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} | 3.265625 | 3 | CC-MAIN-2024-33 | latest | en | 0.824348 |
https://profsonly.com/physics-project-kinematics/ | 1,679,825,082,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00289.warc.gz | 548,875,037 | 29,455 | Get help from the best in academic writing.
# Physics Project Kinematics
ScenarioYou have recently joined the team at A
## I need help drawing a circuit diagram
1, A single cell powers an open circuit with a light bulb, and a resistor, both controlled by a switch. Always show the direction of the current flow. 2. A three-cell battery with two light bulbs wired so that if one burns out, the other will stay on. Include a closed switch in your circuit that will turn both bulbs on and off together. If you could help me quickly that would be greatly appricated, thank you.
## speed questions
Physics Assignment Help 1- A girl cycles for 3hrs at a speed of 40 km/h. What distance did she travel?
2- A car travels a distance of 540km in 6 hours. What speed did it travel at?
3- Lauren walks 100m in half a minute. What must her speed have been to travel this distance?
4- How long does it take to travel a distance of 672km at a speed of 96km/h?
## Physics Question
Identify a current problem in physics by searching for news articles and current events. One reputable source of news in physics is Phys.org. Choose one article, and in two pages, describe how the scientific method is being used to solve the problem mentioned in the article. Identify the initial observations that identified the problem, the hypothesis, tests, and any revisions of the original hypothesis. Cite the article in APA format as well as other references you might use.
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# Responding to the publics fascination with-and sometimes
Author Message
Manager
Joined: 26 Mar 2008
Posts: 96
Schools: Tuck, Duke
Responding to the publics fascination with-and sometimes [#permalink]
### Show Tags
03 Oct 2008, 18:24
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Responding to the public’s fascination with-and sometimes
undue alarm over-possible threats from asteroids, a scale
developed by astronomers rates the likelihood that a particular
asteroid or comet may collide with Earth
.
A. a scale developed by astronomers rates the likelihood
that a particular asteroid or comet may
B. a scale that astronomers have developed rates how
likely it is for a particular asteroid or comet to
C. astronomers have developed a scale to rate how likely
a particular asteroid or comet will be to
D. astronomers have developed a scale for rating the
likelihood that a particular asteroid or comet will
E. astronomers have developed a scale that rates the
likelihood of a particular asteroid or comet that may.
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Manager
Joined: 25 May 2008
Posts: 185
### Show Tags
06 Oct 2008, 07:48
The first must be "astronomers". Among C,D and E i think that D is the best choice.
Manager
Joined: 26 Mar 2008
Posts: 96
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### Show Tags
06 Oct 2008, 08:07
Even I marked D as ans. But given OA is C. This from one of the GMAT SETS test.
Intern
Joined: 15 Aug 2008
Posts: 19
### Show Tags
07 Oct 2008, 16:49
the correct idiom is "likely to be" not "likely be to" as in C ..!!
VP
Joined: 30 Jun 2008
Posts: 1018
### Show Tags
07 Oct 2008, 23:32
"likely + infinitive" is the right usage.
eg. The Zoman Empire is likely to fall.
_________________
"You have to find it. No one else can find it for you." - Bjorn Borg
Intern
Joined: 02 Oct 2008
Posts: 16
### Show Tags
07 Oct 2008, 23:41
IMO E assuming collide with earth is not underlined.
SVP
Joined: 17 Jun 2008
Posts: 1502
### Show Tags
08 Oct 2008, 03:26
"for rating" in D does not seem right....I went for C.
Manager
Joined: 22 Sep 2008
Posts: 119
### Show Tags
08 Oct 2008, 18:12
1
KUDOS
what is correct use of RATING?
For rating
To rate
That rates?
Thanks
Intern
Joined: 24 Sep 2008
Posts: 18
### Show Tags
09 Oct 2008, 03:15
1
KUDOS
to rate the the correct idiom.
VP
Joined: 30 Jun 2008
Posts: 1018
### Show Tags
09 Oct 2008, 04:16
vr4indian wrote:
what is correct use of RATING?
For rating
To rate
That rates?
Thanks
I guess the usage depends on context ....
Joey is a member of the committee that rates employee performance in this company.
Joey loves to rate his female colleagues.He gets special attention from them.
I am not sure if the example is right for the usage of "for rating"
Joey gets no extra salary for rating his colleagues.
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Posts: 832
### Show Tags
05 Dec 2008, 09:06
This is a GMAT Set question. OA is C, but I am not comfortable with it.
SVP
Joined: 29 Aug 2007
Posts: 2452
### Show Tags
05 Dec 2008, 12:26
arorag wrote:
Responding to the public’s fascination with-and sometimes undue alarm over-possible threats from asteroids, a scale developed by astronomers rates the likelihood that a particular asteroid or comet may collide with Earth.
A. a scale developed by astronomers rates the likelihood that a particular asteroid or comet may
B. a scale that astronomers have developed rates how likely it is for a particular asteroid or comet to
C. astronomers have developed a scale to rate how likely a particular asteroid or comet will be to
D. astronomers have developed a scale for rating the likelihood that a particular asteroid or comet will
E. astronomers have developed a scale that rates the likelihood of a particular asteroid or comet that may.
Agree with C.
The choice is among C, D and E.
"Scale to rate" in C is better idiom than "scale for rating" in D.
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Re: SC: Scale [#permalink] 05 Dec 2008, 12:26
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https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=DifferentialGeometry/Tensor/EnergyMomentumTensor | 1,642,955,310,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00197.warc.gz | 413,255,829 | 62,837 | DifferentialGeometry/Tensor/EnergyMomentumTensor - Maple Help
Tensor[EnergyMomentumTensor] - find the energy-momentum tensor for various matter fields
Tensor[MatterFieldEquations] - find the field equations for various matter fields
Tensor[DivergenceIdentities] - check the divergence identities for the energy-momentum tensor field for various matter fields
Calling Sequences
EnergyMomentumTensor(FieldType, g, F1, F2, ...)
MatterFieldEquations(FieldType, g, F1, F2, ...)
DivergenceIdentities(FieldType, g, F1, F2, ... , T, E1, E2,...)
Parameters
FieldType - a string, one of "DiracWeyl", "Dust", "Electromagnetic", "PerfectFluid", "Scalar", "NMCScalar"
g - a metric tensor
F1, F2,.. - scalars, tensors or spinors, defining the fields needed for the field theory designated by FieldType
T - a rank 2 tensor (the energy-momentum tensor)
E1, E2,.. - scalars, tensors or spinors, defining the field equations for the field theory designated by FieldType
Description
• The energy momentum tensor is a symmetric, rank-2 contravariant tensor $T$ which determines the right-hand side of the Einstein field equations.
• If FieldType = "DiracWeyl", then the additional arguments for EnergyMomentumTensor are: a solder form (compatible with the metric $g$), a rank 1 covariant spinor $\mathrm{ψ}$, and the complex conjugate $\stackrel{‾}{\mathrm{ψ}}$.
• If FieldType = "Dust", then the additional arguments for EnergyMomentumTensor are: a vector field $U$, a scalar $\mathrm{μ}$ (energy density).
• If FieldType = "Electromagnetic", then the additional arguments are either: a 1-form $A$ (the electromagnetic 4-potential), or a skew-symmetric rank 2 tensor $F$ (the field strength tensor).
• If FieldType = "PerfectFluid", then the additional arguments for EnergyMomentumTensor are: a vector field $U$, and scalars $\mathrm{μ}$ (energy density) and $p$ (pressure).
• If FieldType = "Scalar", then the additional argument for EnergyMomentumTensor is a scalar $\mathrm{φ}$.
• If FieldType = "NMCScalar", then the additional argument for EnergyMomentumTensor is a non-minimally coupled scalar $\mathrm{φ}$.
• See the Details help page for the explicit formulas used to calculate the various energy-momentum tensors, the matter field equations and the divergence identities.
• These commands are part of the DifferentialGeometry:-Tensor: package, and so can be used in the form EnergyMomentumTensor(...), MatterFieldEquations(...), DivergenceIdentities(...) only after executing the commands with(DifferentialGeometry); with(Tensor); in that order. They can always be used in the long form DifferentialGeometry:-Tensor:-EnergyMomentumTensor, DifferentialGeometry:-Tensor-MatterFieldEquations, DifferentialGeometry:-Tensor:-DivergenceIdentities.
Examples
> $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{Tools}\right):$$\mathrm{with}\left(\mathrm{Tensor}\right):$
Example 1. "DiracWeyl"
First create a vector bundle $N$ with base coordinates $\left(t,x,y,z\right)$ and fiber coordinates $\left(\mathrm{z1},\mathrm{z2},\mathrm{w1},\mathrm{w2}\right)$.
> $\mathrm{DGsetup}\left(\left[t,x,y,z\right],\left[\mathrm{z1},\mathrm{z2},\mathrm{w1},\mathrm{w2}\right],N\right)$
${\mathrm{frame name: N}}$ (2.1)
Define a metric of signature $\left(1,-1,-1,-1\right)$ and an orthonormal tetrad.
N > $\mathrm{g1}≔\mathrm{evalDG}\left({x}^{4}\mathrm{dt}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dt}-\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}-\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}-\mathrm{dz}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dz}\right)$
${\mathrm{g1}}{≔}{{x}}^{{4}}{}{\mathrm{dt}}{}{\mathrm{dt}}{-}{\mathrm{dx}}{}{\mathrm{dx}}{-}{\mathrm{dy}}{}{\mathrm{dy}}{-}{\mathrm{dz}}{}{\mathrm{dz}}$ (2.2)
N > $\mathrm{OTetrad}≔\mathrm{evalDG}\left(\left[\frac{1}{{x}^{2}}\mathrm{D_t},\mathrm{D_x},\mathrm{D_y},\mathrm{D_z}\right]\right)$
${\mathrm{OTetrad}}{≔}\left[\frac{{1}}{{{x}}^{{2}}}{}{\mathrm{D_t}}{,}{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}\right]$ (2.3)
Calculate the solder form.
N > $\mathrm{σ1}≔\mathrm{SolderForm}\left(\mathrm{OTetrad}\right)$
${\mathrm{σ1}}{≔}\frac{{{x}}^{{2}}{}\sqrt{{2}}}{{2}}{}{\mathrm{dt}}{}{\mathrm{D_z1}}{}{\mathrm{D_w1}}{+}\frac{{{x}}^{{2}}{}\sqrt{{2}}}{{2}}{}{\mathrm{dt}}{}{\mathrm{D_z2}}{}{\mathrm{D_w2}}{+}\frac{\sqrt{{2}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{D_z1}}{}{\mathrm{D_w2}}{+}\frac{\sqrt{{2}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{D_z2}}{}{\mathrm{D_w1}}{-}\frac{{I}}{{2}}{}\sqrt{{2}}{}{\mathrm{dy}}{}{\mathrm{D_z1}}{}{\mathrm{D_w2}}{+}\frac{{I}}{{2}}{}\sqrt{{2}}{}{\mathrm{dy}}{}{\mathrm{D_z2}}{}{\mathrm{D_w1}}{+}\frac{\sqrt{{2}}}{{2}}{}{\mathrm{dz}}{}{\mathrm{D_z1}}{}{\mathrm{D_w1}}{-}\frac{\sqrt{{2}}}{{2}}{}{\mathrm{dz}}{}{\mathrm{D_z2}}{}{\mathrm{D_w2}}$ (2.4)
Define a rank 1-spinor field $\mathrm{ψ1}$ and its complex conjugate.
N > $\mathrm{ψ1}≔\mathrm{evalDG}\left(h\left(x\right)\mathrm{dz1}-f\left(x\right)\mathrm{dz2}\right)$
${\mathrm{ψ1}}{≔}{h}{}\left({x}\right){}{\mathrm{dz1}}{-}{f}{}\left({x}\right){}{\mathrm{dz2}}$ (2.5)
N > $\mathrm{barpsi1}≔\mathrm{evalDG}\left(h\left(x\right)\mathrm{dw1}-f\left(x\right)\mathrm{dw2}\right)$
${\mathrm{barpsi1}}{≔}{h}{}\left({x}\right){}{\mathrm{dw1}}{-}{f}{}\left({x}\right){}{\mathrm{dw2}}$ (2.6)
Calculate the Dirac-Weyl energy momentum tensor $T$.
N > $\mathrm{T1}≔\mathrm{EnergyMomentumTensor}\left("DiracWeyl",\mathrm{g1},\mathrm{σ1},\mathrm{ψ1},\mathrm{barpsi1}\right)$
${\mathrm{T1}}{≔}{-}\frac{\sqrt{{2}}{}\left({{f}{}\left({x}\right)}^{{2}}{-}{{h}{}\left({x}\right)}^{{2}}\right)}{{{x}}^{{3}}}{}{\mathrm{D_t}}{}{\mathrm{D_y}}{+}\sqrt{{2}}{}\left({-}{h}{}\left({x}\right){}{f}{\prime }{}\left({x}\right){+}{f}{}\left({x}\right){}{h}{\prime }{}\left({x}\right)\right){}{\mathrm{D_x}}{}{\mathrm{D_y}}{-}\frac{\sqrt{{2}}{}\left({{f}{}\left({x}\right)}^{{2}}{-}{{h}{}\left({x}\right)}^{{2}}\right)}{{{x}}^{{3}}}{}{\mathrm{D_y}}{}{\mathrm{D_t}}{+}\sqrt{{2}}{}\left({-}{h}{}\left({x}\right){}{f}{\prime }{}\left({x}\right){+}{f}{}\left({x}\right){}{h}{\prime }{}\left({x}\right)\right){}{\mathrm{D_y}}{}{\mathrm{D_x}}$ (2.7)
Evaluate the Dirac-Weyl field equations $\mathrm{E1}$ for the given spinor field $\mathrm{ψ}$.
N > $\mathrm{E1}≔\mathrm{MatterFieldEquations}\left("DiracWeyl",\mathrm{g1},\mathrm{σ1},\mathrm{ψ1},\mathrm{barpsi1}\right)$
${\mathrm{E1}}{≔}\frac{\frac{{I}}{{2}}{}\sqrt{{2}}{}\left({f}{\prime }{}\left({x}\right){}{x}{+}{f}{}\left({x}\right)\right)}{{x}}{}{\mathrm{D_w1}}{-}\frac{\frac{{I}}{{2}}{}\sqrt{{2}}{}\left({h}{\prime }{}\left({x}\right){}{x}{+}{h}{}\left({x}\right)\right)}{{x}}{}{\mathrm{D_w2}}{,}{-}\frac{\frac{{I}}{{2}}{}\sqrt{{2}}{}\left({f}{\prime }{}\left({x}\right){}{x}{+}{f}{}\left({x}\right)\right)}{{x}}{}{\mathrm{D_z1}}{+}\frac{\frac{{I}}{{2}}{}\sqrt{{2}}{}\left({h}{\prime }{}\left({x}\right){}{x}{+}{h}{}\left({x}\right)\right)}{{x}}{}{\mathrm{D_z2}}$ (2.8)
Check the divergence identity for the dust energy momentum tensor $T$. The LHS is the covariant divergence of the energy momentum tensor and the RHS is a combination of the field equations.
N > $\mathrm{Div1},\mathrm{RHS1}≔\mathrm{DivergenceIdentities}\left("DiracWeyl",\mathrm{g1},\mathrm{σ1},\mathrm{ψ1},\mathrm{barpsi1},\mathrm{T1},\mathrm{E1}\right)$
${\mathrm{Div1}}{,}{\mathrm{RHS1}}{≔}\frac{\sqrt{{2}}{}\left({-}{x}{}{h}{}\left({x}\right){}{f}{″}{}\left({x}\right){+}{x}{}{f}{}\left({x}\right){}{h}{″}{}\left({x}\right){+}{2}{}{f}{}\left({x}\right){}{h}{\prime }{}\left({x}\right){-}{2}{}{h}{}\left({x}\right){}{f}{\prime }{}\left({x}\right)\right)}{{x}}{}{\mathrm{D_y}}{,}\frac{\sqrt{{2}}{}\left({-}{x}{}{h}{}\left({x}\right){}{f}{″}{}\left({x}\right){+}{x}{}{f}{}\left({x}\right){}{h}{″}{}\left({x}\right){+}{2}{}{f}{}\left({x}\right){}{h}{\prime }{}\left({x}\right){-}{2}{}{h}{}\left({x}\right){}{f}{\prime }{}\left({x}\right)\right)}{{x}}{}{\mathrm{D_y}}$ (2.9)
N > $\mathrm{Div1}-\mathrm{RHS1}$
${0}$ (2.10)
We note that $f\left(x\right)=h\left(x\right)=\frac{1}{x}$ is a solution of the Dirac-Weyl field equations:
N > $\mathrm{map}\left(\mathrm{DGsimplify},\mathrm{eval}\left(\left[\mathrm{E1}\right],\left[f\left(x\right)=\frac{1}{x},h\left(x\right)=\frac{1}{x}\right]\right)\right)$
$\left[{0}{}{\mathrm{D_z1}}{,}{0}{}{\mathrm{D_z1}}\right]$ (2.11)
The covariant divergence of the energy momentum tensor vanishes on this solution:
N > $\mathrm{DGsimplify}\left(\mathrm{eval}\left(\mathrm{Div1},\left[f\left(x\right)=\frac{1}{x},h\left(x\right)=\frac{1}{x}\right]\right)\right)$
${0}{}{\mathrm{D_t}}$ (2.12)
Example 2. "Dust"
First create a manifold $M$ with base coordinates $\left(t,x,y,z\right)$:
N > $\mathrm{DGsetup}\left(\left[t,x,y,z\right],M\right)$
${\mathrm{frame name: M}}$ (2.13)
Define a metric.
M > $\mathrm{g2}≔\mathrm{evalDG}\left(\mathrm{dt}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dt}-{t}^{2}\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}-\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}-\mathrm{dz}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dz}\right)$
${\mathrm{g2}}{≔}{\mathrm{dt}}{}{\mathrm{dt}}{-}{{t}}^{{2}}{}{\mathrm{dx}}{}{\mathrm{dx}}{-}{\mathrm{dy}}{}{\mathrm{dy}}{-}{\mathrm{dz}}{}{\mathrm{dz}}$ (2.14)
Define the normalized 4-vector representing the 4-velocity of the dust.
M > $\mathrm{u2}≔\mathrm{evalDG}\left(\mathrm{cosh}\left(f\left(t\right)\right)\mathrm{D_t}-\frac{\mathrm{sinh}\left(f\left(t\right)\right)}{t}\mathrm{D_x}\right)$
${\mathrm{u2}}{≔}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{D_t}}{-}\frac{{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right)}{{t}}{}{\mathrm{D_x}}$ (2.15)
M > $\mathrm{TensorInnerProduct}\left(\mathrm{g2},\mathrm{u2},\mathrm{u2}\right)$
${1}$ (2.16)
Define the energy density.
M > $\mathrm{μ2}≔h\left(t\right)$
${\mathrm{μ2}}{≔}{h}{}\left({t}\right)$ (2.17)
Calculate the dust energy- momentum tensor $\mathrm{T2}$.
M > $\mathrm{T2}≔\mathrm{EnergyMomentumTensor}\left("Dust",\mathrm{g2},\mathrm{u2},\mathrm{μ2}\right)$
${\mathrm{T2}}{≔}{h}{}\left({t}\right){}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{}{\mathrm{D_t}}{}{\mathrm{D_t}}{-}\frac{{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right)}{{t}}{}{\mathrm{D_t}}{}{\mathrm{D_x}}{-}\frac{{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right)}{{t}}{}{\mathrm{D_x}}{}{\mathrm{D_t}}{+}\frac{{h}{}\left({t}\right){}\left({{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{-}{1}\right)}{{{t}}^{{2}}}{}{\mathrm{D_x}}{}{\mathrm{D_x}}$ (2.18)
Evaluate the dust field equations $\mathrm{E2}$ for the given $\mathrm{u2}$ and $\mathrm{μ2}$.
M > $\mathrm{E2}≔\mathrm{MatterFieldEquations}\left("Dust",\mathrm{g2},\mathrm{u2},\mathrm{μ2}\right)$
${\mathrm{E2}}{≔}\frac{{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){+}\stackrel{{\mathbf{.}}}{{h}}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{t}{+}{h}{}\left({t}\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}}{{t}}{,}\frac{{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{-}{1}{+}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}}{{t}}{}{\mathrm{D_t}}{-}\frac{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}\left({\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){+}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}\right)}{{{t}}^{{2}}}{}{\mathrm{D_x}}$ (2.19)
Check that the following values for $f\left(t\right)$ and solve the dust field equations.
M > $\mathrm{Soln}≔\left[h\left(t\right)=\frac{\mathrm{_C2}}{{\left(1+{t}^{2}{\mathrm{_C1}}^{2}\right)}^{\frac{1}{2}}},f\left(t\right)=\mathrm{arcsinh}\left(\frac{1}{t\mathrm{_C1}}\right)\right]$
${\mathrm{Soln}}{≔}\left[{h}{}\left({t}\right){=}\frac{{\mathrm{_C2}}}{\sqrt{{1}{+}{{t}}^{{2}}{}{{\mathrm{_C1}}}^{{2}}}}{,}{f}{}\left({t}\right){=}{\mathrm{arcsinh}}{}\left(\frac{{1}}{{t}{}{\mathrm{_C1}}}\right)\right]$ (2.20)
M > $\mathrm{simplify}\left(\mathrm{eval}\left(\left[\mathrm{E2}\right],\mathrm{Soln}\right),\mathrm{symbolic}\right)$
$\left[{0}{,}{0}{}{\mathrm{D_t}}{+}{0}{}{\mathrm{D_x}}\right]$ (2.21)
Check the divergence identity for the dust energy-momentum tensor $\mathrm{T2}$. The LHS is the covariant divergence of the energy-momentum tensor and the RHS is a combination of the field equations.
M > $\mathrm{Div2},\mathrm{RHS2}≔\mathrm{DivergenceIdentities}\left("Dust",\mathrm{g2},\mathrm{u2},\mathrm{μ2},\mathrm{T2},\mathrm{E2}\right)$
${\mathrm{Div2}}{,}{\mathrm{RHS2}}{≔}\frac{{2}{}{h}{}\left({t}\right){}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{-}{h}{}\left({t}\right){+}\stackrel{{\mathbf{.}}}{{h}}{}\left({t}\right){}{t}{}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{+}{2}{}{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}}{{t}}{}{\mathrm{D_t}}{-}\frac{{2}{}{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){+}\stackrel{{\mathbf{.}}}{{h}}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){}{t}{+}{2}{}{h}{}\left({t}\right){}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}{-}{h}{}\left({t}\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}}{{{t}}^{{2}}}{}{\mathrm{D_x}}{,}\frac{{2}{}{h}{}\left({t}\right){}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{-}{h}{}\left({t}\right){+}\stackrel{{\mathbf{.}}}{{h}}{}\left({t}\right){}{t}{}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{+}{2}{}{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}}{{t}}{}{\mathrm{D_t}}{-}\frac{{2}{}{h}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){+}\stackrel{{\mathbf{.}}}{{h}}{}\left({t}\right){}{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right){}{\mathrm{sinh}}{}\left({f}{}\left({t}\right)\right){}{t}{+}{2}{}{h}{}\left({t}\right){}{{\mathrm{cosh}}{}\left({f}{}\left({t}\right)\right)}^{{2}}{}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}{-}{h}{}\left({t}\right){}\stackrel{{\mathbf{.}}}{{f}}{}\left({t}\right){}{t}}{{{t}}^{{2}}}{}{\mathrm{D_x}}$ (2.22)
M > $\mathrm{Div2}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&minus\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{RHS2}$
${0}{}{\mathrm{D_t}}$ (2.23)
Example 3. "Electromagnetic"
First create a manifold $M$ with base coordinates $\left(t,x,y,z\right)$.
M > $\mathrm{DGsetup}\left(\left[t,x,y,z\right],M\right)$
${\mathrm{frame name: M}}$ (2.24)
Define a metric.
M > $\mathrm{g3}≔\mathrm{evalDG}\left({x}^{2}\mathrm{dt}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dt}-\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}-\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}-\mathrm{dz}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dz}\right)$
${\mathrm{g3}}{≔}{{x}}^{{2}}{}{\mathrm{dt}}{}{\mathrm{dt}}{-}{\mathrm{dx}}{}{\mathrm{dx}}{-}{\mathrm{dy}}{}{\mathrm{dy}}{-}{\mathrm{dz}}{}{\mathrm{dz}}$ (2.25)
Define an electromagnetic 4-potential $\mathrm{A3}$.
M > $\mathrm{A3}≔\mathrm{evalDG}\left(\mathrm{f1}\left(x\right)\mathrm{dt}+\mathrm{f2}\left(x\right)\mathrm{dy}\right)$
${\mathrm{A3}}{≔}{\mathrm{f1}}{}\left({x}\right){}{\mathrm{dt}}{+}{\mathrm{f2}}{}\left({x}\right){}{\mathrm{dy}}$ (2.26)
Calculate the electromagnetic energy-momentum tensor $\mathrm{T3}$.
M > $\mathrm{T3}≔\mathrm{EnergyMomentumTensor}\left("Electromagnetic",\mathrm{g3},\mathrm{A3}\right)$
${\mathrm{T3}}{≔}{-}\frac{{{\mathrm{f2}}{\prime }{}\left({x}\right)}^{{2}}{}{{x}}^{{2}}{+}{{\mathrm{f1}}{\prime }{}\left({x}\right)}^{{2}}}{{2}{}{{x}}^{{4}}}{}{\mathrm{D_t}}{}{\mathrm{D_t}}{+}\frac{{\mathrm{f1}}{\prime }{}\left({x}\right){}{\mathrm{f2}}{\prime }{}\left({x}\right)}{{{x}}^{{2}}}{}{\mathrm{D_t}}{}{\mathrm{D_y}}{+}\frac{{-}{{\mathrm{f2}}{\prime }{}\left({x}\right)}^{{2}}{}{{x}}^{{2}}{+}{{\mathrm{f1}}{\prime }{}\left({x}\right)}^{{2}}}{{2}{}{{x}}^{{2}}}{}{\mathrm{D_x}}{}{\mathrm{D_x}}$ | 6,831 | 17,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 92, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.545807 |
http://slideplayer.com/slide/2426053/ | 1,544,751,720,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825123.5/warc/CC-MAIN-20181214001053-20181214022553-00096.warc.gz | 263,202,515 | 21,973 | # “How much soil is being lost from the fields in Tarland each year?” This is a very valid question since the soil is the farmer’s principal agricultural.
## Presentation on theme: "“How much soil is being lost from the fields in Tarland each year?” This is a very valid question since the soil is the farmer’s principal agricultural."— Presentation transcript:
“How much soil is being lost from the fields in Tarland each year?” This is a very valid question since the soil is the farmer’s principal agricultural resource. It has taken effort to clear the soil of rocks, to make water conditions and then nutrient levels correct for crop growth. When soil is lost valuable phosphorus fertiliser and seed is lost too. The phosphorus and the soil itself (termed eroded sediment) pollutes waters and damages conditions for fish and other ecology.
“How much soil is being lost from the fields in Tarland each year?” Our calculations from monitoring during 2004-2005 show that 600 tonnes of soil were lost down the river network from the 50 km 2 catchment upstream of Coull
and here’s how we worked it out… The full story can be found in the journal article
“How much soil is being lost from the fields in Tarland each year?” There are 3 ways to answer this: a)Using typical values for soil erosion losses from fields under different land uses taken from scientific literature b)Using detailed field scale experiments, observing volumes of soil lost, which have given us the literature values above c)Using a landscape scale approach measuring the amount leaving the catchment through the river network over a year
How to calculate annual river sediment loads from the upper Tarland catchment We will demonstrate method (c), then compare the average catchment values to possible losses under severe examples of localised erosion
River sampling to determine sediment losses The amount of material moving down a river is referred to as the mass load. The basic calculation is: This is then scaled up to a year using all the sampled time points and the average river flow rate (kg / year) Then this can be expressed on an area basis according to the size of the catchment (kg / hectare / year) Concentration at time of sampling (kg / Litre) Water flow rate when sampling (Litres / second) Mass load (kg / second) ×=
River sampling to determine sediment losses Sediments are sampled by collection of a water sample using an autosampler to fill a 1 litre bottle The autosampler can either take a sample at a set time each day, or be automatically triggered by a rise in river level On return to the laboratory samples are filtered and the mass of dried sediment on each filter paper is recorded
River sampling to determine sediment losses An intensive period of sampling over 2004 to 2005 provides the best data to determine sediment loads During this year 495 sediment samples were obtained between daily and 4 hourly storm sampling River flow was measured every 15 minutes Samples were collected near Coull This defines a contributing catchment area of 50 km 2
What do the final figures mean? During 2004-2005 the average suspended sediment concentrations in the river at Coull ranged from 7 mg / Litre in early summer to 44 mg / Litre in late winter The loss of sediment measured at Coull was 596240 kg (approximately 600 tonnes of soil lost) or 116 kg / hectare / year This carries approximately 600 kg of pure phosphorus equivalent to £1000 of chemical fertiliser each year, This is not much in cost, but is equivalent to the phosphorus pollution from the septic tanks of ~500 family homes each year
Further considerations The fine suspended sediments measured in the river are not the only component of the river mass load. We have not considered here the courser bed material that moves slowly through the river network. So, whilst our method underestimates the true river mass load: it is the finer particles considered here by this method that actually carry the bulk of the nutrient load from the fertilised fields, also the fine particles are most physically damaging to habitats
Further considerations Local field erosion rates can be much greater, dependant on topography and management Much of this soil may not make it to the river network, but instead accumulates in depressions, field edges and buffer strips An example can be seen from our work in the Lunan catchment (Angus). Experiments with a simple 200 m long filter fence at the bottom of a single potato field after harvested trapped 70 tonnes of soil from a single field during two months
Similar presentations | 942 | 4,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.944468 |
https://mathphysicsexpert.com/let-a-be-a-non-empty-set-and-pa-be-the-power-set-of-a/ | 1,669,990,117,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00481.warc.gz | 434,151,450 | 11,247 | # Let A be a non-empty set and P(A) be the power set of A.
Let A be a non-empty set and P(A) be the power set of A. Then P(A) is a Boolean algebra under the usual operations of union, intersection and complementary in P(A). The sets ∅ and A are the zero element and unit element of the Boolean algebra P(A). Observe that if A is an infinite set, then the Boolean algebra P(A) will contain infinite number of elements. | 104 | 418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.812755 |
https://www.enotes.com/homework-help/girl-throws-tennis-ball-upward-an-initial-633319 | 1,670,001,072,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710909.66/warc/CC-MAIN-20221202150823-20221202180823-00797.warc.gz | 757,720,735 | 18,706 | # A girl throws a tennis ball upward with an initial velocity of 4 m/s. What is the maximum displacement of the ball?
For future reference, the formula below can be used to solve similar problems.
The formula is `h=(V^2)/(2g)` where `h` is the maximum height (maximum displacement) and `V` is the initial upward velocity.
Approved by eNotes Editorial Team
Hello!
A tennis ball will go up and its velocity will decrease. When the velocity is zero, the ball will be at its maximum height and will begin to fall down. This maximum height is what we are solving for.
The easiest way to find this is to consider energy. Energy will be conserved, so the energy of the ball at the start and at the point of maximum height will be the same.
At the start a ball has only kinetic energy `(mV^2)/2,` where `m` is the mass and `V` is the initial speed. In contrast, at the maximum height it will have only the potential energy `mgh` where `h` is the maximum height.
So `(mV^2)/2=mgh,` i.e. `V^2/2=gh` and `h=V^2/(2g) approx 16/20` = 0.8 (m). This is the answer.
Note that we ignored air resistance. If we hadn't, the real height would be less.
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https://www.6miu.com/read-3850309.html | 1,709,648,507,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00118.warc.gz | 611,309,931 | 6,112 | # 5873. 小p的属性
xiaoxiao2021-03-01 26
## 程序：
#include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> #define N 2005 #define LL long long using namespace std; struct data{LL x,y,z;}x[N]; LL sum[N][N],f[N][N],b[N]; LL n,m,size,cnt; int main(){ freopen("growth.in","r",stdin); freopen("growth.out","w",stdout); scanf("%lld%lld",&n,&m); for (int i=1;i<=n;i++) { scanf("%lld%lld%lld",&x[i].x,&x[i].y,&x[i].z); } for (int i=1;i<=n;i++) b[++cnt]=x[i].x,b[++cnt]=x[i].y; sort(b+1,b+cnt+1); size=unique(b+1,b+cnt+1)-b-1; for (int i=1;i<=n;i++){ x[i].x=lower_bound(b+1,b+size+1,x[i].x)-b; x[i].y=lower_bound(b+1,b+size+1,x[i].y)-b; sum[x[i].x][x[i].y]+=x[i].z; } for (int i=1;i<=size;i++) for (int j=1;j<=size;j++){ sum[i][j]=sum[i-1][j]+sum[i][j-1]+sum[i][j]-sum[i-1][j-1]; } for (int i=1;i<=size;i++) for (int j=1;j<=size;j++){ if (i==1&&j==1) continue; f[i][j]=max(f[i-1][j]+sum[i-1][j]*(b[i]-b[i-1]-1),f[i][j-1]+sum[i][j-1]*(b[j]-b[j-1]-1))+sum[i][j]; } LL ans=0; for (int i=1;i<=size;i++) for (int j=1;j<=size;j++) if (b[i]+b[j]<=m){ ans=max(ans,f[i][j]+sum[i][j]*(m-b[i]-b[j])); } printf("%lld\n",ans); } | 501 | 1,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.188917 |
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## NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2
Ex 5.2 Class 9 Maths Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
Two distinct intersecting lines cannot be parallel to the same line.
Ex 5.2 Class 9 Maths Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes.
According to Euclid’s fifth postulate when line x falls on line y and z such that ∠1+ ∠2< 180°. Then, line y and line z on producing further will meet in the side of ∠1 arid ∠2 which is less than 180°.
We find that the lines which are not according to Euclid’s fifth postulate. i.e., ∠1 + ∠2 = 180°, do not intersect.
### NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry (युक्लिड के ज्यामिति का परिचय) (Hindi Medium) Ex 5.2
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Comments are closed | 271 | 952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-26 | latest | en | 0.825302 |
http://fliphtml5.com/tags/equation%20xk | 1,582,194,292,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00052.warc.gz | 56,076,304 | 9,624 | Theorem 6.2: Properties of Isomorphisms Acting on Elements
For a fixed integer k and a fixed group element b in G, the equation xk = b has the same number of solutions in G as does the equation xk = φ(b) in G¯.
MA441: Algebraic Structures I - AbstractAlgebra.net: The ...
For a fixed integer k and a fixed group ele- ment b in G1, the equation xk = b has the same number of solutions in G1 as does the equation xk = φ(b) in G2.
MA441: Algebraic Structures I - AbstractAlgebra.net: The ...
For a fixed integer k and a fixed group ele- ment b in G1, the equation xk = b has the same number of solutions in G1 as does the equation xk = φ(b) in G2. | 194 | 643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.899907 |
https://www.lidolearning.com/questions/m-bb-rdsharma9-ch21-ex21p2-q12/a-cylinder-whose-height-is-two/ | 1,680,276,638,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00639.warc.gz | 950,182,846 | 13,128 | # RD Sharma Solutions Class 9 Mathematics Solutions for Surface Area And Volume Of Sphere Exercise 21.2 in Chapter 21 - Surface Area And Volume Of Sphere
Question 5 Surface Area And Volume Of Sphere Exercise 21.2
A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Radius of a sphere (R)= 4 cm (Given)
Height of the cylinder = 2/3 diameter (given)
Let h be the height and r be the base radius of a cylinder, then h = 2/3× (2r) = 4r/3
Volume of the cylinder = Volume of the sphere or r = 4
Therefore, radius of the base of the cylinder is 4 cm.
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Ellipses are necessary but troublesome. Learn to deal with them appropriately.
Català - Castellano
If you have a Boss like mine who loves ellipse shaped Buildings, Drop-offs, Piazzas, etc. you probably have to deal with ellipses more than you wish. Ellipses are quite troublesome.
First of all they don't offset as ellipses, rather when we offset an ellipse the parallel geometry obtained is a SPLINE. Splines have the disadvantage that they can only be trimmed but not extended.
One way to partially solve this problem is to know the PELLIPSE system variable. The default value of this paramter is 0, which means ellipses are drawn as ellipses. If we set it to 1, instead of an ellipse the geometry drawn is a POLYLINE. The advantage of using polylines is that they can be exploded (Ellipses can't) and when offset they remain polylines. Another advantage is that Polylines can be converted to other types of objects (in ACA you can convert them to Walls, Spaces, etc).
The disadvantage of drawing ellipses as Polylines is that since they are an approximation of an ellipse made with multiple curved segments, if we trim part of the ellipse shaped polyline we will probably not going to be able to recover the old geometry by using the extend command.
Another consideration when deciding if you want to set PELLIPSE to 1 or 0 is to think if you will need to modify that ellipse often. If you will be trimming and extending segments of an ellipse my recommendation is to keep it in 0 (the ellipse is a "smart object" you will be able to recover all of its geometry from its smallest segment if is a real ellipse, if it is a polyline yo will have to redraw it) If you will be offsetting the ellipse better set it to 1 so you avoid getting Splines that will leave you with no flexibility.
Another issue I found when dealing with ellipses is when I want to redraw one that is not aligned with the active UCS. For some reason the "UCS Object" feature doesn't allign properly with the axis of ellipses. So, if you want to draw an ellipse using the same alignment for its axis as a previous one this is what you need to do.
You need to know the concept of Quadrant Osnap. In ellipses (and circles) quadrant osnap is a point on the ellipse that intersects the ellipse axis. See the three images below for a clearer explanation (click on the image to enlarge)
The image on the left shows a selected ellipse. The grips we see correspond to the centre and quadrant osnaps of the ellipse that we will use to get the righ UCS alignment. The image in the middle shows how we can access the Osnap contextual menu to select the quadrant osnap. (follow this link for more info on the topic). The image on the right shows the line from quadrant to quadrant drawn. We will use the align UCS to Object to this line to get the correct UCS that will allow us to redraw an ellipse equal or parallel to the original one.(This only works with Ellipses drawn with PELLIPSE = 0).
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Clip your Blocks or XREFs using curved polylines, circles, etc
Català - Castellano - Deutsch
A common problem found when using the XCLIP command is that it doesn't clip properly if the polyline we use has curved segments. The same way we can't use circles to clip Blocks or XREFs. If you try to clip a block using a polyline that contains curved segments you will see that along the curved segment the clipping is not done properly. See the following image. The block was clipped using the yellow polyline. The result is not satisfactory at all.
The first image shows the original block, the second one is the block clipped with the XCLIP command. As you see we need a work around this result. The work around is to use an Express Tool Command.
CLIPIT is an express tool command that will allow us to do what we wanted to do. CLIPIT will ask us to select the new clipping frame, that can actually be a POLYLINE, CIRCLE, ARC, ELLIPSE, or TEXT object. It works the opposite as the XCLIP command. With CLIPIT you first select the frame and then the Block, Image, Wipeout or XREF you want to clip. See the following image with the result.
What CLIPIT does, is turn the curved clipping frame into a set of segments. If you switch on the XCLIPFRAME and select the block, you will see that the frame is actually made of straight lines. See it below.
The only bad part of this express tool, is that so far (till version 2009) it doesn't allow inverted clips. We talked about how useful inverted XCLIPS can be on a previous post. Hopefully in future versions of AutoCAD this will be solved.
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### AutoCAD: LISP Routine for Converting Splines to Polylines
A LISP Routine to Convert some SPlines to Polylines. It has some bugs though.
Català - Castellano - Deutsch
I was recently looking for a way to convert Splines to Polyines and found a LISP routine via Cadalyst. The routine works fine for Splines that we create from scratch. Unfortunately it doesn't work properly for Splines that have been generated when we offset an ellipse. In that case, the resulting polyline does not correspond to the spline we selected. For any other spline, it seems to work properly. You can download it at the cadalyst website.
As I previously explained, remember that you have the option to create ellipses as such or to do it creating an ellipse looking polyline through the PELLIPSE system variable.
Update: Kerry pointed out that there is a much easier way to convert Splines into Polylines. It is as simple as using the FLATTEN command for the spline you want to Convert. Thanks Kerry for the feedback.
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### AutoCAD: AutoLISP to Generate Multiple Hatch Boundaries
Do you need to re-generate the boundaries of multiple hatch patterns? An AutoLISP routine will do it.
Català - Castellano - Deutsch
I had to work on some files converted from ArchiCAD to AutoCAD and needed to use the AEC Space entities to get the areas. Unfortunately, plain AutoCAD doesn't read the are of this objects, and by exploding them, it generated Hatch Patters that would not show their area on the properties palette.
The solution for a single AEC_SPACE, was simple, explode it, regenerate the boundary of the Hatch Pattern, and select the resulting Polyline to see the area (or to extract the area to an excel file as explained on a previous post). The problem came becuase there were a lot of this AEC_SPACE objects,and AutoCAD does not allow to regenerate boundaries of multiple Hatch Patterns at the same time.
The solution was found via DigitalCAD, in the form of a LISP routine called HATCHB.LSP. This routine when used, allows you to select as many Hatch Patterns as you want and obtain their boundaries in the form of polylines. The polylines will be generated on the current layer, and properties.
Some thoughts: This real life situation is a clear case that shows how format incompatibilities makes us waste a lot of time. In this case the lead architect works in ArchiCAD, but we as Contruction Managers have only AutoCAD, so everytime we get files from them there is a lot of information in those files that is wasted, because we can not read it properly, so we have to waste hours on retracing polylines to be able to double check the information we have received...
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### SketchUp Plugin: Draw Polylines, Bezier Curves, Splines and Chamfered PLines
Do you find the default Bezier curve tool hard to use? This plugin will make dealing with Bezier Curves and other 2D geometry much easier.
Fredo6 has been busy again creating another amazing Plugin to draw, and most important, edit bezier curves. With it you can draw bezier curves, polylines, B-splines, Courbettes (multi tangent arc polyline), Catmull Splines, F-Splines... Yeah I don't know either the exact meaning of most of these terms, but they are very useful as different methods to draw curves. The best as I mentioned is the possibility to edit them using their control points. This is the toolbar that will appear when you install the Plugin.
The Plugin can be found here. For an extra feature (drawing chamfered Polylines) you will need the extension posted here. To download both the plugin and the chamfer extension, you will need to be registered at Sketchucation.com. See some shapes generated with the different curves and the option "close loop nicely".
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### AutoCAD: List of Express Tools
A complete list of the Express Tools that make your drafting expierence much easier.
Las week I found this list of the Express Tools in the Autodesk discussion Groups. With no means of just copying it, I though I could reproduce it here and use it as a directory to link to those Posts that I have written already about them. With more time, I will eventually update the list wit more and more links to Posts talking to one or more Express Tools. So here is the list.
ALIASEDIT Allows you to create, modify, and delete AutoCAD command aliases on-the-fly.
ALIGNSPACE Adjusts a viewport's zoom factor and panning position based on the alignment points specified in model space and paper space.
ARCTEXT Places text along an arc
ATTIN Imports block attribute values from an external, tab-delimited ASCII file.
ATTOUT Exports block attribute values to an external file in tab-delimited ASCII format.
BCOUNT Calculates number of blocks
BEXTEND Extends objects to nested objects within blocks or xrefs
BLOCK? Lists block objects
BLOCKREPLACE Allows you to globally replace all inserts of one block with another block.
BLOCKTOXREF Replaces all instances of a standard block with an xref. Unbinds xrefs that are bound.
BREAKLINE Creates a polyline and inserts the breakline symbol.
BSCALE Scales a block insert from its insertion point.BTRIM Trims to entities nested in blocks or xrefs
BURST Explodes block and converts attributes to text
CDORDER Arranges the drawing order of objects by colour number
CHSPACE Seamlessly moves objects from one space to the other while maintaining the appearance of the original objects.
CHURLS Provides a method to change a previously placed URL (Uniform Resource Locator) address.
CLIPIT Xclip command with arc, circle, and polyline capability
CLOSEALL Close all open drawings, asking whether to save changes if the drawing has been modified since the last Save.
COPYM Copies multiple objects with Repeat, Array, Divide and Measure options.
COPYTOLAYER Copies selected objects to a different layer, leaving original objects intact.
DIMEX Saves dimension styles to a file
DIMIM Imports dimension styles saved to a file with
DIMEX command
DIMREASSOC Restores a measurement value to overridden or modified dimension text.
DUMPSHX Converts existing SHX (compiled) files to the equivalent SHP (decompiled) files
DWGLOG Creates and maintains an individual log file for each drawing file as it is accessed.
EXOFFSET This enhanced version of the OFFSET command offers several advantages over the standard command, including layer control, undo, and a multiple option.
EXPLAN An extended version of the PLAN command.
EXTRIM Trim using closed polyline for cookie cutter effect
FASTSEL Creates a selection set of objects that touch the selected object.
FLATTEN Converts 3D geometry to 2D geometry
FULLSCREEN Toggles FULLSCREEN mode
GATTE Globally changes attribute values
GETSEL Collects specific entity types and makes them the current selection
IMAGEEDIT Launches the image-editing program (for example, Paintbrush) for the selected image.
JULIAN Contains a collection of AutoLISP routines for calendar date conversions
LAYCUR Changes the layer of selected objects to the current layer
LAYDEL Permanently deletes layer from drawing, even if it contains objects
LAYFRZ Freezes layers of selected objects
LAYISO Isolates layers of selected objects
LAYLCK Locks layer of selected object
LAYMCH Changes the layer of selected objects to the layer of a selected destination object
LAYMRG Merges two layers, and removes the first layer from the drawing
LAYOFF Turns of layers of selected objects
LAYON Turns on all layers in drawing
LAYOUTMERGE Combines specified layouts into the current layout
LAYTHW Thaws all layers in drawing
LAYULK Unlocks layers of selected object
LAYUNISO Turns on all layers that were turned off by the last LAYISO command.
LAYVPI Isolates an object's layer to the current viewport
LAYVPMODE Controls whether the layer utilities LAYISO, LAYFRZ and LAYOFF use VPFreeze or the standard layer Freeze or Off when used in a floating paper space viewport.
LAYWALK Dynamically displays objects on selected layers
LMAN Layer Manager saves and restores layer settings
LSP Displays a list of AutoLISP commands available at the command prompt
LSPSURF Displays the contents of an AutoLISP file by individual functions.
MKLTYPE Creates linetypes based on selected entities
MKSHAPE Creates shapes based on selected entities
MOCORO Moves, copies, rotates and scales entities
MOVEBAK Changes the destination directory for BAK files
MPEDIT PEDIT for multiple polylines
MSTRETCH Stretches using multiple selection windows
NCOPY Copies entities nested inside blocks and xrefs
OVERKILL Removes unneeded objects by deleting duplicates and combining line and arc segments that overlap.
PACK Starts Pack & Go program
PLT2DWG Imports HPGL files into the current drawing session. All colours are retained.
PROPULATE Update, list or clear Drawing Properties data
PSBSCALE Sets or updates the scale of block objects relative to paper space.
PSTSCALE Paper space text scaling utility
QLATTACH Attaches leader to annotation object
QLATTACHSET Globally attaches leaders to annotation objects
QLDETACHSET Detaches leader from annotation object
QQUIT Close all open drawings and exit.
REDIR Redefines hard-coded directory paths in xrefs, images, shapes, styles, and rtext
REDIRMODE Sets options for the REDIR command by specifying which object types the command should act on.
REVERT Closes and re-opens the current drawing.
RTEDIT Allows editing existing Remote Text (Rtext) objects
RTEXT Inserts or edits remote text entity
RTUCS Rotate the UCS dynamically with your pointing device.
SAVEALL Save all open drawings
SHOWURLS Shows location of embedded URLs S
HP2BLK Creates a new block definition based on the appearance of a shape object.
SSTOOLS A collection of AutoLISP routines that creates exclusionary selection sets SSX returns a selection set either exactly like a selected object or, if the filter list is adjusted, very similar to it.
SUPERHATCH Uses images, blocks, xrefs, and wipeouts as hatch patterns
SYSVDLG Allows you to view, edit and save system variable settings.
TCASE Changes the case of selected text, mtext, attributes and dimension text.
TCIRCLE Places a circle, a slot, or a rectangle around each selected text or mtext object.
TCOUNT Adds sequential numbering to text objects. The numbering can appear as a prefix, suffix or replacement text.
TEXTFIT Fits text between specified points
TFRAMES Toggles the state of frames for Wipeout and image objects. If frames are turned on, this command turns them off, and vice versa.
TJUST Changes a text object's justification without changing its position. Works with text, mtext, and attribute definition objects.
TORIENT Aligns text, mtext and block attribute objects to new orientation
TSCALE Scales text, mtext, attributes and attribute definitions.
TXT2MTXT Converts text entities created with text or Dtext to Mtext
TXTEXP Explodes text or Mtext objects into polylines
VPSCALE Displays a clear translation of the scale of the current or selected viewport.
VPSYNC Synchronizes one or more viewports with a master viewport. All synchronized viewports will take on the zoom factor of the master viewport.
XDATA Attaches extended entity data to any object
XDLIST Lists extended entity data attached to object
XLIST Displays properties of entities nested in external references or blocks
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### AutoCAD: List of 2D Objects
Català - Castellano - Deutsch
There are many different types of Objects in AutoCAD. On an effort to keep updating the directory of CAD-Addict.com here comes a complete list of the 2D Objects and the links to those articles that explain their use. I will try to keep adding lists of objects weekly and at the end I will create a post that works as a main directory for this series of posts.
2D Objects
There are 2D objects that are created through other commands, but regardless when created they are in one of these categories. An example of this would be Polygons, Rectangles or Donuts which are actually Polylines.
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### AutoCAD: PEDIT Command Without Confirmation
A trick to speed up the PEDIT command
Català - Castellano - Deutsch
PEDIT is a command that allows us to edit Polylines. It also allows us to edit lines, arcs and splines, by asking us the question we many hate
"Object selected is not a polyline
Do you want to turn it into one? "
Although this might be useful to remember that the geometry selected is not a polyline, it turns to be an extra click that we would love to skip. We will skip it from now on thanks to Josh from Lazydrafter.
Josh posted recently about the PEDITACCEPT system variable. The variable is by default set to "0", and that is why AutoCAD asks us that annoying question. Well, set it up to "1", and the question is gone. Now it takes no extra step to edit lines, arcs. Spline still ask for the conversion factor, but still you'll be saving couple of clicks. All to work faster!
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How can you trick AutoCAD to make circular Wipeouts?
Català - Castellano
A problem often found is when we want to create a wipeout based on a polyine with curved segments or simply a circular wipeout. AutoCAD won't allow us to do it since wipeouts must be made from polylines containing only linear segments.
There is an easy trick to "cheat". If we want to create a circular wipeout, we wil create a 100 sides polygon instead of creating a circle. That way AutoCAD will accept the polyline and visually it will be like having a circular wipeout.
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### AutoCAD: Select all Connected Geometry.
Wonder how to select with a single click all connected geometry? AutoCAD can do it, you just need to know the right command.
Català - Castellano - Deutsch
Usually my day at work involves designing and drawing both in AutoCAD and SketchUp (and eventually in Rhino depending on the complexity of the geometry of the project). Switching back and forth from one software to another makes you miss on one program some of the functions that the other program has. This was the case of the "Select all Connected" option in SketchUp. It is so useful to select all connected geometry (it can be accessed through a right click menu or by triple clicking any line on the drawing). But can something similar be done in AutoCAD? The answer is yes.
There is an Express Tool for AutoCAD called FASTSEL (accessible also via the FS shortcut) that allows us to select all the geometry that is connected (touching) any element on our drawing. The elements that can be used using this commands are LINE, POLYLINE, LWPOLYLINE, CIRCLE, ARC, ATTDEF, TEXT, MTEXT, ELLIPSE, SPLINE or IMAGE objects. BLOCKS or XREFs containing these objects will also work.
AEC Objects from some of the vertical products such as Walls or AEC Polygons are not supported. In case you have a Block or XREF containing both supported and unsupported elements, the command will only use the supported elements.
The command can be accessed transparently while using other commands by typing 'FS while the command is running. Also, use the FSMODE System Variable to control if the FS command selects all connected geometry (FSMODE = ON) or only the geometry connected to the first object (FSMODE = OFF).
Note that this Express Tool Command is mostly intended for 2D since I tried to use it in 3D and lines that are actually not touching other geometry did get connected. I find it specially useful when I accidentally explode polylines, with this command you can select all the lines that where previously connected with a single click. Select faster than a private jet with this express tool!
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### SketchUp Plugins: Extrude with Rotation
Extrude Edges or faces with a rotation.
Català - Castellano - Deutsch
ExtrudeEdgesByLathe.rb by TIG allows us to Extrude Polylines or Faces following a circular path defined by an axis and a rotation angle. The Plugin is part of a suite of several Extrusion Plugins by the same Author that can be found here. The Plugin Allows you to do the following.
We can set all the properties of the extrusion on a dialog box that appears after we run the Plugin. These are the options for the previous example.
And if we select a face instead of a polyline, the plugin works too!!
Chek here how to Install SketchUp Plugins.
Show me more... | 4,654 | 20,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-23 | latest | en | 0.942457 |
https://help.quickqna.click/2022/10/08/what-is-an-adjacent-angle-easy/ | 1,679,739,356,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00743.warc.gz | 330,277,195 | 23,027 | # What is an adjacent angle easy?
## What is an adjacent angle easy?
Adjacent angles are Two angles that have a common side and a common vertex (corner point) but do not overlap in any way. When you break down the phrase adjacent angles, it becomes easy to visualise exactly what it is; they are two angles that are next to each other.
## What is an adjacent in math?
In geometry, Two angles are adjacent if they have a common side and a common vertex. In other words, adjacent angles are directly next to each other and do not overlap.
## Do triangles have adjacent angles?
In a right triangle, the hypotenuse is the longest side, an “opposite” side is the one across from a given angle, and An “adjacent” side is next to a given angle. We use special words to describe the sides of right triangles. The hypotenuse of a right triangle is always the side opposite the right angle.
## How do i find adjacent?
Adjacent, adjoining, contiguous, juxtaposed mean Being in close proximity. adjacent may or may not imply contact but always implies absence of anything of the same kind in between. a house with an adjacent garage adjoining definitely implies meeting and touching at some point or line.
## What are the adjacent sides?
If two sides share a common angle, then they are called adjacent sides.
: not adjacent: such as. a : Not having a common endpoint or border Nonadjacent buildings/rooms. b of two angles : not having the vertex and one side in common.
## What is an adjacent shape?
Adjacent means “next to”, so in a shape an adjacent sides would be Two sides that are next to each other.
Adjacent means Close to or near something. You may consider the people up and down your street to be neighbors, but your next-door neighbor is the person who lives in the house or apartment adjacent to yours. Adjacent can refer to two things that touch each other or have the same wall or border.
## What is an adjacent in a sentence?
If one thing is adjacent to another, the two things are next to each other. He sat in an adjacent room and waited. The schools were adjacent but there were separate doors.
## What does adjacent look like?
If one thing is adjacent to another, the two things are next to each other. He sat in an adjacent room and waited. The schools were adjacent but there were separate doors.
## What is a good sentence for adjacent?
Adjacent sentence example. A bird hide is situated adjacent to the lake. We moved to adjacent land with a higher elevation in case there was a tsunami after the earthquake. He encouraged adjacent landowners to install some signs. | 546 | 2,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.953059 |
https://shakyradunn.com/slide/lecture-11-context-free-grammar-nxzehv | 1,603,511,378,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00558.warc.gz | 539,962,503 | 9,092 | Lecture 11 Context-Free Grammar
Lecture 11 Context-Free Grammar Definition A context-free grammar (CFG) G is a quadruple (V, , R, S) where V: a set of non-terminal symbols : a set of terminals (V = ) R: a set of rules (R: V (V U )*) S: a start symbol. Example V = {q, f,}
= {0, 1} R = {q 11q, q 00f, f 11f, f } S=q (R= {q 11q | 00f, f 11f | }) How do we use rules? If A B, then xAy xAy derivates xBy. xBy and we say that
If s t, then we write s * t. A string x in * is generated by G=(V,,R,S) if S * x. L(G) = { x in * | S * x}. Example G = ({S}, {0,1}. {S 0S1 | }, S) in L(G) because S .
01 in L(G) because S 0S1 01. 0011 in L(G) because S 0S1 00S11 0011. n n n n 0 1 in L(G) because S * 0 1 . n n L(G) = {0 1 | n > 0} Context-Free Language (CFL) A language L is context-free if there exists a CFG G such that L = L(G).
Theorem For every regular set, there exists a CFG G such that L=L(G). Proof. Let L=L(M) for a DFA M=(Q, , , s, F). Construct a CFG G=(V, , R, S) as follows. V = Q, = , R = { q ap | (q,a) = p } U { f | f in F}, S = s. x1
s xn q1 S f=qn x q x x q x1xnf x1xn 1 1
1 2 2 x in L(M) There is a path associated with x from initial state to a final state. S
*x Therefore, L(M) = L(G). Corollary Every regular language is a CFL. The class of regular languages is a proper subclass of CFLs. CFL Regular Why, proper?
Regular Grammar Regular grammar is a CFG (V, , R, S) such that every rule is in form V *(V+) Example G = ({S, A}, {0, 1}, {S 1A, A 00}, S) Remark: Every regular language can be generated by a regular grammar. Theorem Every regular grammar generates a regular language. Proof.
Consider a regular grammar G=(V, , R, S). Construct a string-labeled digraph with vertex set V U {f} as follows: For each rule A xB, x in * and B in V, x draw an edge A B. x For each rule A x, x in *, draw an edge A f Example 0
S G = ({S,A}, {0,1}, {S0S | 10A, A00}, S) 10 A 00 f This string-labeled digraph with initial state S and a final state f is a state diagram of an NFA M.
S * x in * There is a path associated with x from S to f in M. Therefore, L(G) = L(M). Corollary A language L is regular if and only if L can be generated by a regular grammar.
Right-Linear and Left-Linear The regular grammar is also called a rightlinear grammar. A grammar G=(V, , R, S) is left-linear if every rule is in form V (V+)*. (e.g., ({S,A}, {0, 1}, {SA01, A10}, S) Remark: Every language generated from a left-linear grammar is regular. Why? Why? For left-linear grammar G = (V, , R, S), R
R construct G = (V, , R , S) where R R R = {AW | AW in R}. R R G is right-linear. Hence, L(G ) is regular. R R Therefore, L(G) = L(G ) is regular. Example 1 G = ({S,A}, {0, 1}, {SA01, A10}, S)
R G =({S,A}, {0, 1}, {S 10A, A 01}, S) R NFA accepts L(G ) S R L(G )={1001} L(G)={1001} 10 A
01 Example 2 L(G) = 0*1 R L(G ) = 10* NFA accepts 10* S R 1
A 0 G = ({S,A}, {0,1}, {S 1A, A 0A|}, S) G = ({S,A}, {0,1}, {S A1, A A0|}, S)
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https://physics.stackexchange.com/questions/216083/force-exerted-on-carseats | 1,725,742,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00307.warc.gz | 453,900,195 | 43,662 | # Force exerted on carseats?
Can anyone explain how force is exerted on car seats during a crash?
I am being told in car seat safety forums that $force=mass \times acceleration$. So if a $2000 \;\text{lb}$ vehicle bumps you at even $5 \;\text{mph}$, the force exerted on your car seat equals $10000 \;\text{lb}$, which is good reason to replace the seat.
My husband (an engineer), says it's not that simple and that the mass being used in the above equation should not be the mass of the vehicle but that of the child and car seat combined. So at $5 \;\text{mph}$ the force exerted on the car seat would be something much less, on the order $500 \;\text{lb}$, depending on total weight of the seat and child. He also said something about the energy transfer into the brakes, the bumper, and car seats - therefore not all of that force is being applied to the car seat.
What mass should be used to calculate the force exerted on a car seat during a collision, $m_\text{vehicle}$ or $m_\text{seat+child}$?
Can anyone shed some light on this or explain better what either my husband or the car seat safety forums are trying to say?
• What gets accelerated by the seat is the child, so it's the mass of the child that applies. The car accelerates the seat plus the child, of course, but it's more than likely that the frame of the car can take that. At 5mph even the car seat and the body of the child should be able to absorb the forces that accelerate the child without damage, but I wouldn't push it much beyond that... Commented Nov 2, 2015 at 1:57
• Sorry, none of this makes any difference. The physics of whether or not your car seat is damaged is not something that you're going to be able to determine with a simple application of Newton's 2nd law. The overall amount of force is only one variable that will affect how damaged or not your car seat is. You should follow the manufacturer's guidelines - they're the ones who have done crash tests with your make & model seat to determine when it's safe and when it isn't. Commented Nov 2, 2015 at 2:17
• @Bronius: She asked if we can explain to her what forces are being exerted on the car seat. Yes, we can explain that. Whether the car seat is safe or not depends on a lot more than on the guidelines. If the manufacturer is not honest (and that wouldn't be the first time child car seats are under fire), then those guidelines are meaningless. Irrespective of that the seat only has to hold the child, not the car. Commented Nov 2, 2015 at 2:41
• I really hope that "I am being told in car seat safety forums that force=massXacceleration, so if a 2000lb vehicle bumps you at even 5mph you are looking at 10,000lb of force exerted on your car seat" represents some loss of information along a whisper chain, rather than something that someone serious is actually putting out there. Sheesh, the units don't match up. But, no matter. Your husband is right that this math is badly bogus. Commented Nov 2, 2015 at 3:03
The short answer- The force exerted on a car seat depends on how much your car crushes to absorb the impact. The more your car deforms, the lower the force.
When you are thinking about this type of problem, remember that $$5 \text{mph}$$ is a speed (velocity), not an acceleration. Acceleration describes how quickly an object changes speed ($$a = \frac {v_{f}-v_{i}}{t}$$). As you correctly noted, $$F = ma$$, however, forces are caused by accelerations, not speeds!
**To calculate Force in [$$\text{lbf}$$]: Acceleration has units [$$\frac {ft}{s^2}$$], with velocity in [$$\frac {ft}{s}$$] and time in [$$\text{s}$$]; Mass has units [$$\text{slug} = \frac {\text{lbm}}{\text{g}}$$] where $$g=32.2 \;[\frac {ft}{s^2}]$$.
## Calculate Acceleration:
$$a = \frac {0 \;\text{[ft/s]} - 7.33 \;\text{[ft/s]}}{t}$$
This is where your question becomes a matter of guess work, because we dont know how long the crash takes- it depends on how much your car deforms during a crash. As a ballpark guess, lets assume that the crash takes $$t = 0.25 \;\text{seconds}$$.
$$\therefore a = \frac {0 \;\text{[ft/s]} - 7.33 \;\text{[ft/s]}}{\approx 0.25 \;[\text{s}]} \approx -29 \;[\frac {ft}{s^2}]$$
## Calculate Force:
The mass of the driver and the mass of the car seat are accelerated with the car body. Assuming the driver weighs $$150 \;\text{[lbm]}$$ and the car seat weighs $$50 \;\text{[lbm]}$$, their masses are $$4.6 \;\text{[slug]}$$ and $$1.5 \;\text{[slug]}$$. The force on the car seat is calculated by:
$$F_{seat} = (m_{driver}+m_{seat})a \qquad \Rightarrow \qquad \therefore F_{seat} \approx 182 \;\text{[lbf]}$$
Similarly, to determine force exerted on the driver:
$$F_{driver} = (m_{driver})a \qquad \Rightarrow \qquad \therefore F_{driver} \approx 137 \;\text{[lbf]}$$
In reality, the seat cushion compresses, increasing the impact time (felt by the driver). This decreases the drivers acceleration and reduces the force felt by the driver.
While braking does exert force on the seat and driver, it is calculated separately because it occurs before the collision. For example, you might calculate the force as the car decelerates from $$30 \;\text{[mph]}$$ to $$5 \;\text{[mph]}$$ (in some time interval) before the collision. However, this deceleration is much less than the deceleration/force involved in the collision and can be safely ignored. After all, you dont need to replace a car seat after braking!
The car seat is easily strong enough to handle these forces. However, this is a concern in high speed crashes, where forces are large enough to create stresses in the seat frame that exceed the yield limit, the seat should be replaced.
Can anyone explain how force is exerted on car seats during a crash?
I am being told in car seat safety forums that force=mass×acceleration. So if a 2000lb vehicle bumps you at even 5mph, the force exerted on your car seat equals 10000lb, which is good reason to replace the seat.
The formula is correct. $F=ma$. This is the total force $F$ it takes to accelerate a mass $m$ with the acceleration $a$.
Now, the point is - maybe surprisingly - not the speed of the car! The point is not how fast you are going (or how fast the bumping car is going).
The point is how fast you are decelerating. In other words, how fast you are speeding down! If you drive the car strait into a stone wall, it is a lot more damaging for the child on the back seat than if you drive into a large soft vertical foam matress.
Why? Because the foam matress makes the car (and you) slow down slowlier! It simply takes longer for the speed to go from 5mhp to 0mhp, which means that the deceleration (negative acceleration) $a$ is smaller and therefore the force $F$ is smaller as well.
My husband (an engineer), says it's not that simple and that the mass being used in the above equation should not be the mass of the vehicle but that of the child and car seat combined.
Yes, because $F$ in the formula above is the total force needed to accelerate the mass $m$. And what you are interested in is the force on the child, so the child would be your $m$. The force on the car might be very different and much larger than the force on the child.
I am not quite sure from the question, exactly what object you wish to look at, so I've assumed we are looking at the impact on the child. If you wish to look at the seat alone, we should define what the seat exactly is (or is a childs seat?)
He also said something about the energy transfer into the brakes, the bumper, and car seats - therefore not all of that force is being applied to the car seat.
Yes, here comes a point of car evolution and improvements. This is the reason that we cannot just calculate the impact force on people in a car - it is way too complicated with way too many unknown factors - so crash tests are what is being used.
The point is again that the foam wall is making the impact softer because is gives the car more time to slow down from the 5mhp to 0mhp. We want to avoid a sudden stop, which would cause a large deceleration.
So, if we do hit a stone wall, we must instead imitate the foam somewhere else. For example in the structure of the car before the impact reaches the child on the backseat.
In accidents you very often see extremely damaged and crumbled cars - it looks very violent even when the people inside weren't that injured. That is because many bigger cars are made so their frontend will crumble a lot. This is like a foam sponge that absorbs the energy because it slows the car down more slowly than if the car was totally rigid.
This foam sponge effect is the case in many cases throughout the car chasis - everywhere where the car is not perfectly rigid, energy is absorbed for the car deformation giving the rest of the car and passengers more time to slow down.
The airbag on the frontseats works like a pillow that your head can hit. Then, when the head flies forward, because the car is being stopped suddenly, it is slowly slowed down by this "pillow" and decelerated less just over a longer period of time. If the airbag wasn't there, the head might continue with the same speed until it hits the dashboard and there it would be stopped suddenly and experience a large deceleration. Or, the neck would have to stop the head and maybe the bones and the body is not strong enough the cause this deceleration of the head, so something will break.
I am not a physicist
The force applied to the seat will be the net (difference) of all the following:
1. Force absorbed by the car body.
2. Force absorbed by the brakes/friction.
3. Force dissipated through the frame (on to which the car seat is attached).
4. Force absorbed by the car seat itself.
Finally, whatever may be left will be transferred to the occupant.
The only way you can affect a 10,000 lbs of force onto an object is if you discount all the other forces in effect and exert a much larger force in terms of the momentum of the object.
Keep in mind that vehicles are designed with a certain amount of safety and that there are many mechanisms in place (in terms of the structural design) absorb any force and dissipate it away from the occupants. These design mechanisms include the placements and size/shape of the seats.
At 5 miles per hour - which is nearly idle speed any bump to a vehicle will result in at most a small dent to the part of the body hit by the vehicle; the force exerted would not be sufficient to structurally compromise the car seat such that it would need replacement.
Source: Sadly, experience with car accidents first hand.
Your husband is right. The mass to use in that equation is the mass of the car seat and passenger combined. | 2,613 | 10,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-38 | latest | en | 0.969176 |
https://www.mikescher.de/blog/25/Advent_of_Code_2017/day-05 | 1,639,019,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363659.21/warc/CC-MAIN-20211209030858-20211209060858-00631.warc.gz | 963,985,713 | 5,468 | 2017-12-05
Day 05: A Maze of Twisty Trampolines, All Alike
Description:
--- Day 5: A Maze of Twisty Trampolines, All Alike ---
An urgent interrupt arrives from the CPU: it's trapped in a maze of jump instructions, and it would like assistance from any programs with spare cycles to help find the exit.
The message includes a list of the offsets for each jump. Jumps are relative: -1 moves to the previous instruction, and 2 skips the next one. Start at the first instruction in the list. The goal is to follow the jumps until one leads outside the list.
In addition, these instructions are a little strange; after each jump, the offset of that instruction increases by 1. So, if you come across an offset of 3, you would move three instructions forward, but change it to a 4 for the next time it is encountered.
For example, consider the following list of jump offsets:
0
3
0
1
-3
Positive jumps ("forward") move downward; negative jumps move upward. For legibility in this example, these offset values will be written all on one line, with the current instruction marked in parentheses. The following steps would be taken before an exit is found:
(0) 3 0 1 -3 - before we have taken any steps.
(1) 3 0 1 -3 - jump with offset 0 (that is, don't jump at all). Fortunately, the instruction is then incremented to 1.
2 (3) 0 1 -3 - step forward because of the instruction we just modified. The first instruction is incremented again, now to 2.
2 4 0 1 (-3) - jump all the way to the end; leave a 4 behind.
2 (4) 0 1 -2 - go back to where we just were; increment -3 to -2.
2 5 0 1 -2 - jump 4 steps forward, escaping the maze.
In this example, the exit is reached in 5 steps.
How many steps does it take to reach the exit?
--- Part Two ---
Now, the jumps are even stranger: after each jump, if the offset was three or more, instead decrease it by 1. Otherwise, increase it by 1 as before.
Using this rule with the above example, the process now takes 10 steps, and the offset values after finding the exit are left as 2 3 2 3 -1.
How many steps does it now take to reach the exit?
Input:
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Part 1:
``````#!/usr/bin/env python3
import aoc
import itertools
instructions = list(map(lambda x: int(x), rawinput.splitlines()))
ilen = len(instructions)
pos = 0
for i in itertools.count(1):
v = instructions[pos]
instructions[pos] += 1
pos += v
if pos < 0 or pos >= ilen:
print(i)
exit()
``````
Result: 388611
Part 2:
``````#!/usr/bin/env python3
import aoc
import itertools
instructions = list(map(lambda x: int(x), rawinput.splitlines()))
ilen = len(instructions)
pos = 0
for i in itertools.count(1):
v = instructions[pos]
instructions[pos] += -1 if v >= 3 else 1
pos += v
if pos < 0 or pos >= ilen:
print(i)
exit()
``````
Result: 27763113
made with vanilla PHP and MySQL, no frameworks, no bootstrap, no unnecessary* javascript | 4,006 | 7,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.932152 |
https://edurev.in/course/quiz/attempt/-1_Test-Simple-And-Compound-Interest-Including-Annuit/72c6ac25-bd93-4f1d-bc93-017c2cb9d438 | 1,656,912,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00482.warc.gz | 276,705,305 | 41,068 | # Test: Simple And Compound Interest Including Annuity - 2
## 40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Simple And Compound Interest Including Annuity - 2
Description
Attempt Test: Simple And Compound Interest Including Annuity - 2 | 40 questions in 40 minutes | Mock test for CA Foundation preparation | Free important questions MCQ to study Quantitative Aptitude for CA CPT for CA Foundation Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### A sum of money compounded annually becomes Rs1,140 in two years and Rs.1,710 in three years. Find the rate of interest per annum.
Solution:
QUESTION: 4
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:
Solution:
QUESTION: 5
The S.I. on a sum of money is $\frac{4}{9}$ of the principle and the no. of year is equal to the rate of interest per annum. Find the rate of interest per annum?
Solution:
QUESTION: 6
How much investment is required to yield an Annual income of Rs.420 at 7% p.a. Simple interest.
Solution:
QUESTION: 7
The difference between compound and simple interest on a certain sum of money for 2 years at 4% is Rs.1. The sum (in Rs) is:
Solution:
QUESTION: 8
The partners A and B together lent Rs.3,903 at 4% per annum interest compound annually. After a span of 3 years, A gets the same amount as B gets after 9 years. The share of A in the sum of Rs.3,903 would have been:
Solution:
QUESTION: 9
If the Simple Interest on Rs.1,400 for 3 years is less than the simple interest on Rs.1,800 for the same period by Rs.80, then the rate of interest is
Solution:
QUESTION: 10
In what time will a sum of money double its y at 6.25% p.a. simple interest?
Solution:
QUESTION: 11
Simple interest on Rs.2,000 for 5 months at 16% p.a. is__________.
Solution:
QUESTION: 12
What principle will amount to Rs.370 in 6 years at 8% p.a. at simple interest?
Solution:
QUESTION: 13
On what sum difference between compound interest and simple interest for two years at 7% p.a. interest is Rs.29.4
Solution:
QUESTION: 14
Mr. X invests Rs.90,500 in post office at 7.5% p.a. simple interest. While calculating the rate was wrongly taken as 5.7% p.a.The difference in amounts at maturity is Rs.9,774. Find the period for which the sum was invested
Solution:
QUESTION: 15
A sinking fund is created for redeming debentures worth Rs. 5 lakhs at the end of 25 years. How much provision needs to be made out of profits each year provided sinking fund investments can earn interest at 4% p.a.?
Solution:
QUESTION: 16
A machine costs Rs. 520000 with an estimated life of 25 years. A sinking fund is created to replace it by a new model at 25% higher cost after 25 years with a scrap value realization of Rs. 25000. What amount should be set aside every year if the sinking fund investments accumulate at 3.5% compound interest p.a.?
Solution:
QUESTION: 17
Johnson left Rs. 100000 with the direction that it should be divided in such a way that his minor sons Tom, Dick and Harry aged 9, 12 and 15 years should each receive equally after attaining the age 25 years. The rate of interest being 3.5%, how much each son receive after getting 25 yeas old?
Solution:
QUESTION: 18
A machine worth Rs. 490740 is depreciated at 15% on its opening value each year. When its value would reduce to Rs. 200000?
Solution:
QUESTION: 19
A machine depreciates at 10% of its value at the beginning of a year. The cost and scrap value realized at the time of sale being Rs. 23240 and Rs. 9000 respectively. For how many years the machine was put to use?
Solution:
QUESTION: 20
In how many years will a sum of money double at 5% p.a. compound interest?
Solution:
QUESTION: 21
In how many years a sum of money trebles at 5% p.a. compound interest payable on half yearly basis?
Solution:
QUESTION: 22
A machine worth Rs. 490740 is depreciated at 15% of its opening value each year. When its value would reduce by 90?
Solution:
QUESTION: 23
If the difference of S.I. and C.I. is Rs. 72 at 12% for 2 years. Calculate the amount.
Solution:
QUESTION: 24
If a simple interest on a sum of money at 6% p.a.for 7 years is equal to twice of simple interest on another sum for 9 years at 5% p.a.. The ratio will be:
Solution:
QUESTION: 25
Appu retires at 60 years receiving a pension of 14400 a year paid in half-yearly installments for rest of his life after reckoning his life expectation to be 13 years and that interest at 4% p.a. is payable half-yearly. What single sum is equivalent to his pension?
Solution:
QUESTION: 26
Raja aged 40 wishes his wife Rani to have Rs. 40 lakhs at his death. If his expectation of life is another 30 years and he starts making equal annual investments commencing now at 3% compound interest p.a. how much should he invest annually?
Solution:
QUESTION: 27
The present value of Rs. 10000 due in 2 years at 5% p.a. compound interest when the interest is paid on yearly basis is Rs. ________.
Solution:
QUESTION: 28
Alibaba borrows R.s 6 lakhs Housing Loan at 6% repayable in 20 annual installments commencing at the end of the first year. How much annual payment is necessary.
Solution:
QUESTION: 29
The present value of Rs. 10000 due in 2 years at 5% p.a. compound interest when the interest is paid on half-yearly basis is Rs. __________.
Solution:
QUESTION: 30
If A=Rs. 10000, n=18 yrs., R=4% p.a. C.I, P will be
Solution:
QUESTION: 31
If P = 1000, n = 4yrs., R = 5% p.a. then C.I will be
Solution:
QUESTION: 32
The difference between compound and simple interest at 5% per annum for 4 years. On Rs. 20000 is Rs. _________
Solution:
QUESTION: 33
The time in which a sum of money will be double at 5% p.a. C.I is
Solution:
QUESTION: 34
The time by which a sum of money would treble it self at 8% p.a. C.I is
Solution:
QUESTION: 35
A person bought a house paying Rs. 20000 cash down and Rs. 4000 at the end of each year for 25 yrs. At 5% p.a. C.I. The cash down price is
Solution:
QUESTION: 36
The present value of an annuity of Rs. 80 a years for 20 years at 5% p.a. is
Solution:
QUESTION: 37
The compound interest on half-yearly rests on Rs. 10000 the rate for the first and second years being 6% and for the third years 9% p.a. is Rs. ___________.
Solution:
QUESTION: 38
A person desires to create a fund to be invested at 10% CI per annum to provide for a prize of Rs. 300 every year. Using V=a/I find V and V will be
Solution:
QUESTION: 39
Mr. X borrowed R.s 5120 at 12 ½ % p.a C.I.At the end of 3 yrs, the money was repaid along with the interest accrued. The ammount of interest paid by him is
Solution:
QUESTION: 40
A = Rs. 5200, R = 5% p.a., T = 6 years, P will be
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | 1,882 | 6,843 | {"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-2022-27 | latest | en | 0.921494 |
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## Total circles area
For a demonstration of the algorithm see http://stackoverflow.com/a/1667789/10562
The algorithm itself I described below the Haskell code on the task page, which would probably be easier to follow if you sketch a few circles on a piece of paper and try to verify it step by step. I understand that the Haskell code is kinda hard to follow, despite it having been cleaned up by someone else already, but believe me that it would be even harder to read had I written it in C++, because of, well, me and C++. But once you understand how the algorithm works, it's really nothing more than elementary geometry and bookkeeping, tedious as it may be, rocket science it is not.
That being said, I don't really recommend the algorithm if you need it for serious work. Depending on the circles you have, the algorithm can be error-prone if intersection points are very close to each other. The algorithm depends on the correct construction of an outline for the union of overlapping circles, which in turn depends on correctly intersecting and sorting arc segments, which can get messy when the intersections are dense. The scanline methods on the same task page are much more stable, not necessarily slower or less accurate, and a lot cleaner, so consider using that if at all possible. --Ledrug (talk) 21:00, 5 March 2015 (UTC)
Fortunately, I am deal with numbers of circles typically in the single digits, sometimes in the tens of circles. Your geometric solution certainly seems more satisfying than the scanline methods. I do think you are right that the scanline method would be easier, but I also need to adapt the code to measure perimeter. Your geometric solution would seem to give that inherently, though getting it from the scanline method shouldnt be much more than counting transition from a box in the total area to one that isnt (and vice-versa), right?
I appreciate your fast response to my question. Thank you for your help!
To calculate the perimeter, your idea of tracking transitions between inside and outside should work, provided that you take into account the slopes of the circles at the scanline intersections (segment length ~ scanline spacing / sine of tangential line's angle). I don't know how well it converges, because circles can be a little funny at top and bottom (slope near zero, so 1/sin can get really big), but I think it's worth a try. --Ledrug (talk) 19:41, 6 March 2015 (UTC) | 534 | 2,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.96332 |
https://www.pcorner.com/list/EDUCAT/CNC.ZIP/INFO/ | 1,716,513,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00269.warc.gz | 820,687,768 | 13,570 | File CNC.ZIP from The Programmer’s Corner in
Category Science and Education
Complex Number Calculator Ver. 1.0: evaluates expressions containing real or complex numbers.
File Name File Size Zip Size Zip Type
CNC.DOC 10551 4231 deflated
CNC.EXE 81060 36246 deflated
## Contents of the CNC.DOC file
Complex Number Calculator version 1.0 by Robert G. Kaires on 3/24/90
SHAREWARE
If you find this program useful, please register. Registration is only \$15.
Your payment is appreciated and necessary for the continued support and
upgrade of this product. If you register you will be put on the mailing list
Robert G. Kaires
609 Glover Dr.
Runnemede, NJ 08078
INCLUDE A SELF-ADDRESSED, STAMPED ENVELOPE IF YOU WANT A REPLY AND ARE NOT A
REGISTERED USER.
Anyone wishing a disk mailed to them with the latest version of this program
send \$5 to the above address requesting a copy of CNC (you do not have to
be a registered user).
Other shareware products from R.G.Kaires
Function Plotter 4.0, (plots functions and data) registration \$20, disk \$5
PRELIMINARY
The symbol "i" is used to denote the unit imaginary constant and is equal to
the square root of -1. The symbol "z" denotes a complex variable and
z(t) = x(t) + i * y(t). As you can see x(t) is the real part, and y(t) is the
complex part of the complex number, respectively. The symbol "t" is used to
denote a real parameter and will be useful in the discussion of numerical
integration.
INTRODUCTION
Complex Number Calculator (CNC) evaluates expressions containing real or
complex numbers. Examples of valid expressions are:
1.23e10/.45e-27 ............. numbers can be entered in exponential format
1 + sin(pi/3) ............... same as 1 + sin(180/3) if "degrees" are set
exp( 2 - (1/2) * exp(3) ) ... functions can, of course, be nested
.5 * ln( (1+z)/(1-z) ) ...... you will be prompted for z
(i+1)/(i-1) ................. "i" is the unit imaginary constant
(z+1)/(z-1) ................. enter any real or complex no. at the prompt
(1+i)^(2+i) ................. exponents can be complex
int( 1 / (1+t^2), 4,0,1) .... numerically integrate f(t) = 1 / ( 1+t^2 )
from 0 to 1 using an approx order of 4
int(cos(t)+j*sin(t),4,0,pi).. numerically integrate cos(t) + i * sin(t)
M_m(2.0) .................... convert 2 miles to meters
kW_hp(4.5) .................. convert 4.5 kilo-Watts to horse power
See the end of this document for a full list of functions, constants, and
conversions available in CNC. Be aware that CNC is case sensitive.
CNC "remembers" previously entered expressions (up to 20), they can be
accessed by using the up and down arrow keys. Expressions can be one full line
in length. When the line is full, CNC will not allow any addition characters
to be inserted until some are deleted. All calculations are performed in
double precision with about 15 significant digits of precision. The number of
displayed digits (default is 6) can be changed using the "dd" command.
When responding to prompts in CNC, the default response is given in square
brackets. Simply hit "Enter" (CR) to respond with the default.
At the top of the screen is a message line which lists some information on
valid editing keys and other commands. At the bottom of the screen is the
status line.
OPERATORS, FUNCTIONS AND PRECEDENCE
The valid operators are + - * / ^. The operator ^ (exponentiation) has the
highest precedence, followed by the operators * and / which have equal
precedence, followed by + and - which have equal precedence. Operators with
equal precedence are evaluated left to right. Functions (and conversions) have
higher precedence than operators. Therefore sin(pi/6)^2 is evaluated as
(sin(pi/6) )^2 since sin(pi/6) is evaluated before the exponentiation is
performed. Consider the following two expressions:
sin(5) ^ sqrt(2) and
sin(5) ^ 2^.5
The first yields: (-.958924) ^ 1.41421 ( = -0.250921 - i * 0.90839 )
The second gives: ( (-.958924) ^ 2 ) ^ .5 ( = .958924 )
In the first expression sqrt (which IS a function) and sin are evaluated
first, in the second expression sin is evaluated, followed by the operators ^
and ^, which are evaluated left to right. Enough said. As when using any
calculator, when in doubt, use parenthesis. One final note: do not use
operators side by side with no intervening numbers or parenthesis such as
2*-2, you will be surprised at what you get. Instead use 2*(-2)
PRECISION, ROUND OFF ERROR, MIN AND MAX NUMBERS
All calculations are done in double precision and have about 15 digits of
accuracy. Problems that can occur from this finite accuracy are best
illustrated by the following example session:
>sin(z)^2 + cos(z)^2 ...... (user enters expres at > prompt)
z=?>1 ..................... (user enters expres at z=?> prompt)
1 ......................... (result)
z=?>10
1
z=?>100
1
z=?>i ..................... (of course you may use complex
1 numbers as well)
z=?>10*i
1
z=?>100*i ................. (WHAT HAPPENED HERE!)
6.90175e+71
This last result is not a bug in CNC! It results because of the finite
precision of this (and any) calculator. The functions sin(z)^2 and cos(z)^2
with purely imaginary arguments are -sinh(y)^2 and cosh(y)^2, respectively.
Now the answer is obvious, for large y we are subtracting two very large
numbers. The answer, of course, becomes very inaccurate.
Floating point numbers can range from 1e-308 to 1e308.
FUNCTIONS
sqrt() square root function
exp() inverse natural log function
real() real part of a complex number
imag() imaginary part of a complex number
mag() magnitude of a complex number
ang() angle of a complex number
ln() natural log
log() log base 10
sin(), cos(), tan() trigonometric functions
asin(), acos(), atan() inverse trigonometric functions
sinh(), cosh(), tanh() hyperbolic functions
asinh(), acosh(), atanh() inverse hyperbolic functions
INTEGRATION FUNCTION
int( f(t),N,l,u ) numerical integration of a function of a single real
- f(t) can evaluate to a complex number
- N is the order of the approximation ( 2,4,8,10,12,16,32 )
- l,u lower and upper integration limits (can be a function of T)
The integration method used is Gaussian Quadrature. The numerical procedure
does not put integration points at the integration limits. Therefore the
integration limit points can be singular. Of course, one must be concerned
about convergence. Some singularities are integrable, some are not. Consider
the following two sessions. In both the function is singular at t=0, but the
first function is integrable and the second is not.
>int(ln(t),8,0,1) ......... The ln function is integrable
-0.991239 from 0 to 1
>int(ln(t),16,0,1)
-0.997679
>int(ln(t),32,0,1)
-0.999402
>int(1/t,8,0,1) ............The 1/t function in not integrable
5.43571 from 0 to 1
>int(1/t,16,0,1)
6.76146
>int(1/t,32,0,1)
8.11699
We notice that the first integral converges but that the second does not.
INTEGRATION IN THE COMPLEX PLANE
Integrating a function of a complex variable in the complex plane is possible
with some work. The function must be parameterized as a function of a real
variable as follows: f(z) = f(x+i*y) = f(x(t) + i*y(t)), where t is real.
We then note that dz = ( x'(t) + i*y'(t) ) dt, where ' denote differentiation
with respect to t. Therefore we have
f(z) dz = f( x(t) + i*y(t) ) * ( x'(t) + i*y'(t) ) dt.
Example: We want to integrate 1/z along the unit circle centered at the
origin. The complex variable z can be described by z = cos(t) + i * sin(t).
Therefore we have
1/z dz = ( 1 / (cos(t) + i * sin(t) ) * ( -sin(t) + i*cos(t) ) dt.
The following is from an actual session:
>int( ( 1 / (cos(t) + i * sin(t) ) ) * ( -sin(t) + i*cos(t) ),4,0,2*pi)
3.73223e-18 + i * 6.28319
With the exception of some small round-off error, we obtain approximately
2*pi*i as expected.
BUILT IN CONSTANTS
mathematical constants: pi (3.14159...) gamma (.5772...)
physical constants:
c0 speed of light in vacuum (m/s) N Avagadro's constant (1/mole)
h Planck's constant (J-s) R Universal gas constant (J/K-mole)
qe electron charge (C) k Boltzmann's constant (J/K)
eps permittivity constant (F/m) Me electron mass (kg)
mu permeability constant (H/m) Mn neutron mass (kg)
G Grav constant (N-m^2/kg^2) Mp proton mass (kg)
g standard gravity (m/s^2)
CONVERSION FUNCTIONS (must use real arguments)
M_m(), m_M() miles to meters, meters to miles
M_yd(), yd_M() miles to yards, yards to miles
M_ft(), ft_M() miles to feet, ....
m_in(), in_ft() meters to inches, ....
cm_in(), in_cm() cm to inches, ....
F_C(), C_F() degrees Fahrenheit to degrees Celsius, ....
K_C(), C_K() degrees Kelvin to degrees Celsius, ....
K_F(), F_K() degrees Kelvin to degrees Fahrenheit, ....
lb_kg(), kg_lb() pounds to kilograms, .... (under standard gravity)
lb_N(), N_lb() pounds to Newtons, ....
Btu_J(), J_Btu() Btu's to Joules, ....
kWh_J(), J_kWh() kilo-Watt hours to Joules, ....
kW_hp(), hp_kW() kilo-Watt hours to horse-power, ....
FILES
You can write data to a file only from the z=?> (or T=?>) prompt. Enter a
function of z, at the z=?> prompt press f , you will be prompted for a
file name and whether you want to write the domain and range or just the
range. If you request the domain and range to be written, the real part of z
will be written to the first column, the imaginary part of z to the second
column, the real part of f(z) to the third column and finally the imaginary
part of f(z) to the fourth column. If you request the range (only) to be
written, the real and imaginary parts of f(z) will be written to the first and
second columns, respectively. The file will be close when you press "q" from
the z=?> (or T=?>) prompt. You can plot your data with any standard plotting
package. A shareware plotting package is available from this author if you are
in need (see beginning of this document).
December 15, 2017 | 2,705 | 9,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.768324 |
http://www.ck12.org/book/Basic-Physics-Second-Edition/r6/ | 1,411,468,373,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657138501.67/warc/CC-MAIN-20140914011218-00134-ip-10-234-18-248.ec2.internal.warc.gz | 403,401,184 | 25,611 | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Basic Physics - Second Edition | CK-12 Foundation
You are reading an older version of FlexBook: Basic Physics - Second Edition Go to the latest version.
# Basic Physics - Second Edition
Created by: CK-12
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## Table of Contents
• 1. Basic Physics SE-Units
This chapter covers the basic units used in physics, guidelines for using units, and their importance within physics.
• 2. Basic Physics SE-SHM Wave
This chapter covers objects in harmonic motion, which are defined as objects that return to the same position after a fixed period of time. Objects in harmonic motion have the ability to transfer some of their energy over large distances.
• 3. Basic Physics SE-Light Nature
This chapter covers the nature of light, polarization, and color.
• 4. Basic Physics SE-Optics
This chapter covers reflection, refraction, mirrors and lenses, and these concepts apply to Fermat's principle.
• 5. Basic Physics SE-Motion
This chapter covers the speed and velocity of objects in space. Velocity is speed with a direction, making it a vector quantity. If an object’s velocity changes with time, the object is said to be accelerating.
• 6. Basic Physics SE-Projectile Motion
This chapter covers parabolic motion of a thrown object, known as projectile motion. Motion in one direction is unrelated to motion in other perpendicular directions.
• 7. Basic Physics SE-Force
This chapter covers the basics of force as understood through Newton's 1st, 2nd, and 3rd Law of Motion.
• 8. Basic Physics SE-Momentum
This chapter covers momentum which is the product of mass and velocity. Momentum can be exchanged from one object to another through a collision.
• 9. Basic Physics SE-Energy
This chapter explains energy as a measure of the amount of or potential for movement in something. The total amount of energy in the universe is always the same.
• 10. Basic Physics SE-Heat
This chapter explains heat as a form of energy transfer. Materials do not contain heat. They contain internal energy that can be transferred (i.e. heat) from one body to another.
• 11. Basic Physics SE-Circular Motion
This chapter covers the basics of circular motion. In the absence of a net force, objects move in a straight line. Forces which cause objects to turn around continuously in a circle are known as centripetal forces.
• 12. Basic Physics SE-Gravity
This chapter covers gravity and how it relates to laws of physics governing motion, spcifically the Universal Law of Gravitation.
• 13. Basic Physics SE-Electrostatics
This chapter explores the fourth of the 5 conservation laws in physics, the conservation of charge. There are two charges, + and -, and the symmetry of the electric charge indicates that the total charge in the universe remains the same.
• 14. Basic Physics SE-DC Electric Circuits
This chapter explores electric currents through voltage and resistance as they apply to Ohm's Law, the main equation for electric circuits.
Feb 23, 2012
## Last Modified:
Sep 08, 2014
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ShareThis Copy and Paste | 731 | 3,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | longest | en | 0.87925 |
https://mathexamination.com/lab/mil%C3%BC.php | 1,606,698,583,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00070.warc.gz | 385,489,669 | 8,572 | ## Do My Milü Lab
As discussed over, I used to write a simple and also straightforward math lab with only Milü Nonetheless, the simpler you make your lab, the much easier it becomes to obtain stuck at the end of it, after that at the beginning. This can be very irritating, and all this can occur to you due to the fact that you are using Milü and/or Modular Equations inaccurately.
With Modular Formulas, you are currently making use of the incorrect formula when you get stuck at the start, if not, then you are likely in a stumbling block, as well as there is no feasible way out. This will just become worse as the issue ends up being a lot more complex, yet after that there is the concern of how to wage the trouble. There is no other way to correctly go about solving this sort of math trouble without having the ability to quickly see what is taking place.
It is clear that Milü and Modular Formulas are difficult to learn, as well as it does take technique to establish your very own feeling of instinct. Yet when you want to fix a math trouble, you have to make use of a device, and also the devices for finding out are utilized when you are stuck, as well as they are not used when you make the wrong action. This is where lab Help Solution is available in.
For example, what is wrong with the inquiry is incorrect concepts, such as obtaining a partial value when you do not have sufficient working parts to complete the whole work. There is a great reason that this was wrong, and also it is a matter of logic, not instinct. Reasoning allows you to comply with a detailed procedure that makes good sense, and when you make an incorrect step, you are generally compelled to either attempt to go forward and correct the blunder, or try to go backward and do a backwards step.
Another instance is when the trainee does not comprehend an action of a procedure. These are both rational failures, and there is no chance around them. Also when you are stuck in a location that does not permit you to make any sort of move, such as a triangle, it is still important to understand why you are stuck, to ensure that you can make a far better action and go from the step you are stuck at to the next location.
With this in mind, the best means to fix a stuck scenario is to simply take the progression, as opposed to trying to go backward. Both procedures are various in their method, yet they have some basic similarities. However, when they are tried together, you can quickly inform which one is much better at fixing the issue, and also you can additionally tell which one is extra effective.
Allow's speak about the initial example, which relates to the Milü mathematics lab. This is not too complicated, so let's initial discuss how to start. Take the adhering to process of attaching a part to a panel to be made use of as a body. This would call for three measurements, and also would be something you would certainly require to connect as part of the panel.
Now, you would certainly have an additional dimension, however that doesn't imply that you can simply keep that dimension and go from there. When you made your primary step, you can easily forget the measurement, and afterwards you would certainly need to go back as well as backtrack your actions.
Nevertheless, rather than remembering the additional dimension, you can utilize what is called a "mental shortcut" to help you keep in mind that additional dimension. As you make your first step, imagine on your own taking the dimension and attaching it to the part you wish to attach to, and afterwards see exactly how that makes you feel when you duplicate the process.
Visualisation is a really powerful technique, and is something that you need to not miss over. Picture what it would certainly feel like to really connect the part and also have the ability to go from there, without the measurement.
Now, allow's take a look at the second example. Allow's take the same procedure as previously, but now the student has to remember that they are mosting likely to move back one step. If you tell them that they need to return one step, however after that you get rid of the concept of needing to return one action, then they will not recognize exactly how to wage the trouble, they won't recognize where to try to find that step, and the procedure will certainly be a mess.
Rather, utilize a mental faster way like the psychological diagram to mentally reveal them that they are mosting likely to move back one step. and place them in a position where they can progress from there. without needing to think of the missing an action.
## Pay Me To Do Your Milü Lab
" Milü - Need Help with a Math lab?" Sadly, many trainees have actually had a trouble understanding the principles of straight Milü. Fortunately, there is a brand-new style for straight Milü that can be used to show direct Milü to pupils that have problem with this principle. Students can use the lab Help Service to help them learn new strategies in linear Milü without facing a mountain of troubles and also without needing to take a test on their concepts.
The lab Help Service was created in order to help battling students as they relocate from university and senior high school to the college as well as job market. Many trainees are not able to handle the stress of the understanding procedure as well as can have really little success in understanding the ideas of direct Milü.
The lab Help Service was established by the Educational Testing Service, who provides a variety of various online examinations that students can take and also practice. The Examination Aid Service has aided numerous students enhance their ratings as well as can help you improve your ratings also. As pupils move from university and high school to the university and also job market, the TTS will assist make your pupils' transition much easier.
There are a couple of different manner ins which you can capitalize on the lab Help Solution. The major manner in which trainees use the lab Aid Service is via the Answer Managers, which can aid pupils learn strategies in linear Milü, which they can utilize to help them succeed in their courses.
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The lab Assist Service functions similarly that a teacher performs in regards to assisting pupils grasp the concepts of straight Milü. By giving your students with the devices that they need to discover the important principles of straight Milü, you can make your pupils more successful throughout their studies. Actually, the lab Aid Solution is so reliable that many pupils have actually switched over from traditional mathematics class to the lab Assist Solution.
The Job Manager is designed to help trainees manage their homework. The Job Supervisor can be set up to arrange how much time the trainee has available to finish their assigned homework. You can additionally establish a custom period, which is a wonderful function for pupils who have a hectic routine or an extremely busy high school. This feature can assist students stay clear of sensation bewildered with mathematics projects.
Another helpful attribute of the lab Help Service is the Student Assistant. The Pupil Assistant helps students handle their job and also provides a place to publish their research. The Pupil Aide is valuable for students who don't wish to obtain bewildered with addressing several inquiries.
As pupils get more comfy with their projects, they are motivated to connect with the Task Supervisor as well as the Student Aide to get an on the internet support system. The on the internet support group can assist trainees keep their emphasis as they address their projects.
Every one of the tasks for the lab Assist Service are included in the bundle. Pupils can login as well as complete their appointed job while having the student help readily available behind-the-scenes to help them. The lab Assist Solution can be a great assistance for your pupils as they begin to browse the difficult university admissions and task searching waters.
Students should be prepared to get made use of to their tasks as quickly as possible in order to reach their major objective of getting involved in the college. They need to work hard sufficient to see results that will certainly allow them to stroll on at the following degree of their studies. Getting used to the process of finishing their jobs is really essential.
Trainees have the ability to discover different means to help them learn exactly how to utilize the lab Aid Solution. Discovering how to make use of the lab Aid Service is important to students' success in college as well as task application.
## Hire Someone To Take My Milü Lab
Milü is utilized in a lot of schools. Some teachers, nevertheless, do not use it really efficiently or utilize it improperly. This can have a negative impact on the trainee's discovering.
So, when designating jobs, use an excellent Milü aid solution to aid you with each lab. These solutions give a selection of valuable services, including:
Assignments may need a lot of evaluating as well as browsing on the computer. This is when making use of an assistance service can be an excellent benefit. It allows you to obtain more job done, boost your understanding, and also stay clear of a lot of stress.
These types of homework solutions are a wonderful means to start dealing with the very best kind of aid for your needs. Milü is just one of one of the most tough based on understand for students. Dealing with a service, you can ensure that your needs are satisfied, you are shown correctly, and also you recognize the material correctly.
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Having a homework solution also helps to improve the pupil's qualities. It permits you to include extra credit score in addition to boost your GPA. Obtaining extra credit history is frequently a massive benefit in many colleges.
Pupils who do not take full advantage of their Milü course will end up moving ahead of the remainder of the course. The good news is that you can do it with a quick and easy solution. So, if you intend to continue in your class, utilize a great help service. One point to bear in mind is that if you really want to increase your quality level, your program work requires to obtain done. As much as possible, you require to understand as well as deal with all your issues. You can do this with a great help service.
One advantage of having a homework service is that you can assist on your own. If you do not feel great in your ability to do so, then a great tutor will have the ability to assist you. They will be able to resolve the issues you face and assist you recognize them so as to get a far better quality.
When you graduate from secondary school and also get in college, you will certainly require to strive in order to remain ahead of the various other students. That means that you will need to work hard on your homework. Making use of an Milü solution can assist you get it done.
Maintaining your grades up can be tough because you generally need to study a lot and take a lot of examinations. You don't have time to deal with your grades alone. Having a great tutor can be a great aid because they can help you as well as your research out.
An assistance solution can make it easier for you to manage your Milü class. In addition, you can discover more regarding yourself and also help you do well. Discover the most effective tutoring solution as well as you will have the ability to take your research study skills to the next degree. | 2,437 | 12,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-50 | longest | en | 0.975011 |
http://www.jiskha.com/display.cgi?id=1290565002 | 1,498,388,125,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320489.26/warc/CC-MAIN-20170625101427-20170625121427-00078.warc.gz | 568,956,098 | 3,970 | # Math HELP
posted by .
I have my answers at the bottom but please do these questions so I can check my answers. Check my answers as well.
Subtract these fractions. !Reduce them!
1. 5/7 - 3/10=
2. 11/18 - 5/12=
3. 1/3 - 1/6=
4. 7/12 - 1/16=
5. 7/8 - 3/10=
6. 7/8 - 1/12=
7. 7/10 - 2/5=
8. 3/4 - 1/2=
1. 29/70
2. 7/36
3. 1/6
4.?
5. 46/80
6. 19/24
7. 3/10
8. 1/4
• Math HELP -
4. 7/12 - 1/16 = 28/48 - 3/48 = 25/48
5. 46/80 = 23/40
The rest are correct.
• Math HELP -
Thanks alot PsyDAG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):) | 334 | 721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-26 | latest | en | 0.430048 |
http://conferences2.imfm.si/contributionDisplay.py?contribId=40&confId=1 | 1,679,894,004,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00690.warc.gz | 10,866,999 | 8,778 | # Bled'11 - 7th Slovenian International Conference on Graph Theory
19-25 June 2011
Bled, Slovenia
Europe/Ljubljana timezone
Home > Timetable > Contribution details
PDF | XML
# Rainbow numbers and rainbow connection
Presented by Prof. Ingo SCHIERMEYER
Type: Oral presentation
Track: Coloring, independence and forbidden subgraphs
## Content
\begin{abstract} In this talk we consider edge colourings of graphs. A subgraph $H$ of a graph $G$ is called {\it rainbow subgraph}, if all its edges are coloured distinct. In the first part we will survey the computation of rainbow numbers. For given graphs $G, H$ the rainbow number $rb(G,H)$ is the smallest number $m$ of colours such that if we colour the edges of $G$ with at least $m$ different colours, then there is always a totally multicoloured or rainbow copy of $H.$ For various graph classes of $H$ we will list the known rainbow numbers if $G$ is the complete graph and report about recent progress on the computation of rainbow numbers. Finally, new results on the rainbow numbers $rb(Q_n, Q_k)$ for the hypercube $Q_n$ will be presented. An edge-coloured graph $G$ is called {\it rainbow connected} if any two vertices are connected by a path whose edges have distinct colours. This concept of rainbow connection in graphs was introduced by Chartrand et al.. The {\it rainbow connection number} of a connected graph $G,$ denoted $rc(G),$ is the smallest number of colours that are needed in order to make $G$ rainbow connected. Rainbow connection has an interesting application for the secure communication in networks. In the second part we will show several complexity results and present lower and upper bounds for $rc(G)$. We will discuss $rc(G)$ for graphs with given minimum degree. Finally, we will present some recent Erd\H{o}s-Gallai type results for $rc(G).$ \end{abstract}
## Place
Location: Bled, Slovenia
Address: Best Western Hotel Kompas Bled
More | 453 | 1,926 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.880182 |
http://waiferx.blogspot.com/2012/11/physics-quiz-question-stretching-wire.html | 1,532,021,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00085.warc.gz | 413,635,368 | 20,436 | ## 20121123
### Physics quiz question: stretching a wire
Physics 205A Quiz 6, fall semester 2012
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 10.3
A "000" AWG-gauge copper wire[*] with Young's modulus of 1.2×1011 Pa is 1.0 m long and has a cross-sectional area of 85.0 mm2. If a weight of 300 N is hung from the wire, it will stretch:
(A) 3.6×10–9 m.
(B) 2.1×10–7 m.
(C) 2.8×10–7 m.
(D) 2.9×10–5 m.
[*] wki.pe/American_wire_gauge.
Correct answer (highlight to unhide): (D)
Hooke's law for elastic materials is given by:
(F/A) = Y·(∆L/L),
where the cross-sectional area of the wire is 85.0 mm2 = 8.50×10–5 m2, such that the length that the amount the wire stretches is:
L = (F·L)/(A·Y)
L = ((300 N)·(1.0 m))/((8.50×10–5 m2)·(1.2×1011 Pa) = 2.941176471×10–5 m,
or to two significant figures, ∆L = 2.9×10–5 m.
(Response (A) is F·L·(1.2×10–11 Pa); response (B) is F·L·(85.0 m2)/Y; and response (C) is A/F.)
Sections 70854, 70855
Exam code: quiz06How3
(A) : 2 students
(B) : 5 students
(C) : 9 students
(D) : 30 students
Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.52 | 463 | 1,157 | {"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} | 3.5625 | 4 | CC-MAIN-2018-30 | latest | en | 0.605979 |
https://www.physicsforums.com/threads/need-help-with-thin-film-interference.56184/ | 1,527,122,501,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865863.76/warc/CC-MAIN-20180523235059-20180524015059-00568.warc.gz | 799,223,347 | 14,968 | Homework Help: Need help with thin film interference
1. Dec 9, 2004
pnazari
A student diving in a swimming pool (filled with water with index of refraction
1.33) creates thin films of air. Viewed underwater, what are the first two non-zero thicknesses of an air film for which there will be constructive interference for reflection off the surfaces of the bubble for light of wavelength 655 nm in water
a. 123 nm, 369 nm
b. 218 nm, 653 nm
c. 218 nm, 873 nm
d. 246 nm, 492 nm
e. 436 nm, 871 nm
f. 436 nm, 1091 nm
The answer is B, but I keep getting D. Here is what I have done:
Lambda=2nd/m (m=1,2,3..,n=index,d=thickness)
2. Dec 10, 2004
Andrew Mason
Tricky problem.
It has been a while since I have looked at this area but I do recall from optics that quarter wave coatings minimize reflected glare by making the reflected wave from the air/film surface out of phase by $\pi$ with the reflected wave from the film/glass surface, resulting in these two reflections destructively interfering. Since the light from a lower index of refraction medium meeting a higher one (air to film and film to glass) will reflect with a 180 degree phase shift at both surfaces, we don't have to worry about the phase change on reflections. So the distance of travel through the film (i.e from air/film surface to the glass surface and back to the air/film surface) just has to be a half wavelength longer. So the thickness of the film has to be $\lambda/4$.
When light (or any wave) goes from a higher index to a lower index medium it
does not have a 180 degree phase shift. The reflected wave is in phase with the incident wave. But the reflected wave from the other surface (air to water) does have a 180 degree phase shift because it is going from a low to high index medium. That is the key to this problem.
To provide CONSTRUCTIVE interference (maximum reflection), a phase shift of $2\pi$ is required. A $\pi$ shift is provided by the reflection from the air to water surface. So the two way travel through air has to produce a $\pi$ phase shift. This means that the thickness of air has to produce a $\pi/2, 3\pi/2, 5\pi/2 ...$ phase shift, so the minimum would be a quarter of the wavelength of the light in air and the next 3/4 wavelength. Since $\lambda_{air} = 1.33 \times 655 nm = 871 \text{nm}$ you have your answer.
AM
Last edited: Dec 10, 2004 | 614 | 2,355 | {"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} | 3.53125 | 4 | CC-MAIN-2018-22 | longest | en | 0.914316 |
https://perfectcourseworkhelp.com/a-race-car-starts-from-rest-on-a-circular-track-of-radius-404-m-the-cars-speed-increases-at-the-constant-rate-of-0-400-m-s2/ | 1,696,053,612,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00745.warc.gz | 460,152,893 | 14,833 | # A race car starts from rest on a circular track of radius 404 m. The car’s speed increases at the constant rate of 0.400 m/s2.
A race car starts from rest on a circular track of radius 404 m. The car’s speed increases at the constant rate of 0.400 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following(a.)the speed of the race car(b.)the distance traveled(c.)the elapsed time
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0 replies | 116 | 459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.797558 |
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