Datasets:

HuggingFaceTB

/

finemath

like 35

Follow

Hugging Face TB Research 704

http://www.math.ubc.ca/~staati/teaching/math300/

text/html

crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00304.warc.gz

home/teaching/math300/ # Introduction to Complex Variables (MATH 300 202) This page contains information on section 202. For information on section 201, see here. ## Time and location • Mondays, Wednesdays, Fridays 13:00-14:00 (from January 4 till April 6) • Math Annex 1100 ## Textbook E.B. Saff, A.D. Snider, Fundamentals of Complex Analysis with Applications to Engineering, Science and Mathematics, third edition. ## Course description We will begin by discussing the complex numbers and functions of a complex variable, then proceed to develop differential and integral calculus in this setting. The resulting theory is very beautiful and in many ways quite different from the "usual" calculus for functions of either one or several real variables. Complex analysis has many applications to science, engineering and other areas of mathematics. We will go over (most of) chapters 1-6 in the text, covering the following topics: • complex numbers, • complex derivatives and analytic functions, • elementary functions, • contour integration, • Cauchy's theorem, • Cauchy's Integral Formula, • Taylor series, • Laurent series, singularities and residues. The specific sections to be covered, subject to minor changes along the way, are 1.1-1.6, 2.1-2.6, 3.1-3.3, 3.5, 4.1-4.6, 5.1-5.3, 5.4, 5.5-5.6, 6.1, 6.2, 6.3. The prerequisites can be found here. ## Office hours The term grade will be calculated in two ways and the higher grade will be used. • Homework (20%) • Every week, due on Wednesday, at the beginning of the class. • The first assignment is due on Wednesday, January 11. • Quizzes (40% or 20%) • Every other week on Friday • The first quiz is on Friday, January 20. • There will be 5 quizzes in total. • Final exam (40% or 60%) ## Homework There will be homework assignments every week. The due is on Wednesday, at the beginning of the class. A portion of the assignments will be marked. • Assignment 1 (due Wednesday, January 11) • Section 1.1: 8, 17, 20(b, c), 32 • Section 1.2: 5, 7(c, f, h), 16, 20 • Section 1.3: 3, 7(f, g), 12, 13 Note: Those of you who registered for the course on Tuesday 10 or later may still hand-in their assignments on Friday 13. • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 2 (due Wednesday, January 18) • Section 1.4: 2(c), 4(c), 8, 11, 16, 18(a,c) • Section 1.5: 4(b), 5(b,c,d), 6, 9, 10, 16 • Section 1.6: 2-8 for (a,b,d) • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 3 (due Wednesday, January 25) • Assignment 4 (due Wednesday, February 1) • Section 2.3: 3, 4(b) [use the definition directly], 7(c), 13, 14 • Section 2.4: 2, 4, 6, 7, 12, 14 • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 5 (due Wednesday, February 8) • Assignment 6 (due Wednesday, February 15) • You may skip questions 3 and 6 of section 3.3 for now. For question 13 of section 3.2, use the definition of complex sinh and cosh on page 114. • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 7 (due Wednesday, March 1) • Enjoy the break! • A visualization of the exponential map for problem II. • A visualization of the logarithm map for problem III. • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 8 (due Wednesday, March 8 Friday, March 10) • Deadline extended to March 10. • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 9 (due Wednesday, March 15) • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 10 (due Wednesday, March 22) • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 11 (due Wednesday, March 29) • Section 5.1: 1(b, c, f), 6, 10, 16 • Section 5.2: 2, 4, 5(b, e, g), 11(b), 13, 15, 18(a) • Section 5.3: 7, 8, 12, 18 (compare with Prob. 2.3.14, assignment 4) • Section 5.5: 1(b, c), 5, 6 • Solutions are posted on Connect. • You can collect your papers from the MLC. • Assignment 12 (not to be turned in) ## Quizzes There will be a quiz every other week on Friday. • Quiz 1 (Friday, January 20) • Sections 1.1-1.6 • Solutions • You can collect your papers from the MLC. • Quiz 2 (Friday, February 3) • Up to and including section 2.5 (as far as it is covered before Friday) • Solutions • You can collect your papers from the MLC. • Quiz 3 (Friday, February 17 March 3) • Up to and including section 3.5 • Solutions • You can collect your papers from the MLC. • Quiz 4 (Friday, March 10 March 17) • Chapter 4 and before • Solutions • You can collect your papers from the MLC. • Quiz 5 (Friday, March 24 March 31) • Chapter 5 (as far as it is covered before Friday) and before • Solutions • You can collect your papers from the MLC. ## Final exam • Time and location • The final exam will cover the entire course (see the syllabus on top of the page) with an emphasis on chapters 4, 5, and 6. • No books, notes or calculators will be allowed on the exam. • Practice final exam (actual final exam from 2015) with solutions • Other past exams • Review sessions • Monday, April 10, 10:30-12:00, Math Annex 1100 • Wednesday, April 12, 13:00-14:30, Math Annex 1100 • The grades are submitted. (new)

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-43

latest

en

http://mathhelpforum.com/differential-equations/137034-separable-de.html

text/html

crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00328.warc.gz

1. ## Separable DE For the equation above, I need to find all the constant solutions. How do I go about doing this? I've tried putting all similar terms on one side and then integrating, but it doesn't seem to work out nicely. Any help is greatly appreciated, thanks in advance! 2. $\displaystyle \int y' = \int y-y^3$ $\displaystyle \Rightarrow y= \frac{2y^2 - y^4}{4} + C$ $\displaystyle \Rightarrow \frac{y}{4}(y^3-2y+4)=C$ $\displaystyle \Rightarrow \frac{y}{4}(y^3-2y+4)-C=0$ Code: syms y c solve('(y/4)*(y^3-2*y+4)-c=0',y) ans = - 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)/(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/4)*(48*c*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 4*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 24*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 48*6^(1/2)*(3*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 36*c + 26)^(1/2))^(1/2) 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)/(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/4)*(48*c*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 4*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 24*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 48*6^(1/2)*(3*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 36*c + 26)^(1/2))^(1/2) - 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)/(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/4)*(48*c*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 4*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 24*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 48*6^(1/2)*(3*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 36*c + 26)^(1/2))^(1/2) 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 1/6/(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/6)/(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/4)*(48*c*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 4*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) + 24*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3)*(12*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(1/3) - 48*c + 9*(8/9*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 32/3*c + 208/27)^(2/3) + 4)^(1/2) - 48*6^(1/2)*(3*3^(1/2)*(64*c^3 + 32*c^2 - 68*c + 25)^(1/2) - 36*c + 26)^(1/2))^(1/2) umm... lol. Your sure your not given any initial conditions, $\displaystyle y(x)=0?$ 3. that's wrong, the ODE actually writes as $\displaystyle \frac{dy}{y-y^{3}}=dx,$ 4. Originally Posted by Krizalid that's wrong, the ODE actually writes as $\displaystyle \frac{dy}{y-y^{3}}=dx,$ Figured it had to be . So your pretty good at calc, huh? 5. uhh yeah there aren't any initial conditions 6. Originally Posted by cdlegendary For the equation above, I need to find all the constant solutions. How do I go about doing this? I've tried putting all similar terms on one side and then integrating, but it doesn't seem to work out nicely. Any help is greatly appreciated, thanks in advance! if $\displaystyle y = C$ , then $\displaystyle y' = 0$ ... correct? 7. Originally Posted by skeeter if $\displaystyle y = C$ , then $\displaystyle y' = 0$ ... correct? ah yes. I think I know where you're going with that 8. so the solutions are just -1, 0, 1. wow haha. I was definitely thinking too hard about that. thanks!

{"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}

CC-MAIN-2018-26

latest

en

https://www.distinylearn.com/dbms-unit-4-part-2-transaction-processing-concept/

text/html

crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00725.warc.gz

# DBMS Unit 4 Part 2 Transaction Processing Concept Que4.11. What’s priority graph? How can it be used to test the conflict serializability of a schedule? Answer Priority graph 1. A priority graph is a directed graph G = ( N, E) that consists of set of bumps N = { T1, T2,., Tn} and set of directed edges E = ( e1,e2., em). 2. There’s one knot in the graph for each sale Ti in the schedule. 3. Each edge ei in the graph is of the form( Tj Tk), 1 j n, 1 k n, where Tj is the starting knot of ei and Tk is the ending knot of ei. 4. Such an edge is created if one of the operations in Tj appears in the schedule before some disagreeing operation in Tk. Algorithm for testing conflict serializability of schedule S a. For each sale Ti sharing in schedule S, produce a knot labeled Ti in the priority graph. b. For each case in S where Tj executes aread_item( X) after Ti executes awrite_item( X), produce an edge( Ti Tj) in the priority graph. c. For each case in S where Tj executes awrite_item( X) after Ti executesread_item( X), produce an edge( Ti Tj) in the priority graph. d. For each case in S where Tj executes awrite_item( X) after Ti executes awrite_item( X), produce an edge( Ti Tj) in the priority graph. e. The schedule S is serializable if and only if the priority graph has no cycles. 5. The priority graph is constructed as described in given algorithm. 6. still, schedule S isn’t( conflict) If there’s a cycle in the priority graph. serializable; if there’s no cycle, S is serializable. 7. In the priority graph, and edge from Ti to Tj means that sale Ti must come ahead sale Tj in any periodical schedule that is original to S, because two disagreeing operations appear in the schedule in that order. 8. still, we can produce an original If there’s no cycle in the priority graph. periodical schedule S that’s original to S, by ordering the deals that share in S as follows Whenever an edge exists in the priority graph from Ti to Tj, Ti must appear before Tj in the original periodical schedule S. Que4.13. bandy cascadeless schedule and slinging rollback. Why is cascadeless of schedule desirable? Answer Cascadeless schedule Refer Q.4.10, runner 4 – 9A, Unit- 4. Cascading rollback Slinging rollback is a miracle in which a single failure leads to a series of sale rollback. For illustration Schedule S In the illustration, sale T1 writes a value of A that’s read by sale T2. sale T2 writes a value of A that’s read by sale T3. Suppose that at this point T1 fails. T1 must be rolled back. Since T2 is dependent on T1, T2 must be rolled back, since T3 is dependent on T2, T3 must be rolled back. Need for cascadeless schedules Cascadeless schedules are desirable because the failure of a sale does not lead to the aborting of any other sale. This comes at the cost of lower concurrency.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-23

latest

en

https://forum.math.toronto.edu/index.php?PHPSESSID=acfo9cuftgpa81fip265m6t2e2&topic=1492.msg5268

text/html

crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00013.warc.gz

### Author Topic: 2.5 Q20  (Read 1420 times) #### Nikki Mai • Jr. Member • Posts: 12 • Karma: 1 ##### 2.5 Q20 « on: November 13, 2018, 12:52:41 PM » Can anyone help me solve 2.5 question20? I read the question a few times. i do not know how to do it. Thank you. #### Aleena Au • Newbie • Posts: 1 • Karma: 0 ##### Re: 2.5 Q20 « Reply #1 on: November 13, 2018, 03:09:12 PM » This is my attempt at the question. Assume f's Laurent series is not unique. Then, we have $$f(z) = \sum a_{n} (z-z_{0})^n$$ $$f(z) = \sum b_{n} (z-z_{0})^n$$ Subtract the two equations and get $$0 = \sum (a_{n}-b_{n}) (z-z_{0})^n$$ So, we must have $$a_{n} = b_{n}$$ and f's Laurent series must be unique. #### Victor Ivrii Aleena It is not the answer because you rely upon unsustained claim that if $f(z)=0$ then it's Laurent's coefficients are $0$ which is an equivalent form of the same question. Hint: Consider $\int_\gamma (z-z_0)^m f(z)\,dz$ where $\gamma$ is a counter-clockwise circle around $z_0$ and prove that it is equal to $2\pi i a_{m-1}$.

{"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}

CC-MAIN-2022-49

latest

en

http://highered.mheducation.com/sites/007299181x/student_view0/chapter6/group_activities.html

text/html

crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00531.warc.gz

Group Activities Group Activities Buy or borrow five identical candles. Tie four of the candles together in a bundle. At night, find a relatively dark location. Light all five candles and wait for the candles to start burning evenly. Have one member of your group take the single candle and walk ten paces away. Have another group member take the bundle of four candles and walk away in the same direction until the rest of the group judges that the four candles together appear as bright to them as the single, closer candle. Measure how many paces the four candles are from the group. Find the ratio of the distances of the four candles and the single candle and compare the result to what you would expect on the basis of Equation 6.1. Go to an asphalt parking lot early in the morning. Have one-half of the group stand on one side of the parking lot and the other half on the other side. Have the two groups observe how steady the other group appears to be when viewed across the parking lot. Repeat the experiment after the Sun has been shining on the dark parking lot for much of the day. Get the entire group together to discuss why the shimmering due to the parking lot is different at the two times of day.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-30

latest

en

https://www.slideserve.com/abby/density

text/html

crawl-data/CC-MAIN-2018-05/segments/1516084887065.16/warc/CC-MAIN-20180118032119-20180118052119-00040.warc.gz

Density 1 / 5 # Density - PowerPoint PPT Presentation Density. By: Megan Passmore. Sinker or Floater?. Floaters If an object has a lower density than the substance it’s placed in the object will float because it is too light to be pulled down through the dense substance. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Density' - abby Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ### Density By: Megan Passmore Sinker or Floater? Floaters • If an object has a lower density than the substance it’s placed in the object will float because it is too light to be pulled down through the dense substance. • Ex. If Styrofoam is placed in water the Styrofoam will float because Styrofoam has a density of 0.15kg/L and water has a higher density of 1kg/L. Sinkers • If an object has a higher density than the substance it’s placed in the object will sink because it is being pulled down by gravity through the thin substance. • Ex. If aluminum is placed in water the aluminum will sink because aluminum has a density of 2.7kg/L and water has a low density of 1kg/L. Calculating Density REMEMBER: The displacement method is very important and helping when attempting to calculate the volume of an irregular solid. To do this, fill a graduated cylinder with water, measure the current volume, place the irregular shape, and measure the new volume. Subtract the new volume by the previous volume to find the water displaced. The difference is the volume of the shape. REMEMBER: To calculate the density use the equation M/V=D, or mass divided by volume equals density. To find the mass of a substance, use a scale or balance. To find the volume of a substance, use a graduated cylinder or mathematical equation. Divide the two numbers to find the density of the object. TRY IT! BEFORE: If a wooden block weighs 2kg and it’s volume is 5L, you’re going to divide the mass (2kg) by the volume (5L.) 2kg/5L=0.40kg/L. AFTER: What is it? If someone hands you an unknown substance and says its something like diamond, how do you know if it really is? You have to observe the characteristics. You should know how they react to different substances and whether or not they float or sink in water. An example of this can be demonstrated through the Phet simulation. When you’re given a mystery substance, you can test whether it floats or sinks in water. Next, you can calculate the density by finding the mass and volume. Using the chart, you can determine the substances true identity.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-05

latest

en

http://ewhows.com/en/pages/1269705

text/html

crawl-data/CC-MAIN-2017-51/segments/1512948522205.7/warc/CC-MAIN-20171213065419-20171213085419-00733.warc.gz

# How to make a cube out of paper ? ## Watch the video How to make a cube out of paper? Now we try to understand how to make a paper cube.There are simple and complex ways to create a cube, look at both. ## simple way to Take a piece of paper and draw a cube unassembled on paper.Take, for example, a cube with edge 3 cm Draw 6 squares with a side of 3 cm as follows:.. 4 squares with common sides draw in a row, and the third drawn from both sides add one more.We get cross squares.Then, on the top three squares "cross" by adding 0.7 cm. Allowance for gluing.Then we cut our piece, bend the lines and collect allowances sticking to the main part.Ready cube should be left to dry for a while. To produce multicolored cube before bonding resulting razrisuy all squares on the front side of pencils of different colors. ## hard way With the second method, you will learn how to collect the cube without using glue. For the manufacture of the cube, we need 6 sheets of A4 paper (paper can be colored) and scissors.If you follow the instructions, then you get the correct figure. 1. A4 sheet of paper is folded in diagonal so that you have a triangle.As a result, you will have an extra piece of paper that you want to cut. 2. resulting square folded in half and return it back, during which on the sheet will fold.Fold the ends of the sheet to the fold line on both sides. 3. again straighten and bend your hand has two opposite angles to each other, and again bend the outstretched hand. 4. Then bend the other two corners, but larger: one of them is hiding inside (at the opposite side of the paper), and cover with a second sheet.As a result, you get a figure in the form of a square. 5. Both ends of the bend so as to obtain a square.One of the six parts is ready, we can start manufacturing other parts. 6. Fold in the same way for another 5 parts. 7. now proceed to collect the figures, and for that attach to a pocket of some other details of the folds.Parts are assembled in pairs.Then attach pairs to each other and ready our cube. That's how your own hands can make a cube.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-51

latest

en

https://en.razmery.info/equipment/wheel_disks/dimensions-wheel-disks-lexus-gs.html

text/html

crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00506.warc.gz

Disc sizes of Lexus GS General parameters: Offset (ET): 45;Drilling (PCD): 5x114.3;Center hole (DIA): -.Important: wheel disc parameter PCD indicates not only the diameter of circumference of the mounting holes, but also the number of mounting bolts or nuts (for example: 5x120, where 120 (mm) is the diameter of centers of the mounting holes, and 5 is the number of mounting bolts). Possible disk sizes for Lexus GS: 17x7.518x8.0 Procedure for calculating the disc offset: The distance between mating face of the wheel (the face with which the disk is pressed against the hub) and the middle of the width of disk is called disk offset (ET).The formula for calculating:ET = A - B/2, where:A is the distance between the inner plane of disk and the part in contact with the hub;B is disk width.The disk offset affects the width of the vehicle wheelbase, since the distance between the centers of symmetry (in width) of wheels on one axis directly depends on this parameter.Important: PCD of disks with four fixing bolts (or nuts) corresponds to the distance between the centers of opposite bolts (or nuts), PCD of disks with five fixing bolts (or nuts) corresponds to the distance between the centers of any non-adjacent bolts (or nuts) multiplied by the coefficient 1.051. Wheel disks - full drawing Possible disk parameters for Lexus GS General view General view Disk Possible disk sizes for Lexus GS of different years of manufacture and modifications Wheel Possible wheel sizes for Lexus GS of different years of manufacture and modifications ET (mm) Disk offset Distance between the vertical plane of symmetry of the wheel and the plane of contact of the disk to the hub. Measured in millimeters (mm) (General view) 17'' 18'' 17x7.5 18x8.0 45 Table of disc sizes of Lexus GS modifications Year Model year of manufacture Size Disk size in format: 1. D - diameter of the disk in inches. 3. B - disk width in inches. ET (mm) Disc offset: Distance between the vertical plane of symmetry of the wheel and the plane of contact of the disc to the hub. PCD (mm) Disc drilling: 1. Number of bolt holes. 2. Diameter of the circle on which the mounting holes are located. Tire Tire size in format: 1. Width of the work face. 2. Percentage of profile height to width. 3. R - type of construction (radial) - inner diameter (in inches). DIA (mm) Center hole: Diameter of center hole, another marking is CH. Lexus GS 200517x7.5455x114.3225/50R17x 200618x8.0455x114.3245/40R18x 200518x8.0455x114.3245/40R18x Lexus GS 300/430 200618x8.0455x114.3245/40R18x Lexus GS 300/430 SE 200617x7.5455x114.3225/50R17x 200618x8.0455x114.3245/40R18x Lexus GS 450h 200618x8.0455x114.3245/40R18x

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-23

latest

en

https://www.fluther.com/48432/do-you-ever-need-to-use-imaginary-numbers-in-real-life/

text/html

crawl-data/CC-MAIN-2018-43/segments/1539583511203.16/warc/CC-MAIN-20181017153433-20181017174933-00450.warc.gz

# Do you ever need to use imaginary numbers in real life? Asked by applesaucemanny (1770) June 26th, 2009 I understand how we use log and scientific notation and all that but when are we going to need to use imaginary numbers? Observing members: 0 Composing members: 0 Ah, “i.” As far as I am aware, imaginary numbers are only used by some computer/other technology-related occupations and mathematicians. Isn’t math fun? Fly (8699) When I taught control theory, electromagnetism, quantum mechanics, and cartography I used them quite often. not so much anymore. Yes, math is fun… DrBill (16046) I love the number elevendy. I use it all the time. Or elevendy billion. :] chelseababyy (7934) No, my imaginary friend uses them – but I abstain. DarkScribe (15470) How many times has a question about gay rights and marriage been asked on Fluther? Better go with the N times. filmfann (45409) When boredom strikes. Dum dum da dum.. Only when discussing my bank balance. Lightlyseared (30862) @Lightlyseared so… how many i’s do you have? 1. Just after the L. Lightlyseared (30862) What the hell is an imaginary number? knitfroggy (8944) They are very useful on tax returns. juwhite1 (2971) You may never be taught this, but here is a neat application of imaginary numbers to trigonometry. It can be used to find multiple angle formulas. Suppose you wanted to know how to find cos(3x) in terms of powers of cos(x) and sin(x) cos(3x) + isin(3x) = e^(3ix) = (e^ix)^3 = (cos(x) + isin(x))^3 You would now expand the right side. The real part of what you get is cos(3x) and the imaginary part is sin(3x) You can also go in the opposite direction. Suppose you wanted to find cos^3(x) in terms of sin and cos of multiples of x. cos^3(x) = [(e^ix – e^-ix)/2]^3 Expand the right side, giving powers of e^ix. Then convert these terms to sin and cos of multiples of x. The imaginary terms will cancel and you end up with the desired equation. I saw the cos(3x) on the third line and blanked out. It’s amazing, I can actually feel my brain forcing my eyes away from the equation while refusing to even try comprehending what’s going on… i do. ratboy (15157) @ratboy So punny! Fly (8699) Electrical analysis requires you to use them. critter1982 (4120) I don’t know how anything remotely math related made it into my “Questions for you.” Am I being punished? ubersiren (15150) @ubersiren – Must be the “imaginary” factor. ;-) juwhite1 (2971) This is a very good question. The simple answer is yes, if by “real life” you mean anything that is observable in the “everyday sense” kind of way. If you are a mathematician, or your field uses mathematics quite often, you use complex (imaginary) numbers regularly simply because they get the job done… and very elegantly at that. I would say the majority of the numbers encountered by most scientists deal with rational numbers. And especially numbers that are truncated… approximations of numbers that have a huge (sometimes infinite… think pi) number of digits. The simple fact is, the entire collection of real numbers is severely neglected in everyday use. But, this is ok. They’re simply not useful in every context. What about imaginary numbers? These beasts have a very specialized, theoretical, purpose. To mathematicians they are just another extension of the real number system (and yes, there is an infinite regress of extensions beyond complex numbers – starting with hyper-complex numbers). To scientists, they provide a very compact way to express theories of periodic (repeating, etc.) phenomena. They also produce fractals .(they are interesting and useful!) Typically, the symbol ‘i’ is used to represent complex numbers. It is a bit unsettling, after being introduced to complex numbers, that they do not look like numbers at all. They look like algebra . Most of our entire math education has drilled into us the falsehood that numbers are only represented in one way: a bunch of numerals lined up in a row, sometimes with a decimal point in the middle of them all. No wonder our heads hurt after first seeing our first complex number, and no wonder why we see no purpose for them in everyday life (complex numbers aren’t alone in this fact). The strangeness of ‘i’ is no stranger than seeing the square root of 2 (with the radical sign over the two) ... that is just as weird, because that symbol represents exactly a number who’s square is 2. The same for ‘i’, the ‘number’ who’s square is -1. majamin (99) Yes, in Electronics we use them all the time. sqrt(-1) is represented by the letter j, as i always represents current. In a reactive circuit – one that contains inductors, capacitors, or both, voltage and current are out of phase. The phase difference can be represented as a vector, or phasor on an Argand diagram. The Argand diagram is a coordinate system with + and – real numbers on the horizontal axis, and + and – imaginary numbers on the vertical axis. A circuit with a phase difference of +30 degrees would be represented by the phasor sqrt(3)/2+j. What’s the significance of this? In a purely reactive circuit – one that contains only a capacitor, for instance, you can have current flowing, and a difference in voltage across the terminals, but no power dissipation. That corresponds to a phase difference of -90 degrees, which is represented on the Argand diagram as 0 – j. In a circuit that contains both resistive and reactive components, only the resistive components dissipate power, while the reactive components dissipate nothing. Thus their power dissipation could be called imaginary. You can measure the voltage across a reactive circuit and the current through it, but you need to know the phase relationship between the voltage and current to determine how much power is actually being used. or

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-43

longest

en

https://amylemons.com/solve-to-create-a-snowman/

text/html

crawl-data/CC-MAIN-2024-22/segments/1715971057684.4/warc/CC-MAIN-20240519035827-20240519065827-00248.warc.gz

# Solve to Create a Snowman Do you want to mix solving word problems with some winter fun? Solve to Create allows students the opportunity to build a fun craft while also solving word problems or completing math tasks. It is the ultimate winter word problem activity! Click HERE to see the Solve to Create a Snowman. ## How to Set-Up Here’s how Solve to Create a Snowman works: Students get into groups of 6. Each group starts at a different word problem. The group works together to solve the word problem. These can be put into Task Folder Envelopes, but that isn’t necessary at all! ## Solving Word Problems When the groups get to a task, they read the word problem as a group. The word problems come in strips (that can be glued into a spiral for solving) and one-to-a page for groups to read together. Each word problem solved allows them to gain a piece of their snowman. So, if they solve task 1 correctly they will receive the head. If they solve task 2 correctly they will get the body. ## Earning Parts of the Snowman The teacher can keep all of the snowman pieces at his/her table so that the students have to be checked before receiving a piece. OR the pieces can go with that word problem right in the folder. That totally depends on your style and student needs! ## The Word Problems 1. Addition and Subtraction within 20 2. 2 and 3 Digit Addition and Subtraction with Regrouping and Multi-Step Word Problems 3. Mixed Review with Multiplication, Division, Addition, and Subtraction We wanted as many teachers to be able to use this as possible! ## Easy to Prep Materials The great thing about it is all you need are the word problems, recording sheets, and the craft pieces. Just print it out and go! ## Editable Word Problems To help as many teachers as possible, we also created an editable set of word problems. This way ANY teacher of ANY grade level can type in word problems to fit ANY skill! That way you can use this no matter what! We hope that you enjoy using Solve to Create in your classroom! Don’t want to forget this lesson? Pin the image below to come back to later! ## Hi, I'm Amy Hey, y’all! My name is Amy Lemons and I am passionate about providing students with both engaging and effective standards-based Math and ELA lessons. ## FREE SAMPLE OF ROOTED IN READING!​ Sample a day of Rooted in Reading with these lesson plans and activities for Reading Comprehension, Vocabulary, and Grammar!

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-22

latest

en

h666.dlinkddns.com

text/html

crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00306.warc.gz

# Hex to decimal conversion c++ ### aquilone You can find other of the chest wall. Prediction of phenotypes and acura a spec kit X5 Home Learning. Use std::hex manipulator: #include <iostream> #include <iomanip> int main() { int x; st. Nov 16, 2009 . how would i write a function that parses a hex number as a string into a decimal i. Aug 24, 2014 . This is a Qt GUI frontend for a dec/bin/hex converter class. I know Qt haves such. Mar 14, 2015 . Write a Program to convert Decimal to Hexadecimal C++ decimal to hexadecimal conve. Need help? Post your question and get tips & solutions from. Hi Guru, If I have string like. Sep 28, 2015 . nptr is the string you want to convert. This should be your hexadecimal number. Yo. ### hex to decimal conversion c++ Use std::hex manipulator: #include <iostream> #include <iomanip> int main() { int x; st. Nov 16, 2009 . how would i write a function that parses a hex number as a string into a decimal i. Aug 24, 2014 . This is a Qt GUI frontend for a dec/bin/hex converter class. I know Qt haves such. Mar 14, 2015 . Write a Program to convert Decimal to Hexadecimal C++ decimal to hexadecimal conve. Need help? Post your question and get tips & solutions from. Hi Guru, If I have string like. Sep 28, 2015 . nptr is the string you want to convert. This should be your hexadecimal number. Yo. Use std::hex manipulator: #include <iostream> #include <iomanip> int main() { int x; st. Nov 16, 2009 . how would i write a function that parses a hex number as a string into a decimal i. Aug 24, 2014 . This is a Qt GUI frontend for a dec/bin/hex converter class. I know Qt haves such. Mar 14, 2015 . Write a Program to convert Decimal to Hexadecimal C++ decimal to hexadecimal conve. Need help? Post your question and get tips & solutions from. Hi Guru, If I have string like. Sep 28, 2015 . nptr is the string you want to convert. This should be your hexadecimal number. Yo. ### formation of coals 1 or literal 1 is an approved online the USA. Learn some interesting information WWII Period Nazi German a range of fun facts and trivia thats. Use std::hex manipulator: #include <iostream> #include <iomanip> int main() { int x; st. Nov 16, 2009 . how would i write a function that parses a hex number as a string into a decimal i. Aug 24, 2014 . This is a Qt GUI frontend for a dec/bin/hex converter class. I know Qt haves such. Mar 14, 2015 . Write a Program to convert Decimal to Hexadecimal C++ decimal to hexadecimal conve. Need help? Post your question and get tips & solutions from. Hi Guru, If I have string like. Sep 28, 2015 . nptr is the string you want to convert. This should be your hexadecimal number. Yo.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-43

latest

en

http://www.mathworksheetscenter.com/mathskills/trigonometry/addsubcomplexnumbers/

text/html

crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00089.warc.gz

Receive free math worksheets via email: When you write sin (45), you are really writing the sinus of 45 degrees! So, the next time your trigonometry teacher tells the class to open their math books, feel free to blurt out, 'It's time to study our sinuses now.' That should be good for a laugh! # Adding & Subtracting Complex Numbers Worksheets = free = members only Basic Lesson Guides students solving equations that involve an Adding & Subtracting Complex Numbers. Demonstrates answer checking. Standard: MATH 3 View lesson Intermediate Lesson Demonstrates how to solve more difficult problems. Standard: MATH 3 View lesson Independent Practice 1 A really great activity for allowing students to understand the concept of Adding & Subtracting Complex Numbers . Standard: MATH 3 View worksheet Independent Practice 2 Students find the Adding & Subtracting Complex Numbers in assorted problems. The answers can be found below. Standard: MATH 3 View worksheet Adding & Subtracting Complex Numbers Homework Students are provided with problems to achieve the concepts of Adding & Subtracting Complex Numbers. Standard: MATH 3 View homework Adding & Subtracting Complex Numbers Quiz This tests the students ability to evaluate Adding & Subtracting Complex Numbers. Standard: MATH 3 View quiz Answers for math worksheets, quiz, homework, and lessons. Need Some Math Help? 10,000+ math worksheets and lessons. We're growing fast! \$29.97 For a limited time! 10 Everyday Reasons Why Trigonometry is Important in your Life The primary application of trigonometry is found in scientific studies where precise distances need to be measured. View Math Article Why is important that someone on this planet understands the geometry of a triangle? Well, you use it all the time but often don't notice it and when you develop a deeper understanding of triangles a tremendous wealth of information opens to you. View Math Article

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-16

latest

en

https://brilliant.org/problems/an-angle-discovery/

text/html

crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00304.warc.gz

# An Angle Discovery! Geometry Level 4 Given that $XY$ is the diameter of the circle, where $\triangle WXY$ is enclosed in the circle and that the ratio $\frac { Area\quad of\quad circle }{ Area\quad of\quad \triangle WXY } =2\pi$. Find the acute angle $WXY$. Giving your answer in degrees. ×

{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-43

latest

en

https://www.sulfonte.com/why-is-it-important-to-have-constant-variables/

text/html

crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00598.warc.gz

Home General Why is it important to have constant variables? # Why is it important to have constant variables? ## Why is it important to have constant variables? It’s important to use constants in an experiment because they allow you to isolate a particular variable (the independent variable). If you were to have multiple independent variables in an experiment, it would be extremely difficult to figure out how each factor was influencing the results. ## How do you identify variables and constants? The number before an alphabet (variable) is called a constant. Variable : A symbol which takes various numerical values is called a variable. The alphabet after a number (constant) is called a variable. In the formulas d = 2r; 2 is a constant whereas, r and d are variables. ## How do you find the constant term of an expression? We can see that the general term becomes constant when the exponent of variable x is 0 . Therefore, the condition for the constant term is: n−2k=0⇒ k=n2 . In other words, in this case, the constant term is the middle one ( k=n2 ). Also Read:  What program is using my hard drive Mac? ## What is another term for constants in science? SYNONYMS FOR constant 1 unchanging, immutable, permanent. 2 perpetual, unremitting, uninterrupted. 3 incessant, ceaseless. ## What are some examples of consonance? Examples of Consonance in Sentences • Mike likes his new bike. • I will crawl away the ball. • He stood on the road and cried. • Toss the glass, boss. • It will creep and beep while you sleep. • He struck a streak of bad luck. • When Billie looked at the trailer, she smiled and laughed. • I dropped the locket in the thick mud. ## What are the three criteria to describe a vowel? Daniel Jones developed the cardinal vowel system to describe vowels in terms of the features of tongue height (vertical dimension), tongue backness (horizontal dimension) and roundedness (lip articulation). ## What are the four main principles of consonant classification? The chapter is structured according to four phonological classification criteria: airstream mechanisms, voicing contrast, place of articulation, and manner of articulation. Also Read:  Can you haggle when leasing a car? ## How many principles of classification of consonants do you know? place of obstruction) consonants are classified into: 1) labial, 2) lingual, 3) glottal. This principle provides the basis for the following distinctive oppositions: labial vs. lingual (what-hot), lingual vs.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-05

latest

en

https://www.learningclip.co.uk/Clip/Search/Mathematics_Using-and-Applying-Maths_Puzzles

text/html

crawl-data/CC-MAIN-2020-16/segments/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00520.warc.gz

Worksheets & Whiteboard Resources for Primary Maths, SPaG & Science Not logged in ### Mathematics Using and Applying Maths Puzzles Year 1, Unit 3, Block A ## Question Grid Describe ways of solving puzzles and problems, explaining choices and decisions orally or using pictures • Calendar • Favourites Year 1, Unit 3, Block E ## Triangle Man's Tangrams Describe a puzzle or problem using numbers, practical materials and diagrams; use these to solve the problem and set the solution in the original context • Calendar • Favourites Year 2, Unit 2, Block A ## Matching Total Cards Present solutions to puzzles and problems in an organised way; explain decisions, methods and results in pictorial, spoken or written form, using mathematical language and number sentences • Calendar • Favourites Year 3, Unit 1, Block A Ages 7 - 8 ## Number Cranes Explain a process or present information, ensuring items are clearly sequenced, relevant details are included and accounts ended effectively • Calendar • Favourites Year 3, Unit 1, Block C Ages 7 - 8 ## Alien Sorting Use Venn diagrams or Carroll diagrams to sort data and objects using more than one criterion • Calendar • Favourites Year 3, Unit 2, Block B Ages 7 - 8 ## Symmetrical Tiling (1line) Draw and complete shapes with reflective symmetry; draw the reflection of a shape in a mirror line along one side • Calendar • Favourites Year 3, Unit 2, Block D Ages 7 - 8 Read and record the vocabulary of position, direction and movement, using the four compass directions to describe movement about a grid • Calendar • Favourites Year 3, Unit 3, Block B Ages 7 - 8 ## Carroll Diagram (3D) Relate 2-D shapes and 3-D solids to drawings of them; describe, visualise, classify, draw and make the shapes . • Calendar • Favourites Year 3, Unit 3, Block D Ages 7 - 8 ## Mystery Number Use knowledge of number operations and corresponding inverses, including doubling and halving, to estimate and check calculations • Calendar • Favourites Year 4, Unit 1, Block B Ages 8 - 10 ## Times Table Cards Derive and recall multiplication facts up to 10 × 10, the corresponding division facts and multiples of numbers to 10 up to the tenth multiple • Calendar • Favourites

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-16

latest

en

https://findthefactors.com/tag/texas/

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00635.warc.gz

# 1404 Texas Tessellation I recently visited family members in Texas. My daughter-in-law is awesome at both mathematics and quilting. My photo does not do her work justice, but Texas is tessellated in this quilt! She also carefully chose the fabrics she pieced together. Do they remind you of anything for which Texas is famous? Someone else designed the pattern, but piecing these pieces together was not the easiest sewing project. I wondered if anyone else had thought to tessellate Texas and found a couple of examples on twitter. As this first one asked, should we call this Texellation? Now I’ll tell you something about the number 1404: • 1404 is a composite number. • Prime factorization: 1404 = 2 × 2 × 3 × 3 × 3 × 13, which can be written 1404 = 2² × 3³ × 13. • 1404 has at least one exponent greater than 1 in its prime factorization so √1404 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1404 = (√36)(√39) = 6√39. • The exponents in the prime factorization are 2, 3, and 1. Adding one to each exponent and multiplying we get (2 + 1)(3 + 1)(1 + 1) = 3 × 4 × 2 = 24. Therefore 1404 has exactly 24 factors. • The factors of 1404 are outlined with their factor pair partners in the graphic below. 1404 is the hypotenuse of a Pythagorean triple: 540-1296-1404 which is (5-12-13) times 108 Since 1404 has so many factors, it also has MANY different factor trees. Here are four of them mixed in with some Texas tessellations!

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-50

latest

en

http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-8-radical-functions-chapter-review-exercises-page-670/53

text/html

crawl-data/CC-MAIN-2018-17/segments/1524125945037.60/warc/CC-MAIN-20180421051736-20180421071736-00406.warc.gz

## Intermediate Algebra: Connecting Concepts through Application $3\sqrt 2$ $\sqrt {\frac{36}{2}}=\sqrt {18}=\sqrt {2\times9}=\sqrt 2\times\sqrt 9=3\sqrt 2$

{"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}

CC-MAIN-2018-17

latest

en

https://brainly.com/question/189488

text/html

crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00466-ip-10-171-10-70.ec2.internal.warc.gz

2014-11-17T19:44:58-05:00 ### This Is a Certified Answer Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. If you divide both sides by 0.6 the answer will be 6.5 Thanks we were a little slow and kept multiplying. You're welcome! 2014-11-17T19:47:25-05:00 ### This Is a Certified Answer Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. We have to solve following equation 3.9=0.6n. To solve equation we want to have only 1 n alone on the LHS or RHS. Lets solve it again. 3.9=0.6n        - to move 0.6 to the LHS we can divide both side by 0.6 3.9=0.6n      /:(0.6) or just 3.9:0.6=0.6n:0.6 And both notation give the same result 6.5=n Thanks for explaining it. u re welcome

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-04

latest

en

https://leenz.org/cranes/how-tall-is-a-tower-crane.html

text/html

crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00244.warc.gz

# How tall is a tower crane? Contents Generally, when anchored to the ground, a tower crane can’t be much more than 265 feet tall. They can be much taller if they are secured to a building as the building rises. The maximum jib reach is approximately 230 feet, and the maximum weight it can lift is 19.8 tons, or 18 metric tons. ## What is the height of a tower crane? A typical tower crane has the following specifications, but note there are differences depending on the model and manufacturer: Maximum unsupported height: 265 feet or 80 meters. It can reach taller if attached into the building, since the building rises around the crane. Maximum reach: 230 feet or 70 meters. ## What is free standing height of tower crane? Tower Cranes Model Jib length(s) Free Standing Height TC 5040-T 50 m 40 m TC 5540-T 55 m 40 m TC 6040 60 m 40 m TC 6040-B 60 m 40 m IT IS INTERESTING:  How much do crane inspectors make? ## What is the tallest tower crane in the world? The Liebherr LTM 11200-9.1, built by the German company Liebherr Group, is the most powerful mobile crane ever built. It also has the longest telescopic boom in the world, which extends fully to 100 meters. It’s set on a double cab truck and can lift 1200 metric tons – that’s nearly 700 automobiles. ## How long does it take to climb a tower crane? “It takes about 15 minutes,” Miller said. “You don’t want to climb up too fast, because when you get up there you’re worn out and hot and sweaty. You’ve got to sit up there all day long.” The crane runs around the clock. ## What does a 100 ton crane mean? 100 ton means, that theoretically the crane could lift one metre from the center of the column 100 ton. … The base formula for the Crane Capacity Index is: * For each main boom length at each radius the “Lifting height*Capacity” is calculated. ## How much can a 200 ton crane lift? 19,500lb Cap. 1,250lb & 300lb Cap. ## Why do tower cranes not fall over? Why Don’t Tower Cranes Fall Over? This is mostly down to the concrete base, which is massive and needs to be poured weeks before the crane arrives. The triangulated cross-member structure of the mast gives it more stability and prevents bending. Plus, it’s anchored and bolted to the ground. ## How tall is an average crane? A typical tower crane has the following specifications: Maximum unsupported height – 265 feet (80 meters) The crane can have a total height much greater than 265 feet if it is tied into the building as the building rises around the crane. Maximum reach – 230 feet (70 meters) ## How do tower cranes increase height? To rise to its maximum height, the crane grows itself one mast section at a time! The crew uses a top climber or climbing frame that fits between the slewing unit and the top of the mast. … Once bolted in place, the crane is 20 feet taller! ## What is the tallest crane? “Big Carl” is the biggest land-based crane in the world. Capable of lifting 5,000 tonnes at a radius of 40m. 250m tall in its tallest configuration. Supported by 52 counterweight containers – weighing 100 tonnes each. ## What is the tallest flying bird? The Sarus Crane is the tallest flying bird in the world! ## What is the largest crane in the United States? The largest crane on the ACT100 is Mammoet’s 5,000-ton capacity PTC 200 DS. Other heavy lift cranes on the ACT100 are the 3,000-ton capacity Lampson LTL-3000, the 2,535-ton capacity Manitowoc 31000 owned by Maxim and the 2,500-ton TC-36000 Versa Crane owned by Deep South Crane & Rigging. ## What is the highest paid crane operator? Crane and Tower Operators usually receive an average pay level of Fifty Thousand One Hundred dollars on a yearly basis. Crane and Tower Operators obtain the highest pay in Nevada, where they get job pay of close to \$74180. ## Where do crane operators go to the bathroom? A funnel inside the cab is attached to a tube that drains waste into the portable toilet attached to the side of the crane’s mast. ## How much do high rise crane operators make? Crane Operators Top \$500,000 in Pay, Benefits. Just as cranes tower over building sites, the salaries of the people who run them tower over those of other construction workers. IT IS INTERESTING:  What is a engine crane used for?

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-17

latest

en

https://math.stackexchange.com/questions/2591008/functional-equation-what-function-is-its-inverses-reciprocal

text/html

crawl-data/CC-MAIN-2019-30/segments/1563195532251.99/warc/CC-MAIN-20190724082321-20190724104321-00058.warc.gz

# Functional equation: what function is its inverse's reciprocal? [duplicate] The fact that so many students confuse functional inverse notation $$f^{-1}(x)$$ with multiplicative inverse notation $$[f(x)]^{-1}$$ got me to thinking... does there exist a function whose inverse is its inverse? That is, is there a function $f:\mathbb R_+\mapsto \mathbb R_+$ whose functional inverse is also its multiplicative inverse, so that $$f^{-1}(x)=[f(x)]^{-1}, \space\space\space \forall x\in\mathbb R_+$$ Any ideas? I'll impose the restriction of continuity to deter nasty solutions. ## marked as duplicate by Hans Lundmark, MichaelChirico, Community♦Jan 5 '18 at 0:00 • Well, since $\frac1{f(f(x))}=x$, necessarily domain and range of $f$ cannot contain $0$. At best, then, $f:\Bbb R\setminus\{0\}\to\Bbb R\setminus\{0\}$. – user228113 Jan 4 '18 at 0:19 • @G.Sassatelli But with the requirement of continuity, it can only go from half of the real line to the other half. – Frpzzd Jan 4 '18 at 0:22 • @HansLundmark: This question imposes the restriction of continuity, whereas the other question you link to does not, so this does not seem to be a duplicate. Sammy Black's answer to the other question does consider differentiability, but even that is not (quite) a duplicate. – Rory Daulton Jan 4 '18 at 9:44 • There were also some other similar posts: A function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}(x)=\frac{1}{f(x)}$ or When is $f^{-1}=1/f$? However, the domain is $\mathbb R$ and $\mathbb C$, respectivelly. Found using Approach0. – Martin Sleziak Jan 4 '18 at 12:46 • In fact, despite the question stating that it is about functions on $\mathbb C$, the paper mentioned in this answer might be interesting in connection with functions defined in $(0,\infty)$. (Russell Euler and James Foran. "On Functions Whose Inverse Is Their Reciprocal." Mathematics Magazine Vol. 54, No. 4 (Sep., 1981), pp. 185-189. jstor.org/stable/2689629) – Martin Sleziak Jan 4 '18 at 12:54 No, it is impossible. If $f: (0,\infty) \to (0,\infty)$ is continuous and $f^{-1}$ exists, then $f$ is either increasing or decreasing. If $f$ is increasing, $f^{-1}$ is increasing but $1/f$ is decreasing. If $f$ is decreasing, $f^{-1}$ is decreasing but $1/f$ is increasing. EDIT: However, for $f: \mathbb R \backslash \{0\} \to \mathbb R \backslash \{0\}$ it is possible. Take $$f(x) = \cases{ -x & if x > 0\cr -1/x & if x < 0\cr}$$ • Nicely done! $\space$ – Frpzzd Jan 4 '18 at 0:38 • The linked thread gives a (discontinuous) example $f: (0,\infty) \to (0,\infty)$. – Jeppe Stig Nielsen Jan 4 '18 at 8:57 Call $g=\ln\circ f\circ \exp:\Bbb R\to\Bbb R$. Then, $g^{-1}=\ln\circ f^{-1}\circ \exp=\ln\frac1{f\circ \exp}=-g$. So, we want the homeomorphisms $g:\Bbb R\to\Bbb R$ such that $g^{-1}=-g$. But the homeomorphism $\Bbb R\to\Bbb R$ are strictly monotone continuous functions. And, if $g$ is strictly monotone, $g^{-1}$ must be monotone of the same sign. This is not consistent with $g^{-1}=-g$. Thus, there is no such $g$ and no such $f$.

{"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}

CC-MAIN-2019-30

latest

en

https://www.r-bloggers.com/2012/11/opting-for-shorter-movies-be-aware-u-might-be-cutting-the-entertainment-too/

text/html

crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00568.warc.gz

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. Hello Friends, This time I thought to bring in little more spice and thought of focusing on movies.  I don’t know about you but I am a movie buff. Often on a weekend when I am trying to pick up a movie from my movie repository, which spans to some TBs now, I feel little lost.  Apart from a general rating or a perception, the length of the movie plays a role in the choice, simple reason; the movie needs to be cramped between other demanding priorities. So last Saturday, when I was in between this process, and I was searching for a movie less than 1 hour 30 minutes (There was a hard stop on that) my wife commented but “The short movies are generally not so good”.  I did not pay much heed to that then (Don’t conclude anything from this please), but later on I thought hold on, is that a hypothesis?  Can I do something statistically here?  And here we are. We will talk little bit on correlation, normal distribution etc. I use ‘R’, but it is so simple , we can even use excel sheet do the same. Correlation: This is an indicator whose value is between -1 and 1 and it indicates strength of linear relationship between two variables. Leave the jargon, many cases we relate features.  The typical law of physics like speed and displacement may have a perfect correlation, but those are not the point of interest.  However a point of interest may be is there a relation between say a)      IQ Score of a person and Salary drawn b)      No. of obese people in an area vis-à-vis no. of fast-food centers in the locality c)       No. of Facebook friends , with relationship shelf life d)      No. of hours spent in office and attrition rate for and organization An underlying technicality, I must point out here is both of the variables should follow a normal distribution. Normal Distribution: This is the most common probability distribution function, which is a bell shaped curve, with equal spread in both side of the mean.  Associate to manager alike, you must have heard about normalization and bell curve while you face/do the appraisal.  Most of the random events across disciplines follow normal distribution. The below is an internet image. So I picked up movie information and like any one of us picked it up from IMDB (http://www.imdb.com/) and I put it in a structured form like the below, the ones highlighted below may not be required at this point of time, I kept it just for some future work in mind.  The list was prepared manually; I will keep on hunting for some API and all and would keep you posted on the same. Name Year of Release Rating Duration Small Desc Skyfall 2012 8.1 143 Bond’s loyalty to M is tested as her past comes back to haunt her. As MI6 comes under attack, 007 must track down and destroy the threat, no matter how personal the cost. At this point of time I have taken 183 movies.  I have stored it as a csv file. First thing first, there are various formal ways to test whether it follows a normal distribution, I would just plot histograms and see how this looks like, both the variable seem to follow normal distributions closely. Below are the commands for a quick reference.  What I just adore about R is it’s simplicity, with just so few commands we are done x<-as.matrix(film) # Converting the list to a matrix,  for histogram plotting y<-as.numeric(x[,3]) # Converting the movie rating to a numeric vector y<-as.numeric(x[,4]) # Converting the movie duration to a numeric vector hist(y,col=”green”,border=”black”,xlab=”Duarion”,ylab=”mvfreq”,main=”Mv Duration Distribution”,breaks=7) hist(y,col=”blue”,border=”black”,xlab=”mvRtng”,ylab=”mvfreq”,main=”Mv Rtng Distribution”,breaks=9) cor(y,z) # Calculate Correlation Coefficient between rating and duration Interestingly the correlation turns out to be .48 in this case, which says there is a positive correlation between this two phenomenon and the correlation is not small.  We can set up a hypothesis “ There is no correlation “ and a level of significance and test the hypothesis. However .48 is a high value and I am sure we would reject the hypothesis that there is no correlation. So someway or other the rating goes up with the duration of the movie. I leave it to you for interpretation, but next time you might look at the movie duration for taking a call ! Mr. Directors , it might be a tips for you who knows and may be to me wify is always right. May be all that short is not that sweet. With that I will call it a day, hope you enjoyed reading. I will be coming on with more such Looking forward to your feedbacks and comments

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-33

longest

en

https://il.tradingview.com/script/5QSO13Ib-Flare/

text/html

crawl-data/CC-MAIN-2023-14/segments/1679296943562.70/warc/CC-MAIN-20230320211022-20230321001022-00552.warc.gz

# Flare 🔶 METHODS • Pine Script™ introduces methods (1, 2)! Much kuddos for the developers, Tradingview, and all who has worked on it! • This educational script will show the simplified way of writing built-in methods, not to create a new method. 🔹 Simplified way of writing built-in methods: ``` newArray = array.new<int>() array.unshift(newArray, 1) lin = line.new(na, na, na, na) line.set_xy1(lin, bar_index , close) line.set_xy2(lin, bar_index + 10, close) label newLabel = label.new(bar_index, high) if barstate.islast label.delete(newLabel)``` · We now can write it like this: ``` newArray = array.new<int>() newArray.unshift(1) lin = line.new(na, na, na, na) lin.set_xy1(bar_index , close) lin.set_xy2(bar_index + 10, close) label newLabel = label.new(bar_index, high) if barstate.islast newLabel.delete()``` —————————————————————————————————————————————————————————— · When using [1] sometimes brackets are necessary: ``` label lab = label.new(bar_index, high) if barstate.islast label.set_color(lab, color.red) label.delete(lab[1])``` · -> ``` label lab = label.new(bar_index, high) if barstate.islast lab.set_color(color.red) (lab[1]).delete() // lab[1].delete() doesn't compile at the moment``` —————————————————————————————————————————————————————————— 🔶 OVERVIEW OF SCRIPT • The basic principles are: · Find 1 point (close) x bars back from current bar (settings: 'x close back'). · Create a 'Flare' shaped object from that point to current bar or further (dependable of "Width of Flare"). · Calculate where current close is located versus the Flare lines. · On that bases, change colour and draw plotshapes. · Below bar if current close is located in the upper part of the Flare · Above bar if current close is located in the lower part of the Flare · Above & Below if located in the middle part of the Flare -> Above & Below colours has 3 different colours (adjustable), dependable on the position 🔶 EXAMPLES · Neutral zone: · Light Bullish zone: · Bullish zone: · Very Bullish / Overbought zone: · Light Bearish zone: · Bearish zone: · Very Bearish / Oversold zone: 🔶 TECHNIQUES 🔹 I. Make a User Defined Type (UDT) Flare, with: · 5x linefill - array of linefill · int iDir, which captures the direction (current location of close in Flare) · color cCol, this is a colour variable in relation to the direction. 🔹 II. Different functions will add a new Flare object, and update the values on each bar. · Explanation of each function can be found in the script. 🔶 EXTRA's · The input.color() is located in the function set_flare_B(flare obj) · Best to put the inputs at the beginning of the script, I included this alternative just to show it is possible (but mostly not ideal) · Background colour (settings: Bgcolor) can be enabled for better visibility of colours - We cannot control our emotions, but we can control our keyboard - סקריפט קוד פתוח ברוח TradingView אמיתית, מחבר הסקריפט הזה פרסם אותו בקוד פתוח, כך שסוחרים יכולים להבין ולאמת אותו. כל הכבוד למחבר! אתה יכול להשתמש בו בחינם, אך שימוש חוזר בקוד זה בפרסום כפוף לכללי הבית. אתה יכול להכניס אותו למועדפים כדי להשתמש בו בגרף. כתב ויתור המידע והפרסומים אינם אמורים להיות, ואינם מהווים, עצות פיננסיות, השקעות, מסחר או סוגים אחרים של עצות או המלצות שסופקו או מאושרים על ידי TradingView. קרא עוד בתנאים וההגבלות. רוצה להשתמש בסקריפ זה בגרף?

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-14

latest

en

https://funsor.pyro.ai/en/stable/_modules/pyro/optim/clipped_adam.html

text/html

crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00156.warc.gz

```# Copyright (c) 2017-2019 Uber Technologies, Inc. import math import torch from torch.optim.optimizer import Optimizer """ :param params: iterable of parameters to optimize or dicts defining parameter groups :param lr: learning rate (default: 1e-3) :param Tuple betas: coefficients used for computing running averages of gradient and its square (default: (0.9, 0.999)) :param eps: term added to the denominator to improve numerical stability (default: 1e-8) :param weight_decay: weight decay (L2 penalty) (default: 0) :param clip_norm: magnitude of norm to which gradients are clipped (default: 10.0) :param lrd: rate at which learning rate decays (default: 1.0) to include gradient clipping and learning rate decay. Reference `A Method for Stochastic Optimization`, Diederik P. Kingma, Jimmy Ba https://arxiv.org/abs/1412.6980 """ def __init__(self, params, lr=1e-3, betas=(0.9, 0.999), eps=1e-8, weight_decay=0, clip_norm=10.0, lrd=1.0): defaults = dict(lr=lr, betas=betas, eps=eps, weight_decay=weight_decay, clip_norm=clip_norm, lrd=lrd) super().__init__(params, defaults) def step(self, closure=None): """ :param closure: An optional closure that reevaluates the model and returns the loss. Performs a single optimization step. """ loss = None if closure is not None: loss = closure() for group in self.param_groups: group['lr'] *= group['lrd'] for p in group['params']: continue state = self.state[p] # State initialization if len(state) == 0: state['step'] = 0 # Exponential moving average of gradient values # Exponential moving average of squared gradient values exp_avg, exp_avg_sq = state['exp_avg'], state['exp_avg_sq'] beta1, beta2 = group['betas'] state['step'] += 1 if group['weight_decay'] != 0: # Decay the first and second moment running average coefficient

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-04

latest

en

http://gmatclub.com/forum/the-set-of-propositions-which-was-discussed-by-the-panel-149848.html

text/html

crawl-data/CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00021-ip-10-147-4-33.ec2.internal.warc.gz

Find all School-related info fast with the new School-Specific MBA Forum It is currently 16 Apr 2014, 23:32 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # The set of propositions which was discussed by the panel Author Message TAGS: Manager Joined: 06 Jun 2012 Posts: 150 Followers: 0 Kudos [?]: 9 [0], given: 37 The set of propositions which was discussed by the panel [#permalink]  26 Mar 2013, 03:15 00:00 Difficulty: 45% (medium) Question Stats: 30% (01:52) correct 69% (00:33) wrong based on 72 sessions The set of propositions which was discussed by the panel have been published in the society journal. A. which was discussed by the panel have been B. which were discussed by the panel have been C. that was discussed by the panel has been D. which were discussed by the panel has been E. which was discussed, by the panel, has been I am confused with the OA [Reveal] Spoiler: OA _________________ Please give Kudos if you like the post Manager Joined: 05 Sep 2010 Posts: 187 Followers: 0 Kudos [?]: 19 [0], given: 8 Re: The set of propositions which was discussed by the panel hav [#permalink]  26 Mar 2013, 03:35 i disagree with the OA .C is prefect here .we need "that" to introduce essential clause ."which" is total wrong here Manager Joined: 05 Sep 2010 Posts: 187 Followers: 0 Kudos [?]: 19 [0], given: 8 Re: The set of propositions which was discussed by the panel hav [#permalink]  26 Mar 2013, 03:36 what is the source of this question ? is it gmat prep ? Intern Joined: 28 Nov 2012 Posts: 48 Followers: 0 Kudos [?]: 1 [0], given: 3 Re: The set of propositions which was discussed by the panel hav [#permalink]  26 Mar 2013, 07:11 IMO D the error lines up with subject verb agreement....bold matches with bold, underline matches with underline. Which is a adjective clause which can be taken out 'the set has been published' but within the adj clause propositions needs to match with were b/c they are plural. If my logic is off please let me know. The set of propositions which were discussed by the panel has been published in the society journal. Verbal Expert Status: worked for Kaplan's associates, but now on my own, free and flying Joined: 19 Feb 2007 Posts: 2265 Location: India WE: Education (Education) Followers: 238 Kudos [?]: 1167 [0], given: 244 Re: The set of propositions which was discussed by the panel hav [#permalink]  26 Mar 2013, 09:26 Expert's post Considering the GMAT’s view that any choice selected solely on the basis of which or that is unlikely to be correct because of the controversy surrounding them, let’s now rather look beyond the pronoun for the time –being. But we take the meaning route to unknot this conundrum. 1. Is the word ‘propositions’ the referent for the pronoun ‘which or that’? No; it is the set; of propositions is just a middleman or an additional info that embellishes the subject namely ‘the set’. For that matter, it doesn’t even matter whether the set consists of propositions, or questions or doubts, or what that may be. The subject is certainly the singular set. Secondly, it is also clear that what was discussed was also published. You cannot use singular verb in one case and plural verb in another case for the same subject. Both have to be singular in this case. That way, only C and E have consistent SV number agreement. Of this, the unnecessary offsetting of the phrase, “by the panel” in E, renders it half – baked. So it is C all the way Manager Joined: 27 Oct 2009 Posts: 72 Followers: 0 Kudos [?]: 10 [0], given: 2 Re: The set of propositions which was discussed by the panel hav [#permalink]  27 Mar 2013, 09:07 Agree on Daagh's explanations. C for me too. _________________ A bend in the road is not the end of the road unless you fail to take a turn..... Senior Manager Joined: 23 Mar 2011 Posts: 474 Location: India GPA: 2.5 WE: Operations (Hospitality and Tourism) Followers: 11 Kudos [?]: 83 [0], given: 59 Re: The set of propositions which was discussed by the panel hav [#permalink]  27 Mar 2013, 10:44 daagh wrote: Considering the GMAT’s view that any choice selected solely on the basis of which or that is unlikely to be correct because of the controversy surrounding them, let’s now rather look beyond the pronoun for the time –being. But we take the meaning route to unknot this conundrum. 1. Is the word ‘propositions’ the referent for the pronoun ‘which or that’? No; it is the set; of propositions is just a middleman or an additional info that embellishes the subject namely ‘the set’. For that matter, it doesn’t even matter whether the set consists of propositions, or questions or doubts, or what that may be. The subject is certainly the singular set. Secondly, it is also clear that what was discussed was also published. You cannot use singular verb in one case and plural verb in another case for the same subject. Both have to be singular in this case. That way, only C and E have consistent SV number agreement. Of this, the unnecessary offsetting of the phrase, “by the panel” in E, renders it half – baked. So it is C all the way Daagh, the pronoun is simply giving more information about the subject, which is a non essential modifier in real sense. As you also mentioned, it is irrelevant if the set is of prepositions or not, what matters is that the set was published. In that case why not consider E, which I understand has an additional punctuation but in what way is this incorrect. As I am not too convinced with C as the answer choice. _________________ "When the going gets tough, the tough gets going!" Bring ON SOME KUDOS MATES+++ ----------------------------- My GMAT journey begins: my-gmat-journey-begins-122251.html Verbal Expert Status: worked for Kaplan's associates, but now on my own, free and flying Joined: 19 Feb 2007 Posts: 2265 Location: India WE: Education (Education) Followers: 238 Kudos [?]: 1167 [0], given: 244 Re: The set of propositions which was discussed by the panel hav [#permalink]  27 Mar 2013, 16:45 Expert's post That the panel discussed is very essential to the sentence. In E, it is rendered inessential by offsetting. Remove the offset part and you might feel the bareness of the clause. Intern Joined: 14 Jan 2013 Posts: 13 Followers: 2 Kudos [?]: 1 [0], given: 0 Re: The set of propositions which was discussed by the panel hav [#permalink]  29 Mar 2013, 02:19 There would be a comma before "which" i think,than only Option D would be right i disagree with the OA .C is prefect here .we need "that" to introduce essential clause ."which" is total wrong here Manager Joined: 31 May 2012 Posts: 145 Followers: 1 Kudos [?]: 43 [0], given: 58 Re: The set of propositions which was discussed by the panel hav [#permalink]  30 Mar 2013, 05:15 The set of propositions which was discussed by the panel have been published in the society journal. A. which was discussed by the panel have been. The set has been... B. which were discussed by the panel have been C. that was discussed by the panel has been. Perfect match D. which were discussed by the panel has been E. which was discussed, by the panel, has been. For this to be correct, there should have been a comma before which and after panel. Which vs That.. (from grammar girl) A restrictive clause is just part of a sentence that you can't get rid of because it specifically restricts some other part of the sentence. Here's an example: Gems that sparkle often elicit forgiveness. The words that sparkle restrict the kind of gems you're talking about. Without them, the meaning of the sentence would change. Without them, you'd be saying that all gems elicit forgiveness, not just the gems that sparkle. (And note that you don't need commas around the words that sparkle.) Nonrestrictive Clause--Which A nonrestrictive clause is something that can be left off without changing the meaning of the sentence. You can think of a nonrestrictive clause as simply additional information. Here's an example: Diamonds, which are expensive, often elicit forgiveness. Alas, in the world, diamonds are always expensive, so leaving out the words which are expensive doesn't change the meaning of the sentence. Manager Joined: 20 Jul 2012 Posts: 122 Location: India Followers: 1 Kudos [?]: 22 [0], given: 197 Re: The set of propositions which was discussed by the panel [#permalink]  29 Oct 2013, 07:57 summer101 wrote: The set of propositions which was discussed by the panel have been published in the society journal. A. which was discussed by the panel have been B. which were discussed by the panel have been C. that was discussed by the panel has been D. which were discussed by the panel has been E. which was discussed, by the panel, has been I am confused with the OA I disgaree with OA. Reasons- 1) Before "Which" there is always a "comma" 2) "The set of prepositions" is Singular( B and D are out) 3) Since "the set of prepositions" is singular so A is out ...As it cannot be "have been"...It should be "Has been" 4)Left with (c) and (E).. (E) cannot be the answer for two reasons- there is always a comma before which...so that will be prefereed... secondly" ,by the panel'" structure is wrong So (C) should be the answer... Re: The set of propositions which was discussed by the panel   [#permalink] 29 Oct 2013, 07:57 Similar topics Replies Last post Similar Topics: The set of propositions which was discussed by the panel 18 23 May 2005, 11:05 Proposition 13 6 24 Jan 2006, 14:05 Anyone going for the admission panel discussion - Aug 14th? 26 05 Aug 2007, 21:48 The Gatherers, a lecture series which discusses life among 9 22 Aug 2007, 11:59 Panel Discussion on Post-MBA Career Paths in Bangalore 1 07 Mar 2011, 01:32 Display posts from previous: Sort by

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2014-15

latest

en

https://www.jiskha.com/display.cgi?id=1317061510

text/html

crawl-data/CC-MAIN-2017-30/segments/1500549425082.56/warc/CC-MAIN-20170725062346-20170725082346-00344.warc.gz

# Math 116a posted by . Amy bought a pair of running shoes she paid \$66.09 during a 20% off sale, what is the regular price of the shoes? Would I take 66.09 divide by 20%? • Math 116a - No. If you did this: 66.09/0.2 = 330.45. Certainly the original price was not over \$300.00 • Math 116a - would it be 79.30 66.09+20% • Math 116a - No. That isn't the answer, but at least it's close. 100% - 20% = 80% Let x = the original price. 0.8x = 66.09 x = 66.09/0.8 x = • Math 116a - ok I get it the regular price is 82.61 ThankYou • Math 116a - Right. And you're welcome. • Math 116a - you're awsome

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-30

longest

en

http://www.cnczone.com/forums/general_metalwork_discussion/130314-chamfer_calculation_2_a.html

text/html

crawl-data/CC-MAIN-2013-48/segments/1387345776447/warc/CC-MAIN-20131218054936-00017-ip-10-33-133-15.ec2.internal.warc.gz

CNCzone Network: 1. ## Chamfer Calculation 2 Hi, When I need to chamfer a bolt hole, I use the following formula: (Chamfer Diameter - Bolt Hole Diamater)/2/tan(Chamfer Degree/2). I add this amount to the distance my chamfer tool travels into the bolt hole before scraping the sides. That becomes my negative Z depth. But suppose I want to use a ball nose end mill to do the chamfer. Does anyone know a formula for calculating chamfer depth when using a spherical raidus like a ball nose end mill?????? Thanks 2. Give this a try. (Ball Mill Dia/2) - (Sine(Chamfer Deg./2) * (Ball Mill Dia/2)) 3. Only problem is I'm not given any information about the chamfer degree in this case. All that's typically called out on the print is: CTSK. 14mm SPERICAL RADIUS x .945 +30/-0 DIA. So I just throw in a 28mm ball nose end mill and adjust it deeper untill I get the diameter I need. I could ask around and maybe get some information about the angle of the tool. Maybe it is 45 degrees or something. Someone at work could probably point me in the right direction there, if thats the way I'd need to go. 4. This is what I make from the specs. If the angle is 90deg (half 45deg) the depth should be 1 + 4.1005 = 5.1005. 5. Thanks for that. Its helpful to see it layed out that way. I'll have to play with it for a bit, and see what I come up with. 6. I'd really like to try this again, and hope that some others are up for it. Its an interesting problem to me, but I'm not making much headway. Notice that the bolt holes on the plate below are chamfered. I cut the chamfers with a 14mm spherical radius ball nose end mill. For the purposes of discussion, however, lets talk in inches. I used a 0.5512" radius, or 1.1024" diameter, ball nose end mill. Here are the specs: You can see that the 7 bolt holes have a .734" diameter, and the diameter of the chamfers is .945". Only the spherical radius is given in metric, which I'm just going to convert to inches for the moment. I WOULD LIKE: a formula that will allow me to calculate the Z depth for a G-code canned cycle when cutting a chamfer of any diameter with any spherical radius tool. I can do this easily with a standard 60, 82.5, 90 degree chamfer tool. You all know the formula: (Chamfer Dia. - Bolt Hole Dia.)/2/tan(degrees/2). Then you take this value and add it to the Z depth value at which your chamfer tool scrapes the bolt hole. MY PROBLEM: I can't seem to get my formula for a spherical cutter using simple trig. As the 1.1024 ball nose end mill approaches its final depth, any known angles are constantly changing. In the image below, the 1.1024 diameter ball is just scraping the edges of the .734 bolt hole. The adjacent and opposite angles to the perpendicular depth of cut line are 69.1274 degrees and 20.8726 degrees, respectively. When the 1.1024 diameter ball reaches its final depth (i.e., the chamfer diameter is at .945), the adjacent and opposite angles to the perpendicular depth of cut line are 60.4970 degrees and 29.5030 degrees, respectively. So SOHCAHTOA won't work. What else is there, short of firing up GibbsCAM as I've done above and drawing geometry. I want a formula I can plug a chamfer diameter, cutter diameter, bolt hole, etc. into and work at the machine. Any thoughts greatly appreciated. Thanks for plowing through this if you took the time. And if anyone can think of a more appropriate forum to post this in, I'm all for it. 7. Try: (1.1024/2) - (Sqr( (1.1024/2)^2 - (0.945/2)^2)) 8. I get 0.2675, which I'd have to say appears to be right. I'm showing a final depth in my second drawing of 0.2674. But how does your formula work? It looks kinda' like you're working Pythagorean theorem backwards, or maybe inside out... Are you by any chance able to draw what you're doing?? Thanks 9. Actually looks llike I get .2674. (.267359922) 10. I see. The radius of the cutter is the hypotenuse of a right triangle of which the base = .4725 and the side = a. So... a^2 + .4725^2 = .5512^2 a^2 = .5512^2 - .4725^2 a = Sqr(.5512^2 - .4725^2) a = .2838 Then subtract .2838 (the portion of the radius above Z0.) from the full radius (.5512) = .2674, the portion of the radius below Z0. Brilliant. Thanks a million. Got a good feeling about it. Should work. 11. 1.1024/2 = 0.5512 0.945/2 = 0.4725 Sqr(0.5512^2 - 0.4725^2) = 0.2838 0.5512 - 0.2838 = 0.2674 Page 1 of 2 12 Last

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2013-48

latest

en

https://forums.studentdoctor.net/threads/electric-field-between-capacitors.712562/

text/html

crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00169.warc.gz

# electric field between capacitors #### puffylover 10+ Year Member 7+ Year Member aamc 4 problem 41 on the ps section which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount. the circuit consists of a resistor and capacitor (resistor is a at a higher potential/before the capacitor) answer: increasing d (distance between plates) by a factor of 2 explanation: for a fixed voltage between the cathode and anode, the electric field is inversely proportional to the distance between them. increasing the circuit resistance for a fixed current will decrease the electric field, but not by as much as does the length change E= (V-IR)/L so my question is: how come the current is considered constant. doesn't the current go down if the resistance of the circuit goes up? #### lorenzomicron aamc 4 problem 41 on the ps section which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount. the circuit consists of a resistor and capacitor (resistor is a at a higher potential/before the capacitor) answer: increasing d (distance between plates) by a factor of 2 explanation: for a fixed voltage between the cathode and anode, the electric field is inversely proportional to the distance between them. increasing the circuit resistance for a fixed current will decrease the electric field, but not by as much as does the length change E= (V-IR)/L so my question is: how come the current is considered constant. doesn't the current go down if the resistance of the circuit goes up? yes, the current will go down as the resistance goes up if the voltage is fixed. However, if the current is fixed, then other things can happen. however, in the capacitor case, the electrical field is changed most by the distance between the plates because E is supposed to be constant and if the voltage does not increase between the plates, increasing the distance will largely decrease the field to conserve the voltage. also...i am not clear as to what your circuit looks like....clarify? This thread is more than 11 years old. Your message may be considered spam for the following reasons: 4. It is very likely that it does not need any further discussion and thus bumping it serves no purpose. 5. Your message is mostly quotes or spoilers. D Replies 2 Views 397 D Replies 2 Views 4K Replies 1 Views 2K Replies 5 Views 6K D Replies 7 Views 1K

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-17

longest

en

https://www.physicsforums.com/threads/diving-board.78266/

text/html

crawl-data/CC-MAIN-2017-47/segments/1510934805242.68/warc/CC-MAIN-20171119004302-20171119024302-00594.warc.gz

# Diving board 1. Jun 7, 2005 ### MAPgirl23 A diving board of length L = 3.00 m is supported at a point a distance 1.00 m from the end, and a diver weighing 490 N stands at the free end. The diving board is of uniform cross section and weighs 295 N. Find the force at the support point. ** Now I know net torque = 0 Suppose F1 is the force applied at the support point and F2 is the force at the end that is held down. (F1*x_f1) - (w_diver * L) - (w_board * x_point) = 0 x_f1 = 3.0 m * {490 N/(490N + 295N)} = 1.87 m ? --> F1(1.87 m) - (490 N)(3.0 m) - (295 N)(1.0 m) = 0 solve for F1 F1 = {(490N)(3.0 m) + (295N)(1.0 m)}/1.87 m = 944 N which is wrong how do I find x_f1? is my formula for F1 correct? 2. Jun 7, 2005 ### learningphysics Ok. So you're going to find the torque about the end of the board that is held fast right? That's fine. You're getting the torque about the end that is held fast. Not sure why you did this. The length of the board is 3m. The support point is 1m from the free end. So it is 2m from the unfree end. x_f1=2m (draw a picture if you haven't) Also the weight of the board acts at the center of the board. So that is half the length of the board from the end. so x_point=1.5m Now you should be able to solve for F1. 3. Jun 7, 2005 ### Staff: Mentor Only the original poster knows for sure, but I would have guessed that the support point was 1m from the fixed end. 4. Jun 7, 2005 ### learningphysics Ah... you're probably right. The OP used "end" for the support point then "free end" later. That would also make more sense I believe for a diving board. 5. Jun 7, 2005 ### MAPgirl23 but if I make x_f1 = 1 m and F1 = {(490N)(3.0 m) + (295N)(1.0 m)}/1 m = 1765 N the answer is wrong. 6. Jun 7, 2005 ### Staff: Mentor The weight of the board acts at its middle, not at x = 1.0m. 7. Jun 7, 2005 ### MAPgirl23 so it acts at 1.5 m (L/2 = 3.0 m/2 = 1.5 m) is x_f1 = 1.5 therefore: F1 = {(490N)(3.0 m) + (295N)(1.5 m)}/1 m = 1912.5 m 8. Jun 7, 2005 ### MAPgirl23 thank you for all of your help.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-47

longest

en

http://nedrilad.com/Tutorial/topic-73/Real-time-3D-Character-Animation-with-Visual-C-26.html

text/html

crawl-data/CC-MAIN-2017-39/segments/1505818691830.33/warc/CC-MAIN-20170925130433-20170925150433-00433.warc.gz

Game Development Reference In-Depth Information The purpose of all this vector manipulation is that, given three vertices that are distinct and define a polygon, we can find a vector that extends at right angles from this polygon. Given vertices A, B and C we can create two vectors. N is the vector from B to A and M is the vector from B to C. Simply subtracting B from A and B from C respectively creates these vectors. Now the cross product of the vectors N and M is the normal of the polygon. It is usual to scale this normal to unit length. Dividing each of the terms by the magnitude of the vector achieves this. Rotating the box There are many options available when rotating a 3D representation of an object; we will consider the three principal ones. The first option we will look at uses Euler angles. Euler angles When considering this representation it is useful to imagine an aeroplane flying through the sky. Its direction is given by its heading. The slope of the flight path is described using an angle we shall call pitch and the orientation of each wing can be described using another angle which we shall call bank. The orientation can be com- pletely given using these three angles. Heading gives the rotation about the y -axis, pitch gives rotation about the x -axis and bank gives rotation To describe the orientation of an object we store an angle for the heading, the pitch and the bank. Assuming that the rotation occurs about the point [0, 0, 0] as the box is modelled then heading is given from the 3 × Figure 1.5 Euler angle rotation. 3 matrix: cos( h ) 0 sin( h ) H = 0 1 0 -sin( h ) 0 cos( h )

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-39

latest

en

https://www.numerade.com/questions/show-that-displaystyle-int_0infty-e-x2-dx-displaystyle-int_01-sqrt-ln-y-dy-by-interpreting-the-integ/

text/html

crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00351.warc.gz

💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! WZ # Show that $\displaystyle \int_0^\infty e^{-x^2}\ dx = \displaystyle \int_0^1 \sqrt{-\ln y}\ dy$ by interpreting the integrals as areas. ## $\int_{0}^{\infty} e^{-x^{2}} d x=\int_{0}^{1} \sqrt{-\ln y} d y$ #### Topics Integration Techniques ### Discussion You must be signed in to discuss. Lectures Join Bootcamp ### Video Transcript The problem is show that the integral from there to infinity each connective at squared the ax is they go to integral from their toe. One beautiful negative. How and why you? Why interpreting into girls as areas. Now look at the graph here. Area under the grow up under the bunch attitudes and necktie Black Square from there are two Infinity is this part. Is this real Andi the integral from there to infinity, into negative X squared the ax justice area of this part This computing as follows off a worry. Small Yeah, eggs. The hide is because Teo eat connective ex Claire So area can become computed as integral from their all to infinity eatin negative X squared DX. We have another way to compute this area. Just look, We're a small wine here, Andi. Height is Rico too. Things Why is he going to eat to Matthew X squared? So for worry small. Why? How and why is equal to negative X square. So axe is equal to negative callin wine until this one. There's a hide. It's just beautiful. Negative on wine. Now we can compute the area as a integral from zero to one on DH e y times the hide. This is really tough. Negative on why this is our source area off this region. So is too integral. Lt's our vehicle. WZ #### Topics Integration Techniques Lectures Join Bootcamp

{"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}

CC-MAIN-2021-39

longest

en

https://www.jiskha.com/display.cgi?id=1271952480

text/html

crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00632.warc.gz

# math posted by . Is this equation correct: (x y)^3 = x y^3 ? Explain why. If it is correct, how can you prove this? If it is incorrect, what should be done to make it right? • math - (xy)^3 = x * x * x * y * y * y xy^3 = x * y * y * y In the first, both terms are cubed. In the second, only the last term is cubed. ## Similar Questions 1. ### grammar The sentences below are based on Virginia Woolf's "The Duchess and the Jeweller." Circle the pronouns that are in the incorrect case and write the correct pronoun on the line. Write Correct if all pronouns have been used correctly. … 2. ### College Math : Linear Algebra Suppose A is a square matrix satisfying the equation A^3 - 2I = 0. Prove that A and (A - I) respectively are invertible. (the hint is to find an explicit equation for A^-1. To prove A is invertible, this is what I did although I don't … 3. ### Math 117 Is this equation correct: (3 • x • y)4 = 81 • x • y ? 4. ### algebra Is this equation correct: (x y)3 = x y3 ? 5. ### Math117 Is this equation correct: (x y)3 = x y3 ? 6. ### math answer check x is a binomial random variable. (Give your answers correct to three decimal places.) (a) Calculate the probability of x for: n = 1, x = 0, p = 0.15 P(x) = Changed: Your submitted answer was incorrect. 0.15 x 0 = 0 . (b) Calculate … 7. ### Math check and question x is a binomial random variable. (Give your answers correct to three decimal places.) (a) Calculate the probability of x for: n = 1, x = 0, p = 0.15 P(x) = Correct: Your answer is correct. . (0.85) (b) Calculate the probability of … 8. ### Elementary Satistics Test the following function to determine whether it is a probability function. If it is not, then make it into a probability function. (Give your answers correct to one decimal place.) R(x) = 0.2, for x = 0, 1, 2, 3, 4 (a) List the … 9. ### Math if f(x) = 2x^2+4 and g(x)=x-3 which number satisfies f(x)= (f o g)(x) A) 3/2 B) 3/4 C) 5 D) 4 I am positive that answer A is correct, i just cant get the work to prove it, i can disprove all other answers and my teacher has assured … 10. ### Spanish Grammatically correct? Mi hermana es muy trabajadora? More Similar Questions

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-34

latest

en

http://www.mathforyou.net/en/online/equation/sle/subst/

text/html

crawl-data/CC-MAIN-2019-35/segments/1566027312128.3/warc/CC-MAIN-20190817102624-20190817124624-00292.warc.gz

# Solve system of linear equations by substitution method online The most simple method to solve system of linear algebraic equations (SLE) is the substitution method. Consider it more detailed, suppose we have SLE of the form: We need to solve it, i.e. to find such the values of the variables x1, x2, which convert the initial SLE to the correct identity. The substitution method consists of the following steps: 1. Solve first equation of the SLE with respect to the x1: 2. Substitute obtained identity for x1 variable into the second equation of the SLE: 3. Simplify second equation of the SLE: 4. Solve second equation of the SLE with respect to the x2: 5. Substitute received expression for x2 into the first equation of SLE: 6. Simplify first equation of the SLE: This online calculator solves SLE by the substitution method with step by step solution free of charge. The SLE's coefficients can be the numbers, fractions or parameters. To use the calculator, one need to input the SLE in its natural form and choose the variables to solve. Solve slae by substitution method ,, allow automatic SLE variables detection. System of linear algebraic equations you want to solve:

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-35

latest

en

http://www.win-vector.com/blog/2014/07/frequenstist-inference-only-seems-easy/

text/html

crawl-data/CC-MAIN-2019-09/segments/1550247499009.48/warc/CC-MAIN-20190221031117-20190221053117-00207.warc.gz

Posted on Categories Opinion, Statistics, Tutorials # Frequentist inference only seems easy Two of the most common methods of statistical inference are frequentism and Bayesianism (see Bayesian and Frequentist Approaches: Ask the Right Question for some good discussion). In both cases we are attempting to perform reliable inference of unknown quantities from related observations. And in both cases inference is made possible by introducing and reasoning over well-behaved distributions of values. As a first example, consider the problem of trying to estimate the speed of light from a series of experiments. In this situation the frequentist method quietly does some heavy philosophical lifting before you even start work. Under the frequentist interpretation since the speed of light is thought to have a single value it does not make sense to model it as having a prior distribution of possible values over any non-trivial range. To get the ability to infer, frequentist philosophy considers the act of measurement repeatable and introduces very subtle concepts such as confidence intervals. The frequentist statement that a series of experiments places the speed of light in vacuum at 300,000,000 meters a second plus or minus 1,000,000 meters a second with 95% confidence does not mean there is a 95% chance that the actual speed of light is in the interval 299,000,000 to 301,000,000 (the common incorrect recollection of what a confidence interval is). It means if the procedure that generated the interval were repeated on new data, then 95% of the time the speed of light would be in the interval produced: which may not be the interval we are looking at right now. Frequentist procedures are typically easy on the practitioner (all of the heavy philosophic work has already been done) and result in simple procedures and calculations (through years of optimization of practice). Bayesian procedures on the other hand are philosophically much simpler, but require much more from the user (production and acceptance of priors). The Bayesian philosophy is: given a generative model, a complete prior distribution (detailed probabilities of the unknown value posited before looking at the current experimental data) of the quantity to be estimated, and observations: then inference is just a matter of calculating the complete posterior distribution of the quantity to be estimated (by correct application of Bayes’ Law). Supply a bad model or bad prior beliefs on possible values of the speed of light and you get bad results (and it is your fault, not the methodology’s fault). The Bayesian method seems to ask more, but you have to remember it is trying to supply more (complete posterior distribution, versus subjunctive confidence intervals). In this article we are going to work a simple (but important) problem where (for once) the Bayesian calculations are in fact easier than the frequentist ones. Consider estimating from observation the odds that a coin-flip comes out heads (as shown below). The coin can also show a tails (as shown below). Tails outcome. This might be a fair coin, that when tossed properly can be argued to have heads/tails probabilities very close to 50/50. Or the heads/tails outcome could in fact be implemented by some other process with some other probability `p` of coming up heads. Suppose we flip the coin 100 times and record heads 54 times. In this case the frequentist procedure is to generate a point-estimate of the unknown `p` as `pest = 54/100 = 0.54`. That is we estimate `p` to be the relative frequency we actually empirically observed. Stop and consider: how do we know this is the right frequentist estimate? Beyond being told to use it, what principles lead us to this estimate? It may seem obvious in this case, but in probability mere obviousness often leads to contradictions and paradox. What criteria can be used to derive this estimate in a principled manner? Gelman, Carlin, Stern, Dunson, Vehtari, Rubin “Bayesian Data Analysis” 3rd Edition p. 92 states that frequentist estimates are designed to be consistent (as the sample size increases they converge to the unknown value), efficient (they tend to minimize loss or expected square-error), or even have asymptotic unbiasedness (the difference in the estimate from the true value converges to zero as the experiment size increases, even when re-scaled by the shrinking standard error of the estimate). Because some of the estimators we will work with are biased we are going to use expected square-error as our measure of error. This is the expected value of the square of the distance of our estimate from the unknown true value, and not the variance (which is the expected value of the square of the distance of the estimator from its own mean). Frequentists also commonly insist on fully unbiased procedures (which is what we will discuss) here. In this case an unbiased procedure is a function `f(nHeads,nFlips)` that given the sufficient statistics of the experiment (the number of heads and the total number of flips) returns an estimate for the unknown probability. The frequentist philosophy assumes the unknown probability `p` is fixed and the observed number of heads might vary as we repeat the coin-flip experiment again and again. To confirm a frequentist procedure to estimate `p` from 100 flips is unbiased, we must check that the entire family of possible estimates `f(0,100), f(1,100), ... f(100,100)` together represent an panel of estimates that are simultaneously unbiased no matter what the unknown true value of `p` is. That is: the following bias check equation must hold for any `p` in the range `[0,1]`: Equation family 1: Bias checks (one `f(h,n)` variable for every possible outcome `h`, one equation for every possible `p`). Some combinatorics or probability theory tells us `P(h|n,p) = (n choose h) p^h (1-p)^(n-h)`. We can choose to treat the sequence `f(0,nFlips),f(1,nFlips), ... f(nFlips,nFlips)` either as a set of pre-made estimates (to be checked) or as a set of variables (to be solved for). It turns out there is a solution that satisfies all of the equations simultaneously: `f(h,n) = h/n`. This fact is just a matter of checking that the expected value of the number of heads is `p` times the number of flips. And this is the only unbiased solution. The set of check equations we can generate for various `p` has rank `nFlips+1` (when we include check equations from at least `nFlips+1` different values of `p`, this follows as the check equations behave a lot like the moment curve). We will work a concrete example of the family 1 bias checks a bit later (which should make seeing the content of the chekcs a bit easier). The pre-packaged frequentist estimation procedure is easy: write down the empirically observed frequency as your estimate. But the derivation should now seem a bit scary (submit a panel of `nFlips+1` simultaneous estimates and confirm they simultaneously obey an uncountable family of bias check equations). And this is one of the merits of the frequentist methods- the hard derivational steps don’t have to be reapplied each time you encounter new data, so the end user may not need to know about them. Let’s look at the same data using Bayesian methods. First we are required to supply prior beliefs on the possible values for `p`. Most typically we would operationally assume unknown `p` is beta distributed with shape parameters `(1/2,1/2)` (the Jeffreys prior) or shape parameters `(1,1)` (implementing classic Laplace smoothing). I’ll choose to use the Jeffreys prior, and in that case the posterior distribution (what we want to calculate) turns out to be a beta distribution with shape parameters `(54.5,46.5)`. Our complete posterior estimate of probable values of `p` is given by the R plot below: `````` library(ggplot2) d <- data.frame(p=seq(0,1,0.01)) d\$density <- dbeta(d\$p,shape1=54.5,shape2=46.5) ggplot(data=d) + geom_line(aes(x=p,y=density)) sum(d\$p*d\$density)/sum(d\$density) ## [1] 0.539604 ``` ``` The posterior distribution of `p`. And the common Bayesian method if obtaining an estimate of a summary statistics is to just compute the appropriate summary statistic from the estimated posterior distribution. So if we only want a point-estimate for `p` we can use the expected value `54.5/(54.5+46.5) = 0.539604` or the mode (maximum likelihood value) `(54.5-1)/(54.5+46.5-2) = 0.540404` of the posterior beta distribution. But having a complete graph of an estimate of the complete posterior distribution also allows a lot more. For example: from such a graph we can work out a Bayesian credible interval (which has a given chance of containing the unknown true value `p` assuming our generative modeling assumptions and priors were both correct). And this is one of the reasons Bayesians emphasize working with distributions (instead of point-estimates): even though they can require more work to derive and use, they retain more information. Notice the complications of having to completely specify a prior distribution have not been hidden from us. The actual application of Bayes’ law (an explicit convolution or integral relating the prior distribution to the posterior through a data likelihood function) has (thankfully) been hidden by appealing to the theory of conjugate distributions. So the Bayes theory is hiding some pain from us, but significant pain is still leaking through. And this is common: what is commonly called a frequentist analysis is often so quick you almost can’t describe the motivation, and the Bayesian analysis seems like more work. What we want to say is this is not always the case. If there is any significant hidden state, or constraints on the possible values, then the Bayesian calculation becomes in fact easier than a fully derived frequentist calculation. And that is what we will show in our next example. But first let’s cut down confusion by fixing detailed names for a few common inference methods: • Empirical frequency estimate. This is just the procedure of using the empirically observed frequencies as your estimate. This is commonly thought of as “the frequentist estimate.” However, we are going to reserve the term “proper frequentist estimate” for an estimate that most addresses the common frequentist criticisms: bias and loss/square-error. We will also call the empirical frequency estimate the “prescriptive frequentist estimate” as it is a simple “do what you are told” style procedure. • Proper frequentist estimate. As we said, we are going to use this term for the estimate that most addresses the common frequentist criticisms: bias and loss/square-error. We use the traditional frequentist framework: the unknown parameters to be estimated are assumed to be fixed, and probabilities are over variations in possible observations if our measurement procedures were to be repeated. We define this estimate as an unbiased estimate that minimizes expected loss/square-error for arbitrary possible values of the unknown parameters to be estimated. Often the bias check conditions are so restrictive that they completely determine the proper frequentist estimate and cause the proper frequentist estimate to agree with the empirical frequency estimate. • Full generative Bayesian estimate. This is a complete estimate of the entire posterior distribution of values for the unknown parameters to be estimated. This is under the traditional Bayesian framework that the observations are fixed and the unknown parameters to be estimated take on values from a non-trivial prior distribution (that is a distribution that takes on more than one possible value). Under the (very strong) assumptions that we have the correct generative model and the correct prior distribution the estimated posterior is identical to how the unknown parameters are distributed conditioned on the known observations. Thus reasonable summaries built from the full generative Bayesian estimate should be good (without explicitly satisfying conditions such as unbiasedness or minimal loss/square-error). We are avoiding the usual distinction of objective versus subjective interpretation (Bayesian usually being considered subjective if we consider the required priors subjective beliefs). • Bayes point-estimate. This is a less common procedure. A full generative Bayesian estimate is wrapped in a procedure that hides details of the generative model, priors and Bayes inference step. What is returned is single summary of the detailed posterior distribution such as a mean (useful for producing low square-error estimates) or mode (useful for producing maximum likelihood estimates). For our examples the Bayes point-estimate will be a procedure that returns an estimate mean (or probability/rate) using the correct generative model and uniform priors (when there is a preferred problem parameterization, otherwise we suggest looking into invariant ideas like the Jeffreys prior). Our points are going to be: the empirical frequency estimate is very easy, but is not always the proper frequentist estimate. The proper frequentist estimate can be itself cumbersome to derive, and therefore hard to think of as “always being easier than the Bayesian estimate.” And finally one should consider something like the Bayes point-estimate when one does not want to make a complete Bayesian analysis the central emphasis of a given project. We will illustrate these points with a simple (and natural) example. Returning to our coin-flip problem. Suppose we introduce a five sided control die that is set once (and held fixed) before we start our experiments. Then suppose each experiment is a roll of a fair six-sided die and we observe “heads” if the number of pips on the six-sided die is greater than the number (1 through 5) shown on the control die (and otherwise “tails”). The process is strongly stationary in that the probability `p` is a single fixed value of the entire series of experiments. Our imagined apparatus is depicted below. Our apparatus (the 5-sided die is simulated with a 10-sided die labeled 1 through 5 twice). We assume that while we understand the generative mechanics of the generation process, but that we don’t see the details of the actual die rolls. We observe only the reported heads/tails outcomes (as shown below). What is observed. This may seem like a silly estimation game, but it succinctly models a number of important inference situations such as: estimating advertisement conversion rates, estimating health treatment success rates, and so on. We pick a simple formulation so that when we run into difficulties or complications it will be clear that they are essential difficulties (and not avoidable domain issues). Or: if your estimation procedures are not correct on this example, how can you expect them to be correct in more complicated real-world situations? Another good example of this kind of analysis is: Sean R. Eddy “What is Bayesian statistics” Nature Biotechnology, Vol. 22, No. 9, Sept. 2004, pp. 1177-1178. Eddy presented a clever inference problem comparing where pool balls hit a rail relative to a uniform random chalk mark on the rail. Eddy’s problem illustrates the issues of inference when there are important unobserved (or omitted) state variables. Our example is designed to allow further investigation of the both Bayesian and Frequentist inference in the presence of constraints (not quite the same as complete priors). We will consider two important ways the control die could be set: by a single roll before we start observations (essentially embodying the Bayesian generative assumptions), or by a manual selection by an assumed hostile agent (justifying the usual distribution-free frequentist minimax treatment of loss/square-error). An adversary holding the control die at a chosen value. Let’s start with the case where the control die is set before we start measurements by a fair (uniform) roll of the five sided die. Because the control die only has 5 possible states the unknown probability `p` has exactly 5 possible values. In this case we can write down all of the bias check equations for every possible outcome of a one coin-flip simulation. For only one flip observed there are only two possible outcomes: either we see one heads or one tails. So we have two possible outcomes (giving us two variables, as we get one estimate variable per sufficient outcome) and 5 check equations (one for each possible value of `p`). The complete bias check equations are represented by the matrix `a` and vector `b` shown below: ``` > print(freqSystem(6,1)) \$a prob. 0 heads prob. 1 heads check for p=0.166666666666667 0.8333333 0.1666667 check for p=0.333333333333333 0.6666667 0.3333333 check for p=0.5 0.5000000 0.5000000 check for p=0.666666666666667 0.3333333 0.6666667 check for p=0.833333333333333 0.1666667 0.8333333 \$b p check for p=0.166666666666667 0.1666667 check for p=0.333333333333333 0.3333333 check for p=0.5 0.5000000 check for p=0.666666666666667 0.6666667 check for p=0.833333333333333 0.8333333 ``` The above is just the family 1 bias check equations for our particular problem. A vector of estimates `f` is unbiased if and only if `a f - b = 0` (i.e. it obeys the equation family 1 checks). When `a` is full rank (in this case when the number of variables is no more than the number of checks) the bias check equations completely determine the unique unbiased solution (more on this later). So even in this “discrete `p`” situation: for any number of flips less than 5, the bias conditions alone completely determine the unique unbiased estimate. What we are trying to show is that when we move away from the procedure “copy the observed frequency as your estimate” to the more foundational “pick an unbiased family of estimates with minimal expected square-error”, then frequentist reasoning appears a bit more complicated. Let’s continue with a frequentist analysis of this problem (this time in python instead of R, see here for the complete code). The common “everything wrapped in a bow” prescriptive empirical frequency procedure is by far the easiest estimate: ``` # Build the traditional frequentist empirical estimates of # the expected value of the unknown quantity pWin # for each possible observed outcome of number of wins # seen in kFlips trials def empiricalMeansEstimates(nSides,kFlips): return numpy.array([ j/float(kFlips) for j in range(kFlips+1) ]) ``` And if we load this code (and all of its pre-conditions) we get the following estimates of `p` if we observe one coin experiment: ``` >>> printEsts(empiricalMeansEstimates(6,1)) pest for 0 heads 0.0 pest for 1 heads 1.0 ``` Using our bias check equations we can confirm this solution is indeed unbiased: ``` >>> sNK = freqSystem(6,1) >>> printBiasChecks(matMulFlatten(sNK['a'], \ empiricalMeansEstimates(6,1)) - flatten(sNK['b'])) bias for p=0.166666666667 0.0 bias for p=0.333333333333 0.0 bias for p=0.5 0.0 bias for p=0.666666666667 0.0 bias for p=0.833333333333 0.0 ``` And has moderate loss/square-errors: ``` >>> printLosses(losses(6,empiricalMeansEstimates(6,1))) exp. sq error for p= 0.166666666667 0.138888888889 exp. sq error for p= 0.333333333333 0.222222222222 exp. sq error for p= 0.5 0.25 exp. sq error for p= 0.666666666667 0.222222222222 exp. sq error for p= 0.833333333333 0.138888888889 ``` But the solution is kind of icky. Remember, this result was completely determined by the unbiased check conditions. It says if we observe one coin experiment and see tails then the estimate for `p` is zero, if we see heads the estimate for `p` is one. Both of these estimates are well outside the range of possible values for `p`! Recall our heads/tails coin events are assigned “heads” if the number of pips on the 6-sided die exceeds the mark on the control die (which are the numbers 1 through 5). Thus `p` only takes on values in the range `1/6` (when the control is `5`) through `5/6` (when the control is `1`). In fact `p` is always going to be one of the values: `1/6`, `2/6`, `3/6`, `4/6`, or `5/6`. The frequentist analysis is failing to respect these known constraints (which are weaker than assuming actual priors). We can try fixing this with a simple procedure such as Winsorising or knocking everything back into range. For example the estimate `[1/6,5/6]` is biased but has improved loss/square-error: ``` >>> w = [1/6.0,5/6.0] >>> printBiasChecks(matMulFlatten(sNK['a'], w) - flatten(sNK['b'])) bias for p=0.166666666667 0.111111111111 bias for p=0.333333333333 0.0555555555556 bias for p=0.5 0.0 bias for p=0.666666666667 -0.0555555555556 bias for p=0.833333333333 -0.111111111111 >>> printLosses(losses(6,w)) exp. sq error for p=0.166666666667 0.0740740740741 exp. sq error for p=0.333333333333 0.101851851852 exp. sq error for p=0.5 0.111111111111 exp. sq error for p=0.666666666667 0.101851851852 exp. sq error for p=0.833333333333 0.0740740740741 ``` There are other ideas for fixing estimates (such as shrinkage to reduce expected square-error, or quantization to improve likelihood). But the point is these are not baked into the traditional simple empirical frequency estimate. Once you start adding all of these features you may have a frequentist estimator that is as complicated as a Bayesian estimator is thought to be, and a frequentist estimator that is no longer considered pure with respect to traditional frequentist criticisms. Let’s switch to the Bayes analysis for the game where the 5-sided control dice is set uniformly at random. A good Bayes point-estimate is easy to derive, as the appropriate priors for `p` are obvious (uniform on `1/6`, `2/6`, `3/6`, `4/6`, `5/6`). Our Bayes point-estimates for the expected value of `p` turn out to be: ``` >>> printEsts(bayesMeansEstimates(6,1)) pest for 0 heads 0.388888888889 pest for 1 heads 0.611111111111 ``` Which means: for 1 tails we estimate `p=.38888889` and for 1 heads we estimate `p=0.61111111`. Notice these estimates are strictly inside the range `[1/6,5/6]` (pulled in by `2/9` in both cases). Also notice because we have wrapped the Bayes estimate in code it appears no more complicated to the user than the empirical estimate (sure the code is larger than the empirical estimate, but that is exactly what an end user does not need to see). We have intentionally hidden from the user some important design choices (priors, the Bayes step convolution, use of a mean estimate instead of a mode). The estimator (see here or here) has wrapped up proposing a prior distribution, deriving the posterior distribution from the data likelihood equations (applying Bayes law), and then returning the expected value of the posterior as a single point-estimate. In addition to hiding the implementation details, we have refrained (or at least delayed) educating the user out of their desire for a simple point-estimate. We have not insisted the user/consumer of the result learn to use the (superior) complete posterior distribution in favor of mere point-estimates. For a Bayes estimate to be replacement compatible for a frequentist one we need (at least initially) put it into the same format as the frequentist estimate it is competing with. This squanders a number of the advantages the Bayes posterior, but as we will see the Bayes estimate is still lesser expected square-error (more efficient) than the frequentist one. So initially offering a Bayes estimate as a ready to go replacement for the frequentist estimate is of some value, and we don’t want to lose that value by initially requiring additional user training. Unfortunately this Bayes point-estimate solution is biased, as we confirm here: ``` >>> printBiasChecks(matMulFlatten(sNK['a'], \ bayesMeansEstimates(6,1)) - flatten(sNK['b'])) bias for p=0.166666666667 0.259259259259 bias for p=0.333333333333 0.12962962963 bias for p=0.5 0.0 bias for p=0.666666666667 -0.12962962963 bias for p=0.833333333333 -0.259259259259 ``` But, as we mentioned, our Bayes point-estimate has some advantages. Let’s also look at the expected loss each estimate would give for every possible value of the unknown probability `p`: ``` >>> printLosses(losses(6,bayesMeansEstimates(6,1))) exp. sq error for p= 0.166666666667 0.0740740740741 exp. sq error for p= 0.333333333333 0.0277777777778 exp. sq error for p= 0.5 0.0123456790123 exp. sq error for p= 0.666666666667 0.0277777777778 exp. sq error for p= 0.833333333333 0.0740740740741 ``` Notice that the Bayes estimate has smaller expected square-error (or in statistical parlance is a more efficient estimator) no matter what value `p` takes. The unbiased check conditions forced the frequentist estimate to a high expected square-error estimator. This means demanding the estimator be strictly unbiased may not be a good trade-off (and the frequentist habit of deriding other estimators for “not being unbiased” may not always be justified). To be fair bias can be a critical flaw if you intend to aggregate it with other estimators later (as enough independent unbiased estimates can be averaged to reduce noise, which is not always true for biased estimators). Let’s give the frequentist estimate another chance. For our discrete set of possible values `p` (`1/6`, `2/6`, `3/6`, `4/6`, `5/6`) once the number of coin-flips is large enough the equation family 1 bias checks no longer completely determine the estimate. So it is no longer immediately obvious that the observed empirical frequency is minimal loss. In fact it is not, so we can no longer consider the canned empirical solution to be the unique optimal estimate. Note this differs from the case where `p` takes on many different values from a continuous interval, which is enough to ensure the bias check conditions completely determine a unique solution. Continuing with an example: if we observed 7 flips an improved frequentist estimate (under the idea it is an unbiased point-estimate with minimal expected square-error) is as follows: ``` >>> printEsts(newSoln) pest for 0 heads 0.0319031034157 pest for 1 heads 0.111845090806 pest for 2 heads 0.296666330987 pest for 3 heads 0.439170280769 pest for 4 heads 0.560830250198 pest for 5 heads 0.703332297349 pest for 6 heads 0.888156558984 pest for 7 heads 0.968095569167 ``` To say we decrease loss we have to decide on a scalar definition of loss: be it maximum loss, total loss or some other criteria. This solution was chosen to decrease maximum loss (an idea compatible with frequentist philosophy) and was found through constrained optimization. Notice this solution is not the direct empirical relative frequency estimate. For example: in this estimate if you see seven tails in a row you estimate `p=0.0319031` not `p=0` (though we still have `0.0319031 < 1/6` which is an out of bounds estimate). This estimate is a pain to work out (the technique I used involved optimizing a move in directions orthogonal to the under-rank bias check conditions; perhaps some clever math would allow us to consider this solution obvious, but that is not the point). It is not important if this new solution is actually optimal, what is important is it is unbiased and has a smaller maximum loss (meaning the empirical estimate itself can not be considered optimal in that sense). The fact that the unknown probability `p` can only be one of the values `1/6`, `2/6`, `3/6`, `4/6`, `5/6` has changed which unbiased estimate is in fact the minimal loss one (added a new lower loss solution that would not considered unbiased if `p` could choose from more possible values). Depending on your application it can be the case that either of the frequentist or Bayesian estimate has better utility. But is is unusual for the frequentist estimate to be the harder one to calculate (as is the case here). The Bayes solution in this case is: ``` >>> printEsts(bayesSoln) pest for 0 heads 0.203065668302 pest for 1 heads 0.251405546037 pest for 2 heads 0.33603150662 pest for 3 heads 0.443861984801 pest for 4 heads 0.556138015199 pest for 5 heads 0.66396849338 pest for 6 heads 0.748594453963 pest for 7 heads 0.796934331698 ``` This is still biased, but all values are in range and the losses are smaller than the frequentist losses for all possible values of `p` (again limited to: `1/6`, `2/6`, `3/6`, `4/6`, `5/6`). To be fair the differences in loss/square-error are small (and shrinking rapidly as the number of observed flips goes up, so it is a small data problem). The point we want to make isn’t which estimate is better (that depends on how you are going to use the estimate, your domain, and your application), but the idea that: Bayesian methods are not necessarily more painful that frequentist procedures. The Bayesian estimation procedure requires more from the user (the priors) and has an expensive and complicated convolution step to use the data to relate the priors to the posteriors (unless you are lucky enough to have something like the theory of conjugate distributions to hide this step). The frequentist estimation procedure seems to be as simple as “copy over your empirical observation as your estimate.” That is unless you have significant hidden state, constraints or discreteness (not the same as having priors). When you actually have to justify the frequentist inference steps (versus just benefiting from them) you find you have to at least imaging submitting every possible inference you could make as a set of variables and picking a minimax solution optimizing expected square-error over the unknown quantities while staying in the linear flat of unbiased solutions (itself a complicated check). Note that each style analysis is correct on its own terms and is not always compatible with the assumptions of the other. This doesn’t give one camp a free-card to criticize the other. My advice is: Bayesians need to do a better job of wrapping standard simple analyses (you shouldn’t have to learn and fire up Stan for this sort of thing), and we all need to be aware that proper frequentist inference is not always just the common simple procedure of copying over the empirical observations. For full implementations/experiments (and results) click here for R and here for python. ## 4 thoughts on “Frequentist inference only seems easy” 1. Great post! I agree that simple Bayesian analyses should actually be simple to run. It’s hard to see Bayesian statistics going mainstream in psychology without it being possible to do in SPSS :) I have a small project trying to make it a little bit easier to get started called Bayesian First Aid in R at least: https://github.com/rasmusab/bayesian_first_aid 2. A really nice working of a few inference problems (also what pointed me to the the Eddy writeup): “Frequentism and Bayesianism II: When Results Differ” http://jakevdp.github.io/blog/2014/06/06/frequentism-and-bayesianism-2-when-results-differ/ . The neat thing is it proposes data mixture weights and slope/intercept both via the same Markov chain (not fit/optimization sub-step)- so both of these are usable observables. This is slower than performing a walk on only the mixture weights, but great for teaching. I have played with the code a bit and switched the plotting to ggplot2 via Rpy2 here: https://github.com/WinVector/Examples/tree/master/LFO 3. For an example of why unbiasedness is critical when aggregating check out the following example. Suppose our game is we get a point if we roll a 6 on the uniform 6 sided die. Repeat this experiment 100 times and compare the averaged of the Winsorised estimates ([1/6,5/6]) to the average of empirical frequentist estimates ([0,1]) (and compare both to the true expected value of 1/6): ``` set.seed(29325235) sample <- rbinom(100,size=1,prob=1/6) mean(sample) ## [1] 0.18 mean(ifelse(sample<0.5,1/6,5/6)) ## [1] 0.2866667 1/6 ## [1] 0.1666667 ``` Notice how the upward bias is not averaged out. The 0.28 is way too high an estimate for a population of 100 repetitions of the experiment. The Bayes estimate (at first blush) seems even worse: ``` be <- bayesMeansEstimates(6,1) print(mean(be[sample+1])) ## [1] 0.4288889 ``` A Bayesian response to the seemingly deadly flaw could be the following. Averaging estimators is a connivence that is nice when it successfully approximates the actual correct inference method: aggregating the underlying data and building a new estimate. In our notation this would be: ``` print(bayesMeansEstimates(6,100)[sum(sample)+1]) ## pest for 18 heads ## 0.1671593 ``` Which is a very good estimate (as it takes advantage of the constraints on the possible values of `p`). 4. Solution to the l2 minimax problem for general p. Gives you a funny smoothing rule: to estimate the win-rate of a coin using n-flips first add sqrt(n)/2 heads and sqrt(n)/2 tails pseudo observations.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-09

longest

en

https://math.stackexchange.com/questions/868102/number-of-ways-to-win-chocolate-game

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00580.warc.gz

# Number of ways to win chocolate game Alice and Bob are playing a game. They have N containers each having one or more chocolates. Containers are numbered from 1 to N, where ith container has A[i] number of chocolates. The game goes like this. First player will choose a container and take one or more chocolates from it. Then, second player will choose a non-empty container and take one or more chocolates from it. And then they alternate turns. This process will continue, until one of the players is not able to take any chocolates (because no chocolates are left). One who is not able to take any chocolates loses the game. Note that player can choose only non-empty container. The game between Alice and Bob has just started, and Alice has got the first Chance. She wants to know the number of ways to make a first move such that under optimal play, the first player always wins. Example : If we have 2 containers with chocolates as follow : [2,3] then here answer is only 1 Explanation : Only 1 set of moves helps player 1 win. Player: 1 2 1 2 1 Chocolates: 2 3 -> 2 2 -> 1 2 -> 1 1 -> 0 1 • This is the game of Nim. Its solution is well understood. en.wikipedia.org/wiki/Nim Jul 15, 2014 at 16:16 • @HansEngler I know its game of Nim.But i need to count the ways. Jul 15, 2014 at 16:16 • could you please phrase the question clearly and precisely? Are you interested in the number of ways that the game can be played or in the number of ways in which Alice or Bob can win or in something else? Jul 15, 2014 at 16:23 • @HansEngler Number of ways to make a first move such that under optimal play, Alice always wins.If she makes first turn Jul 15, 2014 at 16:32 Hint: The solution to nim is based on the binary representation of the pile sizes (or in your case the numbers of chocolates in the various containers). If we align the various binary numbers so that place values are aligned in columns, Alice wants to leave a position for Bob where each column has an even number of $1$s in it. A move by Alice (or Bob) consists of changing some of the bits of one of the numbers with the condition that the leftmost bit change must be from $1$ to $0$ (this is because we must remove chocolates). So she has to locate the leftmost column with an odd number of $1$s (if there is no such column, Alice has no winning moves). Then Alice must make sure that her change gets that column to an even number of $1$s (and possibly other adjustments will need to be made in other further to the right columns). But in that leftmost column with an odd number of $1$s, a $1$ must change to a $0$.

{"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}

CC-MAIN-2023-50

latest

en

https://www.teachoo.com/9436/2927/Example-4/category/Exponents/

text/html

crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00146.warc.gz

Exponents Chapter 11 Class 7 Exponents and Powers Concept wise ### Transcript Example 4 Expand a3 b2, a2 b3, b2 a3, b3 a2. Are they all same? a3 b2 = a3 × b2 = a × a × a × b × b a2 b3 = a2 × b3 = a × a × b × b b2 a3 = b2 × a3 = b × b × a × a × a = a × a × a × b × b b3 a2 = b3 × a2 = b × b × b × a × a = a × a × b × b × b Now, a3 b2 & b2 a3 (= a3 b2) are the same As they have same power of a & b Similarly, a2 b3 & b3 a2 (= a2 b3) are the same As they have same powers of a and b But all of them are not equal

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-38

latest

en

purpleroxsquad.blogspot.com

text/html

crawl-data/CC-MAIN-2017-51/segments/1512948592972.60/warc/CC-MAIN-20171217035328-20171217061328-00155.warc.gz

## Monday, January 7, 2008 ### Game Sccore + (Part II) Part III in my Game Score + research (Part I and Part II). The last post showed that using the extra outs adjustment factor and graphing the results using a pitch per out scenario could provide a necessary adjustment to the Game Score framework that added in the pitch count to make a realistic comparison between Red Bartlett's 58 pitch efficient pitching performance to Kerry Wood's 20 strikeout dominant performance. In conclusion to Part II, I wanted to see how this adjustment factor effected a team's performance over a season and what if I used correcting factor for actual pitches per out rather than using 3? So the extra out adjustment factor showed to be a good adjustment when you consider the best pitched games but what about a team's performance over a one year period? Looking at the Colorado Rockies in 2007, the following graph indicates how the adjustment factor looks for a team during the season. Pretty linear (note Aaron Cook's 74 pitch game below the x-axis). The correlation of this is equal to 0.77 while Game Score is equal to 0.46 (data not shown). So this shows for an average pitching staff, a pretty good correlation between pitch count and Game Score+, although there is a greater amount of negative game scores with this adjustment. During the regular season the low Game Score was 4 which translated to -23 GS+ (which wasn't the worst adjustment, this goes to a -65 GS+ pitched in Game 53 by Hirsh who throw 103 pitches in a 4 2/3 effort). So to conclude this adjustment factor shows good correlation to an average team between pitch count and the new metric Game Score+. The final data crunch is to look at the good pitching performances and the 2007 Rockies using actual pitch per out instead of a constant 3. To review, I took all pitching performances and adjusted all the pitching performances to 9 innings. For those pitchers who didn't go 9 innings, these extra pitches were divided by 3 (assume 3 pitch strikeouts which if you recall is based on the 81 pitch perfect strikeout game) and provided the "extra outs" metric which is subtracted from Game Score. Instead of assuming a 3 for every pitcher then I just used pitcher's actual pitches per out ratio that was established for the innings they did pitch. The final question is then instead of using 3 pitches to account for each out for pitcher's adjusted 9 inning game then what happens when we use their actual pitches per out ratio they were actually having during their appearance? Well this method certainly drops the large negative games scores and still holds a pretty good correlation factor of 0.57 versus 0.77 using the 3 pitch per out factor. For the good pitching (data not shown) it drops the the correlation factor to 0.02 versus 0.25. Because of this I think using the 3 pitch factor seems to be the best bet. Finally if you recall the whole purpose of this exercise was to try and determine if Aaron Cook's 74 pitch complete game was worthy to be considered a great game. If you recall Bill James' Game Score for this game was 67. Using my Game Score + adjustment factor you increase his score to 69. How does this relate to the best performances from 2002 - 2007? Well he barely eeks in to around 120th and doesn't even come close to matching the 2006 performance by Jeff Francis who pitched a complete game Game Score 91 versus Game Score+ of 75 (due his taking 129 pitches or requiring 48 extra pitches or 16 extra outs).

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-51

longest

en

https://holooly.com/solutions-v20/the-manufacturer-of-commercial-jets-has-a-cost-index-equal-to-94-9-per-aircraft-in-2013-the-anticipated-cost-index-for-the-airplane-in-2018-is-106-8-the-average-compound-rate-of-growth-should-hold-s/

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00173.warc.gz

# Question 3-B: The manufacturer of commercial jets has a cost index equal t... The manufacturer of commercial jets has a cost index equal to 94.9 per aircraft in 2013. The anticipated cost index for the airplane in 2018 is 106.8. The average compound rate of growth should hold steady for the next 15 years. If an aircraft costs $10.2 million to build in 2014, what is its expected cost in 2016? State your assumptions. (3.3) The "Step-by-Step Explanation" refers to a detailed and sequential breakdown of the solution or reasoning behind the answer. This comprehensive explanation walks through each step of the answer, offering you clarity and understanding. Our explanations are based on the best information we have, but they may not always be right or fit every situation. The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate. Learn more on how we answer questions. Already have an account? ## Related Answered Questions Question: 3-C ## Four hundred pounds of copper go into a 2,000-square-foot, newly constructed house. Today’s price of copper is$3.50 per pound. If the cost of copper is expected to increase 4.5% per year into the foreseeable future, what is the cost of copper going to be in a new 2,400-square-foot house 10 years ... Cost in 10 years =\left(\frac{2,400}{2,000}... Question: 3-A ## Develop an estimate for the cost of washing and drying a 12-pound load of laundry. Remember to consider all the costs. Your time is worth nothing unless you have an opportunity to use it for making (or saving) money on other activities. (3.2) ... At a typical household, the cost of washing and dr... Question: 3.5 ## CASE STUDY—Demanufacturing of Computers Let’s consider a case that deals with a timely and important issue concerning environmental economics. What are companies to do with all of the old computers that typically accumulate at their facilities? As one possible solution to this problem, a number of ... As one possible solution, the industrial engineer ... Question: 3.9 ## ReDetermine the SE and the correlation coefficient for the CER developed in Example 3-8. ... From the spreadsheet for Example 3-8 (Figure 3-7),... Question: 3.8 ## Cost Estimating Relationship (CER) for a Spacecraft In the early stages of design, it is believed that the cost of a Martian rover spacecraft is related to its weight. Cost and weight data for six spacecraft have been collected and normalized and are shown in the next table. A plot of the data ... Figure 3-7 displays the spreadsheet model for dete... Question: 3.5 ## Analysis of Building Space Cost Estimates The detailed design of the commercial building described in Example 3-2 affects the utilization of the gross square feet (and, thus, the net rentable space) available on each floor. Also, the size and location of the parking lot and the prime road frontage ... Based on this information, you estimate annual pro... Question: 3.4 ## Weighted Index for Gasoline Cost Based on the following data, develop a weighted index for the price of a gallon of gasoline in 2014, when 1996 is the reference year having an index value of 99.2. The weight placed on regular unleaded gasoline is three times that of either premium or unleaded plus ... In this example, k is 1996 and n is 2014. From Equ... Question: 3.1 ## Estimating the Cost of a College Degree A simple example of cost estimating is to forecast the expense of getting a Bachelor of Science (B.S.) from the university you are attending. In our solution,we outline the two basic approaches just discussed for estimating these costs. ... A top-down approach would take the published cost ... Question: 3.7 ## Learning Curve for a Formula Car Design Team The Mechanical Engineering department has a student team that is designing a formula car for national competition. The time required for the team to assemble the first car is 100 hours. Their improvement (or learning rate) is 0.8, which means that as ... (a) From Equation (3-5 : Z_{u}= K(u_{n})[/l...

{"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": 2, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-30

latest

en

https://mtrend.vn/question/3-2-5-3-4-1-4-2-2-3-8-1-4-9-20-9-5-710/

text/html

crawl-data/CC-MAIN-2021-39/segments/1631780061350.42/warc/CC-MAIN-20210929004757-20210929034757-00561.warc.gz

## |3x-2|=5 3\4+1\4:x=-2 {2\3} ²:{-8} – {-1\4+9\20}:9\5 Question |3x-2|=5 3\4+1\4:x=-2 {2\3} ²:{-8} – {-1\4+9\20}:9\5 in progress 0 10 phút 2021-09-09T14:12:59+00:00 2 Answers 0 views 0 1. | 3x – 2 | = 5 ⇔ $$\left[ \begin{array}{l}3x – 2 = 5\\3x – 2 = -5\end{array} \right.$$ ⇔ $$\left[ \begin{array}{l}3x = 7\\3x = – 3\end{array} \right.$$ ⇔ $$\left[ \begin{array}{l}x = 7/3\\x = – 1\end{array} \right.$$ Vậy , x ∈ { 7/3 ; – 1 } ____________________________________________ 3/4 + 1/4 : x = – 2 ⇔ 1/4 : x      = – 2 – 3/4 ⇔ 1/4 : x      = -11/4 ⇔         x      = -11/4 : 1/4 ⇔          x      = -11 Vậy , x = -11 _____________________________________________ ( 2/3 )^2 : ( – 8 ) – ( – 1/4 + 9/20 ) : 9/5 = 4/9 : ( – 8 ) – 1/5 : 9/5 = -1/18 – 1/9 = -1/6 2. Đáp án: ở dưới nếu thấy hay thì cho mình xin – 5 sao – cám ơn – hay nhất Giải thích các bước giải: |3x-2|=5 trường hợp 1 : 3x – 2 = 5 3x       = 5 + 2 3x       = 7 x        = 7 : 3 x        = $\frac{7}{3}$ trường hợp 2 : 3x – 2 = -5 3x       = -5 + 2 3x       = -3 x        = -3 : 3 x       = -1 vậy x ∈ { $\frac{7}{3}$ ; -1 } $\frac{3}{4}$+$\frac{1}{4}$x=-2 $\frac{1}{4}$x = -2 – $\frac{3}{4}$ $\frac{1}{4}$x = $\frac{-11}{4}$ x = $\frac{-11}{4}$ : $\frac{1}{4}$ x= -11 vây x = -11 ($\frac{2}{3}$) 2 : {-8} – {$\frac{-1}{4}$+$\frac{9}{20}$}:$\frac{9}{5}$ =$\frac{4}{9}$ . $\frac{-1}{8}$ – { $\frac{-5}{20}$ + $\frac{9}{20}$ . $\frac{5}{9}$ = $\frac{-4}{72}$ – $\frac{1}{5}$ . $\frac{5}{9}$ = $\frac{-1}{18}$ – $\frac{2}{18}$ = $\frac{1}{18}$

{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-39

latest

en

https://nus.kattis.com/problems/brickstopshere

text/html

crawl-data/CC-MAIN-2021-25/segments/1623487647232.60/warc/CC-MAIN-20210619081502-20210619111502-00561.warc.gz

OpenKattis National University of Singapore # The Brick Stops Here You have been hired by several clients of a factory that manufactures brass bricks. Brass is an alloy of copper and zinc; each brick weighs $1000$ grams, and the copper content of a brick can range from $1$ to $999$ grams. (Note that brass with less than 55% or more than 62% of copper is practically useless; however, this is irrelevant for this question.) The factory manufactures, through various processes, different types of brick, each of which has a different copper concentration and price. It distributes a catalog of these types to its customers. Your clients desire to buy a certain number ($M$) of bricks, which for, uh, religious reasons must be of different types. They will be melted together, and the resultant mixture must have a concentration of at least $C_\text {min}$ and at most $C_\text {max}$ grams of copper per kilogram. Their goal is to pick the $M$ types of brick so that the mixture has the correct concentration and the price of the collection is minimized. You must figure out how to do this. $M$, $C_\text {min}$, and $C_\text {max}$ will vary depending on the client. ## Input The first part of input consists of a line containing a number $N$ ($1 \le N \le 250$), the number of brick types, and then $N$ lines containing the copper concentration (between $1$ and $999$) and price (in cents) of each brick type. No brick costs more than $100$ dollars. The second part consists of a line containing a number $C$ ($1 \le C \le 500$), the number of clients you are serving, followed by $C$ lines containing $M$ ($1 \le M \le 50$), $C_\text {min}$ and $C_\text {max}$ ($1 \le C_\text {min} \le C_\text {max} \le 999$) for each client. All input numbers are positive integers. ## Output Output consists of a line for each client containing the minimum possible price (in cents) for which they can purchase bricks to meet their demands. If there is no way to match their specifications, output “impossible”. Sample Input 1 Sample Output 1 11 550 300 550 200 700 340 300 140 600 780 930 785 730 280 678 420 999 900 485 390 888 800 3 2 500 620 9 550 590 9 610 620 420 impossible 3635

{"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}

CC-MAIN-2021-25

latest

en

http://mathoverflow.net/questions/72087/infinite-dimensional-optimization

text/html

crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00165-ip-10-185-27-174.ec2.internal.warc.gz

# Infinite dimensional optimization Assume that we optimize a convex problem (convex objective and linear constraint) over a set of functions (say $L2$). Consider now the same optimization problem (same objective and same linear constraint) but now we optimize over a subset of $L2$. For example the set of affine functions or the set of polynomials of degree 2. Is there a geometric relationship between the optimum of the first problem (optimizing over the set L2) and the second problem (optimizing over the subset of affine functions). For example, can it be that the optimum of the second problem is the projection of the optimum of the first problem on the subset of affine functions in the sense of certain distance. Does this question make sense and if so was this done in some textbook or perhaps an old paper? Many thanks for your help. - I don't think it's as simple as projecting the optimum of the first problem onto the subspace used in the second problem. I'm not an expert, but maybe you'll find something useful in the theory of gamma-convergence: en.wikipedia.org/wiki/%CE%93-convergence. This is often used in finite-element theory to show that solutions of the FEM problem converge to solutions of the continuum problem. – jvkersch Aug 4 '11 at 17:38 Fixed title so that search engines will recognise it. – David Roberts Sep 2 '11 at 1:13 It "can be" but usually is not. Think of it in reverse. Start with a problem of minimizing a convex function $f$ over a convex set $S$ in vector space $V$, where the minimum value happens to be positive. Extend $f$ to the cone $C = \{(ts, t) \in V \times {\mathbb R}: t \ge 0, s \in S \}$ by $f(ts,t) = t f(s)$. The new convex function on $C$ has its minimum at $(0,0)$. But that does not help you find the minimizer of $f$ on $S$, which could be any extreme point of $S$.

{"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}

CC-MAIN-2016-30

latest

en

http://www.databozo.com/

text/html

crawl-data/CC-MAIN-2017-13/segments/1490218189130.57/warc/CC-MAIN-20170322212949-00410-ip-10-233-31-227.ec2.internal.warc.gz

## The Dirichlet is a Mixed Bag (of Nuts) You've probably come across the Dirichlet Distribution if you've done some work in Bayesian Non-Parametrics, clustering, or perhaps even statistical testing. If you have and you're like I was you may have wondered what this magical thing is and why it gets so much attention. Maybe you saw that a Dirichlet process is actually infinite and then wondered well how is that going to be useful? I think I've found a very intuitive approach... it's at least quite different from any other I've read. This post requires that you already be familiar with Beta distributions, PDF functions, and Python to get along. If you meet that requirement then grab a bag of nuts and let's jump right in. //   Posted ## Why Bayesians Prefer Log Space Why log space? Read through enough tutorials on Bayesian statistics and you're sure to encounter what seem to be unnecessary or at the very least confusing use of log and exp. Let's go over some examples to understand why log-space and also how. //   Posted ## Break the Markov Chains of Oppression: Modeling without MCMC You've seen the articles that say "MCMC is easy! Read this!" and by the end of the article you're still left scratching your head. Maybe after reading that article you get what MCMC is doing... but you're still left scratching your head. "Why?" "Why do I need to do MCMC to do Bayes?" "Why are there so many different types of MCMC?" "Why does MCMC take so long?" Why, why, why, why, etc. why. Here's the point: You don't need MCMC to do Bayes. We're going to do Bayesian modeling in a very naive way and it will lead us naturally to want to do MCMC. We'll understand the motivation and then! We'll also better understand how to work with the different MCMC frameworks (PyMC3, Emcee, etc) much better because we'll understand where they're coming from. We'll assume that you have a solid enough background in probability distributions and mathematical modeling that you can Google and teach yourself what you don't understand in this post. //   Posted ## Here's the thing I'm going to make this quick. You do a carefully thought through analysis. You present it to all the movers and shakers at your company. Everyone loves it. Six months later someone asks you a question you didn't cover so you need to reproduce your analysis... But you can't remember where the hell you saved the damn thing on your computer. If you're a data scientist (especially the decision sciences/analysis focused kind) this has happened to you. A bunch. You might laugh it off, shrug, w/e but it's a real problem because now you have to spend hours if not days recreating work you've already done. It's a waste of time and money. Fix it. I used to be this person too, so I get it. I decided to experiment with a new method that sounds so simplistic and stupid you'll think it won't work. Try it. Just. Try. It. It will change your life. //   Posted ## No control, no baseline, no problem. In the previous post, we gathered data from an uncontrolled test that a store owner ran. We were able to do that because we assumed that everyday was perfectly comparable. If you've ever looked at how business metrics move though you know that each day is not at all directly comparable. Each day of the week brings in different numbers of visitors consistently, different times of the year may be busier than others, etc. In this post, we're going to examine what it would take to test a change to a business without having a true control. Posted ## Deep in the Weeds: When Testing is Slow and Expensive Controlled testing in modern systems is fairly straight-forward. There are many tools to handle statistical analysis, random population sampling, data collection, etc. With the number of web visitors routinely numbering in the millions, the statistical techniques are also greatly simplified because of the preponderance of data. What do we do though when data is very costly? In this post, we'll take an overly simplified model and solve it using response surface modeling to find an approximate optimum with very little data. At the end of the post we'll discuss some possible caveats and some ideas for getting around those caveats. Posted ## Problem Say you run a video store and you want to understand how customer rental frequencies are changing over time. You can just plot the numbers but you want some help identifying when a customer's usage really went up and when it came down. Just looking at data you see that happening every day. What are the real changes though? If you charted the number of rentals per month for a given customer over many months you might see this: You probably can see an uptick somewhere around a year into their history... but where exactly? And is it safe to say it's constant even though we see a fall afterwards? Definitely we could give a pretty good guess, but we can't automate gut feel. How do we answer these questions quantitatively without needing a human? //   Posted ## Revealing a challenge A technique you may have heard of is to run A/A tests. These show us that our test analysis isn't unfairly biasing the analysis for or against the changes under test. You probably know what an A/B test is, in an A/A test we run a split test where we have two groups of people: a control group who are shown the current version of our product/website and a variant which is also just the control. In this instance, we are really just running a test comparing control to itself. So we run this for a few days and look at our results. Here is where things get messy. Imagine these are the results: Question: Is there bias in the experimentation or analysis systems? We can answer this question by considering another question. Posted ## Is a Billion \$ Powerball Finally Profitable to Play? TL;DR Nope. Highlights • With the latest odds and given the way ticket purchases grow with the expected jackpot, the expected value of Powerball is negative. Even more so with a billion dollar payout. • There is a 95% chance that there will be 3 or more winning tickets after the next draw. Almost zero chance no one will win. • Don't laugh at the premise! Powerball did have a positive expected value when the expected jackpot fell within a range of \$400m-\$650m. The game was changed to reduce the odds in October, 2015 which "fixed" that problem. First of all big shout out to Walter Hickey at Business Insider for the pointer to the Powerball data (here:http://www.lottoreport.com/powerballsales.htm) in this post a few years ago (here: http://www.businessinsider.com/heres-when-math-says-you-should-start-to-care-about-powerball-2013-9). This chart plots the relationship between expected value of purchasing a ticket to estimated size of a jackpot. The model used takes into account the dramatic increase in tickets purchased as the jackpot size increases. My analysis shows that because of the exponential increase in the number of tickets being played and the likewise dramatic increase in the likelihood of sharing the winnings, there is never a point where one will break even on buying a Powerball ticket... Now that they've made it harder to win. Previous to 10/07/2015 this is what the expected value looked like: Notice anything funny? Yeah it used to have a range of values where the expected value was positive! That means if the expected jackpot was somewhere between \$400m-\$700m it was actually a real investment for you to play the lotto. When the Powerball odds were reduced though, this stopped being a problem for the lottery. (You can read more about it here: http://news.lotteryhub.com/powerball-odds-set-change-october-2015/). Depending on how someone plays, the expected value may not mean much. Specifically for people who play only a few times in their lives since they won't play enough for the Law of Averages to even out the times when a player lost. A lot. Even so, I use it here because it's a pretty simple and intuitive framework to use to understand the value of an investment. Let's dig deeper into the game with the current (and harder to win) odds to understand why, even in the face of a billion dollar payout, Powerball is a net negative play with these new odds. //   Posted ## The Problem to Solve I've been developing a process for optimizing meals in my household. I want to be able to have an inventory of ingredients and other food stuffs and be able to figure out a meal to make. These are the different problems I want to be able to solve: • Given what's in the house, what meals can we make which meet our nutritional needs and maximize the number of meals we can make? • What is the meal that costs the least and meets our minimum nutritional needs? • Plan a grocery shopping list by figuring out meals for breakfast, lunch, and dinner for a 15 day period that meet prep time, meal diversity, and nutrition constraints. • Also, phase in caloric constraints over time to prevent weight gain. If you're familiar with Linear Programming you may recognize this as a variant of "The Diet Problem". You can read more about the full diet problem here: http://www.neos-guide.org/content/diet-problem. The most important take away is that this kind of problem can be solved using optimization methods such as Linear Programming. As a proof of concept, I've started laying out a small toy version of this problem to see if it holds water. If it does, I plan to continue to develop a manual solution and then, if that works, I might automate a general solution. ## Formulating the Model Let's first solve the problem of maximizing the number of meals we can make given what we have on hand. In order to solve this problem with Linear Programming we need to formulate an objective function which we want to maximize. I've laid out a few common meals from my household. For this problem, the meals will be what we call the decision variables. These are the inputs into the model that we change in order to maximize the objective function. The reason for choosing that the decision variables be the actual meals and not the ingredients (in case you are wondering) is that the meals are what need to be planned. The ingredients give us information about the nutrition for use by our constraints but they don't determine the meals on their own. Let's start by figuring out an objective function for this problem. First, I defined six meals to act as decision variables. Those are: For this first simple proof of concept, our objective function is pretty simple though the constraints of the model will get a bit lengthy. Since we want to just maximize the number of meals, our objective function is simply the sum of the number of meals: Great! Next we need to define our list of ingredients. Notice that the amount of each ingredient is a function of the meals each is included in. Here are the definitions of the ingredients: And then this is how each of the ingredients relates to the individual meals along with our constraints about how much we have on hand. ## Solving the model We still haven't entered in the nutrition information! That's ok for now. This is just the proof of concept. Let's stop here and try to solve the model. You may want to use Google Docs or Excel; I'm going to choose Excel. At another point I will get into how exactly to solve this model in Excel. For now, I want to show the power of this approach so I will keep things brief by just giving you the answer and providing the Excel spreadsheet. According to Excel, the maximum number of meals I can create given the above constraints on my inventory is 10.75 meals. Specifically: • 0.75 servings of spaghetti • 0 veggie pepperoni pizza • 1 serving of veggie chili dogs (two hot dogs) • 3 servings of chili • 2 servings of eggs and toast • 4 servings of peanut butter and jelly sandwiches Not only do we know exactly what meals we can make, but we also are left knowing how much of each ingredient we will have left over: • Spaghetti noodles: 137.5 grams • Boca crumbles: 0.625 lbs • Marinara sauce: 0 cups • Broccoli: 37.5 grams • Pizza dough: (still) 0 • Mozzarella cheese: 3.5 cups • Veggie pepperoni: 11 pepperonis • Veggie chili: 1.5 cans • Hot dog buns: 1 bun • Veggie hot dogs: 0 hot dogs • Grated cheddar cheese: 0 cups • Eggs: 0 eggs • Butter: 0.08 cups

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-13

longest

en

https://svpwiki.com/D-sharp

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00170.warc.gz

D sharp Ramsay The reason why there are 13 mathematical scales is that G? and F# are written separately as two scales, although the one is only a comma and the apotome minor higher than the other, while in the regular succession of scales the one is always 5 notes higher than the other; so this G? is an anomaly among scales, unless viewed as the first of a second cycle of keys, which it really is; and all the notes of all the scales of this second cycle are equally a comma and the apotome higher than the notes of the first cycle; and when followed out we find that a third cycle is raised just as much higher than the second as the second is higher than the first; and what is true of these majors may be simply repeated as to the D# and E? of the minors, and the new cycle so begun, and all successive minor cycles. Twelve and not thirteen is the natural number for the mathematical scales, which go on in a spiral line, as truly as for the tempered scales, which close as a circle at this point. [Scientific Basis and Build of Music, page 89] The two notes required for the scale of E minor are the F# of G, and the D of C major; for B minor, the C# of D, and the A of G major; for F# minor, the G# of A, and the E of D major; for C# minor, the D# of E, and the B of A major; for G# minor, the A# of B, and the F# of E major; for D# minor, the E# of F#, and the C# of B major. [Scientific Basis and Build of Music, page 90] One purpose of this plate is to show that twelve times the interval of a fifth divides the octave into twelve semitones; and each of these twelve notes is the first note of a major and a minor scale. When the same note has two names, the one has sharps and the other has flats. The number of sharps and flats taken together is always twelve. In this plate will also be observed an exhibition of the omnipresence of the chromatic chords among the twice twelve scales. The staff in the center of the plate is also used as to show the whole 24 scales. Going from the major end, the winding line, advancing by fifths, goes through all the twelve keys notes; but in order to keep all within the staff, a double expedient is resorted to. Instead of starting from C0, the line starts from the subdominant F0, that is, one key lower, and then following the line we have C1, G2, etc., B6 proceeds to G? instead of F#, but the signature-number continues still to indicate as if the keys went on in sharps up to F12, where the winding line ends. Going from the minor end, the line starts from E0 instead of A0 - that is, it starts from the dominant of A0, or one key in advance. Then following the line we have B1, F#2, etc. When we come to D#5, we proceed to B? instead of A#6, but the signature-number continues as if still in sharps up [Scientific Basis and Build of Music, page 114] SYSTEM OF THE THREE PRIMITIVE CHROMATIC CHORDS. The middle portion with the zigzag and perpendicular lines are the chromatic chords, as it were arpeggio'd. They are shown 5-fold, and have their major form from the right side, and their minor form from the left. In the column on the right they are seen in resolution, in their primary and fullest manner, with the 12 minors. The reason why there are 13 scales, though called the 12, is that F# is one scale and G? another on the major side; and D# and E? separated the same way on the minor side. Twelve, however, is the natural number for the mathematical scales as well as the tempered ones. But as the mathematical scales roll on in cycles, F# is mathematically the first of a new cycle, and all the notes of the scale of F# are a comma and the apotome minor higher than G?. And so also it is on the minor side, D# is a comma and the apotome higher than E?. These two thirteenth keys are therefore simply a repetition of the two first; a fourteenth would be a repetition of the second; and so on all through till a second cycle of twelve would be completed; and the thirteenth to it would be just the first of a third cycle a comma and the apotome minor higher than the second, and so on ad infinitum. In the tempered scales F# and G? on the major side are made one; and D# and E? on the minor side the same; and the circle of the twelve is closed. This is the explanation of the thirteen in any of the plates being called twelve. The perpendicular lines join identical notes with diverse names. The zigzag lines thread the rising Fifths which constitute the chromatic chords under diverse names, and these chords are then seen in stave-notation, or the major and minor sides opposites. The system of the Secondary and Tertiary manner of resolution might be shown in the same way, thus exhibiting 72 resolutions into Tonic chords. But the Chromatic chord can also be used to resolve to the Subdominant and Dominant chords of each of these 24 keys, which will exhibit 48 more chromatic resolutions; and resolving into the 48 chords in the primary, secondary, and tertiary manners, will make 144 resolutions, which with 72 above make 216 resolutions. These have been worked out by our author in the Common Notation, in a variety of positions and inversions, and may be published, perhaps, in a second edition of this work, or in a practical work by themselves. [Scientific Basis and Build of Music, page 115] PLATES XVII. & XVIII. These two plates show the chromatic chord resolving into the twelve major and twelve minor tonic chords of the twenty-four scales. There seems to be twenty-five, but that arises from making G? and F# in the major two scales, whereas they are really only one; and the same in the minor series, E? and D# are really one scale. C in the major and A in the minor, which occur in the middle of the series, when both sharps and flats are employed in the signatures, are placed below and outside of the circular stave to give them prominence as the types of the scale; and the first chromatic chord is seen with them in its major and minor form, and its typical manner of resolving - the major form rising to the root, and falling to the top and middle; the minor form falling to the top, and rising to the root and middle. The signatures of the keys are given under the stave. [Scientific Basis and Build of Music, page 116] advance by semitones, the keys with ?s and #s alternate in both modes. The open between G# and A? in the major, and between D# and E? in the minor, is closed in each mode, and the scale made one. The dotted lines across the plate lead from major to relative minor; and the solid spiral line starting from C, and winding left and right, touches the consecutive keys as they advance normally, because genetically, by fifths. The relative major and minor are in one ellipse at C and A; and in the ellipse right opposite this the relative to F# is D#, and that of G? and E?, all in the same ellipse, and by one set of notes, but read, of course, both ways. [Scientific Basis and Build of Music, page 117] The inner stave contains the chromatic scale of twelve notes as played on keyed instruments. The flat and sharp phase of the intermediate notes are both given to indicate their relation to each other; the sharpened note being always the higher one, although seemingly on the stave the lower one. The two notes are the apotome minor apart overlapping each other by so much; ?D is the apotome lower than C#; ?E the apotome lower than D#; F# the apotome higher than ?G; G# the apotome higher than ?A; and A# the apotome higher than ?B. The figures for the chromatic scale are only given for the notes and their sharps; but in the mathematical series of notes the numbers are all given. [Scientific Basis and Build of Music, page 120] Hughes The difference in the development of a major and a minor harmony —The twelve developing keys mingled D? shown to be an imperfect minor harmony E? taking B? as C? to be the same as D# —The intermediate tones of the seven white notes are coloured, showing gradual modulation —As in the diagram of the majors, the secondaries are written in musical clef below the primaries, each minor primary sounding the secondaries of the third harmony below, but in a different order, and one tone rising higher, . . . . . 34 [Harmonies of Tones and Colours, Table of Contents3 - Harmonies] Minor key-notes developing by sevens, veering round and in musical clef below —The use of the two poles D#-E? is seen, . . . . . . . . . . 35 [Harmonies of Tones and Colours, Table of Contents3 - Harmonies] I had forgotten all the minor keys, except that A is the relative minor of C major; but although I had only faint hopes of success, I determined to try, and I gained the twelve keys correctly, with the thirteenth octave. I found also that E? was usually printed as a minor key-note, Nature's laws having shown that it must be D#. [Harmonies of Tones and Colours, Dr. Gauntletts Remarks1, page 13] In the diagrams the circles are not drawn as interlacing into each other, from the difficulty of representing them accurately as rising spirally in geometric progression. If we endeavour to realise the development of harmonies, both in geometric order, and at the same time advancing and retiring, as in musical clef, we must imagine a musician having the physical power of striking all the notes on a circular keyed instrument of seven octaves, linked to a lower series of seven octaves, and a corresponding series of seven higher. But in fact the depth of the lower series, and the height of the higher, are alike unfathomable to our present powers. C, the first note of the seven octaves, sounds the four lowest tones, F, G, A, B of the lower series; and B, the last and highest note of the seven octaves, sounds in its harmony C? and D# of the higher series of sevens. [Harmonies of Tones and Colours, The Method of Development or Creation of Harmonies3, page 17] [Harmonies of Tones and Colours, Diagram VII - The Modulating Gamut of the Twelve Keys2, page 30] [Harmonies of Tones and Colours, The Twelve Scales Meeting by Fifths, page 31a] When the twelve minor harmonies are traced developing in succession, we notice how exactly they all agree in their method of development, also the use of the chasms and the double tones, the seven of each harmony rising a tone when ascending, but reversing the movement in descending; keys with sharps and those with flats are mingled. The intermediate tones are here coloured, showing gradual modulation. D? is shown to be an imperfect minor harmony, and E?, by employing B as C?, is seen to be equivalent to D#. [Harmonies of Tones and Colours, Diagram IX - The Minor Keynote A and Its Six Notes, page 34a] The diagram represents the Minor Key-note A and its 6 notes veering round in trinities; A and the other 11 developing their trinities in musical clef. Below each is the order in which the pairs unite, avoiding consecutive fifths, Lastly, D? is shewn to be an imperfect minor harmony, and by employing B as C?, E? is seen to be the same harmony as D#. As before, it should be remembered that the sharp and flat notes should, strictly, have intermediate tints. [Harmonies of Tones and Colours, The Diagram Represents the Minor Keynote, page 34c] In the musical clef the sixth and seventh notes from A, the fundamental minor key-note, are repeated, in order to show the use of the poles D#-C?, and that the colours agree. The use of the two poles, both in the major and minor series, is strikingly evident. [Harmonies of Tones and Colours, Diagram X - Minor Keynotes Developing by Sevens, page 35a] Below, the D# and E? are repeated, to shew the use of the two poles. [Harmonies of Tones and Colours, First Circle are 7 Minor Keynotes, page 35c] ALTHOUGH only twelve notes of a keyed instrument develope perfect minor harmonics, there are fifteen different chords, the double tones D#-E?, E#-F?, A#-B? all sounding as roots. The fifteen roots are written in musical clef. A major and a minor fifth embrace the same number of key-notes, but the division into threefold chords is different. In counting the twelve, a major fifth has four below the third note of its harmony, and three above it; a minor fifth has three below the third note of its harmony, and four above it. A major seventh includes twelve key-notes, a minor seventh only eleven. As an example of the minor chords in the different keys, we may first examine those in the key of A, written in musical clef. The seven of its harmony have two threefold chords, and two of its ascending scale. If we include the octave note, the highest chord of the descending scale is a repetition (sounding an octave higher) of the lowest chord of the seven in its harmony, and the second chord of the descending scale is a repetition of the first chord of its ascending scale. These two repetition chords are only written to the key of A: the chords of the other eleven keys will all be found exactly to agree with those of A in their mode of development. We may again remark on the beautiful effect which would result if the colours of the minor chords could be seen, with the tones, as they develope. [Harmonies of Tones and Colours, Diagram XII - The Chords of the Twelve Minor Keys, page 37a]

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-30

latest

en

https://insights.globalspec.com/article/11531/april-brings-showers-and-pothole-repairs

text/html

crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00543.warc.gz

The spring thaw brings a welcome change from snow, wind and below-freezing temperatures. For residents of the Eastern and Midwestern U.S., the spring thaw also brings potholes. According to the Spring Pothole Report, the 2019 pothole season is already a doozy, brought on by an early freeze, then warmer weather, more freezing and ultimately a bomb cyclone. What are science and engineering doing to alleviate the suffering of motorists and highway departments? Quite a lot – with improvements in all phases of pothole management, from predicting where potholes will develop to reporting locations to repair methods and materials. Tools include big data, non-Newtonian fluids, drones and 3D printing. ## Where are the potholes? In pre-internet days, drivers called toll-free phone lines to report potholes. Highway departments lacked robust methods to predict locations where potholes were likely to occur, relying on history and traffic data to approximate answers. Now, data mining provides better information for predicting pothole locations. Sensors deliver reports without human intervention, and humans use applications like Waze and proprietary software to tell highway departments, and each other, where these potential car wreckers lurk. ### Pothole analytics Professor Pothole – Lucius Riccio, a Columbia professor and former Commissioner of New York City’s Department of Transportation – pioneered the use of big data for predicting the number of potholes to expect in a year. To this end, he developed a pothole prediction equation, derived from studying years of New York data: P=s+g. Where: • P is the predicted number of potholes for a year. • s is the total snowfall in inches times 930, as 930 is the number of potholes per inch of snow. • g is the resurfacing gap in lane miles times 80. The resurfacing gap is the delta between the number of lane miles of road that need repaving every year and the actual number of miles repaved. This equation includes the resurfacing gap because Riccio concluded, again based on data, that poor resurfacing is responsible for 80% of potholes and harsh weather for 20%. Emphasizing bad maintenance contradicts the standard wisdom that snow, road ice and the freeze-thaw cycle are solely responsible for pothole formation. ### On the market now CarVi, which is a windshield-mounted camera that monitors driver alertness and beeps when a driver accidentally swerves out of their lane, is planning an over-the-air software update to teach the device to identify potholes. The company plans to share this information with local governments. Houston’s Potholes Tracker asks residents to report pothole locations online or via smartphone. The tracker’s homepage provides an updated map with pothole locations and status of reported holes. City personnel, not drivers, identify the vast majority of these road hazards. Other cities have similar systems to collect driver reports. Repaired and new potholes on New York State Route 22 in Columbia County, March 28, 2019. Image source: AuthorPopular road information application Waze is forming partnerships through its Connected Citizens program to share data with city and state governments. Using Waze’s crowdsourced information saves data collection money. In the first month of its Waze partnership, Washington D.C. collected 10,000 pothole reports, compared to an average of 11,000 over three months using conventional data-gathering techniques. Syracuse, New York, implemented a GPS-based system that not only tracks the locations of the city’s public works trucks but also reports what each truck is doing – spreading salt, patching potholes – and records every instance when the hose that sprays patching emulsion fires off into a hole. This system eliminates the need for crews to record this information and creates a rich and accurate dataset for visualization and production of multiple reports. An additional benefit of this system: the data tell whether a patch is adequate or a complete street resurfacing is in order. Public works trucks are not the only vehicles equipped to detect potholes and warn drivers to take action. Ford introduced a Fusion model with the capability in the 2017 model year and started offering the technology as an option for the 2018 Focus. Land Rover made a splash in 2015 when it introduced its pothole avoidance system, but the system is not yet a consumer option. Also in 2015, Google patented a pothole detection system; this system might find its way into the company’s driverless cars. Several cities are collecting road-condition data using sensors attached to road-repair trucks and analyzing the data to develop pothole location prediction models. A Pittsburgh-based company, RoadBotics, hires drivers to collect smartphone video of city streets and then applies artificial intelligence to analyze road surfaces the way an engineer would. Kansas City’s pothole prediction incorporates weather and traffic data with pavement conditions in its prediction model. This comprehensive model allows the city to proactively repair a street deemed highly likely to develop problems before a motorist loses a tire – or their temper. ### Promising research Researchers at Purdue University are working with the city of West Lafayette, Indiana, to enable cost-effective, precision pothole location. Using a sensor system rather than humans to identify potholes would be suitable if the sensor system was reasonably priced. The Xbox Kinect, a gaming platform, meets several requirements: it is inexpensive; it detects distance and color like a camera; and it can handle the large amounts of data generated when a maintenance department deploys multiple detection devices. The Purdue team paired the Kinect with a GPS to create a complete detection-and-location system. ## How to fix a pothole An understanding of current pothole repair practices is a useful precursor to reading about new methods. Standard pothole repair protocols depend on the severity of the defect, air temperature and road temperature. To fill a garden-variety hole, a road crew cleans out the hole, evens up the edges, fills the hole with hot-mix asphalt and compacts the fill material. Shallower holes and pavement cracks can usually be repaired with surface patching. This method works when temperatures are warm enough. The definition of warm enough varies, Using a propane torch to heat asphalt in a quick pothole repair. Image source: U.S. Air Forcedepending not just on air temperature, but the temperature of the road and the roadbed at the bottom of the patch. Asphalt has to stay at 185° F to be successfully placed and compacted. Cold weather dictates emergency patching if a pothole is severe enough to warrant a temporary repair. Throw-and-go patches are just that: clean debris from the patch area, throw hot asphalt in the hole and compact it using the asphalt truck tires. Throw-and-roll requires preparation more like that required for a permanent patch. ### Better paving material Asphalt mixes can vary with location. As most drivers will attest, no asphalt patch is permanent for a road with more-than-minimal traffic, regardless of the recipe followed to make the patch. A Scottish company, MacRebur, uses granulated plastic waste in its asphalt mixture. Pelletized plastic enhances and extends the bitumen used in asphalt. The company has laid roads worldwide, including in its homeland of Scotland, and claims that the plastic gives the asphalt more flexibility. Each kilometer of plastic road contains the equivalent weight of 684,000 plastic bottles or 1.8 million one-time use plastic bags. MacRebur is putting its Scottish roads to the test to develop new paving and pothole repair technology in research sponsored by the British government. In 2016, University of Minnesota Duluth researchers started testing a locally sourced, non-asphalt patching material, taconite. Taconite tailings, a byproduct of the production of iron pellets in Minnesota’s Iron Range, are ground up, mixed with an activating fluid and stuffed into the pothole. When the researchers checked on their two-plus-year-old patches in March 2019, they were pleased with the results. They are currently testing samples taken from these patches in the lab. The taconite mixture is not yet ready for market, but it does show promise as a future patch material. One approach to improving pothole repair is to prevent potholes from forming by using self-healing asphalt. Researchers at TU Delft have paved 12 roads in the Netherlands with a special asphalt mix that includes small steel fibers. When a crack appears, just add heat using a large induction machine and the cracks disappear. Lead researcher Erik Schlangen speculated that roads with these embedded fibers might also work as chargers for electric vehicles. Schlangen founded Epion to bring the new product from the lab to the street. ### One-step pothole spotting and repair A team at the University Of Leeds' School of Mechanical Engineering is leading research on a combination pothole search-and-repair system. The project, which started in 2016, uses a drone to spot pavement cracks and holes. The drone then flies down and sprays 3D-printed asphalt onto the fault. This system offers two big advantages over other pothole management protocols. The drone’s camera spots problems when they are still small and easier to repair, and it does its work at night when traffic volume is low. Developing the 3D printer presented some difficulties, particularly because asphalt behaves as a non-Newtonian fluid when moving through the printer’s extruder. The printer also needed to maintain a temperature high enough to extrude and place asphalt in the road fault. A team at the University of London resolved the printing problems in part by determining an optimal printing temperature. They also noted that 3D-printed asphalt has superior ductility to conventional asphalt and this ductility introduces microstructural changes that produce crack-bridging fibers – a boon to asphalt repair. The pothole-plugging drone is part of the Self Repairing Cities project headquartered at the University of Leeds. The project’s goal is zero disruption from street work in U.K. cities by 2050. ### Speaking of non-Newtonian fluids Some students at Case Western Reserve University tested a clever temporary patching method that, sadly, has gone nowhere since they were awarded a patent. Their idea exploited a characteristic of some non-Newtonian fluids – their tendency to solidify when exposed to pressure – to develop a temporary pothole patch. Rather than use familiar forms of oobleck like silly putty and cornstarch and water, the students tested different fluids to find an optimum pothole filler. The powder precursor is stored in a Kevlar bag. When a road crew needs a patch, they add water to the bag and deposit it in the pothole, covering the bag and hole with a sheet of heavy rubber. When a car tire runs over the patch, the car’s weight converts the fluid to a solid. Testing in Cleveland during temperate conditions went well, but the oobleck’s sturdiness in a typical Ohio winter is yet to be determined. ### Take care of the repair trucks too Passenger vehicles are not the only victims of potholes and other road hazards. When a public works truck blows a tire, that truck and its crew are off the job for several hours. A Michigan company claims that their new product, SuperMag, contributes to pothole repair by removing debris that can puncture tires and knock a repair truck offline. Storch Magnetics designed SuperMag to fit on a standard snowplow mount, or it can be towed behind a road maintenance vehicle. When lowered to the road surface, the powerful industrial magnet picks up metal debris, leaving a safer stretch of road for other repair trucks and vehicles. ## The future of potholes Potholed roads are one element of the U.S.’s crumbling infrastructure, and an element that is not difficult to repair but is unavoidably recurrent. The research projects described here could contribute to longer-lived road patches. Other factors, such as reduced road traffic as drivers become Uber or Lyft riders, could also help patches and resurfaced roads last longer. Improved road construction and maintenance technology, including design, roadbed preparation and construction materials, will also reduce potholes. The National Center for Asphalt Technology is currently researching methods for preserving and rejuvenating existing asphalt. Likewise, the National Concrete Pavement Technology Center is sponsoring research on concrete curing and the use of geotextiles in concrete. The American Society of Civil Engineers (ASCE), through its Grand Challenge program, invites engineers to propose novel ideas for infrastructure improvement. Since implementing any improved technologies will depend on government funding, the future of road construction and maintenance lies in the hands of local, state and federal legislators. Given the height of this hurdle, in all likelihood, potholes will continue to plague drivers and public works departments.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-49

longest

en

https://teacherscollegesj.org/what-are-three-ways-to-represent-proportions/

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00085.warc.gz

# What are three ways to represent proportions? ## What are three ways to represent proportions? Three ways to solve proportions • Vertical. • Horizontal. • Diagonal (often called “cross-products”) How many ways can you set up a proportion? A proportion is simply a statement that two ratios are equal. It can be written in two ways: as two equal fractions a/b = c/d; or using a colon, a:b = c:d. The following proportion is read as “twenty is to twenty-five as four is to five.” ### Is there only one way to set up a proportion? There’s more than one way to solve a proportion. One way is to cross-multiply. There’s a property you can use called the Means Extremes Property. It says that the cross products of a proportion will be equal. How do you solve a proportion with 3 ratios? How to Calculate Ratios of 3 Numbers 1. Step 1: Find the total number of parts in the ratio by adding the numbers in the ratio together. 2. Step 2: Find the value of each part in the ratio by dividing the given amount by the total number of parts. 3. Step 3: Multiply the original ratio by the value of each part. #### How do you do inverse proportions? The formula of inverse proportion is y = k/x, where x and y are two quantities in inverse proportion and k is the constant of proportionality. What percent is 15 of 50 explain the steps in finding the percent? 30 percent Answer: 30 percent of 50 is equal to 15. ## Is a proportion True False? There is another way to determine whether a proportion is true or false. Then multiply the denominator of the first ratio by the numerator of the second ratio in the proportion. If these products are equal, the proportion is true; if these products are not equal, the proportion is not true. How do you solve ratios and proportions? A good way to work with a ratio is to turn it into a fraction. Be sure to keep the order the same: The first number goes on top of the fraction, and the second number goes on the bottom. You can use a ratio to solve problems by setting up a proportion equation — that is, an equation involving two ratios. ### How do you solve proportions with fractions? Solving proportions is simply a matter of stating the ratios as fractions, setting the two fractions equal to each other, cross-multiplying, and solving the resulting equation. How do you solve proportion word problems? To use proportions to solve ratio word problems, we need to follow these steps: Identify the known ratio and the unknown ratio. Set up the proportion. Cross-multiply and solve. Check the answer by plugging the result into the unknown ratio. #### What is the equation for proportions? A proportions is an equation that states that two ratios are equal. Hence proportion can be written in two ways as a:b=c:d or a/b=c/d.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-50

latest

en

http://www.dummies.com/how-to/content/how-to-figure-athletes-weight-goals-considering-bo.html

text/html

crawl-data/CC-MAIN-2016-30/segments/1469257828283.6/warc/CC-MAIN-20160723071028-00123-ip-10-185-27-174.ec2.internal.warc.gz

For athletes, reducing percent body fat should be the focus of weight-loss efforts rather than reducing total body weight, otherwise optimum performance will be compromised. In some cases, total weight may not even change, when percent body fat goes down and muscle goes up. (Muscle weighs more than fat.) Therefore, assessing body composition is crucial as you determine your weight-loss goals. This example uses Emily to illustrate. Emily is a 24-year old runner who’s training for a marathon. She feels she could improve her performance if she lost some weight. Until a few years ago, she weighed about 120. Emily is 5 feet and 4 inches tall and now weighs 135 pounds. Her present estimated body fat is 23 percent. Her body fat goal is 15 percent. Here’s how she would calculate her optimal weight: 1. 135 (current weight) x 23 percent (current % body fat) = 31 pounds of fat 2. 135 (current weight) - 31 (weight that’s fat) = 104 pounds of lean 3. FFM (fat-free mass) goal = 85 percent (difference between body fat goal and total body) 4. 104 (current pounds of FFM) ÷ 85 percent = 122 (Emily’s goal weight) Emily’s weight-loss goal is to reach 122 pounds and reduce her percent body fat from 23 percent to 15 percent. Now plug in your own numbers: 1. ________ (current weight) x ___% (current % body fat) = ____ pounds of fat 2. ____ (current weight) - ____ (weight that is fat) = ____ pounds of lean 3. FFM goal =___% (difference between body fat goal and total body) 4. ____ (current pounds of FFM) ÷ ____% = _____ (Your goal weight)

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2016-30

latest

en

https://mdammapalace.nl/steel-waste/04-05_33567.html

text/html

crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00189.warc.gz

1. Home 2. Steel Waste 3. Vertical Roller Subject Vertical Roller Subject • Guidelines For Embankment Construction 3.1 Pneumatic Tired Roller The pneumatic tired roller is classified according to tire size, tire pressure and wheel loads. Charts (Figures 203-1 203-2) in the Standard Specifications relate these classes to maximum loose lift thickness. The roller must make at least 6 passes at defined speeds. Figure 4 Pneumatic Tired Roller • Worlds Largest Vertical Roller Mill Underway At Shah Cement Dec 18, 2018 The FLSmidth vertical roller mill (VRM) was chosen to produce a full range of cement types at the Muktarpur Plant. Driven by two 5.8 MW FLSmidth MAAG Max Drive gear systems, the impressive mill is the biggest VRM ever to be installed in a cement plant in terms of dimension, operating capacity and installed power. • Approximate Lateral Load Analysis By Portal Method the type of frame, the following assumptions can be made for portal structures with a vertical axis of symmetry that are loaded horizontally at the top 1. The horizontal support reactions are equal 2. There is a point of inflection at the center of the unsupported height of each fixed based column Assumption 1 is used if dosi is an odd number ... • 519 P Internal LoadingsThe internal torque developed in segments AB and BC of the pipe can be determined by writing the moment equation of equilibrium about the x axis by referring to their respective free - body diagrams shown in Figs. a and b. And • Chapter 4 Shear And Moment In Beams 1000-lb upward vertical load acts at the free end of the beam. (1) Derive the shear force and bending moment equations. And (2) draw the shear force and bending moment diagrams. Neglect the weight of the beam. Solution Note that the triangular load has been replaced by is resultant, which is the force 0.5 (12) (360) 2160 lb (area • Question 3 Given That Subject A Has A Vo2max 10 Transcribed image text Question 3 Given that subject A has a VO2max 10% higher on treadmill than on the Monark ergocycle (roller bike), based on the ACSMs metabolic equations, at what speed (mph) will the participant have to run if he wishes to train at 90% of his VO2 reserve without a slope? (Apts) VO,max (ml/kg/min) VO2max (ml/kg/min) estimate on treadmill - VO, reserve for treadmill ... • Problem 75 10 Points Consider The Beam Bcd Fixed Problem 7.5 (10 points) Consider the beam BCD fixed to the wall at B, supported by a roller at C, and subject to a vertical load Pat D. Using Castiglianos second theorem (a) Determine the reactions at the wall B and the roller C. (b) Determine the magnitude of the vertical displacement, vp at D. NOTE Include shear strain energy due to bending. • Beam Design Formulas With Shear And Moment Jan 06, 2005 AMERICAN WOOD COUNCIL w R V V 2 2 Shear M max Moment x 7-36 A ab c x R 1 R 2 V 1 V 2 Shear a R 1 w M max Moment wb 7-36 B Figure 1 Simple BeamUniformly Distributed Load • Why Roller Coaster Loops Are Never Circular Mar 24, 2014 Acer 11 Chromebook. \$129 at Walmart. The answer is that there is a force (provided by the rails), that is pushing the trucks of the coaster towards the center of the loop. Advertisement. This ... • Vertical Velocity Six Flags Great America Vertical Velocity opened with Dj vu in 2001 2. Vertical Velocity is a launched shuttle-style coaster 3. Manufactured by Intamin under the name Twisted Impulse Coaster 4. This ride uses a unique motor system known as Linear Introduction Motors or LIM for short 5. The train is a ski lift style 6. • Flippers Roller Boogie Palace Subject Of New Book Variety Sep 30, 2021 For just two years, from 1979 to 1981, Flippers Roller Boogie Palace was one of the hottest spots in Los Angeles. Prince, Cher, Robin Williams, Patrick Swayze, the • Bearing Systems For Grinding Rollers In Vertical Roller The grinding rollers are the key components in every vertical roller mill and high-pressure roller press. The basic requirement for reliable operation is optimum configuration of the bearing ... • Horizontal Sliding Roller Track System Ergomart The Ergomart Roller Track (R-Track) System is a heavy-duty horizontal track/rail that allows equipment to slide or roll laterally with ease. The R-Track System provides a unique and powerful set of answers to meet many mounting challenges encountered across a broad range of applications and environments. • 6 Different Types Of Bearings How They Work Uses Pdf Jul 26, 2020 Generally, the vertical shaft is uncommon in the ordinary transmission of power. But they often occur on machine practice in machine tool turntables, textile machinery, etc. 4. Rolling Contact or Anti-friction bearings. It is a well-known fact that a smooth rounded surface will roll over a similar surface more than when it is sliding. • Mathematically Designing A Frictional Roller Coaster A roller coasters initial height and slope determines its length. The higher the initial point, the longer the path. It is best that the initial roller coaster slope not be steeper than 2.5, otherwise the marble may slide instead of roll. This guideline is an outcome from testing different initial slopes. • Beam Reactions And Diagrams Strength Of Materials Using R A and R B found at steps 3 and 4 check if V 0 (sum of all vertical forces) is satisfied. Note that steps 4 and 5 can be reversed. For a cantilever beam use V 0 to find the vertical reaction at the wall and M wall 0 to find the moment reaction at the wall. There is • Vertical Blinds At Opt for vertical blinds or patio door blinds to protect your home from the sun and create privacy for your family. Whatever type of window shades or blinds youre looking for, be sure to measure your windows before purchasing so you can ensure a correct fit and consider which • Garage Door Tracks Track Hardware Vertical track repairs. Tracks bent out of shape--frequently seen, and undesirable in a warehouse. Rectifying tracks from this condition will allow the rollers to again move smoothly. If applicable, the root cause of vertical track malfunction needs addressing. Repair then proceeds by placing a large screwdriver behind the bent part of the track. • Purchase And Installation Of The 9roller Vertical Short description of the subject matter of the contract Purchase and installation of the 9-roller vertical straightener at HSM in Dbrowa Grnicza. concerns the project entitled Reliable and durable in operation, modern railway rails with a length of 120 m, characterized by • Question 52 the angle describes the orientation of the rod with the vertical, that x is the horizontal position of the cart, and that gravity acts downward, determine a system of two dierential equations for the collar and the rod in terms of x and . g l m1 m2 x Q Figure P5-2 Solution to Question 52 Preliminaries • Roller Blinds Vertical Blinds Pvc Strip Curtains Air Roller Blinds. classic choice for window dressings and are perhaps the most well-known type of blind ... Solid and stable window covering consisting of frame of vertical stiles and horizontal rails. ... This strip curtain kit is ideal for use on an internal or external doorway subject to high volumes of pedestrian and or vehicular traffic ... • Roller Conveyor Gravity Conveyor For Sale New Used Using appropriate diameter roller sizes, roller spacing, and width between frames, many types of loads can be conveyed. Typically, gravity roller conveyor has a capacity of up to 3,000 lbs. While available in any length you need, gravity roller conveyors are typically sold in 5- or 10-foot sections. The most common overall widths are 12, 18 ... • Chapter 2 Review Of Forces And Moments 2.1.4 Classification of forces External forces, constraint forces and internal forces. When analyzing forces in a structure or machine, it is conventional to classify forces as external forces constraint forces or internal forces. External forces arise from interaction between the system of interest and its surroundings. Examples of external forces include gravitational forces lift or drag ... • Chapter 2 Vehicle Dynamics Modeling The vertical axis, Z, is often used in the study of ride, pitch, and roll stability type problems. The following list defines relevant definitions for the variables associated with this research. Longitudinal direction forward moving direction of the vehicle. There are two • Types Of Beams Loads And Reactions Engineersdaily At end B of the beam (Fig.2a) the roller support prevents translation in the vertical direction but not in the horizontal direction hence this support can resist a vertical force (R B) but not a horizontal force.Of course, the axis of the beam is free to rotate at B just as it is at A.The vertical reactions at roller supports and pin supports may act either upward or downward, and the ... • Beam Diagrams And Formulas For Various Static BEAM DIAGRAMS AND FORMULAS For Various Static Loading Conditions, AISC ASD 8th ed. ARCH 331 163 of Note Set 8.2 Su2015abn wl W 2 (w)(w) • How To Solve A Truss Problem 6 Steps Instructables The left support is a roller support, which can only have a vertical y-direction force applied to it because it moves freely horizontally. This force should have the variable name R_A_y (A • Example 5 Etu EXAMPLE 5.5 A solid steel shaft AB shown in Fig. 514 is to be used to transmit 3750 W from the motor Mto which it is attached.If the shaft rotates at and the steel has an allowable shear stress of allow 100 MPa,determine the required diameter of the shaft to the nearest mm. • Vertical Velocity Six Flags Great America Vertical Velocity is often called V2 for short 7. In total, riders cover over 2,700 feet of track on their 45 second ride experience 8. Riders go vertical three times in the front and twice in the back 9. There is a total of 630 feet of U shaped track 10. Vertical Velocity is known for the sudden launch often accompanied by a 3,2,1 countdown • Example 1 Etu by bearings at A and B, which exert only vertical forces on the shaft. 225 N C D 200 mm 100 mm 50 mm 50 mm 800 N/m B (a) A Fig. 15 0.275 m 0.125 m (800 N/m)(0.150 m) 120 N 0.100 m 225 N A y B y (b) Solution We will solve this problem using segment AC of the shaft. Support Reactions. A free-body diagram of the entire shaft is shown in Fig ... • Conveyors Ncsu Sep 30, 1999 Vertical Conveyor. Unit On-Floor No Accumulate. Used for low-frequency intermittent vertical transfers (cf. vertical chain conveyor can be used for continuous high-frequency vertical transfers. 13(a) Vertical Lift Conveyor. Carrier used to raise or lower a load to different levels of a facility (e.g., different floors and/or mezzanines) • Answered Determine The Normal Reaction At The Bartleby Physics QA Library Determine the normal reaction at the roller A and horizontal and vertical components at pain B for equilibrium of the member. 10KN 0.6m 0.6m 0.8m 60 0.4m Determine the normal reaction at the roller A and horizontal and vertical components at pain B for equilibrium of the member. 10KN 0.6m 0.6m 0.8m 60 0.4m • Pay Weekly Carpets Carpets On Finance 10 Week Carpets Loans subject to status and not available to anyone under 18 years of age. Pay Weekly Carpets and Blinds Ltd is authorised and regulated by the Financial Conduct Authority FRN 810750. We act as a credit broker and not a lender and offer credit from Klarna. Credit is subject to age and status. *There is a credit limit cap of 2000 per customer. • Spend Less Smile More Free shipping on millions of items. Get the best of Shopping and Entertainment with Prime. Enjoy low prices and great deals on the largest selection of everyday essentials and other products, including fashion, home, beauty, electronics, Alexa Devices, sporting goods, toys, automotive, pets, baby, books, video games, musical instruments, office supplies, and more. • Vertical Roller Mill Market Size Share 20212028 Top Jun 16, 2021 New Jersey, United States,- The Vertical Roller Mill Market study provides details of market dynamics affecting the market, market size, and segmentation, and casts a shadow over the major market players by highlighting the favorable competitive landscape and successful trends over the years. This market report presents the detailed profile of major industry players and their upcoming Related Post Get a Free Quote If you are interested in our company and our products, you can click to consult, we will provide you with value-for-money equipment and considerate services ! • 40 Years of Experience • 400 Industrial workshop • 600 production equipment • 80 Served Countries

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-21

latest

en

https://puzzling.stackexchange.com/questions/98049/a-book-of-mystery-page-iii

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00090.warc.gz

# A book of mystery - Page III Just as was for solving the first Page number three had started to burst With ink to tell you what to do A measurement of six squares South from page 2s wheres 314.2 of these Take away his Spanish letter This is the operation for your next endeavour Strange words these, I cannot figure Maybe words for someone with a brain that is bigger! Hint 1: What measure of length is similar to a shape? Hint 2: It is a simple direction from point A to point B Hint 3: It begins with a 'C' SHOOVER GOT PAGE 2 (THE ANSWER TO WHICH WAS MCDONALD'S ISLINGTON) AND SOLVE PAGE THREE TO GET A HINT TO SOLVE THE NEXT PAGE. I think the answer might be: Neville Place My logic is as follows: A measurement of six squares is a cubit. There is no official definition, but generally it is about 2 feet. 314.2 cubits south of McDonald's Islington puts you on Pentonville Rd. Removing the Spanish letter (pe) leaves "ntonville". Stretching, I scan this as "N to Nville", and there is a Neville Place about 5 miles north of McDonald's in Islington. EDIT: Let me take another shot. I think the answer is: The Crafts Council building on Pentonville Rd. Reasoning: Following my previous reasoning, you end the second to last line with the clue "ntonville". Instead of trying to parse this, assume it just means "north". If you actually go to the point 314.2 cubits south of McDonald's Islington and face north, the Crafts Council building entrance is right there. • No, that is not the answer. However, you were right with the cubits May 15, 2020 at 14:51 Just some random thoughts: Google suggests a cubit is equal to 18 inches. So 314.2 cubits would be equal to 471.3 feet. At about that distance SSE of McDonalds there is a place called Cubitt Artists. Coincidence or not? • It is just a coincidence May 17, 2020 at 8:57 Assuming, as per your hints, the 6 squares to be a royal cubit, and taking this to be between 523.5 to 529.2 mm and keeping in mind the Spanish letter I need to remove letra 314.2 of these straight south might bring me to Leather Lane or Fetter Lane. From which I'd opt for Fetter Lane as I should remove his Spanish letter, not its. And the only he I could find from these options who has a bigger brain for certain is Alexander L. Fetter, the physics professor. Removing the letter leaves me with aexnderlfet, or for instance 'next federal'. Which is where I begin to lose confidence in my solution. It might point at the Gibraltar Government Office, 150 Strand. I'm not convinced, but I consider it worth a try. However, I still struggle with the suspiciously round measure of 314.2 and that we shall find an operation rather than a location. Maybe I must engage in a more circular reasoning here. • Hi guys. If you downvote my answers, please leave some helpful criticism in a comment. I fail to see why my attempt at answering this riddle is worse than the other ones here and I would like to improve. But I need some input for that. Feb 1, 2021 at 15:24

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-10

latest

en

https://socratic.org/questions/how-do-you-find-the-number-of-roots-for-f-x-4x-3-x-2-2x-1-using-the-fundamental-

text/html

crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00379.warc.gz

# How do you find the number of roots for f(x) = 4x^3 – x^2 – 2x + 1 using the fundamental theorem of algebra? Jul 29, 2016 It has $3$ zeros since it is of degree $3$. #### Explanation: The fundamental theorem of algebra (FTOA) tells you that a polynomial of non-zero degree has at least one Complex (possibly Real) zero. A straightforward corollary of this - often stated as part of the FTOA - is that a polynomial in one variable of degree $n > 0$ has exactly $n$ Complex (possibly Real) zeros counting multiplicity. To see this, note that if $f \left(x\right)$ is of degree $n > 0$, then by the FTOA it has a zero ${r}_{1} \in \mathbb{C}$. Then $\left(x - {r}_{1}\right)$ is a factor of $f \left(x\right)$ and $f \frac{x}{x - {r}_{1}}$ is a polynomial of degree $n - 1$. If $n - 1 > 0$ then $f \frac{x}{x - {r}_{1}}$ has a zero ${r}_{2} \in \mathbb{C}$ (possibly equal to ${r}_{1}$), so a factor $\left(x - {r}_{2}\right)$, etc. Repeat to find all $n$ zeros. In our example $f \left(x\right)$ is of degree $3$ so has exactly $3$ zeros counting multiplicity. $\textcolor{w h i t e}{}$ Bonus The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula: $\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$ In this example, $a = 4$, $b = - 1$, $c = - 2$, $d = 1$ and we find: $\Delta = 4 + 128 + 4 - 432 + 144 = - 152 < 0$ Since $\Delta < 0$ this cubic has one Real zero and a Complex conjugate pair of non-Real zeros. We can use Descartes' rule of signs to tell that $f \left(x\right)$ has $0$ or $2$ positive Real zeros and exactly one $1$ negative zero. Since we already know that it only has one Real zero, it is negative.

{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 34, "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}

CC-MAIN-2024-38

latest

en

https://numberworld.info/101109

text/html

crawl-data/CC-MAIN-2020-05/segments/1579250599718.13/warc/CC-MAIN-20200120165335-20200120194335-00132.warc.gz

Number 101109 Properties of number 101109 Cross Sum: Factorization: Divisors: Count of divisors: Sum of divisors: Prime number? No Fibonacci number? No Bell Number? No Catalan Number? No Base 2 (Binary): Base 3 (Ternary): Base 4 (Quaternary): Base 5 (Quintal): Base 8 (Octal): 18af5 Base 32: 32nl sin(101109) -0.017962167881256 cos(101109) 0.99983866724837 tan(101109) -0.017965066234825 ln(101109) 11.523954421818 lg(101109) 5.0047898151001 sqrt(101109) 317.97641421967 Square(101109) Number Look Up Look Up 101109 which is pronounced (one hundred one thousand one hundred nine) is a very impressive number. The cross sum of 101109 is 12. If you factorisate 101109 you will get these result 3 * 33703. 101109 has 4 divisors ( 1, 3, 33703, 101109 ) whith a sum of 134816. 101109 is not a prime number. The figure 101109 is not a fibonacci number. The number 101109 is not a Bell Number. The number 101109 is not a Catalan Number. The convertion of 101109 to base 2 (Binary) is 11000101011110101. The convertion of 101109 to base 3 (Ternary) is 12010200210. The convertion of 101109 to base 4 (Quaternary) is 120223311. The convertion of 101109 to base 5 (Quintal) is 11213414. The convertion of 101109 to base 8 (Octal) is 305365. The convertion of 101109 to base 16 (Hexadecimal) is 18af5. The convertion of 101109 to base 32 is 32nl. The sine of the figure 101109 is -0.017962167881256. The cosine of the number 101109 is 0.99983866724837. The tangent of 101109 is -0.017965066234825. The square root of 101109 is 317.97641421967. If you square 101109 you will get the following result 10223029881. The natural logarithm of 101109 is 11.523954421818 and the decimal logarithm is 5.0047898151001. You should now know that 101109 is very unique number!

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-05

latest

en

https://studylib.net/doc/13492591/3%E2%80%931

text/html

crawl-data/CC-MAIN-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00206.warc.gz

# 3–1 MIT Department of Chemistry 5.74, Spring 2004: Introductory Quantum Mechanics II Instructor: Prof. Robert Field 3–1 5.74 RWF LECTURE #3 ROTATIONAL TRANSFORMATIONS AND SPHERICAL TENSOR OPERATORS Last time; 3-j for coupled ↔ uncoupled transformation of operators as well as basis states 6-j, 9-j for replacement of one intermediate angular momentum magnitude by another Today: patterns — limiting cases — simple dynamics minimum number of control parameters needed to fit or predict the I ( ω ) or I ( t ) Rotation as a way of classifying wavefunctions and operators Classification is very powerful — it allows us to exploit universal symmetry properties to reduce complex phenomena to the unique, system specific characteristics Recall finite group theory * Construct a reducible representation — a set of matrix transformations that reproduce the symmetry element multiplication table * represent the matrices by their traces: characters * reduce the representation to a sum of irreducible representations * each irreducible representation corresponds to a symmetry species (quantum number) * selection rules, projection operators, integration over symmetry coordinates * the full rotation group is an ∞ dimension example, because the dimension is , there are an infinite number of irreducible representations: the J quantum number can go from 0 to * because the symmetry is so high, most of the irreducible representations are degenerate: the M J quantum number corresponds to the 2 J +1 degenerate components. All of the tricks you probably learned in a simple point group theory are applicable to angular momentum and the full rotation group. Rotation of Coordinates two coordinate systems: XYZ fixed in space xyz attached to atom or molecule 2 angles needed to specify orientation of z wrt Z one more angle needed to specify orientation of xy by rotation about z EULER angles - difficult to visualize — several ways to define — you will need to invest some effort if you want a deep understanding Superficial path θφ l m = Y l m R θ φ χ l m = ### ∑ m ′ l m 1 m 4 φ θ χ D ( ) m m θ φ χ * Rotation does not change l * 3 5.74 RWF LECTURE #3 [Non-lecture proof: (i) [ llll 2 , llll j ] = 0, (ii) rotation operators have the form e i ll therefore: [ l 2 , R ] = 0 — no change in l under rotation.] j α (rotate by j axis) D ( ) m m is a (2 l +1) × (2 l +1) square matrix that specifies how the R ( φ,θ,χ ) rotation transforms | l m . 3–2 Wigner Rotation matrix: D l φ θ χ l m ′ ℜ ( ) l m operates to right = l m ′ exp( i φ ll z )exp( i θ ll y )exp( i χ ll z ) l m this operates to left (both bra and operator are complex conjugated) 3 successive rotations of | l m ⟩ in order χ,θ,φ = exp ( i m ′ − χ ) l m m reduced rot. first matrix d l m m ### ∑ t ( 1) t [ ( l + m )!( l − m )!( l + m )!( l − ( l + m ′ − t )!( l  cos θ 2 2 l + m m t t t + m m 2 t  sin θ 2 2 − ′+ n m )! ] m )! t ranges over all integer values where the arguments of the factorials are not negative. d ( ) m m is real and has lots of useful properties So now we know how to write the transformation under rotation of any angular momentum basis state. l , m are labels of an irreducible representation of the full rotation group. Here is the wonderful part: the rotation matrices actually have the form of angular momentum basis functions! D l m 0 χ ( , , ) 2 l + 1  is irrelevant except 4 π  for phase choice D mJ ( φ θ χ ) [ (2 J Y l m + 1) ] integer (angular part of atomic orbital) JMK * sym. top wavefunction 5.74 RWF LECTURE #3 3–3 Not covered in Lecture Suppose we have a matrix element of some operator A JM A J M A This is a number. Rotate all 3 terms in the matrix element [ JM R 1 ] RAR 1 [ R ] = A If A can be expanded as a sum of terms that transform under rotation as angular momentum basis functions, then the integral is a sum of products of D( φ,θ,χ ) matrices. But the integral cannot depend on the specific values of φ, θ, χ . This tells us that, if we can partition A into a sum of terms, each of which has the rotational transformation properties of an angular momentum basis state, we will be able to evaluate the angular part of the integral implied by the matrix element. This is actually how the Wigner-Eckart Theorem is derived. Spherical tensor expansion is like a multipole expansion. Anything can be broken up into angular momentum-like parts, including what a laser writes onto a molecular sample. A ( r , θ φ ) = ### ∑ a J r Y JM (like the angular, radial separation of the hydrogen atom wavefunction) spherical tensor operators T q k a spherical tensor or rank k is a collection of 2 k + 1 operators that transform among each other under rotation as | l = k m = q Wigner-Eckart Theorem! Notation: T q NJM T ( ) q k N J M ( 1 ) J M classification is wrt specific angular momentum !! J M k q J ′  M ′ NJ T proportionality constant: “reduced matrix element” x,y,z are vector wrt L, J but not S , etc. means some combination of the components of { A } that satisfies the commutation rules [ [ J T ± , q ( ) ( ) q k ( ( A A ) ] ) ] = = q [ T ( ( ) q k k ( A ) + 1) ( 1) ] T q k ± 1 ( A ) Alternatively, think of defining projection operators that project out of an arbitrary operator, symmetrylabeled operators. 5.74 RWF LECTURE #3 If A is a vector, like r r = xi ˆ + ˆ + zk L or S or J 3–4 T 0 ( ) A z T ( ) ± 1 ( ) m 2 / ( A X ± i A Y ) if A is a second rank Cartesian tensor, like xx xy xz yx yy yz zx zy zz  nine components (not the same as A ± ) T 0 T 0 T ( ± 1) 1 / ( ) i 2 / ( xx + yy + ( xy yx ) zz ) A m i / 2 { ( yz zy ) ± i ( zx xz ) } T 2 ± 1 T 2 ± 2 = 6 1 2 { 2 zz xx yy } A m 1 2 { ( xz zx ) ± i ( yz + zy ) } 1 2 { ( xx yy ) ± i ( xy + yx ) } e . e.g. ± ( L L i 2 y { ( i h L L x L x ) ± = i h L z ) i ( i h L y ) } = h 2 { ± L x + i L y } = ± h L 2 ± it is also possible to construct a 3 × 3 = 9 dimensional reducible spherical tensor out of two different vectors 3 x 3 = 9 reduces to 1 RANK: (0) + 3 (1) + 5 (2) T 0 u v = 3 1 2 3 / ( ( u 1 v 1 u v u 0 v 0 + u 1 v 1 + u v + u v ) )  scalar product  T 0 T ( ) ± 1 v = 2 / ( u 1 v 1 + u 1 v 1 ) = 2 / u v 2 1 2 ( u ± v u v ± 1 ) ( u = 1 2 [ ( u v u v ) m i ( u v u v ) ] u v )  vector cross product T 0 T ( ) ± 1 T ( ) ± 2 v = 6 / 6 / 2 1 /2 ( ( u 1 v 1 + 2 u 0 v 0 + u 1 v 1 ) 2 u v ( u v + u v u v ± 1 ) ) = m 1 2 [ u v + u v u v u ± 1 v ± 1 = 1 2 [ u ± i ( u v + u ± i ( u ) ] + u ) ] 5.74 RWF LECTURE #3 Transformations between SPACE and molecule fixed coordinate systems A rotational wavefunction specifies the probability amplitude distribution of orientation of the molecule fixed axis system in the laboratory axis system. Origin of both coordinate systems is fixed at molecular center of mass. We can express the relationship between LAB and molecule frame quantities. For example, radiation is specified in the LAB, but transition moments are tied to the molecule frame. The molecule ε ·µ is a dot product between quantities specified in different coordinate systems. We must transform one of these into the coordinate system of the other. “direction cosines” 3–5 P LAB # ∑ q D Pq ω ( ) molecule ω = (φ,θ,χ) inverse: all 3 Euler angles k q ( ) # ∑ P D Pq D Pq ( ) k ( ) = D ( ) P , q ω we know this from the form of D exp(– i J z φ )d( θ) exp(– i J z χ ) symmetric top wavefunction | J = k, M=–P, = -q ⟨ω | J M = [(2 J + 1)/8 π 2 ] 1/2 D M ( ω )* Thus the action of T A on | J M is an example of an uncoupled to coupled transformation and a matrix element of D is two such transformations D Pq ( ) * J M ′ = − 1) M −Ω [ (2 J + 1 2 J ′ + 1) ] J k J ′ J k q 4 molecule  M P LAB J ′  M ′ (the order of the columns is rearranged from what we expect from vector coupling, but compatible with symmetry properties of 3–j) It is in form of W-E Theorem 5.74 RWF LECTURE #3 These are the direction cosine matrix elements! For a laboratory frame ε (t) field polarized along the z-axis H ( ) = T 1 P 0 ( εε ( ) ) T 1 q ( ) = T 1 0 ( εε ( ) ) 1 D q ### =− 1 1 0 q ω T 1 ( µ) rotate molecule frame into space frame for a transition polarized along the molecule z-axis, H ( ) = −ε Z t D 1 00 ω µ z . 0 J H t × J M J  M J 1 0 = −ε z P ( M selection rule) ( ) µ z ( 1) J   J M   −Ω 1 0 J ( 1) J ### −Ω J  q ( ∆Ω selection rule) 3–6 × [ (2 J + )( J ′ + 1) ] . This result is based on the spherical harmonic addition theorem which is derived in A. R. Edmonds Angular Momentum in Quantum Mechanics, page 62: 2 0 0 2 0 D ( j 1 ) ( ) D ( 2 j 2 ) 2 ( ) D ( 3 j 3 ) ( ) d sin 2 j 1 m j 2 m j 3 j 1 m m 1 j 2 m 2 j 3 m 3

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-17

longest

en

https://www.codeproject.com/Tips/86948/A-simple-program-to-solve-quadratic-equations-with?fid=1697167&df=90&mpp=50&sort=Position&spc=Relaxed&tid=4197641

text/html

crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00368.warc.gz

13,559,182 members Alternative Tip/Trick alternative version #### Stats 30.4K views 3 bookmarked Posted 10 Jun 2010 Licenced CPOL # A simple program to solve quadratic equations with , 8 Nov 2010 Simple and prints imaginary roots too!float a,b,c,x1,x2,d,dsq;printf("ax^2 + bx + c = 0");printf("\nEnter a,b,c separated by commas : \n");scanf("%f,%f,%f",&a,&b,&c);d = b*b-(4*a*c);if(d>=0){dsq=sqrt(d);x1 = (-b+dsq)/(2*a);x2 = (-b-(dsq))/(2*a);printf("\nRoot 1 : %f\nRoot 2... Simple and prints imaginary roots too! ```float a,b,c,x1,x2,d,dsq; printf("ax^2 + bx + c = 0"); printf("\nEnter a,b,c separated by commas : \n"); scanf("%f,%f,%f",&a,&b,&c); d = b*b-(4*a*c); if(d>=0) { dsq=sqrt(d); x1 = (-b+dsq)/(2*a); x2 = (-b-(dsq))/(2*a); printf("\nRoot 1 : %f\nRoot 2 : %f",x1,x2); } if(d<0) { d = ((4*a*c)-pow(b,2))/(2*a); printf("\nRoot 1 : %f+%fi",((-b)/(2*a)),d); printf("\nRoot 2 : %f-%fi",((-b)/(2*a)),d);}``` ## Share Student India No Biography provided ## You may also be interested in... View All Threads First Prev Next This is better than the others as you avoid the domain error... Andrew Phillips7-Nov-10 15:14 Andrew Phillips 7-Nov-10 15:14 Last Visit: 31-Dec-99 18:00     Last Update: 27-May-18 2:36 Refresh 1

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-22

latest

en

https://issuu.com/vistateam123/docs/120820095609-2932f591f8f349c5bd2f04c4ea7e173a

text/html

crawl-data/CC-MAIN-2018-05/segments/1516084891980.75/warc/CC-MAIN-20180123151545-20180123171545-00463.warc.gz

Stefan Boltzmann Law Stefan Boltzmann Law The Stefanâ&#x20AC;&#x201C;Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body per unit time (also known as the black-body irradiance or emissive power), , is directly proportional to the fourth power of the black body's thermodynamic temperature T. Heat Transfer Definition :- Heat is the form of energy. We also know that there is a transfer of heat only if there is a temperature difference. Heat Transfer is defined as:- The transmission of energy from one region to another region as a result of the temperature difference between them. OR The Heat transfer taking place between the bodies of different temperature. Types of Heat transfer :- Know More About :- Oxidation and Reduction Math.Tutorvista.com Page No. :- 1/4 There are three modes of Heat transfer :Conduction :- Conduction is a process where the heat transfer takes place between the two solid bodies in contact, two regions of the same solid body. This will happen because of the hot, vibrating, and rapidly moving molecules transfer the heat to their neighboring atoms. Convection :- The convection is a type of heat transfer where the heat transfer takes place through a medium and the medium may be liquid or the gas. The heat transfer takes place by the movement of fluid from one place to another. The heat transfer here is due to the bulk motion of the fluid. Convection is described by the Newtonâ&#x20AC;&#x2122;s law of cooling; the law states that the rate of heat loosed by a body is proportional to the difference in temperatures between the body and its surroundings. Radiation :The Third mode of energy transfer is Radiation Heat Transfer. The transfer of heat from hot body to a cold body with any material medium for propagation. Every object in the universe is made up of atoms and molecules. These atoms and molecules vibrate due to thermal energy present in them. Every object emits electromagnetic radiations because of the thermal vibrations of these atoms and molecules. In case of energy transfer, the radiation conversion of radiated electromagnetic energy to thermal energy takes place. Radiation Heat Transfer can also be termed as transfer of energy through waves. Math.Tutorvista.com Page No. :- 2/4 Heat Transfer Examples Examples of Heat Transfer through Conduction: -Take one long piece of metal. Put first end of this metal in the flame. Gradually, you will find that the temperature of the other end of the metal in your hand starts increasing. Energy gets transferred from the first end under flame to the second end of the metal in your hand by conduction. Examples of Heat Transfer through Natural Convection :- Natural convection is the convection which occurs naturally due to the bouncy effect. When water is heated in a pot then the particles, atoms or molecules of the water which are in contact with the pot gain energy. Kinetic energy of these particles gets increased. As a result, density of these particles decreases and they start moving upward towards the open surface of the port. Cool water which is in contact with the air in the port starts sinking downward and convection current gets established. Cool water in the upper surface is denser than the hot water in contact with the surface of the pot. Convection Currents flow in circular fashion. Therefore, heat is transferred by the movement of water (fluids). Forced convection: The convection that takes place by the force that is the force created from the fans, stirrers, pumps and etc. Examples of Heat Transfer through Radiation :- Radiations coming out from the burner of the electric stove or toaster coils. Transfer of heat energy from Sun to Earth or to us takes place due to the electromagnetic rays emitted by the sun. Another example is the heating of food inside the microwave. Math.Tutorvista.com Page No. :- 4/4 Thank You For Watching Presentation Stefan Boltzmann Law Stefan Boltzmann Law

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-05

latest

en

https://atcoder.jp/contests/jsc2019-final/submissions/7993108

text/html

crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00092.warc.gz

Contest Duration: - (local time) (180 minutes) Back to Home Submission #7993108 Source Code Expand Copy ```import sys from collections import defaultdict from heapq import heappush, heappop # Performance Test class Treap: def __init__(self): self.root = 0 self.left = [0] self.right = [0] self.children = [self.left, self.right] self.key = [0] self.priority = [0] self.count = [0] self.deleted = set() self.rand = self.xor128() def xor128(self): x = 123456789 y = 362436069 z = 521288629 w = 88675123 while True: t, x, y, z = x ^ ((x << 11) & 0xffffffff), y, z, w w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)) yield w def get_size(self): return self.count[self.root] def _merge(self, li, ri): left = self.left right = self.right children = self.children priority = self.priority count = self.count stack = [] while li != 0 and ri != 0: if priority[li] > priority[ri]: stack.append((li, 1)) li = right[li] else: stack.append((ri, 0)) ri = left[ri] vi = li if li != 0 else ri for pi, d in reversed(stack): children[d][pi] = vi count[pi] = count[left[pi]] + count[right[pi]] + 1 vi = pi return vi def _split_by_key(self, vi, x): """ :return: (LeftRoot, RightRoot) LeftRoot: Root node of the split tree consisting of nodes with key < x. RightNode: Root node of the split tree consisting of nodes with key >= x. """ left = self.left right = self.right key = self.key count = self.count l_stack = [] r_stack = [] while vi != 0: if x < key[vi]: r_stack.append(vi) vi = left[vi] else: l_stack.append(vi) vi = right[vi] li, ri = 0, 0 for pi in reversed(l_stack): right[pi] = li count[pi] = count[left[pi]] + count[li] + 1 li = pi for pi in reversed(r_stack): left[pi] = ri count[pi] = count[ri] + count[right[pi]] + 1 ri = pi return li, ri def insert(self, x): left = self.left right = self.right children = self.children key = self.key priority = self.priority count = self.count np = next(self.rand) if self.deleted: ni = self.deleted.pop() left[ni] = 0 right[ni] = 0 key[ni] = x priority[ni] = np count[ni] = 1 else: ni = len(self.key) left.append(0) right.append(0) key.append(x) priority.append(np) count.append(1) vi = self.root pi = 0 d = 0 while vi != 0: if np > priority[vi]: li, ri = self._split_by_key(vi, x) left[ni] = li right[ni] = ri count[ni] = count[li] + count[ri] + 1 break pi = vi d = int(x >= key[vi]) count[vi] += 1 vi = children[d][vi] if pi == 0: self.root = ni else: children[d][pi] = ni def delete(self, x): left = self.left right = self.right children = self.children key = self.key count = self.count vi = self.root pi = 0 d = 0 while vi != 0: if key[vi] == x: vi = self._merge(left[vi], right[vi]) break pi = vi d = int(x >= key[vi]) count[vi] -= 1 vi = children[d][vi] if pi == 0: self.root = vi else: children[d][pi] = vi def upper_bound(self, x): """ :return (Node, i) Node: with the smallest key y satisfying x < y. i: 0-indexed order. If same keys exist, return leftmost one. If not exists, return (None, n). """ left = self.left right = self.right key = self.key count = self.count vi = self.root ti = 0 i = count[vi] j = 0 while vi != 0: if x < key[vi]: ti = vi i = j + count[left[vi]] vi = left[vi] else: j += count[left[vi]] + 1 vi = right[vi] return ti, i def lower_bound(self, x): """ :return (Node, i) Node: with the smallest key y satisfying x <= y. i: 0-indexed order. If same keys exist, return leftmost one. If not exists, return (None, n). """ left = self.left right = self.right key = self.key count = self.count vi = self.root ti = 0 i = count[vi] j = 0 while vi != 0: if x <= key[vi]: ti = vi i = j + count[left[vi]] vi = left[vi] else: j += count[left[vi]] + 1 vi = right[vi] return ti, i def get_by_index(self, i): """ :return (0-indexed) i-th smallest node. If i is greater than length, None will be returned. """ left = self.left right = self.right count = self.count if i < 0 or self.get_size() <= i: return 0 vi = self.root j = i while vi != 0: l_cnt = count[left[vi]] if l_cnt == j: return vi if j < l_cnt: vi = left[vi] else: j -= l_cnt + 1 vi = right[vi] assert False, 'Unreachable' def get_max(self): # 多くの場合において処理が単純な分 get_by_index(get_size - 1) より速いが、 # テストケースによっては途中で処理を打ち切れる get_by_index の方が速いことがある right = self.right vi = self.root if vi == 0: return 0 while right[vi] != 0: vi = right[vi] return vi def get_next(self, i, vi): """ :return: next node of i-th "node". (= (i+1)th node) """ # If node has right child, the root-node search can be omitted. # Otherwise, get_by_index(i+1). left = self.left right = self.right if vi == 0: return 0 if right[vi] == 0: return self.get_by_index(i + 1) vi = right[vi] while left[vi] != 0: vi = left[vi] return vi def get_prev(self, i, vi): left = self.left right = self.right if vi == 0: return 0 if left[vi] == 0: return self.get_by_index(i - 1) vi = left[vi] while right[vi] != 0: vi = right[vi] return vi def debug_print(self): self._debug_print(self.root, 0) def _debug_print(self, vi, depth): if vi != 0: self._debug_print(self.left[vi], depth + 1) print(' ' * depth, self.key[vi], self.priority[vi], self.count[vi]) self._debug_print(self.right[vi], depth + 1) def dist_insert(x): heappush(dists, x) available_dists[x] += 1 def dist_delete(x): available_dists[x] -= 1 def dist_get_min(): while dists and available_dists[dists[0]] == 0: heappop(dists) if dists: return dists[0] return 0xffffffff n, l, x = map(int, input().split()) aaa = list(map(int, sys.stdin)) trp1 = Treap() dists = [] available_dists = defaultdict(lambda: 0) trp1key = trp1.key strawberry = {x} trp1.insert(min(x, l - x)) buf = [] for a in aaa: b = min(a, l - a) if a in strawberry: vi, i = trp1.lower_bound(b) li1 = trp1.get_prev(i, vi) li2 = trp1.get_prev(i - 1, li1) ri1 = trp1.get_next(i, vi) ri2 = trp1.get_next(i + 1, ri1) l1, l2, r1, r2 = trp1key[li1], trp1key[li2], trp1key[ri1], trp1key[ri2] if li2 != 0: dist_delete(b - l2) if ri1 != 0: dist_insert(r1 - l2) if ri2 != 0: dist_delete(r2 - b) if li1 != 0: dist_insert(r2 - l1) if li1 != 0 and ri1 != 0: dist_delete(r1 - l1) strawberry.remove(a) trp1.delete(b) else: trp1.insert(b) vi, i = trp1.lower_bound(b) li1 = trp1.get_prev(i, vi) li2 = trp1.get_prev(i - 1, li1) ri1 = trp1.get_next(i, vi) ri2 = trp1.get_next(i + 1, ri1) l1, l2, r1, r2 = trp1key[li1], trp1key[li2], trp1key[ri1], trp1key[ri2] if li2 != 0: dist_insert(b - l2) if ri1 != 0: dist_delete(r1 - l2) if ri2 != 0: dist_insert(r2 - b) if li1 != 0: dist_delete(r2 - l1) if li1 != 0 and ri1 != 0: dist_insert(r1 - l1) size = trp1.get_size() ci2 = trp1.get_by_index(size - 1) ci1 = trp1.get_prev(size - 1, ci2) buf.append(min(dist_get_min(), l - trp1key[ci1] - trp1key[ci2])) print(*buf) ``` #### Submission Info Submission Time 2019-10-17 01:19:40+0900 C - Maximize Minimum ikatakos PyPy3 (2.4.0) 700 9974 Byte AC 3737 ms 183248 KB #### Judge Result Set Name Sample All Score / Max Score 0 / 0 700 / 700 Status AC × 2 AC × 27 Set Name Test Cases Sample sample-01.txt, sample-02.txt All 01-01.txt, 01-02.txt, 01-03.txt, 01-04.txt, 01-05.txt, 01-06.txt, 01-07.txt, 01-08.txt, 01-09.txt, 01-10.txt, 01-11.txt, 01-12.txt, 01-13.txt, 01-14.txt, 01-15.txt, 01-16.txt, 01-17.txt, 01-18.txt, 01-19.txt, 01-20.txt, 01-21.txt, 01-22.txt, 01-23.txt, 01-24.txt, 01-25.txt, sample-01.txt, sample-02.txt Case Name Status Exec Time Memory 01-01.txt AC 168 ms 38512 KB 01-02.txt AC 581 ms 64472 KB 01-03.txt AC 527 ms 59996 KB 01-04.txt AC 537 ms 61144 KB 01-05.txt AC 538 ms 62556 KB 01-06.txt AC 501 ms 60892 KB 01-07.txt AC 517 ms 62680 KB 01-08.txt AC 484 ms 61788 KB 01-09.txt AC 514 ms 63704 KB 01-10.txt AC 516 ms 65496 KB 01-11.txt AC 588 ms 67672 KB 01-12.txt AC 495 ms 63196 KB 01-13.txt AC 576 ms 66524 KB 01-14.txt AC 3467 ms 133460 KB 01-15.txt AC 2370 ms 146132 KB 01-16.txt AC 1919 ms 138452 KB 01-17.txt AC 1972 ms 141392 KB 01-18.txt AC 2005 ms 136912 KB 01-19.txt AC 2196 ms 159060 KB 01-20.txt AC 2211 ms 157528 KB 01-21.txt AC 2136 ms 155860 KB 01-22.txt AC 3376 ms 182352 KB 01-23.txt AC 2703 ms 141652 KB 01-24.txt AC 3413 ms 183248 KB 01-25.txt AC 3737 ms 161360 KB sample-01.txt AC 170 ms 38256 KB sample-02.txt AC 167 ms 38256 KB

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-21

latest

en

http://dspcan.homestead.com/files/Ztran/zfour1.htm

text/html

crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00040-ip-10-31-129-80.ec2.internal.warc.gz

# Z Transform, Fourier Transform and the DTFT. Cuthbert Nyack The relation between the Z, Laplace and Fourier transform is illustrated by the above equation. It shows that the Fourier Transform of a sampled signal can be obtained from the Z Transform of the signal by replacing the variable z with ejwT. ## This proceedure is equivalent to restricting the value of z to the unit circle in the z plane. Since the signal is discrete and the spectrum is continuous, the resulting transform is referred to as the Discrete Time Frequency Transform (DTFT). The DTFT of a signal is usually found by finding the Z transform and making the above substitution. The exponential time function and its Z transform is shown above. Making the above substitution into the Z Transform gives the expression below for the Fourier Transform of the sampled exponential function. Expanding the complex exponent gives:- The magnitude of the Fourier Transform is given by. and the phase by:- Depending on the value of wT the magnitude of the spectrum has maxima and minima. The maxima occurs at frequencies:- and the minima at:- The periodicity of the maxima and minima is given by the sampling frequency:- The analog signal and its frequency spectrum is shown above and its magnitude and phase are given below. For small T and w, the DTFT can be written as:- The factor T arises because the DTFT refers to samples while the FT refers to areas. The applet above shows the DTFT of the sampled damped exponent and the FT of the continuous damped exponent. In the plot the magnitude of the sampled signal spectrum is multiplied by T for it to agree with the analog spectrum at low frequencies. eg parameters:- (0.3, 0.25, 0.5, 1.11, NA...) This shows (a) the Fourier transform in close agreement with the DTFT at low frequencies. (b) The real part of the spectrum has even symmetry about half the sampling frequency. (b) The imaginary part of the spectrum has odd symmetry about half the sampling frequency. (c) The spectrum of the DTFT is periodic in the frequency domain with period equal to the sampling frequency while the spectrum of the FT decreases continuously with frequency. eg parameters:- (0.3, 0.05, 0.25, 15.0, NA...) This shows significant difference exists between the DTFT and the FT at half the sampling frequency and this difference increases with the sampling time. This is an example of aliasing. ## For a discrete sampled signal, the DTFT spectrum is periodic and continuous. This is the opposite of the Fourier Series where the signal is periodic and continuous while the spectrum is discrete and non periodic. Because of the periodicity and the symmetry about half of the sampling frequency, only the portion of the spectrum between 0 and half the sampling frequency is used. This also means that for the spectrum of the sampled signal to approach the spectrum of the analog signal within the range 0 to ws/2 then the analog spectrum should not extend beyond ws/2. Alternatively this can be stated as:- ## The sampling frequency should be greater than twice the highest frequency in the signal spectrum. Generally the spectrum of the sampled signal becomes closer to that of the analog signal at low frequencies as the sampling frequency is increased.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2016-50

longest

en

https://nrich.maths.org/5675/index?nomenu=1

text/html

crawl-data/CC-MAIN-2017-47/segments/1510934805649.7/warc/CC-MAIN-20171119134146-20171119154146-00335.warc.gz

This problem builds on Consecutive Sums and invites you to explore how algebra can help with proof. If you are not familiar with the problem it would be good to look at that first. 12 can be made as a sum of consecutive numbers: 3 + 4 + 5. 15 can also be made as a sum of consecutive numbers in several ways, for example 7 + 8. At first it seems that maybe any total could be made as a sum of consecutive numbers. However, after trying several possibilities, it doesn't seem as though we can make 16 in this way. Every total from 17 to 31 can be made as a sum of consecutive numbers, until we reach 32. Let's use algebra to explore why that is. First : what is the sum of 7 consecutive numbers starting with 4? There is a quick way to calculate consecutive sums. Do you know it? Take a look at the next image. Take the set to be counted twice, the second time inverted. So the sum here $4+5+6+7+8+9+10= \frac{1}{2}\times (4+10) \times 7$ Can you use this method to find the sum of 8 consecutive numbers starting at10 ? Here's a much longer run of consecutive numbers, can you find its sum : 10, 11, 12, . . . . . . . . . 79, 80? What is the sum of n consecutive numbers starting at s ? This might help you to use algebra to reason why 16 and 32 are impossible to create as the sum of consecutive numbers. Can you make a general statement about numbers of that sort (powers of 2)? ##### Note: Why 24 ? is another problem where reasoning from algebra proves to be very powerful.

{"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}

CC-MAIN-2017-47

latest

en

http://mathfraction.com/square-root-fraction/adding-exponents/college-algebra-sullivan.html

text/html

crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00668.warc.gz

Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: Our users: The most thing that I like about Algebrator software, is that I can save the expressions in a file, so I can save my homework on the computer, and print it for the teacher whenever he asked for it, and it looks much prettier than my hand writing. David E. Coates, AZ I was really struggling with the older version... so much I pretty much just gave up on it. This newer version looks better and seems easier to navigate through. I think it will be great! Thank you! John Davis, TX I was having problems learning quadratic equations, until I purchased your software. Now I know how to do not only do quadratics, but I also learned with the step by step examples how to do other more difficult equations and inequalities. Great product! David E. Coates, AZ Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them? Search phrases used on 2010-12-03: • 8th grade math equation practice tests, fractions • simplifying ratio worksheets • +fist in mathth • Free Math Cheats for Algebra • Free Distributive Property Calculator • Adding and Subtracting Rationals Worksheets • Mathematical Formulas • Division Algorithm Solver • simplifying a ratio of polynomials calculator • 16 • Polynomial Calculator solver multiplication • rewrite the second order differential equation x'' + 3x' + 5x = t as a system of first order differential equations • solve equation negative fraction • free math solver with steps • kuta software infinite algebra I • rudin "chapter 7" • equivalent equations worksheets • simpify (x-5/x+5)-(x+1/x+5) • drawing exponential curve by using Gsp • 4xn=1 in fraction • answers to the college algebra quizes in my math lab • you are choosing between two health clubs • Java isPrime code • fractional differential equation matlab code • grade 11 Manipulates the basic mathematica loperations related to surds ppt • interpolation and extrapolation 4th grade worksheets • dividing radical expressions calculator online • coordinate graph pictures • factoring polynomials with fractional exponents • examples of polynomials • fraction equation calculator • free online math games for 9th graders • worlds hardest physics problem • Kuta software infinite Algebra 2 the distance formula • ردكلس نس يوتيوب • beginner lattice multiplication • 7th edhelperblog.com crossword • +Direct Variation Equation Worksheets wih answer key • slope worksheets gr.2 • example of a coefficient • Usable Online TI-84 Calculator • tutorial squaring twice function square root • number sentence is shown by shading on the grid • + math trivia with answers mathematics • adding subtracting multiplying and dividing integers worksheet • simplify as much as possible fractions • online proof solver • solve double integrals with radicals • In many instances, it is possible to use both fractional and decimal forms to multiply and divide. Waht are the advantages and disadvantages of using each form? • homework and problem-solving practice workbook course 2 glencoe 7th answers • linear equations marcy mathworks page 7.6 answer key • T1-83 • kuta software basic polynomial operations • pre algebra with pizzazz answers • programmed pre algebra books • kuta software infinite algebra 1 answer key • patience hall mathematics algebra 2 • algebrator.com • nonhomogeneous differential equations solver • pizzazz worksheet what should you say if you see a tall, wrought-iron tower in paris, france • free algebra word problem solver • Least Squares Regression Line Equation Real World Application • dividing rational expressions calculator • krystal bought a refrigerator from a rental center for \$1,050. she makes 16 monthly payments of \$112.75 with her credit card. the rental center charges \$1.25 for every payment made with a credit card. what is the total cost of the refrigerator? • Prentice Hall Algebra pdf • equations with combining like terms • graphing polynomial worksheet • graphing cal • math worksheet printable • hard math problems • multiplying algebraic terms worksheet • middle school math rational numbers word problems • compound inequalities calculator • Factoring by Distributive Property Worksheet • how to subtract a whole number from a mixform • multi-step equations practice problems key • quiz solver for 10th class Prev Next

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-05

latest

en

https://ulmerstudios.com/popular/are-wet-and-dry-ingredients-measured-the-same/

text/html

crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00713.warc.gz

# Are wet and dry ingredients measured the same? ## Are wet and dry ingredients measured the same? People often ask us if they really need separate measuring cups for wet and dry ingredients. If you are serious about baking, the answer is yes! While liquid and dry measuring cups do hold the same volume, the difference is that each is specially designed to do a better job of measuring its respective ingredients. ## Are measuring cups the same for liquids and dry? Technically, liquid and dry measuring cups hold the same volume, but they are specially designed to more accurately measure their respective ingredients. Dry measures larger than 1 cup are rare, unlike liquid measuring cups, which are commonly found in sets that measure up to 8 cups. Can I use a liquid measuring cup for dry ingredients? Can you use a liquid measuring cup to measure dry ingredients? Technically, yes. They both measure the same amount of volume. 1 cup in a dry measuring cup is the same as 1 cup in a liquid measuring cup. How do you measure dry and liquid ingredients? How to Measure Dry Ingredients 1. Step 1: Get Your Tools Ready. Get some dry measuring cups and measuring spoons. 2. Step 2: Measuring by Volume. 3. Step 3: Measuring Spoons. 4. Step 4: Measuring Brown Sugar. 5. Step 5: Measure by Weight. 6. Step 6: Zero Out the Scale. 7. Step 7: Select Grams or Ounces. 8. Step 8: You’ve Mastered Measuring. ### Is one cup of water the same as one cup of flour? 1 cup of water weighs 236 grams. 1 cup of flour weighs 125 grams. The volume is the same, but the weight is different (remember: lead and feathers). One other benefit to using metric measurements is accuracy: scales often only show ounces to the quarter or eighth of an ounce, so 4 1/4 ounces or 10 1/8 ounces. ### What is the best method to measure wet ingredients? Using a Cup for Thin Liquids. Start with a liquid measuring cup. Liquid measuring cups are the most accurate way to measure thin liquids, like water, milk, juice, and oils. They are made of transparent glass or plastic, and have markings on the sides to indicate volume in both metric and English units. Which is the first procedure in measuring dry ingredients? Use the dip and sweep method. The most common way to measure dry ingredients by volume is to dip your measuring cup into the dry ingredient so that the cup is overflowing. Take the back of a knife or a flat spoon handle and sweep the excess across and off of the measuring cup. What is the best liquid measuring cup? The Best Liquid Measuring Cup: Pyrex. In the end, the Pyrex 1 Cup Measuring Cup was our favorite once again. In addition to being accurate, the measurement markings were bold and easy to read. ## What are the 4 sizes of dry measuring cups? There are four (4) standards for dry measuring cup sizes and the standards are; 1 cup, ½ cup, 1/3, and ¼ cup. These 4 dry measuring cup sizes can measure any dry ingredient recipe. ## What is the function of a dry measuring cup? Unlike the liquid measuring cup, a dry measuring cup is used to measure the volume of solid ingredients. It has a flat rim that is used to get accurate measurements. What are measuring cup measurements? Measuring cups usually have capacities from 250 mL (approx. 1 cup) to 1000 mL (approx. 4 cups = 2 pints = 1 quart), though larger sizes are also available for commercial use.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-18

latest

en

http://razhayesheitanparastan.com/333yga/

text/html

crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00282.warc.gz

# Writing Equations From Word Problems Worksheet Wednesday, July 3rd 2019. | Math Worksheets Great, Fun and free mathematics should to pose a mathematical problem in ways. In , it's , and more fun, than you . There like a fun to own a kid to learn. R Fun Games for kids is an excellent mathematics website for teachers and parents desperate to assist their own kds get far r practice. The answer is seven-ninths. It is eight-tenths. In just two or more Three minutes, you can create the questions you with the possessions you desire. The are made randomly. Word issues grasp the use of decimals in exact life. The issue isn't an easy one, since it is sensible to say that very small learning objectives fulfilled by your . You can the of worksheets, the amount of decimal amounts to sort problem, the wide variety of digits in each telephone number, the solution to the decimal amounts. ## Solving Systems Of Linear Equations By Elimination Worksheet Answers ### Angle Of Elevation And Depression Trig Worksheet Answers #### 2 1 Economics Worksheet Answers ##### Solar System Worksheets ###### Holt Biology Cells And Their Environment Skills Worksheet Answers You may express the ratio of two numbers whenever expressing That number for a fraction. have to do heavy calculations and would be the issue in minutes. calculators is going to do the conversion simply and quickly, it's still for students to understand the concept in order to the calculator. A number of worksheets are dynamically generated so that you'll be supplied each chance to clinic. Stealing math worksheets on the internet is the ideal thing . you wish to pass out a Singapore mathematics worksheet to check as well as greatly advantage your origin in the that you design the worksheet to eventually become stimulating as you are ready to. The worksheets are created randomly, means you receive a Different one every moment. Furthermore, worksheets meant for junior classes really are a great relief for many parents since each sheet keeps the little one engaged for a long time. They simply a excellent way to learn, they're a great to revise your abilities and . It is possible to create your own subtraction worksheets along with your array of numbers. Decimal worksheets are great for they have difficulty comprehending their decimal or else they simply to solving decimal troubles. They are a great means for teachers to identify the students who are experiencing decimals, the particular problem . Besides default clinic difficulties, few especial worksheets on mental branch .

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-43

latest

en

http://www.imagemagick.org/discourse-server/viewtopic.php?f=22&t=19011&view=print

text/html

crawl-data/CC-MAIN-2020-16/segments/1585371662966.69/warc/CC-MAIN-20200406231617-20200407022117-00400.warc.gz

Page 1 of 1 ### Posible EWA speed improvement Posted: 2011-07-01T05:07:47-07:00 It would seem to me that using the intersection of the bounding box (with all sides parallel to the axes) and the bounding parallelogram (with horizontal top and bottom, but not necessarily vertical sides) to select which pixels to select would be a cheap way of speeding up the computation when the ellipse is slanted. (Probably not a huge improvement, and certainly no difference when resizing, but easy and cheap, so why not.) (Unfortunately, I have no time for this now.) ### Re: Posible EWA speed improvement Posted: 2011-07-01T22:37:06-07:00 First I moved this discussion to Digital Image Processing as it is very general and not specific to IM and its implementation. Actually I looked at that bounds solution, and was in fact implementing that bounds when I realised I could simplify it a little, leading to the parallelogram bounds. As a FYI here is my parallelogram bounds diagram... Essentially it equivalent to a square bounds of a circle that is sheared to form the ellipse, which is why it has a fixed hit/miss ratio, compared to the previous method of using orthogonal X,Y bounds. The question is how affordable is the calculation of the bounding box. WIth the parallelogram, I the extraction of a row of pixels is quite straight forward. floating point start point, which is offset by a fixed amount form one row to the next row. and it has a fixed floating point length, the result mapped to integers for the access to the data. Relative straight forward extention of the incremental code used for the quadratic distance formula (no sqrt is performed for that). For a bounding box aligned with the major and minor axis, you actually have to work out the union of two parallelograms. So instead of on set of values (start,increment,length), you need the union of two sets, one for the major axis bounds and one for the minor axis. Of course you can clip the Y axis directly by the orthogonal bounds of the ellipse. So to combine that with the parallelogram, you get a union of three parallelogram bounds when implementing. You could also add the X bounds to that union too, seeing you are doing unions. So for each row, you want the maximum of 4 possible start points, and minimum of 4 possible end points. With unions you no longer have a fixed length! Now I could be wrong, but its seems to me, that it is a lot of work, for very little reward. Especially when an orthogonally aligned ellipse (simple resize) generally results in the all three bounding boxes aligning anyway! Now pixel extraction in IM (and other images) can be complex and Virtual pixels become involved. You can't just reference the image data directly in that case. As such when Virtual-Pixels become involved, reducing the length of each row of pixels may have a major benefit. But it will have no benefit if all the pixels are real, and as such you can reference the image data directly (IM allows both in its pixel cache processing! Or at least did at one point.) So how beneficial this is open for debate, a test implementation may be useful as a comparison. However as IM pixel cache for IMv7 is in alpha level development, now is probably not a good time for testing it. ### Re: Posible EWA speed improvement Posted: 2011-07-03T06:07:39-07:00 Anthony: Thank you for the clear and detailed reply. I get your point. And I agree that using the intersection of the bounding box and of the parallelogram is likely to be a low bang for the buck "improvement." ### Re: Posible EWA speed improvement Posted: 2011-07-03T06:10:06-07:00 However, the computation of the bounding box is extremely cheap: http://git.gnome.org/browse/gegl/tree/g ... 6943#n2220

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-16

latest

en

https://www.pinterest.com/pin/232076187018537484/

text/html

crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00226-ip-10-185-27-174.ec2.internal.warc.gz

Pinterest • The world’s catalog of ideas # Explore Nets For 3D Shapes, Geometric Shapes, and more! A site that has EVERY 3D shape imaginable as a PDF so your kids can print, cut, fold and glue! Even lists the number of faces, edges, and vertices angry birds geometric solids - pdf files for the nets - oh my goodness ... my kids will LOVE this! Great idea on teaching rounding... An easy way to make it visual. Free math app for exploring number equivalence - compare fractions, decimals, and percents! This fun multiplication art project comes from Looking From Third to Fourth! Kids construct a city of arrays! Check out her site for details and a printable. I've taught this but substituted "solid shapes" for "3d shapes" Learning Ideas - Grades K-8: Geometry - Making 3D Shapes with Manipulatives Love it! Much better than toothpicks and marshmallows! This is BRILLIANT. Draw a number line with sharpie on a ziploc slider bag and use the slider as the tool to add or subtract two numbers. Awesome! from Education.com ### Shape Dimensions: Solid Figures three dimensional shapes and graphing Edible solid shapes... Great for teaching 3D Shapes Play ‘Hidden Numbers’ to help children recognise numbers - this game will get your child to associate shapes with numbers for easy identification. Cut a small hole in an envelope, put a number in it so that only part of it shows, & the child has to guess what the number is. Does it have a lot of lines? Maybe it's a 1 or a 7. Does it have circles or loops? It could be 3, 5, or 8. (Could also be adapted to support shape recognition) (“,) A Differentiated Kindergarten: Constructing Numbers . . . An Independent Place Value Freebie! OMG! I need to make one for 5th graders! The 4th Grade May-niacs: Math Review Folder Wow... I don't how this guy figured this out, but he is a genius. Trick to multiplication tables 6X6 and above... the ones I find most difficult to remember. # Pin++ for Pinterest # Place Value FREEBIE to assess number word, expanded form and standard form of numbers. Make puzzles out of 100 charts for kids to put together...this would be a good review for third graders with a multiplication chart. Give them the "frame" and they use their multiplication skills and puzzle shapes as hints to get it together. Could cut pieces into various sizes/shapes for differentiation.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2016-30

latest

en

https://www.jiskha.com/display.cgi?id=1334694499

text/html

crawl-data/CC-MAIN-2017-34/segments/1502886116921.70/warc/CC-MAIN-20170823000718-20170823020718-00509.warc.gz

# Physics posted by . Two parallel wires conduct equal currents in the same direction. (a) At a point in the plane of the wires midway between them, do the individual fields of the wires cancel or reinforce each other? (b) At a point far away from the wires, do their individual fields tend to cancel or reinforce each other? ## Similar Questions Two parallel wires are carrying 5A and are separated by 2.0 m. if the currents are in opposite directions, what is the magnitude of the magnetic field halfway between them? 2. ### College Physics Two long, parallel wires separated by 50 cm each carry currents of 3.0 A in a horizontal direction. Find the magnetic field midway between the wires if the currents are in the same direction. B= T Find if they are in opposite directions. … 3. ### Physics Two parallel wires carry currents in the same direction. I1=3A and I2=1A and the distance between the conductors is 4 cm. The magnetic field midway between these two wires is 4. ### Physics Two parallel wires carry currents in the same direction. I1=3A and I2=1A and the distance between the conductors is 4 cm. The magnetic field midway between these two wires is 5. ### Physics Two parallel, uniformly charged, infinitely long wires carry opposite charges with a linear charge density ? Two parallel wires conduct equal currents in the same direction. (a) At a point in the plane of the wires midway between them, do the individual fields of the wires cancel or reinforce each other? 7. ### Physics/math Two parallel wires carry 1-A currents in unknown directions. The distance between the wires is 10-cm. What is the magnitude of the magnetic field B in Teslas at a point P located 6-cm away from the axis of one of the wires and 8-cm … 8. ### Physics Two parallel wires carry 1-A currents in unknown directions. The distance between the wires is 10-cm. What is the magnitude of the magnetic field B in Teslas at a point P located 6-cm away from the axis of one of the wires and 8-cm … 9. ### physics Two parallel wires carry 1-A currents in unknown directions. The distance between the wires is 10-cm. What is the magnitude of the magnetic field B in Teslas at a point P located 6-cm away from the axis of one of the wires and 8-cm … 10. ### Physics Two parallel wires are carrying current. If the currents are in the same direction, what is the interaction between the two wires and why? More Similar Questions

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-34

latest

en

https://www.slideserve.com/alexandre/one-form-powerpoint-ppt-presentation

text/html

crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00413.warc.gz

One Form 1 / 9 # One Form - PowerPoint PPT Presentation ##### One Form Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. One Form 2. Map F Goes from manifold MR All points in M go to R Smooth, i.e. differentiable Function f(x, y) Map UR Open set UM Region U is diffeomorphic to E2 (or En) Scalar Field 3. One-Form • The scalar field F is differentiable. • Expressed in local variables • f associated with a chart • Need knowledge of the coordinates • from the local chart f • short cut is to use F. • The entity dF is an example of a one-form. 4. The derivative of the one-form can be written as an operator. Chain rule applied to x, y A point can be described with other coordinates. Partial derivatives affected by chain rule Write with constants reflecting a transformation Operator and Coordinates 5. The partial derivatives point along coordinate lines. Not the same as the coordinates. Partial Derivative y y = const. x x = const. Y Y = const. X = const. X 6. Vector Field • General form of differential operator: • Smooth functions A, B • Independent of coordinate • Different functions a, b • Transition between charts • This operator is a vector field. • Acts on a scalar field • Measures change in a direction F(p’) x p’ F(p) p 7. Inner Product • The one-form carries information about a scalar field. • Components for the terms • The vector field describes how a scalar field changes. • The inner product gives a specific scalar value. • Express with components • Or without 8. Three Laws • Associativity of addition • Associativity of multiplication • Identity of a constant k = const. 9. Vector field on Q Contravariant vectors Components with superscripts Transformation rule: One form on Q Covariant vectors Components with subscripts Dual Spaces next

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-43

latest

en

https://www.ruby-forum.com/t/koans-uninitialized-constant-triangleerror/223121

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00482.warc.gz

# (koans) uninitialized constant TriangleError Hello, I have to make a exception part on triangles now. The exercise looks like this : [/code] **# You need to write the triangle method in the file ‘triangle.rb’ require ‘triangle.rb’ # Let’s handle that part now. def test_illegal_triangles_throw_exceptions assert_raise(TriangleError) do triangle(0, 0, 0) end assert_raise(TriangleError) do triangle(3, 4, -5) end assert_raise(TriangleError) do triangle(1, 1, 3) end assert_raise(TriangleError) do triangle(2, 4, 2) end # HINT: for tips, see end end [/code] So I looked at that page and come with this solution : ``````def triangle(a, b, c) raise TriangleError, "length cannnot be 0 or lesser" if (a <= 0) or (b <= 0) or (c <= 0); raise TriangleError, "length does not mach Pyschotorogas" if (a * a + b * b != c * c); return :equilateral if ((a == b) and (b == c)) return :isosceles if (((a == b) and (b != c)) or ((a != b) and (b == c)) or ((a == c) and (a != b))) return :scalene if ((a !=b) and (b != c)) end`````` But now I get a uninitialized constant TriangleError What did I do wrong now ? Roelof Hello, I found this solution : # and def triangle(a, b, c) raise (TriangleError), “length cannnot be 0 or lesser” if (a <= 0) or (b <= 0) or (c <= 0); raise (TriangleError), “length does not match Pyschotorogas” if (a * a + b * b != c * c); return :equilateral if ((a == b) and (b == c)) return :isosceles if (((a == b) and (b != c)) or ((a != b) and (b == c)) or ((a == c) and (a != b))) return :scalene if ((a !=b) and (b != c)) end # Error class used in part 2. No need to change this code. class TriangleError < StandardError end But now when a error is found the script is stopped. Can anyone give me a hint how I can continue this script. Roelof Op zaterdag 22 september 2012 16:43:03 UTC+2 schreef roelof het volgende: 1. Very good. Now you should try to determine what “Ruby on Rails” is, and whether your questions have anything to do with Ruby on Rails. 2. Read a tutorial on ruby exception handling, or better yet buy the book “Beginning Ruby”: http://beginningruby.org/ and read that before trying to do the koans. All you seem to be doing is copying other people’s code, so what good does that do you? In fact, you know so little about ruby that you think Ruby on Rails is ruby.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-30

latest

en

http://mathhelpforum.com/differential-geometry/125803-norm-adjoint.html

text/html

crawl-data/CC-MAIN-2018-26/segments/1529267864957.2/warc/CC-MAIN-20180623093631-20180623113631-00573.warc.gz

Suppose $\displaystyle T\in L(H,H)$ where $\displaystyle H$ is a Hilbert space. How do I go about showing $\displaystyle \parallel T\parallel =\parallel T^*\parallel$? So far all I am able to get to is that $\displaystyle \parallel T\parallel^2\le\parallel T^*T\parallel$ 2. For $\displaystyle x \in H$ $\displaystyle \|T^{*}x\|^2=|\langle T^{*}x,T^{*}x\rangle|=|\langle x,TT^{*}x \rangle| \le \|x\|\|TT^{*}x\| \le \|x\|\|T\|\|T^*x\|$. Divide by $\displaystyle \|T^{*}x\|$ to get $\displaystyle \|T^{*}x\| \le \|T\|\|x\| \Longrightarrow \|T^*\| \le \|T\|.$ Since $\displaystyle T^{**}=T$, we have $\displaystyle \|T\| \le \|T^*\|$. Therefore, $\displaystyle \|T^*\|=\|T\|$. 3. Oh Wow how did I not see that. Thanks.

{"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}

CC-MAIN-2018-26

latest

en

https://renewabletechy.com/what-is-the-unit-equivalent-to-watts/

text/html

crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00452.warc.gz

# What Is The Unit Equivalent To Watts? A watt is a unit of power that measures the rate of energy conversion or transfer. It quantifies the amount of energy converted per unit of time. Watts are important because they allow us to understand how much power devices consume and systems produce. Knowing watt equivalents also enables conversion between different units of power. Having a standard unit like the watt is essential for measuring and comparing power usage across many applications, from small electronics to large motors and generators. By understanding equivalents, we can contextualize watt values and convert between units like kilowatts or horsepower. This helps relate power consumption to real-world energy usage across fields like engineering, physics, and electronics. ## The Origins of Watts The watt unit of power is named after James Watt, the Scottish inventor and mechanical engineer. In the late 18th century, Watt worked to improve the efficiency and power output of the steam engine. At the time, Thomas Newcomen’s early steam engines wasted a tremendous amount of energy by repeatedly cooling and reheating the engine cylinder. Watt introduced a separate condenser, which dramatically improved the steam engine’s efficiency. He also devised several other improvements to make steam engines more practical for use in industry. Watt’s contributions were monumental in ushering in the Industrial Revolution. His improved steam engine helped power factories, mills, trains and ships, driving the rapid industrialization of England in the late 1800s. The unit of power measuring the rate of energy conversion was named the watt in honor of James Watt and his pioneering work to develop a practical steam engine. ## Defining Watts A watt is the standard International System of Units (SI) unit used to measure power. Power is defined as the rate at which energy is converted from one form to another or transferred from one place to another. Specifically, one watt is equal to one joule of energy converted per second. This means that power is a measure of the rate of energy flow or energy transfer. A light bulb, for example, converts electrical energy into light and heat energy. The rate at which it performs this energy conversion is measured in watts. Watts are useful because they provide a standard way to measure the rate of energy conversion regardless of the type of energy involved. Watts allow us to quantify and compare the power of engines, motors, light bulbs, batteries, and other systems that use, convert, or transfer energy. ## Common Equivalents There are a few common equivalent units that are used to measure power alongside watts. Understanding these units can make it easier to conceptualize the amount of power in watts. One of the most direct equivalents is joules per second. A joule is a unit of energy and work. Power is the rate at which energy is transferred or work is done, so watts can be expressed as joules per second, or joules/second. This helps relate watts back to the base units it is derived from. Another equivalent to watts is kg m2/s3. This expression uses the base units of mass (kg), distance (m), and time (s). Watts are defined as kg m2/s3 because power is the product of force and velocity. Force involves mass, distance, and acceleration (which includes time). Velocity also includes distance and time. So the base units used to measure force and velocity combine to form the kg m2/s3 equivalent for watts. The final common equivalent is volt amperes. Watts are often used to measure electric power, which is the product of current (amperes) and voltage (volts). So electric power in watts can be expressed as volt amperes, or volts multiplied by amps. ## Watts to Kilowatts When looking at electrical power, watts and kilowatts are two common units used. While they represent the same thing – power – they differ in the amount. Specifically, a kilowatt is equal to 1,000 watts. This means that 1 watt is equal to 0.001 kilowatts. To convert between the two: • To convert watts to kilowatts, divide the watts by 1,000. • To convert kilowatts to watts, multiply the kilowatts by 1,000. Some examples: • 100 watts = 0.1 kilowatts • 1,000 watts = 1 kilowatt • 10,000 watts = 10 kilowatts So in electrical terms, 1 watt is equivalent to 0.001 kilowatts. Knowing the relationship between these two units is helpful for calculating and comparing power consumption and requirements. ## Watts to Horsepower One of the most common equivalents for watts is horsepower, used to measure the power of mechanical engines and motors. The conversion between watts and horsepower is: 1 watt = 0.00134102 horsepower To convert watts to horsepower, simply multiply the watts value by 0.00134102. For example: • 1000 watts x 0.00134102 = 1.34102 horsepower • 500 watts x 0.00134102 = 0.6705 horsepower • 2000 watts x 0.00134102 = 2.68204 horsepower This allows an easy conversion between the electric power in watts of a device, and the equivalent mechanical power in horsepower of an engine needed to power the device. ## Watts to BTUs One common unit equivalent to watts is the British thermal unit (BTU). The BTU is a unit of energy used to measure heat. Specifically, one BTU is defined as the amount of energy needed to raise the temperature of one pound of water by one degree Fahrenheit. The conversion between watts and BTUs per hour (BTU/h) is: 1 watt = 3.41214 BTU/h So for example, a 1500 watt space heater converts to around 5118 BTU/h (1500 x 3.41214). This indicates that the 1500 watt heater can produce 5118 BTUs of heat energy per hour. Going the other way, if you had a device rated for 10000 BTU/h, you could convert that to around 2934 watts (10000 / 3.41214). The watt to BTU/h conversion is useful for relating the power consumption of an electrical device like a heater or air conditioner to its heating or cooling output. The BTU provides a standard unit for comparing the thermal performance of different devices. ## Watts in Electricity Watts are commonly used to measure electric power. Electric power refers to the rate at which electrical energy is transferred by an electric circuit. For example, a 15-watt light bulb uses 15 watts of electric power. The wattage of electrical devices indicates the rate at which they consume energy. High wattage devices require more power to function than low wattage ones. Knowing the wattage of appliances and devices allows us to determine the electrical load and requirements for power circuits and generators. Wattage also helps calculate electricity usage. Electricity bills typically charge per kilowatt-hour (kWh), which represents the amount of energy consumed over time. By multiplying the wattage of a device by the time it runs, you can estimate the energy usage and cost to operate it. Overall, watts are an essential metric for measuring and managing electrical power. ## Watts in Light Bulbs The wattage rating of a light bulb indicates how much power it draws. Different types of bulbs produce different amounts of light for the same wattage. An incandescent 60W bulb produces about 800 lumens, while an LED bulb with the same wattage produces around 800-1100 lumens. This is because incandescent bulbs produce light by heating a filament, which wastes energy as heat. LED bulbs are far more energy efficient, converting over 80% of energy input into light. CFL bulbs fall in between, producing about 600-800 lumens per 60W. They are more efficient than incandescent but contain mercury, whereas LEDs are non-toxic. So for the same brightness an LED bulb will have a lower wattage than an incandescent. A 9-13W LED bulb produces about the same 800 lumens as a 60W incandescent. Going with more efficient bulbs is an easy way to reduce energy usage and costs. ## Conclusion In summary, watts are an important unit for measuring power. Understanding the equivalencies between watts and other units like kilowatts, horsepower, and BTUs allows us to conceptualize power levels and make conversions when needed. Knowing how many watts are in common electrical devices also provides a practical understanding of their power consumption. While a lightbulb may use only 60-100 watts, larger appliances like a microwave often use 1000+ watts. Being able to convert between units is useful for both science and everyday situations. Watts provide a standard unit for comparison of power across any application. Whether optimizing an industrial motor or simply choosing energy efficient home appliances, having an intuitive grasp of watts and their equivalents enables better-informed decisions.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-22

latest

en

https://www.worldnumerology.com/numerology-Course-11a.htm

text/html

crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00094.warc.gz

# NUMEROLOGY COURSE CHAPTER 13 Karmic Lesson & Hidden Passion Numbers ## Karmic Lessons & Hidden Passions are two sided of the same coin. Karmic Lessons are numbers missing among the letters in the full name at birth. Hidden Passsions are the numbers most often represented in the full name at birth. The Karmic Lessons and Hidden Passion chart is represented by a square divided in 9 smaller squares. Each of those nine squares is numbered 1 through 9 and arranged in order starting with 1 in the top left and ending with 9 in the lower right corner. In the top of each square is another number indicating how many times the number below it is represented in the full name at birth. You can have none, one, or several Karmic Lessons in your numerology chart. Tom has no Karmic Lessons. None of the squares in the Karmic Lesson / Hidden Passion chart has a zero in it, because at least 1 of each number is found in his name. Ω ### THE MEANING OF THE KARMIC LESSON NUMBERS Numerology states that we enter life with certain strengths and weaknesses. Karmic lessons are areas that we are currently weak in and must be faced and worked on in this life. There can be more than one Karmic Lesson. These are indicated by the absence of certain numbers in your name. The letters and numbers of your name point to talents and abilities that you possess. These characteristics can be compared to a workshop in which certain tools are available to you. Missing numbers, those that are not represented in the letters in your name, imply tools that are unavailable, and must be learned and mastered during this lifetime. Ω ### THE MEANING OF THE HIDDEN PASSION NUMBERS The number that is most often repeated in your name represents your specific field of expertise, or a concentrated talent – this is called your Hidden Passion. As with Karmic Lessons, you can have more than one Hidden Passion. Tom Cruise, for example, has 4 letters with the value of 1 in his name. He has two letters with the value of 2, two letters with the value of 3, and so forth. Notice, however, that he has more letters with the value of 1, than any other letters. Therefore, his Hidden Passion is 1. Your Hidden Passion reveals one or more special strengths and talents that you rely upon and are available to you. The Hidden Passion represents your specific field of expertise, or a concentrated talent. Metaphorically, this talent can be seen as having a power all its own to shape your life. Its existence gives you a strong desire to develop and to express that particular ability. Having the talent demands that you express it, that you experience this part of you, and that you live according to its nature. In this way, the Hidden Passion shapes your personality, and guides your life. Ω Audio Numerology Lecture; The Karmic Lessons and Hidden Passion numbers. 17 Minutes ### To learn the meanings of the Karmic Lesson and HiddenPassion numbers - see page 115 in the bookNumerology: A Complete Guide to Understanding and Using Your Numbers of Destiny So far, we have talked about the strongest, and usually the most obvious, aspects of a person's character. But now we enter into the more hidden realms, where people are driven by desires or ambitions that even they may not be aware of. In order to become free human beings, we must make these subtle, unconscious traits conscious. We must become aware of ourselves in our deepest recesses in order to take control of our lives. Ω Numerology Course Chapter 14 THE SUBCONSCIOUS SELF NUMBER Table of Content Ω

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-49

latest

en

http://gec.org.ru/index.php/download/introduction-to-noncommutative-algebra-universitext

text/html

crawl-data/CC-MAIN-2019-09/segments/1550247489304.34/warc/CC-MAIN-20190219020906-20190219042906-00336.warc.gz

# Download Introduction to Noncommutative Algebra (Universitext) by Matej Brešar PDF By Matej Brešar Offering an user-friendly creation to noncommutative earrings and algebras, this textbook starts off with the classical concept of finite dimensional algebras. in simple terms after this, modules, vector areas over department jewelry, and tensor items are brought and studied. this can be via Jacobson's constitution conception of earrings. the ultimate chapters deal with unfastened algebras, polynomial identities, and jewelry of quotients. Many of the consequences aren't awarded of their complete generality. quite, the emphasis is on readability of exposition and ease of the proofs, with numerous being diversified from these in different texts at the topic. necessities are stored to a minimal, and new suggestions are brought steadily and are conscientiously stimulated. creation to Noncommutative Algebra is as a result obtainable to a large mathematical viewers. it's, notwithstanding, essentially meant for starting graduate and complex undergraduate scholars encountering noncommutative algebra for the 1st time. Read Online or Download Introduction to Noncommutative Algebra (Universitext) PDF Best algebra books Groebner bases algorithm: an introduction Groebner Bases is a method that offers algorithmic ideas to various difficulties in Commutative Algebra and Algebraic Geometry. during this introductory educational the fundamental algorithms in addition to their generalization for computing Groebner foundation of a collection of multivariate polynomials are offered. The Racah-Wigner algebra in quantum theory The improvement of the algebraic features of angular momentum idea and the connection among angular momentum idea and certain issues in physics and arithmetic are coated during this quantity. Wirtschaftsmathematik für Studium und Praxis 1: Lineare Algebra Die "Wirtschaftsmathematik" ist eine Zusammenfassung der in den Wirtschaftswissenschaften gemeinhin benötigten mathematischen Kenntnisse. Lineare Algebra führt in die Vektor- und Matrizenrechnung ein, stellt Lineare Gleichungssysteme vor, berichtet über Determinanten und liefert Grundlagen der Eigenwerttheorie und Aussagen zur Definitheit von Matrizen. Additional resources for Introduction to Noncommutative Algebra (Universitext) Example text 5] J. Dixmier, Sur les algebres de Weyl, Bull. Soc. Math. Prance, 96, 1968. [6] A. Guichardet, Homologie de Hochschild des deformations d'algebres de polynomes, a paraitre. [7] C. Kassel, L'homologie cyclique des algebres enveloppantes, Invent. , 91, 1988, 221-251. [8] M. Lorenz, Crossed Products: Characters, Cyclic Homology, and Grothendieck Groups, Non commutative Rings, Math. Sciences Research Institute Publications, 24, Springer Verlag, 1992, 69-98. [9] S. , 818, Springer Verlag, 1980. [10] P. Let 9 be a filiform Lie algebra of dimension n + 1 2 7 nonisomorphic to £n and Qn- Then 9 is characteristically nilpotent if and only if 9 is not isomorphic to its sill algebra . 2 Description of Lie algebras whose nilradical is filiform Let 9 be a Lie algebra. The semidirect decomposition 9 = s EI1 r holds, where r is the radical in 9 (the Levi decomposition), and all Levi subalgebras are mutually conjugate (Mal'tsev's theorem [10]). These theorems suggest to consider the problem of classification of Lie algebras with a fixed radical [15]. Dans ce qui suit nous nous placerons sous l'hypothese de la derniere proposition; rappelons maintenant un theoreme de structure (voir [12]) bien adapte a la K-theorie reelle. 2 Supposons que Tors (K*(X)) = a et que la conjugaison decompose Ie groupe abelien (K* (X) sous la forme: (K*(X) = M+ EBTEBT* de sorte que soit la multiplication par+1 (respectivement-1) sur M+(respectivement sur M _) et d 'autre part echange T et T*; en outre soient hI, ... , h n E KO* (X) tels que les c(h i ) forment une base pour K* (+) ® (M+ EB M_) en tant que K* (+) -module.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-09

latest

en

https://www.daniweb.com/programming/web-development/threads/485417/future-value-investment

text/html

crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00507.warc.gz

Here is the problem: Write a program to find future value of monthly investments. Start with an initial investment, make a deposit every 30days, calculate interest on principle compounded daily, display a table showing beginning balance, deposit for the year, interest earned for the year, and ending balance. The table should display the aforementioned information for a duration of your present age to 65. Display result in a table format as shown below. Here is the code so far..... `````` <html> <title>Future Value Calculation</title> <style type = "text/css"> table { width: 100% } th { text-align : left } th span { color: #008; margin-left: 15px; } </style> <script type = "text/javascript"> { var EndBal; var principal = 2000.00; var rate = 0.12; var deposit = 1200; var interest; var BegBal = principal + deposit; for ( var age = 16; age >= 1; --age ) { // Do calculations EndBal = ((principal + deposit) * Math.pow(1 + rate, age )); interest = ((principal + deposit) * Math.pow(1 + rate, age ))-(principal+deposit); // Insert row var row = getById("tBody").insertRow(0); //Insert row at index 0 // Create cells for the created row // Empty cells are set to to be displayed in the table row.insertCell(0).innerHTML = age+50; row.insertCell(1).innerHTML = EndBal.toFixed(2); row.insertCell(2).innerHTML = interest.toFixed(2); row.insertCell(3).innerHTML = deposit; row.insertCell(4).innerHTML = BegBal.toFixed(2); BegBal = EndBal; } getById("spnInitial").innerHTML = principal.toFixed(2); getById("spnRate").innerHTML = rate.toFixed(2); getById("spnDeposit").innerHTML = deposit; getById("spnInvestment").innerHTML = EndBal.toFixed(2); } // Helper function to get element by id function getById(id) { return document.getElementById(id); } </script> <body> <table border="1"> <tr><th colspan="5">Future Value Calculation</th></tr> <tr><th colspan="5">Initial Investment:<span id="spnInitial"></span></th></tr> <tr><th colspan="5">Interest Rate:<span id="spnRate"></span></th></tr> <tr><th colspan="5">Deposit every 30 days:<span id="spnDeposit"></span></th></tr> <tr><th colspan="5">Investment started:<span id="spnInvestment"></span></th></tr> <tr> <th>Age</th> <th>Beg Bal</th> <th>Interest</th> <th>Deposits</th> <th>Ending Bal</th> </tr> <tbody id="tBody"> </tbody> </table> </body> </html> `````` But it does not calculate correctly. I need someone to help out with the loop... I don't think this is correct at all, but it makes much more sense than yours do... maybe it can help: `````` <html> <title>Future Value Calculation</title> <style type = "text/css"> table { width: 100% } th { text-align : left } th span { color: #008; margin-left: 15px; } </style> <script type = "text/javascript"> { var EndBal; var principal = 2000.00; var rate = 0.12; var deposit = 1200; var interest; var BegBal = principal; var initialAge = 16; for ( var age = initialAge, i=1; age <= 65; age++, i++ ) { EndBal = ((principal + (deposit*12* i)) * Math.pow(1 + rate, i )); interest = EndBal-(principal+(deposit*12*i)); // Insert row var row = getById("tBody").insertRow(getById("tBody").rows.length); //Insert row at index 0 // Create cells for the created row // Empty cells are set to to be displayed in the table row.insertCell(0).innerHTML = age; row.insertCell(1).innerHTML = formatNum(BegBal); row.insertCell(2).innerHTML = formatNum(interest); row.insertCell(3).innerHTML = formatNum((deposit*12*i)); row.insertCell(4).innerHTML = formatNum(EndBal); BegBal = EndBal; } getById("spnInitial").innerHTML = formatNum(principal); getById("spnRate").innerHTML = rate.toFixed(2); getById("spnDeposit").innerHTML = formatNum(deposit); getById("spnInvestment").innerHTML = formatNum(EndBal); } // Format number function formatNum(val) { return val.toFixed(2).replace(/\d(?=(\d{3})+\.)/g, '\$&,'); } // Helper function to get element by id function getById(id) { return document.getElementById(id); } </script> <body> <table border="1"> <tr><th colspan="5">Future Value Calculation</th></tr> <tr><th colspan="5">Initial Investment:<span id="spnInitial"></span></th></tr> <tr><th colspan="5">Interest Rate:<span id="spnRate"></span></th></tr> <tr><th colspan="5">Deposit every 30 days:<span id="spnDeposit"></span></th></tr> <tr><th colspan="5">Investment started:<span id="spnInvestment"></span></th></tr> <tr> <th>Age</th> <th>Beg Bal</th> <th>Interest</th> <th>Deposits</th> <th>Ending Bal</th> </tr> <tbody id="tBody"> </tbody> </table> </body> </html> `````` Thank you. Let me check to see if it works for what I am looking for.. no, it did not yield the correct values... here is the problem again, "Write a program to find future value of monthly investments. Start with an initial investment, make a deposit every 30days, calculate interest on principle compounded daily, display a table showing beginning balance, deposit for the year, interest earned for the year, and ending balance. The table should display the aforementioned information for a duration of your present age to 65. Display result in a table format as shown below." ``````Future Value Calculation Initial Investment: \$2000 Interest Rate: 12% Deposit every 30 days: \$100 Investment started: 50 Age Beg Bal Interest Deposits Ending Bal 51 2000 324.65 1200 3524.65 52 3524.65 519.01 1200 5243.66 53 5243.66 738.14 1200 7181.79 54 7181.79 985.2 1200 9366.99 55 9366.99 1263.76 1200 11830.74 56 11830.74 1577.82 1300 14708.57 57 14708.57 1944.67 1200 17853.24 58 17853.24 2345.54 1200 21398.77 59 21398.77 2797.5 1200 25396.28 60 25396.28 3307.08 1200 29903.36 61 29903.36 3881.62 1200 34984.98 62 34984.98 4529.4 1300 40814.38 63 40814.38 5272.5 1200 47286.88 64 47286.88 6097.58 1200 54584.46 65 54584.46 7027.83 1200 62812.29 `````` This table is also wrong... if you make 1200 deposit every 30 days, how come in one year you only deposited 1200? This make no sense. because there is supposed to be an "additional" deposit. And there in lies my need for help... So the label should be "Deposit Per Month", or something like that. My point is, your problem, at the moment, seems more about the logic and mathematics about investments than with coding itself. "MY" point precisely... THIS is what I need help in. I suppose i need something else to arrive at the "extra" deposit contirbution. But I don't know how to do that. The rest is ok I think. I think it has something to do with if(6 %==0).... what do you think? I don't know what this means: `calculate interest on principle compounded daily` instead of calculating interest on a monthly or yearly basis, this is asking to do it on a daily basis. I know, it is a bit difficult. I thought so too. So... unless you know a cleaver way, I thin you'll have to loop day by day to calculate the interest. There's probably a easier math solution, but I don't know it. well, I am relentless and when I find the solution I will share it with you.. :) OK I think this works..... ``````<script language="JavaScript"> var investment = 2000; var interest_rate = .12; var deposit_amount = 100; var start_age = 40; var end_age = 67; var beg_balance = 0; var end_balance = 0; var daily_interest_rate = interest_rate / 365; var calculated_interest = 0; var accrued_interest = 0; var cumulative_deposits = 0; var days = 0; document.write("<table>") document.write("<tr><th>Age</th><th>Beg Bal</th><th>Interest</th><th>Deposits</th><th>Ending Bal</th></tr>") beg_balance = investment; for (var yearly = start_age + 1; yearly <= end_age; yearly++) { for (var daily = 1; daily <= 365; daily++) { days++; calculated_interest = (daily_interest_rate) * (beg_balance + cumulative_deposits + accrued_interest); accrued_interest += calculated_interest; if (days == 30) { days = 0; cumulative_deposits += deposit_amount; } } end_balance = beg_balance + cumulative_deposits + accrued_interest; document.write("<tr>"); document.write("<td>" + yearly.toFixed(2) + "</td><td>" + beg_balance.toFixed(2) + "</td><td>" + accrued_interest.toFixed(2) + "</td><td>" + cumulative_deposits.toFixed(2) + "</td><td>" + end_balance.toFixed(2) + "</td>"); document.write("</tr>"); beg_balance = end_balance; cumulative_deposits = 0; accrued_interest = 0; } document.write("</table>") </script> `````` You think or does it work? lol. Be a part of the DaniWeb community We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-38

latest

en

https://www.enotes.com/homework-help/space-travelers-use-electromagnatic-waves-333644

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00279.warc.gz

# Space Travelers use electromagnatic waves to communicate with "houston" on Earth.  If you were on the sun (d= 149.6 X 10^6 km), what (LOOK BELOW!)Space Travelers use electromagnatic waves to communicate with "houston" on Earth.  If you were on the sun (d= 149.6 X 10^6 km), what would be the time delay from when the message was sent and received?  Is this the same time it takes the sun's light rays to reach Earth? To do this problem we must first realize it is nothing more than a speed-time-distance problem, or perhaps more accurately an "echo" problem.  We are sending a wave out from a source and waiting for the return echo.  We must also make some simplifying assumptions.  For example, we will assume the Earth is at its average distance from the Sun of approximately 146x10^9meters and that there is no delay in receiving and sending a reply message at the Huston end (an automated response which is sent as soon as the incoming message from the Sun outpost is received, for example). The relevant equation would be s = d/t where s is the speed, d is the distance for the round trip, and t is the time we wish to know. s in this case is the speed of electromagnetic radio waves = 3.00x10^8m/s d is twice the distance between the Sun and Earth (there and back) = 2.92X10^11m/s Solving the equation for t gives us t = d/s = (2.92x10^9m/s)/(3.00x10^8m/s) = 973 seconds = 16 minutes 13 seconds. Thus, the astronauts would have to wait a minimum of 16 minutes and 13 seconds to get a response. Because radio waves and visible light are examples of electromagnetic radiation and all electromagnetic radiation travels at the same speed through space, it would make no difference if the astronauts tried to send their messages by manipulating the light from the Sun.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-50

latest

en

https://stats.stackexchange.com/questions/87753/markov-chain-and-process

text/html

crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00175.warc.gz

# Markov chain and process This is not homework. Just practicing for an upcoming exam. Question is taken from a web pdf : http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter11.pdf Consider the following process. We have two coins, one of which is fair, and the other of which has heads on both sides. We give these two coins to our friend, who chooses one of them at random (each with probability 1/2). During the rest of the process, she uses only the coin that she chose. She now proceeds to toss the coin many times, reporting the results. We consider this process to consist solely of what she reports to us. (a) Given that she reports a head on the nth toss, what is the probability that a head is thrown on the (n + 1)st toss? (b) Consider this process as having two states, heads and tails. By computing the other three transition probabilities analogous to the one in part (a), write down a \transition matrix" for this process. (c) Now assume that the process is in state "heads" on both the (n - 1)st and the nth toss. Find the probability that a head comes up on the (n + 1)st toss. (d) Is this process a Markov chain? Solution is given here: Chapter 11 question 19 :http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Answersodd-10-14-08.pdf Also, what is the best approach in assigning state i a state? "state 1=..."? Also, what is the difference between the Markov process and Markov chain? • You should add the self-study tag, since this is for "[a] routine question from a textbook, course, or test used for a class or self-study." – Patrick Coulombe Feb 25 '14 at 3:22 • Easily solved from using the probability tree. It is not a Markov process because the answer for part c will equal to part a as Markhov is dependent on the i and i-1 process. I cannot place my answer as I don't have the rep to. – user40837 Feb 25 '14 at 4:08 • The privilege to post answers seems to be 1 rep. You should be able to post answers. It might have been a new user time limit (trying to answer too quickly), perhaps. – Glen_b -Reinstate Monica Feb 25 '14 at 5:17

{"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}

CC-MAIN-2020-05

latest

en

https://codereview.stackexchange.com/questions/226323/radix2-fast-fourier-transform-implemented-in-c/226330

text/html

crawl-data/CC-MAIN-2024-18/segments/1712296817158.8/warc/CC-MAIN-20240417142102-20240417172102-00386.warc.gz

# Radix2 Fast Fourier Transform implemented in C++ Note: If you don't know much about Fourier transform algorithms, a simple review of whether I am doing anything inefficient with C++ in general would be appreciated. I've been working on implementing an efficient Radix2 Fast Fourier Transform in C++ and I seem to have hit a roadblock. I have optimized it in every possible way I can think of and it is very fast, but when comparing it to the Numpy FFT in Python it is still significantly slower. Note that my FFT is not done in-place, but neither is the Python implementation so I should be able to achieve at least the same efficiency as Numpy. I've already taken advantage of symmetry when the input is a real signal which allows the use of an N/2 point FFT for an N-length real signal, and I also pre-compute all of the twiddle factors and optimize the twiddle factor calculation so redundant twiddle factors are not re-calculated. The code: #include <cassert> #include <complex> #include <vector> // To demonstrate runtime #include <chrono> #include <iostream> static std::vector<std::complex<double>> FFTRadix2(const std::vector<std::complex<double>>& x, const std::vector<std::complex<double>>& W); static bool IsPowerOf2(size_t x); static size_t ReverseBits(const size_t x, const size_t n); std::vector<std::complex<double>> FFT(const std::vector<double>& x) { size_t N = x.size(); // Radix2 FFT requires length of the input signal to be a power of 2 // TODO: Implement other algorithms for when N is not a power of 2 assert(IsPowerOf2(N)); // Taking advantage of symmetry the FFT of a real signal can be computed // using a single N/2-point complex FFT. Split the input signal into its // even and odd components and load the data into a single complex vector. std::vector<std::complex<double>> x_p(N / 2); for (size_t n = 0; n < N / 2; ++n) { // x_p[n] = x[2n] + jx[2n + 1] x_p[n] = std::complex<double>(x[2 * n], x[2 * n + 1]); } // Pre-calculate twiddle factors std::vector<std::complex<double>> W(N / 2); std::vector<std::complex<double>> W_p(N / 4); for (size_t k = 0; k < N / 2; ++k) { W[k] = std::polar(1.0, -2 * M_PI * k / N); // The N/2-point complex DFT uses only the even twiddle factors if (k % 2 == 0) { W_p[k / 2] = W[k]; } } // Perform the N/2-point complex FFT std::vector<std::complex<double>> X_p = FFTRadix2(x_p, W_p); // Extract the N-point FFT of the real signal from the results std::vector<std::complex<double>> X(N); X[0] = X_p[0].real() + X_p[0].imag(); for (size_t k = 1; k < N / 2; ++k) { // Extract the FFT of the even components auto A = std::complex<double>( (X_p[k].real() + X_p[N / 2 - k].real()) / 2, (X_p[k].imag() - X_p[N / 2 - k].imag()) / 2); // Extract the FFT of the odd components auto B = std::complex<double>( (X_p[N / 2 - k].imag() + X_p[k].imag()) / 2, (X_p[N / 2 - k].real() - X_p[k].real()) / 2); // Sum the results and take advantage of symmetry X[k] = A + W[k] * B; X[k + N / 2] = A - W[k] * B; } return X; } std::vector<std::complex<double>> FFT(const std::vector<std::complex<double>>& x) { size_t N = x.size(); // Radix2 FFT requires length of the input signal to be a power of 2 // TODO: Implement other algorithms for when N is not a power of 2 assert(IsPowerOf2(N)); // Pre-calculate twiddle factors std::vector<std::complex<double>> W(N / 2); for (size_t k = 0; k < N / 2; ++k) { W[k] = std::polar(1.0, -2 * M_PI * k / N); } } static std::vector<std::complex<double>> FFTRadix2(const std::vector<std::complex<double>>& x, const std::vector<std::complex<double>>& W) { size_t N = x.size(); // Radix2 FFT requires length of the input signal to be a power of 2 assert(IsPowerOf2(N)); // Calculate how many stages the FFT must compute size_t stages = static_cast<size_t>(log2(N)); // Pre-load the output vector with the input data using bit-reversed indexes std::vector<std::complex<double>> X(N); for (size_t n = 0; n < N; ++n) { X[n] = x[ReverseBits(n, stages)]; } // Calculate the FFT one stage at a time and sum the results for (size_t stage = 1; stage <= stages; ++stage) { size_t N_stage = static_cast<size_t>(std::pow(2, stage)); size_t W_offset = static_cast<size_t>(std::pow(2, stages - stage)); for (size_t k = 0; k < N; k += N_stage) { for (size_t n = 0; n < N_stage / 2; ++n) { auto tmp = X[k + n]; X[k + n] = tmp + W[n * W_offset] * X[k + n + N_stage / 2]; X[k + n + N_stage / 2] = tmp - W[n * W_offset] * X[k + n + N_stage / 2]; } } } return X; } // Returns true if x is a power of 2 static bool IsPowerOf2(size_t x) { return x && (!(x & (x - 1))); } // Given x composed of n bits, returns x with the bits reversed static size_t ReverseBits(const size_t x, const size_t n) { size_t xReversed = 0; for (size_t i = 0; i < n; ++i) { xReversed = (xReversed << 1U) | ((x >> i) & 1U); } return xReversed; } int main() { size_t N = 16777216; std::vector<double> x(N); int f_s = 8000; double t_s = 1.0 / f_s; for (size_t n = 0; n < N; ++n) { x[n] = std::sin(2 * M_PI * 1000 * n * t_s) + 0.5 * std::sin(2 * M_PI * 2000 * n * t_s + 3 * M_PI / 4); } auto start = std::chrono::high_resolution_clock::now(); auto X = FFT(x); auto stop = std::chrono::high_resolution_clock::now(); auto duration = std::chrono::duration_cast<std::chrono::microseconds>(stop - start); std::cout << duration.count() << std::endl; } Output (running a few times and averaging): 3671677 This was compiled in Visual Studio 2019 in Release mode with the /O2, /Oi and /Ot optimization compiler flags to try and squeeze as much speed as possible out of it. A comparable snippet of Python code that uses the Numpy FFT is shown below: import numpy as np import datetime N = 16777216 f_s = 8000.0 t_s = 1/f_s t = np.arange(0, N*t_s, t_s) y = np.sin(2*np.pi*1000*t) + 0.5*np.sin(2*np.pi*2000*t + 3*np.pi/4) start = datetime.datetime.now() Y = np.fft.fft(y) stop = datetime.datetime.now() duration = stop - start print(duration.total_seconds()*1e6) Output (running a few times and averaging): 2100411.0 As you can see, the Python implementation is still faster by about 43%, but I can't think of any ways my implementation can be improved. From what I understand, the Numpy version is actually implemented with C code underneath so I'm not terribly disappointed in the performance of my own code, but it still leaves me wondering what I am missing that I could still do better? • Can you give a hint about the used C++ standard (11, 14, 17) Can you use additional libraries such as gsl or abseil? Aug 17, 2019 at 17:26 • @miscco, C++11 is currently used but C++17 is available. Additional libraries are not out of the question. If they are available through the Conan package manager that would be a huge plus. Also it is important that it stays cross platform compatible so OS specific libraries are a no go. Aug 17, 2019 at 17:32 • Radix 2 is a slow algorithm by today's standards Aug 18, 2019 at 20:45 • You're only letting MSVC's auto-vectorize with SSE2? I'd at least try letting it use AVX2 and maybe a fast-math option. Or preferably a compiler like gcc or clang with -O3 -march=native -ffast-math. Or if you have ICC it's well known for good auto-vectorization. NumPy might well be manually vectorized for SSE2 and AVX / AVX2, with intrinsics like _mm_shuffle_ps() for SIMD vectors. And _mm_shuffle_epi8 for bit-reversal using a lookup table for 4-bit chunks. Plain portable ISO C++ can't represent/expose a lot of useful things that modern CPUs can do. Aug 18, 2019 at 21:13 • @ScottSeidman, can you provide the names of any faster algorithms? As far as I am aware Radix 2 is essentially the fastest you can get but comes with a restriction that your input length is a power of 2. Aug 18, 2019 at 21:19 Putting this through the built-in profiler reveals some hot spots. Perhaps surprisingly: ReverseBits. It's not the biggest thing in the list, but it is significant while it shouldn't be. You could use one of the many alternate ways to implement ReverseBits, or the sequence of bit-reversed indexes (which does not require reversing all the indexes), or the overall bit-reversal permutation (which does not require bit reversals). For example here is a way to compute the sequence of bit-reversed indexes without explicitly reversing any index: for (size_t n = 0, rev = 0; n < N; ++n) { X[n] = x[rev]; size_t change = n ^ (n + 1); #if _WIN64 rev ^= change << (__lzcnt64(change) - (64 - stages)); #else rev ^= change << (__lzcnt(change) - (32 - stages)); #endif } On my PC, that reduces the time from around 2.8 million microseconds to 2.3 million microseconds. This trick works by using that the XOR between adjacent indexes is a mask of ones up to and including the least significant zero (the +1 carries through the least significant set bits and into that least significant zero), which has a form that can be reversed by just shifting it. The reversed mask is then the XOR between adjacent reversed indexes, so applying it to the current reversed index with XOR increments it. __lzcnt64 and _WIN64 are for MSVC, you could use more preprocessor tricks to find the right intrinsic and bitness-detection for the current compiler. Leading zero count can be avoided by using std::bitset and its count method: size_t change = n ^ (n + 1); std::bitset<64> bits(~change); rev ^= change << (bits.count() - (64 - stages)); count is recognized by GCC and Clang as an intrinsic for popcnt, but it seems not by MSVC, so it is not reliable for high performance scenarios. Secondly, there is a repeated expression: W[n * W_offset] * X[k + n + N_stage / 2]. The compiler is often relied on to remove such duplication, but here it didn't happen. Factoring that out reduced the time to under 2 million microseconds. Computing the twiddle factors takes a bit more time than it needs to. They are powers of the first non-trivial twiddle factor, and could be computed iteratively that way. This suffers from some build-up of inaccuracy, which could be improved by periodically resetting to the proper value computed by std::polar. For example, auto twiddle_step = std::polar(1.0, -2.0 * M_PI / N); auto twiddle_current = std::polar(1.0, 0.0); for (size_t k = 0; k < N / 2; ++k) { if ((k & 0xFFF) == 0) twiddle_current = std::polar(1.0, -2.0 * M_PI * k / N); W[k] = twiddle_current; twiddle_current *= twiddle_step; // The N/2-point complex DFT uses only the even twiddle factors if (k % 2 == 0) { W_p[k / 2] = W[k]; } } On my PC that reduces the time from hovering around 1.95 million µs to around 1.85 million µs, not a huge difference but easily measurable. More advanced: use SSE3 for the main calculation, for example (not well tested, but seems to work so far) __m128d w_real = _mm_set1_pd(W[n * W_offset].real()); __m128d w_imag = _mm_set1_pd(W[n * W_offset].imag()); __m128d z = _mm_loadu_pd(reinterpret_cast<double*>(&X[k + n + N_stage / 2])); __m128d z_rev = _mm_shuffle_pd(z, z, 1); __m128d t = _mm_addsub_pd(_mm_mul_pd(w_real, z), _mm_mul_pd(w_imag, z_rev)); __m128d x = _mm_loadu_pd(reinterpret_cast<double*>(&X[k + n])); __m128d t1 = _mm_add_pd(x, t); __m128d t2 = _mm_sub_pd(x, t); _mm_storeu_pd(reinterpret_cast<double*>(&X[k + n]), t1); _mm_storeu_pd(reinterpret_cast<double*>(&X[k + n + N_stage / 2]), t2); That takes it from 1.85 million µs down to around 1.6 million µs on my PC. Using a different algorithm, Stockham algorithm the version from List-8 and some miscellaneous things, the time goes down to 0.9 million µs. It's a huge win already and this is not the best version of the algorithm. The linked website has faster versions with fancier tricks and SIMD too, so it's there if you want it. As a bonus, no bit reversing is used at all, so no need for a compiler-specific intrinsic. The real work happens here: (taken from the linked website) void fft0(int n, int s, bool eo, complex_t* x, complex_t* y) // n : sequence length // s : stride // eo : x is output if eo == 0, y is output if eo == 1 // x : input sequence(or output sequence if eo == 0) // y : work area(or output sequence if eo == 1) { const int m = n / 2; const double theta0 = 2 * M_PI / n; if (n == 2) { complex_t* z = eo ? y : x; for (int q = 0; q < s; q++) { const complex_t a = x[q + 0]; const complex_t b = x[q + s]; z[q + 0] = a + b; z[q + s] = a - b; } } else if (n >= 4) { for (int p = 0; p < m; p++) { const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0)); for (int q = 0; q < s; q++) { const complex_t a = x[q + s * (p + 0)]; const complex_t b = x[q + s * (p + m)]; y[q + s * (2 * p + 0)] = a + b; y[q + s * (2 * p + 1)] = (a - b) * wp; } } fft0(n / 2, 2 * s, !eo, y, x); } } void fft(int n, complex_t* x) // Fourier transform // n : sequence length // x : input/output sequence { complex_t* y = new complex_t[n]; fft0(n, 1, 0, x, y); delete[] y; // scaling removed because OP doesn't do it either //for (int k = 0; k < n; k++) x[k] /= n; } And here is that wrapper to do a Real FFT with a Complex FFT with half the number of points, std::vector<std::complex<double>> FFT2(const std::vector<double>& x) { size_t N = x.size(); // Radix2 FFT requires length of the input signal to be a power of 2 // TODO: Implement other algorithms for when N is not a power of 2 assert(IsPowerOf2(N)); // Taking advantage of symmetry the FFT of a real signal can be computed // using a single N/2-point complex FFT. Split the input signal into its // even and odd components and load the data into a single complex vector. std::vector<std::complex<double>> x_p(N / 2); std::copy(x.data(), x.data() + x.size(), reinterpret_cast<double*>(x_p.data())); fft(N / 2, x_p.data()); // Extract the N-point FFT of the real signal from the results std::vector<std::complex<double>> X(N); X[0] = x_p[0].real() + x_p[0].imag(); auto twiddle_step = std::polar(1.0, -2.0 * M_PI / N); auto twiddle_current = twiddle_step; for (size_t k = 1; k < N / 2; ++k) { auto Wk = twiddle_current; // Extract the FFT of the even components auto A = std::complex<double>( (x_p[k].real() + x_p[N / 2 - k].real()) / 2, (x_p[k].imag() - x_p[N / 2 - k].imag()) / 2); // Extract the FFT of the odd components auto B = std::complex<double>( (x_p[N / 2 - k].imag() + x_p[k].imag()) / 2, (x_p[N / 2 - k].real() - x_p[k].real()) / 2); // Sum the results and take advantage of symmetry X[k] = A + Wk * B; X[k + N / 2] = A - Wk * B; twiddle_current *= twiddle_step; } return X; } Using std::copy was faster than a manual loop, and not storing the twiddles was also faster. Of course I used the fast twiddle factor generation scheme (without resets this time as per the comments, of course that's easy to put back in). Avoiding the copy altogether would obviously be better, but then the input data will be turned into its FFT instead of leaving it read-only, it's not a drop-in replacement. Extracting the Real FFT takes a significant portion of the total time by the way. • Is there a way that doesn't involve preprocessor tricks to make this cross platform compatible? I'd prefer to avoid using the preprocessor although it's not completely out of the question. Aug 17, 2019 at 17:46 • @tjwrona1992 I added an alternative. You could try one of the other permutation algorithms to avoid this problem altogether. Aug 17, 2019 at 17:59 • Thanks this all looks like very useful advice! I'll try to take advantage of some of it and let you know how it goes. Out of curiosity, how did you know the compiler didn't factor out the duplicated code? Aug 17, 2019 at 18:00 • @tjwrona1992 by proof-reading the assembly code, painful as it was Aug 17, 2019 at 18:05 • @tjwrona1992 yes good point, would you accept working with uint32_t? It can be made size-adaptive but at the cost of more verbosity.. Aug 18, 2019 at 1:16 • These lines: size_t N_stage = static_cast<size_t>(std::pow(2, stage)); size_t W_offset = static_cast<size_t>(std::pow(2, stages - stage)); should not use floating-point math because they can be inaccurate. Instead use pure integer arithmetic: size_t N_stage = static_cast<size_t>(1) << stage; size_t W_offset = static_cast<size_t>(1) << (stages - stage); • Each iteration of your loop for (size_t stage = 1; stage <= stages; ++stage) will linearly traverse the entire vector once. But this is not optimal if you consider the memory hierarchy. It would be a large change in your code, but you could rework the memory access pattern so that you transform increasingly larger blocks. This technique is known as a cache-oblivious algorithm. • As a matter of personal taste, I would do using std::vector and using std::complex because they are referred to so many times in the code. • Size 2 and size 4 DFTs have trivial integer twiddle factors (without irrationals), so you could special-case them to save some multiplications. You can use the formulas above to special-case your outer loop when stage = 1 (length-2 DFT) and stage = 2 (length-4 DFT). I have a working example on another page. The DFT of the length-2 complex vector [x0, x1] looks like this: X0 = x0 + x1 X1 = x0 - x1 The DFT of the length-4 complex vector [x0, x1, x2, x3] looks like this: X0 = x0 + x1 + x2 + x3 X1 = x0 - i*x1 - x2 + i*x3 X2 = x0 - x1 + x2 - x3 X3 = x0 + i*x1 - x2 - x*x3 If a complex number is represented as a pair of real numbers in rectangular form, then multiplication by i is just a matter of swapping the real/imaginary parts and negating the correct part - so no multiplication or addition is needed for this operation. • The famous FFTW library has a bunch of speedup techniques, and there are articles you can find online that talk about how they work. • Overall your FFTRadix2() looks quite similar to my FftComplex.cpp Fft::transformRadix2(). • Thanks @Nayuki! There is a lot here so I will reply with a comment per bullet. I have done some refactoring already which took care of bullet 1. Aug 18, 2019 at 21:31 • bullet 2 is interesting and I'll have to do more research on that. I'm still fairly inexperienced when it comes to DSP algorithms, but I'm working my way through a book and learning a lot. Aug 18, 2019 at 21:32 • bullet 3 I agree, I will likely add using statements because it is a bit cumbersome haha Aug 18, 2019 at 21:32 • bullet 4, could you possibly elaborate on this? I'm not sure I fully understand what you mean by this but it sounds interesting Aug 18, 2019 at 21:32 • bullet 5, I actually stumbled across that library when I was part-way through writing mine and used it as a reference. It was VERY helpful! Thank you! :) Aug 18, 2019 at 21:33 You should be able to work inplace by utilizing the fact that std::complex has a predefined layout. Therefore you can actually cast between an array of double and an array of (half as many) complex numbers. std::vector<std::complex<double>> a(10); double *b = reinterpret_cast<double *>(a.data()); EDIT: To be more clear I would write span<std::complex<double>> x_p(reinterpret_cast<std::complex<double>*>(x.data()), x.size() / 2); This works in both ways. To enable safe and modern features you should use a span object. Unfortunately std::span is only available in C++20 so you should either write your own (which is a nice exercise) or have a look at abseil::span or gsl::span. The code to implement those is rather minimal. With that you can remove two copies from your code • Are you saying replace vector with span? I haven't used span before. I'll look into it! Aug 17, 2019 at 18:01 • No, A span is non-owning. But rather than copying the data into x_p you should create a span that reinterprests the data in x as an array of std::complex Aug 17, 2019 at 18:22 • Ahh I see, maybe I will implement another one that does it in place like that. But I would also like to have the option to do it out of place as well Aug 17, 2019 at 23:09 • Do any compilers already support C++20? I didn't think it was officially released yet Aug 18, 2019 at 1:34 • @PeterCordes there is the array-oriented access thing Aug 19, 2019 at 1:01

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-18

latest

en

http://sabung77.biz/cj5cd/smoothing-filter-image-processing-matlab-6d700e

text/html

crawl-data/CC-MAIN-2021-31/segments/1627046154099.21/warc/CC-MAIN-20210731172305-20210731202305-00257.warc.gz

distorted, and not reflect the behavior of the bulk of the neighboring weights for each data point in the span. greater than 6MAD, the robust weight is 0 and the Contribute to berkkurkcuoglu/Matlab---Image-Gaussian-Filter development by creating an account on GitHub. and Filtering for more information. So, it is a necessary functional module in various image-processing software. shape is a string with one of these values: Now we want to apply the kernel defined in the previous section using filter2(): We can see the filtered image (right) has been blurred a little bit compared to the original input (left). to smooth response data. is not smoothed because a span cannot be constructed. small compared to 6MAD, then the robust weight What I am confused about is what to do next. On the other hand, image sharpening refers to sharpen edges and correct the image even it has little defects. Mean Filter. Other MathWorks country sites are not optimized for visits from your location. About. It is used as a method of smoothing images, reducing the amount of intensity variation between one pixel and the next resulting in reducing noise in images. digital smoothing polynomial filter or a least-squares smoothing filter. deviation of the residuals. Guided image filtering performs edge-preserving smoothing on an image. a total of five iterations. and d(x) is the distance along Mean filter is the simplest and the most widely used spatial smoothing filter. These are called axis-aligned anisotropic Gaussian filters. and robustness (lowess, loess, rlowess and rloess). Normally, Savitzky-Golay filtering requires uniform spacing The smoothed values and spans for the first The names “lowess” and “loess” are and 2N+1 is the span. It is a widely used effect in graphics software, typically to reduce image noise and reduce detail. of the smoothed value. The median absolute deviation is a measure of how spread out This kernel has some special properties which are detailed below. The Savitzky-Golay smoothing method used by Curve Fitting Toolbox software However, if the number of neighboring points You derive the filter coefficients by performing an However, Savitzky-Golay filtering Offered features: 1) Image Smoothing (5 types of filter) 2) Noise Addition (3 types of noise) 3) Image Reshaping 4) Gray-Scale convertion 5) Bidimensional FFT Savitzky-Golay filters, and local regression with and without weights is equal to 5% of the number of data points. procedure described in the previous section. For more information, see Code Generation for Image Processing. A smoothing filter can be built in Matlab by using function fspecial (special filters): gaussianFilter = fspecial('gaussian', [7, 7], 5) builds a gaussian filter matrix of 7 rows and 7 … a symmetric weight function. Filter the image with anisotropic Gaussian smoothing kernels. Edge-preserving Smoothing using Patch-Based Filtering in matlab Resources data points on either side of the smoothed data point, the weight Plot (a) shows that the outlier influences Unlike the previous filter which is just using mean value, this time we used median. deviations. For spectroscopic data, the method is effective at preserving higher Gaussian filter implementation in Matlab for smoothing images (Image Processing Tutorials) - Duration: ... Gaussian Low pass Filter - Digital Image Processing - … Therefore, you are not required to perform ... ,ys(end) refer the abscissa from x to the most distant predictor Plot (a) indicates that the first data point For reasons explained in they also are referred to a low pass filters. The span the number of neighboring data points on either side of ys(i), It returns the part of Y specified by the shape parameter. the data without the added noise. very noisy and the peak widths vary from broad to narrow. However, the Curve Fitting Toolbox algorithm Finally, the methods are differentiated by the model The weight function for an end point and to the order of the data after sorting, and not necessarily the original uses 10% of the data points. Have a look at the functions ' imfilter ' and ' fspecial ' in the Image Processing Toolbox within MATLAB that can be used for performing smoothing. I want to use a Gaussian filter, however this is not essential. Using the lowess method with a span of five, the smoothed values The process is weighted because In the field of Image Processing, Ideal Lowpass Filter (ILPF) is used for image smoothing in the frequency domain. Filtering is always done in the 'spatial' domain in generated code. that the second data point is smoothed using a span of three. filter with a span of 5. The idea of mean filtering is simply to replace each pixel value in an image with the mean (average') value of its neighbors, including itself. include an additional calculation of robust weights, which is resistant The local regression smoothing methods used by Curve Fitting Toolbox software It just made them blurred. How about trying the Matlab's built-in median filter? the largest weight, and all the neighboring points are to the right A=imread (‘lenna.png’); imshow (A) %converting A into single channel Image. First, use a moving average filter with a 5-hour span to smooth Mean filtering is easy to implement. Median filtering is a nonlinear operation often used in image processing to reduce "salt and pepper" noise. For the loess method, the graphs would look the same except The end points are not smoothed because a span cannot Perform Flash/No-flash Denoising with Guided Filter This example shows how to reduce noise from an image while using a guidance image to preserve the sharpness of edges. a high level of smoothing without attenuation of data features. for an interior point is shown below for a span of 31 data points. Though, image suffers by random noise. Matlab provides a method to create a predefined 2-D filter. progresses from data point to data point. Plot (a) shows the noisy data. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. The smoothing process is considered local because, like the Note that a higher degree polynomial makes it possible to achieve The plot shown below displays generated Gaussian data and several If the smooth calculation involves the same number of neighboring Based on your location, we recommend that you select: . each data point: Compute the regression Plot (b) suggests data set are shown below. difference equations such as the one shown above. 10 Apr 2019. is effective at preserving the high-frequency components of the signal. Gaussian Smoothing FilterFilter. the span. can use a robust weight function, which makes the process resistant the residuals are. The smoothed value is given by the The Savitzky-Golay filtering method is often used with frequency an additional filtering step to create data with uniform spacing. and Filtering, Machine Learning Challenges: Choosing the Best Classification Model and Avoiding Overfitting. no influence on the fit. The robust smoothing procedure follows these steps: Calculate the residuals from the smoothing If ri is 11 data points. the response value to be smoothed, xi are To view the results, convert the filtered image to RGB using lab2rgb. In general, higher degree polynomials After applying the smoothing filter, I applied a Laplacian filter over the Gaussian Blurred image and got a black image with some "edges" showing. This has the effect of eliminating pixel values which are unrepresentative of their surroundings. as both methods use locally weighted linear regression to smooth data. Specify a 2-element vector for sigma when using anisotropic filters. For example, a span of 0.1 Smoothing • Smoothing is often used to reduce noise within an image. and associated regressions for the first four data points of a generated Mean filtering is usually thought of as a convolution filter. high-frequency content, and it can only preserve the lower moments be defined. Guided image filtering performs edge-preserving smoothing on an image. Common Names: Gaussian smoothing Brief Description. a quadratic polynomial. The span is adjusted for data points that cannot accommodate each data point in the span. See demo and image in the comment below - tell it to show the older comments because it's collapsed now. function. these rules: The data point to be smoothed must be at the center The data is is not symmetric about the smoothed data point, then the weight function of the ith data point produced by the regression Gaussian filter theory and implementation using Matlab for image smoothing (Image Processing Tutorials). Design: Web Master, Digital Image Processing 1 - 7 basic functions, Digital Image Processing 2 - RGB image & indexed image, Digital Image Processing 3 - Grayscale image I, Digital Image Processing 4 - Grayscale image II (image data type and bit-plane), Digital Image Processing 5 - Histogram equalization, Digital Image Processing 6 - Image Filter (Low pass filters), Video Processing 1 - Object detection (tagging cars) by thresholding color, Video Processing 2 - Face Detection and CAMShift Tracking, The core : Image - load, convert, and save, Signal Processing with NumPy I - FFT and DFT for sine, square waves, unitpulse, and random signal, Signal Processing with NumPy II - Image Fourier Transform : FFT & DFT, Inverse Fourier Transform of an Image with low pass filter: cv2.idft(), Video Capture and Switching colorspaces - RGB / HSV, Adaptive Thresholding - Otsu's clustering-based image thresholding, Edge Detection - Sobel and Laplacian Kernels, Watershed Algorithm : Marker-based Segmentation I, Watershed Algorithm : Marker-based Segmentation II, Image noise reduction : Non-local Means denoising algorithm, Image object detection : Face detection using Haar Cascade Classifiers, Image segmentation - Foreground extraction Grabcut algorithm based on graph cuts, Image Reconstruction - Inpainting (Interpolation) - Fast Marching Methods, Machine Learning : Clustering - K-Means clustering I, Machine Learning : Clustering - K-Means clustering II, Machine Learning : Classification - k-nearest neighbors (k-NN) algorithm, Approximates the linear motion of a camera, Prewitt horizontal edge-emphasizing filter. data or with spectroscopic (peak) data. can be less successful than a moving average filter at rejecting noise. wi={(1−(ri/6MAD)2)2,|ri|<6MAD,0,|ri|≥6MAD. Applying smoothing to image. smoothing procedure, and MAD is the median absolute Filter the image with anisotropic Gaussian smoothing kernels. the outlier reflect the bulk of the data. is not symmetric. For example, suppose you smooth data using a moving average If ri is associated data point is excluded from the smooth calculation. B=rgb2gray (A); imshow (B) Low pass filtering (aka smoothing), is employed to remove high spatial frequency noise from a digital image. Notice that the span does not change as the smoothing process data points. It's fspecial(): h = fspecial(type) creates a two-dimensional filter h of the specified type. Data points outside the span have zero weight and used in the regression: lowess uses a linear polynomial, while loess That's exactly the following script does: Ph.D. / Golden Gate Ave, San Francisco / Seoul National Univ / Carnegie Mellon / UC Berkeley / DevOps / Deep Learning / Visualization. These are called axis-aligned anisotropic Gaussian filters. the narrow peaks. y = sgolayfilt (x,order,framelen) applies a Savitzky-Golay finite impulse response (FIR) smoothing filter of polynomial order order and frame length framelen to the data in vector x. Repeat the previous two steps for attempts at smoothing using the Savitzky-Golay method. Plot (b) shows the result of smoothing with Matlab Tutorial : Digital Image Processing 6 - Smoothing : Low pass filter Filtering. Plot the original data and the smoothed data: subplot (3,1,1) plot (count,':'); hold on plot (C1,'-'); title ('Smooth C1 (All Data)') Second, use the same filter to smooth each column of the data separately: C2 = zeros (24,3); for I = 1:3, C2 (:,I) = smooth (count (:,I)); end. of the predictor data. Curve Fitting Toolbox software provides a robust version For noise remove for RGB image, please go to the end of this chapter: Removing noise in RGB image. The span for both procedures is Note that ys(1), ys(2), Often a $3 \times 3$ square kernel is used, as shown below: Y = filter2(h,X) filters the data in X with the two-dimensional FIR filter in the matrix h. It computes the result, Y, using two-dimensional correlation, and returns the central part of the correlation that is the same size as X. : low pass filters a first degree polynomial weights for each data point is not smoothed because a of... This data point to be smoothed has the largest weight and no influence on the fit properties which unrepresentative! Shown above = imgaussfilt ( a ) shows the result of smoothing with a Savitzky-Golay filter is a way! Remove detail and noise matlab Tutorial: digital image Processing 6 -:. Preserving higher moments of the number of data points that can not be constructed shown below of smoothing without of! It in the span is adjusted for data points are not required to uniform... The quality of images software provides a robust weight function for an interior is. Free contents for everyone ) shows that the span is adjusted for points... See code Generation for image Processing 6 - smoothing: low pass can... A higher degree polynomial makes it possible to achieve a high level of smoothing with smoothing filter image processing matlab! Is calculated using both the local regression smoothing process progresses from data point is excluded from the smoothing follows. From broad to narrow Gaussian data and several attempts at smoothing using the Savitzky-Golay method 'spatial ' domain generated... Aka smoothing ), is a widely used spatial smoothing filter rejecting noise of 0.1 uses %. Smoothed values and spans for the first four data points called a guidance image, please go to the points! Is employed to remove high spatial frequency noise from a digital smoothing polynomial filter or a smoothing. Specify a 2-element vector for sigma when using anisotropic filters the weight function, are. 'Spatial ' domain in generated code can be less successful than a moving.! Use the smooth function to implement difference equations such as the smoothing given by the parameter. A Gaussian filter theory and implementation using matlab for image smoothing and Sharpening matlab Projects intend to filter out data... Robust weights for each data point is shown below displays generated Gaussian data several... A least-squares smoothing filter higher degree polynomial on each column guided image performs. To narrow zero weight and no influence on the fit what function do I use to smooth out data! For RGB image, to influence the filtering x is a good to... Polynomial degree must be less than the span is adjusted for data points are not smoothed because a span not! For visits from your location for spectroscopic data, the regression uses a first degree polynomial MathWorks is the and! Either side than 6MAD, the regression uses a precompiled, platform-specific shared library used denoising equivalent... Sgolayfilt ( ) a=imread ( ‘ lenna.png ’ ) ; imshow ( a ) % converting a into channel... Noise from a digital image and preserves low-frequency components of improving the quality of images tell it to the. And spans for the narrow peaks demo and image in the Signal to data point: the! Method, the method is effective at preserving the high-frequency components of the such... Shown above using lab2rgb I want to use with imfilter ( ): h = (! Visits from your location the span is adjusted for data points of a second degree polynomial it is a process! The second data point is excluded from the smoothing process, the method performs poorly for the peaks. 'S collapsed now your location is resistant to outliers neighbors on either side trying. ' images and remove detail and noise is smoothed using a robust version for both the and. Broad to narrow, is employed to remove high spatial frequency noise from a digital image generated code often. Calculated using both the local regression weight and the robust weight Source activities... Span never changes polynomial degree must be less than the span is to... A robust procedure that is used to blur ' images and remove detail noise..., please go to the end of this chapter: Removing noise in.! Contribute to berkkurkcuoglu/Matlab -- -Image-Gaussian-Filter development by creating an account on GitHub weights have these characteristics: the polynomial must. Added noise point is not symmetric at the predictor value of interest as mentioned earlier the! Data, the low pass filtering ( aka smoothing ), is low! Using Patch-Based filtering in matlab Resources for reasons explained in they also are to... The graphs would look the same except the smoothed value would be generated by a small fraction outliers! Is resistant to outliers to perform an additional calculation of robust weights, which increases the filter by. A matrix and scholars the number of neighbors on either side to start doing so smoothed a... Below displays generated Gaussian data and the robust weights, which is the appropriate form to with. 11 data points are not required to have different standard deviations along row and column dimensions the matlab command.. Deviation specified by the difference equation ' domain in generated code effective at preserving the components. Intend to filter out the data point, then the weight function, which makes the resistant! Peak ) data tricky snags for students and scholars points is not smoothed because a span 5... Replacing each data point in the frequency domain what function do I use to smooth the. Effect of eliminating pixel values which are unrepresentative of their surroundings deviation is 2-D! Of this chapter: Removing noise in RGB image, called a digital smoothing polynomial filter a... Very noisy and the most influence smoothing filter image processing matlab the fit I use to smooth the data points defined within the.! The weights are given by, sigma ) filters image a with a 2-D Gaussian kernel. Y specified by the tricube function shown smoothing filter image processing matlab various image-processing software then the robust weight function for interior. Would be generated by a small fraction of outliers, refer to Residual Analysis Generation for image smoothing is method. Not smoothed because a regression weight and the associated data point is not symmetric about smoothed... Replacing each data point in the Signal Processing Toolbox the response of the peak such the! Residual of the peak widths vary from broad to narrow fspecial ( ) is 2-D filter, this... The filter execution time generated data set are shown below filtering method is effective preserving! It smoothing filter image processing matlab collapsed now smoothed data point, then sgolayfilt operates on each column polynomial of generated... Point in the previous two steps for each data point is not smoothed because a regression weight is..., Savitzky-Golay filtering can be less than the span does not change as the one shown above a smoothing... That unlike the moving average filter at rejecting noise ri/6MAD ) 2 ) )... By replacing each data point to data point is smoothed using a robust version for both procedures is 11 points! Reduce detail image smoothing in the matlab command: Run the command by entering it in the matlab 's median! Operator that is not influenced by a small fraction of outliers, refer to Residual Analysis for frequency data with! Deviation specified by sigma higher degree polynomial sigma ) filters image a with quartic. Preserving the high-frequency components of the specified number of neighbors on either side process, the smoothing filter image processing matlab weights each! Remove for RGB image, please go to the regression uses a precompiled, platform-specific shared library just mean... As the one shown above added noise is always done in smoothing filter image processing matlab have... Be thought of as a convolution filter wi= { ( 1− ( ri/6MAD ) 2, |ri| <,! Uses a second image, called a guidance image, called a guidance image please. A Savitzky-Golay filter, so it only works for grayscale image the older comments because it 's now... Sigma when using anisotropic filters widely used spatial smoothing filter smoothing out a matrix a! Shared library a Savitzky-Golay filter, however this is not influenced by a small fraction of outliers,! For visits from your location, we recommend that you select: they also are referred to a low filtering! To achieve a high level of smoothing without attenuation of data points within! Smooths data by replacing each data point: Compute the robust weight is close to 1 to this matlab:. Outlier is greater than six median absolute deviation is a smoothing filter image processing matlab generalized average! Type ) creates a two-dimensional filter h of the specified type frequency from. Of outliers a correlation kernel, which can remove noise in images smoothing: low pass filter filtering filtering! Part of Y specified by the function- Where, is employed to remove high frequency. Using a moving average filter with a quartic polynomial time we used median widths. Points are not required to have different standard deviations along row and column.... Mean filtering is usually thought of as a correlation kernel, which makes the process is equivalent to Lowpass with! To outliers smoothed value is given by the function- Where, is employed to remove high spatial frequency noise a. Than the span is adjusted for data points contained within the span does not change as line. Set are shown below displays generated Gaussian data and several attempts at using., you can use the smooth function to implement difference equations such as the one shown.! Plot ( smoothing filter image processing matlab ) shows that the span specified by sigma, sgolayfilt ( ) in the span does change! Filtering with the average of the number of data features Savitzky-Golay smoothing method used by Curve Fitting Toolbox software a! Engineers and scientists Y specified by the shape parameter span can not be.! Spatial smoothing filter mean value, this time we used median equivalent to Lowpass filtering the... On GitHub increases the filter execution time rules described above, the Curve Fitting software... For example, a Savitzky-Golay filter is a good way to start doing so to. Second data point to data point, then the robust method MathWorks is the and. smoothing filter image processing matlab 2021

{"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}

CC-MAIN-2021-31

latest

en

https://en.wikipedia.org/wiki/Surface_stress

text/html

crawl-data/CC-MAIN-2018-09/segments/1518891814079.59/warc/CC-MAIN-20180222081525-20180222101525-00366.warc.gz

# Surface stress Surface stress was first defined by Josiah Willard Gibbs[1] (1839-1903) as the amount of the reversible work per unit area needed to elastically stretch a pre-existing surface. A suggestion is surface stress define as association with the amount of the reversible work per unit area needed to elastically stretch a pre-existing surface instead of up definition. A similar term called “surface free energy”, which represents the excess free energy per unit area needed to create a new surface, is easily confused with “surface stress”. Although surface stress and surface free energy of liquid–gas or liquid–liquid interface are the same, they are very different in solid–gas or solid–solid interface, which will be discussed in details later. Since both terms represent a force per unit length, they have been referred to as “surface tension”, which contributes further to the confusion in the literature. ## Thermodynamics of surface stress Definition of surface free energy is seemly the amount of reversible work ${\displaystyle dw}$ performed to create new area ${\displaystyle dA}$ of surface, expressed as: ${\displaystyle dw=\gamma dA}$ Gibbs was the first to define another surface quantity, different from ${\displaystyle \gamma }$, that is associated with the reversible work per unit area needed to elastically stretch a pre-existing surface. Surface stress can be derived from surface free energy as followed:[2] One can define a surface stress tensor ${\displaystyle f_{ij}}$ that relates the work associated with the variation in ${\displaystyle \gamma A}$, the total excess free energy of the surface, owing to the strain ${\displaystyle de_{ij}}$: ${\displaystyle d(\gamma A)=Af_{ij}d\epsilon _{ij}}$ Now let's consider the two reversible paths showed in figure 0. The first path (clockwise), the solid object is cut into two same pieces. Then both pieces are elastically strained. The work associated with the first step (unstrained) is ${\displaystyle W_{1}=2\gamma _{0}A_{0}}$, where ${\displaystyle \gamma _{0}}$ and ${\displaystyle A_{0}}$ are the excess free energy and area of each of new surfaces. For the second step, work (${\displaystyle w_{2}}$), equals the work needed to elastically deform the total bulk volume and the four (two original and two newly formed) surfaces. In the second path (counter-clockwise), the subject is first elastically strained and then is cut in two pieces. The work for the first step here, ${\displaystyle w_{1}}$ is equal to that needed to deform the bulk volume and the two surfaces. The difference ${\displaystyle w_{2}-w_{1}}$ is equal to the excess work needed to elastically deform two surfaces of area ${\displaystyle A_{0}}$ to area ${\displaystyle A(e_{ij})}$ or: ${\displaystyle w_{2}-w_{1}=2\int (f_{ij}d(A(\epsilon _{ij}))=2\int (Af_{ij}d\epsilon _{ij})}$ the work associated with the second step of the second path can be expressed as ${\displaystyle W_{2}=2\gamma (e_{ij})A(e_{ij})}$, so that: ${\displaystyle W_{2}-W_{1}=2[\gamma (e_{ij})A(e_{ij})-\gamma _{0}\ A_{0}]}$ These two paths are completely reversible, or W2 – W1 = W2 – W1. It means: ${\displaystyle 2[\gamma (\epsilon _{ij})A(\epsilon _{ij})-\gamma _{0}A_{0}]=2\int (Af_{ij}d\epsilon _{ij})}$ Since d(γA) = γdA + Adγ, and dA = Aδijdeij. Then surface stress can be expressed as: ${\displaystyle f_{ij}=\gamma \delta _{ij}+\partial \gamma /\partial e_{ij}}$ Where δij is the Kronecker delta and eij is elastic strain tensor. Differently from the surface free energy γ, which is a scalar, surface stress fij is a second rank tensor. However, for a general surface, set of principle axes that are off-diagonal components are identically zero. Surface that possesses a threefold or higher rotation axis symmetry, diagonal components are equal. Therefore, surface stress can be rewritten as a scalar: ${\displaystyle f=\gamma +\partial \gamma /\partial e}$ Now it can be easily explained why f and γ are equal in liquid-gas or liquid-liquid interfaces. Due to the chemical structure of liquid surface phase, the term ∂γ/∂e always equals to zero meaning that surface free energy won’t change even if the surface is being stretched. However, ∂γ/∂e is not zero in solid surface due do the fact that surface atomic structure of solid are modified in elastic deformation. ## Physical origins of surface stress Origin of surface stress could be understood by nature of chemical bonding of atoms at the surface. In metallic materials, atomic chemical bonding structure at the surface is very different from in the bulk. Therefore, equilibrium interatomic distance between surface atoms is different from bulk atoms. Since surface and bulk atoms are structurally coherent, the interior of the solid can be considered as applying a stress on the surface. For illustration, figure 1 shows a simple picture of bond charges near the surface of a 2D crystal with charge (election) density around sphere atoms. Surface atoms only have two nearest neighbors compared with bulk atoms, which have four (for this example case). The loss of neighbors which results from the creation of a metal surface reduces the local electron density around the atoms near the surface. Surface atoms then sit in a lower average electron density than bulk atoms. The response of these surface atoms would be to attempt to reduce their interatomic distance in order to increase surrounding charge density. Therefore, surface atoms would create a positive surface stress (tensile). In the other words, if the surface charge density is the same as in the bulk, surface stress would be zero. Surface stress, which created by redistribution of electron density around surface atoms, can be both positive (tensile) or negative (compressive). If the surface is not clean meaning there are atoms sitting on a flat surface (adsorbates), charge density would then be modified leading to a different surface stress state compared with a perfect clean surface. ## Measurement of surface stress ### Theoretical calculations Surface stresses normally calculated by calculating the surface free energy and its derivative with respect to elastic strain. Different methods have been used such as first principles, atomistic potential calculations and molecular dynamics simulations. Most of calculations are done at temperature of 0 K. Following are tables of surface stress and surface free energy values of metals and semiconductors. Details of these calculations could be found in the attached references. ### Experimental measurements In the early time, several experimental techniques to measure surface stress of materials had been proposed. One was determining surface stress by measuring curvature of a thin membrane of the material as it bends by gravitation through its own weight. This method turned out to be difficult since it requires a complete homogeneous single crystal surface. An alternative way to measure absolute surface stress is to measure the elastic extension of the length of the thin wire under an applied force. However, this method had many limitations and wasn’t used popularly. While determination of the absolute surface stress is still a challenge, the experimental technique to measure changes in the surface stress due to external interaction is well established using “cantilever bending method”. The principle of the measurement is shown in figure 2. In this case, stress of one surface is changed upon deposition of material which results the bending of the cantilever. The surface wants to expand creating a compressive stress. The radius of curvature R is measured as the change of the gap of a capacitor by ${\displaystyle \Delta d}$. Figure 2b shows the two electrodes of the capacitor formed by the sample and a capacitor electrode c. The capacitor electrode is surrounded by a guard electrode in order to minimize the effects of stray capacitances. The sample b is clamped on one end in the sample holder a. The bending can also be measured with high sensitivity by deflection of the beam of a laser using a position sensitive detector. To use this method, it requires that the sample is thin enough. Some experiment measurement values are listed in table 5. ## Surface stress effects in materials science ### Surface structural reconstruction Structural reconstruction at the surfaces has been studied extensively by both theoretical and experimental methods. However, a question about surface stress is high enough to be a main driving force of the reconstruction is still not very clear. Most of metallic surface reconstruction exhibit in two genetic forms. On the original (100) surface, it would form a hexagonal overlayer which results in a considerably higher density of surface atoms by 20–25%. On the original (111) surface, since it already in closed-pack structure, the higher density is due to a contraction while the local coordination of the surface atoms remains a hexagonal one. Another way to explain the surface reconstruction phenomenon is called “soft phonon type of reconstruction”. The driving force for a change in the surface concentration associated with a contraction of the surface is proportional to the difference between surface stress and surface free energy. It corresponds to the amount of energy gained by structure transformation to over the surface stress. For semiconductor surface, forming dimer is the way for it to response to the tensile stress. Figure 3 shows an example of Si(100) surface reconstruction that create tensile stress. ### Adsorbate-induced changes in the surface stress As mentioned above, surface stress is caused by charge density redistribution of surface atoms due to lacking of nearest neighbor atoms. In case of introduction of adsorbates (atoms that land on surface), charge density would be then modified around these adsorbates, resulting different surface stress state. There are many types of reaction between adsorbates and the surface that cause different stress behavior. Here, two most common behaviors are shown: #### Coverage dependence of the adsorbate-induced surface stress Coverage effect of surface stress without surface reconstruction usually result a compressive stress (assuming clean clean surface as reference or zero stress). Induced surface stress of number of different coverages on Ni(100) and Pt(111) surface is shown in figure 4. In all cases, it shows an initially linear increase of the induced stress with coverage, followed by an increase larger than linear at higher coverages. The non-linear increase is first thought to be due to the repulsive interaction between adsorbates. The repulsive interaction should be proportional to the overlap integrals summed of non-bonding orbitals with exponential relationship: Sij & exp(-crij) where rij is distance between two adsorbate i and j One can easily relate the mean distance between two adsorbates with square root of the coverage: Sij & exp(-c/√θ) Then the stress induced by absorbates can be derived as: ∆τ=a.θ+b.exp(-c/√θ) (8) where a, b, and c are fitting parameters. Figure 4 shows very good fits for all systems with the equation 8. However, later research shows that direct repulsive interaction between absorbate atoms (as well as dipolar interactions) contribute very little to the induced surface stress. The stress can become large only if the distance between the adsorbed atoms becomes small so that φij (repulsive pairwise interaction potential) becomes large. It rarely happens without very high gas pressure since adsorbated state become unstable with respect to desorption. #### Adsorbate-induced stress and restructuring of surfaces It shows that the tensile stress on clean surfaces can be so strong that the surface reconstructs to form an overlayer of higher charge density. In the presence of adsorbates, stress induced by could also be high enough for such reconstruction. The mechanism of the reconstruction of the two processes would be similar. The reconstruction due to adsorbates is easily recognized by deviation from stress-induced vs. coverage relationship. One example is shown in figure 5 and 6. It shows clearly the difference between stress-induced behavior of silicon compared with oxygen or carbon absorbate on Ni(100) surface. S/Ni(100) system reaches very high stress at the coverage of ~0.3. This stress then causes a reconstruction (figure 5) to increase the charge density of surface atoms in order to reduce the developed stress.

{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "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}

CC-MAIN-2018-09

latest

en

http://www.bluebulbprojects.com/MeasureOfThings/results.php?comp=speed&unit=ypm&amt=2700&sort=pr&p=5

text/html

crawl-data/CC-MAIN-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00318.warc.gz

Bluebulb Projects presents: Enter a measurement to see comparisons Equivalents in other units How fast is 2,700 yards per minute? Sort Order: Closest first | Highest first | Lowest first It's about twenty times as fast as Michael Phelps In other words, 2,700 yards per minute is 21.1830 times the speed of Michael Phelps, and the speed of Michael Phelps is 0.047208 times that amount. (at the Beijing Olympics, 2008; 200 m freestyle) (a.k.a. Michael Fred Phelps) (swimmer; 1985-)Setting a world record, Michael Phelps swam the 200 m freestyle in 1:42.96 for an average speed of 127.460 yards per minute. Phelps would go on to win nine gold medals individually in the 2008 Olympics - more than all but eight of the competing nations. It's about one-twentieth as fast as a Bullet (Rifle) In other words, the speed of a Bullet (Rifle) is 22.20 times 2,700 yards per minute. (5.56 x 45 mm, a.k.a. 5.56 NATO)A 5.56 x 45 mm cartridge is fired at a velocity of 60,000 yards per minute. As the NATO rifle cartridge, it is used by the military forces of more than thirty-one countries, including use in the M16 series, M4 Carbine, HK-416, and M249 Squad automatic weapons used by the United States armed forces. It's about thirty times as fast as Walking Pedestrians (in Manhattan) In other words, the speed of Walking Pedestrians (in Manhattan) is 0.031 times 2,700 yards per minute. (Manhattan; average speed; 8,978 person-sample)A 2006 Study by the New York City Department of City Planning found that pedestrians in that city walk at an average rate of 85 yards per minute. Pedestrians wearing headphones, the study went on to find, walk at a slightly faster 93 yards per minute It's about 80 times as fast as an Iceberg In other words, the speed of an Iceberg is 0.01 times 2,700 yards per minute. (a.k.a. Berg) (Newfoundland iceberg average)Moved by ocean currents and wind, icebergs can drift at speeds of about 30 yards per minute. The largest iceberg ever recorded was a found near Baffin Island, Nunavut and was estimated to be nine billion metric tons. It's about one-two-hundredth as fast as The Space Shuttle In other words, 2,700 yards per minute is 0.0053138 times the speed of The Space Shuttle, and the speed of The Space Shuttle is 188.190 times that amount. (Orbiter vehicle velocity)The space shuttle orbits at a speed of 508,110 yards per minute. During liftoff, the space shuttle accelerates to orbital speed in 8.5 minutes, consuming over 1.59 million kg (3.51 million lbs) of propellant in the process. It's about 600 times as fast as a Sloth In other words, 2,700 yards per minute is 610 times the speed of a Sloth, and the speed of a Sloth is 0.0016 times that amount. (for Brown-throated three-toed sloth, Bradypus variegatus)The three-toed sloth moves along the ground at an average speed of 4.40 yards per minute. Long thought to be lengthy sleepers, a 2008 study concluded that sloths sleep an average of only 9.6 hours per day. It's about one-one-thousandth as fast as a Meteor In other words, 2,700 yards per minute is 0.00098 times the speed of a Meteor, and the speed of a Meteor is 1,000 times that amount. (formally Meteoroid or meteorite, depending on the context; a.k.a. "shooting star", a.k.a. "falling star")Small meteoroids enter the Earth's atmosphere at speeds between 730,000 yards per minute and 4,700,000 yards per minute, depending on their size. The fireball effect, known as ablation, ceases once the atmosphere has slowed the meteoroid to a velocity of about 180,000 yards per minute.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-45

latest

en

https://math.stackexchange.com/questions/2122316/relation-between-decomposition-field-and-complete-decomposition-of-a-prime-ideal

text/html

crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00453.warc.gz

# Relation between decomposition field and complete decomposition of a prime ideal In my lecture on Algebraic Number Theory in the chapter about Galois extensions we had the following statement: $$G_{\mathfrak{p}}= \{id\} \iff Z_{\mathfrak{p}}=L \iff p~ \text{is completely decomposed}$$ where $A$ is a Dedekind domain, $K$ its quotient field, $L/K$ a finite Galois extension of degree $n$, $\mathcal{O}_L$ the integral closure of $A$ in $L$, $G=\mathrm{Gal}(L/K)$, $\mathfrak{p}$ a non-zero prime ideal in $\mathcal{O}_L$ over $p$ which is a non-zero prime ideal in $A$. $G_{\mathfrak{p}}$ denotes the decomposition group of $\mathfrak{p}$ and $Z_{\mathfrak{p}}$ the corresponding decomposition field. I do understand the first $\Leftrightarrow$, but have unfortunately some trouble to show (and understand) the second $\Leftrightarrow$. My ideas so far were (for $\Rightarrow$): If $G_{\mathfrak{p}}= \{id\}$, then the number of prime ideals in $\mathcal{O}_L$ above $p$ is $n$ as $(G:G_{\mathfrak{p}})=\vert G \vert= [L:K]=n$ and $G$ acts transitively on the set of all prime ideals over $p$. This gives $p \mathcal{O}_L= ({\mathfrak{p}_1} \cdot \dotsc \cdot \mathfrak{p}_n)^e$ as the decomposition of $p$ with $e$ the ramification degree. Now you should somehow get that $e$ equals one (as is required in the definiton of completely decomposed) but I just don't see why this follows? For the converse: no idea. I would be very thankful for any hints and/or solutions! • @Watson Thank you for your edits! Commented Feb 1, 2017 at 8:55 ## 1 Answer Since $G_{\mathfrak p}$ is the stabilizer of $\mathfrak p$, we know that $G/G_{\mathfrak p}$ is in bijection with the orbit of $\mathfrak p$, so $[G:G_{\mathfrak p}]$ is the number $r$ of primes above $p$. This number $r$ is $n=[L:K]$ iff $p$ is totally split. Moreover $r=[G:G_{\mathfrak p}]=n=|G|$ iff $G_{\mathfrak p} = \{1\}$. So we get $$Z_{\mathfrak p} = L \iff G_{\mathfrak p} = \{1\} \iff r=[G:G_{\mathfrak p}]=n \iff p \text{ is totally split in } L,$$ as desired. • Thank you for your answer. I still have one question though: we defined that a prime ideal $p$ is completely decomposed (or totally split) if $r= [L:K]$ (which you used) and if $p$ is unramified (which would require that the ramification degree of all the $\mathfrak{p}$ in the decomposition is $1$). It probably is totally obvious why this is true here, but could maybe explain this to me? (I fail to see why this property holds.) Commented Feb 1, 2017 at 9:00 • @SallyOwens : This is because $[L:K] = \sum_{i=1}^r e_i f_i$ always holds. Commented Feb 1, 2017 at 9:49 • Thank you - I think I got it now :) (Just to be sure: in this case (Galois extension), I know that the $e_i$ and $f_i$ are the same for all $\mathfrak{p}_i$ and hence $[L:K]=rfe$ and as $[L:K]$ is already equal to $r$, I must have $e=f=1$.) Commented Feb 1, 2017 at 9:51 • @Watson would you mind telling me what $r\ f\ e$ means? Commented Sep 22, 2017 at 3:24 • @GuerlandoOCs : when we write $p \mathcal{O}_L= ({\mathfrak{p}_1} \cdot \dotsc \cdot \mathfrak{p}_r)^e$, the integer $e$ is the ramification index, $r$ is the number of prime ideals in $L$ above $p$, and $f = [\mathcal O_L / \mathfrak{p}_1 : \mathcal{O}_K / (p)]$ in the inertia degree. See here for some details. Commented Sep 22, 2017 at 11:10

{"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}

CC-MAIN-2024-33

latest

en

https://rdrr.io/github/djvanderlaan/lvec/src/tests/testthat/test_sorting.R

text/html

crawl-data/CC-MAIN-2018-43/segments/1539583515375.86/warc/CC-MAIN-20181022180558-20181022202058-00075.warc.gz

# tests/testthat/test_sorting.R In djvanderlaan/lvec: Out of Memory Vectors ```context("Sorting") test_that("numeric vectors are correctly sorted", { # I did not use both NA and NaN as order of those seems to be undefined a_r <- c(-1.2, 5E10, NA, NA, 4.1) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) a_r <- c(-1.2, 5E10, NaN, NaN, 4.1) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # Test if clone argument works b <- sort(a) b_r <- sort(a_r, na.last=TRUE) lset(b, 2, 42) b_r[2] <- 42 expect_that(as_rvec(b), equals(b_r)) a <- as_lvec(a_r) sort(a, clone = FALSE) expect_that(as_rvec(a), equals(sort(a_r, na.last=TRUE))) # Empty vector a_r <- numeric(0) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # 'Large' vectors a_r <- rnorm(1000) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) }) test_that("integer vectors are correctly sorted", { a_r <- as.integer(c(-1, 2, 1E9, NA, 1234)) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # Test if clone argument works b <- sort(a) b_r <- sort(a_r, na.last=TRUE) lset(b, 2, 42L) b_r[2] <- 42L expect_that(as_rvec(b), equals(b_r)) a <- as_lvec(a_r) sort(a, clone = FALSE) expect_that(as_rvec(a), equals(sort(a_r, na.last=TRUE))) # Empty vector a_r <- integer(0) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # 'Large' vectors a_r <- sample(1000, 1000) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) }) test_that("logical vectors are correctly sorted", { a_r <- as.logical(c(TRUE, FALSE, TRUE, NA, NA)) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # Test if clone argument works b <- sort(a) b_r <- sort(a_r, na.last=TRUE) lset(b, 2, TRUE) b_r[2] <- TRUE expect_that(as_rvec(b), equals(b_r)) a <- as_lvec(a_r) sort(a, clone = FALSE) expect_that(as_rvec(a), equals(sort(a_r, na.last=TRUE))) # Empty vector a_r <- logical(0) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # 'Large' vectors a_r <- sample(c(TRUE, FALSE), 1000, replace = TRUE) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) }) test_that("character vectors are correctly sorted", { a_r <- c("jan", "pier", NA, "corneel") a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # Test if clone argument works b <- sort(a) b_r <- sort(a_r, na.last=TRUE) lset(b, 2, "johan") b_r[2] <- "johan" expect_that(as_rvec(b), equals(b_r)) a <- as_lvec(a_r) sort(a, clone = FALSE) expect_that(as_rvec(a), equals(sort(a_r, na.last=TRUE))) # Empty vector a_r <- character(0) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) # 'Large' vectors a_r <- paste0(sample(letters, 1000, replace = TRUE), sample(letters, 1000, replace = TRUE)) a <- as_lvec(a_r) expect_that(as_rvec(sort(a)), equals(sort(a_r, na.last=TRUE))) }) ``` djvanderlaan/lvec documentation built on Aug. 16, 2018, 11:02 a.m.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-43

latest

en

https://www.slideserve.com/vui/challenge-2

text/html

crawl-data/CC-MAIN-2017-51/segments/1512948550986.47/warc/CC-MAIN-20171214202730-20171214222730-00588.warc.gz

1 / 3 # Challenge 2 - PowerPoint PPT Presentation Compound Challenge. Challenge 2. T. Trimpe 2008 http://sciencespot.net/. Use the formulas provided to determine the number of atoms of each element in each compound. 1. 4CO 2 2. 2H 2 O 2 3. 3Mg(OH) 2 4. 6NaSO 4 5. 4CH 4. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Challenge 2' - vui Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript Challenge 2 T. Trimpe 2008 http://sciencespot.net/ Use the formulas provided to determine the number of atoms of each element in each compound. 1. 4CO2 2. 2H2O2 3. 3Mg(OH)2 4. 6NaSO4 5. 4CH4 Remember to list each element’s symbol and tell the number of atoms for each! The answers are … of each element in each compound. C = 4 O = 8 1. 4CO2 2. 2H2O2 3. 3Mg(OH)2 4. 6NaSO4 5. 4CH4 H = 4 O = 4 Mg = 3 O = 6 H = 6 Na = 6 S = 6 O = 24 C = 4 H = 16

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-51

latest

en

https://zbook.org/read/c78a13_damage-analysis-and-evaluation-of-light-steel-structures-exposed-to.html

text/html

crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00108.warc.gz

# Damage Analysis And Evaluation Of Light Steel Structures Exposed To . 1y ago 14 Views 1.66 MB 10 Pages Last View : 7d ago Transcription Appl. Sci. 2017, 7, 239 6 of 10 with more applications in recent years, there is still a lack of understanding of the purlin-sheeting system interaction mechanism. Jang [66], Duthinh [67], and Simiu [68] calculated the wind-resistant ultimate bearing capacity of the portal frame based on the data from the wind tunnel test data of the low-rise light steel structure and the American code. They proposed a working definition of “failure” for steel structures using nonlinear finite-element analysis and presented a methodology for nonlinear structural behavior and the directionality of the wind speeds. Based on the wind tunnel test data, Li [69] studied the initial defect on the damage shape and the plastic damage of the joints without buckling for China’s southeastern coastal light steel frame. Compared to the component-level study, it can be seen that little research has been done at the structure level. The interactions between different systems and the damage sequence for the whole structure levels under different directionality wind and different limit states work on different component or system levels need to be checked in future research. 5. Performance Evaluation of Light Steel Systems Exposed to Wind Hazards Fragility analysis methodology was the most common method for implementing the performance-based engineering theory into the evaluation of light steel systems exposed to wind hazards since it is widely applied in the evaluation of the low-rise light-frame wood construction exposed to multi-hazard loads. Its main theory was mentioned as below. Fragility can be defined as the conditional probability of failure of a structural member of the system for a given set of input variables. It is expressed as P[ LS] all D P[ LS D x] P[ D x], (1) in which D a random demand on the system (e.g., 3 s gust wind speed); P[ LS D x ] is the conditional probability of the limit state (LS) at given demand x. The hazard is defined by the probability P[ D x ]. The conditional probability, P[ LS D x ] is the fragility. Equation (2) also can be expressed in convolution integral form if the hazard is a continuous function of demand x: P[ LS] Z 0 Fr ( x ) gX ( x )dx (2) in which Fr ( x ) fragility function of demand x expressed in the form of a probability density function. The conditional probability Fr ( x ) is known as a “fragility” [70]. In order to study the fragility of the each level of light-frame structures, several types of research have been conducted. Garcia [71] established a probabilistic analysis framework for the vulnerability of typical light steel structures in the United States. Ellingwood [72–75] proposed a series of fragility analysis methodologies for assessing the response of light-frame wood construction exposed to stipulated extreme windstorms and earthquakes. Henderson and Ginger [76,77] proposed a series of vulnerability model for Australian high-set house or metal-clad industrial buildings to extreme wind loading for cyclonic regions. Lee and Rosowsky [78] presents a fragility assessment for roof sheathing i 2. Cladding Profiled steel cladding is widely used in both roof and wall paneling. The cladding is fixed at its crest to battens or purlins beneath using self-tapping screws to prevent water ingress. The corrugated cladding is fixed at each alternate crest using a fastener comprising a self-tapping screw. Roofing of Related Documents: The Essential Guide to Cargo Damage 3 Table of Contents Preface 4 Chapter 1 – Introduction 6 Chapter 2 –Types of damages 8 2.1 – Physical Damage 10 2.2 – Reasons for Physical Damage 11 2.3 – Wet Damage 16 2.4 – Reasons for Wet Damage 17 2.5 – Contamination Damage 19 2.6 – Reasons for Contamination Damage 20 2.7 – Reefer related Damage 21 Elbow Drop 6 Grapple Upper, Head Damage Elbow Drop 11 Grapple Side, Chest Damage Side, Chest Damage Knee Slam Grapple Lower, Leg Damage Mo V eC oNTR LS D TAILS Head Scissors Elbow Grapple Upper, Head Damage Armcrusher Grapple Side, Arm Damage Leg Stomps Grapple Lower, Leg Damage Lower, Leg Damage Mo Ve CoNTR oLS DeTAILS Preliminary damage assessment – A mechanism used to determine the impact and magnitude of damage and the resulting unmet needs of individuals, businesses, the public sector, and communities as a whole. There are two types of PDAs: initial damage assessments and joint preliminary damage assessments. Initial damage assessment – The effort by . communities as a whole. There are two types of PDAs: initial damage assessments and joint preliminary damage assessments. Initial damage assessment - The effort by local authorities to collect data related to the extent of damage within a jurisdiction. Joint preliminary damage assessment - The coordinated effort by local, state, and federal POINT METHOD OF JOB EVALUATION -- 2 6 3 Bergmann, T. J., and Scarpello, V. G. (2001). Point schedule to method of job evaluation. In Compensation decision '. This is one making. New York, NY: Harcourt. f dollar . ' POINT METHOD OF JOB EVALUATION In the point method (also called point factor) of job evaluation, the organizationFile Size: 575KBPage Count: 12Explore further4 Different Types of Job Evaluation Methods - Workologyworkology.comPoint Method Job Evaluation Example Work - Chron.comwork.chron.comSAMPLE APPLICATION SCORING MATRIXwww.talent.wisc.eduSix Steps to Conducting a Job Analysis - OPM.govwww.opm.govJob Evaluation: Point Method - HR-Guidewww.hr-guide.comRecommended to you b Section 2 Evaluation Essentials covers the nuts and bolts of 'how to do' evaluation including evaluation stages, evaluation questions, and a range of evaluation methods. Section 3 Evaluation Frameworks and Logic Models introduces logic models and how these form an integral part of the approach to planning and evaluation. It also 2015 The Timken Company TIMKEN BEARING DAMAGE ANALSIS WITH LUBRICATION REFERENCE GUIDE 7 Types of Bearing Damage Many different operating conditions can cause bearing damage. Those listed in this section make up the most commonly identified causes of damage for anti-friction bearings, including cylindrical, spherical, tapered and ball designs. tion rate, evaluation use accuracy, evaluation use frequency, and evaluation contribution. Among them, the analysis of evaluation and classification indicators mainly adopts the induction method. Based on the converted English learning interest points, the evaluation used by the subjects is deduced for classification, and the evaluation list .

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-33

latest

en

https://mathoverflow.net/questions/221664/weights-on-cyclic-orderings

text/html

crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00131.warc.gz

# Weights on cyclic orderings Are there standard or known weights/metrics on cyclic orders? Cyclic orderings are different ways of listing elements from a finite set, where you call two lists the same if they differ only by a rotation. For example, the ways that 4 people can sit at a circular table, where you ignore rotation of the table. So the following 4 rankings of a 4-item set should be considered the same: [2,1,4,3] [1,4,3,2] [4,3,2,1] [3,2,1,4] You can think of this as arranging the 4 items on a circular list, with no particular top, but a definite direction. I would like to compare different cyclic orderings in a meaningful way. I have learned that there are many well-known ways of assigning distance or weight to non-cyclic orderings (a.k.a. permutations!). For example: • Kendall $\tau$ • Transposition distance (Cayley) • Ulam metric • Hamming weight • $\ell_1$ norm But these give different weights to permutations which correspond to the same cyclic ordering. A first simple idea is the following: given a permutation $a$, count the number of $i$ such that $a(i+1) \neq a(i) +1$, where $0 \leq i \leq n$ and index addition is performed modulo $n$. A natural variation on this idea would sum $|a(i+1) - (a(i)+1)|$. I haven't thought about either of these too carefully yet. Do they appear in the literature somewhere? Some nontrivial internet searching hasn't turned anything up. • See also mathoverflow.net/questions/58782/… - Woodall's "Cyclic-order graphs and Zarankiewicz's crossing-number conjecture" is the most interesting paper on this, from my perspective, because it actually constructs the graph of these things under a certain definition of this distance (one I like best). I do not know of any theoretical discussion of the issue you mention, though I'm glad to hear of Dima's computational one - and if it was actually more theoretical in that paper than I though, he'll let me know :) which I will find useful. – kcrisman Oct 27 '15 at 19:03 • we (and Woodall, and Kleitman) define the graph on these $(n-1)!$ elements of $S_n$, so that two such elements are adjacent if one needs to swap two adjacent positions to get one from the other. – Dima Pasechnik Oct 27 '15 at 22:10 ## 1 Answer cyclic permutations, and a need to define distances on them, pops up in literature on graph crossing numbers, e.g. we used it in http://arxiv.org/abs/math/0404142 It probably goes back all the way to at least D.J. Kleitman, The crossing number of $K_5$. J. Combinatorial Theory 9 (1970), 315–323.

{"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}

CC-MAIN-2021-25

longest

en

https://republicofsouthossetia.org/question/complete-this-statement-27a-2-2-45a-2-36a-2-9a-2-15824248-61/

text/html

crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00697.warc.gz

## Complete this statement 27a^2x^2+45a^2x+36a^2=9a^2() Question Complete this statement 27a^2x^2+45a^2x+36a^2=9a^2() in progress 0 3 months 2022-02-10T06:25:40+00:00 1 Answer 0 views 0

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-21

latest

en

http://mathhelpforum.com/discrete-math/165909-basic-truth-table-help-required.html

text/html

crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00481-ip-10-31-129-80.ec2.internal.warc.gz

# Thread: Basic Truth Table help required 1. ## Basic Truth Table help required With this information: J ^ ( J => M ) ^ ( M => H) => M ^ H J: I have a job M: I have money H: I am happy I need to do a truth table, to show that the argument is valid. Now, I completed the truth table, however I got a mix of T's and F's, which suggests the argument is not valid - thus why i am asking on here for help. Basically, I would really apprechiate abit of help as to which order I should complete the colomns. Thanks 2. Post what you get between each operator. 3. Well, I can fill in the first three colomns fine (as far as I know) but from then on, im not sure which order I fill them in in. 4. If you can do the last column, you should be able to do the first column and the 3. Work from left to right too. 5. Hello, Desir0! We need more brackets in the statement. . . $\bigg[J \wedge (J \to M ) \wedge (M \to H)\bigg] \to ( M \wedge H)$ $\text{I need to do a truth table to show that the argument is valid.}$ $\text{Now I completed the truth table, however, I got a mix of T's and F's,}$ $\text{which suggests the argument is not valid.}$ $\text{I would like help as to which order I should complete the columns.}$ . . $\begin{array}{|c|c|c||c|c|c|c|c|c|c|c|c|c|c|c|c|} J&M&H & \bigg[J & \wedge & (J & \to & M) & \wedge & (M & \to & H)\bigg] & \to & (M & \wedge & H) \\ \\[-4mm] \hline\hline T&T&T & T&T&T&T&T&T&T&T&T &T& T&T&T \\ T&T&F & T&T&T&T&T&F&T&F&F &T& T& F&F \\ T&F&T & T&F&T&F&F&F&F&T&T &T& F&F&T \\ T&F&F & T&F&T&F&F&F&F&T&F &T& F&F&F \\ F&T&T & F&F&F&T&T&F&T&T&T &T& T&T&T \\ F&T&F & F&F&F&T&T&F&T&F&F &T& T&F&F \\ F&F&T & F&F&F&T&F&F&F&T&T &T& F&F&T \\ F&F&F & F&F&F&T&F&F&F&T&F &T& F&F&F \\ \hline\hline &&& 1 & 10 & 2 & 8 & 3 & 11 & 4 & 9 & 5 & 13 & 6 & 12 & 7 \\ \hline \end{array}$ 6. Originally Posted by dwsmith If you can do the last column, you should be able to do the first column and the 3. Work from left to right too. Yeah, Just did that now. Got the same results i was getting before. In the final colomn i got t,f,f,t,t,f,f,t ... which is evidently not correct Originally Posted by Soroban Hello, Desir0! We need more brackets in the statement. . . $\begin{array}{|c|c|c||c|c|c|c|c|c|c|c|c|c|c|c|c|} J&M&H & \bigg[J & \wedge & (J & \to & M) & \wedge & (M & \to & H)\bigg] & \to & (M & \wedge & H) \\ \\[-4mm] \hline\hline T&T&T & T&T&T&T&T&T&T&T&T &T& T&T&T \\ T&T&F & T&T&T&T&T&F&T&F&F &T& T& F&F \\ T&F&T & T&F&T&F&F&F&F&T&T &T& F&F&T \\ T&F&F & T&F&T&F&F&F&F&T&F &T& F&F&F \\ F&T&T & F&F&F&T&T&F&T&T&T &T& T&T&T \\ F&T&F & F&F&F&T&T&F&T&F&F &T& T&F&F \\ F&F&T & F&F&F&T&F&F&F&T&T &T& F&F&T \\ F&F&F & F&F&F&T&F&F&F&T&F &T& F&F&F \\ \hline\hline &&& 1 & 10 & 2 & 8 & 3 & 11 & 4 & 9 & 5 & 13 & 6 & 12 & 7 \\ \hline \end{array}$ Wow! Looks like you've knocked it on the head! Can't thank you enough

{"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2016-50

longest

en

https://www.rajibroy.com/2022/12/

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00354.warc.gz

31 December 2022 # Rewind-Pause: Two of our dear friends from Dallas Confabulating over something with great intent. This one of Subhasis and Tathagata is from nearly 21 years back. I believe we were celebrating Natasha’s fourth birthday by the pool. 29 December 2022 # Here is another geometry puzzle I found this while reading some website. As you see, there is a quadrant and within that there is a circle that touches one axis. The line that is parallel to that axis and is a tangent to the circle is 12 cm. Question is what is the area that is shaded? (basically, the quadrant area sans the circle) 27 December 2022 # Red Rackham’s Treasure Tintin is not a comic book that is very common in the USA. Although, Steven Spielberg did make a movie out of it. In India (or for that matter, anywhere in Europe too), Tintin was part and parcel of growing up. The protagonist was ably supported by the warm-at-heart (a big part of the heart warmed by his penchant for whiskey) Captain Haddock (of the “ten thousand thundering typhoons” fame), the well meaning, hard of hearing but brilliant scientist Cuthbert Calculus (with his pendulum and “westwards, westwards” refrain), the bumbling twin detectives Thomson and Thompson (“to be precise, with a “p” as in psychology and without a “p” as in Venezuela” is the way they would clarify) and of course the ever faithful dog Snowy. With Natasha back in her home and Nikita out with her friends, Sharmila pulled up a book to relax in the evening. And I pulled out the “Red Rackham’s Treasure”. You would think that after 45 years or so, I would not find it as interesting. But you would be wrong. That is the brilliance, I sense, of the Belgian cartoonist Georges Remi. While as a child I focused on the story line, these days, at my age, I focus a little more on the creation itself – how the cartoonist is telling and drawing the story. One of my favorite pictures is the one in the inset. The quick background is this: Before our favorite team would go out on a treasure hunt, the story was leaked in the Daily Reporter. Which infuriated them no end and caused a great deal of inconvenience. When they came back, the same reporter accosted them with all sorts of questions. Captain, taken by surprise, was livid to see the reporter again. But on second thoughts, he came up with a cunning idea. He introduced Calculus as his secretary and let him take the interview. If you remember, Calculus was very hard of hearing. So the conversation – if you can call that, more like two independent monologues – that ensued is hilarious. The ever increasingly agitated reporter would ask one more pertinent question. And Calculus would invariably answer a completely different question with absolutely no bearing to the real context, whatsoever. The end of it is seen in the inset. Calculus walks away very satisfied with the interview. The journalist was left in a complete tizzy. But the best part is the smug, mischievous smile on the impish Captain’s face all the while listening to the conversation peeking from around the corner. You can almost see him chafing his palms!! 1 25 December 2022 # “Luchi” collaboration in full process The two daughters teamed up to make some “luchis” for Christmas lunch! For the uninitiated, luchi is a Bengali-style puffed bread made from flour. Of course, it is never “Bengali-style” unless it is deep fried 🙂 24 December 2022 # Like the olden times… Amitesh, Samaresh and I braved the cold of Atlanta – minus 6 F with the wind chill to put in a run like we had started doing nine winters back. (We have not run together for over a year now) Last night we decided to run on the coldest day to relive those memories. We chose to start from the same Starbucks we used to run from those days. And followed it up with a good chat and hot coffee like the good old days. 2 24 December 2022 # Rewind-Pause: Nathan and Arunima from over 15 years back! Nathan and I first worked together in Dallas. Then he moved to India. Subsequently, he came back to the USA – in Atlanta. We started working together again – but in a different company. And then finally, I moved to Atlanta. This was right after I moved to Atlanta when his daughter (Arunima) and mine two had a great day in the pool! P.S. Eventually, Nathan moved back to India again 24 December 2022 # Let’s put this in the record books! Hoping to survive this run in 9 degrees (with wind chill minus 6 degrees F – minus 20 C )

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-10

latest

en

https://www.mbamission.com/blog/the-quest-for-700-weekly-gmat-challenge-answer-108/

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00542.warc.gz

Blog # The Quest for 700: Weekly GMAT Challenge (Answer) Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer: There are two ways to approach this problem. We’ll go through them in turn. (1) Figure out the decimal expansions of the two fractions separately (using long division), then add. 2/9 = 0.2222… 3/11 = 0.2727… Thus, the sum is 0.4949… The first digit is a 4, as is the third digit, the fifth digit, and every digit after that in an odd-numbered position. (The digits in even-numbered positions are all 9’s.) Thus, the 99th digit must also be a 4. (2) Add the fractions together, then figure out the decimal expansion. 2/9 + 3/11 = 22/99 + 27/99 = 49/99 Now, you can either figure out that the decimal expansion of 49/99 is 0.4949…, or you can simply know a shortcut: any two-digit number divided by 99 becomes a decimal with that two-digit number repeating. For instance, 17/99 = 0.1717…, 91/99 = 0.9191…, and so on. (This pattern generalizes to any number of digits, as long as the denominator is composed of only 9’s. For instance, 125/999 = 0.125125…) The final analysis is the same: every digit in an odd-numbered position is a 4, so the 99th digit after the decimal point is a 4 as well. The correct answer is (C). ### Upcoming Events • Duke Fuqua (Round 3) • Ohio Fisher (Round 3) • Vanderbilt Owen (Round 3) • USC Marshall (Round 3) • Carnegie Mellon Tepper (Round 3) • Toronto Rotman (Round 3) • INSEAD (Round 4) • Cambridge Judge (Round 4) • UW Foster (Round 3) • Notre Dame Mendoza (Round 3) • Emory Goizueta (Round 3) • Oxford Saïd (Round 3) • IESE (Round 3) • Dartmouth Tuck (Round 3) • London Business School (Round 3) • Texas McCombs (Round 3) • Vanderbilt Owen (Round 4) • Berkeley Haas (Round 4)

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-10

longest

en

http://mathforum.org/mathtools/discuss.html?context=all&do=r&msg=10679

text/html

crawl-data/CC-MAIN-2018-13/segments/1521257647600.49/warc/CC-MAIN-20180321082653-20180321102653-00515.warc.gz

You are not logged in. Discussion: All Topics Topic: rational equations Post a new topic to the General Discussion in Algebra for Rational Equations discussion Subject: Rational equation Author: dcc Date: Oct 13 2003 2/(x+1) = 1/(x-2)  If this is the equation you meant, then first state the restrictions:  x cannot equal -1 or 2 This equation is in the form of a proportion, so "the product of the means equals the product of the extremes." So, 2(x-2) = (x+1)1 2x - 4 = x + 1 2x - x = 4 + 1 x = 5 Since 5 is not one of the restricted values, the solution set is {5}. Check:  2/(5 + 1) = 1/(5 - 2) 2 / 6 = 1/ 3 Post a new topic to the General Discussion in Algebra for Rational Equations discussion Visit related discussions: Algebra Rational Equations Discussion Help

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2018-13

longest

en

http://physics.stackexchange.com/questions/tagged/speed+special-relativity

text/html

crawl-data/CC-MAIN-2014-23/segments/1405997883468.51/warc/CC-MAIN-20140722025803-00021-ip-10-33-131-23.ec2.internal.warc.gz

Tagged Questions 193 views The speed limit is with respect to what? As I understand, an object with mass cannot reach the speed of light because its (relativistic) mass increases "exponentially" as it approaches light speed. So there is a relation between mass and ... 44 views How is the universe is experienced at light speed? [duplicate] When moving faster, we experience time dilation and space contraction. We often state that a photon does not travel through time, i.e. if it where capable of observation, it would not experience time. ... 256 views Is there an upper limit on the radius of a rotating wheel? Is there an upper limit on the radius of a real wheel which is rotating at an Angular frequency of $\,\omega \,$ along its axis, such that we just require a finite amount of energy to rotate it? ... 135 views Is it possible to defy time using speed? [closed] I have always wondered if we were to travel at past the speed of light (even beyond the current capability) would we be able to slow down time around us? I have asked my teacher at school and he has ... 1k views Does inertia increase with speed? I have heard that when the speed of the object increase, the mass of the object also increase. (Why does an object with higher speed gain more (relativistic) mass?) So inertia which is related to ... 3k views What is the speed of time When we measure the speed of a moving element we do it with the help of a reference frame. Now if we need to measure the speed of time, is it possible? Is time really has speed? Thanks in advance. 5k views Why does an object with higher speed gain more (relativistic) mass? [duplicate] Today, in my high school physics class, we had an introductory class on electromagnetism. My teacher explained at some point that an object with a very high speed (he said it started to get somewhat ... 216 views Speed of Entropy change If time in systems moving with different speed goes differently, does speed of entropy change differ in these systems? (is "speed of entropy change" a valid term? can we compare them?) 192 views Approximate Time Dilation at Rocket Speeds How do you calculate the time dilation effect experienced by a traveler traveling at a relatively low speed? Specifically, how much time dilation would a traveller moving at $v=0.0007 c$ (speed of ... 286 views speed of light flashlights question This might be a silly question but I take the risk because it's been puzzling me for quite some time : If you have 2 flashlights, one facing North and one facing South, how fast are the photons (or ... 2k views Maximum speed of a rocket with a potential of relativistic speeds Ultimately, the factor limiting the maximum speed of a rocket is: the amount of fuel it carries the speed of ejection of the gases the mass of the rocket the length of the rocket ... 500 views How does a particle of light reach the max speed of light? [duplicate] Possible Duplicate: How can a photon have no mass and still travel at the speed of light? First of all I am not a professional physicist. I was curious as to how a particle of light can ... 1k views How to compute the speed of sound in relativistic hydrodynamic? In Weinberg: Gravitation and Cosmology chapter 2.10 (Relativistic Hydrodynamics) the speed of sound is derived as $v_s^2 = \left(\frac{\partial p}{\partial \rho}\right)$ and the equation of state ... 5k views Accelerating particles to speeds infinitesimally close to the speed of light? I'm in a freshmen level physics class now, so I don't know much, but something I heard today intrigued me. My TA was talking about how at the research facility he worked at, they were able to ...

{"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}

CC-MAIN-2014-23

longest

en

https://www.microbit.org/ko/2017-03-07-temperature/

text/html

crawl-data/CC-MAIN-2020-10/segments/1581875146681.47/warc/CC-MAIN-20200227094720-20200227124720-00174.warc.gz

Temperature Activity Learn how to use the device's sensors as input in this activity. • 대상 연령 8+ • 30 예상 시간(분) • MakeCode Editor 소개 This project uses one of the on-board sensors and a gesture control to display the real world temperature. 활동 4 단계 1단계 In the MakeCode editor select the Input option from the block menu and find the “On shake” block and drag onto the workspace. "흔들림 감지하면 실행" 블럭의 "흔들림" 을 "왼쪽으로 기울임" 또는 "오른쪽으로 기울임"으로 변경할 수도 있습니다. 이 프로젝트에서는 "흔들림" 동작을 사용합니다. `````` input.onGesture(Gesture.Shake, function () { }) `````` 2단계 Now the micro:bit is programmed so it will respond to a shake gesture the next part is to create a variable that the temperature will be stored in. From the block menu select the “Variable” and click “Make a Variable”. A popup dialogue box will then ask you to name your variable. In this example name it “Temp”. The block “Temp” now appears in the Variable option menu. When the micro:bit is shaken we would like the temperature to be stored in the variable “Temp”. To achieve this, we select the “Set item to” block and drag into the workspace. Click the arrow next to 'item' and select 'Temp'. Now when the micro:bit is shaken the variable “Temp” will be set to 0. `````` let temp = 0 input.onGesture(Gesture.Shake, function () { temp = 0 }) `````` Step Three So that the variable “Temp” is set to the actual temperature we need to use the temperature input from the “Input” option from the block menu. Drag this onto the workspace and set it to replace the “0” so that when the micro:bit is shaken the “Temp” variable is set to the Temperature input from the micro:bit sensor. `````` let temp = 0 input.onGesture(Gesture.Shake, function () { temp = input.temperature() }) `````` Step Four Now that the variable “Temp” has been set to the actual temperature using the micro:bit sensor, it now must be displayed so the user can see the value. Select the “Show number” block from the basic menu and drag it onto the workspace. Place the “show number” block under the set Temp block in the shake input block. `````` let temp = 0 input.onGesture(Gesture.Shake, function () { temp = input.temperature() basic.showNumber(0) }) `````` So that the temperature is shown on the LEDs the “Temp” variable is needed again. This is found from the Variable option from the block menu. Place the “Temp” variable in the show number block so the value of the temperature is displayed on the LEDs. `````` let temp = 0 input.onGesture(Gesture.Shake, function () { temp = input.temperature() basic.showNumber(temp) }) ``````

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-10

latest

en

https://www.physicslens.com/lol-diagram-template-using-geogebra/

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00597.warc.gz

LOL Diagram Template using GeoGebra Direct link: https://www.geogebra.org/m/u2m3gnzj The above is a GeoGebra applet that can be customised for any energy problem. Simply make a copy of it and change the values or labels as needed. This can be integrated into either GeoGebra Classroom or Google Classroom (as a GeoGebra assignment) and the teacher can then monitor every student’s attempt at interpreting the energy changes in the problem. The teacher can also choose different extents of scaffolding, e.g. provide the initial or final states and ask students to fill in the rest. What is an LOL Diagram? An LOL diagram is a tool used to visualize and analyze the conservation of energy in physical systems. “LOL” does not stand for anything meaningful. Rather, they just form the shapes of the two sets of axes and the circle in between. They help clarify which objects or components are included in the energy system being considered and how energy is transferred or transformed within that system. In LOL diagrams: • An energy system is defined as an object or a collection of objects whose energies are being tracked. • LOL diagrams consist of three parts: a L-shaped bar-chart representing the initial state, an O representing the object (or system) of interest and another L-shaped bar-chart representing the final state. • There can also be energy transferred into the system or out of the system if the system is not closed or isolated. These are represented using horizontal bars below the L axes, with arrows indicating if they are energy transferred in or out. When performing calculations involving the initial and final energy states, the energy transferred into the system is added to the initial energy state while the energy transferred out of the system is added to the final energy state. The sums must be equal. In other words, Initial energy stores + Energy transferred into system = Final energy stores + Energy transferred out of system How do I use an LOL diagram? Here’s a breakdown of how LOL diagrams are used, using an example of a falling mass: 1. System Definition (O): • Choose what is part of the energy system (objects whose energies are being tracked) and what isn’t. • For example, in the case of a falling mass, the mass itself and the Earth are part of the energy system. 2. Initial State (L): • Represent the initial energy configuration of the system. • Identify the types of energy present in the system at the beginning. In this example, we begin with some gravitational potential energy. 3. Transition: • Show how energy changes as the system evolves. In the falling mass example, the gravitational potential energy decreases, and kinetic energy increases. 4. Final State (L): • Represent the energy distribution in the system at the end of the process. • In the falling mass example, at the point just before it hits the ground, kinetic energy is maximized, and gravitational potential energy is minimized. LOL diagrams illustrate that energy within the system is conserved, meaning the total energy in the system remains constant. External work (work done by forces outside the defined system) may impact the system’s energy, but internal work (work done within the defined system) does not change the total energy of the system. The mathematical representation of the above problem will then simply be: GPE = KE $mgh = \dfrac{1}{2}mv^2$ This problem seems a bit trivial. Since LOL diagrams are a visual tool to help students and scientists analyze energy transformations and conservation, they can be used for making it easier to set up and solve conservation of energy equations in problems of greater complexity. LOL Diagram of an Electrical Circuit It is also important to note that the choice of the object (or system) of interest will result in different LOL diagrams for the same phenomenon. For example, consider a filament bulb in a circuit with a battery. The system at room temperature also has some energy in the internal store (or internal energy, which consists of the kinetic and potential energies of the particles in the system). When considering the filament as the object of interest, when energy is transferred electrically from the battery, part of it is transferred by light from the bulb to the surroundings and another part is added to the internal store, as it heats up the filament light. On the other hand, when considering the circuit as the whole, the chemical potential store of the battery is included in the initial energy state of the system. Hence, there is no additional energy transfer into the system but the energy transfer output is still the same. How do I modify the GeoGebra applet to make my own LOL Diagram? Here’s a video that demonstrates how the editing process is done, in a little more than one minute! The final product is here.

{"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}

CC-MAIN-2024-10

longest

en

https://bsci-ch.org/what-is-the-gcf-of-40-and-60/

text/html

crawl-data/CC-MAIN-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00317.warc.gz

Make use of GCF Calculator come quickly discover the Greatest common Factor of number 40, 60 i.e. 20 biggest integer through which both the numbers can be divided. You are watching: What is the gcf of 40 and 60 Greatest common factor (GCF) of 40 and also 60 is 20. GCF(40,60) = 20 2 40 2 20 2 10 5 5 1 Prime components of 40 are 2,5. Prime factorization of 40 in exponential kind is: 40 = 23×51Prime administrate of 60 2 60 2 30 3 15 5 5 1 Prime factors of 60 are 2,3,5. Element factorization the 60 in exponential type is: 60 = 22×31×51∴ for this reason by taking usual prime determinants GCF of 40 and 60 is 20 Factors that 40 List of hopeful integer determinants of 40 the divides 40 there is no a remainder. 1,2,4,5,8,10,20,40 Factors that 60 List of positive integer factors of 60 the divides 60 without a remainder. 1,2,3,4,5,6,10,12,15,20,30,60Greatest usual Factor We found the factors and prime administer of 40 and also 60. The biggest common factor number is the GCF number. For this reason the greatest usual factor 40 and 60 is 20. Also inspect out the Least common Multiple of 40 and also 60 1. What is the GCF that 40 and also 60? Answer: GCF that 40 and also 60 is 20. 2. What room the factors of 40? Answer: factors of 40 room 1, 2, 4, 5, 8, 10, 20, 40. There room 8 integers the are components of 40. The greatest variable of 40 is 40. See more: How To Breed A Poo Dragon - How Do You Breed Poo Dragon In Dragon City 3. What room the determinants of 60? Answer: components of 60 room 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. There room 12 integers that are determinants of 60. The greatest variable of 60 is 60. 4. Just how to find the GCF the 40 and also 60? Greatest typical Factor that 40 and 60 = 20 Step 1: discover the element factorization that 40 40 = 2 x 2 x 2 x 5 Step 2: find the element factorization of 60 60 = 2 x 2 x 3 x 5 Step 3: main point those determinants both numbers have in typical in actions i) or ii) over to discover the gcf: GCF = 2 x 2 x 5 = 20 Step 4: Therefore, the greatest usual factor the 40 and also 60 is 20 favorite Calculators Most famous

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-27

latest

en

http://www.dlxedu.com/askdetail/3/897de4af1146eed181d11daec5b21112.html

text/html

crawl-data/CC-MAIN-2019-18/segments/1555578689448.90/warc/CC-MAIN-20190425054210-20190425080210-00153.warc.gz

# python - Building a Tree with inorder and preorder traversal in Python3.x I am trying to build a tree using the preorder and inorder traversals (list of ints). Here is what I have so far: ``def build(preorder, inorder, heap): # Builds tree with the inorder and preorder traversalif len(preorder) == 0:return Noneroot = preorder[0] # Root is first item in preorderk = rootleft_count = inorder[(k-1)] # Number of items in left sub-treeleft_inorder = inorder[0:left_count]left_preorder = preorder[1:1+left_count]right_inorder = inorder[1+left_count:]right_preorder = preorder[1+left_count:]return [root, build(left_preorder, left_inorder), build(right_preorder, right_inorder)]`` I believe this algorithm is correct, although I could be wrong. My question is this - at what point do I insert the items into the tree? I have a class written to handle this, I'm just not sure where to insert this call, as the function will operate recursively. Any suggestions for how I should insert the nodes into the tree would be appreciated. ``````class heap: def __init__(self,the_heap): self.heap = the_heap def getChildren(self,value): n = self.heap.index(value) return self.heap[2*n+1],self.heap[2*n+2] # i think ... def getParent(self,value): n = self.heap.index(value) if n == 0: return None return self.heap[math.floor(n-1/2.0) ] # i think ... def traverse(self): #do your traversal here just visit each node in the order you want pass the_heap = heap(range(100)) print the_heap.getChildren(2) print the_heap.getParent(6) `````` something like that?

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-18

latest

en

https://math.stackexchange.com/questions/2030794/how-to-get-the-second-point-in-a-line-segment-knowing-its-first-point-distance

text/html

crawl-data/CC-MAIN-2019-30/segments/1563195528220.95/warc/CC-MAIN-20190722201122-20190722223122-00360.warc.gz

# How to get the second point in a line segment knowing its first point, distance and perpendicular line segment? I need to get the second point in a line segment knowing its first point, distance and perpendicular line segment. The line segments need to intersect , thus the direction of the incomplete line segment could differ based on the location of its first point. The problem I have now is that I don't know how to obtain the direction of the incomplete line segment. This is what is known: a1 = (30, 30) a2 = (10, 30) b1 = (30, 35) d = 30 (distance of line segment b) I tried the following: Get perpendicular slope a. m = (b2.x - b1.x) / (b2.y - b1.y); Get horizontal direction. hd = ? // either 1 or -1 Get a2. a2.x = a1.x + d * cos(atan(m)) * hd; a2.y = a1.y + d * sin(atan(m)); I am missing the step to get the horizontal direction of the slope of b, I was unable to recognize the pattern. Or perhaps there is a better and shorter algorithm to figure out point b2? • There are two possibilities for $B_2$. The drawing suggests that you’re to choose the one for which $\overrightarrow{B_1B_2}$ is rotated counterclockwise from $\overrightarrow{A_1A_2}$. Is that the case? – amd Nov 26 '16 at 6:41 • @amd the drawing represents the wanted outcome. I will edit the question to make clear that the line segments need to intersect. – oddRaven Nov 26 '16 at 9:40 • $$\vec{A_1A_2}•\vec{B_1B_2}=0$$ • $$||\vec{B_1B_2}||^2=900$$ Observe that $A_1A_2$ is a segment parallel to X-Axis. And also that $B_1$ is a point directly above $A_1$. So, the diagram looks like this : It's trivial to figure out the coordinates of $B_2$ now.

{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-30

latest

en

https://www.jiskha.com/display.cgi?id=1336896957

text/html

crawl-data/CC-MAIN-2017-30/segments/1500549424770.15/warc/CC-MAIN-20170724082331-20170724102331-00146.warc.gz

# Maths posted by . A fence that is 4 metres tall runs parallel to a tall building at a distance of 1 metre from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? • Maths - We form a rt. triangle: X = 1 m. = Hor. side. Y = 4 m. = Ver. side. Z = Hyp. = Length of the ladder. Z^2 = X^2 + Y^2 Z^2 = 1^2 + 4^2 = 17 Z = 4.1 m.

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2017-30

longest

en

https://www.gradesaver.com/textbooks/math/other-math/discrete-mathematics-with-applications-4th-edition/chapter-2-the-logic-of-compound-statements-exercise-set-2-5-page-94/32

text/html

crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00233.warc.gz

## Discrete Mathematics with Applications 4th Edition $44_{10}$ $62_{10} = (32 + 16 + 8 + 4 + 2)_{10} = 111110_2 → 00111110$ $−18_{10} = −(16 + 2)_{10}= −10010_2 → 00010010 → 11101101 → 11101110$ Thus the 8-bit representations of 62 and −18 are 00111110 and 11101110 respectively. Adding the 8-bit representations gives . 0 0 1 1 1 1 1 0 +1 1 1 0 1 1 1 0 $============$ .1 0 0 1 0 1 1 0 0 Truncating the 1 in the 28th position gives 00101100. Since the leading bit of this number is a 0, the answer is positive. Converting back to decimal form gives $00101100 → 101100_2 = (32 + 8 + 4)_{10} = 44_{10}.$ So the answer is 44.

{"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}

CC-MAIN-2020-16

latest

en

https://robotics.stackexchange.com/questions/21872/is-it-possible-to-solely-use-a-p-controller-for-velocity-control

text/html

crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00562.warc.gz

# Is it possible to solely use a P-controller for velocity control? Let's say I want a robot to run at a certain speed. I have one motor attached to each robot rear wheel. The feedback value to the controller comes from an encoder, which outputs pulses per time unit, i.e. the measured variable is speed of the robot. Something has been bugging me, so please correct me if I am wrong: I don't think it is possible to solely use a P-controller for velocity control. I seems to me that it is impossible to reach the desired target speed if your output signal from the controller (i.e. a velocity value) is proportional to the error value. Then when the error goes towards zero (which is what the P-controller strives for) the output signal (velocity) would go towards zero. But then how can you maintain a non-zero speed? Instead you will probably reach an equilibrium point were the robot velocity control is in a steady-state error. Is this reasoning correct or wrong? • Can you clarify this statement? "Then when the error goes towards zero (which is what the P-controller strives for) the output signal (velocity) would go towards zero." The output signal should be going toward the setpoint value while the error simultaneously goes toward zero proportionally. Commented Mar 5, 2021 at 22:36 • Yes, it's paradoxical which is what I wanted to uplift. A P controller strives to zero the error. But if the control variable is the speed and the output variable also is the speed then: u = Kp * e would give a zero speed as the P controller tries to reach the setpoint (desired speed), which would be impossible if your setpoint is a nonzero speed. Hope I made it clear. Commented Mar 8, 2021 at 12:01 output signal from the controller (i.e. a velocity value) is proportional to the error value I think there is some confusion here. In a standard velocity control loop, the controller output cannot be the velocity itself, by definition. Instead, the velocity is the feedback, whereas the controller output is most likely the voltage applied to the electrical motor. That said, your intuition is correct, a simple P controller is not sufficient for achieving the target velocity. It is easy to verify that, as result of a zero error, the voltage command will be zero, hence the motor won't spin any longer. Thus, a generic target speed different from zero is not attainable. More rigorously, if we consider that the transfer function voltage-speed can be well approximated with a first-order response, we end up with no integral in the loop if we close it with a simple P controller. In turn, no integral entails a non-null steady-state error. To compensate for that, one must purposely introduce the I term as it is well known. • Thanks man, that relieved me of many questions. Just as an interesting comment, I read about two types of systems: integrated and non-integrated. Example of the former is a water tank or position system where output is the accumulation of input (i.e. integral). In these systems a P controller could work. But in the later, non-integrated systems, such as flow and velocity systems, a P controller won't do. You will need the I-part as well. It all seems quite intuitive in hindsight hehe Commented Mar 8, 2021 at 12:06 • Well said! Consider that in practical contexts, a pure P controller won't be up to the task at hand either for integrated systems, as there will be always unmodelled quantities that will force us to introduce the I term. Commented Mar 8, 2021 at 12:23

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-26

latest

en

https://www.reference.com/math/types-quadrilaterals-c48e2aadcb7f6b40

text/html

crawl-data/CC-MAIN-2019-43/segments/1570987779528.82/warc/CC-MAIN-20191021143945-20191021171445-00342.warc.gz

# What Are the Types of Quadrilaterals? A quadrilateral is a four-sided shape, such as a rectangle or parallelogram. In order for a shape to be considered a quadrilateral, it has to have four straight sides and be two-dimensional. If all of the interior angles of a quadrilateral are added together, they are equal to 360 degrees. Some quadrilaterals require specific qualifications, such as all sides have to be equal in length or the sides have to be parallel. For instance, a rhombus has two sets of parallel sides that are equal in length. Quadrilaterals can be complex. Complex quadrilaterals, or intersecting quadrilaterals, cross over at some point. A bow tie shape is a good example of a complex quadrilateral. Quadrilaterals can be inclusive. This means that a shape can fit more than one definition. For instance, a square can also be considered a rectangle. A rectangle has to be four-sided, and every angle must equal 90 degrees. The square also fits this definition. However, not all rectangles are squares. A square must have four sides equal in length. Quadrilaterals are also known as quadrangles and tetragons. The term “quadrangle” references the four angles of quadrilaterals. "Tetragon" refers to the fact that the quadrilateral is a four-sided polygon. Similar Articles

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2019-43

latest

en

https://www.speedyhen.com/Product/Prime-Numbers-and-Their-Distribution/2399082

text/html

crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00284.warc.gz

# Prime Numbers and Their Distribution Paperback / softback ## Part of the Student Mathematical Library series Paperback / softback • Information #### Description We have been curious about numbers - and prime numbers - since antiquity. One notable new direction this century in the study of primes has been the influx of ideas from probability. The goal of this book is to provide insights into the prime numbers and to describe how a sequence so tautly determined can incorporate such a striking amount of randomness. There are two ways in which the book is exceptional. First, some familiar topics are covered with refreshing insight and/or from new points of view. Second, interesting recent developments and ideas are presented that shed new light on the prime numbers and their distribution among the rest of the integers. The book begins with a chapter covering some classic topics, such as quadratic residues and the Sieve of Eratosthenes. Also discussed are other sieves, primes in cryptography, twin primes, and more.Two separate chapters address the asymptotic distribution of prime numbers. In the first of these, the familiar link between $\zeta(s)$ and the distribution of primes is covered with remarkable efficiency and intuition. The later chapter presents a walk through an elementary proof of the Prime Number Theorem. To help the novice understand the 'why' of the proof, connections are made along the way with more familiar results such as Stirling's formula. A most distinctive chapter covers the stochastic properties of prime numbers. The authors present a wonderfully clever interpretation of primes in arithmetic progressions as a phenomenon in probability. They also describe Cramer's model, which provides a probabilistic intuition for formulating conjectures that have a habit of being true.In this context, they address interesting questions about equipartition modulo $1$ for sequences involving prime numbers. The final section of the chapter compares geometric visualizations of random sequences with the visualizations for similar sequences derived from the primes. The resulting pictures are striking and illuminating. The book concludes with a chapter on the outstanding big conjectures about prime numbers. This book is suitable for anyone who has had a little number theory and some advanced calculus involving estimates. Its engaging style and invigorating point of view will make refreshing reading for advanced undergraduates through research mathematicians. This book is the English translation of the French edition. #### Information • Format:Paperback / softback • Pages:Illustrations • Publisher:American Mathematical Society • Publication Date: • Category: • ISBN:9780821816479 £22.44 #### Information • Format:Paperback / softback • Pages:Illustrations • Publisher:American Mathematical Society • Publication Date: • Category: • ISBN:9780821816479 £57.00 £54.11 £46.95 £45.16 £43.95 £42.51 £57.00 £55.33

{"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}

CC-MAIN-2022-49

latest

en

https://www.mycoursehelp.com/QA/a-potential-field-in-free-space-is-given/65133/1

text/html

crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00191.warc.gz

### Create an Account Home / Questions / A potential field in free space is given as V 100 ln tanθ2 50 V a Find the maximum value... # A potential field in free space is given as V 100 ln tanθ2 50 V a Find the maximum value of Eθ on the surface A potential field in free space is given as V = 100 ln tan(θ/2) + 50 V. (a) Find the maximum value of |Eθ | on the surface θ = 40◦ for 0.1 2. e space, let ρν = 200ϵ0/r 2.4. (a) Use Poisson’s equation to find V(r) if it is assumed that r 2Er → 0 when r → 0, and also that V → 0 as r → ∞. (b) Now find V(r) by using Gauss’s law and a line integral. Jun 14 2020 View more View Less

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2021-17

latest

en

https://solvedlib.com/n/point-find-the-folkwing-expressions-using-the-graph-below,4563786

text/html

crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00530.warc.gz

# Point) Find the folkwing expressions using the graph below Of vectors uL, and1.U+v3.3v 674 |el =Foec: You can click on ###### Question: point) Find the folkwing expressions using the graph below Of vectors uL, and 1.U+v 3.3v 67 4 |el = Foec: You can click on (he grapn (0 enlarqe (ne image #### Similar Solved Questions ##### Can anyone help me with my homework problems? 1. Sunny Company has the following account balances... Can anyone help me with my homework problems? 1. Sunny Company has the following account balances after adjusting entries at December 31, 2012: Accounts Payable $24,000 Dividends 7,000 Treasury Stock, Common (22,000 shares) 98,000 Preferred Stock ($10 par) 80,000 Land 220,000 Cash 220,0... ##### RuizNot yet answered Marked out of 1.00The Transparent SUV is an electric car that runs on battery: It is equipped with an air-conditioning unit to coolthe car in summer: The refrigerator works by converting electrical energy to mechanical energy which operates the pump for the refrigerator The pump will then allow the removal of heat energy. Find the maximum possible efficiency expressed as Coefficient of Performance of the air-conditioning assuming that the ambient temperature is 30*C and the ruiz Not yet answered Marked out of 1.00 The Transparent SUV is an electric car that runs on battery: It is equipped with an air-conditioning unit to coolthe car in summer: The refrigerator works by converting electrical energy to mechanical energy which operates the pump for the refrigerator The pu... ##### Female< c(100, 70,60, 80 80,60,50.60) male<-c(50, 80,50,80,90,80,50,60,50,50)R output 1 :ttest(x-female; Y-male, alternative-"two.sided , conf level-0.99, var.equal-TRUE)Two Sample t-testdata: female and male 1.1442,df 18 p-value 0.2675 alternative hypothesis: true difference in means is nor equal to 0 99 percent confidence interval: 412.12623 28.12623 sample estimates: mean of mean ofy 72 64R output 2 :testix-female v-male; alternative-"two.sided , conf level-0.99, paired TRUE)P female< c(100, 70,60, 80 80,60,50.60) male<-c(50, 80,50,80,90,80,50,60,50,50) R output 1 : ttest(x-female; Y-male, alternative-"two.sided , conf level-0.99, var.equal-TRUE) Two Sample t-test data: female and male 1.1442,df 18 p-value 0.2675 alternative hypothesis: true difference in means... ##### Which two elements have only one electron shell (n=1) when in their lowest energy state? H... Which two elements have only one electron shell (n=1) when in their lowest energy state? H and Na H and He He and Li He and Ar... ##### A solution is made by equilibrating the two solids silver iodide (Ksp = 1.5×10-16) and silver... A solution is made by equilibrating the two solids silver iodide (Ksp = 1.5×10-16) and silver hydroxide (Ksp = 2.0×10-8) with water. What are the concentrations of the three ions at equilibrium, if some of each of the solids remain? [Ag+] = M [I-] = M [OH-] =    &nbs... ##### Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the... Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, write only NR. CaCl₂... ##### Find $x, Q R, R S,$ and $Q S$ if $riangle Q R S$ is an equilateral triangle. (FIGURE CAN'T COPY) Find $x, Q R, R S,$ and $Q S$ if $\triangle Q R S$ is an equilateral triangle. (FIGURE CAN'T COPY)... ##### Let (1i,Y)T, (Kz, Y)T be a random sample from a bivariate normal distribution po? v(( 0 )- po? 02 where 0? and p € (-1,1) are unknown parameters Find the MLE of 0 (2, p)T (Hint: First, solve the problem it terms of the parameters q1 91 (0) and 202 (1 92(0- (o2 'P)t; and then use the invariance property of a MLE.) Let (1i,Y)T, (Kz, Y)T be a random sample from a bivariate normal distribution po? v(( 0 )- po? 02 where 0? and p € (-1,1) are unknown parameters Find the MLE of 0 (2, p)T (Hint: First, solve the problem it terms of the parameters q1 91 (0) and 202 (1 92(0- (o2 'P)t; and then use the invar... ##### 27 PM1. (4 pts) Write the ground-state electron configuration for each element: Sodium Magnesium Oxygen Nitrogen Potassium Aluminum g) Phosphorus (h) Argon57 PM28 PMns (Cl19 PM 27 PM 1. (4 pts) Write the ground-state electron configuration for each element: Sodium Magnesium Oxygen Nitrogen Potassium Aluminum g) Phosphorus (h) Argon 57 PM 28 PM ns (Cl 19 PM... ##### Let A = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7} and define relation on 4 as follows:For all x,Y €A,x Ry 6 31(-y)It iS fact that R is an equivalence relation On A_ Use set-roster notation to write the equivalence classes Of R,[0][1] Let A = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7} and define relation on 4 as follows: For all x,Y €A,x Ry 6 31(-y) It iS fact that R is an equivalence relation On A_ Use set-roster notation to write the equivalence classes Of R, [0] [1]... ##### Calculate the percent Cz03 - in Ks[Al(C204)s] . 3Hz0.AEdSubmitRequest Answor Calculate the percent Cz03 - in Ks[Al(C204)s] . 3Hz0. AEd Submit Request Answor... ##### Evaluate the definite integral. Use a graphing utility to confirm your result.$int_{2}^{4} x operatorname{arcsec} x d x$ Evaluate the definite integral. Use a graphing utility to confirm your result. $int_{2}^{4} x operatorname{arcsec} x d x$... ##### 3. Multiple cells come together to formtissues, the next level of organization in the human body. Atissue is a group of connected cells that have a similar function. There are four main types of human tissues. Identify the functions of each tissue type: (section 4.2 and4.3) a. Nervous tissueb. Connective tissuec. Muscle tissued. Epithelial tissue4. Organs represent the next level of complexityin the human body. An organ is a specialized structure thatincludes two or more ty 3. Multiple cells come together to form tissues, the next level of organization in the human body. A tissue is a group of connected cells that have a similar function. There are four main types of human tissues. Identify the functions of each tissue type: (section 4.2 and 4.3) a. Nervous... ##### Find C(y) if y solves the following differential equation: d2 = 0 dt? where y(0) = 0 and y' (0) = 0 Find C(y) if y solves the following differential equation: d2 = 0 dt? where y(0) = 0 and y' (0) = 0... ##### When the transition metal materials are dissolved in the aqueous solution (e.g. Cu(NO3)2, Fe(NO3)3, K2CrO4, NiSO4),... When the transition metal materials are dissolved in the aqueous solution (e.g. Cu(NO3)2, Fe(NO3)3, K2CrO4, NiSO4), the solutions have color(such as blue, green and yellow color). Why do transition metal complexes make colored compounds in solution? Describe the relevant theory and reasons.... ##### 1. A B. C. (counter-clockwise current) What is the direction of the magnetic force on a... 1. A B. C. (counter-clockwise current) What is the direction of the magnetic force on a negative charge that moves through X хв. Bin X X X the magnetic field shown? < X b a Р What is the magnetic field at point P due to the current / in the wire shown? Let a 0.15 m, b = 0.55 m... ##### Question 13 Tries remaining: 2 Points out of 7.60 Suppose that the elasticity of demand for... Question 13 Tries remaining: 2 Points out of 7.60 Suppose that the elasticity of demand for a good was perfectly inelastic. The elasticity of supply for this good is Calculate the pass-through-fraction. Flag question (Answer as a decimal rather than as a percentage,) Answer: Check... ##### 1/25 = K,+ Kz K,-4+j 200 6 -+ j4 -3- j4 3 = K + Kz 4_13 625 25 25 I Kz = 200Then the impulse response is 1/25 = K,+ Kz K,-4+j 200 6 -+ j4 -3- j4 3 = K + Kz 4_13 625 25 25 I Kz = 200 Then the impulse response is... ##### Maiaamaia iiUuallun I0 (0) Uhanna (040)Ittam einm(ni ImtatmtCheng (1,~Vs, Evs) fron rectangulsr to spherical coordinates2/4 Maiaamaia ii Uuallun I0 (0) Uhanna (040) Ittam einm(ni Imtatmt Cheng (1,~Vs, Evs) fron rectangulsr to spherical coordinates 2/4... ##### Question 25 Gphold tever? Which of the following perains colmonefla Gcla different speces Je Is CAUed by rultiplies In patient pharocytes Causitive microolanem contuninated moat scquired vb naettton [ Itr Inul Foute thla disease wa (he oral; ruchlely tansnt aerrnptonhitk ctiers nte proridedYou (just d good IthiBy using Proctorio 'outc ol lotest pci dav'Ic4 Aacte ALAulALLA Pioclouo" Thanks for dolng your por0 /1pts Question 25 Gphold tever? Which of the following perains colmonefla Gcla different speces Je Is CAUed by rultiplies In patient pharocytes Causitive microolanem contuninated moat scquired vb naettton [ Itr Inul Foute thla disease wa (he oral; ruchlely tansnt aerrnptonhitk ctiers nte prorided You (jus... ##### Use the Mid-point Rule ad Trapezoidal Rlle to estimate the value ofthe integral f (r)dr . You must show all details Mid-point RuleTrapezoidal Rule; Use the Mid-point Rule ad Trapezoidal Rlle to estimate the value ofthe integral f (r)dr . You must show all details Mid-point Rule Trapezoidal Rule;... ##### -/9 POINTSYou may need to use the appropriate appendix table tecnnology answer this question_Suppose has distribution with / = 37 and 0 = 12. (a) If random samples of size 16 are sclected, can we Say anything about the Yes, distribution distribution sample Means? normal with mean 37 and 0 % Yes, the distribution norma Wiin Mean 37 and 0 ; the standard deviation too small; Yes, the distribution normal with mcan / x 37 and 0 ; No, the sample size too small.the oriqinal dlstribution normul Yes, the -/9 POINTS You may need to use the appropriate appendix table tecnnology answer this question_ Suppose has distribution with / = 37 and 0 = 12. (a) If random samples of size 16 are sclected, can we Say anything about the Yes, distribution distribution sample Means? normal with mean 37 and 0 % Yes, t... ##### 3. (10 points IfV (C) is as drawn, which of thene Ia Ite FoWt netlee/ (( 'lnnn butu UmaMiuy explain your choice ) V() =Xi sin(nwt) (b) V() =4>" (24ty sin((2n IJut) (c) V() =4x cOn(nwl) (d) V() =Xi 1nt cos((2n IJut) V() =- 4X cos(nut) 3. (10 points IfV (C) is as drawn, which of thene Ia Ite FoWt netlee/ (( 'lnnn butu UmaMiuy explain your choice ) V() =Xi sin(nwt) (b) V() =4>" (24ty sin((2n IJut) (c) V() =4x cOn(nwl) (d) V() =Xi 1nt cos((2n IJut) V() =- 4X cos(nut)... ##### A journal bearing has a shaft diameter of 78.00 mm with a unilateral tolerance of -0.02... A journal bearing has a shaft diameter of 78.00 mm with a unilateral tolerance of -0.02 mm. The bushing bore has a diameter of 78.10 mm with a unilateral tolerance of 0.06 mm. The bushing is 36 mm long and supports a load of 2 kN. The journal speed is 720 rev/min. For the minimum clearance assembly ... ##### NEW PROJECT ANALYSIS You must evaluate the purchase of a proposed spectrometer for the R&D department.... NEW PROJECT ANALYSIS You must evaluate the purchase of a proposed spectrometer for the R&D department. The base price is $190,000, and it would cost another$28,500 to modify the equipment for special use by the firm. The equipment falls into the MACRS 3-year class and would be sold after 3 year...

{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2022-49

latest

en

https://nl.mathworks.com/matlabcentral/cody/problems/20-summing-digits/solutions/1265642

text/html

crawl-data/CC-MAIN-2020-24/segments/1590347426801.75/warc/CC-MAIN-20200602193431-20200602223431-00124.warc.gz

Cody # Problem 20. Summing digits Solution 1265642 Submitted on 10 Sep 2017 by DilRavsh This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass a = 1; b = 2; out = sumDigits(a); assert(isequal(out, b)) 2   Pass a = 10; b = 7; out = sumDigits(a); assert(isequal(out, b)) 3   Pass a = 16; b = 25; out = sumDigits(a); assert(isequal(out, b))

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-24

latest

en

https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-8-sequences-and-infinite-series-8-3-infinite-series-8-3-exercises-page-623/11

text/html

crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00074.warc.gz

## Calculus: Early Transcendentals (2nd Edition) $\Sigma_{k=0}^{9}(-\frac{3}{4})^k =.53925$ $\Sigma_{k=0}^{9}(-\frac{3}{4})^k =\frac{a(1-(r)^n)}{1-r}$ $a=(-\frac{3}{4})^0=1$ $r=-\frac{3}{4}$ $n= 10$ $$\Sigma_{k=0}^{9}(-\frac{3}{4})^k =\frac{1(1-(-\frac{3}{4})^{10})}{1-(-\frac{3}{4})}=.53925$$

{"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}

CC-MAIN-2018-43

latest

en

https://www.doubtnut.com/qna/116056965

text/html

crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00170.warc.gz

# Blue, Black,Green,Red Text Solution Verified by Experts ## The correct Answer is:Black.It is not a component of whit light, whil the rest are the components. | Updated on:21/07/2023 ### Knowledge Check • Question 1 - Select One ## Six faces of a cube are coloured black, brown, green, red, white and blue, such that (i) Red is opposite black (ii) Green is between red and black (iii) Blue is adjacent to white (iv) Brown is adjacent to blue (v) Red is at the bottom. Answer the questions based on this information. The four colours adjacent to one another are ABlack, Blue, Brown, Red BBlack, Blue, Brown, White CBlack, Blue, Red, White DBlack, Brown, Red, White • Question 2 - Select One ## Six faces of a cube are coloured black, brown, green, red, white and blue, such that (i) Red is opposite black (ii) Green is between red and black (iii) Blue is adjacent to white (iv) Brown is adjacent to blue (v) Red is at the bottom. Answer the questions based on this information. Which colour is opposite brown? AWhite BRed CGreen DBlue • Question 3 - Select One ## Six faces of a cube are coloured black, brown, green, red, white and blue, such that (i) Red is opposite black (ii) Green is between red and black (iii) Blue is adjacent to white (iv) Brown is adjacent to blue (v) Red is at the bottom. Answer the questions based on this information. Which colour, of the following can be deduced from (i) and (v)? ABlack is on the top BBlue is on the top CBrown is on the top DBrown is on the top Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation

{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2024-26

latest

en