url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.teachoo.com/5059/1287/Ex-7.6--14---Integrate-x-(log-x)2---Chapter-7-Class-12/category/Integration-by-parts/ | 1,686,396,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00074.warc.gz | 1,135,252,975 | 35,654 | Integration by parts
Chapter 7 Class 12 Integrals
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 7.6, 14 γπ₯(logβ‘π₯)γ^2 β«1βγπ₯(logβ‘π₯ )^2.ππ₯ " " γ β΄ β«1βγπ₯(logβ‘π₯ )^2.ππ₯γ=β«1βγ(logβ‘π₯ )^2 π₯ .ππ₯γ = (logβ‘π₯ )^2 β«1βγπ₯ .γ ππ₯ββ«1β((π(logβ‘π₯ )^2)/ππ₯ β«1βγπ₯ .ππ₯γ) ππ₯ = (logβ‘π₯ )^2 . π₯^2/2ββ«1β(2(logβ‘π₯ ) 1/π₯ β«1βγπ₯ .ππ₯γ) ππ₯ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x and g(x) = (log x)2 = π₯^2/2 (logβ‘π₯ )^2β2β«1βγlogβ‘π₯/π₯ . π₯^2/2γ ππ₯ = π₯^2/2 (logβ‘π₯ )^2ββ«1βγπ₯ logβ‘π₯ γ ππ₯ Solving I1 I1 = β«1βγπ₯ logβ‘π₯ γ ππ₯ β«1βγπ₯ logβ‘π₯ γ ππ₯=β«1β(logβ‘π₯ )π₯ ππ₯ =logβ‘π₯ β«1βπ₯ ππ₯ββ«1β(π(logβ‘π₯ )/ππ₯ β«1βγπ₯.ππ₯γ)ππ₯ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x and g(x) = log x =logβ‘π₯ (π₯^2/2)ββ«1βγ1/π₯ . π₯^2/2. ππ₯γ =γπ₯^2/2 logγβ‘γ π₯γβ1/2 β«1βγπ₯. ππ₯γ =γπ₯^2/2 logγβ‘π₯β1/2 . π₯^2/2 +πΆ =γπ₯^2/2 πππγβ‘γ π₯γβ π₯^2/4 +πΆ Putting value of I1 in (1), β«1βγπ₯(logβ‘π₯ )^2.ππ₯γ=π₯^2/2 (logβ‘π₯ )^2ββ«1βγ π .πππβ‘π π
πγ =π₯^2/2 (logβ‘π₯ )^2β((π₯^2 (logβ‘π₯ ))/2 β π₯^2/4 +πΆ1) =π₯^2/2 (logβ‘π₯ )^2β (π₯^2 (logβ‘π₯ ))/2 + π₯^2/4 βπΆ1 =π^π/π (πππβ‘π )^πβ (π^π (πππβ‘π ))/π + π^π/π+πͺ " " | 1,685 | 1,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | longest | en | 0.517065 |
https://www.nonviolenta.org/united-states-cjua/1a7dee-area-of-triangles-worksheet-pdf | 1,618,717,755,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464146.56/warc/CC-MAIN-20210418013444-20210418043444-00473.warc.gz | 1,021,646,550 | 18,497 | area of triangles worksheet pdf
8.4 Area of Triangles 435 15. 9) A=6 sq. Free trial available at KutaSoftware.com Area of a Triangle Worksheets, Geometry Worksheets for 4th grade, 5th grade and middle school Geometry worksheets: Area of right triangles. This section contains worksheets on area at a 5th and 6th grade level. Geometry - Area and Perimeter Math Worksheets|Printables PDF for kids. With this worksheet generator, you can make free worksheets for calculating the area of triangles, parallelograms, other quadrilaterals, and polygons (pentagons/hexagons) in the coordinate grid. Geometry: Triangle perimeter area grade 7 worksheet pdf. Find the Area and Perimeter of Triangles Worksheet ... find the area and perimeter of each triangle. Area of Triangles 1 1) A=12 squares 2) A=17.5 squares 3) A=14.5 squares 4) A=21 squares 5) A=9 squares 6) A=10 squares 7) A=10.5 sq. Find the missing measure. 7 ft 3 ft 6 in 6 in 7.1 in 7.1 in m 4 cm 5 cm 5 ft 4.5 ft 3. area worksheets with answers pdf.perimeter word problems worksheet.perimeter of polygons worksheet.area of irregular shapes worksheet pdf.area and perimeter worksheets pdf kuta.area and perimeter of rectangles.area and perimeter word problems.irregular polygon perimeter worksheet.area missing side worksheet.area and perimeter problems with solutions.area of triangles worksheet pdf.area … View Key - Area of Triangles Worksheet (Day 10).pdf from MATH 315 at University of Richmond. 4. These worksheets are pdf files. The area of a triangle is half that of a quadrilateral. Use the buttons below to print, open, or download the PDF version of the Calculating the Perimeter and Area of Triangles (A) math worksheet. FREE (12) Popular paid resources. Find the area of triangles, worksheet #1. It has an answer key attached on the second page. The Corbettmaths Practice Questions on finding the Area of a Triangle Areas of Triangles Worksheet 1 – Here is a nine problem worksheet that will allow your students to practice calculating the area of a triangle. Area of Rectangles and Triangles - I : Worksheet for Fifth Grade Math Practice calculating the area of a square, triangle , or rectangle using formulas in this worksheet. This pdf file includes 4 worksheets about the area of triangles. Other contents: Area of the triangle Add to my workbooks (4) Download file pdf Embed in my website or blog Add to Google Classroom By the time your class has completed this extensive series, they will undoubtedly be experts at finding the area of a triangle. This is a math PDF printable activity sheet with several exercises. Take a square sheet of paper, ask the students how to measure its area (b x h), then fold it in half to show a triangle. This page has printable geometry PDFs on angle types. Similar: Area of right triangles Word Doc PDF. Prepared by teachers of the best CBSE schools in India. The base and height of the triangle is shown in all questions. Find the area of triangles, worksheet #2. Helping Students learn Pre-Algebra through lessons, worksheets, step by step examples, and interactive practice. Find the area of the triangle. in. Geometry worksheets: Area of triangles. cm. The size of the PDF … In the example, the base is 9m and the perpendicular height is 4m. Sale. A 5 63 cm 2 18. Draw three different triangles that each have an area of 24 square units. That’s why we chose to create area of triangle worksheets and printables. The above Practice worksheets for Class 7 Triangles and Its Properties have been designed as per latest NCERT CBSE and KVS guidelines and 2021 syllabus. Area of a triangle = 45 km 9 km 6 km m 12 m 14 cm 9 cm 16 in 15 yd 10 yd 1 5 m 12 m 1 11 mm 10 k m 9 km . Alea — Perimeter = Alea — Perimeter = z 5. Word Doc PDF. Most worksheets require students to identify or analyze acute, obtuse, and right angles. Using Algebra In Exercises 16–18, A gives the area of the triangle. (Double the height)-2-Create your own worksheets like this one with Infinite Geometry. Download PDF. Series of question on area of triangle, when height and length are provided. Units are left out due to formatting and the nature of the worksheet. These worksheets are pdf files. Perimeter of Triangles Worksheet B. Perimeter of Triangles Worksheet B Answers. 16. Problem 4 : area of 100 cm2. Problem 3 : Find the area of the scalene triangle whose side lengths are 12 cm, 18 cm and 20 cm. Finding the Area of a Trapezoid - Integers | Worksheet #2. Notch up your skills on finding the area of a trapezoid with this pdf worksheet. The focus is on calculating the area of triangles and different quadrilaterals. These sheets help students recognize the formula for the area of a rectangle, triangle, octagon, and more. The problems give the coordinates of the vertices of the shapes and ask to calculate the area. Word Doc PDF. Using these sheets will help your child to: know how to calculate the area of a triangle; know how to calculate the area of a … right triangle area worksheets, area of triangle worksheet.pdf and area of triangle worksheet.pdf are some main things we will show you based on the post title. Find the area of parallelograms. ©This area and/or perimeter worksheet is from www.teach-nology.com Find the Area and Perimeter of Triangles Using the given measurements, find the area and perimeter of each triangle. Therefore students will be asked to find the perimeter and area. One of the four activities provides practice dealing with triangle dimensions that contain decimals. Contact Us. How Long? emmavgriffin Year 1 3D shapes: 10 Worksheets : 8 Minutes Standards Met: Area and Perimeter in Application with Triangles Instructions for Printing the Worksheet or Answer Key. About Us. The more advanced worksheets include straight and reflex angles too. Download PDF. Area of Triangles Worksheet (Day 10) 1) 17.75 units2 2) … In addition to that, they might try to figure out the answers mentally, without having to write down intermediary steps. The height of the triangle is shown in all cases, so trigonometry is not requited to answer the questions. Student s can use math worksheets to master a math skill through practice, in a study group or for peer tutoring. I used this with a top set year 4 class. Bellow, you find the triangle perimeter area grade 7 worksheet pdf for free. Each exercise provides a drawing of the triangle as well as the length of its base and height.. A 5 80 m 2 19. Area of a Triangle Worksheet 2. ... Find the area of triangles, parallelograms, trapezoids, and circles, long workshet. These worksheets are prepared for class practice / additional practice assignments / practice from home. Find the area of triangles, parallelograms, trapezoids, and circles. So the area of right triangle is ½ x 8 x 5 = ½ x 40 = 20 square cm or 20 cm 2; Example 2) Find the area of the right triangle below. Find the area of the equilateral triangle having the side length 10 cm. Download PDF. Print or download free pdf printable worksheet and teach students about Area of Triangle. So the area is ½ x 9 x 4 = ½ x 36 = 18 square cm or 18 cm 2; Example 3) Find the area of the triangle … Below are six versions of our grade 6 math worksheet on determining the area of right triangles. CBSE Class 9 Mathematics Worksheet - Triangles - Practice worksheets for CBSE students. When we talk related with Triangle- area Worksheet, scroll the page to see several related pictures to add more info. 717 896 0006. shane@mathvine.com. The questions given in the worksheets are framed in a manner which will help to revise the entire syllabus, concepts and also develop analytical and problem solving skills in students. Finding the Height A triangle has an area of 78 square inches and Alea — Perimeter = Alea — Do Worksheet 41.1. After students have learned how to find the area of different shapes then they can practice these mixed problems of finding areas. Triangle perimeter area worksheet for 7th grade children. Area of Compound Shapes (including triangles) (Chris Duncan) Area Assessment (Ginny Dorrington) PDF; Area: Compound Shapes (Craig Stevens) PDF; Area of shapes made from rectangles (5 sheets) (Michele Zylstra) PDF; Area of Triangles (Tracey West) PDF; Compound Shapes - Area and Perimeter (Alistair Johnson) Area of Compound Shapes (Helen Newton) DOC Many answers 22) Change one number in the diagram you drew for the last question so that the area is now 200 cm2. Each question lists the given information with a, b, c, A, B, or C. All answers are rounded to the tenth place. O Worksheet by Kuta Software LLC Advanced Geometry ID: 1 Name_____ Date_____ Period____ ©f w2q0s1 83F YKnu5tOan ASFoFfpt3w8a5rQe w EL ILMCv.i 6 LAxl Xlv Rroi8gnhwtMs6 orKePsNeNrPvje hd m.W Area of triangles using Trigonometry Find the area of each triangle to the nearest tenth. FREE (40) michaelgrange Dictionary worksheet. ... Area of triangle worksheet. Given the area of a square, find the side length. This worksheet focuses on finding the area of oblique triangles using SAS (A=1/2absinC) and SSS (Heron's Formula). Learn to find area of different shapes like circle, square, rectange, parallelogram and triangle with math area worksheets for kids. ft. 8) A=18 sq. Our area of triangle worksheets and printables use easy-to-understand language to encourage independent learning. Similar: Area of triangles Area of triangles, parallelograms, trapezoids Common Core State Standard 6.G.1 This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 … Do #1 together and let the students work independently on the rest. 0. The worksheets can be made in PDF or html formats. This worksheet is a supplementary seventh grade resource to help teachers, parents and … Problem 2 : The altitude drawn to the base of an isosceles triangles is 8 cm and the perimeter is 32 cm. These perimeter area of triangle worksheets all come with a corresponding printable answer page. 5. by Christopher Rudolph. First, add up the bases and divide the sum by two and multiply the answer by the height to compute the area of the given trapezoid. Worksheet by Kuta Software LLC Geometry HW 70 Area: Rectangles, Triangles, Parallelograms, Trapezoids Name_____ ©Z h2[0D1J6o fKvuvtcaK `SioEfctWwqaDrGeS gLZL_Ct.w k aAulQlc arDiwg[hwt\sJ erXeEswemrkvoeydL.-1-Find the area of each. Below are six versions of our grade 6 math worksheet on area of triangles, only some of which will be right triangles. Explain why the area formula for a triangle is ½ times base times height. ID: 980479 Language: English School subject: Math Grade/level: 5 Age: 12-14 Main content: Find out the Area. Teachers, parents, and students can print these out and make copies. PDF printable worksheet for children of different grades to practice math topic - Area of Triangle. A 5 22 ft 2 17. Area of a Triangle Worksheet 3. Area of triangle Math Worksheet for kids with answer key. Students must free download and practice these worksheets to gain more marks in exams.CBSE Class 9 Mathematics Worksheet - Triangles This worksheet is a PDF document. 1) 8 cm 11 cm C A B 113° 2) 9.9 ft | 2,666 | 11,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | longest | en | 0.797101 |
http://tex.stackexchange.com/questions/tagged/graphs?sort=faq&pagesize=15 | 1,469,521,855,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.62/warc/CC-MAIN-20160723071024-00237-ip-10-185-27-174.ec2.internal.warc.gz | 236,138,348 | 26,017 | # Tagged Questions
Graphs are (1) graphs of functions or (2) sets of nodes and edges.
6k views
### What is the clearest way to graph a piecewise function?
I have a function defined as: How can I graph that using functions instead of coordinates? I'm currently working with pgfplots but I'm open to something else if it's better.
18k views
### Typesetting a directed, weighted graph with TikZ
I just started doing things with TikZ today and I run into a problem: there is just no example code snippets for typesetting directed, weighted graphs. Can anyone supply one simple example in an ...
13k views
### How to draw Bar & Pie Chart
I'm want to include two graphs mentioned below in a document. I'm able to draw pie charts using pgf-pie. Is there any package which is capable of generating BAR & PIE Chart both? [Also which ...
2k views
### How to plot a graph from its adjacency matrix and coordinates of vertices?
A similar question is here. If we give all the data of the vertices's coordinates and the graph's adjacency matrix, how to plot it with tikz, pstrick or other tool in tex? here are the data of ...
6k views
### Bipartite graphs
I want to draw something similar to this in latex. How can I do it? I want it to be a directed graph and want to be able to label the vertices.
4k views
### TikZ - Two Colored Circle Split
I need to draw a two-colored circle split, exactly as is specified in the section 39.6 of the manual for rectangle splits, meaning that one half of the circle split has one fill color and the other ...
8k views
### How do I develop a complex time line?
I currently have a resume I've made in Illustrator with a time line of my work with academics on the lower part and jobs on the upper section: The trouble with it being in Illustrator of course, is ...
2k views
### Package for drawing RDF graphs
Is there some package to draw nice RDF graphs, such as the one below (taken from there)? Can such a graph (with the same style) be produced in TikZ?
4k views
### Drawing a hypergraph
What is the proper way to draw a hypergraph? Is there a package to do this, or is there an easy way to draw such diagrams with a common drawing method like tikz? I'm trying to draw something like
4k views
### Generating dozens of graphs
I am revising old lecture notes on microeconomics and they contain a lot of graphs. As I am rewriting it in LaTeX, I would like to recreate the old graphs, but I need to select the right tool for the ...
3k views
### How to create a gaussian curve with pgfplot?
What I want: This is what I have so far in terms of code: \newcommand\gauss[2]{1/(#2*sqrt(2*pi))*exp(-((x-#1)^2)/(2*#2^2))} % Gauss function, parameters mu and sigma \begin{tikzpicture} \begin{...
23k views
### Is it possible to put MS Excel graphs into a LaTeX document?
I will be using Microsoft Excel to produce graphs and I was wondering if there was a simple way to put those graphs into a LaTeX document?
2k views
### Why does PGF/TikZ 3.0 draw my simple layered graph as non-planar by default?
I have two questions: If I’m reading §30.4 “Crossing Minimization” of the PGF/TikZ 3.0 manual correctly, the layered graph drawing algorithm should have crossing minimization enabled by default. So ...
1k views
### By which pkg to visualize a hypercube graph?
I am trying to visualise hypercube like the below. I have asked help here but getting nodes on top of each other with the packages there. The N-th hypercube has 2^n nodes and each vertex with the ...
15k views
### Basic bar chart with text as x axis labels
I am trying to create a very basic bar chart using pgfplots. I need four bars representing percentages with text as labels under the x axis. Text labels contain spaces and accented, non-English ...
872 views
### Numbering nodes in a for loop
I've just started using TikZ to make basic graphs, and I'm trying to reproduce the Petersen graph. I've managed to make it without using for loops, but this obviously isn't a great method for bigger ...
6k views
### Graph into graph with arrows
I'm using tikzpicture and I have some graphs, but I want some graphs with arrows. Is there any simple syntax change for that? My Graph: \begin{tikzpicture}[thick] \begin{scope} \...
40k views
### Drawing graphs in LaTeX
Is it possible to draw graphs like in image below in LaTeX? It doesn't have to look exactly the same. I need arrows with numbers and circles (or dots or other symbol) with text in it (or next to it).
2k views
### How can I produce the history graph of a Git repository in LaTeX?
Is there a relatively simple way to produce the history graph of a git repository in LaTeX? I'm looking for something nicer-looking than the verbatim output of git log --graph --oneline
3k views
### How can I display an array (as in the data structure from Computer Science, not the mathematical version) in LaTeX?
I'm finding this extraordinarily difficult to search for, as the mathematical meaning of "array" is much more common in LaTeX documentation. I essentially want a single row of boxes, into which I can ...
2k views
### Which LaTeX package is needed to draw radar-like diagrams?
Can you please suggest some LaTeX package of drawing charts in the format mentioned below? It should offer the possibility to put text as "Text A" as well as "Text B" . Note: Currently I'm using ...
6k views
### How do I change the aspect ratio of gnuplot output?
How do I change the aspect ratio of the canvas of a gnuplot plot in LaTeX using the gnuplottex package? The following changes the aspect ratio of the output but not the canvas size, thus leaving a ...
1k views
### TikZ - Four Colored Circle Split
Following @Alain Matthes's reply on this question, wa can get a picture like this: I am not good enough to adapt the code to split the circle into 4 quarters. Can you help me? \documentclass[11pt]{...
709 views
### How to plot a graph from its adjacency matrix (coordinates manually specified)? [duplicate]
For example, I want to plot a graph whose adjacency matrix is [0,1,0;1,0,1;0,1,0]. I first plot three points labeled by p1,p2,p3, and link the points p1,p2, p2,p3. For a general case, how to link the ...
4k views
### Bar plot labels and grids
\documentclass[a4paper,11pt]{report} \usepackage{tikz} \usepackage{verbatim} \usepackage{pgfplots} \begin{document} \begin{tikzpicture} \begin{axis}[ title = {Optimalization based upon co-occurences}, ...
153 views
### Use of forest/tree&box, how can I get result as attached picture,use forest drawing table
1.I want to change the forest height dynamic,like this for tree={ grow=0,reversed, % tree direction parent anchor=east,child anchor=west, % edge anchors edge path={none}, minimum width=...
1k views
### Diagramming workflow in LaTeX
I'm trying to put together some diagrams in LaTeX. My dilemma is as follows. I have a technical report that contains a lot of images from MATLAB. I need to be able to freehand / draw other plots (...
1k views
### Drawing a tree of grids
I need to draw a game tree that is similar to the image below. Is there any way to do this using the Tikz package so that the grids can simply be added as nodes?
1k views
### Simplest way to create a grid with colored squares and labels
I am trying to create a grid with colored squares, multi-line labels within the squares, and labels on the axes. Is there a simple and easy way to achieve this? This question is the closest I could ...
355 views
### Graphs in commutative diagrams in TikZ
I want to draw a diamond shape commutative diagram using tikz. Instead of just letters I want to have simple graphs (just edges and vertices) in the diagram. I tried some stuff but the diagram still ...
439 views
### Ellipsis positioning when drawing large graphs in Tikz
I tried to use Tikz to draw large graphs with specific toologies and I am running into some problems. Since not all nodes can be represented explicitly, I am forced to use an ellipsis. I tried to ...
153 views
### How to draw the graph of function x+y=100
Following is the code for drawing the graph of x+y=4. Now i am thinking to draw the graph of x+y=100 for that what are the necessary changes in the given codes. \documentclass[12pt,a4paper]{article} \...
182 views
### Graphing y=x^2 and y=x
I want to draw the region bounded by y=x^2 and y=x. Any ideas? I was hoping this would be somewhat simple. I've tried used a similar, with \begin{axis}..\end{axis}, but it doesn't like that. Did I not ...
769 views
### Drawing Tapered Edges in Graph
I'd like to achieve the following effect in a TikZ picture when drawing a straight edge from one node to another. How can I achieve this tapering effect? Notice what happens when an edge exists from ...
2k views
### Family tree with multiple marriage ties
I'm trying to build a family tree, and among the many examples I've found online this one has been the most useful: How can I improve this family tree in TikZ? However, there are two things I cannot ...
485 views
### Flip x and y axis of ternary graph
I am trying to reproduce the USDA Soil Triangle. To do so, I would need to flip the x and y axis of the default pgfplots-ternary. I understand, that I could switch my inputs (don't really want to do ...
702 views
### Use “layered layout” with user-specified node ordering (i.e., without crossing minimization)
If and when someone answers this question in a way that demonstrates lua programming for TikZ graph algorithms, I will guarantee a 500-point bounty to that person. According to the PGF 3.0 manual, ...
4k views
### Chart with long x axis text labels looks ugly
Thanks to Caramdir and Alan Munn I am now able to create basic bar charts that use text as x axis labels. Functionally they are perfect, but some of them look kind of ugly. See for yourself: \...
750 views
### Typesetting a directed hypergraph in tikz
Similar to several other questions, I am trying to draw hypergraphs in tikz. I want to draw a directed hypergraph that looks something like those in this image from here: As opposed to other ...
4k views
### 2D and 3D vectors in Tikz
I would like to draw some 2D and 3D vectors in Tikz with certain properties. The following primitive graph sums what I desire: I just need the first quadrant of the Cartesian coordinate system for ...
2k views
### How to add a single axis label to multiple graphs
Please I need someone to help me address this problem. I have three graphs that I created using excel and I have been able to save them as eps files. I've been able to use the \raisebox command to ...
437 views
### drawing two curves in the same figure
Please i need someone who can help me to draw two curves into one figure. I need to draw the two curves Y1 and Y2 in the same figure. By the way, i tried to write the code below: \documentclass{...
894 views
### How to plot integral as summation, as pictured?
I'm new to TikZ and pgfplots but would like to know how to plot the square as well as how to plot the arrows + delta x as well as arrow + delta A: I'd be most grateful if some of you had some ...
3k views
### Is there a pure LaTeX or TeX package for chart drawing?
We are creating an easy-to-use LaTeX template for our institution. I'd like to have a pure LaTeX way to create simple bar graphs and pie charts. Does anything exist? I can certainly throw the students ...
302 views
### Matrix from graph description
Are there any LaTeX package that draws/plots/generates a matrix from a "graph" description (node relations). Any type of description/language could be useful. The trick is to generate automaticly a ...
1k views
### Petersen graph with new tikz graph library
I'm considering moving from tkz-berge to the new tikz graph library for drawing my graphs (in the sense of graph theory). I have produced the Petersen graph, but is there is a more elegant way of ...
1k views
### How to automatically draw a graph?
Is there a simple package for which I can input an abstract graph and it returns a drawing? A similar question was asked in 2010, it did not receive any answers relevant to my question. I want no ...
157 views
### Is there a TeX-native way to generate a readable graph from an adjacency matrix?
related questions This question goes the opposite way from the conversion I would like: Matrix from graph description This question is almost what I want, but the vertex coordinates are manually ... | 2,958 | 12,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-30 | latest | en | 0.9448 |
https://getrevising.co.uk/forums/topics/how_to_find_the_coeficient_of_x3_in_the_expansion_1_05x21plus2x7 | 1,590,623,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00544.warc.gz | 369,850,131 | 12,136 | How to find the coeficient of x^3 in the expansion (1-0.5x)^2(1+2x)^7?
Would much appreciate any help. I have an exam in 2 days.
Thank you
Posted Mon 14th May, 2012 @ 18:25 by divya.k
x^3 can be made when X^2 * X or when X^3 times any number
so binomially expand the first part to give (1-X + 0.25X^2)
expand the second part to to the X^3 term. this gives (1+ 14X + 84X^2 +280X^3 ...)
multiply the relevant terms together. so 1*280X^3 + X * 84X^2 + 0.25X^2 * 14X
this gives 367.5X^3
so the coefficient is 367.5
**
Answered Mon 14th May, 2012 @ 19:48 by Pui Pui
is there something missing in the question?.. in-between the squared and the (1+2x)^7 ?
Answered Mon 14th May, 2012 @ 19:19 by Lalloo19
Thank you so much Pui Pui, your method helped so much, and you made it really simple.
The answer you got was incorrect, probably a small mistake but I used your method to get the correct answer: 199.5
Really good explanation.
Thank you again for your help, I really appreciate it :D x
Answered Mon 14th May, 2012 @ 20:31 by divya.k
Edited by divya.k on Mon 14th May, 2012 @ 20:38 | 381 | 1,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-24 | latest | en | 0.921063 |
http://hackage.haskell.org/package/modular-arithmetic | 1,618,918,853,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00604.warc.gz | 48,505,348 | 6,893 | # modular-arithmetic: A type for integers modulo some constant.
[ bsd3, library, math ] [ Propose Tags ]
A convenient type for working with integers modulo some constant. It saves you from manually wrapping numeric operations all over the place and prevents a range of simple mistakes. Integer Mod 7 is the type of integers (mod 7) backed by Integer. We also have some cute syntax for these types like ℤ/7 for integers modulo 7.
Versions [faq] 1.0.0.0, 1.0.0.1, 1.0.1.0, 1.0.1.1, 1.1.0.0, 1.2.0.0, 1.2.1.0, 1.2.1.1, 1.2.1.2, 1.2.1.3, 1.2.1.4, 1.2.1.5, 2.0.0.0, 2.0.0.1 (info) CHANGELOG.md base (>4.9 && <5), typelits-witnesses (<0.5) [details] BSD-3-Clause Tikhon Jelvis Tikhon Jelvis Math https://github.com/TikhonJelvis/modular-arithmetic https://github.com/TikhonJelvis/modular-arithmetic/issues head: git clone git://github.com/TikhonJelvis/modular-arithmetic.git by TikhonJelvis at 2020-08-31T02:28:11Z NixOS:2.0.0.1 8138 total (18 in the last 30 days) 2.0 (votes: 1) [estimated by Bayesian average] λ λ λ Docs uploaded by userBuild status unknown
## Modules
[Index] [Quick Jump]
#### Maintainer's Corner
For package maintainers and hackage trustees
[back to package description]
# Modular Arithmetic
This package provides a type for integers modulo some constant, usually written as ℤ/n.
Here is a quick example:
>>> 10 * 11 :: ℤ/7
5
It also works correctly with negative numeric literals:
>>> (-10) * 11 :: ℤ/7
2
Modular division is an inverse of modular multiplication. It is defined when divisor is coprime to modulus:
>>> 7 div 3 :: ℤ/16
13
>>> 3 * 13 :: ℤ/16
7 | 538 | 1,587 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-17 | latest | en | 0.725191 |
https://qifulefuvotyv.holidaysanantonio.com/write-an-exponential-function-that-passes-through-given-points-32804nj.html | 1,618,916,669,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00066.warc.gz | 569,237,794 | 5,957 | # Write an exponential function that passes through given points
For one example, the big bang theory places demands on the amount of past time there must be.
The philosopher of time Craig Callender has joked that time is a big invisible thing that will kill you. Something powerful and angry and determined. His "poor man's R-bomb" is constrained by a particular exhaust velocity, and the question is how to squeeze the maximum kinetic energy into the payload.
You can substitute this value for b in either equation to get a. But the theory places no limit on the maximum amount of past time. They have emergent space but not time. Newton also assumed that, if you are five feet tall in one reference frame, then you are that tall in other frames.
A million tasks that had been running inside the system froze in the confusion as directives coming down from the highest operating levels of the nucleus redefined the whole structure of priority assignments to force an emergency analysis of the new data.
In many cases, a default value of the token will be taken for optional arguments if the token is omitted, but these cases are not always denoted with braces around the angle brackets.
This is a surprising implication of the theory of relativity, and it undermines a key ingredient of the manifest image. Unfortunately, the word "time" probably is not a member of the large class that Wittgenstein is speaking of, and even if it were, most philosophers want to know much more than what "time" means.
Calls to computer operating systems Colors in images As the basis of most approaches to image compression City populations a small number of large cities, a larger number of smaller cities Wealth distribution a small number of people have large amounts of money, large numbers of people have small amounts of money Company size distribution Artificial intelligence in particular, "chat bots" that can chat with humans relies on the limited number of questions and statements that people actually write in chats.
Because the x-value of the first point is zero, we can easily find a. You already need a radio telescope to see the far end of this ship's wrap sheet. One reason why many people believe time exists is that they notice time by noticing a leaf fall. Divide all space into three disjoint regions, called region 3, region 4, and region 5.
Could the second hand just as well go counterclockwise to lower numbers instead of clockwise. It cannot, says Newton. Its agents - not even human equivalent on this primitive hardware - raced through the ship's automation, shutting down, aborting.
Cut your drives and prepare to be boarded.
After transmitting a lengthy statement of intent — by all accounts quite stirring, if rabid carbon chauvinism is to your taste — every ship of the True Life Alliance fleet fired its mass drivers and flushed its missile tubes simultaneously at the lone processing node. Note that it would have been much easier to factor this quadratic, but, like the quadratic equation, we can use the completing the square method for any quadratic.
This presupposition in the article is itself philosophically controversial. Then, looking east, she saw it coming -- at least her eyes began to register it -- but her optic nerves did not last long enough to transmit what the eyes had seen.
A clock's proper time depends on the clock's history, its history of speed and gravitational influence. The undamaged processing node returned to the Methizar Traverse with its freshly acquired escort fleet and missile cloud, which unsubstantiated rumor claims were broken down for raw materials upon arrival.
Remember that the sign of a term comes before it, and pay attention to signs. An Example from the Real World Sincehuman population growth has been exponential, and by plotting a growth curve, scientists are in a better position to predict and plan for the future.
And since multiplication is Commutative, we can do these operations in any order we choose. Why would someone reject a feature of the manifest image in favor of the scientific image.
The manifest image is a collection of commonsense beliefs, and it is an important part of our implicit model of the world.
There is a plan to be put into effect here. How to Find an Exponential Equation With Two Points By Chris Deziel; Updated March 13, If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points.
Then it just turns out that we can factor using the inverse of Distributive Property. For example, if I am walking along the road and you drive by me toward the traffic signal ahead, then we will very nearly agree on the time at which the traffic signal changed color, but if we want to know what event on a planet in the Andromeda Galaxy is simultaneous with the traffic signal's color change, we will correctly choose Andromeda events that differ from each other by several weeks.
With these cautions in mind, here is a brief summary of speculations throughout the centuries about whether time has a beginning or end. Visualize the exponential function that passes through two points, which may be dragged within the x-y plane.
Also see the resulting equation. Built-in Types¶. The following sections describe the standard types that are built into the interpreter.
The principal built-in types are numerics, sequences, mappings, classes, instances and exceptions. We are looking for an exponential function. That kind of equation takes the form `y = ab^x` We need to solve for a and b. We will use the points given to create two equations with two unknowns and.
Write an exponential function for the graph that passes through the given points.
(0, ) and (3, ) 62/87,21 Substitute for y and for a and 3 for x into an. You can see a lot more detail for smaller values of `x` and `y`. Notice that the graph of an exponential function on a semi-log graph is a straight line.
Notice also that the numbers along the x axis are evenly spaced, while along the y-axis, we have powers of 10 evenly spaced.
Finally, here's the graph of `y=5^x` on lin-log (linear vertical axis, logarithmic horizontal axis). How do I find an appropriate exponential model of given data points on a TI?
How do I find an exponential model of the form #y = ae^(kt)# on a TI? How do I create a logistic model of population growth on a TI?
Write an exponential function that passes through given points
Rated 0/5 based on 82 review
Log-Log and Semi-log Graphs | 1,336 | 6,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-17 | latest | en | 0.93783 |
https://careericons.com/quantitative-aptitude/compound-interest/basic-problems-using-formula/823-1/ | 1,720,888,572,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00548.warc.gz | 127,675,846 | 8,149 | # model 1 basic compound interest using formula Practice Questions Answers Test with Solutions & More Shortcuts
#### compound interest PRACTICE TEST [6 - EXERCISES]
Question : 1 [SSC CGL Tier-I 2013]
The time in which Rs.80,000 amounts to Rs.92,610 at 10% p.a. compound interest, interest being compounded semi annually is :
a) 3 years
b) 1\$1/2\$ years
c) 2\$1/2\$ years
d) 2 years
Using Rule 1 and 2,
Time = t half year
and R = 5% per half year
A = P\$(1 + R/100)^T\$
\$92610/80000 = (1 + 5/100)^T\$
\$9261/8000 = (21/20)^T\$
T = 3 half years or 1\$1/2\$ years
\$(21/20)^3 = (21/20)^T\$
Question : 2 [SSC CPO S.I.2006]
If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of Rs.10,000 for 3 years will be
a) Rs.2,000
b) Rs.1,600
c) Rs.1,575.20
d) Rs.1,625.80
Using Rule 3,
If there are distinct 'rates of interest' for distinct time periods i.e.,
Rate for 1st year → \$r_1\$%
Rate for 2nd year → \$r_2\$%
Rate for 3rd year → \$r_3\$% and so on
Then A = P\$(1 + r_1/100)(1 + r_2/100)(1 + r_3/100)\$...
C.I. = A - P
Amount = P\$(1 + R_1/100)(1 + R_2/100)(1 + R_3/100)\$
= 10000\$(1+ 4/100)(1 + 5/100)(1 + 6/100)\$
= \$10000 × 26/25 × 21/20 × 53/50\$
A = Rs.11575.2
C.I. = Rs.(11575.2–10000) = Rs.1575.2
Question : 3 [SSC CGL Tier-I 2010]
In how many years will a sum of Rs.800 at 10% per annum compound interest, compounded semi-annually becomes Rs.926.10 ?
a) 2\$1/2\$ years
b) 1\$1/2\$ years
c) 2\$1/3\$ years
d) 1\$2/3\$ years
Using Rule 1 and 2,
Rate = 10% per annum = 5% half yearly
A = P\$(1 + R/100)^T\$
926.10 = 800\$(1 + 5/100)^T\$
\$9261/8000 = (21/20)^T\$
\$(21/20)^3 = (21/20)^T\$
Time = 3 half years = 1\$1/2\$ years
Question : 4 [SSC CGL Tier-I 2010]
At what rate per cent per annum will a sum of Rs.1,000 amounts to Rs.1,102.50 in 2 years at compound interest ?
a) 6.5%
b) 5%
c) 6%
d) 5.5%
Using Rule 1,
A = P\$(1 + R/100)^T\$
Let rate be 'r'
\${1102.50}/1000 = (1 + r/100)^2\$
\$11025/10000 = (1 + r/100)^2\$
\$(105/100)^2 = (1 + r/100)^2\$
1 + \$r/100 = 105/100\$
\$r/100 = 5/100\$ = 5%
Question : 5 [SSC CGL Prelim 2003]
In what time will Rs.1000 amounts to Rs.1331 at 20% per annum, compounded half yearly ?
a) 2\$1/2\$ years
b) 1\$1/2\$ years
c) 1 year
d) 2 years
Using Rule 1 and 2,
Let the required time be t years. Interest is compounded half yearly.
Time = 2t half years and rate
= \$20/2\$ = 10%
1000\$(1 + 10/100)^{2t}\$ = 1331
\$(11/10)^{2t} = 1331/1000\$
\$(11/10)^{2t} = (11/10)^3\$ ⇒ 2t = 3
t = \$3/2\$ years or 1\$1/2\$ years
## Recently Added Subject & Categories For All Competitive Exams
#### 100+ One Word Substitution Practice Test, PDF for SSC
Learn more One Word Substitution questions and answers, PDF for SSC. The most important General English MCQ Practice exercise for all competitive exams
11-Jul-2024 by Careericons
#### Top Para Jumbled Sentences Questions PDF for SSC Exams
New Para-Jumbles or Jumbled-Sentences Questions and Answers with PDF for SSC. General English Rearrangement of Sentences model exercise for all Bank Exams
11-Jul-2024 by Careericons
#### Top Permutation and Combination Practice Questions PDF
Top Permutation and Combination Questions and Answers with Solutions, PDF for SSC. Best Aptitude MCQ Practice Exercise for all competitive, Govt exams
10-Jul-2024 by Careericons
#### Top 100+ Probability MCQ Questions and Answers PDF: SSC
Recent Probability Multiple Choice Questions and Answers with Detailed Solution, download PDF for SSC. Aptitude practice exercise for all Government exams
10-Jul-2024 by Careericons | 1,325 | 3,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.773229 |
http://www.numbersaplenty.com/43867 | 1,596,601,958,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735909.19/warc/CC-MAIN-20200805035535-20200805065535-00287.warc.gz | 149,251,737 | 3,205 | Search a number
43867 is a prime number
BaseRepresentation
bin1010101101011011
32020011201
422231123
52400432
6535031
7241615
oct125533
966151
1043867
112aa5a
1221477
1316c75
1411db5
15cee7
hexab5b
43867 has 2 divisors, whose sum is σ = 43868. Its totient is φ = 43866.
The previous prime is 43853. The next prime is 43889. The reversal of 43867 is 76834.
It can be divided in two parts, 438 and 67, that added together give a palindrome (505).
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-43867 is a prime.
It is not a weakly prime, because it can be changed into another prime (43067) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21933 + 21934.
It is an arithmetic number, because the mean of its divisors is an integer number (21934).
243867 is an apocalyptic number.
43867 is a deficient number, since it is larger than the sum of its proper divisors (1).
43867 is an equidigital number, since it uses as much as digits as its factorization.
43867 is an evil number, because the sum of its binary digits is even.
The product of its digits is 4032, while the sum is 28.
The square root of 43867 is about 209.4445033893. Note that the first 3 decimals coincide. The cubic root of 43867 is about 35.2678765186.
The spelling of 43867 in words is "forty-three thousand, eight hundred sixty-seven". | 420 | 1,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-34 | latest | en | 0.901927 |
http://www.popflock.com/learn?s=Spherical_geometry | 1,611,535,614,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00621.warc.gz | 168,585,626 | 19,254 | Spherical Geometry
Get Spherical Geometry essential facts below. View Videos or join the Spherical Geometry discussion. Add Spherical Geometry to your PopFlock.com topic list for future reference or share this resource on social media.
Spherical Geometry
The sum of the angles of a spherical triangle is not equal to 180°. A sphere is a curved surface, but locally the laws of the flat (planar) Euclidean geometry are good approximations. In a small triangle on the face of the earth, the sum of the angles is only slightly more than 180 degrees.
A sphere with a spherical triangle on it.
Spherical geometry is the geometry of the two-dimensional surface of a sphere. In this context the word "sphere" refers only to the 2-dimensional surface and other terms like "ball" or "solid sphere" are used for the surface together with its 3-dimensional interior.
Long studied for its practical applications to navigation and astronomy, spherical geometry bears many similarities and relationships to, and important differences from, Euclidean plane geometry. The sphere has for the most part been studied as a part of 3-dimensional Euclidean geometry (often called solid geometry), the surface thought of as placed inside an ambient 3-d space. It can also be analyzed by "intrinsic" methods that only involve the surface itself, and do not refer to, or even assume the existence of, any surrounding space outside or inside the sphere.
Because a sphere and a plane differ geometrically, (intrinsic) spherical geometry has some features of a non-Euclidean geometry and is sometimes described as being one. However, spherical geometry was not considered a full-fledged non-Euclidean geometry sufficient to resolve the ancient problem of whether the parallel postulate is a logical consequence of the rest of Euclid's axioms of plane geometry. The solution was found instead in hyperbolic geometry.
## Overview
In plane (Euclidean) geometry, the basic concepts are points and (straight) lines. In spherical geometry, the basic concepts are point and great circle.
In the extrinsic 3-dimensional picture, a great circle is the intersection of the sphere with any plane through the center. In the intrinsic approach, a great circle is a geodesic; a shortest path between any two of its points provided they are close enough. Or, in the (also intrinsic) axiomatic approach analogous to Euclid's axioms of plane geometry, "great circle" is simply an undefined term, together with postulates stipulating the basic relationships between great circles and the also-undefined "points". This is the same as Euclid's method of treating point and line as undefined primitive notions and axiomatizing their relationships.
Great circles in many ways play the same logical role in spherical geometry as lines in Euclidean geometry, e.g., as the sides of (spherical) triangles. This is more than an analogy; spherical and plane geometry and others can all be unified under the umbrella of geometry built from distance measurement, where "lines" are defined to mean shortest paths (geodesics). Many statements about the geometry of points and such "lines" are equally true in all those geometries provided lines are defined that way, and the theory can be readily extended to higher dimensions. Nevertheless, because its applications and pedagogy are tied to solid geometry, and because the generalization loses some important properties of lines in the plane, spherical geometry ordinarily does not use the term "line" at all to refer to anything on the sphere itself. If developed as a part of solid geometry, use is made of points, straight lines and planes (in the Euclidean sense) in the surrounding space.
In spherical geometry, angles are defined between great circles, resulting in a spherical trigonometry that differs from ordinary trigonometry in many respects; for example, the sum of the interior angles of a spherical triangle exceeds 180 degrees.
## Relation to similar geometries
Spherical geometry is closely related to elliptic geometry.
An important geometry related to that of the sphere is that of the real projective plane; it is obtained by identifying antipodal points (pairs of opposite points) on the sphere. Locally, the projective plane has all the properties of spherical geometry, but it has different global properties. In particular, it is non-orientable, or one-sided, and unlike the sphere it cannot be drawn as a surface in 3-dimensional space without intersecting itself.
Concepts of spherical geometry may also be applied to the oblong sphere, though minor modifications must be implemented on certain formulas.
Higher-dimensional spherical geometries exist; see elliptic geometry.
## History
### Greek antiquity
The earliest mathematical work of antiquity to come down to our time is On the rotating sphere (? ? ?, Peri kinoumenes sphairas) by Autolycus of Pitane, who lived at the end of the fourth century BC.[1]
Spherical trigonometry was studied by early Greek mathematicians such as Theodosius of Bithynia, a Greek astronomer and mathematician who wrote the Sphaerics, a book on the geometry of the sphere,[2] and Menelaus of Alexandria, who wrote a book on spherical trigonometry called Sphaerica and developed Menelaus' theorem.[3][4]
### Islamic world
The Book of Unknown Arcs of a Sphere written by the Islamic mathematician Al-Jayyani is considered to be the first treatise on spherical trigonometry. The book contains formulae for right-handed triangles, the general law of sines, and the solution of a spherical triangle by means of the polar triangle.[5]
The book On Triangles by Regiomontanus, written around 1463, is the first pure trigonometrical work in Europe. However, Gerolamo Cardano noted a century later that much of its material on spherical trigonometry was taken from the twelfth-century work of the Andalusi scholar Jabir ibn Aflah.[6]
### Euler's work
Leonhard Euler published a series of important memoirs on spherical geometry:
• L. Euler, Principes de la trigonométrie sphérique tirés de la méthode des plus grands et des plus petits, Mémoires de l'Académie des Sciences de Berlin 9 (1753), 1755, p. 233-257; Opera Omnia, Series 1, vol. XXVII, p. 277-308.
• L. Euler, Eléments de la trigonométrie sphéroïdique tirés de la méthode des plus grands et des plus petits, Mémoires de l'Académie des Sciences de Berlin 9 (1754), 1755, p. 258-293; Opera Omnia, Series 1, vol. XXVII, p. 309-339.
• L. Euler, De curva rectificabili in superficie sphaerica, Novi Commentarii academiae scientiarum Petropolitanae 15, 1771, pp. 195-216; Opera Omnia, Series 1, Volume 28, pp. 142-160.
• L. Euler, De mensura angulorum solidorum, Acta academiae scientiarum imperialis Petropolitinae 2, 1781, p. 31-54; Opera Omnia, Series 1, vol. XXVI, p. 204-223.
• L. Euler, Problematis cuiusdam Pappi Alexandrini constructio, Acta academiae scientiarum imperialis Petropolitinae 4, 1783, p. 91-96; Opera Omnia, Series 1, vol. XXVI, p. 237-242.
• L. Euler, Geometrica et sphaerica quaedam, Mémoires de l'Académie des Sciences de Saint-Pétersbourg 5, 1815, p. 96-114; Opera Omnia, Series 1, vol. XXVI, p. 344-358.
• L. Euler, Trigonometria sphaerica universa, ex primis principiis breviter et dilucide derivata, Acta academiae scientiarum imperialis Petropolitinae 3, 1782, p. 72-86; Opera Omnia, Series 1, vol. XXVI, p. 224-236.
• L. Euler, Variae speculationes super area triangulorum sphaericorum, Nova Acta academiae scientiarum imperialis Petropolitinae 10, 1797, p. 47-62; Opera Omnia, Series 1, vol. XXIX, p. 253-266.
## Properties
Spherical geometry has the following properties:[7]
• Any two great circles intersect in two diametrically opposite points, called antipodal points.
• Any two points that are not antipodal points determine a unique great circle.
• There is a natural unit of angle measurement (based on a revolution), a natural unit of length (based on the circumference of a great circle) and a natural unit of area (based on the area of the sphere).
• Each great circle is associated with a pair of antipodal points, called its poles which are the common intersections of the set of great circles perpendicular to it. This shows that a great circle is, with respect to distance measurement on the surface of the sphere, a circle: the locus of points all at a specific distance from a center.
• Each point is associated with a unique great circle, called the polar circle of the point, which is the great circle on the plane through the centre of the sphere and perpendicular to the diameter of the sphere through the given point.
As there are two arcs determined by a pair of points, which are not antipodal, on the great circle they determine, three non-collinear points do not determine a unique triangle. However, if we only consider triangles whose sides are minor arcs of great circles, we have the following properties:
• The angle sum of a triangle is greater than 180° and less than 540°.
• The area of a triangle is proportional to the excess of its angle sum over 180°.
• Two triangles with the same angle sum are equal in area.
• There is an upper bound for the area of triangles.
• The composition (product) of two reflections-across-a-great-circle may be considered as a rotation about either of the points of intersection of their axes.
• Two triangles are congruent if and only if they correspond under a finite product of such reflections.
• Two triangles with corresponding angles equal are congruent (i.e., all similar triangles are congruent).
## Relation to Euclid's postulates
If "line" is taken to mean great circle, spherical geometry obeys two of Euclid's postulates: the second postulate ("to produce [extend] a finite straight line continuously in a straight line") and the fourth postulate ("that all right angles are equal to one another"). However, it violates the other three: contrary to the first postulate, there is not a unique shortest route between any two points (antipodal points such as the north and south poles on a spherical globe are counterexamples); contrary to the third postulate, a sphere does not contain circles of arbitrarily great radius; and contrary to the fifth (parallel) postulate, there is no point through which a line can be drawn that never intersects a given line.[8]
A statement that is equivalent to the parallel postulate is that there exists a triangle whose angles add up to 180°. Since spherical geometry violates the parallel postulate, there exists no such triangle on the surface of a sphere. The sum of the angles of a triangle on a sphere is , where f is the fraction of the sphere's surface that is enclosed by the triangle. For any positive value of f, this exceeds 180°.
## Notes
1. ^ Rosenfeld, B.A (1988). A history of non-Euclidean geometry : evolution of the concept of a geometric space. New York: Springer-Verlag. p. 2. ISBN 0-387-96458-4.
2. ^
3. ^
4. ^ "Menelaus of Alexandria Facts, information, pictures". HighBeam Research. Retrieved 2015.
5. ^ School of Mathematical and Computational Sciences University of St Andrews
6. ^ Victor J. Katz-Princeton University Press
7. ^ Merserve, pp. 281-282
8. ^ Gowers, Timothy, Mathematics: A Very Short Introduction, Oxford University Press, 2002: pp. 94 and 98.
## References
This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0. | 2,674 | 11,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-04 | latest | en | 0.938947 |
https://tuxdoc.com/category/Validity/2 | 1,674,887,455,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00827.warc.gz | 571,053,206 | 4,704 | # Validity
Leyes y Equivalencias de Conectivos
Determinism / If And Only If / Validity / Logical Consequence / Metalogic
Matematica Basica UTP
Set (Mathematics) / Proposition / Rational Number / Validity / Engineering
G127401
Proposition / Bit / Mathematical Logic / Validity / Argument
Key to all exercises.doc
Inductive Reasoning / Validity / Deductive Reasoning / Argument / Relativism
Logica Matematica
Validity / Proposition / Argument / Logical Consequence / If And Only If
Que Es La Logica
Proposition / Validity / Logic / Inference / Argument
Tugendhat. Propedéutica Lógico-semántica - Caps. 1-2-3
Logic / Validity / Sentence (Linguistics) / Truth / Inductive Reasoning
Matemática - Noções Elementares de Lógica
Validity / Argument / Logic / Semantics / Logical Consequence
Tipos de razonamientos .pdf
Validity / Argument / Inductive Reasoning / Proposition / Mathematical Proof
10 DIAGRAMAS LOGICOS
Validity / Logic / Mathematics / Physics & Mathematics / Psychology & Cognitive Science
Comanducci, Goznalez, Ahumada, Positivismo Juridico y Neoconstitucionalismo
Theory Of Justification / Judge / Validity / Argument / Legislation
Logica Proposicional
Proposition / Validity / Logic / If And Only If / Argument
Luis Camacho N: Lectura Inteligente (2009 ,II edición)
Validity / Logic / Inductive Reasoning / Argument / Don Quixote
Guía Lógica Clásica
Logic / Validity / Reason / Science / Aristotle
Filosofia 11ºAno Sebenta do Aluno.pdf
Argument / Validity / Logic / Fallacy / Knowledge
Explicacion Cientifica - W. C. Salmon Traducción Juan Duran
Validity / Inductive Reasoning / Argument / Science / Rotation | 433 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.242864 |
fleschutz.droppages.com | 1,713,255,335,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00228.warc.gz | 235,994,595 | 4,862 | (Photo: pixabay.com)
# SHORTY
SHORTY is a complete public key algorithm, it is suitable for both encryption and digital signatures. The ciphertext is hardly larger than the plaintext and the modular multiplication used is much less computationally intensive than modular exponentiation as used in other algorithms.
... By Markus Fleschutz 🕓 April 12, 2000.
## Description
To prepare, divide the message into sub-blocks of fixed size nbits and choose the four values a, b, c and m so that holds:
• c and m must not have a common divisor (relatively prime)
• m must be greater than 2 ^ nBits
• a * b * c mod m must be 1
The values a and b form the private key, c and m the public key. The larger these values are chosen, the stronger the security. Then calculate the number sequence K as follows:
For i = 0, 1, 2, ... to nBits - 1 K[i] = (2 ^ i) * a mod m
For encryption set the ciphertext ("cipher") to 0.
For i = 0, 1, 2... to nBits - 1 If in the message the bit 2 ^ i is set, then: cipher = cipher + K[i]
cipher = cipher * b mod m
For decryption calculate:
Plaintext = cipher * c mod m
## Example
We divide the message into sub-blocks of 6 bits and choose a = 7, b = 2, c = 9 and m = 125.
The number sequence K is calculated from this as follows:
``````1 * 7 mod 1301 = 7
2 * 7 mod 1301 = 14
4 * 7 mod 1301 = 28
8 * 7 mod 1301 = 56
16 * 7 mod 1301 = 112
32 * 7 mod 1301 = 99
``````
Thus, the sequence of numbers K consists of: { 7, 14, 28, 56, 112, 99 }.
If the message is 010110, the encryption proceeds as follows:
010110 equals 14 + 28 + 112 = 154. The ciphertext is then: 154 * 2 mod 125 = 58
The decryption is done quite simply via:
58 * 9 mod 125 = 22, which corresponds to 10110
The recovered plaintext is then: 010110
## Derivation
SHORTY is derived from the knapsack algorithm of Merkle-Hellman [1], which is defined as follows:
1. choose a knapsack sequence with the superincreasing property.
2. choose a multiplier n and a modulus m with the following properties:
m > sum of all numbers in the knapsack sequence, n and m have no common divisors.
1. calculate the normal knapsack sequence from: superKnapsack[i] * n mod m
2. add for each set bit in the message the corresponding element in the normal knapsack sequence to the ciphertext
knapsack sequence to the ciphertext.
1. determine nn so that: n * nn = 1 (mod m)
2. calculate: ciphertext * nn mod m
3. gain the plaintext over the knapsack sequence with the superincreasing property.
For SHORTY, we take the power-of-two knapsack sequence (1, 2, 4, 8, 16, 32, ...) in step (1), which satisfies the superincreasing property, so step (7) can then be omitted.
Step (6) is split into two modular multiplications, since holds: ab * x mod m equals: (a * x mod m) * b mod m
nn is split into the 2 multipliers b and c, n corresponds to a in SHORTY.
## Example key
For 80 bits:
``````private a = 1b538a (24 bit)
private b = abd93 (20 bit)
public mod = dcffdb7769a382d02f69 (80 bit)
``````
For 96 bits:
``````private a = 1a4824 (24 bit)
private b = 219a9c (24 bit)
public mod = c3c694bcb70d5dcfb3d5f239 (96 bit)
``````
For 100 bit:
``````private a = 1fdb48 (24 bit)
private b = 2457c6 (24 bit)
public c = 181e0b1397d13baa66e861d0feae1734b418db2e1aa5efc74a3c1a2fe920dacc26adca8b1dd579bb2bc51ef204b778e8dd04d0f3f02ed8edb254582f8a99ecb26aab5a58c433a3332ca82f5c44f64a5265fb611d1fc0a3b200fdc21f42836ece72fb20060d03b30f9416cbc82f8fe9f4ea0fc15de1e8084d518a6b (984 bit)
public mod = 8ee82a56a73549610d9dbf8db (100 bit)
`````` | 1,153 | 3,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.829784 |
https://math.stackexchange.com/questions/436134/testing-sin-theta-and-cos-theta-without-referring-to-the-trigonometric-fun | 1,716,641,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00189.warc.gz | 332,164,018 | 35,646 | # Testing $\sin\theta$ and $\cos\theta$ without referring to the trigonometric functions
This is very much not my area so apologies if this is an obvious no.
Suppose values have been calculated for $\sin\theta$ and $\cos\theta$. Is it possible to test their correctness, without referring to the trigonometric functions, imaginary numbers, or performing an infinite number of operations?
• How have these values been calculated? (The method in which they were calculated can give insight on how to test for correctness) Jul 4, 2013 at 14:58
• There are "well known" formulas for trigonometric functions of multiplied angles. So, for example, if you are given $\sin \theta$ and $\cos \theta$, you can compute $\cos k \theta$ in a very elementary way. If $\theta$ is a rational multiple of $\pi$, say $\theta = \frac{a}{b} \pi$, then you can use this to compute, say $\cos a \pi$ and $\sin a \pi$. A possible test is to verify if these values come out right ($\pm 1$ and $0$ respectively). Jul 4, 2013 at 14:59
• @anorton: if I understand correctly, the idea is that $\sin \theta$ and $\cos \theta$ are provided by some "black box" that we don't know much about, and the objective is to verify if the answers the "box" produced are correct. Alec: is my understanding correct? Jul 4, 2013 at 15:01
• @Feanor Yep, you are correct.
– Max
Jul 4, 2013 at 15:02
• Do we also know the angle in question? Jul 4, 2013 at 15:03
1. Assign $t\leftarrow\frac{2\theta}\pi$, $s\leftarrow\sin\theta$, $c\leftarrow\cos\theta$
2. Test if $s^2+c^2=1$ (within desired precision).
3. Let $r= \lfloor t \rfloor$. If $r\equiv 0\pmod 4$, verify that $s\ge 0, c\ge 0$. If $r\equiv 1\pmod 4$, verify that $s\ge 0, c\le 0$. If $r\equiv 2\pmod 4$, verify that $s\le 0, c\le 0$. If $r\equiv 3\pmod 4$, verifiy that $s\le 0,c\le 0$.
4. Assign $t\leftarrow 2t$, $s\leftarrow 2sc$, $c\leftarrow 2c^2-1$ and go back to step 3 | 615 | 1,894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-22 | latest | en | 0.818746 |
https://electronics.stackexchange.com/questions/169904/how-is-speed-torque-curve-of-an-induction-motor-plotted-experimentally | 1,623,721,595,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487614006.8/warc/CC-MAIN-20210614232115-20210615022115-00447.warc.gz | 223,502,502 | 38,437 | # How is speed torque curve of an induction motor plotted experimentally?
I'm wondering how is obtained a 3-phase induction motor's speed-torque curve obtained. I couldnt find any procedure.
Are they first applying fixed voltages with fixed freq. to stators. and then gradually increasing the torque or they speed up the motor with a variable torque such as wind fan?
• When I was at university, you connected the motor to a leather belt driving a stationary (fixed) wheel, with the wheel mounted on a spring. All energy "generated" by the motor was dissipated by belt slipping on the wheel, and the torque was read off the spring. Yes, you could do the same driving a fan. – david May 11 '15 at 2:00
• But for diffetent torques different loads needed right? Otherwise we cannot plot anything. How do they adjust the load? – user16307 May 11 '15 at 6:42
• By tightening the belt. (See screw wheels at top of picture posted by @raj) The belt is slipping, and the tighter it gets, the more energy is used by friction – david May 12 '15 at 7:04
• I see, so the readings will show the torque? why there are two readings? – user16307 May 12 '15 at 7:31
Usually in college laboratory they connect brake drum and spring balance meters with motor's pulley to measure the force. (you may refer this image below i attached)
While running test you need to measure the speed using tachometer and spring balance force.
And need to find the radius of brake drum .
Torque= force*distance
Torque= (difference between spring balance )* (radius of brake drum)
In industires they may use test benches they can easily do this
• +1 Neat approach. Could probably use cheap digital luggage scales. – Spehro Pefhany May 11 '15 at 2:16
• Thanks for your comment. Yes. We can use its depends on accuracy, How long u are going to use. @ spehro – Photon001 May 11 '15 at 2:23
• Incidentally, the term for these type of devices in general is a dynamometer. There are several different designs, this is one of the simplest I've seen. – helloworld922 May 11 '15 at 2:37
• But how does the rotor rotate if it is connected to break drums? – user16307 May 11 '15 at 6:38
• The brake is a progressive brake - it slows the motor but doesn't stop it. (The brake will get hot!) You increase its pressure slowly with the handwheels at top. As the motor is pulling the brake one way, the torque is the difference between gauge readings * the radius. – user_1818839 May 11 '15 at 9:02 | 618 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.944636 |
https://www.scribd.com/presentation/205661900/SAPM-Risk-Aversion | 1,548,261,938,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584334618.80/warc/CC-MAIN-20190123151455-20190123173455-00300.warc.gz | 922,835,174 | 34,819 | You are on page 1of 13
# Risk and Risk Aversion
Chapter 6
293 6-2 1-2 .4(80=122)2 = 1.176.Risk .6 (150-122)2 + .4(80) = 122 s2 = p[W1 .E(W)]2 + (1-p) [W2 .4 W2 = 80 Profit = -20 E(W) = pW1 + (1-p)W2 = 6 (150) + .E(W)]2 = .000 s = 34.Uncertain Outcomes W1 = 150 Profit = 50 W = 100 1-p = .
4 Risk Free T-bills W2 = 80 Profit = -20 Profit = 5 Risk Premium = 17 6-3 1-3 .Risky Investments with Risk-Free Risky Inv. W1 = 150 Profit = 50 100 1-p = .
.Risk Aversion & Utility Investor’s view of risk Risk Averse Risk Neutral Risk Seeking Utility Utility Function U = E ( r ) .005 A s 2 A measures the degree of risk aversion 6-4 1-4 .
.22 .005 A (34%) 2 Risk Aversion A Value High 5 -6.005 A s 2 = .90 3 4..66 Low 1 16.Risk Aversion and Value: U = E ( r ) .22 T-bill = 5% 6-5 1-5 .
Dominance Principle Expected Return 4 2 1 Variance or Standard Deviation 3 • 2 dominates 1. has a higher return 6-6 1-6 . has a higher return • 2 dominates 3. has a lower risk • 4 dominates 3.
Utility and Indifference Curves Represent an investor’s willingness to tradeoff return and risk.0 25.9 2 2 6-7 1-7 .005As2 20. Example Exp Ret 10 15 St Deviation U=E ( r ) .5 2 2 20 25 30.0 33..
Indifference Curves Expected Return Increasing Utility Standard Deviation 6-8 1-8 .
Expected Return Rule 1 : The return for an asset is the probability weighted average return in all scenarios. E (r ) = P( s )r ( s ) s 6-9 1-9 .
Variance of Return Rule 2: The variance of an asset’s return is the expected value of the squared deviations from the expected return. P( s)[ r ( s) E (r )] s = s 2 2 6-10 1-10 .
with the portfolio proportions as weights.Return on a Portfolio Rule 3: The rate of return on a portfolio is a weighted average of the rates of return of each asset comprising the portfolio. r p = W 1r 1 + W 2r 2 W1 = Proportion of funds in Security 1 W2 = Proportion of funds in Security 2 r1 = Expected return on Security 1 r2 = Expected return on Security 2 6-11 1-11 .
Portfolio Risk with Risk-Free Asset Rule 4: When a risky asset is combined with a risk-free asset. the portfolio standard deviation equals the risky asset’s standard deviation multiplied by the portfolio proportion invested in the risky asset. s p = wriskyasset s riskyasset 6-12 1-12 .
are combined into a portfolio with portfolio weights w1 and w2. respectively. the portfolio variance is given by: sp2 = w12s12 + w22s22 + 2W1W2 Cov(r1r2) Cov(r1r2) = Covariance of returns for Security 1 and Security 2 6-13 1-13 .Portfolio Risk Rule 5: When two risky assets with variances s12 and s22. respectively. | 857 | 2,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-04 | latest | en | 0.794183 |
https://www.studysmarter.us/textbooks/math/calculus-1st/limits/q-11-write-each-of-the-inequalities-in-interval-notation/ | 1,685,889,808,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649986.95/warc/CC-MAIN-20230604125132-20230604155132-00653.warc.gz | 1,086,164,134 | 22,379 | • :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon
Suggested languages for you:
Americas
Europe
Q 11.
Expert-verified
Found in: Page 107
### Calculus
Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861
# Write each of the inequalities in interval notation:$\left|f\left(x\right)-L\right|<\in$
Answer is $\left(L-\in ,L+\in \right)$
See the step by step solution
## Step 1. Given infromation
An inequality is $\left|f\left(x\right)-L\right|<\in$
## Step 2. Explanation
$\begin{array}{rcl}\left|f\left(x\right)-L\right|& <& \in \\ L-& \in &
Solution in interval notation is $\left(L-\in ,L+\in \right)$ | 231 | 682 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-23 | latest | en | 0.487682 |
https://www.basicsofelectricalengineering.com/2018/01/formula-to-find-current-in-single-phase.html | 1,618,154,665,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00240.warc.gz | 763,257,053 | 27,935 | Basics of Electrical Engineering
Learn the basics of Electrical Engineering.
The basic formula for calculating current in the single-phase electrical motor is:
Let's solve an example for it.
Example: Find the current through the single phase, 2 HP motor at 120. The motor has an efficiency of 90% and has the power factor of 85%.
Solution: I = (2 HP * 746) / (120 * 0.9 * 0.85) = 16.25 A | 103 | 391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-17 | longest | en | 0.903069 |
https://www.asterclasses.com/chapter-3-algebra-ex-3-1/ | 1,628,065,748,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00661.warc.gz | 627,339,419 | 15,871 | # Chapter 3 -Algebra – Ex 3.1
10 Dec 2020 1:57 pm
Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 3, Algebra, Ex 3.1,
Question 1.
Solve the following system of linear equations in three variables
(i) x + y + z = 5
2x – y + z = 9
x – 2y + 3z = 16
x + y + z = 5 ….(1)
2x – y + z = 9 ….(2)
x – 2y + 3z = 16 ….(3)
Substituting z = 4 (4)
3x + 2(-4) = 14
3x – 8 = 14
3x = 14 – 8
3x = 6
x = 63 = 2
Substituting x = 2 and z = 4 in (1)
2 + y + 4 = 5
y + 6 = 5
y = 5 – 6
= -1
∴ The value of x = 2, y = -1 and z = 4
(ii) 1x – 2y + 4 = 0, 1y – 1z + 1 = 0, 2z + 3x = 14
Let 1x = p, 1y = q and 1z = r
p – 2q + 4 = 0
p – 2q = -4 ……(1)
q – r + 1 = 0
q – r = -1 ……(2)
3p + 2r = 14 ……(3)
Substituting the value of p = 2 in (1)
2 – 2q = -4
-2q = – 4 – 2
-2q = -6
q = 62 = 3
Substituting the value of q = 3 in (2)
3 – r = 1
– r = – 1 – 3
r = 4
The value of x = 12, y = 13 and z = 14
(iii) x + 20 = 3y2 + 10 = 2z + 5 = 110 – (y + z)
x + 20 = 3y2 + 10
Multiply by 2
2x + 40 = 3y + 20
2x – 3y = -40 + 20
2x – 3y = -20 ……(1)
3y2 + 10 = 2z + 5
Multiply by 2
3y + 20 = 4z + 10
3y – 4z = 10 – 20
3y – 4z = -10 ……(2)
2z + 5 = 110 – (y + z)
2z + 5 = 110 – y – z
y + 3z = 110 – 5
y + 3z = 105 ….(3)
Substitute the value of z = 25 in (2)
3y – 4(25) = -10
3y – 100 = – 10
3y = – 10 + 100
3y = 90
y = 903 = 30
∴ The value of x = 35, y = 30 and z = 25
Substitute the value of y = 30 in (1)
2x – 3(30) = -20
2x – 90 = -20
2x = -20 + 90
2x = 70
x = 702 = 35
Question 2.
Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6, – 3x – 2y + 5z = -12 , x – 2z = 3
x + 2y – z = 6 …..(1)
-3x – 2y + 5z = -12 …..(2)
x – 2z = 3 ……(3)
Adding (1) and (2) we get
Adding (3) and (4) we get
The above statement tells us that the system has an infinite number of solutions.
(ii) 2y + z = 3(- x + 1) ,-x + 3y – z = -4, 3x + 2y + z = –12
2y + z = 3 (- x + 1)
2y + z = -3x + 3 ……(1)
3x + 2y + z = –12
-x + 3y – z = – 4
x – 3y + z = 4 …..(2)
3x + 2y + z = – 12
Hence we arrive at a contradiction as 0 ≠ 7
This means that the system is inconsistent and has no solution.
(iii) y+z4 = z+x3 = x+y2, x + y + z = 27
y+z4 = z+x3
3y + 3z = 4z + 4x
-4x + 3y + 3z – 4z = 0
-4x + 3y – z = 0
4x – 3y + z = 0 ………(1)
z+x3 = x+y2
2z + 2x = 3x + 3y
-3x + 2x – 3y + 2z = 0
-x – 3y + 2z = 0
x + 3y – 2z = 0 ………(2)
Substituting the value of x in (5)
6 + 5z = 81
5z = 81 – 6
5z = 75
z = 755 = 15
Substituting the value of x = 3
and z = 15 in (3)
3 + y + 15 = 27
y + 18 = 27
y = 27 – 18
= 9
The value of x = 3, y = 9 and z = 15
This system of equations have unique solution.
Question 3.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
x+y+z3 = 53
x + y + z = 159 ….(1)
By the given 2nd Condition.
12 z + 13y + 14x = 65
Multiply by 12
6z + 4y + 3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
– 4x + z = – 16 + 4
4x – z = 12 ……(3)
Vani age = 24 years
Vani’s father age = 51 years
Vani grand father age = 84 years
Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51
Question 4.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?
Let the hundreds digit be x and the tens digit be ”y” and the unit digit be “z”
∴ The number is 100x + 10y + z
If the digits are reversed the new number is 100z + 10y + x
By the given first condition
x + y + z = 11 ….(1)
By the given second condition
100z + 10y + x = 5 (100x + 10y + z) + 46
= 500x + 50y + 5z + 46
x – 500x + 10y – 50y + 100z – 5z = 46
– 499x – 40y + 95z = 46
499x + 40y – 95z = -46 ….(2)
By the given third condition
x + 2y = z
x + 2y – z = 0 ….(3)
∴ The number is 137
Subtituting the value of y = 3 in (5)
2x + 3(3) = 11
2x = 11 – 9
2x = 2
x = 22 = 1
Subtituting the value of x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 – 4
= 7
Question 5.
There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. But when first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Let the number of ₹5 currencies be “x”
Let the number of ₹10 currencies be “y”
and the number of ₹20 currencies be “z”
By the given first condition
x + y + z = 12 ………(1)
By the given second condition
5x + 10y + 20z = 105
x + 2y + 4z = 21 (÷5) ……….(2)
By the given third condition
10x + 5y + 20z = 105 + 20
10x + 5y + 20z = 125
2x + y + 4z = 25 ………..(3)
Substituting the value of x = 7 in (5)
7 – y = 4 ⇒ -y = 4 – 7
-y = -3 ⇒ y = 3
Substituting the value of x = 7, y = 3 in …. (1)
7 + 3 + z = 12
z = 12 – 10 = 2
x = 7, y = 3, z = 2
Number of currencies in ₹ 5 = 7
Number of currencies in ₹ 10 = 3 Number of currencies in ₹ 20 = 2
#### VISITORS COUNT
Users Today : 400
Total Users : 122622
Views Today : 964
Total views : 476952 | 2,529 | 5,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-31 | latest | en | 0.571186 |
https://mathhelpforum.com/threads/multiplying-matrices.183880/ | 1,581,998,460,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143505.60/warc/CC-MAIN-20200218025323-20200218055323-00160.warc.gz | 481,065,975 | 16,578 | # Multiplying matrices
#### Googl
How do you multiply matrices with:
2 rows and 3 columns
#### TKHunny
Same as any other. Why do you ask?
#### Googl
Same as any other. Why do you ask?
What do you mean the same way as any other? In what pattern?
#### Plato
MHF Helper
How do you multiply matrices with:
2 rows and 3 columns
Suppose that we have $$\displaystyle A_{2,3}~\&~B_{2,3}$$ then it impossible to form $$\displaystyle A\cdot B$$.
But $$\displaystyle B^T$$, the transpose is a $$\displaystyle 3\times 2$$ matrix.
Thus $$\displaystyle A\cdot B^T$$ is a $$\displaystyle 2\times 2$$ matrix.
mr fantastic
#### TKHunny
What do you mean the same way as any other? In what pattern?
Well, that would be why I asked, "Why do you ask?" Just trying to make sure we have a good question. If matrices are not conformable, you DON'T multiply them.
Last edited:
#### Googl
Thanks for the reply. I have managed to solve this one. | 266 | 934 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-10 | longest | en | 0.856009 |
https://mathisfunforum.com/viewtopic.php?pid=440901 | 1,726,036,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00755.warc.gz | 345,837,855 | 3,895 | # Math Is Fun Forum
Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
You are not logged in.
## #1 2023-11-28 00:36:02
larry04
Novice
Registered: 2023-11-27
Posts: 1
### Calculadora de Alicia: A Calculator That Makes Math Easy
In today's world, calculators are everywhere. We use them to balance our checkbooks, solve complex math problems, and even just to do basic arithmetic. But have you ever wished there was a calculator that could show you how to solve a problem step by step? Well, now there is! Calculadora de Alicia is a revolutionary new calculator that does just that.
Unlike traditional calculators that just show you the answer, Calculadora de Alicia shows you each step of the way. This makes it easy to understand how the problem was solved and to learn from your mistakes.
I remember when I was a kid, I loved to solve math problems by hand. I found it so satisfying to figure out how to get the answer. Calculadora de Alicia brings back that feeling. It's like having a math teacher right there with you, guiding you through each step of the problem.
I think Calculadora de Alicia could be a game-changer for math education. It could help students of all ages learn math in a more fun and engaging way. I encourage you to check it out and see for yourself!
Offline
## #2 2023-11-28 01:23:28
Bob
Registered: 2010-06-20
Posts: 10,542
### Re: Calculadora de Alicia: A Calculator That Makes Math Easy
hi larry04
Welcome to the forum.
I had a quick look and it just looked like any other calculator. Would you be able to post an example that shows it's superior features?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
## #3 2024-05-27 06:57:58
robertscott56
Novice
Registered: 2024-05-27
Posts: 1
### Re: Calculadora de Alicia: A Calculator That Makes Math Easy
Bob wrote:
hi larry04
Welcome to the forum.
I had a quick look and it just looked like any other calculator. Would you be able to post an example that shows it's superior features?
Bob
Here is the look for the Calcualdora Alicia which has more unique funstions as mentioned above for quick math solutions.
Offline | 615 | 2,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.911951 |
https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/ | 1,624,516,115,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00633.warc.gz | 958,342,531 | 18,268 | # Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample
Type of paper: Report
Pages: 2
Words: 550
Published: 2021/02/17
Find the biggest directory of over
1 million paper examples!
## Graphs, Tables, and Calculations
Line graph showing mean annual air temperature for the period between 1975 and 2013
Line graph showing mean annual water temperature for the period between 1975 and 2013
Scatterplot showing the relationship between mean air temperature and mean water temperature
Summary output table from your linear regression analysis of air temperature through time
Summary output table from your linear regression analysis of water temperature
B. Questions
Question 1
The hypothesis of interest is to test whether there is a relationship between the water temperature and the air temperature. The null hypothesis states that there is no significant relationship between the water temperature and the air temperature. However, the regression analysis rejects the null hypothesis by showing the there is a positive correlation between the air temperature and water temperature.
## Question 2
The air temperature and water temperature are correlated because the scatter plot shows a positive correlation. As the number of the year increases the two variable increases. This was expected because the study aims to prove that there is a relationship between air and water temperatures. The relationship strength is measured by the correlation coefficient or multiple r. The correlation coefficient of air temperature is 0.097768, implying that the relationship is weak. On the other hand, the correlation coefficient on the water temperature is 0.407109, implying that the relationship is relatively stronger.
## Question 3
The line graphs indicate that the temperature trend in Lake Michigan is significantly increasing as the years elapse. According to the trend lines, the air temperature and water temperature increases with the years. According to the slopes of the trend lines, the average temperature increases over time. The slope of the air temperature trend line is 0.0132 while the slope of the air temperature trend line is 0.0596. This implies that the air temperature increases at a higher rate than the air temperature over the years.
## Question 4
The R-squared analysis or mean air temperature was 0.009559 indicates that the model was not acceptable because it is less than 0.3. The R-square implies that the only 0.95 percent of the variables fit the model. On the other hand, the R-squared of mean water temperature was 0.165738 implies that 16.57 percent of the variables fit into the model. The p-value of both mean water and mean air temperature is 0.000 indicating that both variables are statistically significant at 95 percent confidence level hence null hypothesis is rejected. This supports the conclusion in question 3.
## Question 5
The linear regression provides the qualitative analysis of data the helps the researcher to effectively predict the future using the forecasting techniques. For instance, using the regression analysis, one can predict that the air and water temperature are anticipated to change.
## Question 6
The use of the subset of the raw date might have a significant effect on the conclusion regarding the average temperature changes. Although for the period of 32 years all years were included, the omission of some moths of the year could have a significant impact on the results. For instance, in 1975 the moths of January to June were omitted. During these months that temperatures could have a significant effect on the mean temperature. However, according to the p-values of 0.0000, the model is statistically significant indicating that there is a positive relationship between the air temperature and water temperature.
Choose cite format:
• APA
• MLA
• Harvard
• Vancouver
• Chicago
• ASA
• IEEE
• AMA
WePapers. (2021, February, 17) Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample. Retrieved June 24, 2021, from https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/
"Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample." WePapers, 17 Feb. 2021, https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/. Accessed 24 June 2021.
WePapers. 2021. Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample., viewed June 24 2021, <https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/>
WePapers. Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample. [Internet]. February 2021. [Accessed June 24, 2021]. Available from: https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/
"Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample." WePapers, Feb 17, 2021. Accessed June 24, 2021. https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/
WePapers. 2021. "Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample." Free Essay Examples - WePapers.com. Retrieved June 24, 2021. (https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/).
"Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample," Free Essay Examples - WePapers.com, 17-Feb-2021. [Online]. Available: https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/. [Accessed: 24-Jun-2021].
Are The Great Lakes (Lake Michigan) Being Influenced By The Global Greenhouse Effect? Report Sample. Free Essay Examples - WePapers.com. https://www.wepapers.com/samples/are-the-great-lakes-lake-michigan-being-influenced-by-the-global-greenhouse-effect-report-sample/. Published Feb 17, 2021. Accessed June 24, 2021.
Copy
Share with friends using:
Please remember that this paper is open-access and other students can use it too.
If you need an original paper created exclusively for you, hire one of our brilliant writers!
GET UNIQUE PAPER
Some topics are tougher than others.
If these samples aren’t enough, order your unique customized essay from us.
It’s okay to ask for help! | 1,495 | 6,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-25 | latest | en | 0.918972 |
https://www.reference.com/science/heron-s-formula-78dff34de05e0406 | 1,487,902,768,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171271.48/warc/CC-MAIN-20170219104611-00390-ip-10-171-10-108.ec2.internal.warc.gz | 889,885,527 | 20,577 | Q:
# What is Heron's formula?
A:
Heron's formula is used to calculate the area of a triangle. The formula finds the area by taking the square root of s times a, multiplied by the difference between s and b, and multiplied by the distance difference between s and c; it can also be expressed as A = square root of (s-a)(s-b)(s-c).
## Keep Learning
Credit: Liane Cary age fotostock Getty Images
In the formula, "A" is the symbol for area, and "a," "b" and "c" are the symbols for each side of the triangle. "S" is the symbol for the semiperimeter as determined by the formula dividing the sum of a, b and c by two. Integrating the two formulas can result in a version without the variable s. The formula is named after the Greek mathematician, Hero of Alexandria.
Sources:
## Related Questions
• A: The formula is Area = ½ times (Base times Height) or Area = (Base times Height) divided by 2. Finding the area of an isosceles or right triangle is easy as... Full Answer >
Filed Under:
• A: To find the height of a scalene triangle, the formula for the area of a triangle is necessary. The equation is area = 1/2hb, where h is the height and b is... Full Answer >
Filed Under:
• A: The common formula to find the area of an isosceles triangle is area equals 1/2 * base * height of triangle. The height is the line bisecting the two equal... Full Answer >
Filed Under:
• A: The area of a triangle can be calculated from its vertices by using a formula to find the distances between them. This allows for the calculation of the ar... Full Answer >
Filed Under:
PEOPLE SEARCH FOR | 390 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-09 | longest | en | 0.920064 |
https://www.homeworklib.com/question/2094808/a-240103-resistor-and-a-capacitor-are-connected | 1,726,244,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00627.warc.gz | 737,529,796 | 9,759 | Question
# A 2.40×103 Ω resistor and a capacitor are connected in series and then a 2.00 V...
A 2.40×103 Ω resistor and a capacitor are connected in series and then a 2.00 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 1.300 V in 4.70 μs. Calculate the time constant of the circuit. Tries 0/10 Find the capacitance of the capacitor.
#### Earn Coins
Coins can be redeemed for fabulous gifts.
Similar Homework Help Questions
• ### A(n) 14.0 kΩ resistor and a capacitor are connected in series and then a(n) 17.0 V...
A(n) 14.0 kΩ resistor and a capacitor are connected in series and then a(n) 17.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 11.0 V in 1.35 μs. (a) Calculate the time constant of the circuit. ???μs (b) Find the capacitance of the capacitor. ???pF
• ### A 11.4 kΩ resistor and a capacitor are connected in series and then a 12.0 V...
A 11.4 kΩ resistor and a capacitor are connected in series and then a 12.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 4.82 V in 1.48 µs. (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor. A 11.4 k2 resistor and a capacitor are connected in series and then a 12.0 V potential difference is suddenly applied across them. The potential difference across the capacitor...
• ### A 21.1 kΩ resistor and a capacitor are connected in series and then a 12.0 V...
A 21.1 kΩ resistor and a capacitor are connected in series and then a 12.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 4.48 V in 1.32 µs. (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.
• ### a) A 19.1 kΩ resistor and a capacitor are connected in series and then a 12.0...
a) A 19.1 kΩ resistor and a capacitor are connected in series and then a 12.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 3.51 V in 1.49 µs. (i) Calculate the time constant of the circuit. (ii) Find the capacitance of the capacitor. b) The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) 2.2 µF capacitor drops to one-fourth its initial...
• ### A 13.0 kO resistor and a capacitor are connected in series and then a 28.0 V...
A 13.0 kO resistor and a capacitor are connected in series and then a 28.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 4.0 Vin 2.80 us. Calculate the time constant of the circuit. (Your result must be in units of us's. Include 2 digit after the decimal point and maximum of 5% of error is accepted in your answer.)
• ### An uncharged capacitor and a resistor are connected in series to a source of EMF. If...
An uncharged capacitor and a resistor are connected in series to a source of EMF. If ε = 6.48 V, C = 24.6 μF, and R = 125 Ω, calculate the time constant τ of the circuit. 3.08 ms You are correct. Previous Tries Calculate the maximum charge on the capacitor. Incorrect. Tries 1/20 Previous Tries Calculate the charge on the capacitor after one time constant.
• ### The discharge curve for a capacitor connected in series with a resistor is shown below, If...
The discharge curve for a capacitor connected in series with a resistor is shown below, If the capacitance of the capacitor is 22.50 farads, what is the resistance of the resistor? Please calculate the time constant for this circuit. After two time constants what is the potential across capacitor? Please calculate the amount of charge on the capacitor 25.00 minutes after discharge has started, Once capacitor is fully discharged, the charging process begun. How much time, after charging begins, is...
• ### A resistor and capacitor are connected in series to a power supply of 50 Volts. The...
A resistor and capacitor are connected in series to a power supply of 50 Volts. The resistor has a value of 30000 ohms and the Capacitor is initially uncharged. After a time of 16.0 seconds, the capacitor is charged 72% fully charged. a) What is the capacitance of capacitor in this circuit? b) What is the time constant of this system? c) What is the current through the resistor at t= 20.0 seconds? d) What is the potential difference across...
• ### A 502.6 Ω resistor is connected in series with a capacitor. What must be the capacitance...
A 502.6 Ω resistor is connected in series with a capacitor. What must be the capacitance of the capacitor to produce a time constant of 2.26 s ?
• ### A 503.7 Ω resistor is connected in series with a capacitor. What must be the capacitance...
A 503.7 Ω resistor is connected in series with a capacitor. What must be the capacitance of the capacitor to produce a time constant of 2.17 s? | 1,222 | 4,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-38 | latest | en | 0.889148 |
https://www.enotes.com/homework-help/solve-x-y-x-log-base-3-4-x-2log-base-3y-8-320034 | 1,516,654,621,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00598.warc.gz | 891,545,008 | 9,825 | # solve for x,y `x*log_3 y = 4` `x^2*log_3 y = 8`
sciencesolve | Certified Educator
You need to solve the system of simultaneous equations such that:
`x*log_3 y = 4`
`x^2*log_3 y = 8`
You should express `log_3 y` in terms of x such that:
`log_3 y = 4/x`
Substituting `4/x` for `log_3 y` in the equation `x^2*log_3 y = 8` yields:
`x^2*(4/x)= 8`
You need to subtract 8 both sides such that:
`4x^2/x - 8 = 0 =gt (4x^2 - 8x) = 0`
You need to factor out 4x such that:
`4x(x - 2) = 0 =gt 4x = 0 =gt x = 0`
`x - 2 = 0 =gt x = 2`
You need to substitute 0 for x in equation `x*log_3 y = 4` , hence `0*log_3 y = 4 =gt 0 != 4`
Notice that if x=0, the equation `x*log_3 y = 4` does not hold, hence 0 is not a solution to equation.
You need to substitute 2 for x in equation `x*log_3 y = 4` , hence `2*log_3 y = 4 =gt log_3 y = 2 =gt y = 3^2 =gt y = 9` .
Hence, the solution to simultaneous equation is (2;9).
justaguide | Certified Educator
The equations to be solved for x and y are:
`x*log_3 y = 4` ...(1)
`x^2*log_3 y = 8` ...(2)
(2)/(1)
=> `(x^2*log_3 y)/(x*log_3 y)` = `8/4` = 2
=> x = 2
Substitute in (1)
=> `2*log_3 y = 4`
=> `log_3 y = 2`
=> `y = 3^2 = 9`
The solution of the system of equations is x = 2 and y = 9 | 516 | 1,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-05 | latest | en | 0.797423 |
https://socratic.org/questions/a-spring-with-a-constant-of-2-kg-s-2-is-lying-on-the-ground-with-one-end-attache-6 | 1,708,675,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00596.warc.gz | 555,740,974 | 5,967 | # A spring with constant 2 kgs^-2 is lying on the ground with one end attached to a wall. An object with a mass of 1/2 kg and speed of 3/5 ms^-1 collides with and compresses the spring until it stops moving. How much will the spring compress?
The kinetic energy of the moving mass, ${E}_{k} = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times \frac{1}{2} \times {\left(\frac{3}{5}\right)}^{2} = 0.09$ $J$, will be converted into spring potential energy in the spring, ${E}_{p} = \frac{1}{2} k {x}^{2}$. Rearranging, $x = \sqrt{\frac{2 {E}_{p}}{k}} = \sqrt{\frac{2 \times 0.09}{2}} = 0.30$ $m$ = $30$ $c m$. | 222 | 600 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-10 | latest | en | 0.818864 |
https://www.teachersource.com/product/magnetic-accelerator/electricity-magnetism?utm_source=newsletter&utm_medium=newsletter&utm_campaign=forces-and-motion | 1,696,449,776,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00580.warc.gz | 1,101,593,089 | 62,224 | # Magnetic Accelerator
(8 reviews)
• Magnetic Accelerator
Item #: MAG-400
• - +
• Magnetic Accelerator - 6 additional spheres
Item #: MAG-415
• 1 magnetic sphere, 5 non-magnetic.
• - +
Great for open-ended discovery!
Description
Place three seemingly identical spheres on a 3-foot horizontal track touching each other. Slowly roll another sphere slowly into the spheres at rest. Wow! All of a sudden one sphere takes off with tremendous velocity. Has the system gained energy? Could this be the basis of a perpetual motion machine? Hmm... Great for open-ended discovery! Colors may vary. Also available, six additional spheres for additional experimentation.
Read more on our Blog - The Magnetic Accelerator
Video
Lesson Ideas
Place three seemingly identical spheres on a horizontal track touching each other. Roll another sphere slowly into the spheres at rest. Wow! All of a sudden the last sphere takes off with tremendous velocity. The system seems to have gained energy!
Materials:
Track
Two wooden blocks
Five metal spheres (2.0 cm)
One magnetic metal sphere (2.0 cm)
(Safety glasses recommended, but not provided.)
Note: The track can be used in several different orientations: straight; raised 4.0 cm on either or both ends; and raised 6.0 cm on either or both ends.
Procedure A. Student Investigation
1. With the track raised 2.5 cm on each end, release one of the five metal spheres from one end of the track. What did you observe? At which point on the track does the sphere have the most Potential Energy? The most Kinetic?
(Expected observations: How high did it roll up the opposite side? How many changes in direction occurred before it stopped? How long was it in motion?)
2. Place two of the spheres at the bottom and release a third one from the end. What did you observe?
3. Experiment with different numbers of the five spheres at the bottom and releasing different numbers of spheres from the end of the curved track. What do you observe?
Teacher Note: As with Newton's cradle (NEW-100) both the kinetic energy and the momentum of the initial moving spheres are transferred to the final moving spheres. In this case, in order for both the Law of Conservation of Momentum (mv) and the Law of Conservation of Kinetic Energy (1/2 mv2) to hold, the initial moving spheres must equal the final moving spheres 'What comes in is what goes out.' Deviations are due to friction, and the slight magnetism of the spheres, causing them to stick together slightly and not roll well.
4. With two of the five spheres at the bottom of the curved track, release the special sphere given to you by your instructor. What did you observe?
5. With two of the five spheres and the one special one at the bottom of the curved track, release another sphere into the special one. What did you observe?
Teacher Note: As the released sphere gets closer to the stationary magnetic sphere at the bottom of the track, it becomes more and more attracted to the magnetic sphere. This causes a great increase in velocity. As a result, the sphere on the other end shoots out quickly. Notice that the final moving sphere is initially separated by another sphere from magnetic sphere. Consequently, it is attracted less to the magnetic sphere.
6. With only one end of the track raised and the other end near the edge of the table, set up the magnetic accelerator so that the final moving sphere travels the furthest. You can mark on floor with masking tape the final point of contact. Caution: the strong magnetic sphere is ceramic and may crack if dropped onto a hard surface.
7. With the track horizontal, place two of the five spheres and the one special one at the center as shown. With a pencil, slowly push a sphere from the end toward the special sphere. Observe.
Procedure B. Teacher Demo
Repeat steps 1 thru 5 from Procedure A in front of the class. Before each step, ask students to predict the result.
Procedure C. Open Ended
Provide students with all of the materials listed at the top. Ask them to experiment with the materials, carefully recording what they do and the results.
How does this activity demonstrate the Law of Conservation of energy?
Reviews
8 reviews
Associate Prof. Emeritus
Oct 31, 2015
This magnetic accelerator is great, even for quantitative measurements. I took a 120 fps video of the colliding spheres, from which I was able to determine the amount of magnetic potential energy that was converted to kinetic energy. I taped the magnetic and one steel sphere to the track so they would not recoil on the collision. Highly recommended.
Rich
2 0
Oct 2, 2015
It's great, but the magnetic marble begins to chip after a while.
Brian Laird
1 0
very useful for conservation of energy and not!
Mar 25, 2015
Easy to set up and rugged (no strings to tangle) can clearly demonstrate conservation of energy and when that concept is beginning to sink in just roll the magnetic sphere and astound your students with what looks like a serious exception.
Michael Thuot
2 0
replacements
Feb 18, 2015
So glad to have replacement parts
Blair Cochran
1 0
Useful demonstration - not a toy
Mar 22, 2014
This device is an excellent demonstration of Newton's Laws of Motion provided you keep to the non-magnetic balls. Once you introduce the magnetic ball (1) into the system all the 'laws' seem to go out the window (literally depending on your location!). The magnetic ball introduces a different energy source into the system - it's still Newtonian but the magnetic attraction between the steel balls and the magnetic ball can be very confusing - it is a secondary source of potential energy that increases in strength with the number of steel balls in the system. This can be very difficult to explain to students w/o a familiarity with the basic laws of magnetism. The system with five steel balls and a single magnetic ball appears to violate the conservation of energy principle - until the magnetic attraction effect is brought into the equations. This is, unfortunately, a complex function that is beyond most elementary Newtonian expressions and can reduce the teacher to hand waving (not a good situation despite the spectacular results of the fifth ball being shot off the end of the apparatus). Unless you are ready to explain magnetism to your students I would steer clear of this device - despite its wonderful simplicity. On the other hand, if you want to keep to strictly Newtonian physics (w/o magnets) this is an excellent device - it functions as a pendulum and momentum transfer demonstration that works at a low enough speed to allow students to observe and record data that reinforces the Newtonian laws to a 'T.' Worth the cost - and then some. wb
Warren Buckles
3 0
magnetic accelerator
Mar 13, 2013
What a great tool for teaching so many concepts of physics, including energy, and conservation of energy
Don Stokes
0 0
Crowd Pleasing Demonstration
May 30, 2012
The students in Science Club love to watch this demonstration. I do have a parent or older student helper closely monitor this activity since the steel balls can get lost and to make sure they aren't launched towards breakable objects.
Cynthia House
0 0
Great product
May 29, 2012
Great item.
Aaron Geery
0 0
### NGSS
This product will support your students' understanding of the Next Generation Science Standards (NGSS)*, as shown in the table below.
Elementary Middle School High School K-PS2-1 Students can use the Magnetic Accelerator to plan and conduct an investigation to compare the effects of different strengths or different directions of pushes and pulls on the motion of an object. K-PS2-2 Students can use the Magnetic Accelerator in an investigation to analyze data to determine if a design solution works as intended to change the speed or direction of an object with a push or a pull. 3-PS2-2 Students can use the Magnetic Accelerator in an investigation to understand motion. Students can make observation and/or measurements of an object's motion to provide evidence that a pattern can be used to predict future motion. 4-PS3-4 Students can use the Magnetic Accelerator to design, test, and refine a device that converts energy from one form to another. MS-PS3-5 Students can use the Magnetic Accelerator to construct, use and present arguments or experiments to support the claim that when the motion energy of an object changes, energy is transferred to or from the object. HS-PS2-1 Students can use the Magnetic Accelerator in an investigation to analyze data to support the claim that Newton's Second Law of Motion describes the mathematical relationship among the net force on a macroscopic object, its mass, and its acceleration. HS-PS2-2 Students can use the Magnetic Accelerator in an investigation and use mathematical representations to support the claim that the total momentum of a system of objects is conserved when there is no net force on the system. HS-PS3-4 Students can use the Magnetic Accelerator design, build, and refine a device that works within given constraints to convert one form of energy into another form of energy.
Suggested Science Idea(s)
K-PS2-1
K-PS2-2
3-PS2-2
4-PS3-4
MS-PS3-5
HS-PS2-1
HS-PS2-2
HS-PS3-4
Students can use the Magnetic Accelerator in a number of investigations to demonstrate and teach Newton's Laws of Motion. This science tool creates a dramatic demonstration of energy transfer, and much more.
* NGSS is a registered trademark of Achieve. Neither Achieve nor the lead states and partners that developed the Next Generation Science Standards were involved in the production of, and do not endorse, this product.
Q & A | 2,090 | 9,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-40 | longest | en | 0.910794 |
https://physicscatalyst.com/article/how-to-find-acceleration-with-velocity-and-distance/ | 1,718,391,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00822.warc.gz | 416,946,959 | 25,292 | Home » Tips and tricks » How to find acceleration with velocity and distance
# How to find acceleration with velocity and distance
In this article learn how to find acceleration with velocity and distance. We will look at the cases for motion with uniform acceleration and non-uniform acceleration in one dimension.
We know that acceleration is defined as a change in velocity per unit time. So given velocities and time, we can calculate acceleration. Now here we will be checking how to find acceleration with velocity and distance
## How to find acceleration with velocity and distance
### For Uniform acceleration motion in one dimension
In uniformly acceleration motion, acceleration can be found with velocity and distance using the formula
$v^2= u^2 + 2as$
Where v is the final velocity
$u$ is the initial velocity
$a$ is the acceleration
$s$ is the distance traveled
Rearranging the formula we get
$a=\frac{v^2-u^2}{2s}$
Example 1
An object slows downs from 50 m/s to rest in a distance of 100m. what is the acceleration of the object assuming constant?
Solution
We know that
$v^2= u^2 + 2as$
or $a = \frac {v^2 -u^2}{2s} = -50/200 =- .5 m/s^2$
the negative sign is present as it is deceleration or retardation
### For Non Uniform acceleration motion in one dimension
For an object whose velocity varies with position, the instantaneous acceleration can be obtained as
$a = \frac {dv}{dx}$
Example 2
An object moves in a straight line such as $v= x+ x^2 + 5$ where $v$ is the velocity and $x$ is the position Find the acceleration equation and acceleration at $x=0$.
Solution
Given
$v= x+ x^2 + 5$
Now
$a = v \frac {dv}{dx}$
$a = (x+ x^2 + 5) \frac {d}{dx} (x+ x^2 + 5)$
$a=(x+ x^2 + 5)(2x+ 1) = 2x^3+ 3x^2 +11x + 5$
At $x=0$
a= 5 m/s2
Practice Problems
1. An object velocity increase from 30 m/s to 40 m/s uniformly while covering the distance 10 m. Find the acceleration
2. A ball is thrown vertically upwards with the initial velocity 25 m/s. What height does the ball reach. Take g=10 m/s2
3. A charged particle in a uniform electric field is accelerated in a straight line from rest to 1000 km/s in a distance 1 m.What is the acceleration
Hope you like the article on how to find acceleration with velocity and distance. Please do provide the feedback
Related Articles
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 640 | 2,380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.843252 |
https://www.studyrankers.com/2020/08/mcq-questions-for-class-9-maths-constructions.html | 1,607,025,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732696.67/warc/CC-MAIN-20201203190021-20201203220021-00064.warc.gz | 874,635,475 | 51,963 | ## MCQ Questions for Class 9 Maths: Ch 11 Constructions
1. Which of the following angles can be constructed using ruler and compasses?
(a) 35°
(b) 45°
(c) 95°
(d) 55°
► (b) 45°
2. With the help of a rular and compass, it is not possible to construct an angle of
(a) 35°
(b) 67.5°
(c) 82.5°
(d) 7.5°
► (a) 35°
3. With the help of a ruler and compass, it is possible to construct an angle of
(a) 40°
(b) 37.5°
(c) 65°
(d) 50°
► (b) 37.5°
4. To construct a ΔABC in which BC = 10 cm and ∠B= 60 degrees and AB + AC = 14 cm, then the length of BD used for construction.
(a) 7 cm
(b) 14 cm
(c) 20 cm
(d) 10 cm
â–º(b) 14 cm
5. An angle whose measure is more than 180° and less than 360° is called a
(a) Reflex angle
(b) Acute angle
(c) Straight angle
(d) Complete angle
â–º (a) Reflex angle
6. If two circles touches internally then distance between their centres is equal to
(b) not possible to determine
(d) none
7. On a ray AB with initial point A, Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E. Draw the ray AC passing through E. Then, the measure of ∠CAB is
(a) 30°
(b) 60°
(c) 45°
(d) 15°
► (b) 60°
8. The bisector of an angle lies in its
(a) Interior
(b) On the arms of the angle
(c) Any where in the plane
(d) Exterior
â–º (a) Interior
9. If two circles touches internally then distance between their centres is equal to
(c) not possible to determine
(d) none
10. The internal and external bisectors of an angle form a
(a) Acute angle
(b) Straight angle
(c) Right angle
(d) Reflex angle
â–º (c) Right angle
11. Two radii of same circle are always :
(a) may inchired at any angle
(b) perpendicular
(c) parallel
(d) parallel and may inchired at any angle
â–º (d) parallel and may inchired at any angle
12. Which of these angles cannot be constructed using ruler and compasses?
(a) 120°
(b) 60°
(c) 140°
(d) 135°
► (c. 140°
13. Which of the following angles can be constructed using ruler and compass?
(a) 35°
(b) 40°
(c) 90°
(d) 50°
► (c) 90°
14. On a ray AB with initial point A, Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E. Draw the ray AC passing through E. Then, the measure of ∠CAB is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
► (d) 60°
15. The construction of △ABC, given that BC = 5 cm, ∠B = 600 is not possible when the difference of AB and AC is equal to
(a) 4.2 cm
(b) 5.9 cm.
(c) 4 cm.
(d) 3 cm.
â–º (b) 5.9 cm.
16. The construction of a triangle ABC with AB = 4 cm and ∠A = 60° is not possible when difference of BC and AC is equal to
(a) 3.5 cm
(b) 4.5 cm
(c) 2.5 cm
(d) 3 cm
â–º (b) 4.5 cm
17. With the help of a ruler and a compass, it is possible to construct an angle of
(a) 40°
(b) 37.5°
(c) 47.5°
(d) 35°
► (b) 37.5°
18. The point of concurrence of the three angle bisectors of a triangle, is called
(a) Centroid
(b) Incentre
(c) Circumcentre
(d) Orthocentre
â–º (b) Incentre
19. In Δ ABC, which of the following information is needed to construct it if it is known that measure of ∠B = 60 and BC = 6 cm :
(a) AB + BC
(b) CA + AB
(c) BC + CA
(d) All of the above
â–º (d) All of the above
20. An external bisector of an angle measuring 70° will divide the angle into two angles measuring
(a) 35°
(b) 55°
(c) 70°
(d) 110°
► (b) 55°
21. Two radii of the same circle are always:
(a) may inchired at any angle
(b) parallel
(c) parallel and may inchired at any angle
(d) perpendicular
â–º (c) parallel and may inchired at any angle
22. In Δ ABC if ∠B = ∠C = 300, which of the following is the longest side?
(a) BC
(b) AC
(c) AB
(d) none
â–º (a) BC
X | 1,394 | 3,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-50 | longest | en | 0.773312 |
https://mail.python.org/pipermail/python-ideas/2017-January/044300.html | 1,660,890,068,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00642.warc.gz | 364,952,747 | 2,963 | Juraj Sukop juraj.sukop at gmail.com
Sun Jan 15 12:25:59 EST 2017
```Hello!
Fused multiply-add (henceforth FMA) is an operation which calculates the
product of two numbers and then the sum of the product and a third number
with just one floating-point rounding. More concretely:
r = x*y + z
The value of `r` is the same as if the RHS was calculated with infinite
precision and the rounded to a 32-bit single-precision or 64-bit
double-precision floating-point number [1].
Even though one FMA CPU instruction might be calculated faster than the two
the increased precision of numerical computations that involve the
accumulation of products. Examples which benefit from using FMA are: dot
product [2], compensated arithmetic [3], polynomial evaluation [4], matrix
multiplication, Newton's method and many more [5].
C99 includes `fma` function to `math.h` [6] and emulates the calculation if
the FMA instruction is not present on the host CPU [7]. PEP 7 states that
"Python versions greater than or equal to 3.6 use C89 with several select
C99 features" and that "Future C99 features may be added to this list in
the future depending on compiler support" [8].
This proposal is then about adding new `fma` function with the following
signature to `math` module:
math.fma(x, y, z)
'''Return a float representing the result of the operation `x*y + z` with
single rounding error, as defined by the platform C library. The result is
the same as if the operation was carried with infinite precision and
rounded to a floating-point number.'''
There is a simple module for Python 3 demonstrating the fused multiply-add
operation which was build with simple `python3 setup.py build` under Linux
[9].
Any feedback is greatly appreciated!
Juraj Sukop
[1] https://en.wikipedia.org/wiki/Multiply%E2%80%93accumulate_operation
[2] S. Graillat, P. Langlois, N. Louvet. Accurate dot products with FMA.
2006
[3] S. Graillat, Accurate Floating Point Product and Exponentiation. 2007.
[4] S. Graillat, P. Langlois, N. Louvet. Improving the compensated Horner
scheme with a Fused Multiply and Add. 2006
[5] J.-M. Muller, N. Brisebarre, F. de Dinechin, C.-P. Jeannerod, V.
Lefèvre, G. Melquiond, N. Revol, D. Stehlé, S. Torres. Handbook of
Floating-Point Arithmetic. 2010. Chapter 5
[6] ISO/IEC 9899:TC3, "7.12.13.1 The fma functions", Committee Draft -
Septermber 7, 2007
[7] https://git.musl-libc.org/cgit/musl/tree/src/math/fma.c
[8] https://www.python.org/dev/peps/pep-0007/
[9] https://github.com/sukop/fma
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.python.org/pipermail/python-ideas/attachments/20170115/e7708c56/attachment.html>
``` | 751 | 2,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-33 | latest | en | 0.889655 |
http://www.markedbyteachers.com/international-baccalaureate/maths/math-hl-portfolio.html | 1,601,167,476,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00070.warc.gz | 196,226,661 | 29,759 | • Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
19. 19
19
20. 20
20
21. 21
21
22. 22
22
23. 23
23
24. 24
24
25. 25
25
26. 26
26
27. 27
27
28. 28
28
29. 29
29
30. 30
30
31. 31
31
32. 32
32
33. 33
33
34. 34
34
35. 35
35
36. 36
36
37. 37
37
38. 38
38
# Math HL portfolio
Extracts from this document...
Introduction
Math HL
Portfolio
Patrick Vollmer
Description:
In this task you will investigate the patterns in the intersection of parabolas and the lines y=x and y=2x. Then you will be asked to prove your conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials.
The main aim of my investigation is to conclude an answer for the relations between the graph intersections of the graphs y=x and y=2x
Method
1. Consider the parabola y=(x-3)²+2=x²-6x+11 and the lines y=x and y=2x.
Using Technology find the four intersections illustrated on the right.
Using the Microsoft based program autograph® the four intersections between the three given graphs were found.
Label all the x-values of these intersections as they appear from left to right on the x-axis as , , and. ( Label on the actual graph)
Find the values of - and - and name them respectievlyand .
= {1,764} - = 2,382 - 1,764 = 0,618 =
= {2,382} - = 6,236 – 4,618= 1,618 =
= {4,618}
= {6,236}
Finally, calculate D = -
D= - = 0,618 – 1,618 = – 1= 1 D = 1
2. Find Values for D for other parabolas of the form y=ax²+bx+c, a > 0, with vertices in quadrant 1, intersected by the lines y=x and y=2x. Consider various values of a, beginning with a=1. Make a conjecture about the value of D for these parabolas.
I used several parabolas to derive to a relation or formula.
The first parabola I used in the form of y=ax²+bx+c is
y=2x²-5x+4
= {0,719} - = 1 - 0.719 = 0.281 =
= {1} - = 2,781 – 2 = 0.781 =
= {2}
= {2,781} D= - = ( - ) - (- ) = 0.2808 – 0.781= 0.5
The third parabola I used in the form of y=ax²+bx+c is
Y= 0.5x²-3x+5
= {1,127}
= {1,5505}
= {6,4495}
= {8,873}
D= - =( - ) - (- ) = 0.4235 – 2.4235= 2
The fourth parabola I used in the form of y=ax²+bx+c is
Y=2x²-3x+1.2
= {0,26893}
= {0,36754}
= {1,6325}
= {2,2311}
D= - =( - ) - (- ) = 0.09861 – 0.59861= 0.5
Middle
1
1
2
0.5
2.5
0.4
Π
0.31833
This conjecture is only possible if the vertex is in the 1st quadrant and if the values of b and c are bounded to limitations.
B has to be smaller than zero in mathematical terms b<0
For example see the same quadratic equation just the variable b is changed
Y= x²-4x+5 (a=1, b= -4, c=5)
And Y= x²+4x+5 (a=1, b= 4, c=5)
You can complete the square to convert ax2 + bx + c to vertex form, but it's simpler to just use a formula (derived from the completing-the-square process) to find the vertex.
For a given quadratic y = ax2 + bx + c, the vertex (h, k) is found by computing h = b/2a, and then evaluating y at h to find k where k = (4ac – b2) / 4a.
So the vertex for the first one is is h = =
K = =
So the Vertex is (2,1)
The Vertex of the second quadratic does not lie in the first quadrant
Y= x²+4x+5 (a=1, b= 4, c=5)
h = =
K = =
That the vertex is not in the first quadrant can also be proved by using technology (autograph) which is shown in the graph below:
3. Investigate your conjecture for any real value of a and any placement of the vertex. Refine your conjecture as necessary and prove it. Maintain the labelling convention used in parts 1 and 2 by having the intersections of the first line to be and , and the intersections with the second line to be and .
This Question implies that I have to investigate and modify my conjecture for any real value for “a” and any placement of the vertex. This means that I am going to change not only “a”, but also the variables “b” and “c” in the parabola. The lines of y = x and y = 2x are still going to be the same. Similarly, I will note the intersections of those lines with the parabola and also calculate D again.
I used several parabolas to derive to a modified conjecture to the case of a vertex in different quadrants.
Y=−x²+4x+5
= {-1,449}
= {-1.193}
= {4,193}
= {3.449}
D= - =( - ) - (- ) = 0.256 – (-0.744)= 1
= -1 D-1
My previously conjecture has to be modified from = D
To = D
In this case it would be = -1 = 1 = D
The second parabola I used in the form of y=ax²+bx+c is
y=-13x²-4x+7
= {-1}
= {-0,9509}
= {0,5663}
= {0,5385}
D= - =( - ) - (- ) = 0.0491 – (-0.0278)= 0.0769
= = 0.0769 = D
Y = log(6)x² + 3x – 2
= -2.3697
= -3.3398
= 0.7696
= 1.0846
D =- =( - ) - (- ) = | (-0.9701) – (0.3150) | = | (-1.2851) | = 1.2851
D= = = 1.2851
Y = 5x² - 3x – 2
= -0.3062
= -0.3483
= 1.1483
= 1.3062
D =- =( - ) - (- ) = | (-0.0421) – (0.1579) | = | -0.2000 | = 0.2
D= = = 0.2
Y = -18x² + 5x +6
= -0.5
= -0.4768
= 0.6991
= 0.6667
D =- =( - ) - (- ) = | 0.0232 – (-0.0324) | = | -0.0556 | = 0.0556
D= = = 0.0556 regarding only 4 significant figures.
My modified conjecture works so far which is shown in table of results below:
Value for “a”-18 Value for “D”0.0556 -13 0.0769 -1 1 Log(6) 1.2851 5 0.2
Conclusion
Therefore my new general conjecture for any polynomial is: D = |d2-d1| / |(power-1)*a|,
Where d1 and d2 are the gradients of the lines intersecting the polynomial and a is the gradient of the polynomial and the power is the order of the polynomial.
Now I want to summarize my findings form question 6 and check whether my conjecture is true.
Polynomial Lines D Calculated value for D Y=(x-1)(x-3)(x-4)(x-5) Y=x-6 and y=0.5x-6 0.1367 0.166 Y=2(x-1)(x-3)(x-4)(x-5) Y=4x-19 and y=3x-12 0.2047 0.166 Y=0.7(x-1)(x-3)(x-4)(x-5) Y=2x-7 and y=x-3 0.439 0.952 Y=(x-1)(x-3)(x-4)(x-5)(x-6) Y=3x and y=5x 0.1449 0.5
Again the calculated value for D did not come out to be exactly the same as the value for D that I got from drawing the polynomials and intersecting it with the lines. This is due to the fact that the gradient of the curve does not stay constant and so is only approximated. Therefore, my conjecture can only come out to be close, but not exactly the same.
Conclusion
After now having investigated in the patterns of the intersections of parabolas with two lines, I want to conclude that the general conjecture for any polynomial is:
D = (d2-d1) / ((power-1)*a), where d1 and d2 are the gradients of the lines intersecting the polynomial and a is the gradient of the polynomial and the power is the order of the polynomial.
The difficulties in this Project were that it was impossible to find an exact conjecture for D. It was always only an approximate. This is due to the fact that the gradient of the line, a, is approximated and changed throughout the whole curve. Therefore the calculated values for D will always vary slightly from the actual value for D.
However, the conjecture that I found comes out to be very close.
Parabola Investigation
Patrick Vollmer
Math Higher Level
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
## Found what you're looking for?
• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month
Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
# Related International Baccalaureate Maths essays
1. ## This portfolio is an investigation into how the median Body Mass Index of a ...
rural girls in Shaanxi has the data after the min point growing exponentially while the data before the min point decreasing in almost a linear fashion. Therefore, the sinusoidal model cannot be adjusted to fit the data unless the sinusoidal function is abandoned completely for linear function with a negative slope followed by an exponential growth function.
2. ## Extended Essay- Math
G`= 6�\ �)-l�E �0��W�3�1�'"2�-��4�*�'a�Sz���'A�*"v&(c)�)IF"��"���:�0]-J�q�P��L�[ÅF� 3�����L� *iA�-[k> �D>/B�z�|^,Ï�f" �xèH-9�'�c1�D�">��u�(tm)��p����H-�cA�?.�--�1/4Q3��1kÇ"g�:C�,(é°3/4Y� K2�Af�"�+l�(c)�YX9�%�x-/K��(c)y&��K'w�o�\$� Äv�N�'2wꦺrÞ��,o-��1/4]-���y;W�y�5V���1/4�6��mh�N�j���j���^�(�(tm)�8�e���Ñ|vÑW��rb�×�Ü^w�{}�a�q�o\����8����F�},�_6�W�Mc|T*6�o�N ��W~���{���j�u��-�zh��m��T=�:�-���|���-�GÅ���Ҹvv�X?<�<ET*�t�~��~--�"�=Aw���s�v��y"�"��n�j�\�fé±ï¿½ï¿½3/4��\(c)](c)y���W���� g %�(r)y���W��nc*�c9e��Q�{..."�X;vZ�{�?z�:8�2w{��Z��h��7�gw'�v�R�F�]�جÞ���@r�=>��(r)Ú%[s��(tm)a"O=�Um���@�"6����E��wc[�ak(I c�1/4��:cwx�/��W��{x' EV���1/2�յ73/4?h+ǵ�����ެ�}È¡Y��;��S��k��֬�Ì1/2��3/4���k�W�'O���'�(r)x�_�Î��q��[� /1/4����N���_?���ը5><*�3/4k�T���ɸv|\1/2�*�~R�9��Ó�5j߶�����Q[w��< :"�i���_�["-����Y�<U""�Q�92�1/2��Q?_1�~�1/4Y{"tÖ�H-��Ç1/2'1/2�s�Y�����p�@H1/2n��uq�m f}ct�?1/29T"�`�s�g�n��6��Û�T[~�bɬ�Vǹz<�yê���}�rr`�n���+� n��-��Ũzq�6N�"�'1/2�1/2.�Þ��{�(��l�6k�N(c)_.�e�2V� gv�")��y����\�^ ëµï¿½o�jO����Sk�h� [HI�t-���a�Q{~3jaÙEy"����Ó2� G�y"q��ů���{]���T"(��� �"'"��CI����pok��'�ky;�l��,��y�����1/2�"4���[:�ã¡ï¿½Nz�� �/-�K���p����Ӭ��>���{�|V4vNO�1/2��p-����*5�׵}P�@9�R�#��_R��}Ó°w=�7��7 �;��ã}3ê¤ y�"��YT�V5P���c{�xq�8����"#�~�3/4j�z�"���Ö�s"Õ´[e43/4(]���-w��s�BF�]1/2�6��nG...�A]�� }Vb�&�S��Tj(�1/41/2Ò¥k�T��\>]��Z����lT-� -�Pjk���3��1/2��...y��(r)�qK�<� �]c"�!+ØR3/4�t�p3�d-�+'[��H�*G�(c)̶�C�s&��.&�0?� <1 t� �.���rvY�2%" �j�]- ���4��� ��w0�O�8�`\(�� 8\O"�A0D 0�,'pÑ\$'4��6\$m�\$v�)�#@H '��2<j\$F7B~����Sn�R�6�(r)C��� f Q��l �CR�9���<bnZñ %C� r�D����"XG�_�H�k3z���'�o M "+J�A�� ,;�1ÒvB�IfܤL+ -I�FYU��".��"����1/4�7�'��z��O��+�>�O� :�"0}(r),U,>�� �3(c)�r �����3/4��"�`4��1O�-)P���z��-�EE�<yZ�n��":���� ,)
1. ## MAths HL portfolio Type 1 - Koch snowflake
16 14 805306368 2.09075E-07 168.3695562 0.6928173 17 15 3221225472 6.96917E-08 224.4927416 0.6928190 18 16 12884901888 2.32306E-08 299.3236555 0.6928197 19 17 51539607552 7.74352E-09 399.0982074 0.6928201 20 18 2.06158E+11 2.58117E-09 532.1309432 0.6928202 21 19 8.24634E+11 8.60392E-10 709.5079242 0.6928203 22 20 3.29853E+12 2.86797E-10 946.0105656 0.6928203 To obtain the above table, various formulas were used in the spreadsheet software (Office '07).
2. ## Investigating Parabolas
To find the conjecture (a = 2) i. Equation: y = ax2 + bx + c ii. let a =2, b = -10, and c = 14 iii. y = 2x2-10x+14 Process - x2 - x1 = 2 - 1.58 = 0.41 = SL - x4 - x3 =
1. ## Math SL Circle Portfolio. The aim of this task is to investigate positions ...
0 The 0 is rejected, because according to the graph shown before, the point P' does not lie on the origin. The coordinates of P' is ( , 0). Again, by using the distance formula, the length of can be calculated.
2. ## Ib math SL portfolio parabola investigation
Situation 8: A concave up parabola in the third quadrant works. Situation 9: Anthat is concave down in the third quadrant works. Situation 10: Intersections in both 1st and 3rd quadrants work as long as the intersections of one line are x2 and x3, and the intersections of the other
1. ## Ib math HL portfolio parabola investigation
Hence: ,where a >0 . We also know that for the turning point (h,k) to be in the first quadrant both h and k must be greater than zero . We want to find four intersection points. This means that the parabola MUST intersect with y=x.
2. ## Math IA Type 1 Circles. The aim of this task is to investigate ...
Therefore there is no point A and circle C3 does not exist because the center of C3 is the point of intersection of C1 and C2; and by extension OP? cannot be determined. Similarly, OP? does not exist when r=4.01, but it does exist when r = 4, see Figure 11.
• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to | 6,437 | 13,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-40 | latest | en | 0.736411 |
https://www.mathworks.com/matlabcentral/cody/problems/753-solitaire-cipher/solutions/116155 | 1,496,092,498,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612553.95/warc/CC-MAIN-20170529203855-20170529223855-00264.warc.gz | 1,121,318,373 | 11,793 | Cody
# Problem 753. Solitaire Cipher
Solution 116155
Submitted on 20 Jul 2012 by Rijk
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
deck = [1 4 7 10 13 16 19 22 25 28 3 6 9 12 15 18 21 24 27 2 5 8 11 14 17 20 23 26]; n = 10; out = [8 26 1 4 7 23 8 8 4 24]; assert(isequal(solitaire(deck, n),out))
``` out = 8 out = 8 26 out = 8 26 1 out = 8 26 1 4 out = 8 26 1 4 7 out = 8 26 1 4 7 23 out = 8 26 1 4 7 23 8 out = 8 26 1 4 7 23 8 8 out = 8 26 1 4 7 23 8 8 4 out = 8 26 1 4 7 23 8 8 4 24 ```
2 Pass
%% deck = 1:28; n = 10; out = [3 22 9 4 23 7 25 16 14 14]; assert(isequal(solitaire(deck, n),out))
``` out = 3 out = 3 22 out = 3 22 9 out = 3 22 9 4 out = 3 22 9 4 23 out = 3 22 9 4 23 7 out = 3 22 9 4 23 7 25 out = 3 22 9 4 23 7 25 16 out = 3 22 9 4 23 7 25 16 14 out = 3 22 9 4 23 7 25 16 14 14 ```
3 Pass
%% deck = 1:28; n = 9; out = [3 22 9 4 23 7 25 16 14]; assert(isequal(solitaire(deck, n),out))
``` out = 3 out = 3 22 out = 3 22 9 out = 3 22 9 4 out = 3 22 9 4 23 out = 3 22 9 4 23 7 out = 3 22 9 4 23 7 25 out = 3 22 9 4 23 7 25 16 out = 3 22 9 4 23 7 25 16 14 ``` | 645 | 1,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-22 | longest | en | 0.500931 |
http://www.velocityreviews.com/forums/t953742-how-to-apply-qsort-with-a-parameter.html | 1,386,654,112,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164009894/warc/CC-MAIN-20131204133329-00047-ip-10-33-133-15.ec2.internal.warc.gz | 595,514,426 | 11,984 | Velocity Reviews > How to apply qsort() with a parameter?
# How to apply qsort() with a parameter?
none
Guest
Posts: n/a
10-22-2012
How does one apply qsort() if the comparison criterion
depends on a parameter?
Specifically, suppose I wish to sort an array of int,
according to distance from a prescribed value `n':
int compar(const void *aa, const void *bb)
{
int a = *(int *)aa;
int b = *(int *)bb;
return abs(a - n) < abs(b - n);
}
One way to make this work, is to declare n as a global
(file scope) variable. Is there a clever way to do this
without bringing in a global variable?
--
Rouben Rostamian
ImpalerCore
Guest
Posts: n/a
10-22-2012
On Oct 22, 1:51*pm, rouben@shadow.(none) (Rouben Rostamian) wrote:
> How does one apply qsort() if the comparison criterion
> depends on a parameter?
>
> Specifically, suppose I wish to sort an array of int,
> according to distance from a prescribed value `n':
>
> int compar(const void *aa, const void *bb)
> {
> * * * * int a = *(int *)aa;
> * * * * int b = *(int *)bb;
> * * * * return abs(a - n) < abs(b - n);
>
> }
>
> One way to make this work, is to declare n as a global
> (file scope) variable. *Is there a clever way to do this
> without bringing in a global variable?
The best method is to reparameterize the qsort function with a
comparison that takes an external parameter.
void qsort_ext( void* base,
size_t num,
size_t size,
int (*cmp_ext_fn)( const void*, const void*, void* ),
void* ext_data );
Then you write your modified comparison function in this fashion.
int compar(const void *aa, const void *bb, void *nn)
{
int a = *(const int *)aa;
int b = *(const int *)bb;
int n = *(int *)n;
return abs(a - n) < abs(b - n);
}
If you have the source of a 'qsort' you want to use, you can create
'qsort_ext' by simply adding the external parameter to the comparison
function pointer call.
Best regards,
John D.
ImpalerCore
Guest
Posts: n/a
10-22-2012
On Oct 22, 2:18*pm, ImpalerCore <(E-Mail Removed)> wrote:
> int compar(const void *aa, const void *bb, void *nn)
> {
> * int a = *(const int *)aa;
> * int b = *(const int *)bb;
> * int n = *(int *)n;
> * return abs(a - n) < abs(b - n);
> }
The comparison should probably be 'abs(a - n) - abs(b - n)', not '<'
which is a boolean expression.
Best regards,
John D.
tom st denis
Guest
Posts: n/a
10-22-2012
On Oct 22, 1:51*pm, rouben@shadow.(none) (Rouben Rostamian) wrote:
> How does one apply qsort() if the comparison criterion
> depends on a parameter?
>
> Specifically, suppose I wish to sort an array of int,
> according to distance from a prescribed value `n':
>
> int compar(const void *aa, const void *bb)
> {
> * * * * int a = *(int *)aa;
> * * * * int b = *(int *)bb;
> * * * * return abs(a - n) < abs(b - n);
>
> }
>
> One way to make this work, is to declare n as a global
> (file scope) variable. *Is there a clever way to do this
> without bringing in a global variable?
The most portable way is to normalize your input first based on what
you are measuring the distance from then perform your sort.
Tom
Eric Sosman
Guest
Posts: n/a
10-22-2012
On 10/22/2012 1:51 PM, none Rouben Rostamian wrote:
> How does one apply qsort() if the comparison criterion
> depends on a parameter?
>
> Specifically, suppose I wish to sort an array of int,
> according to distance from a prescribed value `n':
>
> int compar(const void *aa, const void *bb)
> {
> int a = *(int *)aa;
> int b = *(int *)bb;
> return abs(a - n) < abs(b - n);
Aside: This can't be right, as it doesn't define
a total ordering.
> }
>
> One way to make this work, is to declare n as a global
> (file scope) variable. Is there a clever way to do this
> without bringing in a global variable?
No, nothing clever. You could write multiple comparators,
each hard-wiring a different `n' value:
int compar10(const void *aa, const void *bb) {
... use n=10 ...
}
int compar20(const void *aa, const void *bb) {
... use n=20 ...
}
Or, you could subtract n from all the array elements,
use the equivalent of compar0(), and then add n back again.
(Sounds wasteful, but it only adds O(N) time to an operation
that's probably O(N log N) already.)
Neither alternative strikes me as "clever." You either
live with the qsort() interface or write your own sort.
--
Eric Sosman
http://www.velocityreviews.com/forums/(E-Mail Removed)d
Eric Sosman
Guest
Posts: n/a
10-22-2012
On 10/22/2012 2:41 PM, Eric Sosman wrote:
> On 10/22/2012 1:51 PM, none Rouben Rostamian wrote:
>> How does one apply qsort() if the comparison criterion
>> depends on a parameter?
A further observation: Don't go overboard in parameterizing
your comparators. A comparator may be called a large number of
times during a sort, so it's "in the inner loop" and should be
kept short and swift. A comparator that examines parameters
P1,P2,P3,... in addition to the items it's comparing starts to
stray from the "short and swift" ideal.
A particular abuse that always seems to crop up is trying
to use one comparator to sort on multiple keys:
struct s {
int k1;
char *k2;
double k3;
};
int which_key;
int compare(const void *aa, const void *bb) {
const struct s *a = aa;
const struct s *b = bb;
switch (which_key) {
case 1:
return (a->k1 > b->k1) - (a->k1 < b->k1);
case 2:
return strcmp(a->k2, b->k2);
case 3:
return (a->k3 > b->k3) - (a->k3 < b->k3);
default:
assert(false);
}
}
...
which_key = 2;
qsort(array, count, sizeof array[0], compare);
This setup repeats the switch decision -- which is "just
overhead" -- for every comparison, and there may be a lot of
them. It would almost surely be better to have three short
comparators than one omnibus comparator.
--
Eric Sosman
(E-Mail Removed)d
Ben Bacarisse
Guest
Posts: n/a
10-22-2012
ImpalerCore <(E-Mail Removed)> writes:
> On Oct 22, 1:51Â*pm, rouben@shadow.(none) (Rouben Rostamian) wrote:
>> How does one apply qsort() if the comparison criterion
>> depends on a parameter?
>>
>> Specifically, suppose I wish to sort an array of int,
>> according to distance from a prescribed value `n':
>>
>> int compar(const void *aa, const void *bb)
>> {
>> Â* Â* Â* Â* int a = *(int *)aa;
>> Â* Â* Â* Â* int b = *(int *)bb;
>> Â* Â* Â* Â* return abs(a - n) < abs(b - n);
>>
>> }
>>
>> One way to make this work, is to declare n as a global
>> (file scope) variable. Â*Is there a clever way to do this
>> without bringing in a global variable?
>
> The best method is to reparameterize the qsort function with a
> comparison that takes an external parameter.
>
> void qsort_ext( void* base,
> size_t num,
> size_t size,
> int (*cmp_ext_fn)( const void*, const void*, void* ),
> void* ext_data );
>
> Then you write your modified comparison function in this fashion.
>
> int compar(const void *aa, const void *bb, void *nn)
> {
> int a = *(const int *)aa;
> int b = *(const int *)bb;
> int n = *(int *)n;
> return abs(a - n) < abs(b - n);
> }
>
> If you have the source of a 'qsort' you want to use, you can create
> 'qsort_ext' by simply adding the external parameter to the comparison
> function pointer call.
Many C libraries already have such a thing. GNU libc has qsort_r and
Microsoft champions qsort_s -- to the extent that it's in C11. The BSD
C library also has qsort_r but with a different argument order (for both
qsort_r and the comparison function). I'll bet there are others.
A mess of #ifdefs can probably render such code reasonably portable, but
maybe the OP wants the code to work in only one of these environments.
--
Ben.
tom st denis
Guest
Posts: n/a
10-22-2012
On Oct 22, 2:59*pm, Eric Sosman <(E-Mail Removed)>
wrote:
> On 10/22/2012 2:41 PM, Eric Sosman wrote:
>
> > On 10/22/2012 1:51 PM, none Rouben Rostamian wrote:
> >> How does one apply qsort() if the comparison criterion
> >> depends on a parameter?
>
> * * *A further observation: Don't go overboard in parameterizing
> your comparators. *A comparator may be called a large number of
> times during a sort, so it's "in the inner loop" and should be
> kept short and swift. *A comparator that examines parameters
> P1,P2,P3,... in addition to the items it's comparing starts to
> stray from the "short and swift" ideal.
>
> * * *A particular abuse that always seems to crop up is trying
> to use one comparator to sort on multiple keys:
>
> * * * * struct s {
> * * * * * * int k1;
> * * * * * * char *k2;
> * * * * * * double k3;
> * * * * };
>
> * * * * int which_key;
>
> * * * * int compare(const void *aa, const void *bb) {
> * * * * * * const struct s *a = aa;
> * * * * * * const struct s *b = bb;
> * * * * * * switch (which_key) {
> * * * * * * case 1:
> * * * * * * * * return (a->k1 > b->k1) - (a->k1 < b->k1);
> * * * * * * case 2:
> * * * * * * * * return strcmp(a->k2, b->k2);
> * * * * * * case 3:
> * * * * * * * * return (a->k3 > b->k3) - (a->k3 < b->k3);
> * * * * * * default:
> * * * * * * * * assert(false);
> * * * * * * }
> * * * * }
>
> * * * * ...
> * * * * which_key = 2;
> * * * * qsort(array, count, sizeof array[0], compare);
>
> * * *This setup repeats the switch decision -- which is "just
> overhead" -- for every comparison, and there may be a lot of
> them. *It would almost surely be better to have three short
> comparators than one omnibus comparator.
But that code is not thread safe and it's not really any simpler than
just calling qsort with a variable as the compare callback that is
determined locally on the stack.
Tom
Eric Sosman
Guest
Posts: n/a
10-22-2012
On 10/22/2012 4:25 PM, tom st denis wrote:
> On Oct 22, 2:59 pm, Eric Sosman <(E-Mail Removed)>
> wrote:
>>[...]
>> This setup repeats the switch decision -- which is "just
>> overhead" -- for every comparison, and there may be a lot of
>> them. It would almost surely be better to have three short
>> comparators than one omnibus comparator.
>
> But that code is not thread safe and it's not really any simpler than
> just calling qsort with a variable as the compare callback that is
> determined locally on the stack.
Since qsort() itself need not be thread-safe (it need not
of a comparator scarcely matters. A thread-safe comparator
could be called by qsort() in one thread while another called
it via bsearch(), but that's a bit of a stretch.
My personal parser recognizes all the words in your second
and third lines, but is unable to make sense of them.
In any event, the code I showed was an illustration of
what I consider an inferior pattern; the suggested alternative
--
Eric Sosman
(E-Mail Removed)d
Öö Tiib
Guest
Posts: n/a
10-22-2012
On Monday, 22 October 2012 20:51:55 UTC+3, none wrote:
> How does one apply qsort() if the comparison criterion
> depends on a parameter?
Others have already replied that most platforms offer qsort variants
Also think over why you need to sort with dynamic criteria.
Maybe you only need to find few items with highest order before
On that case building an heap
http://en.wikipedia.org/wiki/Heap_(data_structure)
can be more efficient solution. | 3,229 | 10,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-48 | longest | en | 0.62298 |
https://pedagogue.app/3-desmos-tricks-you-might-not-know-and-will-want-to-use-pronto/ | 1,708,874,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00699.warc.gz | 447,578,158 | 26,608 | # 3 Desmos Tricks You Might Not Know (And Will Want To Use Pronto)
Desmos, an advanced graphing calculator implemented as a web application, has become a staple tool for students and teachers alike. It offers remarkable features for graphing and analyzing mathematical functions, equations, and tables. But despite its popularity, there are some tricks that users might not know about. In this article, we will unveil three Desmos tricks that you might not know but will want to use pronto.
Regression lines play an essential role in understanding the relationship between two variables. With the correct syntax, Desmos can help you create regression models with ease. For instance, if you have a set of points and want to find the linear regression model (y = mx + b), type “y=a(x-h)^2+k” into the expression list. You will be prompted to manually adjust the parameters ‘a’, ‘h’, and ‘k’ until you find the best-fit curve.
However, if you want to take a shortcut, type “y_1~mx_1+b” where x_1 and y_1 represent your data table’s column names. Once you input this formula into Desmos, it will automatically find the optimal values of ‘m’ and ‘b’ for your regression line.
2. Solving Systems of Equations Effortlessly
Desmos can also solve systems of equations quickly using its built-in feature. Type in your system of equations and add curly braces “{}” around each equation. Make sure that you are using your variables (like integers or decimals) consistently within each equation when entering them into the calculator. After typing both x and y in the calculator workspace, the solution to your system of equations will be displayed as an ordered pair (x,y).
For example:
{x+y = 5,
x-y = 3}
After entering these equations, Desmos will provide you with the solution (x=4, y=1) as an ordered pair.
3. Desmos Geometry Tool
While many people use Desmos for graphing functions and equations, it can also create geometric figures. To access the geometry tool, go to “desmos.com/geometry” and start constructing lines, circles, polygons, angles, and more using the tools provided.
With Desmos Geometry, you can perform various geometric transformations like translations, rotations, and reflections. Plus, you can measure lengths and angles directly on the screen. Exporting your created figures as images or sharing them with others is also a breeze thanks to the tool’s sharing features.
In conclusion, it’s clear that Desmos offers a plethora of useful tricks to bolster your mathematical endeavors. From performing regression analysis to effortlessly solving systems of equations and even exploring geometrical figures, this powerful tool has something for everyone. So if you haven’t tried these tricks yet, give them a go and elevate your math game. | 586 | 2,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-10 | longest | en | 0.893747 |
https://www.physicsforums.com/threads/limit-as-x-approaches-infinity-involves-sinx-and-cosx.454185/ | 1,624,631,131,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00607.warc.gz | 854,782,869 | 13,343 | # Limit as x approaches infinity, involves sinx and cosx
y=1/2(sinx-cosx+e(^pi-x))
question: if x approaches infinity, which term or terms will dominate?
from my understanding, sinx and cosx will oscillate and the e term will approach zero. so would the answer be sinx-cosx?
plz and ty | 80 | 288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | longest | en | 0.908384 |
https://www.mrexcel.com/board/threads/workday-schedule-less-holidays.175759/ | 1,701,175,366,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00796.warc.gz | 1,020,158,727 | 22,352 | WORKDAY SCHEDULE LESS HOLIDAYS
Tandem_2
New Member
We are using the WORKDAY formula to create a construction schedule returning dates for each task.
Start Date 12/22/05 (Keyed Date) Cell B1
Slab Date =workday(B1,2,Holidays)
Formula Returns 12/27/05 Cell B2
Inspection =workday(B2,1,Holidays)
Formula Returns 12/28/05 Cell B3
This is perfect...EXCEPT...now we want to use a 6 day work week. How can this formula be modified to acknowledge Saturday as a working day?
Appreciate the help!!!
Excel Facts
Enter current date or time
Ctrl+: enters current time. Ctrl+; enters current date. Use Ctrl+: Ctrl+; Enter for current date & time.
6 day Work Week
OK...I have noticed that a number of people have viewed this question but no one has replied.
Have I worded the question properly? Or, is there just not a formula or an easy way to extend the work week to include Saturday and return work day dates accordingly?
If I need to better explain the question....please let me know.
Thanks again.
Schedule Generator using 6 Day Work Week excluding Holidays
Thanks for the replys. I had looked at the TOPIC referenced below, but it did not address holidays.
http://www.mrexcel.com/board2/viewtopic.php?t=146713&highlight=
The second posting dealt with Holidays but it calculated NET DAYS...thus it delivers a count.
http://www.mrexcel.com/board2/viewtopic.php?t=72395
BUT, my problem is HOW to convert this into a schedule that returns dates....like the Workday function does. Thus you start with a given date, provide a number of days for task completion, which is then added to the prior date excluding ONLY Sundays and Holidays as work days.
I can't seem to figure out how to make a SCHEDULE work.
I believe the following will work for you, although I have no experience with the Workday function so some adjustment may be needed. I'm assuming you want to skip all Sundays and all holidays as possible End dates, but allow Saturdays. The example below shows the behavior of the formula on intervals containing holidays.
workdays.xls
ABCDEFG
1Start DateWork DaysEnd DateHolidays
2Thu, 13 Jan 20051Fri, 14 Jan 2005Sun, 01 Jan 1900dummy start date
3Thu, 13 Jan 20052Sat, 15 Jan 2005Mon, 17 Jan 2005Dr. Martin Luther King Day (3rd Monday in January)
4Thu, 13 Jan 20053Tue, 18 Jan 2005Mon, 21 Feb 2005Presidents Day (3rd Monday in February)
5Thu, 13 Jan 20054Wed, 19 Jan 2005Mon, 30 May 2005Memorial Day (last Monday in May)
6Thu, 13 Jan 20055Thu, 20 Jan 2005Mon, 04 Jul 2005Independence Day
7Thu, 13 Jan 20056Fri, 21 Jan 2005Mon, 05 Sep 2005Labor Day (first Monday in September)
8Thu, 13 Jan 20057Sat, 22 Jan 2005Mon, 10 Oct 2005Columbus Day (2nd Monday in October)
9Thu, 13 Jan 20058Mon, 24 Jan 2005Fri, 11 Nov 2005Veterans Day
10Thu, 13 Jan 20059Tue, 25 Jan 2005Thu, 24 Nov 2005Thanksgiving Day (4th Thursday in November)
11Fri, 25 Nov 2005Thanksgiving Friday (day after Thanksgiving)
12Mon, 26 Dec 2005Christmas Day (observed)
13Tue, 22 Nov 20051Wed, 23 Nov 2005
14Tue, 22 Nov 20052Sat, 26 Nov 2005
15Tue, 22 Nov 20053Mon, 28 Nov 2005
16Tue, 22 Nov 20054Tue, 29 Nov 2005
17
6 Day Work Week
The array formula (entered with CTRL-SHIFT-ENTER) in C2, and copied down, is:
=A2+MATCH(TRUE,ROW(\$1:\$1000)-INT((ROW(\$1:\$1000)-1+WEEKDAY(A2))/7)-MATCH(A2+ROW(\$1:\$1000),E\$2:\$E\$12,1)+MATCH(A2,E\$2:\$E\$12,1)=B2,0)
Some notes and caveats:
In order for the MATCH usage to work correctly, the Holidays list (E2:E12 in the example) must be sorted in ascending order and a dummy date for the first holiday (smaller than any start date you'll be using) is needed, as I've shown.
The 1000 number is an arbitrary choice -- it must be large enough to cover the number of total days needed in any of your task periods. If you wish to adjust this, do so in all instances of ROW(\$1:\$1000).
If you must allow for possible deletions or insertions of rows above any cell containing the formula, then all instances of ROW(\$1:\$1000) (or whatever you choose) should be replaced by
ROW(INDIRECT("1:1000"))
The formula isn't efficient, but making it so would make it longer, and besides, I've struggled long enough with this as it is.
Adapting Fairwinds formula from the thread posted above by NVBC
=SMALL(IF(WEEKDAY(ROW(INDIRECT(A1&":"&WORKDAY(A1,B1,holidays))))<>1,IF(ISNA(MATCH(ROW(INDIRECT(A1&":"&WORKDAY(A1,B1,holidays))),holidays,0)),ROW(INDIRECT(A1&":"&WORKDAY(A1,B1,holidays))))),B1+(AND(WEEKDAY(A1)>1,ISNA(MATCH(A1,holidays,0)))))
confirmed with CTRL+SHIFT+ENTER
where A1 contains a date B1 the number of days to be added and holidays is a named range containing holiday dates
Replies
4
Views
376
Replies
3
Views
363
Replies
4
Views
336
Replies
5
Views
167
Replies
34
Views
1K
1,206,711
Messages
6,074,465
Members
446,071
Latest member
gaborfreeman
We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
Allow Ads at MrExcel
Which adblocker are you using?
Follow these easy steps to disable AdBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
Follow these easy steps to disable AdBlock Plus
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back | 1,665 | 5,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.927231 |
http://www.vttoth.com/CMS/physics-notes/261-tensor-calculus-cheat-sheet | 1,719,153,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00357.warc.gz | 51,447,235 | 6,548 | Once and for all, I need a cheat sheet for tensor calculus, to avoid having to recalculate things from scratch every time I deal with a new (or for that matter, old) theory.
First, differentiation of the metric tensor with respect to the metric. Obviously,
\begin{align}
{\dfrac{\partial g_{\alpha\beta}}{\partial g_{\mu\nu}}=\delta^\mu_\alpha\delta^\nu_\beta.}
\end{align}
Now, given that $g^{\alpha\beta}g_{\beta\gamma}=\delta^\alpha_\gamma$ and the derivative of the Kronecker-delta is identically zero, we can compute the derivatives of the contravariant metric tensor:
\begin{align}
\frac{\partial}{\partial g_{\mu\nu}}g^{\alpha\beta}g_{\beta\gamma}=0&=\frac{\partial g^{\alpha\beta}}{\partial g_{\mu\nu}}g_{\beta\gamma}+g^{\alpha\beta}\delta^\mu_\beta\delta^\nu_\gamma,\\
\frac{\partial g^{\alpha\beta}}{\partial g_{\mu\nu}}g_{\beta\gamma}&=-g^{\alpha\mu}\delta^\nu_\gamma,\\
\frac{\partial g^{\alpha\beta}}{\partial g_{\mu\nu}}g_{\beta\gamma}g^{\gamma\kappa}&=-g^{\alpha\mu}\delta^\nu_\gamma g^{\gamma\kappa},\\
\frac{\partial g^{\alpha\kappa}}{\partial g_{\mu\nu}}&=-g^{\alpha\mu}g^{\kappa\nu}.
\end{align}
Therefore,
\begin{align}
{\dfrac{\partial g^{\alpha\beta}}{\partial g_{\mu\nu}}=-g^{\alpha\mu}g^{\beta\nu}.}
\end{align}
Next, derivatives of the Christoffel-symbols with respect to the metric. Given
\begin{align}
{\Gamma^\gamma_{\alpha\beta}=\frac{1}{2}g^{\gamma\delta}\left(\partial_\alpha g_{\beta\delta}+\partial_\beta g_{\alpha\delta}-\partial_\delta g_{\alpha\beta}\right),}
\end{align}
we obtain
\begin{align}
{\dfrac{\partial\Gamma^\gamma_{\alpha\beta}}{\partial g_{\mu\nu}}=g^{\mu\gamma}\Gamma^\nu_{\alpha\beta}.}
\end{align}
When dealing with Lagrangians, we also need to differentiate with respect to $\partial_\kappa g_{\mu\nu}$:
\begin{align}
\frac{\partial\Gamma^\gamma_{\alpha\beta}}{\partial(\partial_\kappa g_{\mu\nu})}=\frac{1}{2}g^{\gamma\delta}\left(\delta^\kappa_\alpha\delta^\mu_\beta\delta^\nu_\delta+\delta^\kappa_\beta\delta^\mu_\alpha\delta^\nu_\delta-\delta^\kappa_\delta\delta^\mu_\alpha\delta^\nu_\beta\right),
\end{align}
or
\begin{align}
{\dfrac{\partial\Gamma^\gamma_{\alpha\beta}}{\partial(\partial_\kappa g_{\mu\nu})}=\dfrac{1}{2}\left(\delta^\kappa_\alpha\delta^\mu_\beta g^{\gamma\nu}+\delta^\kappa_\beta\delta^\mu_\alpha g^{\gamma\nu}-\delta^\mu_\alpha\delta^\nu_\beta g^{\gamma\kappa}\right).}
\end{align}
The coordinate derivative of the contravariant metric tensor can be expressed in terms of the covariant metric tensor's derivative in a similar way:
\begin{align}
{\partial_\mu g^{\alpha\beta}=-g^{\alpha\gamma}g^{\beta\delta}\partial_\mu g_{\gamma\delta}.}
\end{align}
Finally, let us consider coordinate derivatives of the Christoffel symbols:
\begin{align}
{\partial_\kappa\Gamma^\gamma_{\alpha\beta}=-\frac{1}{2}g^{\gamma\epsilon}\partial_\kappa g_{\epsilon\zeta}\Gamma^\zeta_{\alpha\beta}+\frac{1}{2}g^{\gamma\delta}\left(\partial_\alpha\partial_\kappa g_{\beta\delta}+\partial_\beta\partial_\kappa g_{\alpha\delta}-\partial_\delta\partial_\kappa g_{\alpha\beta}\right),}
\end{align}
\begin{align}
{\dfrac{\partial}{\partial g_{\mu\nu}}\partial_\kappa\Gamma^\gamma_{\alpha\beta}=\frac{1}{2}g^{\gamma\mu}g^{\epsilon\nu}\partial_\kappa g_{\epsilon\zeta}\Gamma^\zeta_{\alpha\beta}-\frac{1}{2}g^{\gamma\epsilon}\partial_\kappa g_{\epsilon\zeta}g^{\mu\zeta}\Gamma^\nu_{\alpha\beta}-\frac{1}{2}g^{\mu\gamma}g^{\nu\delta}\left(\partial_\alpha\partial_\kappa g_{\beta\delta}+\partial_\beta\partial_\kappa g_{\alpha\delta}-\partial_\delta\partial_\kappa g_{\alpha\beta}\right),}
\end{align}
\begin{align}
{\dfrac{\partial}{\partial(\partial_\rho g_{\mu\nu})}\partial_\kappa\Gamma^\gamma_{\alpha\beta}=-\frac{1}{2}g^{\gamma\mu}\delta^\rho_\kappa\Gamma^\nu_{\alpha\beta}-\frac{1}{4}g^{\gamma\epsilon}\partial_\kappa g_{\epsilon\zeta}\left(\delta^\rho_\alpha\delta^\mu_\beta g^{\zeta\nu} + \delta^\rho_\beta\delta^\mu_\alpha g^{\zeta\nu}-\delta^\mu_\alpha\delta^\nu_\beta g^{\zeta\rho}\right),}
\end{align}
\begin{align}
{\dfrac{\partial}{\partial(\partial_\rho\partial_\sigma g_{\mu\nu})}\partial_\kappa\Gamma^\gamma_{\alpha\beta}=\frac{1}{2}\delta^\sigma_\kappa\left(\delta^\rho_\alpha\delta^\mu_\beta g^{\gamma\nu}+\delta^\rho_\beta\delta^\mu_\alpha g^{\gamma\nu}-\delta^\mu_\alpha\delta^\nu_\beta\ g^{\gamma\rho}\right).}
\end{align}
Last but not least, the derivative of the covariant volume element:
\begin{align}
{\dfrac{\partial}{\partial g_{\mu\nu}}\sqrt{-g}=\frac{1}{2}g^{\mu\nu}\sqrt{-g}.}
\end{align}
This is all we need to compute derivatives of the Ricci-tensor that are relevant for variational calculus. We define the Riemann curvature tensor as
\begin{align}
{R_\mu{}^\alpha{}_{\nu\beta}=\partial_\beta\Gamma^\alpha_{\mu\nu}-\partial_\nu\Gamma^\alpha_{\mu\beta}+\Gamma^\kappa_{\mu\nu}\Gamma^\alpha_{\kappa\beta}-\Gamma^\kappa_{\mu\beta}\Gamma^\alpha_{\kappa\nu},}
\end{align}
while the Ricci tensor is given by
\begin{align}
{R_{\mu\nu}=R_\mu{}^\alpha{}_{\nu\alpha}=\partial_\alpha\Gamma^\alpha_{\mu\nu}-\partial_\nu\Gamma^\alpha_{\mu\alpha}+\Gamma^\kappa_{\mu\nu}\Gamma^\alpha_{\kappa\alpha}-\Gamma^\kappa_{\mu\alpha}\Gamma^\alpha_{\kappa\nu},}
\end{align}
and the Ricci scalar is given by $R=g^{\mu\nu}R_{\mu\nu}$.
As it turns out, for a general Lagrangian density that depends on the metric and its first and second derivatives (and other fields), the Euler-Lagrange equation with respect to the metric is given by
\begin{align}
{-\frac{1}{2}g^{\mu\nu}{\cal L}-\frac{2}{3}\dfrac{\partial{\cal L}}{\partial(\partial_\gamma\partial_\mu g_{\alpha\beta})}R_\beta{}^\nu{}_{\alpha\gamma}-\nabla_\lambda\nabla_\kappa\dfrac{\partial {\cal L}}{\partial(\partial_\lambda\partial_\kappa g_{\mu\nu})}=0.}
\end{align}
This indicates that the only derivative of the Lagrangian we need is with respect to second derivatives of the metric tensor. Specifically, for the Ricci-scalar multiplied by the volume element, this derivative is given by
\begin{align}
{\dfrac{\partial R\sqrt{-g}}{\partial(\partial_\lambda\partial_\kappa g_{\mu\nu})}=-\dfrac{\sqrt{-g}}{2}(2g^{\mu\nu}g^{\kappa\lambda}-g^{\mu\kappa}g^{\nu\lambda}-g^{\mu\lambda}g^{\nu\kappa}).}
\end{align}
Lovelock, David and Rund, Hanno: Tensors, Differential Forms, and Variational Principles, Dover Publications, 1989. | 2,197 | 6,266 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-26 | latest | en | 0.590213 |
https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Electric_Circuits_IV_-_Digital_Circuitry_(Kuphaldt)/02%3A_Binary_Arithmetic/2.04%3A_Binary_Subtraction | 1,721,377,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514900.59/warc/CC-MAIN-20240719074314-20240719104314-00172.warc.gz | 544,719,866 | 30,184 | # 2.4: Binary Subtraction
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
We can subtract one binary number from another by using the standard techniques adapted for decimal numbers (subtraction of each bit pair, right to left, “borrowing” as needed from bits to the left). However, if we can leverage the already familiar (and easier) technique of binary addition to subtract, that would be better. As we just learned, we can represent negative binary numbers by using the “two’s complement” method and a negative place-weight bit. Here, we’ll use those negative binary numbers to subtract through addition. Here’s a sample problem:
If all we need to do is represent seven and negative five in binary (two’s complemented) form, all we need is three bits plus the negative-weight bit:
Since we’ve already defined our number bit field as three bits plus the negative-weight bit, the fifth bit in the answer (1) will be discarded to give us a result of 00102, or positive two, which is the correct answer.
Another way to understand why we discard that extra bit is to remember that the leftmost bit of the lower number possesses a negative weight, in this case equal to negative eight. When we add these two binary numbers together, what we’re actually doing with the MSBs is subtracting the lower number’s MSB from the upper number’s MSB. In subtraction, one never “carries” a digit or bit on to the next left place-weight.
Let’s try another example, this time with larger numbers. If we want to add -2510 to 1810, we must first decide how large our binary bit field must be. To represent the largest (absolute value) number in our problem, which is twenty-five, we need at least five bits, plus a sixth bit for the negative-weight bit. Let’s start by representing positive twenty-five, then finding the two’s complement and putting it all together into one numeration:
Essentially, we’re representing negative twenty-five by using the negative-weight (sixth) bit with a value of negative thirty-two, plus positive seven (binary 1112).
Now, let’s represent positive eighteen in binary form, showing all six bits:
Since there were no “extra” bits on the left, there are no bits to discard. The leftmost bit on the answer is a 1, which means that the answer is negative, in two’s complement form, as it should be. Converting the answer to decimal form by summing all the bits times their respective weight values, we get:
Indeed -710 is the proper sum of -2510 and 1810.
This page titled 2.4: Binary Subtraction is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform. | 2,356 | 7,057 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.196497 |
https://cs.nyu.edu/courses/fall03/V22.0002-001/morn2.html | 1,516,248,385,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887065.16/warc/CC-MAIN-20180118032119-20180118052119-00135.warc.gz | 655,435,877 | 1,464 | SECOND EXAM will bw on Dec 1, 2003 .
1. Write a program that reads a number and then prints the sum of its divisors. The divisor of a number is one that when the number is divided by it, the remainder is zero. Thus if the number read is 16, the program would print the sum of 1 + 2 + 4 + 8, or 15.
2. Write a program that reads characters until it encounters a period. It then prints the character read before the period. Thus if the input is {\tt asd5f.}, the program prints
3. What does the following program produce?
```public class Prog3
{
public void plot(int j)
{
for(int i = 0; i < j; i++)
{
for(int k =0; k <= j; k++)
System.out.print(' ');
System.out.println(j);
}
public static void main(String[] asd)
{
int digit;
for(char number = '0'; number <= '3'; number++)
{
digit = number - '0';
plot(digit);
}
}
}
```
4. Evaluate the following (you may use the table on the board):
```(a) (7 >= 3) || ! 9 == 8) (b) !(4 >= 5) && !(3 < 3)
(c) !((5 == 9) || !(3 > 2)) (d) (6 <= 8) && (3 < 5) && (3 = 3)
```
5. What does the following code produce?
```public class WhatDoIDo
{
public static void main (String [] args)
{
print(8);
print(3);
}
public static void print(int number)
{
int index, uScores, numToShow;
numToShow = formula(number);
for(index = 1; index <= number; index++)
{
for(uScores = 1; uScores <= index; uScores ++)
System.out.print(``_'');
System.out.println(numToShow);
}
}
public static int formula(int value)
{
return (value % 3) + 1;
}
}
```
6. Write a method that takes three integers as parameters and returns the largest of the three. | 473 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-05 | latest | en | 0.638888 |
https://alfasoft.com/en/alfasoft/news/mathcad-news/1198-mathcad-news-15.html | 1,652,891,701,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00466.warc.gz | 129,323,581 | 7,983 | [8.Mar.2022] What’s new in Mathcad Prime 8
PTC has recently released a new Mathcad version - Prime 8. So let’s take a look at the improvements.
1. Redefinition Warnings
With the redefinition warnings functionality in Mathcad Prime 8, warnings automatically appear on redefinitions in regions that contain them. You can turn warnings on and off for all or individual categories of identifier, so you can more closely manage your identifier use.
2. Show Frame (header, footer, page body)
Turn borders on for the header, footer, and page body calculation area of your worksheet. These borders appear in page view, printed documents, and worksheets saved in PDF format.
3. Partial Derivative Operator
Insert and use a partial derivative operator in your worksheet. From the Operators menu, use the new operator to write expressions in terms of partial derivatives.
4. Copy/paste multiple values to combo-box
Mathcad Prime 7 included a combo-box input control so you could control calculations in your worksheets with easy to create, easy-to-use drop-down menus of variable definitions.
Mathcad Prime 8 adds the ability to paste multiple variable values directly into the combo-box edit mode. Copy a selection of values from a matrix or a table, select that same size of data in a combo box, and paste. Or you can paste into one value box and the data will paste from that location down and to the right. In both cases, the combo-box automatically resizes to accommodate the pasted data.
5. Symbolics Engine Enhancements
PTC Mathcad Prime 8 continues to enhance the functionality of the new symbolics engine, which was introduced in PTC Mathcad Prime 6. In PTC Mathcad Prime 8 you’ll find:
• Symbolics integral transform enhancements. Significant improvement and expansion in use and performance, including added supported use cases, have been added for the following symbolic integral transforms:
• fourier and invfourier
• laplace and invlaplace
• ztrans and invztrans
• New nfact and ofact modifiers for use with fourier keyword, which allows you to define normalization and oscillatory factors.
• Improved performance and added supported use cases for calculus operators, including limits, derivative, and integrals.
• Improved performance and added supported use cases for keywords, including solve, rewrite, series, simplify, assume.
• Improvement in undefined variable handling in symbolic results, including better auto-labelling of free undefined variables and variables in lambda expressions.
6. Numeric Engine Enhancements
• Updated physical constants. In alignment with NIST and CODATA, we've updated several physical constants, including the following:
• Planck’s constant
• Boltzmann constant
• Reduced Plank`s constant
• Atomic mass unit
• Molar gas constant
• Fine structure constant
• Elementary charge
• Stefan–Boltzmann constant
• Magnetic flux quantum
• Rydberg constant
• Clearer error messages related to behavior of functions statespace and odesolve when not properly configured.
7. Usability Enhancements
The following usability enhancements have been made:
• Ctrl/Wheel zoom can now be used to zoom in and out of a worksheet.
• Zoom slider now defaults to 5 % steps when you use the + and – controls.
• With the Close button on worksheet tabs, you can now close worksheets directly from the worksheet tab.
• Unsaved worksheet indicator: for worksheets that have yet to be saved, an asterisk will appear on the worksheet tab.
• Context menu on worksheet tabs: A new context menu is now available on all worksheet tabs providing access to useful actions, including:
• Calculate
• Close (this worksheet, all worksheets, other worksheets, worksheets to the left/right, unchanged worksheets)
• Move tab
• Copy full file path
• Open containing folder
• Print
• Save as
• Drag and move worksheet tab: When multiple worksheets are open, a selected worksheet tab can now be dragged and dropped to arrange the order of worksheet tabs to the user’s preference.
• Clear text format command: You can now clear text formatting (font, font size, color, etc.) for text selections back to the default text format.
8. Miscellaneous
Finally, you’ll find these updates available, so your work remains useful moving into the future.
Save legacy worksheets as HTML | 904 | 4,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | longest | en | 0.823329 |
https://brainly.com/question/243655 | 1,484,928,893,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280835.60/warc/CC-MAIN-20170116095120-00561-ip-10-171-10-70.ec2.internal.warc.gz | 783,803,408 | 9,081 | 2015-01-07T09:00:25-05:00
72 = x% from 24
72 = * 24
72 = * 6
72 = / * 25 (both sides)
1800 = 6x / ÷ 6 (both sides)
x = 300
Answer: 72 is 300% of 24.
You can also do it easily in head as 72 is just as much as 3 times 24, so it has to be 300%. | 120 | 283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-04 | latest | en | 0.858345 |
https://muster-themes.net/questions/Trigonometry/1081042 | 1,709,056,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00383.warc.gz | 410,158,648 | 5,590 | # Find Amplitude, Period, and Phase Shift y=4sin((8x)/5-pi/2)
Find Amplitude, Period, and Phase Shift y=4sin((8x)/5-pi/2)
Use the form to find the variables used to find the amplitude, period, phase shift, and vertical shift.
Find the amplitude .
Amplitude:
Find the period of .
The period of the function can be calculated using .
Replace with in the formula for period.
is approximately which is positive so remove the absolute value
Multiply the numerator by the reciprocal of the denominator.
Cancel the common factor of .
Factor out of .
Factor out of .
Cancel the common factor.
Rewrite the expression.
Combine and .
Move to the left of .
Find the phase shift using the formula .
The phase shift of the function can be calculated from .
Phase Shift:
Replace the values of and in the equation for phase shift.
Phase Shift:
Multiply the numerator by the reciprocal of the denominator.
Phase Shift:
Multiply .
Multiply and .
Phase Shift:
Multiply by .
Phase Shift:
Phase Shift:
Move to the left of .
Phase Shift:
Phase Shift:
List the properties of the trigonometric function.
Amplitude:
Period:
Phase Shift: ( to the right)
Vertical Shift: None
Do you know how to Find Amplitude, Period, and Phase Shift y=4sin((8x)/5-pi/2)? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name one billion three hundred sixty million three hundred ninety-seven thousand seven hundred ninety-two
### Interesting facts
• 1360397792 has 128 divisors, whose sum is 11125179576
• The reverse of 1360397792 is 2977930631
• Previous prime number is 13
### Basic properties
• Is Prime? no
• Number parity even
• Number length 10
• Sum of Digits 47
• Digital Root 2
### Name
Name one billion one hundred seventy-nine million nine hundred twenty-three thousand seven hundred seventy-eight
### Interesting facts
• 1179923778 has 32 divisors, whose sum is 3432506112
• The reverse of 1179923778 is 8773299711
• Previous prime number is 3
### Basic properties
• Is Prime? no
• Number parity even
• Number length 10
• Sum of Digits 54
• Digital Root 9
### Name
Name four hundred fifty-three million seven hundred eighty-seven thousand four hundred ninety-one
### Interesting facts
• 453787491 has 16 divisors, whose sum is 626402304
• The reverse of 453787491 is 194787354
• Previous prime number is 47
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 9
• Sum of Digits 48
• Digital Root 3 | 624 | 2,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-10 | latest | en | 0.743214 |
https://www.doubtnut.com/question-answer-physics/as-shown-in-fig-ab-is-rod-of-length-30-cm-and-area-of-cross-section-10cm2-and-thermal-conductivity-3-644110289 | 1,643,073,232,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00319.warc.gz | 766,178,612 | 83,054 | HomeEnglishClass 11PhysicsChapterCalorimetry
<img src="https://d10lpgp6xz60...
# <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V06_C01_E01_139_Q01.png" width="80%"> As shown in Fig. AB is rod of length 30 cm and area of cross section 1.0cm^2 and thermal conductivity 336 SI units. The ends A and B are maintained at temperatures 20^@C and 40^@C, respectively .A point C of this rod is connected to a box D, containing ice at 0^@C through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is (assume latent heat of fusion for ice L_f=80 cal//g)
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 25-5-2021
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,
84mg//s84 g//s20 mg//s40 mg//s
<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V06_C01_E01_139_S01.png" width="80%"> <br> Thermal resistance of AC=(L)/(KA)=(0.1)/(336xx10^-4)=(10^3)/(336) <br> =R(say) <br> Thermal resistance of BC=(0.2)/(336xx10^-4)=2R <br> Heat flow rates are <br> H_1=(20)/(R ),H_2=(40)/(2R)=(20)/(R ) <br> H=H_1+H_2=(40)/(R )=(40xx336)/(10^3) <br> =(13440)/(10^3)=13.44W <br> Rate of melting of ice <br> =(H)/(L_f)=(13.44//4.2)/(80)g//s=40 mg//s | 469 | 1,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-05 | latest | en | 0.742678 |
https://quiz.jagranjosh.com/josh/quiz/index.php?attempt_id=545834 | 1,521,684,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647758.97/warc/CC-MAIN-20180322014514-20180322034514-00577.warc.gz | 693,030,757 | 13,474 | # Time, Distance & Speed
Are you preparing for campus placements,Banking,SSC, IAS, Insurance,Defence and other competitive exams? Then, make sure to take some time in practicing the Speed,time and disctance questions and answer in Quantitative Aptitude. Moreover, only those questions are included that are relevant and likely to be asked in any competitive exam. So, take these questions and answer, brush up your skills and practice to stay fully prepared for any your exam.
• Q1.How long a train of length 300 m running at 240 kmph will take to cross a motor van moving in the same direction at 24 kmph?
• Q2.Speeds of four buses are in the ratio 3:4:1:5. What is the ratio of the times taken by them?
• Q3.A worker reaches his factory 8 minutes late, if his speed from the house to the factory is 5 km/hr. If he walks at a speed of 10 km/hr, then he reaches the factory 7 minutes early. The distance of the factory from his house is:
• Q4.Four vans are hired at the rate of Rs. 5 per km plus the cost of petrol at Rs. 50 a litre. In this context, consider the details given in the following table:
Which van maintained the maximum average speed?
• Q5.A train travels at a certain average speed for a distance of 45 km and then travels a distance of 54 km at an average speed of 3 km/hr more than its original speed. If it takes 6 hours to complete the total journey, what is the original speed of the train in km/hr?
• Q6.A group of men decided to do a job in 10 days. But since 20 men dropped out every day, the job completed at the end of the 14th day. How many men were there at the beginning?
• Q7.A man rows 15 kmph in still water and it takes twice as long in upward stream as to downward stream. What is the rate of water current?
• Q8.A boy started cycling at a point in a square field and reached the diagonally opposite point after 15 minutes riding at a speed of 4 kmph. What is the area of the square field? | 475 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-13 | latest | en | 0.958378 |
http://mathhelpforum.com/math-topics/3639-correctness-sample-function.html | 1,524,759,037,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00211.warc.gz | 203,765,020 | 9,481 | # Thread: Correctness of a sample function
1. ## Correctness of a sample function
function f: {a, b, c} -> {1, 2 ,3} defined by f(a) = 1, f(b) = 1, f(c) = 2
if ~f(y) represents inverse function
is ~f({1,3}) = {a, b} correct? or should it be ~f({1}) = {a, b}
the former is printed in the book and i'm kinda confused why ~f({1, 3}) gives {a,b} since ~f({3}) gives NIL.
Appreciate any clarification
Thank you
2. Originally Posted by terencetham
function f: {a, b, c} -> {1, 2 ,3} defined by f(a) = 1, f(b) = 1, f(c) = 2
if ~f(y) represents inverse function
is ~f({1,3}) = {a, b} correct? or should it be ~f({1}) = {a, b}
the former is printed in the book and i'm kinda confused why ~f({1, 3}) gives {a,b} since ~f({3}) gives NIL.
Appreciate any clarification
Thank you
$\displaystyle f^{-1}(\{1,3\})=f^{-1}(\{1\})\cup f^{-1}(\{3\})=\{a,b\}\cup \emptyset=\{a,b\}$
where $\displaystyle \emptyset$ denotes the empty set, and so for any set $\displaystyle A$, $\displaystyle A \cup \emptyset=A$
RonL | 371 | 1,007 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-17 | latest | en | 0.736663 |
http://openstudy.com/updates/55eee517e4b03567e112993c | 1,495,798,325,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00486.warc.gz | 337,143,199 | 8,314 | • anonymous
Assume a and b are nonzero rational numbers and c and d are irrational numbers. For each of the following expressions, determine whether the result is irrational, rational, or both. Justify your answers. Part A: ac + d Part B: four b square root of three end square root plus c Part C: b2(c + d) what are the answers
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 322 | 1,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-22 | longest | en | 0.383763 |
http://www.mywordsolution.com/question/commissioner-of-the-national-hockey-league-wants/91611 | 1,527,052,215,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00580.warc.gz | 430,207,386 | 9,288 | +1-415-315-9853
info@mywordsolution.com
## Statistics
Applied Statistics
Data Set A: problems 1-3.
SAT GPA SES
SAT Pearson Correlation 1 0.778 0.232
Sig. (2-tailed) . 0.002 0.71
N 145200 145200 145200
GPA Pearson Correlation 0.778 1 0.424
Sig. (2-tailed) 0.002 . 0.081
N 145200 145200 145200
SES Pearson Correlation 0.232 0.424 1
Sig. (2-tailed) 0.71 0.081 .
N 145200 145200 145200
1. Find r xy if SAT = x and GPA = y ?
a. 0.778
b. 0.232
c. 0.002
d. 0.71
2. Which of the following correlations is important(p < .05)?
a. SAT and GPA
b. GPA and SES
c. SAT and SES
d. All the above
3. What is the variance accounted for in SAT due to GPA?
a. 40%
b. 60.5%
c. 5.4%
d. Less than 1%
Research Design A: problems 4-8
A researcher was interested in the effects of sexual arousal on the ability to concentrate, and also wondered if gender and age are important factors or not. The researcher had participants read passages which were low, medium, or high in sexual arousal content. The participants included both males and females and were divided into three age categories (18-24, 25-35, and 36-50 years). After reading the passage, concentration was measured by the proofreading task; researcher measured the number of errors detected on task.
4. What kind of statistical procedure must the researcher use to answer the problems?
a. Standard Deviation
b. Independent Sample t test
c. Correlation
d. Factorial Anova
5. Which of the following are the Independent Variables:
a. Gender, Sexual Arousal, Age
b. Gender, Age, Concentration
c. Sexual Arousal, Age, Concentration
d. Sexual Arousal, Concentration, Gender
6. Which of the following is Dependent Variable
a. Sexual Arousal
b. Gender
c. Concentration
d. Age
7. Which of the following represents design?
a. 3 x 3 x 2
b. 3 x 3
c. 2 x 3
d. 2 x 3 x 1
8. Which of the following represents the possible interaction?
a. Gender affects concentration
b. Sexual arousal affects concentration
c. Sexual arousal affects concentration but only for males
d. Only high levels of sexual arousal affects concentration
Data Set 2: problems 9-10
Commissioner of the National Hockey League wishes to know if offense (Goals_F), defense (Goals_A) and penalties (Pen_Min) predict winning (Tier).
Model Summary
Model R R Square Adjusted R Square Std. Error of the Estimate
1 .581(a) .338 .314 .966
2 .728(b) .531 .496 .828
3 .734(c) .538 .485 .837
a Predictors: (Constant), Goals_F
b Predictors: (Constant), Goals_F, Goals_A
c Predictors: (Constant), Goals_F, Goals_A, Pen_Min
ANOVA(d)
Model Sum of Squares df Mean Square F Sig.
1 Regression 13.336 1 13.336 14.290 .001(a)
Residual 26.131 28 .933
Total 39.467 29
2 Regression 20.945 2 10.473 15.267 .000(b)
Residual 18.521 27 .686
Total 39.467 29
3 Regression 21.245 3 7.082 10.104 .000(c)
Residual 18.222 26 .701
Total 39.467 29
a Predictors: (Constant), Goals_F
b Predictors: (Constant), Goals_F, Goals_A
c Predictors: (Constant), Goals_F, Goals_A, Pen_Min
d Dependent Variable: Tier
9. If he wants to describe maximum variability (Adjusted R squared) in winning then which one must he choose?
a. Model 1
b. Model 2
c. Model 3
10. How much variability in winning is NOT described by the best model (coefficient of alienation converted to a percentage)?
a. 31.4 %
b. 49.6%
c. 68.6%
d. 50.4%
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M91611
Have any Question?
## Related Questions in Statistics and Probability
### Case study mikhail gorbachevs 1988 un speechif the pace of
Case Study: Mikhail Gorbachev's 1988 UN SpeechIf the pace of improving US-Soviet relations seemed rapid, Mikhail Gorbachev's speech to the United Nations General Assembly would shift the process into overdrive. In this ...
### Case study mikhail gorbachevs 1988 un speechif the pace of
Case Study: Mikhail Gorbachev's 1988 UN SpeechIf the pace of improving US-Soviet relations seemed rapid, Mikhail Gorbachev's speech to the United Nations General Assembly would shift the process into overdrive. In this ...
### Consider rolling a fair 6-faced die twice let a be the
Consider rolling a fair 6-faced die twice. Let A be the event that the sum of the two rolls is at most 6, and B be the event that the first one is an even number. (a) What is the probability that the sum of the two rolls ...
### Within a forensic investigation a sample was independently
Within a forensic investigation, a sample was independently analyzed for lead content by two different laboratories, Lab 1 and Lab 2. The analysis that was performed on this sample by Lab 1 yielded a coefficient of varia ...
### Why is a z score a standard score why can standard scores
Why is a z score a standard score? Why can standard scores be used to compare scores from different distributions?
### What is the variation in lsquofalse rejects that is not
What is the variation in ‘False rejects' that is NOT attributable to the Inspectors (as Sum of squares, not as percentage or mean)? 349.4 84.9 264.5 10.58 21.2
### Probabilities of graduation and publicationprofessors have
Probabilities of Graduation and Publication Professors have hundreds of students in their classes each year. Some professors teach only upper division courses to students who are in their major course of study. Of this g ...
### Question the volume of soft drink in plastic bottles is a
Question: The volume of soft drink in plastic bottles is a normal random variable with mean 16 ounces and standard deviation 0.6 ounces. a.) If a bottle is selected at random, find the probability that it contains more t ...
### The borda and condorcet methods- the following example was
The Borda and Condorcet Methods :- The following example was presented by Condorcet, in a critique of the Borda Method. A committee composed of 81 members is to choose a winner from among three candidates, A, B, and C. T ...
### Consider the following gameyou pay 100 and a fair 6-sided
Consider the following game. You pay \$1.00 and a fair 6-sided die is rolled. If the roll is 1 or 3, then you get your dollar back and the game ends. If the roll is 5, you get \$2.00 back (your dollar plus an additional do ...
• 13,132 Experts
## Looking for Assignment Help?
Start excelling in your Courses, Get help with Assignment
Write us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.
### WalMart Identification of theory and critical discussion
Drawing on the prescribed text and/or relevant academic literature, produce a paper which discusses the nature of group
### Section onea in an atwood machine suppose two objects of
SECTION ONE (a) In an Atwood Machine, suppose two objects of unequal mass are hung vertically over a frictionless
### Part 1you work in hr for a company that operates a factory
Part 1: You work in HR for a company that operates a factory manufacturing fiberglass. There are several hundred empl
### Details on advanced accounting paperthis paper is intended
DETAILS ON ADVANCED ACCOUNTING PAPER This paper is intended for students to apply the theoretical knowledge around ac
### Create a provider database and related reports and queries
Create a provider database and related reports and queries to capture contact information for potential PC component pro | 2,119 | 8,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.581877 |
https://www.wiziq.com/online-tests/5496-iit-jee-pmt-module-test-5-chemistry-bias-some-basic-concepts-of-chemistry | 1,653,817,289,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00035.warc.gz | 1,235,409,788 | 36,175 | # IIT JEE / PMT Module Test 5: Chemistry : State of Matters
Normal type of MCQ with one correct option Kinetic energy of a gas depends upon its
Molecular mass
Atomic mass
Equivalent mass
None of these
Normal type of MCQ with one correct option In the ideal gas equation, the gas constant R has the dimensions of
mole-atm/degree K
litre mole
litre-atm/degree K/mole
erg/degree K
Normal type of MCQ with one correct option Volume occupied by a gas at one atmospheric pressure and 0°C is V mL. Its volume at 273 K will be
V mL
V/2 mL
2 V
None of these
Normal type of MCQ with one correct option In deriving the kinetic gas equation, use is made of the root mean square velocity of the molecules because it is
The average velocity of the molecules
The most probable velocity of the molecules
Unaffected by change in temperature
The most accurate form in which velocity can be used in these calculations.
/
100(C
<100(C
>100(C
0(C
/
100(C
>100(C
<100(C
25(C
/
V.P ( Oext.
V.P is independent of external pressure.
V.P = Pext.
Both A and R are true and R is correct explanation of A
Both A and R are true and R is not correct explanation of A
A is true, R is false
A is false, R is true
/
(A) ((R); , (B) ((Q); , (C) ((P); ,(D) ((S);
(A) ((S); , (B) ((Q); , (C) ((P); ,(D) ((R);
(A) ((S); , (B) ((R); , (C) ((P); ,(D) ((Q);
(A) ((P); , (B) ((Q); , (C) ((S); ,(D) ((R);
MCQ with more than one correct option To raise the volume of the gas by four times, which of the following method are correct? (a) T is doubled and P is also doubled (b) Keeping P constant T is raised by four times (c) Temperature is doubled and pressure is halved (d) Keeping temperature constant pressure is educed to of its initial value
A,B,C
B, C ,D
C,D
A,C & D
Join the PMT Medical Entrance ExamCommunity
Description:
This is the fifth in the series of FREE Tests; it's a module test of Chemistry. Topic is State of Matter, in this the syllabus of IIT JEE and PMT both are covered. Students are requested to keep a check for the Paid Test Series at WiZiQ.com at a nominal rate.
Discussion
Really!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
2896 days 18 hours 58 minutes ago
3963 days 6 hours 15 minutes ago
SPARKX Chemistry
Chemistry in Chandigarh for Engineering & Medical entrance
6 Members Recommend
309 Followers
### More Tests By Author
IIT JEE / PMT Test Series : Module 7 : +1 : Chemistry : Chemical Equilibrium
10 Questions | 1550 Attempts
IIT JEE / PMT Test Series : Module 4 : +1 : Physics : Work, Power & Energy
10 Questions | 3419 Attempts
IIT JEE / PMT Test Series : Module 3 : +1 : Chemistry : Classification of Elements
10 Questions | 1077 Attempts
Price:\$10.61
Price:\$175
Price:\$100 | 763 | 2,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.910509 |
https://www.coursehero.com/file/6478833/20-Chapter-25-HomeworkCH25-Optical-Instruments/ | 1,524,812,483,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949489.63/warc/CC-MAIN-20180427060505-20180427080505-00521.warc.gz | 768,863,777 | 55,758 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
20_Chapter 25 HomeworkCH25 Optical Instruments
# 20_Chapter 25 HomeworkCH25 Optical Instruments - 356...
This preview shows page 1. Sign up to view the full content.
356 CHAPTER 25 25.43 A fringe shift occurs when the mirror moves distance 4 λ . Thus, the distance moved (length of the bacterium) as 310 shifts occur is 9 5 650 10 m 310 5.04 10 m 50.4 m 4 4 shifts L N λ µ × = = = × = 25.44 A fringe shift occurs when the mirror moves distance 4 λ . Thus, the distance the mirror moves as 250 fringe shifts are counted is 9 5 632.8 10 m 250 3.96 10 m 39.6 m 4 4 shifts L N λ µ × = = = × = 25.45 When the optical path length that light must travel as it goes down one arm of a Michelson’s interferometer changes by one wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change in path length). The number of wavelengths (in a vacuum) that fit in a distance equal to a thickness t is vac N t λ = . The number of wavelengths that fit in this thickness while traveling through the transparent material is ( ) n n N t t n nt λ λ = = = ( ) λ . Thus, the change number of wavelengths that fit in the path down this arm of the interferometer is
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]} | 362 | 1,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-17 | latest | en | 0.843608 |
http://euler-math-toolbox.de/Programs/Examples/Geometry%20on%20the%20Earth.html | 1,553,072,045,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202324.5/warc/CC-MAIN-20190320085116-20190320111116-00109.warc.gz | 70,830,084 | 3,834 | Euler Math Toolbox - Examples
# Geometry on the Earth
by R. Grothmann
In this notebook, we want to do some spherical computations. The functions are contained in the file "spherical.e" in the examples folder.
We need to load that file first.
```>load "spherical.e";
```
To enter a geographical position, we use a vector with two coordinates in radians (north and east, negative values for south and west). The following are the coordinates for the Campus of the Catholic University in Eichstätt.
```>EI=[rad(48,53.371),rad(11,11.330)]
```
```[0.853283, 0.195282]
```
You can print this position with sposprint (spherical position print).
```>sposprint(EI)
```
```N 48°53.371' E 11°11.330'
```
```>IN=[rad(48,45.854),rad(11,25.055)]; ND=[rad(48,44.248),rad(11,10.712)];
>sposprint(IN), sposprint(ND),
```
```N 48°45.854' E 11°25.055'
N 48°44.248' E 11°10.712'
```
First we compute the vector from one to the other on an ideal ball. This vector is [heading,distance] in radians. To compute the distance on the earth, we multiply with the earth radius at a latitude of 48°.
```>br=svector(EI,IN); deg(br[1]), br[2]*rearth(48°)->km
```
```129.671788237
21.766496146
```
This is a good approximation. The following routines use even better approximations. On such a short distance the result is almost the same.
```>esdist(EI,IN)->" km"
```
```21.7654976508 km
```
There is a function for the heading, taking the elliptical shape of the earth into account. Again, we print in an advanced way.
```>sdegprint(esdir(EI,IN))
```
``` 129.67°
```
The angle of a triangle exceeds 180° on the sphere.
```>asum=sangle(IN,EI,ND)+sangle(EI,IN,ND)+sangle(IN,ND,EI); deg(asum)
```
```180.000207484
```
This can be used to compute the area of the triangle. Note: For small triangles, this is not accurate due to the subtraction error in asum-pi.
```>(asum-pi)*rearth(48°)^2->" km^2"
```
```146.771319034 km^2
```
There is a function for this, which uses the mean latitude of the triangle to compute the earth radius, and takes care of rounding errors for very small triangles.
```>esarea(IN,EI,ND)->" km^2"
```
```146.757696278 km^2
```
We can also add vectors to positions. A vector contains the heading and the distance, both in radians. To get a vector, we use svector. To add a vector to a position, we use saddvector.
```>v=svector(EI,IN); sposprint(saddvector(EI,v)), sposprint(IN),
```
```N 48°45.854' E 11°25.055'
N 48°45.854' E 11°25.055'
```
These functions assume an ideal ball. The same on the earth.
```>sposprint(esadd(EI,esdir(EI,IN),esdist(EI,IN))), sposprint(IN),
```
```N 48°45.854' E 11°25.055'
N 48°45.854' E 11°25.055'
```
Let us turn to a larger example, which I found in the Internet, Brandenburger Tor in Berlin, and the Tejo bridge in Lissabon. For Lissabon, take a negative value for the west coordinate.
```>Berlin=[52.5164°,13.3777°]; Lissabon=[38.692668°,-9.177944°];
>sposprint(Berlin), sposprint(Lissabon)
```
```N 52°30.984' E 13°22.662'
N 38°41.560' W 9°10.677'
```
According to Google Earth, the distance is 2315.75km. We get a good approximation.
```>esdist(Lissabon,Berlin)->" km"
```
```2313.67485976 km
```
The heading is the same as the one computed in Google Earth.
```>degprint(esdir(Lissabon,Berlin))
```
```41°3'5.08''
```
However, we do no longer get the exact target position, if we add the heading and distance to the orginal position. This is so, since we do not compute the inverse function exactly, but take an approximation of the earth radius along the path.
```>sposprint(esadd(Berlin,esdir(Berlin,Lissabon),esdist(Berlin,Lissabon)))
```
```N 38°41.557' W 9°10.680'
```
The error is not large, however.
```>sposprint(Lissabon),
```
```N 38°41.560' W 9°10.677'
```
Of course, we cannot sail with the same heading from one destination to another, if we want to take the shortest path. Imagine, you fly NE starting at any point on the earth. Then you will spiral to the north pole. Great circles do not follow a constant heading!
The following computation shows that we are way off the correct destination, if we use the same heading during our travel.
```>dist=esdist(Berlin,Lissabon); hd=esdir(Berlin,Lissabon);
```
Now we add 10 times one-tenth of the distance, using the heading to Lissabon, we got in Berlin.
```>p=Berlin; loop 1 to 10; p=esadd(p,hd,dist/10); end;
```
The result is far off.
```>sposprint(p), skmprint(esdist(p,Lissabon))
```
```N 41°0.566' W 12°5.348'
357.810km
```
As another example, let us take two points on the earth at the same lattitude.
```>P1=[30°,10°]; P2=[30°,50°];
```
The shortest path from P1 to P2 is not the circle of lattitude 30°, but a shorter path starting 10° further north at P1.
```>sdegprint(esdir(P1,P2))
```
``` 79.69°
```
But, if we follow this compass reading, we will spiral to the north pole! So we must adjust our heading along the way. For rough purposes, we adjust it at 1/10 of the total distance.
```>p=P1; dist=esdist(P1,P2); ...
loop 1 to 10; dir=esdir(p,P2); sdegprint(dir), p=esadd(p,dir,dist/10); end;
```
``` 79.69°
81.67°
83.71°
85.78°
87.89°
90.00°
92.12°
94.22°
96.29°
98.33°
```
The distances are not right, since we will add a bit off error, if we follow the same heading for too long.
```>skmprint(esdist(p,P2))
```
``` 0.203km
```
We get a good approximation, if we adjust out heading after each 1/100 of the total distance from Berlin to Lissabon.
```>p=Berlin; dist=esdist(Berlin,Lissabon); ...
loop 1 to 100; p=esadd(p,esdir(p,Lissabon),dist/100); end;
>skmprint(esdist(p,Lissabon))
```
``` 0.057km
```
For navigational purposes, we can get a sequence of GPS position along the great circle to Lissabon with the function navigate.
```>load spherical; v=navigate(Berlin,Lissabon,10); ...
loop 1 to rows(v); sposprint(v[#]), end;
```
```N 52°30.984' E 13°22.662'
N 51°21.596' E 10°34.058'
N 50°8.382' E 7°53.992'
N 48°51.693' E 5°22.083'
N 47°31.853' E 2°57.902'
N 46°9.158' E 0°40.991'
N 44°43.881' W 1°29.117'
N 43°16.269' W 3°32.890'
N 41°46.547' W 5°30.784'
N 40°14.916' W 7°23.240'
N 38°41.560' W 9°10.677'
```
We write a function, which plots the earth, the two positions, and the positions in between.
```>function testplot ...
useglobal;
plotearth;
plotpos(Berlin,"Berlin"); plotpos(Lissabon,"Lissabon");
plotposline(v);
endfunction
```
Now plot everything.
```>plot3d("testplot",>own,>user,zoom=4):
```
Or use plot3d to get an anaglyph view of it. This looks really great with red/cyan glasses.
```>plot3d("testplot",own=1,anaglyph=1,zoom=4):
```
Examples | 2,196 | 6,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-13 | latest | en | 0.815485 |
http://www.knowingpains.com/tag/adwords-expert/ | 1,534,370,646,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210362.19/warc/CC-MAIN-20180815220136-20180816000136-00043.warc.gz | 521,247,156 | 11,865 | ## Ecpm FormulaEcpm Formula
Finance Formulas / July 29, 2018 / Avalynn Orr
Business valuation is a process and a set of procedures used to estimate the economic value of an owner's interest in a business. Valuation is used by financial market participants...
## Income Approach FormulaIncome Approach Formula
Finance Formulas / August 5, 2018 / Kenley Hopper
For example, if a company had \$150,000 in revenues and \$50,000 in explicit costs, its accounting profit would be \$100,000. The same company also had \$25,000 in implicit, or opportunity...
## Coupon Rate FormulaCoupon Rate Formula
Finance Formulas / August 4, 2018 / Alia Marquez
It's important to note that the CFS is distinct from the income statement and balance sheet because it does not include the amount of future incoming and outgoing cash that...
## Break Even Point FormulaBreak Even Point Formula
Finance Formulas / August 5, 2018 / Avalynn Orr
Instead of calculating interest on a finite number of periods, such as yearly or monthly, continuous compounding calculates interest assuming constant compounding over an infinite number of periods. Even with...
### Cpc FormulaCpc Formula
Finance Formulas / August 4, 2018 / Aniyah Booth
Common shareholders expect to obtain a certain return on their equity investment in a company. The equity holders' required rate of return is a cost from the company's perspective because...
#### Solvency Ratio FormulaSolvency Ratio Formula
Finance Formulas / August 4, 2018 / Alia Marquez
Since bonds are an essential part of the capital markets, investors and analysts seek to understand how the different features of a bond interact in order to determine its intrinsic...
##### Annual Growth Rate FormulaAnnual Growth Rate Formula
Finance Formulas / August 5, 2018 / Alia Marquez
A lower debt ratio usually implies a more stable business with the potential of longevity because a company with lower ratio also has lower overall debt. Each industry has its...
###### Contribution Margin Per Unit FormulaContribution Margin Per Unit Formula
Finance Formulas / August 5, 2018 / Briana Leonard
So how do you know if you’re spending the right amount? You need some numbers. First, you need to know how long the average customer sticks with you before they... | 496 | 2,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-34 | longest | en | 0.90899 |
https://ptcouncil.net/how-many-ml-is-40-oz/ | 1,656,340,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00166.warc.gz | 516,844,079 | 6,583 | Converting 40 oz come ml have to be easy, however it’s really not that simple. That’s because “oz” describes the English measurement ounce, and that’s a ax that have the right to actually mean various things in different contexts. It’s part of the weirdness and unwieldy nature the English dimensions that has actually caused many of the world to transform to the metric system. A liter is 1,000 ml. That’s simple enough come understand. However you need 437.5 grains to do an ounce, and also 16 ounces to make a pound. Therefore if you’re walk to market or desire to convert 40 oz come ml, many people approximately the human being may no have an exact idea the what you’re talk about. So v this internet page, you’ll recognize the milliliter (ml) identical of 40 fl oz.
You are watching: How many ml is 40 oz
## 40 OZ to ML Converter
How lot is 40 oz in ml? right here you will uncover the converter routine to changer your systems from oz to ml (ounces come milliliters). The easy sufficient to use. Enter 40 (which is the amount you have actually in fluid ounces) in the field significant OZ.
Then pick the proper volume unit. You have actually 3 options here:
the us customary liquid ounce,the US liquid ounce because that food and nutrition labeling, andthe UK (Imperial) version.
The converter will automatically process the 40 oz to ml conversion and also you’ll obtain the wanted milliliter equivalent showing up on the ML field. It’s the simple.
Click the “Reset” button if you desire to do another conversion. You may have a different amount of liquid in liquid ounces that you desire to convert to milliliters (fl oz to ml conversion).
If you in a hurry and you want the answer right now, then right here they are:
If she using us customary liquid ounces, your 40 fl oz is equal to 1182,94 ml when rounded off.If you’re placing this ~ above a food label, you need to say the your 40 fl oz is the tantamount of 1200 ml.If you’re making use of UK or imperial ounces, then 40 oz amounts to 1136,52 ml rounded off.
Fluid ounces abbreviations: “fl oz”, “fl. Oz.”, “oz. Fl.”.
## 40 OZ come ML – Unit Definition
What is one Ounce? For this net page, we’re talking around fluid ounces (fl oz). We’re no talking around ounces that are provided to measure solids prefer metals. A liquid ounce is just to measure the volume that a liquid.
So what are those various fluid oz versions every about? The difficulty is that you an initial need to know what type of liquid ounce you’re using. If you a united state citizen, then you’re more than likely using the us customary liquid ounce (US fl oz).
If she in the us or putting in the milliliter identical in a food or nutrition label, yes sir a various ml equivalent. If you’re making use of UK or imperial fluid ounces (UK fl oz), girlfriend again have actually a slightly different ml equivalent.
What is a Milliliter? This one is simpler. As soon as you know how much fluid there is in a liter, friend only need to divide the by the conversion aspect which equates to 1,000. That results in 1 milliliter. It is what “milli-” means—1/1000. When you have 1,000 ml, that’s in reality 1 liter.
## 40 OZ come ML conversion Chart
If you don’t desire to use converter, below is quick handy chart you can use in instance you have to memorize switch factors. In case you want even much easier solution feel cost-free to bookmark this page.
QuantityUS Customary Fl ozUK (Imperial) Fl ozUS Nutrition labeling
40 fl oz1182,94 ml1136,52 ml1200 ml
## 40 OZ to ML
So just how did we obtain our results? how do you get the same outcomes using your very own calculator? It’s yes, really easy. Every you need to do is to first determine the type of fluid ounce (fl oz) measurement you have. Each type of fluid ounce (fl oz) has actually its own identical of ml. So girlfriend only need to multiply by the ml equivalent.
Thus, you begin by start the amount of liquid ounce you have. Here, we’re presume you have 40 ounces for this reason we have the right to use that as an example. Yet the same formulas use regardless that the number of fluid ounces girlfriend have.
### 40 us FL OZ come ML
Most of the moment if she an American, you’re using the conventional or customary US fluid ounce (fl oz). This method that 1 fluid ounce (fl oz) is specifically 29.5735295625 milliliters (ml). So main point 40 by 29.5735295625 and you obtain 1182,94 ml rounded off.
### 40 us FL OZ to ML for Food (Nutrition) Labeling
The dominion for US liquid ounces (US fl oz), however, changes when you’re placing the tantamount ml ~ above a food label. This time, 1 liquid ounce (fl oz) is same to 30 ml. Therefore you just multiply 40 by 30, and also you get 1200 ml. Sure, there’s a difference, yet for many purposes, it is inconsequential.
### 40 UK (Imperial) FL OZ to ML
Now the UK citizens have actually their own idea that what a fluid ounce is. For them, it’s the equivalent of 28.4130625 ml. Therefore you have to multiply 40 by 28.4130625 and also you’ll acquire 1136,52 ml.
## Convert 40 OZ come ML
So, why do you need to convert 40 oz to ml? There space several good reasons for that. One is that you can current the amount much more accurate for those who usage the metric system rather than the English system. No everyone understands the idea that ounces, pints, and also gallons.
The second reason is the you have the right to then make certain that everyone is top top the right page. As soon as you use liquid ounce, there might some confusion regarding what girlfriend mean. Room you making use of the customary US liquid once (US fl oz), the version for food labels, or the UK/Imperial ounce? when you’ve convert the figure to ml, it i do not care clear. A milliliter is a milliliter nevertheless of whereby your leader is based.
Most popular OZ come ML (ounces to milliliters) conversions:
## How numerous ML is 40 OZ?
Basically that somewhere in between 1136,52 come 1200 ml. For many situations, the difference may not mean all that much. However it may be a various story for a chemistry experiment or a an extremely exact food recipe.
In any kind of case, 40 fl oz in ml is:
1182,94 ml if you talking about US customary fluid ounce (US fl oz).1200 ml once you want to placed this equivalent on a food label.1136,52 ml if the data is indigenous a UK publishing (Imperial UK fl oz).
See more: What Is 3% Of 20 ? What Is 3/20 As A Percentage
You can always round off the figures if you want to do it more practical. So in the united state a 40-ounce petrol amount is yes, really 1182,94 ml. If the milk, put it at 1200 ml. If you in the UK, it yes, really is 1136,52 ml. In case you might need to perform one more oz come ml conversion, don’t forget to bookmark this page. | 1,658 | 6,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-27 | latest | en | 0.88886 |
https://www.jiskha.com/display.cgi?id=1286931107 | 1,516,504,336,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889917.49/warc/CC-MAIN-20180121021136-20180121041136-00151.warc.gz | 981,669,873 | 3,524 | posted by .
how do you draw a centimeter box with the dimensions 3 by 4 by 12 centimers which = 144 cubes?
make it 3 high, 4 wide, and 12 long.
## Similar Questions
1. ### math
so if my question is find the dimensions of a box that will hold twice as many cubes as a box that is 2 by 6 by 4 a] volume of original box: 48 b]volume of new box is :96 c] dimensions of new box is 12 by 4 by 2
how many boxes con you find that will hold two times as many cubes as a 2 by 3 by 4 box.
find two boxes that will hold half as many centimeter cubes as the 3 centimeters by 4 centimeters by 6 centimers box. Thank you
find two boxes that will hold HALF as many centimeter cubes as the 3 centimeters by 4 centimeters by 6 centimeters box. what would the dimensions of each box be. dimensions of the first box dimensions of the second box.
The factory wants you to build a box that will hold twice as many cubes.What are the dimensions of a box that contains two times as many cubes as a box that is 2 by 3 by 4 ?
how many boxes can you find that will hold two times as many cubes as a 2 by 3 by4 box.record each of the dimensions
7. ### math 3
elena packed 48 cubes into this box. each cube has edges that are 1 centimeter. how many layes of cubes did elena make?
8. ### math
A clerk is packing cubes into a box each cube has edges of 6 inches the dimensions of the box are 15 inches by 30 i.ches by 18 inches find how many cubes will fit in the box ? | 391 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-05 | latest | en | 0.973572 |
nenk.orispring.it | 1,575,908,581,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00048.warc.gz | 100,683,302 | 12,835 | Gave helpful tips to remember steps. Grading Assignments Inline Grading for Assignments ()When reviewing assignment submissions, you can provide a grade and feedback. The questions consist of algebra and trigonometry problems. 5/23/2019 MyOpenMath Assessment 2/10 #3 Points possible: 1. HOMEWORK 1 ANSWERS TO MOST PROBLEMS PEYAM RYAN TABRIZIAN Section 1. He drops your lowest quiz score. Online learning institutions such as MyopenMath has revolutionized the entire education system, where students can access MyopenMath solutions online. Problem 1 : Write the exponential equation 2 5 = 32 in logarithmic form. Note: Do not use the “Immediately” option if you are allowing multiple attempts per problem. A health assessment is a wide range of survey type question that you are asked to answer. Studying MATH 117 Elements of Statistics at Montgomery College? On StuDocu you find all the study guides, past exams and lecture notes for this course. Maack Scenario analysis has been used by the private sector for the last 25 years to manage risk and develop robust strategic plans in the face of an uncertain future. The book is also in paperback for \$20. Stuck with an online statistics class, online statistics exam, or online statistics test? Welcome to PayForMathHomework. 6970 Here's how to ftwimalaltn: cos @t23¥t ') pmortu and tasks) fat!. The Accuplacer Test is an adaptive test. At of the test they will see all of the questions with their answers and the correct answers marked, and they can print this out to use as a study guide. Pay attention to test dates! *Special very important note--This is how you could possibly make up a missed test: The final exam score may be used to replace one lowest or missing test score at the end of the semester. Geometry Calendar. It is primarily a web-based math assessment tool for delivery and automatic grading of math homework and tests. com: We will also use the on-line homework and help system at MyOpenMath. Make money helping your fellow students. No (by the vertical line test) 1. The course shell was built for use in WAMAP. Questions are algorithmically generated and numerical and math expression answers can be computer graded. Enroll In a New Course. Percents and Applications Meaning of Percents Percent means per hundred or parts of one hundred. Studies can include measures within the Assessment Center library as well as custom instruments entered by the researcher. Types of studies (experimental vs. Get myopenmath help at Studyfull. You will also be able to share this result on Facebook or by email. A researcher needs to estimate the mean height of corn in a eld containing 10,000 corn plants. Many questions in the MyOpenMath libraries are randomized, algorithmic questions. MyOpenMath 2 Module quizzes Individual 20 1 hour each On the day of the lecture of Weeks 3, 7, 9 & 11 To be completed online out of class 3 Final Exam Individual 50 2 hours Centrally scheduled exam period At centrally scheduled exam in Weeks 15-16 100% Assessment Task 1: Mathematical Investigations. The TestNav app needs to be restarted due to network connectivity issues. Anyone can see both the problem and it's answers INCREDIBLY clearly on the android. Just to be clear, we are not a publisher. You Try It! EXAMPLE 1. Student Help Skip to start of help topic. Student solutions shows all odds worked out step by step. He self-funded the site for the first year, then joined forces with Lumen Learning, who had been using MyOpenMath with the Kaleidoscope Project. An answer key is provided so that you may check your answers. Chemistry 121A General Chemistry Syllabus, national study on blended learning and carry through with math pre-test and post-test. Exam 1 Information Sheet. To become a Studypool Tutor, please fill out the form below and click APPLY NOW to start the application process. On this page you can find the answer to Mislay (4) clue from the Crossword Climber iOS/Android Game from AppyNation. Ideally, the sampling frame perfectly coincides with the target population. Loading Unsubscribe from Alaina Wright? GET THEM NOW / HOMEWORK ANSWER KEYS / FREE APPS - Duration: 5:02. The test can be taken at either of our testing centers. The score of the nal exam will not be dropped regardless of the result. The best way to explain how to cheat on homework is by offering a professional online writing & editing services established by the top UK & US writers who wish to help students with their performance. com is the place to go to get the answers you need and to ask the questions you want Describe three examples of how the medical assistant plays an important role in patient assessment. MyOpenMath, "a free, open source, online course management system for mathematics and other quantitative fields," is an exceptional example of a tool for sharing open ancillary materials. Instead, the answers used simpler terms like slower and faster, which seemed to be a pattern amongst students on IEPs. com - a premium homework service where we connect highly experienced experts and students looking for pay someone to take my online statistics class. A bicycle Wheel has radius R_ Let p be a point on the spoke Of a Wheel at a distance d the Center Of the Wheel. The app is actually designed to answer any kind of school question — science, history, etc. 7/5/2019 MyOpenMath Assessment 2/18 #2 Points possible: 1. 8047 tanecsce — —. Internet Math Assessment System lets you create and administer auto-graded math assessments to students. MyOpenMath 2 Module quizzes Individual 20 1 hour each On the day of the lecture of Weeks 3, 7, 9 & 11 To be completed online out of class 3 Final Exam Individual 50 2 hours Centrally scheduled exam period At centrally scheduled exam in Weeks 15-16 100% Assessment Task 1: Mathematical Investigations. MyOpenMath Home | My Classes Home | My Classes | Help | Help Course Forums Calendar Gradebook Home > M1001 Fall 2019_1st_8wk > Assessment Quiz #5 100 points possible 0/10 answered Question 8 Find the odds for rolling a 6-sided die and getting a 4. It is essential to lay a solid foundation in mathematics if a student is to be competitive in today's global market. We guide our students on how to get the authentic and best MyOpenMath answer key which is answer generator enabled to provide best assessment answers for your online math test, quiz and exams. AP Statistics Chapter 7 Practice FR Test: Sampling Distributions Show all work for the following on the answer sheet. I will collect the work you show on paper for all homework problems on test days. You will have to click on your answer. 9/6/2016 Classwork: Went over functions - deck of cards function activity with partner Homework: Determine if each set of ordered pairs is a function!. If you're behind a web filter, please make sure that the domains *. All comments are approved before they are posted. Answer Key (homework) Chapter Reviews. b) k-combinations from a set with n elements (with repetition) k-combinations from a set of n elements (without repetition) is an unordered collection of k not necessarily distinct elements taken from a given set. Take my online test? Yes we do. The most common measures of position are percentiles, quartiles, and standard scores (aka, z-scores). Thanks for all of the answers posted. Together, they help assess ecological risk and environmental and economic injury from contamination and ship groundings. We are talented experts in mathematics, science and technology. Answer to MyOpenMath Assessment hetps://www. Study groups: Many students find that working with classmates is the best way to learn and understand the material. We make no money on these printed editions. We are a team that strives to see our clients succeed in their. * Open a text editor. Decryption Decryption by the intended recipient of a ciphertext received that has been encrypted using the Shift Cipher is also very simple. Because this review was originally written for my Calculus students to use as a test of their algebra and/or trig skills it is generally in the form of a problem set. Most of the applications of derivatives are in the next chapter however there are a couple of reasons for placing it in this chapter as opposed to putting it into the next chapter with the other applications. This can help your child grow with his studies. org Our online math tutors are good in mathematics that involves Statistics and Calculus. Tests: There will be two midterms and a cumulative final test. A _ point-estimate _ is a single number that is used to predict a population _parameter _? Suppose the weight of a bag of carrots at the grocery is normally distributed with mean. XYZ Homework is an easy-to-use, feature-packed online teaching tool that helps students learn online. Because you are enrolled in Phys 111, you will be invited to participate in a research. Each week, students will complete the homework portion of this class in MyOpenMath. So, for example, if a student gets a question wrong they would get an easier one. On this page you can find the answer to Mislay (4) clue from the Crossword Climber iOS/Android Game from AppyNation. Sometimes the frame will be larger or smaller, depending on practical ways of getting in touch with each member of the sample. You will also be able to share this result on Facebook or by email. Myopenmath Homework help - post Assignments, Research Papers and Homework Questions. You can hire a professional math homework answers provider to help you with myopenmath answers key. Clicking just Practice Problems gives problems and answers. If you choose to enter your answers without using the Math Quill, always preview what it looks like before submitting your answer (if available--see #2 above), and also make sure you enter the correct letters. I just wanted to let you know what I did: I added it straight into a Module. Ask your myopenmath homework questions and get answers from verified experts. Use this type of question when more than one answer is correct. Grading: Online Homework 40% In Class Exercise 10% Midterm 20% Final 30% Overall Percentage Grade 90 100 % A 80 89 % B 70 79 % C 60 69 % D Cheating: Cheating is prohibited. With just your email and a couple clicks, any instructor can start evaluating a zyBook right now. Because "x-4" could be positive or negative we don't know if we should change the direction of the inequality or not. WAMAP provides ; algorithmically generated online homework and practice problems with automated grading of numerical and algebraic answers. In other words, each correct answer will add to a total score, however selecting a wrong answer has no impact on that score. Completing class notes and activities and the online homework is the best way to prepare for the tests. Students get immediate feedback after they submit an answer. Our guide on finding Mathxl answer key and test answer key are proficient in providing the official online quiz answers with ease. The test can be taken at either of our testing centers. Take the assessments without the use of the book or your notes and then check your answers. If you choose to enter your answers without using the Math Quill, always preview what it looks like before submitting your answer (if available--see #2 above), and also make sure you enter the correct letters. Online learning institutions such as MyopenMath has revolutionized the entire education system, where students can access MyopenMath solutions online. Many questions in the MyOpenMath libraries are randomized, algorithmic questions. Classroom Use. The test is the same for all jobs that require an assessment, but answers may be weighted differently depending on the job you applied for. This class was a hard class. The best way to understand the power and flexibility of zyBooks is to get in and work with a free evaluation zyBook. In 2011, David started MyOpenMath as a way to share with the world this software system and all the great content that had been built around open textbooks. If the revenue R is R(p) =. Math Course Reviews Reviews and answers below are available in PDF format. More Info about IMathAS. The questions consist of algebra and trigonometry problems. Engage students with immersive content, tools, and experiences. Sampling methods review. We are a team that strives to see our clients succeed in their. What is the area of a circle with circumference 18π ? (a) 9π (b) 324π (c) 18π (d) 81π (e) 36π. It is a course/learning management system and testing system featuring algorithmic questions, similar to WebWork, Webassign, and publisher systems like iLrn, MathXL, etc. Hence,tothenearesttenth: 113. Once you are playing, a problem will appear followed by three multiple choice answers. Note: Do not use the “Immediately” option if you are allowing multiple attempts per problem. Welcome and thank you for attending the 2019 Online Teaching Conference! Please help us improve your conference experience by… • Rating and reviewing our sessions and speakers within our OTC. Filip Mori c Question 1. Many questions in the MyOpenMath libraries are randomized, algorithmic questions. Designed by educators, Lumen Online Homework Manager (OHM) draws from thousands of interactive assessment questions to help students master math and other quantitative skills. The Math Quill makes it MUCH EASIER to enter you answers. In fact, the power rule is valid for any real number n and thus can be used to differentiate a variety of non-polynomial functions. Chapter 11. Tonnette Austin Home > Finite Math > Assessment Section 4. This is one way to make sure that your child get his daily dose of study time. To write an exponential function given a rate and an initial value, start by determining the initial value and the rate of interest. Math in Society is a free, open textbook. sure the correct answer is not easily predicted by the relationship to the distractors. You can always take the test as early as you would like. Consider the following set of numbers as a population: {1, 3, 5, 7, 9} a) Show all possible samples of size 2 from this population (there are 10). k-combinations from a set of n elements (without repetition) is an unordered collection of k distinct elements taken from a given set. Parent Guide to Google Play. Its use is provided free to Washington State public educational institution students and instructors. What is Lumen OHM? Forgot Password Forgot Username. Thank you Adam and Leah for putting in so much work to make this happen!. Question 1057775: Please Answer myOpenmath questions? Click on my name and answer the ones that are open. Whatever your goal, Learning Express Library's resources will help you succeed. com and in every module in the workbook. online learning institutions such as myopenmath has revolutionized the entire education system, where students can access myopenmath solutions online. Webassign Answers. Assessment Center SM is a online data collection tool that enables researchers to create study-specific websites for capturing participant data securely online. To write an exponential function given a rate and an initial value, start by determining the initial value and the rate of interest. Because you are enrolled in Phys 111, you will be invited to participate in a research. The test is the same for all jobs that require an assessment, but answers may be weighted differently depending on the job you applied for. Enroll In a New Course. Test digit Thetestdigitisgreaterthan5. However, I am always happy to answer any questions you have regarding grades. A variable is a symbol (usually a letter) that stands for a value that may vary. A rectangular piece of cardboard measuring 12 inches by 18 inches is to be made into a box w/ an open top by cutting equal-sized squares from each corner and folding up the sides. Because of the open license on the libraries of questions in MyOpenMath, you are also free to edit and create your own questions with the question writing tools. So, for example, if a student gets a question wrong they would get an easier one. ©a 62L0N1 c2t oK 9u tJaT lS GomfNtYwRadr9e a oLLfCY. One service for quick answers is answers. asked • 10/22/18 The grocery store has bulk pecans on sale, which is great since you're planning on making 9 pecan pies for a wedding. Integral from -7 to 7 of sqrt(49 - x^2) dx. When you click on a student's score in the gradebook, you will see the specific assessment that they received and the last answer they provided, as well as their score on each question and the attempts used. WAMAP/MyOpenMath. Completing class notes and activities and the online homework is the best way to prepare for the tests. com in cooperation with Lumen. HOMEWORK 1 ANSWERS TO MOST PROBLEMS PEYAM RYAN TABRIZIAN Section 1. MATHEMATICS TEST FOR STD/GRADE 10 [ CBSE , IGCSE , ICSE , GCSE ETC] 6 Questions | 930 Attempts Trigonometry, Statistics, Pythagoras theorem, Similar figures, Simultaneous equations, Polynomials, PYTHAGORAS THEOREM Contributed By: IGNATIUS GEORGE. NO real solutions because absolute value is always positive. looking at the solution steps and the answer. You should use them simply to help you determine at what skill level in math you should begin study. A researcher needs to estimate the mean height of corn in a eld containing 10,000 corn plants. Exam 1 Information Sheet. Sometimes the frame will be larger or smaller, depending on practical ways of getting in touch with each member of the sample. org are unblocked. com in cooperation with Lumen. The course shell was built for use in WAMAP. @JeffE the fact that the textbook is interactive is not irrelevant at all. See the calendar in Bb for specific dates. A bicycle Wheel has radius R_ Let p be a point on the spoke Of a Wheel at a distance d the Center Of the Wheel. Studying MATH 117 Elements of Statistics at Montgomery College? On StuDocu you find all the study guides, past exams and lecture notes for this course. Tonnette Austin Home > Finite Math > Assessment Section 4. If testing anxiety or math anxiety are a concern for you, we have campus resources to help. Graphic displays are useful for seeing patterns in data. September 3, 2019 Answer Keys; Make Notes About a Question. In fact, the power rule is valid for any real number n and thus can be used to differentiate a variety of non-polynomial functions. sure the correct answer is not easily predicted by the relationship to the distractors. sav file, select females (gender = 1) who give more than the average amount of. You will have multiple. The hallmark of authentic assessment is that the student's 'voice' or principles and standards as a practitioner is heard throughout the work that is done. View Your. Test digit Thetestdigitisgreaterthan5. 6970 Here's how to ftwimalaltn: cos @t23¥t ') pmortu and tasks) fat!. Attempt Statistics is in the drop down menu. Text-based alternatives are available for graph entry. You can select between a random sample and a stratified sample directly below the population and then generate a sample of ten animals. I will try to keep your averages current by updating them regularly. Please talk with me early in the term. For printed copies of our textbooks, click on the link: Stitz Zeager Textbooks on Lulu. Doing so would allow a student to complete the test, view the answers, then reenter the test (since more attempts are allowed) and put in the correct answers. by making our site your first choice to get myopenmath answers, you will not regret it. I will also use MyOpenMath to post the syllabus, my schedule, handouts, test information, grades, and announcements. We guide our students on how to get the authentic and best MyOpenMath answer key which is answer generator enabled to provide best assessment answers for your online math test, quiz and exams. The Accuplacer Test is an adaptive test. com is the place to go to get the answers you need and to ask the questions you want Describe three examples of how the medical assistant plays an important role in patient assessment. I will collect the work you show on paper for all homework problems on test days. Not sure if I have my answer right or wrong? Either way the program will not accept my answer (MyOpenMath). VitalSource Bookshelf is the world’s leading platform for distributing, accessing, consuming, and engaging with digital textbooks and course materials. Hi, my name is Heidi Ahern. To write an exponential function given a rate and an initial value, start by determining the initial value and the rate of interest. Stuck with an online statistics class, online statistics exam, or online statistics test? Welcome to PayForMathHomework. If you choose to enter your answers without using the Math Quill, always preview what it looks like before submitting your answer (if available--see #2 above), and also make sure you enter the correct letters. 243)(20) cos A 4 Aa 25. Each of our Learning Centers offers the practice tests, exercises, skill-building courses, eBooks, and information you need to achieve the results you wantat school, at work, or in life. Total attempts: o A stor. The hallmark of authentic assessment is that the student's 'voice' or principles and standards as a practitioner is heard throughout the work that is done. There is also a forum available on MyOpenMath where you can post questions to the whole class and answer each. Together, they help assess ecological risk and environmental and economic injury from contamination and ship groundings. Myopenmath Homework help - post Assignments, Research Papers and Homework Questions. Clicking just Practice Problems gives problems and answers. You can hire a professional math homework answers provider to help you with myopenmath answers key. Another great OER I've been using for my math classes is My Open Math. 7/4/2017 MyOpenMath Assessment Assignment 3 Name: Nakia Brown #1 Points possible: Total attempts: 3 The following transactions occurred for the month… Please help with the attached. Jones & Bartlett Learning 5 Wall Street Burlington, MA 01803 USA U. Comparing an example of each task-type reveals the difference between the information that can be gained from closed and open tasks. The Assessment and Restoration Division comprises NOAA biologists, toxicologists, ecologists, policy analysts, information specialists, attorneys, geologists, environmental engineers, and economists. All of the videos are closed captioned and ADA compliant. Graph Theory Spring 2013 Prof. 00, and whose percentages for the class are greater than 85. One of the limitations of using MyOpenMath is that the question libraries are smaller than those of its commercial competitors. If you already have a MyOpenMath account: 1) Log into myopenmath. PDF | In this paper, effect on the homework and exam performance of a set of online homework developed for use of Mathematics of Accounting and Finance students was explored. Parent Guide to Google Play. What does your body convert into energy? F Water G Food H Hair J Skin 3. Trignometry resources--video tutorials, interactive lessons and free calculators. KEEP IN MIND - • These sheets are not intended to be a short course. If you're behind a web filter, please make sure that the domains *. Before starting work on a website or application:. There are two due dates each week: Thursdays and Sundays at 11:59 PM Eastern Daylight Time. $$x^{3}=x\cdot x\cdot x$$ There are a couple of operations you can do on powers and we will introduce them now. I clicked the Add button and then clicked the External Tools option. View our entire zyBooks Catalog. Resources Aops Wiki 2018 AMC 8 Answer Key Page. Classroom Use. It will be 10 questions designed solely to test your differentiation skills without a calculator. If testing anxiety or math anxiety are a concern for you, we have campus resources to help. 7 Comparing Decimals We can compare two positive decimals by comparing digits in each place as wemove fromleft to right,place by place. — but the math thing is the slickest part. * Write the following snippet: [code] Your text</figcaption> </figure. The dates on the schedule are the LAST day to take the test. Home; Calculus 1 WebAssign Answers; Calculus 2 Webassign Answers; Calculus 3 Webassign Answers. WAMAP is a web based mathematics assessment and course management platform, primarily used for online homework. assignment and quizzes on myopenmath. 8 CHAPTER 1. Due date is Fri. How to prepare. Answer Key (homework) Chapter Reviews. Because of the open license on the libraries of questions in MyOpenMath, you are also free to edit and create your own questions with the question writing tools. Please close and relaunch it. Through this platform, we became integral parts being our willing test subjects. Clicking a section title downloads the entire section with examples, practice problems and answers. Below is a general guide for improving the accessibility of your web-based project. plagiarize. Once you open the assessment, the timer will start. - MAT 1010 - Intro. The third link is to the online homework system for the textbook. Students had the option to turn in homework assignments or use test grade to replace homework grade. In some cases, there is no tool, and it may not be obvious how you should type a math symbol. Online tutoring available for math help. How to copy homework questions from your book or website and email them to a tutor:. The results from the completed assessments should lead to change in. — but the math thing is the slickest part. What is the area of a circle with circumference 18π ? (a) 9π (b) 324π (c) 18π (d) 81π (e) 36π. COLLEGE LEVEL MATHEMATICS PRETEST This pretest is designed to give you the opportunity to practice the types of problems that appear on the college-level mathematics placement test. To view and print these files you must have Acrobat Reader. Answers to your most common questions for conducting your library research. MAT 167: Statistics, Test I: Chapters 1-4 SOLUTIONS p. * Write the following snippet: [code] Your text</figcaption> </figure. Sometimes the frame will be larger or smaller, depending on practical ways of getting in touch with each member of the sample. Are you an instructor who wants to adopt an open textbook, who feels online interactive homework is valuable, but doesn't want their students to have to pay an additional fee? Then read more about using MyOpenMath in the classroom. At of the test they will see all of the questions with their answers and the correct answers marked, and they can print this out to use as a study guide. Pay Someone to Guide Me take MyOpenMath Class. Exam 1 Practice Problems. One of the best things about the experience was thatI could learn and adjust what & how I wanted to gradeI could learn and adjust what & how I wanted to gradeas I viewed students' answers and they came up with things I had not anticipated. Students can receive immediate feedback on algorithmically generated questions with numerical or algebraic expression answers. What's statistics The term of statistics is commonly used in two senses. Text-based alternatives are available for graph entry. One can either use the table already created above, and find each letter of the ciphertext in the bottom row, and replace with the corresponding plaintext letter directly above it, or the recipient could create the inverse table, with the ciphertext. Therefore, this is a graph of a function. Ask your myopenmath homework questions and get answers from verified experts. MyOpenMath does not provide direct support for instructors or students. Tests: There will be two midterms and a cumulative final test. Calculus Help Please? 1) For an object whose velocity in ft/sec is given by v(t) = -t2 + 4, what is its displacement, in feet, on the interval t = 0 to t = 3 secs? (4 points) A) 7. TestGen is a test generator program from Pearson Education designed to help you create paper handouts, quizzes or tests for your students. AP Statistics Chapter 7 Practice FR Test: Sampling Distributions Show all work for the following on the answer sheet. Adam Gilbert created a template course on MyOpenMath. You can always take the test as early as you would like. Toll-Free: 800-832-0034 Switchboard: +1-978-443-5000 Fax: +1-978-443-8000 [email protected] Students get immediate feedback after they submit an answer. If you are working through this material on your own, use these assessments to test your understanding of the lesson concepts. It is primarily a web-based math assessment tool for delivery and automatic grading of math homework and tests. Self-Assessment. Completing class notes and activities and the online homework is the best way to prepare for the tests. Exams/Quizzes are timed, so you will have to complete them in one sitting. Students can receive immediate feedback on algorithmically generated questions with numerical or algebraic expression answers. Descriptive Statistics Calculator - Find Arithmetic mean, mode, median, minimum, maximum of a data set. 9 27) 7 f2 9 29) g f2 e7 g e5 31) t f10 f6 t e1 5 f9 33) i f2 i f i 35) 9 j1 e4 10 e3 37) 8 f4 ; 8 e32 39) 8 : j e9 ; 8 6 e72 41) 67 : f g e6 7 6 e42. CHAPTER 7: PERCENTS AND APPLICATIONS This means that an 85% on a test represents 85 out of 100 This is the answer as an improper fraction percent. Online learning institutions such as MyopenMath has revolutionized the entire education system, where students can access MyopenMath solutions online. assessment, including the amount of time you have to complete, will be in your weekly notes and will be listed on MyOpenMath. It is a widely popular application among WA CTC faculty, used by over 16,000 students during 2014 winter quarter alone. Carbon Dioxide: A Greenhouse Gas Showdown 5 years ago. Through this platform, we became integral parts being our willing test subjects. I would greatly appreciate it. 9166 Sentence: Sample 1 (after) was 2. Our online math tutors are good in mathematics that involves Statistics and Calculus. Automation at Grading Tests and HW’s. HANDOUTS: Unit 1. com in cooperation with Lumen. 7 hundredths = _____ % (56. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. looking at the solution steps and the answer. 1: Graphing Linear Equations and Inequalities. 7 Comparing Decimals We can compare two positive decimals by comparing digits in each place as wemove fromleft to right,place by place. How to Cheat On a Math Test. Completing class notes and activities and the online homework is the best way to prepare for the tests. MyOpenMath Assessment #1 Alaina Wright. | 6,719 | 30,650 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-51 | latest | en | 0.911683 |
https://www.airmilescalculator.com/distance/sbn-to-cmh/ | 1,603,958,135,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00634.warc.gz | 615,971,569 | 19,509 | # Distance between South Bend, IN (SBN) and Columbus, OH (CMH)
Flight distance from South Bend to Columbus (South Bend International Airport – John Glenn Columbus International Airport) is 215 miles / 346 kilometers / 187 nautical miles. Estimated flight time is 54 minutes.
Driving distance from South Bend (SBN) to Columbus (CMH) is 254 miles / 408 kilometers and travel time by car is about 5 hours 3 minutes.
## Map of flight path and driving directions from South Bend to Columbus.
Shortest flight path between South Bend International Airport (SBN) and John Glenn Columbus International Airport (CMH).
## How far is Columbus from South Bend?
There are several ways to calculate distances between South Bend and Columbus. Here are two common methods:
Vincenty's formula (applied above)
• 214.787 miles
• 345.666 kilometers
• 186.645 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 214.489 miles
• 345.187 kilometers
• 186.386 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A South Bend International Airport
City: South Bend, IN
Country: United States
IATA Code: SBN
ICAO Code: KSBN
Coordinates: 41°42′31″N, 86°19′2″W
B John Glenn Columbus International Airport
City: Columbus, OH
Country: United States
IATA Code: CMH
ICAO Code: KCMH
Coordinates: 39°59′52″N, 82°53′30″W
## Time difference and current local times
There is no time difference between South Bend and Columbus.
EDT
EDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 57 kg (125 pounds).
## Frequent Flyer Miles Calculator
South Bend (SBN) → Columbus (CMH).
Distance:
215
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
215
Round trip? | 484 | 1,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-45 | latest | en | 0.813572 |
https://math.stackexchange.com/questions/1325167/proving-or-disproving-a-function-that-is-onto-itself-is-one-to-one | 1,571,863,124,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00060.warc.gz | 576,629,362 | 31,060 | # Proving or Disproving a function that is onto itself is one to one.
I'm having some trouble formulating a proof for this following problem:
A is a finite set and f a function with f : X → X. Suppose that f is onto.
Now Prove or Disprove: f is one to one.
This makes sense to me logically, and I believe it to be true. I've been trying to directly prove it, but maybe proof by contradiction is better?
Thanks
Suppose $|Y| = n$. Let $F:Y \to Y$ be onto. Suppose that $F$ is not injective. Then, there exists some $t \in Y$ such that $a, b \in Y$, $a \neq b$ and $F(a) = F(b) = t$. Then, at most $n -1$ elements can be mapped to (Since $F$ is a function, at most $n$ elements can be mapped to. Since two elements are mapped to the same element, only $n -1$ can be mapped to) $\implies$ $F$ is not onto. Contradiction. | 243 | 822 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | latest | en | 0.940788 |
https://www.numbersaplenty.com/15003100466219 | 1,624,615,875,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00476.warc.gz | 828,613,751 | 3,460 | Search a number
15003100466219 = 97521296873587
BaseRepresentation
bin1101101001010010111001…
…1110010100110000101011
31222010021121212102101022202
43122110232132110300223
53431302322204404334
651524155035210415
73105636546545525
oct332245636246053
958107555371282
1015003100466219
1148648630a0758
12182384bba6a0b
1384aa2b67ba88
1439c222590a15
151b03e995117e
hexda52e794c2b
15003100466219 has 8 divisors (see below), whose sum is σ = 15186865267728. Its totient is φ = 14819929413120.
The previous prime is 15003100466213. The next prime is 15003100466227. The reversal of 15003100466219 is 91266400130051.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 15003100466219 - 236 = 14934380989483 is a prime.
It is a super-2 number, since 2×150031004662192 (a number of 27 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (15003100466213) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 148386257 + ... + 148487330.
It is an arithmetic number, because the mean of its divisors is an integer number (1898358158466).
Almost surely, 215003100466219 is an apocalyptic number.
15003100466219 is a deficient number, since it is larger than the sum of its proper divisors (183764801509).
15003100466219 is an equidigital number, since it uses as much as digits as its factorization.
15003100466219 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 296874205.
The product of its (nonzero) digits is 38880, while the sum is 38.
The spelling of 15003100466219 in words is "fifteen trillion, three billion, one hundred million, four hundred sixty-six thousand, two hundred nineteen". | 598 | 1,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-25 | latest | en | 0.833434 |
https://danaernst.com/teaching/mat526f22/ | 1,701,872,242,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00566.warc.gz | 222,892,357 | 6,725 | ## Welcome
Welcome to the course web page for the Fall 2022 manifestation of MAT 526: Topics in Combinatorics at Northern Arizona University.
## Course Info
Title: MAT 526: Topics in Combinatorics
Semester: Fall 2022
Credits: 3
Section: 1
Time: 10:20-11:10AM MWF
Location: AMB 207
## Instructor Info
Dana C. Ernst, PhD
AMB 176
11:15-12:15PM MWF, 10:15-12:15 Th
dana.ernst@nau.edu
928.523.6852
danaernst.com
The mathematician does not study pure mathematics because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
## What is This Course All About?
This course focuses on enumerative combinatorics with an introduction to generating function techniques. Here are the proposed topics:
• Eulerian numbers
• Binomial coefficients
• Generating functions
• Classical Eulerian numbers
• Eulerian polynomials
• Two important identities
• Exponential generating function
• Narayana numbers
• Catalan numbers
• Pattern-avoiding permutations
• Narayana numbers
• Dyck paths
• Planar binary trees
• Noncrossing partitions
• Partially ordered sets
• Basic definitions and terminology
• Labeled posets and P-partitions
• The shard intersection order
• The lattice of noncrossing partitions
• Absolute order and Noncrossing partitions
• Weak order, hyperplane arrangements, and the Tamari lattice
• Inversions
• The weak order
• The braid arrangement
• Euclidean hyperplane arrangements
• Products of faces and the weak order on chambers
• Set compositions
• The Tamari lattice
• Rooted planar trees and faces of the associahedron
• Refined enumeration
• The idea of a $q$-analogue
• Lattice paths by area
• Lattice paths by major index
• Euler-Mahonian distributions
• Descents and major index
• $q$-Catalan numbers
• $q$-Narayana numbers
• Dyck paths by area
Don’t fear failure. Not failure, but low aim, is the crime. In great attempts it is glorious even to fail.
# Dana C. Ernst
Mathematics & Teaching
Northern Arizona University
Flagstaff, AZ
Website
928.523.6852
Instagram
Strava
GitHub
arXiv
ResearchGate
Mendeley
Impact Story
ORCID | 556 | 2,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.723766 |
http://www.dreamincode.net/forums/topic/302927-problem-how-to-input-values-of-the-list-into-the-program-to-check/page__pid__1761958__st__0 | 1,464,078,815,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049270513.22/warc/CC-MAIN-20160524002110-00172-ip-10-185-217-139.ec2.internal.warc.gz | 469,497,966 | 20,706 | Problem how to input "values of the list" into the program to
Page 1 of 1
5 Replies - 699 Views - Last Post: 06 December 2012 - 02:01 PMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=302927&s=c53f88935f74bb8d3d9740cde57b546a&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
#1 abbey21
Reputation: 0
• Posts: 18
• Joined: 05-December 12
Problem how to input "values of the list" into the program to
Posted 05 December 2012 - 11:27 PM
I am trying to run this program and I am stuck on how to input the values of the lists in rows and columns. The program is suppose to check whether a 2D list has 4 consecutive numbers of the same value, either horizontally, vertically or diagonally. I get an Exception error anytime I try to input values therefore I cant fully run the program.
```
import java.util.Scanner;
public class ConsecutiveFour {
private final int MATCH_COUNT = 4;
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter number of rows: ");
int rows = input.nextInt();
System.out.print("Enter number of columns: ");
int columns = input.nextInt();
int[][] matrix = new int[rows][columns];
System.out.println("Enter " + matrix.length + " rows and " + matrix[0].length + " columns: ");
for (int row = 0; row < matrix.length; row++) {
for (int column = 0; column < matrix[row].length; column++){
matrix[row][column] = input.nextInt();
int value = matrix[row][column];
}
}
System.out.print(isConsecutiveFour(matrix));
}
public static boolean isConsecutiveFour(int[][] values) {
boolean cons = false;
int columns = values.length;
int rows = values[0].length;
//tests horizontally
for (int r=0; r < rows; r++) {
for (int c=0; c < columns - 3; c++){
if (values[c][r] == values[c+1][r] &&
values[c][r] == values[c+2][r] &&
values[c][r] == values[c+3][r]) {
cons = true;
}
}
}
//tests vertically
for (int r=0; r < rows - 3; r++) {
for (int c=0; c < columns; c++){
if (values[c][r] == values[c][r+1] &&
values[c][r] == values[c][r+2] &&
values[c][r] == values[c][r+3]) {
cons = true;
}
}
}
//tests diagonally (going down and to the right)
for (int r=3; r < rows; r++) {
for (int c=0; c < columns - 3; c++) {
if (values[c][r] == values[c+1][r-1] &&
values[c][r] == values[c+2][r-2] &&
values[c][r] == values[c+3][r-3]) {
cons = true;
}
}
}
//tests diagonally (going down and to the left)
for (int r=0; r < rows - 3; r++) {
for (int c=0; c < columns - 3; c++) {
if (values[c][r] == values[c+1][r+1] &&
values[c][r] == values[c+2][r+2] &&
values[c][r] == values[c+3][r+3]) {
cons = true;
}
}
}
return cons;
}
}
```
Is This A Good Question/Topic? 0
Replies To: Problem how to input "values of the list" into the program to
#2 raghav.naganathan
• Perfectly Squared ;)
Reputation: 410
• Posts: 1,449
• Joined: 14-September 12
Re: Problem how to input "values of the list" into the program to
Posted 05 December 2012 - 11:31 PM
abbey21, on 06 December 2012 - 11:57 AM, said:
I get an Exception error anytime I try to input values therefore I cant fully run the program.
Can you please tell us what is the exact error you get so that it will be helpful to find out where your program is throwing that exception.
regards,
Raghav
#3 abbey21
Reputation: 0
• Posts: 18
• Joined: 05-December 12
Re: Problem how to input "values of the list" into the program to
Posted 05 December 2012 - 11:36 PM
at java.util.Scanner.throwFor(Scanner.java:909)
at java.util.Scanner.next(Scanner.java:1530)
at java.util.Scanner.nextInt(Scanner.java:2160)
at java.util.Scanner.nextInt(Scanner.java:2119)
at ConsecutiveFour.main(ConsecutiveFour.java:20)
#4 raghav.naganathan
• Perfectly Squared ;)
Reputation: 410
• Posts: 1,449
• Joined: 14-September 12
Re: Problem how to input "values of the list" into the program to
Posted 06 December 2012 - 01:35 AM
Ok...I see what your problem is.
Try this...use input.nextLine() immediately after your input.nextInt()
What that does is that it not only takes your number, but also the \n character.
regards,
Raghav
#5 abbey21
Reputation: 0
• Posts: 18
• Joined: 05-December 12
Re: Problem how to input "values of the list" into the program to
Posted 06 December 2012 - 08:34 AM
Thank you that worked!! But it takes 14 lines of numbers before it outputs "false". Is there anyway I can change my code to fix this? Or am I inputing the "Values of Rows and Columns" wrong?
When "Enter the Rows and Columns" I go:
123456
123456
345566
546786
432545
321367
Then 14 lines liner through hitting enter i get a false for "No Four Consecutive numbers"
#6 pbl
• There is nothing you can't do with a JTable
Reputation: 8370
• Posts: 31,956
• Joined: 06-March 08
Re: Problem how to input "values of the list" into the program to
Posted 06 December 2012 - 02:01 PM
123456 this is a single number that will be rea d by Scanner.nextInt() and will be store in one cell of your array not in the six columns of a row. You will have to enter
1 2 3 4 5 6
Also in your method to test... don't test everything more than once instead of
cons = true; // found
simply
return true; // nop need to test the rest
ad the end of the method, simply return false; if you haven't encounter a return true; before | 1,634 | 5,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-22 | latest | en | 0.518888 |
http://www.damon4.com/2016/02/why-is-aliasing-important-to-ad.html | 1,656,267,350,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00764.warc.gz | 66,352,613 | 13,869 | Saturday, February 20, 2016
Why is aliasing important to A/D converters?
To minimize aliasing an anti-aliasing filter is needed before an A/D converter. How do you design this pre-filter? What is this aliasing? Shannon, and Nyquist later, said that the maximum signal we can reconstruct is half of the sampling rate. If the signal we sample has components higher than 1/2 sampling rate then we get a phenomena called 'aliasing. If I were to send an aliased signal back out of a computer to a D/A converter it would look like the signal was 'shifted' in frequency.
Suppose a signal of Fs was put into an A/D converter and the sampling rate was Fs. It might appear on a D/A converter as a DC signal .
As an everyday example of aliasing, have you ever seen a moving fan appear still under a strobe light? When the strobe is blinking at the same rpm's as the fans rotational speed, the fan will appear motionless. The aliasing phenomena can occur if the sampled signal is at or above 1/2 the sampling rate OR IF IT HAS (Fourier) COMPONENTS AT OR ABOVE THIS RATE. In this case just those components above the Nyquist rate would be aliased, but the total signal, if reconstructed, will be distorted.
How do we design to this? Typically we use an anti-aliasing filter on the front end of the A/D. This filter is designed to attenuate signal above the Nyquist rate to below some noise level. It might be as simple as an RC or more complicated as a multi-pole active filter in a Chebychev or Butterworth configuration. The key is to understand how much attenuation we need at the stop band. Suppose our A/D is 12 bits. The ideal SNR is 72dB (= 20* log(4096)). The filter should be designed to provide 72 db of attenuation at the stop band. Now we note that a single pole filter provides 6db per octave of attenuation.
A single pole low pass filter designed to attenuate a signal to this noise specification would be designed to have a cutoff 12 octaves below half the A/D sampling rate! | 461 | 1,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | longest | en | 0.95175 |
https://cstheory.stackexchange.com/questions/5572/relation-between-fixed-parameter-and-approximation-algorithm/5573#5573 | 1,642,817,458,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00492.warc.gz | 245,747,543 | 37,712 | Relation between fixed parameter and approximation algorithm
Fixed parameter and approximation are totally different approaches to solve hard problems. They have different motivation. Approximation looks for faster result with approximate solution. Fixed parameter looks for exact solution with time complexity in terms of the exponential or some function of k and polynomial function of n where n is the input size and k is parameter. Example $2^kn^3$.
Now my question, is there any upper or lower bound result based on the relationship between fixed parameter and approximation approaches or they totally do not have any relationship.For example for a problem $P$ is said to be $W[i]$ hard for some $i>0$ is nothing to do with having c-approximation algorithm or PTAS. please provide some references
• Related, possibly duplicate ?: cstheory.stackexchange.com/questions/4906/… Mar 20 '11 at 17:49
• @suresh venkat That question is about difference in understand NP-complete and fixed parameter. when we talk in terms of NP-hardness only, then independent set and vertex cover are literally same, but when we talk in terms of fixed parameter they have huge difference. vertex cover has good fpt whereas independent set is W[1] hard Mar 20 '11 at 17:59
• but here i am looking for a relationship between approximation and fixed parameter. Mar 20 '11 at 18:03
• I think there is no real relation between them, but by using fixed parameter we may have a good approximation, for example in bin packing (makespan scheduling) you can see this relation, or for example in bounded Treewidth graphs we have approximations on some problems. Mar 20 '11 at 19:08
There are several connections between parameterized complexity and approximation algorithms.
First, consider the so-called standard parameterization of a problem. Here, the parameter is what you would optimize in the optimization version of the problem (the size of the vertex cover for the Vertex Cover problem, the width of the tree decomposition for the Treewidth problem, etc.). Let us concretely look at Vertex Cover. Any kernel with a linear number of vertices for Vertex Cover implies a constant factor polynomial-time approximation algorithm: into the approximate solution, put all the vertices that have been forced into the solution by the kernelization algorithm, and all the vertices of the kernelized instance. On the other hand, lower bounds on the approximation factor imply lower bounds on the size of a kernel. For example, under the Unique Games Conjecture, Khot and Regev (JCSS 2008) rule out approximation algorithms for Vertex Cover with a ratio of any $c<2$, which rules out a kernel for Vertex Cover with at most $ck$ vertices, $c<2$, as well.
EDIT: The argumentation for the kernel lower bound in the previous paragraph is very informal, and to the best of my knowledge it is open whether such lower bounds on the kernel size can be proven, even for Vertex Cover. As @Falk points out in the comments, the argument holds for most (all?) known kernels. However, I don't see how one could exclude the existence of kernelization algorithms where a feasible solution of the kernelized instance has a different approximation ratio than the corresponding solution in the initial instance.
Then, there is the issue of PTAS versus FPTAS. If we want to find a solution within $(1+\epsilon)$ from optimal, we can parameterize by $1/\epsilon$. Then, a PTAS corresponds to an XP-algorithm in the parameterized setting, whereas an FPTAS corresponds to an FPT-algorithm. For an approximation lower bound, we may not expect an EPTAS for any problem whose standard parameterization is W[1]-hard: running the EPTAS with $\epsilon=1/(k+1)$ would solve the problem exactly in FPT time.
Finally, an FPT approximation algorithm is an algorithm with FPT running time and an approximation ratio which may depend on the parameter. For example, the standard parameterization of the Cliquewidth problem has an FPT approximation algorithm with approximation ratio $(2^{3k+2}-1)/k$ (Oum, WG 2005), whereas the standard parameterization of Independent Dominating Set has no FPT approximation algorithm with performance ratio $g(k)$ for any computable function $g$, unless FPT=W[2] (Downey et al., IWPEC 2006). See (Marx, The Computer Journal 2008) for a survey on FPT approximation.
• @Gasper Can you please see the question "Finding a maximum acyclic sub-tournament given two acyclic sub-tournaments" . I still have doubt with my answer. As you have worked with related problem, you can help me out Mar 20 '11 at 14:57
• IS the first paragraph of Serge's answer correct? Does the lower bound on approximability yield lower bound on the size of the kernel? The similar statement is in Niedermeier's book but is this statement correct? Mar 20 '11 at 19:41
• @XXYYXX: In Serge's answer, he wrote "Any kernel with a linear number of vertices for Vertex Cover implies a constant factor polynomial-time approximation algorithm" with a short proof. More precisely, his argument shows if there exists a kernel with ck vertices for some constant c, then there exists a factor-c approximation algorithm. The contrapositive is: if no factor-c approximation algorithm exists, then no kernel with ck vertices exists. Mar 21 '11 at 6:22
• @Prabu: I commented on your answer to the other question. @Yoshio: Thanks for answering @XXYYXX's question. Mar 21 '11 at 13:35
• In fact for probably all known kernelizations, the argument holds. However, I see no reason why there shouldn't be one that e.g. first reduces to another problem, kernelizes there, and then reduces back to Vertex Cover, so that the resulting instance has no vertex correspondence with the initial one. So it seems to me that the only thing we can really show is that kernels that are subgraphs will probably not be smaller than 2k. Mar 28 '11 at 10:36
There is known theorem [1,Theorem 3.1], characterizing approximation class $$FPTAS$$ through parameterized class $$PFPT$$:
Let $$Q=(I_Q, S_Q, f_Q, opt_Q)$$ be a scalable $$NP$$ optimization problem. Then $$Q$$ has an $$FPTAS$$ if and only if $$Q$$ is in $$PFPT$$.
In turn, $$PFPT$$ is defined as:
An $$NP$$ optimization problem $$Q$$ is polynomial fixed-parameter tractable ($$PFPT$$) if its parameterized version is solvable in time $$O(|x|^{O(1)}k^{O(1)})$$, where $$|x|$$ - the size of the input instance $$x$$.
Another characterizations for two approximation classes are proposed in [2,Theorem 6.5].
A problem is
• in $$PTAS$$ if and only if it has a $$ptas^{\infty}$$ and its standard parameterization is in $$XP^w$$.
• in $$FPTAS$$ if and only if it has an $$fptas^{\infty}$$ with a polynomially-bounded threshold function and its standard parameterization is in $$PFPT^w$$.
Here $$(f)ptas^{\infty}$$ means asymptotic (fully) polynomial approximation scheme, standard parametrization - decision version of an optimization problem, $$(XP)PFPT^w$$ - corresponding classes of decision problems for which an algorithm which decides them returns a witness if the answer is Yes, threshold function - function depending on error $$\frac{1}{\epsilon}$$ and bounding from below an optimum value.
1. Polynomial time approximation schemes and parameterized complexity. J. Chen et al. / Discrete Applied Mathematics 155 (2007) 180 – 193.
2. Structure of Polynomial-Time Approximation. E. J. van Leeuwen et al. Technical Report UU-CS-2009-034, December 2009. | 1,728 | 7,424 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.904545 |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=1189&pid=8804 | 1,656,936,493,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00292.warc.gz | 445,338,800 | 11,984 | • 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
Math overflow question on fractional exponential iterations sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/28/2018, 07:57 PM (This post was last modified: 03/28/2018, 09:33 PM by sheldonison.) Dmytro Taranovsky asks in his post, https://mathoverflow.net/questions/28381...rent-bases ... " ... Let a and b be real numbers above e^{1/e}, and c exp(1/e), then my conjecture is that the limit holds. Unfortunately, Peter Walker's solution is also conjectured to be nowhere analytic; see: https://math.stackexchange.com/questions...lytic-slog (2) That for Kneser's solution, there are counter examples and the limit does not hold, even with the restriction that c\;\exp_b^d(x)$ (3) Given any analytic solution for base(a), then if we desire a tetration solution for base(b) which has the desired property then I conjecture that the tetration base b is nowhere analytic! The conjecture is also that the slog for b would be given by a modification of Peter Walker's h function. - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/29/2018, 10:13 PM (This post was last modified: 03/29/2018, 10:18 PM by sheldonison.) (03/28/2018, 07:57 PM)sheldonison Wrote: ... Let a and b be real numbers above e^{1/e}, and c\;\exp_b(x)\;\;\;$ but only if x is big enough when a1, and helps one understand why Walker's solution works. - Sheldon JmsNxn Ultimate Fellow Posts: 754 Threads: 104 Joined: Dec 2010 03/30/2018, 07:37 PM (This post was last modified: 03/30/2018, 08:47 PM by JmsNxn.) It seems upsetting that the only tetrations that could satisfy this are non-analytic. I'm not prone to believe this, only because it doesn't look nice... I'm wondering if we can look at it the following way $\exp_b^{d}(x) = \exp_b^{c}(\exp_b^{\delta}(x))$ Then the question boils into whether, for all $\delta, \delta' >0$ then $\exp^c_{b+\delta}(x) = o(\exp_b^{c+\delta'}(x))$ I think we can show this is true when $c \in \mathbb{N}$. By induction, first, for $c = 0$ the induction step is obvious. Namely $x = o (\exp_b^{\delta'}(x))$. Suppose the result holds for $c = n$ then (taking $<$ to mean asymptotically less than): $\exp_{b+\delta}^{n+1}(x) =\exp^{n}_{b+\delta}(\exp_{b+\delta}(x))<\exp_{b}^{n+\delta'/2}(\exp_{b+\delta}(x)) < \exp_{b}^{n+\delta'/2}(\exp_b^{1+\delta'/2}(x)) = \exp_b^{n+\delta'+1}(x)$ Sadly, I can't think of anyway to generalize this to non-integral $c$.... I'm thinking, a nice way to look at it from here is to look at root functions of the $\exp$ function. But then we'd need an implication $f^{\circ n}(x) = o(g^{\circ n}(x)) \Rightarrow f(x) = o(g(x))$, which looks like it could be true for monotonically growing unbounded functions. But that's probably too easy, we'd probably need a nice condition on the root functions for that to be true. EDIT: It appears I made a fruitful mistake in the above proof. The base induction step would have to be (1) $\exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x)$, not the obvious one $x < \exp^{\delta'}(x)$. This is the base step I should have used. The proof then says if this base step (1) holds the result holds for all natural $c$, namely $\exp_{b+\delta}^n(x) < \exp_b^{n+\delta'}(x)$. And I think with some finesse we can show that this implies it's true for root functions of $\exp_b$, which should leave for a proof where $c \in \mathbb{Q}$. Then perhaps a density argument may work on non rational $c$. I'll work on this more later, but I think maybe we can reduce this entire problem into the condition that if for all $\delta,\delta'>0$ we have $\exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x)$ then it follows that $\exp_{b+\delta}^c(x) < \exp_b^{c+\delta'}(x)$. ...We'll probably have to assume that $\exp_b^c(x)$ is monotone non-decreasing in $x$ and unbounded, or at least, eventually monotone non-decreasing. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/31/2018, 04:19 AM (This post was last modified: 03/31/2018, 04:29 PM by sheldonison.) (03/30/2018, 07:37 PM)JmsNxn Wrote: It seems upsetting that the only tetrations that could satisfy this are non-analytic. I'm not prone to believe this, only because it doesn't look nice... There's an old thread http://math.eretrandre.org/tetrationforu...hp?tid=236 I started the thread in 2009, before I had written generic programs for analytic tetration for any base, and I was using an excel spreadsheet to approximate analytic tetration. I estimated that as x gets arbitrarily large $\text{slog}_2(x)-\text{slog}_e(x)\approx1.1282$ The 2009 thread continues on to discuss what I called "the wobble"... Credit needs to go to William Paulsen and Samuel Cowgill in their upcoming paper which discusses these issues more rigorously than I can. But it was quickly clear in the 2009 thread that there is an inherent wobble when comparing tetration bases; this was apparent for bases a little bit bigger than eta=exp(1/e) using straightforward techniques on an excel spreadsheet. The limit as x get arbitrarily large does not converge to a simple number like the 1.1282 estimate, but instead converges to a 1-cyclic function near that value. For base(2) and for base(e), if you use Kneser's construction; then the 1-cyclic limit is graphed below. $\text{slog}_e(\text{tet}_2(x))-x$ On this forum, other ideas like "the base change function" were discussed, where you use Peter Walker's idea to define tetration base(a) from tetration base(b). For example, you could define tetration base(2) from Kneser's tetration base(e). The relevant equations might look something like this. But the "h" function below is conjectured to be nowhere analytic, even though Walker proved it is $C^{\infty}$ for the case in his paper. Walker defined the base(e) slog from the Abel function for iterating $x\mapsto\exp(x)-1$. This is mathematically conjugate (or exactly equivalent) to iterating base eta. $y\mapsto\eta^y\;\;\eta=\exp(1/e)\;\;x=\frac{y}{e}-1$ $h_n (x)=\ln_b^{[n]}(\exp_a^{[n]}(x))$ $h(x)=\lim_{n\to\infty}h_n (x)$ $\text{slog}_b(x)=\text{slog}_a(h(x))-\text{slog}_a(h(1)));$ /* constant to guarantee slog_b(1)=0 */ - Sheldon JmsNxn Ultimate Fellow Posts: 754 Threads: 104 Joined: Dec 2010 04/01/2018, 03:09 AM (This post was last modified: 04/01/2018, 06:22 AM by JmsNxn.) Okay, so the question and your intuition relates to the old base change formula and that it failed to be analytic. That makes sense, but is disappointing to think we're going to lose this property if we choose an analytic solution. So what this slog limit is saying is that for ''good'' analytic tetrations: $f(x)=\exp_{b+\delta}^{c}(x) - \exp^{c+\delta'}_b(x)$ changes sign infinitely often (given $\delta,\delta'<\epsilon$)? This reminds me of something. I've dealt with those limits before and felt discouraged at an ability to prove uniform convergence. Given holomorphic $f,g : \mathbb{D} \to \mathbb{D}$ where $f(0) = g(0) = 0$, when trying to find a function $\Psi:\mathbb{D}\to\mathbb{D}$ such that $\Psi(f(z)) = g(\Psi(z))$, the natural choice is $\Psi(z) =\lim_{n\to\infty} g^{-n}(f^{n}(z))$ (which never seems to work). But it sure does look nice. The only way this works, I found, is to assume $f'(0) = g'(0) = \lambda$ and take the Schroder function of both functions $h_0, h_1$ where $h_0(f(z)) = \lambda h_0(z)$ and $h_1(g(z)) = \lambda h_1(z)$ and then $\Psi(z) = h_1^{-1}(h_0(z))$ which works locally. Then the above limit for $\Psi$ is convergent. But we had to sacrifice a lot to get there. Of course if we're working on a non simply connected set $H$ instead of $\mathbb{D}$ and we assumed that $f,g$ had no fixed points on this set, this could work. But tetration takes $\mathbb{C}/\{z \in (-\infty,-2)\}\to \mathbb{C}$, so it probably has fixed points (maybe this is provable). Which should guarantee a base change function $h$ is non extendable to $\mathbb{C}/\{z \in (-\infty,-2)\,\}$. This is kinda' helping me understand why these functions fail to be analytic. No conjugation can change the multiplier value and clearly $^ze$ will have a different multiplier at its fixed point as $^z 2$ will have at its fixed point. I'll have to read Walker's paper. The only work around I had to this was working with Schroder functions and when dealing with the real line where there are no fixed points I can't imagine a manner of getting a nice uniform convergence. I'm still wondering if I can prove that if $\exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x)$ then $\exp_{b+\delta}^c(x) < \exp_b^{c + \delta'}(x)$ which could then be a condition for tetration to be non-analytic. Still seems like a lot of this is up in the air though. I apologize if this has me a bit scatter brained. « Next Oldest | Next Newest »
Possibly Related Threads… Thread Author Replies Views Last Post Qs on extension of continuous iterations from analytic functs to non-analytic Leo.W 11 315 07/01/2022, 12:13 AM Last Post: MphLee Slog(Exponential Factorial(x)) Catullus 17 503 06/28/2022, 02:03 PM Last Post: tommy1729 A random question for mathematicians regarding i and the Fibonacci sequence. robo37 1 3,878 06/27/2022, 12:06 AM Last Post: Catullus Question about tetration methods Daniel 17 435 06/22/2022, 11:27 PM Last Post: tommy1729 Fractional iteration of x^2+1 at infinity and fractional iteration of exp bo198214 17 28,567 06/11/2022, 12:24 PM Last Post: tommy1729 A question concerning uniqueness JmsNxn 4 9,521 06/10/2022, 08:45 AM Last Post: Catullus Consistency in the composition of iterations Daniel 9 323 06/08/2022, 05:02 AM Last Post: JmsNxn Has anyone solved iterations of z+Γ(z)? Leo.W 5 1,836 01/07/2022, 08:15 AM Last Post: JmsNxn Master of Nuclear Math Daniel 0 780 05/18/2021, 11:30 PM Last Post: Daniel Math.Stackexchange.com question on extending tetration Daniel 3 2,087 03/31/2021, 12:28 AM Last Post: JmsNxn
Users browsing this thread: 1 Guest(s) | 3,026 | 9,876 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 61, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-27 | longest | en | 0.844294 |
https://www.coursehero.com/file/6811919/MIT2-71S09-gquiz1/ | 1,519,597,498,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817437.99/warc/CC-MAIN-20180225205820-20180225225820-00607.warc.gz | 895,984,565 | 41,539 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
MIT2_71S09_gquiz1
# MIT2_71S09_gquiz1 - shown above where lenses L1 and L2 have...
This preview shows pages 1–4. Sign up to view the full content.
2.710 Quiz 1 50 min 8:05–8:55am EST 9:05–9:55pm SST
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Spring ’09 QUIZ 1 Wednesday, March 9 th , 2009 α CL EL EP R 15 50 distances shown in mm (not to scale) CL =corrective lens; EL =eye lens; EP =eye pupil; R =retina 1. Eye correction The schematic above is a grossly simplifed model a person’s eye who suffers ±rom myopia. The unaccommodated ±ocal length EL is f e = 45mm, whereas the distance ±rom EL to R is longer, as shown. The purpose this problem is to study the corrective action CL ±or objects at infnity. We model both CL and EL as thin lenses. a) (15%) First consider an on–axis object, i.e. α = 0. Calculate CL’s ±ocal length f c such that the combination CL and EL ±ocus properly on R. b) (15%) Locate the 2 nd Principal Plane and the Effective Focal Length the combination CL and EL. c) (15%) Now consider an off–axis object, i.e. α = 0. Which elevation on R is the image ±ormed at? d) (15%) Does this person’s EP appear smaller or larger than its natural size when viewed through CL by an observer? By how much?
S1 f 1 f 2 object plane 2 a 1 a f 1 /2 f 1 /2+ f 2 S2 L2 L1 2. Aperture and feld stops in a telescope with fnite conjugates For the telescope confguration
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: shown above, where lenses L1 and L2 have ±ocal lengths f 1 , f 2 , respectively, the object plane and two stops S1 and S2 o± hal±–sizes a 1 , a 2 , respectively, are at the locations shown, a) (10%) identi±y the Aperture Stop and the Field Stop, and trace the Chie± Ray and Marginal Ray ±or a sample off-axis point object o± your choice; b) (10%) locate the Entrance Pupil, Exit Pupil, Entrance Window, and Exit Window; c) (10%) calculate the Numerical Aperture and Field o± View; and d) (10%) critique whether stops S1, S2 are optimally located and, i± not, sug-gest better location(s). GOOD LUCK! MIT OpenCourseWare http://ocw.mit.edu 2.71 / 2.710 Optics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms ....
View Full Document
{[ snackBarMessage ]}
### Page1 / 4
MIT2_71S09_gquiz1 - shown above where lenses L1 and L2 have...
This preview shows document pages 1 - 4. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 809 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-09 | latest | en | 0.834378 |
https://www.ceder.net/def/alterthewave.php?level=master&language=usa | 1,719,081,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00018.warc.gz | 601,713,589 | 6,187 | Definitions of Square Dance Calls and Concepts
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
Alter The Wave [C1]
(Lee Kopman 1970)
From a Wave or Facing Couples.
1. Arm Turn 1/2;
2. Centers Cast Off 3/4 as Ends U-Turn Back;
3. Counter Rotate the Diamond 1/2;
4. Flip The Diamond.
Ends in a Wave (of opposite handedness). Alter The Wave is a 4-part call.
beforeAlter The Wave afterArm Turn 1/2;Centers Cast Off 3/4 as Ends U-Turn Back afterDiamond Counter Rotate 1/2 afterFlip The Diamond (done)
Notes:
• The Diamond Counter Rotate 1/2 is often referred to as Star 1/2.
• Dancers who Cast Off 3/4 become the Ends of the resulting Wave no matter how far the Diamond is turned. Dancers who U-Turn Back always become the Centers.
• The Diamond Counter Rotate 1/2 is danced by turning the whole Diamond 180 degrees around its center of gravity. Centers retain handholds throughout the Diamond Counter Rotate and the Points should NOT place their hands into the 'Star', but instead should point as they would when identifying Diamonds. This is very important, because if there are 3 or more hands are in the Center, dancers tend to forget who is to flip whom for the Flip The Diamond, especially if the caller modifies the amount to 'turn the Star'.
• From a Wave, Ends remain Ends as Centers remain Centers.
Alter The Wave, Star 1/4:
beforeAlter The Wave, Star 1/4 afterArm Turn 1/2;Centers Cast Off 3/4 as Ends U-Turn Back afterDiamond Counter Rotate 1/4 afterFlip The Diamond (done)
Spin Back [C4] (Norm Wilcox 1972): From Facing Couples or a Wave. Arm Turn 1/2; Centers Cast Off 3/4 as Ends U-Turn Back. Ends in a Diamond. This is the first 1/2 of Alter The Wave. | 557 | 1,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.805406 |
https://www.lmfdb.org/L/ModularForm/GL2/Q/holomorphic/670/2/e/d/171/1/ | 1,579,470,588,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250595282.35/warc/CC-MAIN-20200119205448-20200119233448-00268.warc.gz | 960,605,915 | 7,210 | # Properties
Degree 2 Conductor $2 \cdot 5 \cdot 67$ Sign $-0.978 - 0.205i$ Motivic weight 1 Primitive yes Self-dual no Analytic rank 0
# Related objects
## Dirichlet series
L(s) = 1 + (0.5 + 0.866i)2-s + 3-s + (−0.499 + 0.866i)4-s − 5-s + (0.5 + 0.866i)6-s + (−1 + 1.73i)7-s − 0.999·8-s − 2·9-s + (−0.5 − 0.866i)10-s + (−3 + 5.19i)11-s + (−0.499 + 0.866i)12-s + (−1 − 1.73i)13-s − 1.99·14-s − 15-s + (−0.5 − 0.866i)16-s + (−3 − 5.19i)17-s + ⋯
L(s) = 1 + (0.353 + 0.612i)2-s + 0.577·3-s + (−0.249 + 0.433i)4-s − 0.447·5-s + (0.204 + 0.353i)6-s + (−0.377 + 0.654i)7-s − 0.353·8-s − 0.666·9-s + (−0.158 − 0.273i)10-s + (−0.904 + 1.56i)11-s + (−0.144 + 0.249i)12-s + (−0.277 − 0.480i)13-s − 0.534·14-s − 0.258·15-s + (−0.125 − 0.216i)16-s + (−0.727 − 1.26i)17-s + ⋯
## Functional equation
\begin{aligned} \Lambda(s)=\mathstrut & 670 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & (-0.978 - 0.205i)\, \overline{\Lambda}(2-s) \end{aligned}
\begin{aligned} \Lambda(s)=\mathstrut & 670 ^{s/2} \, \Gamma_{\C}(s+1/2) \, L(s)\cr =\mathstrut & (-0.978 - 0.205i)\, \overline{\Lambda}(1-s) \end{aligned}
## Invariants
$$d$$ = $$2$$ $$N$$ = $$670$$ = $$2 \cdot 5 \cdot 67$$ $$\varepsilon$$ = $-0.978 - 0.205i$ motivic weight = $$1$$ character : $\chi_{670} (171, \cdot )$ primitive : yes self-dual : no analytic rank = 0 Selberg data = $(2,\ 670,\ (\ :1/2),\ -0.978 - 0.205i)$ $L(1)$ $\approx$ $0.110885 + 1.06793i$ $L(\frac12)$ $\approx$ $0.110885 + 1.06793i$ $L(\frac{3}{2})$ not available $L(1)$ not available
## Euler product
$L(s) = \prod_{p \text{ prime}} F_p(p^{-s})^{-1}$ where, for $p \notin \{2,\;5,\;67\}$, $$F_p(T)$$ is a polynomial of degree 2. If $p \in \{2,\;5,\;67\}$, then $F_p(T)$ is a polynomial of degree at most 1.
$p$$F_p(T)$
bad2 $$1 + (-0.5 - 0.866i)T$$
5 $$1 + T$$
67 $$1 + (-2.5 + 7.79i)T$$
good3 $$1 - T + 3T^{2}$$
7 $$1 + (1 - 1.73i)T + (-3.5 - 6.06i)T^{2}$$
11 $$1 + (3 - 5.19i)T + (-5.5 - 9.52i)T^{2}$$
13 $$1 + (1 + 1.73i)T + (-6.5 + 11.2i)T^{2}$$
17 $$1 + (3 + 5.19i)T + (-8.5 + 14.7i)T^{2}$$
19 $$1 + (-2 - 3.46i)T + (-9.5 + 16.4i)T^{2}$$
23 $$1 + (-3 - 5.19i)T + (-11.5 + 19.9i)T^{2}$$
29 $$1 + (-3 + 5.19i)T + (-14.5 - 25.1i)T^{2}$$
31 $$1 + (2.5 - 4.33i)T + (-15.5 - 26.8i)T^{2}$$
37 $$1 + (-3.5 - 6.06i)T + (-18.5 + 32.0i)T^{2}$$
41 $$1 + (1.5 - 2.59i)T + (-20.5 - 35.5i)T^{2}$$
43 $$1 - 8T + 43T^{2}$$
47 $$1 + (3 - 5.19i)T + (-23.5 - 40.7i)T^{2}$$
53 $$1 + 3T + 53T^{2}$$
59 $$1 + 6T + 59T^{2}$$
61 $$1 + (4 + 6.92i)T + (-30.5 + 52.8i)T^{2}$$
71 $$1 + (-7.5 + 12.9i)T + (-35.5 - 61.4i)T^{2}$$
73 $$1 + (-2 - 3.46i)T + (-36.5 + 63.2i)T^{2}$$
79 $$1 + (4 - 6.92i)T + (-39.5 - 68.4i)T^{2}$$
83 $$1 + (-7.5 - 12.9i)T + (-41.5 + 71.8i)T^{2}$$
89 $$1 - 9T + 89T^{2}$$
97 $$1 + (-5 - 8.66i)T + (-48.5 + 84.0i)T^{2}$$
\begin{aligned} L(s) = \prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1} \end{aligned} | 1,599 | 2,855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-05 | latest | en | 0.363834 |
giorginikolaishvili.com | 1,695,656,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00379.warc.gz | 314,910,186 | 5,245 | When undergraduate students begin learning econometrics, they’re required to master the following skills:
1. Probabilistic and statistical thinking
2. Calculus and (linear) algebra
3. Programming in R (at UO – the language of choice may vary by school)
In my experience, students tend to struggle or neglect programming when starting out, mainly due to a complete lack of familiarity and “fear” of the unknown.
This page provides a list of links to carefully-picked free online resources that should ease students’ transition into econometric programming using R.
## General guides/tutorials
R is a large and diverse statistical programming language. In an introductory econometrics course, there usually isn’t enough time to cover all of the fundamentals and nuances that are necessary to achieve full competence as an R-user – you’re expected to dive into statistical analysis pretty early on instead of mastering data types, loops etc. The following beginner-friendly tutorials should help with this problem:
1. The best thorough step-by-step tutorial I could find: https://r-coder.com/learn-r/
2. Brief document. I recommend skipping Ch. 4: https://cran.r-project.org/doc/contrib/Paradis-rdebuts_en.pdf
3. Not as great of a tutorial as r-coder, but still fine as an additional reference: https://www.statmethods.net/r-tutorial/index.html
## Manipulating data
One of the most universally practical skills you can gain while learning R, in my opinion, is data manipulation using `dplyr`. I highly recommend the following set of tutorials by Susan Baert that take you through basic `dplyr` concepts step-by-step:
## Plotting
The following resources are excellent for learning how to plot using `ggplot2`. I recommend sticking with `ggplot2` for any kind of plot-making in R.
1. Step-by-step plotting tutorial: https://cedricscherer.netlify.app/2019/08/05/a-ggplot2-tutorial-for-beautiful-plotting-in-r/
2. Another step-by-step tutorial, but less thorough: https://cedricscherer.netlify.app/2019/08/05/a-ggplot2-tutorial-for-beautiful-plotting-in-r/
3. Any time you want to make some sort of graph, and you don’t know how – simply go to this website, select the relevant plot type, select `R gallery`, select the most relevant example plot that is provided, and follow the instructions: https://cedricscherer.netlify.app/2019/08/05/a-ggplot2-tutorial-for-beautiful-plotting-in-r/
4. Learn how to make 50 different types of plots through examples: https://cedricscherer.netlify.app/2019/08/05/a-ggplot2-tutorial-for-beautiful-plotting-in-r/ | 595 | 2,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.854547 |
http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/parabolic | 1,432,976,189,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930916.2/warc/CC-MAIN-20150521113210-00175-ip-10-180-206-219.ec2.internal.warc.gz | 84,872,817 | 33,652 | # Fractals/Iterations in the complex plane/parabolic
"Most programs for computing Julia sets work well when the underlying dynamics is hyperbolic but experience an exponential slowdown in the parabolic case." ( Mark Braverman )[1]
In other words it means that one can need days for making a good picture of parabolic Julia set with standard / naive algorithms.
There are 2 problems here :
• slow or lazy dynamics ( in the neighbourhood of a parabolic fixed point )
• some parts are very thin ( hard to find using standard plane scanning)
# Planes
## Dynamic plane
Discrete dynamic in case of complex quadratic polynomial. exterior = white, interior = gray, unknown=red;
How image is changing with various Iteration Max
Dynamic plane = complex z-plane $\mathbb{C} = J_f \cup F_f$ :
• Julia set $J_f \subset \mathbb{C}$ is a common boundary : $J_f\, = \partial A_f(p) =\partial A_{f}(\infty)$
• Fatou set $F_f \subset \mathbb{C}$
• exterior of Julia set = basin of attraction to infinity : $A_{f}(\infty) \ \overset{\underset{\mathrm{def}}{}}{=} \ \{ z \in \mathbb{C} : f^{(k)} (z) \to \infty\ as\ k \to \infty \}.$
• interior of Julia set = basin of attraction of finite, parabolic fixed point p : $A_f(p) \ \overset{\underset{\mathrm{def}}{}}{=} \ \{ z \in \mathbb{C} : f^{(k)} (z) \to p\ as\ k \to \infty \}.$
• immediate basin = sum of componets which have parabolic fixed point p on it's boundary ; the immediate parabolic basin of p is the union of periodic connected components of the parabolic basin.
• attracting Lea-Fatou flower = sum of n attracting petals = sum of 2*n attracting sepals
• petal = part of the flower. Each petal contains 2 sepals,
• sepals ( Let 1 be an invariant curve in the first quadrant and L 1 the region enclosed by 1 ∪ {0}, called a sepal. ) [2]
$\mathbb{C} \supset F(f) \supset A_{f}(p)$
• Filled Julia set[3] $K_f$
# Key words
• parabolic chessboard
• parabolic implosion
• germ[4][5] [6]
• germ of the function : Taylor expansion of the function
• multiplicity[7]
• Julia-Lavaurs sets
• flower / petal / sepal
• The Leau-Fatou flower theorem[8]
• The horn map
• Blaschke product
• Inou and Shishikura's near parabolic renormalization
• complex polynomial vector field [9]
## Numbers
• "a positive integer ν, the parabolic degeneracy with the following property: there are νq attracting petals and νq repelling petals, which alternate cyclically around the fixed point." [10]
• combinatorial rotation number
## Ecalle cilinder
Ecalle cylinders[11] or Ecalle-Voronin cylinders ( by Jean Ecalle [12])
"... the quotient of a petal P under the equivalence relation identifying z and f (z) if both z and f (z) belong to P. This quotient manifold is called the Ecalle cilinder, and it is conformally isomorphic to the infinite cylinder C/Z"[13]
## eggbeater dynamics
Hand Egg beater
Here is real model of what happens in parabolic case
It is a dynamic plane for fc(z)=z^2 + 1/4. It is zoom around parabolic fixed point z=0.5. Orbits of some points inside Julia set are shown (white points)
Contunuus model of dynamics inside a 2 petals flower - dipol
Model of dynamics inside a 4 petals flower - quadrupole
Physical model : the behaviour of cake when one uses eggbeater.
The mathematical model : a 2D vector field with 2 centers ( second-order degenerate points ) [14][15]
The field is spinning about the centers, but does not appear to be diverging.
Fatou coordinate
## germ
Germ : [16]
• z+z^2
• z+z^3
• z+z^{k+1}
• z+a_{k+1}z^{k+1}
• z+a_{k+1}z^{k+1}
## Petal
repelling petals around fixed point and its preimages
There is no unified definition of petals.
Petal of a flower can be :
• attracting / repelling
• small/alfa/big/ ( small attracting petals do not ovelap with repelling petals, but big do)
Each petal is invariant under f^period. In other words it is mapped to itself by f^period.
Attracting petal P is a :
• Each petal is invariant under $f^n$. In other words it is mapped to itself by $f^n$: $f^n(\overline{P} ) \subset P \cup \left\{ p \right\}$
• domain (topological disc ) containing :
• parabolic periodic point p in its boundary $\overline{P} \ni \left\{ p \right\}$ ( precisely in its root , which is a coomon points of all attracting and repelling petals = center of the Lea-Fatou flower)
• critical point or it's n=period images on the other side ( only small ?? )
• trap which captures any orbit tending to parabolic point [17]
• set contained inside component of filled-in Julia set. The attracting petals of parabolic fixed point are contained in it's basin of attraction
Petals $P_j$ are symmetric with respect to the d-1 directions :
$arg(z) = \frac{2 \Pi l}{d - 1}$
where :
• d is (to do)
• l is from 0 to d-2
Petals can have different size.
If $\lambda^n = 1$ then Julia set should approach parabolic periodic point in n directions, between n petals. [18]
### 0/1
How the target set is changing along an internal ray 0
Cpp code by Wolf Jung see function parabolic from file mndlbrot.cpp ( program mandel ) [19][20]
To see effect :
• run Mandel
• (on parameter plane ) find parabolic point for angle 0, which is c=0.25. To do it use key c, in window input 0 and return.
C code :
// in function uint mndlbrot::esctime(double x, double y)
if (b == 0.0 && !drawmode && sign < 0
&& (a == 0.25 || a == -0.75)) return parabolic(x, y);
// uint mndlbrot::parabolic(double x, double y)
if (Zx>=0 && Zx <= 0.5 && (Zy > 0 ? Zy : -Zy)<= 0.5 - Zx)
{ if (Zy>0) data[i]=200; // show petal
else data[i]=150;}
Gnuplot code :
reset
f(x,y)= x>=0 && x<=0.5 && (y > 0 ? y : -y) <= 0.5 - x
unset colorbox
set isosample 300, 300
set xlabel 'x'
set ylabel 'y'
set sample 300
set pm3d map
splot [-2:2] [-2:2] f(x,y)
### 1/2
Cpp code by Wolf Jung see function parabolic from file mndlbrot.cpp ( program mandel ) [21] To see effect :
• run Mandel
• (on parameter plane ) find parabolic point for angle 1/2, which is c=-0.75. To do it use key c, in window input 0 and return.
C code :
// in function uint mndlbrot::esctime(double x, double y)
if (b == 0.0 && !drawmode && sign < 0
&& (a == 0.25 || a == -0.75)) return parabolic(x, y);
// uint mndlbrot::parabolic(double x, double y)
if (A < 0 && x >= -0.5 && x <= 0 && (y > 0 ? y : -y) <= 0.3 + 0.6*x)
{ if (j & 1) return (y > 0 ? 65282u : 65290u);
else return (y > 0 ? 65281u : 65289u);
}
### Number of petals
In parabolic point child period coincides with parent period
Multiplicity = ParentPeriod + ChildPeriod
NumberOfPetals = multiplicity - ParentPeriod
It is because in parabolic case fixed point coincidence with periodic cycle. Length of cycle ( child period) is equal to number of petals
f(z) number of petals explanation
$z^d+z$ d-1 for $z^d+z=z$ point z=0 has multiplicity d
$z^d-z$ d+2 (?)for $f^2(z)$ a root z=0 has multiplicity d+3
For f(z)= -z+z^(p+1) parabolic flower has :
• 2p petals for p odd
• p petals for p even [22]
... ( to do )
### Sepal
A sepal is the intersection of an attracting and repelling petal.
## Flower
Lea-Fatu flower
Flower with four petals and parabolic point in the center
Critical orbit for internal angle 1/5 showing 5 attracting directions
Sum of all petals creates a flower with center at parabolic periodic point.[23]
"... an attracting petal is a set of points in a sufficient small disk around the periodic point whose forward orbits always remain in the disk under powers of return map. " ( W P Thurston : On the geometry and dynamics of Iterated rational maps)
## Parabolic chessboard
Parabolic chessboard = parabolic checkerboard
See :
• Tiles: Tessellation of the Interior of Filled Julia Sets by T Kawahira[24]
• coloured califlower by A Cheritat [25]
Color points according to :[26]
• the integer part of Fatou coordinate
• the sign of imaginary part
## Cauliflower
Cauliflower or broccoli :[27]
• empty ( its interior is empty ) for c outside Mandelbrot set. Julia set is a totally disconnected (
• filled cauliflower for c=1/4 on boundary of the Mandelbrot set. Julia set is a Jordan curve ( quasi circle).
Pleae note that :
• size of image differs because of different z-planes.
• different algorithms are used so colours are hard to compare
### Bifurcation of the Cauliflower
How Julia set changes along real axis ( going from c=0 thru c=1/4 and futher ) :
Perturbation of a function $f(z)$ by complex $\epsilon$ :
$g(z) = f(z) + \epsilon$
When one add epsilon > 0 ( move along real axis toward + infinity ) there is a bifurcation of parabolic fixed point :
• attracting fixed point ( epsilon<0 )
• one parabolic fixed point ( epsilon = 0 )
• one parabolic fixed point splits up into two conjugate repelling fixed points ( epsilon > 0 )
"If we slightly perturb with epsilon<0 then the parabolic fixed point splits up into two real fixed points on the real axis (one attracting, one repelling). "
See :
• demo 2 page 9 in program Mandel by Wolf Jung
# Local dynamics
Local dynamics :
• in the exterior of Julia set
• on the Julia set
• near parabolic fixed point ( inside Julia set )
## Near parabolic fixed point
Orbits near parabolic fixed point and inside Julia set
Why analyze f^p not f ?
Forward orbit of f near parabolic fixed point is composite. It consist of 2 motions :
• around fixed point
• toward / away from fixed point
# How to compute parabolic c values
Description
Parabolic points of period 1 component of Mandelbrot set (parameter plane)
n Internal angle (rotation number) t = 1/n The root point c = parabolic parameter Two external angles of parameter rays landing on the root point c (1/(2^n+1); 2/(2^n+1) fixed point $z_{\alpha}$ external angles of dynamic rays landing on fixed point $z_{\alpha}$
1 1/1 0.25 (0/1 ; 1/1) 0.5 (0/1 = 1/1)
2 1/2 -0.75 (1/3; 2/3) -0.5 (1/3; 2/3)
3 1/3 0.64951905283833*%i-0.125 (1/7; 2/7) 0.43301270189222*%i-0.25 (1/7; 2/7; 3/7)
4 1/4 0.5*%i+0.25 (1/15; 2/15) 0.5*%i (1/15; 2/15; 4/15; 8/15)
5 1/5 0.32858194507446*%i+0.35676274578121 (1/31; 2/31) 0.47552825814758*%i+0.15450849718747 (1/31; 2/31; 4/31; 8/31; 16/31)
6 1/6 0.21650635094611*%i+0.375 (1/63; 2/63) 0.43301270189222*%i+0.25 (1/63; 2/63; 4/63; 8/63; 16/63; 32/63)
7 1/7 0.14718376318856*%i+0.36737513441845 (1/127; 2/127) 0.39091574123401*%i+0.31174490092937 (1/127; 2/127, 4/127; 8/127; 16/127; 32/127, 64/127)
8 1/8 0.10355339059327*%i+0.35355339059327 0.35355339059327*%i+0.35355339059327
9 1/9 0.075191866590218*%i+0.33961017714276 0.32139380484327*%i+0.38302222155949
10 1/10 0.056128497072448*%i+0.32725424859374 0.29389262614624*%i+0.40450849718747
For internal angle n/p parabolic period p cycle consist of one z-point with multiplicity p[29] and multiplier = 1.0 . This point z is equal to fixed point $z_{alfa}$
## Period 1
One can easily compute boundary point c
$c = c_x + c_y*i$
of period 1 hyperbolic component ( main cardioid) for given internal angle ( rotation number) t using this cpp code by Wolf Jung[30]
t *= (2*PI); // from turns to radians
cx = 0.5*cos(t) - 0.25*cos(2*t);
cy = 0.5*sin(t) - 0.25*sin(2*t);
or this Maxima CAS code :
/* conformal map from circle to cardioid ( boundary
of period 1 component of Mandelbrot set */
F(w):=w/2-w*w/4;
/*
circle D={w:abs(w)=1 } where w=l(t,r)
t is angle in turns ; 1 turn = 360 degree = 2*Pi radians
*/
ToCircle(t,r):=r*%e^(%i*t*2*%pi);
(
[w],
/* point of unit circle w:l(internalAngle,internalRadius); */
w:ToCircle(angle,radius), /* point of circle */
float(rectform(F(w))) /* point on boundary of period 1 component of Mandelbrot set */
)$compile(all)$
/* ---------- global constants & var ---------------------------*/
Numerator :1;
DenominatorMax :10;
/* --------- main -------------- */
for Denominator:1 thru DenominatorMax step 1 do
(
InternalAngle: Numerator/Denominator,
display(Denominator),
display(c),
/* compute fixed point */
alfa:float(rectform((1-sqrt(1-4*c))/2)), /* alfa fixed point */
display(alfa)
)$ ## Period 2 // cpp code by W Jung http://www.mndynamics.com t *= (2*PI); // from turns to radians cx = 0.25*cos(t) - 1.0; cy = 0.25*sin(t); ## Periods 1-6 /* batch file for Maxima CAS computing bifurcation points for period 1-6 Formulae for cycles in the Mandelbrot set II Stephenson, John; Ridgway, Douglas T. Physica A, Volume 190, Issue 1-2, p. 104-116. */ kill(all); remvalue(all); start:elapsed_run_time (); /* ------------ functions ----------------------*/ /* exponential for of complex number with angle in turns */ /* "exponential form prevents allroots from working", code by Robert P. Munafo */ GivePoint(Radius,t):=rectform(ev(Radius*%e^(%i*t*2*%pi), numer))$ /* gives point of unit circle for angle t in turns */
GiveCirclePoint(t):=rectform(ev(%e^(%i*t*2*%pi), numer))$/* gives point of unit circle for angle t in turns Radius = 1 */ /* gives a list of iMax points of unit circle */ GiveCirclePoints(iMax):=block( [circle_angles,CirclePoints], CirclePoints:[], circle_angles:makelist(i/iMax,i,0,iMax), for t in circle_angles do CirclePoints:cons(GivePoint(1,t),CirclePoints), return(CirclePoints) /* multipliers */ )$
/* http://commons.wikimedia.org/wiki/File:Mandelbrot_set_Components.jpg
Boundary equation b_n(c,P)=0
defines relations between hyperbolic components and unit circle for given period n ,
allows computation of exact coordinates of hyperbolic componenets.
b_n(w,c), is boundary polynomial ( implicit function of 2 variables ).
*/
GiveBoundaryEq(P,n):=
block(
if n=1 then return(c + P^2 - P),
if n=2 then return(- c + P - 1),
if n=3 then return(c^3 + 2*c^2 - (P-1)*c + (P-1)^2),
if n=4 then return( c^6 + 3*c^5 + (P+3)* c^4 + (P+3)* c^3 - (P+2)*(P-1)*c^2 - (P-1)^3),
if n=5 then return(c^15 + 8*c^14 + 28*c^13 + (P + 60)*c^12 + (7*P + 94)*c^11 +
(3*P^2 + 20*P + 116)*c^10 + (11*P^2 + 33*P + 114)*c^9 + (6*P^2 + 40*P + 94)*c^8 +
(2*P^3 - 20*P^2 + 37*P + 69)*c^7 + (3*P - 11)*(3*P^2 - 3*P - 4)*c^6 + (P - 1)*(3*P^3 + 20*P^2 - 33*P - 26)*c^5 +
(3*P^2 + 27*P + 14)*(P - 1)^2*c^4 - (6*P + 5)*(P - 1)^3*c^3 + (P + 2)*(P - 1)^4*c^2 - c*(P - 1)^5 + (P - 1)^6),
if n=6 then return( c^27+
13*c^26+
78*c^25+
(293 - P)*c^24+
(792 - 10*P)*c^23+
(1672 - 41*P)*c^22+
(2892 - 84*P - 4*P^2)*c^21+
(4219 - 60*P - 30*P^2)*c^20+
(5313 + 155*P - 80*P^2)*c^19+
(5892 + 642*P - 57*P^2 + 4*P^3)*c^18+
(5843 + 1347*P + 195*P^2 + 22*P^3)*c^17+
(5258 + 2036*P + 734*P^2 + 22*P^3)*c^16+
(4346 + 2455*P + 1441*P^2 - 112*P^3 + 6*P^4)*c^15 +
(3310 + 2522*P + 1941*P^2 - 441*P^3 + 20*P^4)*c^14 +
(2331 + 2272*P + 1881*P^2 - 853*P^3 - 15*P^4)*c^13 +
(1525 + 1842*P + 1344*P^2 - 1157*P^3 - 124*P^4 - 6*P^5)*c^12 +
(927 + 1385*P + 570*P^2 - 1143*P^3 - 189*P^4 - 14*P^5)*c^11 +
(536 + 923*P - 126*P^2 - 774*P^3 - 186*P^4 + 11*P^5)*c^10 +
(298 + 834*P + 367*P^2 + 45*P^3 - 4*P^4 + 4*P^5)*(1-P)*c^9 +
(155 + 445*P - 148*P^2 - 109*P^3 + 103*P^4 + 2*P^5)*(1-P)*c^8 +
2*(38 + 142*P - 37*P^2 - 62*P^3 + 17*P^4)*(1-P)^2*c^7 +
(35 + 166*P + 18*P^2 - 75*P^3 - 4*P^4)*((1-P)^3)*c^6 +
(17 + 94*P + 62*P^2 + 2*P^3)*((1-P)^4)*c^5 +
(7 + 34*P + 8*P^2)*((1-P)^5)*c^4 +
(3 + 10*P + P^2)*((1-P)^6)*c^3 +
(1 + P)*((1-P)^7)*c^2 +
-c*((1-P)^8) + (1-P)^9)
)$/* gives a list of points c on boundaries on all components for give period */ GiveBoundaryPoints(period,Circle_Points):=block( [Boundary,P,eq,roots], Boundary:[], for m in Circle_Points do (/* map from reference plane to parameter plane */ P:m/2^period, eq:GiveBoundaryEq(P,period), /* Boundary equation b_n(c,P)=0 */ roots:bfallroots(%i*eq), roots:map(rhs,roots), for root in roots do Boundary:cons(root,Boundary)), return(Boundary) )$
/* divide llist of roots to 3 sublists = 3 components */
/* ---- boundaries of period 3 components
period:3$Boundary3Left:[]$
Boundary3Up:[]$Boundary3Down:[]$
for m in CirclePoints do (
P:m/2^period,
eq:GiveBoundaryEq(P,period),
roots:bfallroots(%i*eq),
roots:map(rhs,roots),
for root in roots do
(
if realpart(root)<-1 then Boundary3Left:cons(root,Boundary3Left),
if (realpart(root)>-1 and imagpart(root)>0.5)
then Boundary3Up:cons(root,Boundary3Up),
if (realpart(root)>-1 and imagpart(root)<0.5)
then Boundary3Down:cons(root,Boundary3Down)
)
)$--------- */ /* gives a list of parabolic points for given : period and internal angle */ GiveParabolicPoints(period,t):=block ( [m,ParabolicPoints,P,eq,roots], m: GiveCirclePoint(t), /* root of unit circle, Radius=1, angle t=0 */ ParabolicPoints:[], /* map from reference plane to parameter plane */ P:m/2^period, eq:GiveBoundaryEq(P,period), /* Boundary equation b_n(c,P)=0 */ roots:bfallroots(%i*eq), roots:map(rhs,roots), for root in roots do ParabolicPoints:cons(float(root),ParabolicPoints), return(ParabolicPoints) )$
compile(all)$/* ------------- constant values ----------------------*/ fpprec:16; /* ------------unit circle on a w-plane -----------------------------------------*/ a:GiveParabolicPoints(6,1/3); a$
# How to draw parabolic Julia set
All points of interior of filled Julia set tend to one periodic orbit ( or fixed point ). This point is in Julia set and is weakly attracting. [31] One can analyse only behevior near parabolic fixed point. It can be done using critical orbits.
There are two cases here : easy and hard.
If the Julia set near parabolic fixed point is like n-th arm star ( not twisted) then one can simply check argument of of zn, relative to the fixed point. See for example z+z^5. This is an easy case.
In the hard case Julia set is twisted around fixed.
## Estimation from exterior
Description
### Long iteration method
Long iteration method [32]
### Dynamic rays
Parabolic Julia set for internal angle 1 over 15 - made with use of external rays as a aproximation of Julia set near alfa fixed point
One can use periodic dynamic rays landing on parabolic fixed point to find narrow parts of exterior.
Let's check how many backward iterations needs point on periodic ray with external radius = 4 to reach distance 0.003 from parabolic fixed point :
period Inverse iterations time
1 340 0m0.021s
2 55 573 0m5.517s
3 8 084 815 13m13.800s
4 1 059 839 105 1724m28.990s
One can use only argument of point z of external rays and its distance to alfa fixed point. ( see code from image) It works for periods up to 15 ( maybe more ... )
## Estimation from interior
Julia set is a boundary of filled-in Julia set Kc.
• find points of interior of Kc
• find boundary of interior of Kc using edge detection
If components of interior are lying very close to each other then find components using :[33]
color = LastIteration % period
For parabolic components between parent and child component :[34]
periodOfChild = denominator*periodOfParent
color = iLastIteration % periodOfChild
where denominator is a denominator of internal angle of parent comonent of Mandelbrot set.
### Angle
"if the iterate zn of tends to a fixed parabolic point, then the initial seed z0 is classified according to the argument of zn−z0, the classification being provided by the flower theorem " ( Mark McClure [35])
### Attraction time
Various types of dynamics
Interior of filled Julia set consist of components. All comonents are preperiodic, some of them are periodic ( immediate basin of attraction).
In other words :
• one iteration moves z to another component ( and whole component to another component)
• all point of components have the same attraction time ( number of iteration needed to reach target set around attractor)
It is possible to use it to color components. Because in parabolic case attractor is weak ( weakly attracting) it needs a lot of iterations for some points to reach it.
Here are some example values :
iWidth = 1001 // width of image in pixels
PixelWidth = 0.003996
AR = 0.003996 // Radius around attractor
denominator = 1 ; Cx = 0.250000000000000; Cy = 0.000000000000000 ax = 0.500000000000000; ay = 0.000000000000000
denominator = 2 ; Cx = -0.750000000000000; Cy = 0.000000000000000 ax = -0.500000000000000; ay = 0.000000000000000
denominator = 3 ; Cx = -0.125000000000000; Cy = 0.649519052838329 ax = -0.250000000000000; ay = 0.433012701892219
denominator = 4 ; Cx = 0.250000000000000; Cy = 0.500000000000000 ax = 0.000000000000000; ay = 0.500000000000000
denominator = 5 ; Cx = 0.356762745781211; Cy = 0.328581945074458 ax = 0.154508497187474; ay = 0.475528258147577
denominator = 6 ; Cx = 0.375000000000000; Cy = 0.216506350946110 ax = 0.250000000000000; ay = 0.433012701892219
denominator = 1 ; i = 243.000000
denominator = 2 ; i = 31 171.000000
denominator = 3 ; i = 3 400 099.000000
denominator = 4 ; i = 333 293 206.000000
denominator = 5 ; i = 29 519 565 177.000000
denominator = 6 ; i = 2 384 557 783 634.000000
where :
C = Cx + Cy*i
a = ax + ay*i // fixed point alpha
i // number of iterations after which critical point z=0.0 reaches disc around fixed point alpha with radius AR
denominator of internal angle ( in turns )
internal angle = 1/denominator
Note that attraction time i is proportional to denominator.
Attraction time for various denominators
Now you see what means weakly attracting.
One can :
• use brutal force method ( Attracting radius < pixelSize; iteration Max big enough to let all points from interior reach target set; long time or fast computer )
• find better method (:-)) if time is to long for you
## Estimation from interior and exterior
Julia set is a common boundary of filled-in Julia set and basin of attraction of infinity.
• find points of interior/components of Kc
• find escaping points
• find boundary points using Sobel filter
It works for denominator up to 4.
## Inverse iteration of repelling points
Inverse iteration of alfa fixed point. It works good only for cuting point ( where external rays land). Other points still are not hitten. | 7,107 | 21,676 | {"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": 26, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2015-22 | latest | en | 0.714283 |
https://devsenv.com/example/-2682-beecrowd-online-judge-solution-2682-fault-detector-solution-in-c,-c++,-java,-js-and-python | 1,726,435,244,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00512.warc.gz | 176,894,438 | 38,547 | ## Algorithm
Problem Name: 2 AD-HOC - beecrowd | 2682
# Fault Detector
By Emilio Wuerges, UFFS Brazil
Time limit: 1
There is a machine, that produces an increasing sequence of numbers. That is, every number of this sequence is larger than its predecessor.
However, this machine is starting to break. When it starts, everything is OK. However, after some time, it starts producing wrong results.
Your task is, whenever the machine produces the first wrong number or if it turns off, ignore all following results and produce the next smallest valid number.
Since we are just checking the machine, we cannot turn it off. We have to wait it turn it of on its own. That is, we must keep reading numbers until the machine turns of on its own.
## Input
The input consists of 0 < N < 104 lines, and ends with EOF.
Each line consists of a single integer 0 < X < 230.
## Output
A single line, with a single integer Y, the solution for the problem.
Input Sample Output Sample 1 2 41 5 2 1 42
## Code Examples
### #1 Code Example with C Programming
```Code - C Programming```
``````
#include <stdio.h>
#include <stdbool.h>
int main (void)
{
unsigned numeroAtual = 1, numeroAnt = 0, numero = 0;
bool temfalha = false;
while (scanf("%u", &numeroAtual) != EOF)
{
if ((!temfalha) && (numeroAtual < numeroAnt))
{
temfalha = true;
numero = numeroAnt;
}
numeroAnt = numeroAtual;
}
if (temfalha)
printf("%u\n", numero + 1);
else
printf("%u\n", numeroAtual + 1);
}
``````
Copy The Code &
Input
cmd
1
2
41
5
2
1
Output
cmd
42
### #2 Code Example with C++ Programming
```Code - C++ Programming```
``````
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
int max=0;
bool wrong=false;
while(cin>>n)
{
if(n>max&&!wrong)max=n;
else
{
wrong=true;
}
}
cout<<max+1<<endl;
}
``````
Copy The Code &
Input
cmd
1
2
41
5
2
1
Output
cmd
42
### #3 Code Example with Javascript Programming
```Code - Javascript Programming```
``````
"use strict"
const { readFileSync } = require("fs")
const input = readFileSync("/dev/stdin", "utf8").split("\n").map(line => Number.parseInt(line, 10))
input.pop() // Remove EOF
const error = input.find((value, index, arr) => value > arr[index + 1]) || input[input.length - 1]
console.log(error + 1)
``````
Copy The Code &
Input
cmd
1
2
41
5
2
1
Output
cmd
42
### #4 Code Example with Python Programming
```Code - Python Programming```
``````
maior = 0
ant = 0
while True:
try:
n = int(input())
if ant > n and maior == 0:
maior = ant + 1
ant = n
except EOFError:break
if maior == 0:maior = ant + 1
print(maior)
``````
Copy The Code &
Input
cmd
1
2
41
5
2
1
Output
cmd
42 | 780 | 2,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.691416 |
http://wcipeg.com/problem/boi09p5 | 1,540,312,838,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516480.46/warc/CC-MAIN-20181023153446-20181023174946-00356.warc.gz | 404,416,297 | 3,764 | ### 2009 Bulgarian Olympiad in Informatics
Consider the positive integers whose squares contain only (and all) the digits 0,4,9. Let's call them "special". For example, 2120 is special, because 21202 = 4494400 and the square contains only (and all) of 0,4,9. 97 is also special: 972 = 9409. 13 and 7 are not special - 132 = 169 (1 and 6 aren't allowed) and 72 = 49 (there's no 0)
Consider the sequence of special numbers, in order:
{70, 97, 700, 970, 997, 2120, 3148, 7000, 9700, 9970, 9997, 20102, 21200, 31480, 70000, 97000, ...}.
Write a program to find the Nth number in this sequence.
### Input
The positive integer N ≤ 250, on a single line.
### Output
The Nth number in the special sequence (starting from 1).
`12`
### Sample Output
`20102`
Point Value: 30 (partial)
Time Limit: 1.00s
Memory Limit: 32M | 275 | 819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-43 | longest | en | 0.791449 |
https://goprep.co/a-copper-wire-having-resistance-0-01-ohm-in-each-meter-is-i-1nm2mk | 1,642,866,728,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00141.warc.gz | 368,984,174 | 32,602 | # A copper wire hav
Given:
Resistance per unit length(R0) = 0.01 ohm
Total number of turns(N) = 400
Radius of the wire(r) = 1 cm = 0.01 m
Length of the wire(l) = 20 cm = 0.2 m
Magnetic field near the center(B) = 1.0×10–2 T
Formula used:
Magnetic field at the center of a solenoid, B = μ0ni,
Where
μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1,
n = number of turns per unit length,
i = current carried by the wire
Emf(E) = iR,
where i = current, R = resistance
Now, n(number of turns per unit length) = Total number of turns(N)/Length of wire(l) = = 2000
Total resistance (R of all turns) = R0 x 2πr x 400 = (0.01 x 2π x 0.01 x 400) = 0.25 ohm
Now, substituting the given values:
1.0×10-2 = 4π x 10-7 x 2000 x i => i = 3.98 A
Therefore, Emf(E) = iR = (3.98 X 0.25) = 0.995 V which is almost equal to 1 V.
Emf of the battery = 1 V (Ans)
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :
<span lang="EN-USHC Verma - Concepts of Physics Part 2
A tightly-wound, HC Verma - Concepts of Physics Part 2
An electron is prPhysics - Exemplar
The magnetic fielHC Verma - Concepts of Physics Part 2
A long solenoid iHC Verma - Concepts of Physics Part 2
A capacitor of caHC Verma - Concepts of Physics Part 2
A tightly-wound, HC Verma - Concepts of Physics Part 2
A tightly-wound, HC Verma - Concepts of Physics Part 2
A toroid has a coNCERT - Physics Part-I | 544 | 1,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.801567 |
http://www.scienceu.com/geometry/articles/tiling/symmetry.html | 1,586,547,410,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00129.warc.gz | 269,891,041 | 7,124 | Symmetric Tilings
For most people, the idea of symmetry conjures up thoughts of Art and Nature, as well as mathematics. Our ideas of beauty are closely tied up with symmetry, and principles of symmetry manifest themselves in subtle ways throughout the natural world. Books like Design and Color in Islamic Architecture : Eight Centuries of the Tile-Maker's Art and Structure in Nature Is a Strategy for Design pursue these connections in greater detail.
### What is Symmetry?
Even though we all understand and recognize symmetry intuitively, it is a little harder to say just what it is. However, in the plane, the basic idea is clear enough:
A figure in the plane is symmetric if you could pick up a copy of it, move it around to a new location somehow, and set it back down on the original figure so that it exactly matches up again.
The animations below illustrate this definition of symmetry in the case of a checkerboard tiling. On the left, the tiling is just being shifted or translated is such a way that it matches up with itself again. On the right, the tiling is being rotated a quarter turn to make it match up again.
The checkerboard tiling is a very symmetric tilings. There are lots and lots of ways of moving it on top of itself so that it matches up again in addition to the two shown above. Mathematicians say that different way of moving a tiling onto itself is a "symmetry of the tiling". How many symmetries can you think of for the checkerboard tiling?
### Kinds of Symmetry
One of the first things to notice about symmetry is that there are several different kinds. This is because there are different ways of moving something in the plane. One way is to just translate it a little. Another is to rotate it. Yet another is to turn it over. As a consequence, there are different kinds of symmetry.
Consider the pictures above. The one on the left exhibits rotational symmetry. You could pick it up, rotate it a third of a circle, and set it back down so it would exactly match up. The figure on the right has mirror symmetry. In this case, you could pick it up, turn it over along the axis of the dotted line, and set it back down on top of itself.
There are actually four distinct kinds of symmetry, corresponding to four basic ways of moving a tile around in the plane, illustrated to the right. In mathematical language, these different ways of moving things in the plane are called isometries.
To get a quick idea about how isometries work, you can look at some animations of translations, rotations, reflections, and glide reflections, or you can read about isometries in more depth in Introduction to Isometries in the Science U library.
### Symmetry Groups
If a tiling has any symmetries at all, it usually has lots of them. This is simply because if we do one symmetry followed by another, then we could have just move the tiling directly from its initial postion to its final position, and it would still match up. In other words, the net effect of doing two symmetries one after the other is itself a new symmetry. In mathematical language, we say that the composition of two isometries (i.e. symmetries) is again an isometry. For example, as you can see on the left, if we flip the checkerboard tiling twice, one in the x-direction, and once in the y-direction, it has the same net effect as rotating the tiling half a turn. (Focus on the purple square for a landmark.)
This pleasant fact makes it much easier to think about the symmetry group for a tiling. The symmetry group of a tiling is just the collection of all its symmetries. If we think of the checkerboard tiling continuing on forever in all directions, it is clear that it has infinitely many symmetries. Just for starters, we could tranlate any even number of squares in the x-direction or y-direction and the tiling would still match up. This doesn't even count all the rotations and other isometries which take it back onto itself.
In general, the more symmetric a tiling is, the larger its symmetry group is. Indeed, as our checkboard example shows, symmetry groups can be huge, infinite things. However, because we can compose symmetries to get other symmetries, instead of thinking about all symmetries, we need only think about a few basic ones that we can compose with each other to get all the rest!
In the case of the checkerboard tiling, for example, instead of thinking about all of the possible ways of translating it back on top of itself, we can concentrate on two basic translations, one in the x-direction and one in y-direction, as show above on the left. We can get any other translational symmetry just by doing these two enough times. For example, the diagonal translation on the right is the same as the net effect of the two vertical and one horizontal basic translations shown in the middle.
Mathematicians call such a collection of basic symmetries the generators of the symmetry group. As you will see, generators play an important role in classifying different kinds of symmetry groups.
Info Center | Geometry Center | Library | Observatory | Studio | Store | Science Me
Page last updated Wed Oct 19 23:11:18 CDT 2005 | 1,113 | 5,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-16 | longest | en | 0.954664 |
http://www.reddit.com/r/learnprogramming/comments/k63j4/c_for_loop_question/ | 1,406,361,070,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997895170.20/warc/CC-MAIN-20140722025815-00089-ip-10-33-131-23.ec2.internal.warc.gz | 1,049,353,866 | 16,493 | [–] 1 point2 points
sorry, this has been archived and can no longer be voted on
all the powers from 1 to the user's integer
This should tell you all you need to know about the variable to use in the for loop, i.e. what to initialize to and when to stop looping.
As for the sum, initialize a variable outside the loop, then add to it inside the loop with `+=`.
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
What do you mean by "total the sum of all the powers from 1 to the user's input"? The powers of what?
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
the powers of the user integer, i'm assuming.
[–][S] 0 points1 point
sorry, this has been archived and can no longer be voted on
say the user enters 10. Then the program would add 12 22 32 etc. Then the program would cout the total of all powers.
[–] 1 point2 points *
sorry, this has been archived and can no longer be voted on
oh all the squares, not powers. The pseudo code would be:
``````int x; //users input
int sum; //tally of the sum of the squares
int i; //counting variable
input x;
for(from i = 0 to the users input){
sum += i^i;
}
output sum;
``````
try to code it up yourself, tell me if you have any problems changing it to c++.
Edit: read your post thing wrong. i saw 12 + 22 and assumed they were all squared. i have changed my pseudo to fit.
[–][S] 0 points1 point
sorry, this has been archived and can no longer be voted on
Thank you I did not know about the += operator. I have one question, how do you properly initialize sum. I just used zero and it works, but is there a better way?
[–] 2 points3 points
sorry, this has been archived and can no longer be voted on
The += operator isn't actually needed, it is just convenient. You also could have done:
`````` sum = sum + i^i;
``````
It is basically just a quick for to save useless typing. As for initializing sum, you can do it at sum's declaration, or do it explicitly later. either works. but yes, initialize it to zero.
[–][S] 0 points1 point
sorry, this has been archived and can no longer be voted on
Thank you, you really helped me out a lot.
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
sum = sum + ii
This would actually perform a XOR operation, resulting in nothing being added to sum.
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
Oh my bad, totally spaced on that. Yeah the ^ is the XOR operator. It was half pseudo code I was writing, I should have done that better and used the power operator.
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
C++ and C are void of a exponent operator :P
[–] 0 points1 point *
sorry, this has been archived and can no longer be voted on
Isn't exp() the power operator?
Edit: oh its in the standard library, not actually built in. My bad.
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
sum = sum + i
Is the same as
sum += i
and if i == 1, then it is the same as
sum++
[–] -1 points0 points *
sorry, this has been archived and can no longer be voted on
adding the powers up to an integer? That could be done a few ways, starting with the absolute easiest:
``````#include <cmath>
``````
Then anywhere in your loop you'll need a subloop, so something like this
``````int sum=0;
for (int i=0; i< usr; i++){
sum+= pow(i, i);
}
``````
The only other way you could do this is a little more contrived:
``````int sum=0;
for(int i=1; i<=usr; i++){
int temp=i;
for (int j=1; j<i; j++){
temp *= i;
}
sum+= temp;
}
``````
Edit 1: Cleaned up code
Edit 2: rehosted code so you can see it better: https://ideone.com/9LXBd
Edit 3: Made the code posted in part 1, apparently pow only accepts doubles, so we used a run-time cast to convert it: https://ideone.com/nXLV3
Edit 4: As a sidenote, a program like this would be better suited to use double or long, as the max value for int won't make it very high in this kind of program.
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
long in many things is just going to be an integer. You probably want long long or __int64 instead.
You can also double the range by using and unsigned integer (or long long/__int64)
Working with floating point numbers can be very frustrating due to the error. I wouldn't recommend using double for math that doesn't need fractional values (or floating point values in general, unless you REALLY do need them.)
[–] 0 points1 point
sorry, this has been archived and can no longer be voted on
I was thinking u long long or something like that would be more appropriate, but I hadn't taught that type to my class at that point, so I was just going with "largest simple data storage that I've covered", and that's what it was =P | 1,293 | 4,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2014-23 | latest | en | 0.959158 |
http://www.kylesconverter.com/frequency/radians-per-hour-to-terahertz | 1,521,629,565,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647612.53/warc/CC-MAIN-20180321102234-20180321122234-00340.warc.gz | 403,926,899 | 5,314 | # Convert Radians Per Hour to Terahertz
### Kyle's Converter > Frequency > Radians Per Hour > Radians Per Hour to Terahertz
Radians Per Hour (rad/h)* Terahertz (THz) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Terahertz to Radians Per Hour
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Radian per Hour:
1 Radian per hour is comparative to 1/7200π Hertz. Radians per hour is a measure of angular frequency, it can be compared to Hertz or other angular units. Formula uses an approximation of PI. 1 rad/h is approximately 0.000044209706414415 Hz.
1 Terahertz:
1 Terahertz is exactly one trillion Hertz. 1 THz = 1 x 1012 Hz. 1 THz = 1000000000000 Hz.
Link to Your Exact Conversion
Conversions Table
1 Radians Per Hour to Terahertz = 070 Radians Per Hour to Terahertz = 0
2 Radians Per Hour to Terahertz = 080 Radians Per Hour to Terahertz = 0
3 Radians Per Hour to Terahertz = 090 Radians Per Hour to Terahertz = 0
4 Radians Per Hour to Terahertz = 0100 Radians Per Hour to Terahertz = 0
5 Radians Per Hour to Terahertz = 0200 Radians Per Hour to Terahertz = 0
6 Radians Per Hour to Terahertz = 0300 Radians Per Hour to Terahertz = 0
7 Radians Per Hour to Terahertz = 0400 Radians Per Hour to Terahertz = 0
8 Radians Per Hour to Terahertz = 0500 Radians Per Hour to Terahertz = 0
9 Radians Per Hour to Terahertz = 0600 Radians Per Hour to Terahertz = 0
10 Radians Per Hour to Terahertz = 0800 Radians Per Hour to Terahertz = 0
20 Radians Per Hour to Terahertz = 0900 Radians Per Hour to Terahertz = 0
30 Radians Per Hour to Terahertz = 01,000 Radians Per Hour to Terahertz = 0
40 Radians Per Hour to Terahertz = 010,000 Radians Per Hour to Terahertz = 0
50 Radians Per Hour to Terahertz = 0100,000 Radians Per Hour to Terahertz = 0
60 Radians Per Hour to Terahertz = 01,000,000 Radians Per Hour to Terahertz = 0 | 614 | 1,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-13 | longest | en | 0.698696 |
https://web2.0calc.com/questions/need-help_23761 | 1,603,350,625,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00001.warc.gz | 581,085,442 | 6,374 | +0
# need help
0
82
1
James combines 7.28 liters of red paint with 8.06 liters of blue paint to make purple paint. He pours the paint equally into 4 containers and has 1.58 liters of paint leftover. How many liters of paint are in each container?
Sep 25, 2020
#1
+232
+1
7.28 + 8.06 = 15.34
15.34 - 1.58 = 13.76
13.76 / 4 = 3.44
Sep 25, 2020 | 164 | 350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-45 | longest | en | 0.277083 |
http://nrich.maths.org/1351 | 1,438,266,168,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987228.91/warc/CC-MAIN-20150728002307-00206-ip-10-236-191-2.ec2.internal.warc.gz | 177,680,617 | 6,125 | ### Good Approximations
Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.
### There's a Limit
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Not Continued Fractions
Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?
# Continued Fractions I
##### Stage: 4 and 5
Published November 2009,May 1999,October 2009,December 2011,February 2011.
Continued fractions are written as fractions within fractions which are added up in a special way, and which may go on for ever. Every number can be written as a continued fraction and the finite continued fractions are sometimes used to give approximations to numbers like $\sqrt 2$ and $\pi$.
To see how to work out a continued fraction let $$X = {1\over\displaystyle 2\;+\; {\strut 3\over \displaystyle 4 }}.$$ Adding the fractions in the denominator of $X$ we see the denominator is $11/4$. So $X= 1/(11/4)=4/11$.
Let us work out the slightly longer continued fraction $$Y = {1\over\displaystyle 2\;+\; {\strut 3\over \displaystyle 4\;+\; {\strut 5\over \displaystyle 6 }}}.$$ We can calculate $Y$ as follows: $$Y = {1\over\displaystyle 2\;+\; {\strut 3\over \displaystyle 4\;+\; {\strut 5\over \displaystyle 6 }}} = {1\over\displaystyle 2\;+\; {\strut 3\over \displaystyle 29/6}} = {1\over\displaystyle 2\;+\; {\strut 18\over \displaystyle 29}} = {1\over\displaystyle 76/29}={29\over 76}.$$ Can you show that $${1\over\displaystyle 2\;+\; {\strut 2\over \displaystyle 2\;+\; {\strut 2\over \displaystyle 2\;+\; {\strut 2\over \displaystyle 3 }}}}\quad = \quad {11\over 30}\quad ?$$ Now we have got the idea we are in for some surprises! Watch out for some patterns in the numbers that come up. Work out the values of the five continued fractions: $$1,\quad {1\over 1+1},\quad {1\over\displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; {\strut 1 }}},\quad {1\over\displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; {\strut 1 }}}},\quad {1\over\displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; {\strut 1 }}}}}.$$ Check your answers
Did you find the easiest way to calculate these? For example, you should be able to see that the last one is $${1\over\displaystyle 1\;+\; {\strut 3\over \displaystyle 5 }}\quad = \quad {5\over 8}.$$ In this sequence of continued fractions you can always calculate one quickly by using the previous answer. The next fraction in this sequence is $${1\over\displaystyle 1\;+\; {\strut 5\over \displaystyle 8 }}\quad = \quad {8\over 13}.$$ The numbers we get in order are $1, 2, 3, 5, 8, 13$. What do you think the next number is? Yes, these are the Fibonacci numbers. What do you think the next continued fraction in the sequence is?
Now let us find out what happens if the continued fraction goes on for ever. We write this as $$f = {1\over\displaystyle 1+ {\strut 1\over \displaystyle 1\;+\; {\strut 1\over \displaystyle 1\;+\; \cdots }}}.$$ Can you see why we have $$f = {1\over 1+f}\quad ?$$ This gives the quadratic equation $f^2 + f -1 = 0$. Because $f$ is positive we get the one solution $$f = {\sqrt{5}-1\over 2},$$ the ratio of the shorter to the longer side of the Golden Rectangle!
Now investigate the continued fraction $${6\over\displaystyle 1\;+\; {\strut 6\over \displaystyle 1\;+\; {\strut 6\over \displaystyle 1\;+\; {\strut 6\over \displaystyle 1\;+\; \cdots }}}}.$$ The answer is a small whole number (which is obviously less than 6).
You may like to try a problem on continued fractions and its solution or the problem Fibonacci numbers and its solution.
You can find out more about the many contexts in which Fibonacci numbers appear by exploring a website which is devoted entirely to them. This site has practical activities for work on Fibonacci numbers and references to books giving many more.
The next part of the series is here. See also the article Approximations, Euclid's Algorithm and Continued Fractions to find out how continued fractions are used to give very quickly better and better rational approximations to numbers like pi. | 1,296 | 4,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2015-32 | longest | en | 0.787903 |
https://books.google.com/books/about/40_Fun_Tabulous_Puzzles_for_Multiplicati.html?id=F0oVo92Ev8QC&hl=en | 1,495,546,460,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00362.warc.gz | 724,925,191 | 10,934 | # 40 Fun-Tabulous Puzzles for Multiplication, Division, Decimals, Fractions, & More!
Scholastic Inc., Nov 1, 2000 - Education - 48 pages
The world lies devastated after the massive oil crisis that was described in LAST LIGHT. Human society has more or less entirely broken down and millions lie dead of starvation and disease. There are only one or two beacon communities that have managed to fashion a new way of living.Jenny Sutherland runs one of these groups. Based on a series of decaying offshore oil-rigs - for safety - a few hundred people have rebuilt a semblance of normality in this otherwise dead world.But as her and her people start to explore their surroundings once again, they start to realise not every survivor has the same vision of a better future than their catastrophic past. There are people out there who would take everything they have. War is coming, and the stakes are truly massive...
### What people are saying -Write a review
User Review - Flag as inappropriate
User Review - Flag as inappropriate
helps on things. thank-you!
### Contents
Rewriting words as numbers place value 5 CrossNumber Puzzle 6 Regrouping 7 Break the Code 3 digits4addends 8 Sum Number Search 4 digits4 addends 9 Regrouping 10 Last NumberFirst Number 1 5 digits 11 Solve the Mystery 5 digits 12
99s 2 digit divisor4digit quotient 27 More Did You Hear? Riddles 2digit divisor4digit quotient 28 Addition subtraction multiplication and division review 29 Links 30 Shapely Math 1 31 Using Variables 32 Riddle Time addition and multiplication 33 FRACTIONS 34
Addition and subtraction review 13 Cross Them Out 1 14 Regrouping 15 What a MixUp factors and products 16 Solve the Riddle 1 digit x 3 or 4 digits 17 Cross Them Out 2 1 digit x 4 digits 18 Match It 1 2 digits x 2 digits 19 Secret Code Time 2 digits x 2 digits 20 Monster Mystery 2 digits x 3 digits 21 Last NumberFirst Number 2 2 digits x 3 digits 22 Addition subtraction and multiplication review 23 Long division and remainders 24 Remainders 1digit divisor3digit quotient 25 Match It 2 1digit divisor4digit quotient 26
A Sharp Riddle addition with unlikedenominators 35 Wrestle the Code order of operations 36 Decimal Match equivalent fraction and decimals 37 DECIMALS 38 Coded Riddle 2 x 3digit multiplication 39 Decimal Fun multiplestep operations 40 Shapely Math 2 order of operations 41 No Kidding order of operations 42 Locating ordered pairs 43 Hidden Question and Answer 2 44 Converting hours minutes and seconds 45 ANSWERS 46 Copyright | 609 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-22 | longest | en | 0.81517 |
https://ktane.timwi.de/HTML/Tax%20Returns%20optimized%20(Asew54321).html | 1,632,387,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00276.warc.gz | 403,955,997 | 2,458 | ## On the Subject of a Transparent Reading of Tax Returns
Turn off that veto!
• If you ever get a decimal for any of your calculations, disregard the decimal completely and just use the whole number. The HMRC does not concern themselves with pence, after all.
• Follow all instructions on this manual page in the order they are listed.
1. Add together all of the turnovers on the module. This number is your gross turnover (GT).
2. Add together all of the expenses on the module. This number is your gross expenses (GE).
3. Calculate your pension contributions (PC) based on the amount of lit and unlit indicators on the bomb
• If you have no indicators on the bomb, your PC is 0.
• Otherwise, if you have more lit indicators on the bomb than unlit indicators, your PC is your GT multiplied by 0.05.
• Otherwise, if you have more unlit indicators on the bomb than lit indicators, your PC is your GT multiplied by 0.1.
• Otherwise, if you have an equal amount of lit and unlit indicators on the bomb, your PC is your GT multiplied by 0.15.
4. Calculate your tax-free investments (TFI) using the information on the piece of paper behind the list of turnovers and expenses. Use the flowchart on the next page to find your base value.
• Multiply your the number you get from this flowchart by the amount of distinct port types on the bomb. This number is your TFI.
1. Calculate your gross profit (GP) with the following formula:
• GT - GE - PC - TFI = GP
2. Your tax-free allowance (TFA) is 11850. However, in the rare case that your GP is over 100000, then use this formula to find your TFA:
• 11850 - floor((GP - 100000) / 2) = TFA
3. Calculate your taxable income (TI) with the following formula:
• GP - TFA = TI
4. Calculate your basic rate (BR), higher rate (HR), and additional rate (AR) with these formulae:
• Your BR is 6900.
• If your TI is less than 34500, your BR is your TI * 0.2. If you use this formula, skip directly to step 9.
• Your HR is calculated with this formula:
• (TI - 34500) * 0.4
• If your TI is less than 138500, skip directly to step 9.
• Your AR is calculated with this formula:
• (TI - 138000) * 0.45
1. Add up your BR, HR, and AR values. This number is your income tax (IT).
2. Calculate your taxable National Insurance Contributions (TNIC) with the following formula:
• GT - GE = TNIC
3. Calculate your standard rate (SR) and excess rate (ER) with these formulae:
• Your SR is 3413.
• If your TNIC is under 46350, use the following formula, and then skip directly to step 12:
• (TNIC - 8424) * 0.09
• Your ER is calculated with this formula:
• (TNIC - 46350) * 0.02
4. Add your SR and ER values. This number is your National Insurance Contributions (NIC).
5. Add up your IT and NIC values. This number is your total tax bill (TTB).
6. Press the switch on the bottom-right corner of the module, and enter your TTB to disarm the module. | 755 | 2,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-39 | longest | en | 0.856031 |
https://brainly.com/question/49053 | 1,485,161,459,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282202.61/warc/CC-MAIN-20170116095122-00415-ip-10-171-10-70.ec2.internal.warc.gz | 790,759,441 | 8,543 | ## Answers
2014-05-08T12:30:15-04:00
Perimeter is sum of all sides , suppose you have to insert fence on the sides of house , so at that situation we can say that we have to find the perimeter............
here are sum formulae -
square - 4 * sides
rectangle - 2(lenght + breath(width)
triangle - side1+side2+side3 = perimeter
perimeter of a circle - 2r | 102 | 353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-04 | latest | en | 0.796774 |
https://math.stackexchange.com/questions/2798206/prove-limit-using-epsilon-delta-definition | 1,568,918,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00123.warc.gz | 579,864,324 | 32,987 | Prove limit using epsilon-delta definition
How do you prove this using the epsilon-delta definition? I'm unsure of using the min = { } function.
$\lim \limits_{x \to \infty}\frac{2x+1}{1-x}$
These are my steps:
$|f(x) - L| < \epsilon => |\frac{2x+1}{1-x} +2|< \epsilon$
$\qquad \qquad \; \; \; \; \; =>|\frac{3}{1-x} | < \epsilon$
$\qquad \qquad \; \; \; \; \; =>|\frac{-3}{x-1} | < \epsilon$
$\qquad \qquad \; \; \; \; \; =>|-3||\frac{1}{x-1} | < \epsilon$
$\qquad \qquad \; \; \; \; \; =>|\frac{1}{x-1} | < \frac{\epsilon}{3}$
$\qquad \qquad \; \; \; \; \; =>|{x-1} | < \frac{\epsilon |x-1|}{3}$
How do I continue from here?
• I got stuck at |1/(1-x)|<ε, I can't figure how to achieve |x-a|<δ – G.L. May 27 '18 at 16:59
• Achieve $|x-a| < \delta$? How are you approaching this? – JuliusL33t May 27 '18 at 17:02
• |f(x) - L| < ε => |x-1| < ε|x-1|/3 – G.L. May 27 '18 at 17:04
A few pointers:
• The $\varepsilon$-$\delta$ definition of a limit pertains to when you have a limit as $x \to a$, where $a$ is a finite quantity. For a limit as $x \to \infty$, you need an $\varepsilon$-$M$ definition, along the lines of, for all $\varepsilon > 0$, there exists some $M$ such that $$x > M \implies |f(x) - L| < \varepsilon.$$
• Your last line has an error. You'll want to be reciprocating both sides, to get $|x - 1| > \frac{3}{\varepsilon}$. Note the change in sign, as the function $\frac{1}{x}$ is decreasing for positive $x$.
• This means, you are looking for values of $x$ that are a greater distance than $\frac{3}{\varepsilon}$ from $1$. This suggests a that $M = 1 + \frac{3}{\varepsilon}$ is a good choice of $M$. Note that no maximums or minimums are required!
• Beware the flow of logic here! You've written in implications $\implies$ between each step. While this is true, what you really need for the definition is backwards implications: $\impliedby$. This is because you're starting with $|f(x) - L| < \varepsilon$, the thing you need to conclude! This is ok, provided each step you perform implies the previous step, not the next step as we often tend to do. The other way is to simply rewrite the entire proof backwards, starting with assuming $x > 1 + \frac{3}{\varepsilon}$, and working logically to the desired conclusion.
EDIT: Regarding the solution you posted as an answer, it's definitely looking better, though I have a couple more pointers:
• You only really need the last block of implications. Your entire solution could be presented by simply starting with the line that begins $$|x - 1| > \frac{3}{\varepsilon} \iff \ldots$$ and going from there. Sure, it doesn't explain where you got $\frac{3}{\varepsilon}$ from, but then again, it doesn't have to! The logic shows that such a choice was a good choice, no matter how you came by it.
• You still haven't come from the starting point of $x > M$, which you should, as this is part of the definition of a limit as $x \to \infty$. So, I would just add to your solution, \begin{align*} x > 1 + \frac{3}{\varepsilon} &\implies x - 1 > \frac{3}{\varepsilon} > 0 \\ &\implies |x - 1| > \frac{3}{\varepsilon} \\ &\implies \ldots \end{align*} Note that the above implication works because $x - 1 \ge 0$, and so $|x - 1| = x - 1$.
Thanks, Theo Bendit. I re-did my answer, not sure if it's correct.
$|f(x) - L| = |\frac{2x+1}{1-x} +2|< \delta$
$\qquad \quad \; \; \; =|\frac{3}{1-x} | < \delta$
$\qquad \quad \; \; \; =|\frac{-3}{x-1} | < \delta$
$\qquad \quad \; \; \; =|-3||\frac{1}{x-1} | < \delta$
$\qquad \quad \; \; \; =|\frac{1}{x-1} | < \frac{\delta}{3}$
$\qquad \quad \; \; \; =|{x-1} | > \frac{3}{\delta}$
$\frac{3}{\delta} = \epsilon <=> \delta = \frac{3}{\epsilon}$
$|x-1| > \frac {3}{\epsilon} <=> 0 < |\frac{1}{x-1}|<\frac{\epsilon}{3}$
$\qquad \qquad \; \; <=> 0 < |\frac{3}{x-1}|<\epsilon$
$\qquad \qquad \; \; <=> 0 < |-3||\frac{1}{x-1}|<\epsilon$
$\qquad \qquad \; \; <=> 0 < |\frac{3}{1-x}|<\epsilon$
$\qquad \qquad \; \; <=> 0 < |\frac{2x+1}{1-x} + 2|<\epsilon$
$\qquad \qquad \; \; => 0 < |f(x) - L|<\epsilon$
• Ok, I'M confused looking at your answer. You're overthinking this, Theo Bendit.more or less gave you an outline of the solution. Hint: Consider that $\frac {3}{\epsilon}$ > 0 for any value of x, so you can simply drop the absolute value in | x- 1| . Then just proceed with the algebra to get his suggested value of M. – Mathemagician1234 May 28 '18 at 2:48
• Hi G.L., I've edited my answer with a bit of feedback about your answer. Hope you find it helpful. – Theo Bendit May 28 '18 at 4:13 | 1,590 | 4,518 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-39 | latest | en | 0.785326 |
http://www.transtutors.com/homework-help/computer-science/digital-electronics/arithmetic-circuits/adder/ | 1,511,113,888,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00701.warc.gz | 512,992,482 | 16,178 | An adder or summer is a digital circuit used in electronics user for addition of numbers. In computing devices, adders are used in the arithmetic logic units. They are also used in processors to calculate table indices and addresses. They are used in counting circuits.
Adders can be constructed for different types of numerical representations. The most common type of adders is the ones used for binary calculations or in excess-3 equations. Adders can sometimes be modified for adder-subtractor. In instances where two's complement or one's complement is used for representing negative numbers, such a modification is unnecessary. Complex numbers may require more complex adders like a Carry Look-ahead Adder, a Look-ahead Carry Unit or a Ripple Carry Adder.
Binary Adders are made up from standard AND and Ex-OR gates. They allow us to add single bit binary numbers, a and b to produce two outputs, the SUM of the addition and a CARRY called the Carry-out bit.
The 1-bit Full Adder circuit shown above consists of two half adders connected together and has three Ex-OR gates, two AND gates and an OR gate, six logic gates altogether. The truth table for the full adder takes into account an additional column to take into account the Carry-in input as well as the summed output and carry-output. 4-bit full adder circuits are available as standard IC packages in the form of the TTL 74LS83 or the 74LS283 which has the ability to add together two 4-bit binary numbers and produce a SUM and a CARRY output. Now if we want to add two n-bit numbers, then 1-bit full adders have to be connected to each other in order to produce Ripple Carry Adder.
The principal difference between the Full Adder and the previous seen Half Adder is that a full adder has three inputs, the same two single bit binary inputs A and B as before plus an additional Carry-In (C-in) input.
A Ripple Carry Adder can be constructed to perform complex adding circuits. In a Ripple Carry Adder, a logical circuit is used that has multiple full adders to add N-bit numbers with each full adder inputs a Cin, which is the Cout of the previous adder.
Transtutors.com gives you email-based homework assistance with brilliant understanding and simulations that assist in making the subject practical and relevant for any assignment help.
We are able to provide homework help at a reasonable cost which answers all your Computer Science queries in details and enable you to acquire more knowledge about your assignments. We have tutors who have years of experience assisting students and are remarkably qualified to help you with Adder Homework Help or Assignment Help.
## Related Topics
All Computer Science Topics
More Q&A | 587 | 2,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-47 | latest | en | 0.863441 |
https://www.unitconverters.net/pressure/megapascal-to-decipascal.htm | 1,713,128,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00481.warc.gz | 964,180,584 | 3,339 | Home / Pressure Conversion / Convert Megapascal to Decipascal
# Convert Megapascal to Decipascal
Please provide values below to convert megapascal [MPa] to decipascal [dPa], or vice versa.
From: megapascal To: decipascal
### Megapascal to Decipascal Conversion Table
Megapascal [MPa]Decipascal [dPa]
0.01 MPa100000 dPa
0.1 MPa1000000 dPa
1 MPa10000000 dPa
2 MPa20000000 dPa
3 MPa30000000 dPa
5 MPa50000000 dPa
10 MPa100000000 dPa
20 MPa200000000 dPa
50 MPa500000000 dPa
100 MPa1000000000 dPa
1000 MPa10000000000 dPa
### How to Convert Megapascal to Decipascal
1 MPa = 10000000 dPa
1 dPa = 1.0E-7 MPa
Example: convert 15 MPa to dPa:
15 MPa = 15 × 10000000 dPa = 150000000 dPa | 259 | 683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.489033 |
https://reviews.llvm.org/D152278?id=531192 | 1,722,838,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00512.warc.gz | 400,538,354 | 40,056 | [SCEV] Compute SCEV for ashr(add(shl(x, n), c), m) instr tripletClosedPublic
Authored by vedant-amd on Jun 6 2023, 8:14 AM.
Details
Reviewers
fhahn nikic mkazantsev efriedma
Commits
rG5a9a02f67b77: [SCEV] Compute SCEV for ashr(add(shl(x, n), c), m) instr triplet
Summary
%x = shl i64 %w, n
%y = add i64 %x, c
%z = ashr i64 %y, m
The above given instruction triplet is seen many times in the generated
LLVM IR, but SCEV model is not able to compute the SCEV value of AShr
instruction in this case.
This patch models the two cases of the above instruction pattern using
the following expression:
> sext(add(mul(trunc(w), 2^(n-m)), c >> m))
1. when n = m the expression reduces to sext(add(trunc(w), c >> n))
as n-m=0, and multiplying with 2^0 gives the same result.
1. when n > m the expression works as given above.
It also adds several unittest to verify that SCEV is able to compute
the value.
Comparing the snippets of the result of SCEV analysis:
• SCEV of ashr before change ----------------------------
%idxprom = ashr exact i64 %sext, 32
```--> %idxprom U: [-2147483648,2147483648) S: [-2147483648,2147483648)
Exits: 8 LoopDispositions: { %for.body: Variant }```
• SCEV of ashr after change ---------------------------
%idxprom = ashr exact i64 %sext, 32
```--> {0,+,1}<nuw><nsw><%for.body> U: [0,9) S: [0,9)
Exits: 8 LoopDispositions: { %for.body: Computable }```
LoopDisposition of the given SCEV was LoopVariant before, after adding
the new way to model the instruction, the LoopDisposition becomes
LoopComputable as it is able to compute the SCEV of the instruction.
Diff Detail
Event Timeline
vedant-amd created this revision.Jun 6 2023, 8:14 AM
Herald added a project: Restricted Project. Jun 6 2023, 8:14 AM
vedant-amd requested review of this revision.Jun 6 2023, 8:14 AM
Herald added a project: Restricted Project. Jun 6 2023, 8:14 AM
vedant-amd edited the summary of this revision. (Show Details)Jun 6 2023, 8:19 AM
vedant-amd edited the summary of this revision. (Show Details)
vedant-amd edited the summary of this revision. (Show Details)
vedant-amd edited the summary of this revision. (Show Details)
vedant-amd added a comment.EditedJun 6 2023, 8:31 AM
Hey @fhahn @nikic @mkazantsev can you please review this patch, I stumbled upon this while working on a bug in LSR, this patch does fail one of the unit test-case (scaling-factor-incompat-type.ll) in LSR. I believe that might be an effect of SCEV of this AShr pattern being computed, as LSR generally does do some optimizations based on the Shl + Add + AShr instructions.
Hey @fhahn @nikic @mkazantsev can you please review this patch, I stumbled upon this while working on a bug in LSR, this patch does fail one of the unit test-case (scaling-factor-incompat-type.ll) in LSR. I believe that might be an effect of SCEV of this AShr pattern being computed, as LSR generally does do some optimizations based on the Shl + Add + AShr instructions.
ping ! Could you please take a look at this patch once.
llvm/lib/Analysis/ScalarEvolution.cpp
7882
Is there some reason to expect that "c" fits into an int64_t?
7883
I guess the reason AddOperandCI has to be a constant is that this shift needs to constant-fold?
7886
This could be extended to cases where n is greater than m? You can skip that for the initial patch, of course.
I don't really see any reason to restrict the shift amounts like this; the transform is pretty restricted even without that. What effect are you worried about?
38
Try to avoid "undef" in testcases where it isn't relevant. I don't think it really has much effect here, but still.
llvm/lib/Analysis/ScalarEvolution.cpp
7882
Right, I need to add a check like this I guess ?
`if (CI->getValue().uge(BitWidth))`
7883
Theoretically, it doesn't have to be a constant. But, I don't see where such a IR will be emitted. I can handle this case in a future patch.
This IR is usually emitted for the following C code. InstCombine does the optimization.
```int64_t a = 10;
int32_t b = a - 1;
printf("%d", arr[b]);```
vedant-amd marked an inline comment as not done.Jun 8 2023, 9:40 PM
llvm/lib/Analysis/ScalarEvolution.cpp
7882
That doesn't look like the right check.
If it were necessary, you could use isIntN or something like that. But you should just be able to just do the math in APInt: AddOperandCI->getValue().ashr(AShrAmt) or something like that.
vedant-amd updated this revision to Diff 529822.Jun 8 2023, 9:55 PM
Updated unit test to remove ret undefs
vedant-amd marked an inline comment as done.Jun 8 2023, 9:58 PM
38
Sure, will keep this in mind.
vedant-amd marked an inline comment as done.Jun 8 2023, 9:58 PM
llvm/lib/Analysis/ScalarEvolution.cpp
7886
This could be extended to cases where n is greater than m? You can skip that for the initial patch, of course.
I could do that, but the original goal was to handle sext(trunc) expression that are expanded into these statements by instcombine.
I don't really see any reason to restrict the shift amounts like this; the transform is pretty restricted even without that. What effect are you worried about?
We only wanted to support the data types supported by C/C++, and also since these instr are transformed from sext(trunc) it makes sense to just support standard integer types.
llvm/lib/Analysis/ScalarEvolution.cpp
7883
I guess the reason AddOperandCI has to be a constant is that this shift needs to constant-fold?
This can be implemented for non-constants as well, maybe in a future patch. But, it involves coming up with complex SCEV expression involving div.
Added check to make sure AddConstant isn't nullptr and few other changes.
vedant-amd marked 3 inline comments as done.Jun 8 2023, 10:48 PM
llvm/lib/Analysis/ScalarEvolution.cpp
7882
Hey @efriedma , I have made the requested changes (and replied to the questions). Please let me know if anything more needs to be done. Thanks !
updated using clang-format
nikic added a comment.Jun 9 2023, 12:44 AM
I assume the motivation for this is the following InstCombine transform? https://llvm.godbolt.org/z/zd63TWKbW I don't think we can change that canonicalization, so undoing it in SCEV is fine.
llvm/lib/Analysis/ScalarEvolution.cpp
7871–7900
We already have code to handle the general shl+ashr pattern here, including for the case where n and m are not the same. We should reuse the same code. For your pattern, we'd just have an extra add at the start.
llvm/lib/Analysis/ScalarEvolution.cpp
7871–7900
That could be done, but it would be a major refactoring. I will post a patch anyways.
I assume the motivation for this is the following InstCombine transform? https://llvm.godbolt.org/z/zd63TWKbW I don't think we can change that canonicalization, so undoing it in SCEV is fine.
Yup, that's correct. This canonicalization is used several times in LSR, so having it's SCEV might be beneficial.
vedant-amd updated this revision to Diff 529876.Jun 9 2023, 2:52 AM
Refactored code to reuse ashr + shl handling code
I assume the motivation for this is the following InstCombine transform? https://llvm.godbolt.org/z/zd63TWKbW I don't think we can change that canonicalization, so undoing it in SCEV is fine.
Hey @nikic, I did refactor the change, please take a look, all the SCEV test cases do pass with it. It look correct functionally.
I assume the motivation for this is the following InstCombine transform? https://llvm.godbolt.org/z/zd63TWKbW I don't think we can change that canonicalization, so undoing it in SCEV is fine.
Hey @nikic, I did refactor the change, please take a look, all the SCEV test cases do pass with it. It look correct functionally.
ping ! could you please give a +1 ? @nikic @efriedma
xbolva00 added a subscriber: xbolva00.EditedJun 11 2023, 4:35 AM
You should update checks in loopstrengthreduce::scaling-factor-incompat-type.ll
You should update checks in loopstrengthreduce::scaling-factor-incompat-type.ll
I was going to send another patch for that, testcase failure. Sure, will do so.
Ping ! I didn't add the LSR failing testcase fix here, because I honestly don't know if it uncovered a bug, or it's a valid transformation in LSR.
You have to update that file anyway. Pre-commit builds are clearly failing.
You have to update that file anyway. Pre-commit builds are clearly failing.
Right, is it fine if I mark the test case as XFAIL for now ?
nikic added a comment.Jun 13 2023, 6:07 AM
You have to update that file anyway. Pre-commit builds are clearly failing.
Right, is it fine if I mark the test case as XFAIL for now ?
No, you need to regenerate the test by running the update_test_checks.py script.
llvm/lib/Analysis/ScalarEvolution.cpp
7884
There should be no restriction on the shift amount.
7888
We should verify that the add constant has lowest n bits unset, to make reassociation valid. It doesn't matter if n==m, but it matters if n!=m.
7898
You can keep the truncate in the comment code by making this trunc(add()) instead of add(trunc()). Unless I'm missing something, they're equivalent in this context.
7911
Stray newline
7
You can test these cases with basic patterns using just the shl+add+ashr sequence, without a loop (though it's also ok to keep a loop test as a motivating case). Please also test the case where the shift amounts are different.
You have to update that file anyway. Pre-commit builds are clearly failing.
Right, is it fine if I mark the test case as XFAIL for now ?
No, you need to regenerate the test by running the update_test_checks.py script.
Will do so, I was not sure if the testcase will be still valid after the updated CHECKs.
updated failing testcase in LSR
vedant-amd marked an inline comment as done.Jun 13 2023, 11:12 PM
removed restriction on shift amount and the corresponding test case
vedant-amd marked an inline comment as done.Jun 13 2023, 11:19 PM
vedant-amd marked an inline comment as done.
llvm/lib/Analysis/ScalarEvolution.cpp
7888
Do you mean to say check for unset bits from nth bit to the mth bit ? because anything lower than mth bit will anyways be irrelevant once we right shift by m amount.
vedant-amd marked an inline comment as not done.Jun 14 2023, 12:20 AM
llvm/lib/Analysis/ScalarEvolution.cpp
7888
I tried to understand this, I am not able to make sense of it. Can you explain further what you mean to say ?
updated code to shift right the AddConstant by ShlAmt and flip add, trunc exprs
llvm/lib/Analysis/ScalarEvolution.cpp
7898
So, it should look like this ?
`AddTruncateExpr = getTruncateExpr(getAddExpr(ShlOp0SCEV, AddConstant), TruncTy);`
updated code to shift right the AddConstant by ShlAmt and flip add, trunc exprs
vedant-amd added a comment.EditedJul 31 2023, 2:50 AM
ping !! @nikic can you take a look at this, thanks !
Updated LSR testcase and added new testcases for AShr SCEV model
Hey @nikic and @efriedma Sorry for the ping again, but I have addressed all the comments (added the testcases as well), can you please review once again, it's been stuck since a long time. Thanks for your time.
12 ↗(On Diff #545909)
How is this equivalent? Say %a is zero; the original function returns 1, this SCEV expression returns 0. (I think maybe the "ashr" of the constant is shifting by 10, instead of shifting by 8?)
12 ↗(On Diff #545909)
Yeah, the SCEV is wrong. I will fix this. This is what is should have been, but it's a bit opposite.
```= (a*2^10 + 256)/2*8
= a*4 + 1```
So, we need SCEV like this:
2^(shl_amt - ashr_amt) * a + c >> ashr_amt
I have updated my code, here's the correct SCEV:
(1 + (sext i56 (4 * (trunc i64 %a to i56)) to i64))<nuw><nsw> U: [1,-2) S: [-36028797018963967,36028797018963966)
Does it seem correct ?
vedant-amd updated this revision to Diff 546769.Aug 3 2023, 2:37 AM
Fixed issue with generated SCEV for ShlAmt > AshrAmt
Hey @efriedma Thanks for spotting the bug, I updated the SCEV for case when m > n, it seems to be correct now. Can you please review the patch, thanks !
CC: @nikic
12 ↗(On Diff #545909)
I suspect the updated code isn't right... can you add an example with a larger operand to the add? The add needs to happen before the sign-extend, but in this specific example, that doesn't matter because the two expressions are equivalent.
12 ↗(On Diff #545909)
I will add one, but I believe we can move the add out of the sext as the add constant is already shifted right by Ashr amount.
12 ↗(On Diff #545909)
Also, I am confused about one thing, the Type of MulExpr turns out to be i56 and that of the addConstant is i64, I am able to add them before the Sign extend. Adding the m > n part to this ashr model seems to buggy at this point, as I had proposed can this patch just address the case where m=n ? and I submit, the m > n thing later with thorough thinking. But, I think @nikic wants them in the same patch.
12 ↗(On Diff #545909)
Adding to the previous comment, the Ashr + Add + Shl with shlamt > ashramt quite rarely occur in LLVM IR. But, the case where the shlamt = ashramt is quite common in LLVM IR, notably because instcombine like passes do the transformation. I propose that this patch just address the latter and I send a patch for the m > n case at a later point. This patch is stuck from a long time, and just taking care of m=n case won't negatively affect as it'll safely exit in the m > n case.
I'm not sure why you're having so much trouble with the unequal shift math. The case of unequal shift amount is really just treating the shl as two shifts: one shift by an arbitrary amount, followed by one shift with shift amount equal to the ashr.
That said I'd be okay with a patch that bailed out on that case.
Update the SCEV code and add a new testcase
I'm not sure why you're having so much trouble with the unequal shift math. The case of unequal shift amount is really just treating the shl as two shifts: one shift by an arbitrary amount, followed by one shift with shift amount equal to the ashr.
I am having trouble with keeping the addition part inside the SextExpr thing, because it always returns back i64 int, but now came up with a better implementation. It seems correct to me, please take a look.
That said I'd be okay with a patch that bailed out on that case.
I'm not sure why you're having so much trouble with the unequal shift math. The case of unequal shift amount is really just treating the shl as two shifts: one shift by an arbitrary amount, followed by one shift with shift amount equal to the ashr.
That said I'd be okay with a patch that bailed out on that case.
updated with clang-format
added empty line at end of file in a test case
@efriedma I have fixed the SCEV part, it mostly seems correct. Also added the testcase with large add constant. Please take a look once ! thanks !
The revised logic makes sense.
llvm/lib/Analysis/ScalarEvolution.cpp
7899
Instead of checking if (ShlAmt > AShrAmt) here, can you just unconditionally do AddTruncateExpr = getTruncateExpr(ShlOp0SCEV, TruncTy);, then add an if (L->getOpcode() != Instruction::Shl) check to the equal shift amount case?
Or better, just unify the ShlAmt > AShrAmt and ShlAmt == AShrAmt cases; the logic for the ShlAmt > AShrAmt case should just work for the ShlAmt == AShrAmt case (it ends up multiplying by a constant 1, which simplifies to exactly the same thing as the existing ShlAmt == AShrAmt code).
vedant-amd marked an inline comment as done.
Simplified expression handling code, and cleaned up other things
llvm/lib/Analysis/ScalarEvolution.cpp
7899
I have implemented the second suggestion, refactored and cleaned up the comments. Now the code looks in good shape. Please give a final review for the same.
This revision is now accepted and ready to land.Aug 24 2023, 9:51 AM
vedant-amd edited the summary of this revision. (Show Details)
Updated the commit message
This revision was landed with ongoing or failed builds.Aug 24 2023, 10:47 PM
This revision was automatically updated to reflect the committed changes. | 4,233 | 15,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.73853 |
https://us.metamath.org/mpeuni/cringcatALTV.html | 1,718,287,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00536.warc.gz | 549,742,888 | 6,144 | Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cringcatALTV Structured version Visualization version GIF version
Theorem cringcatALTV 44647
Description: The restriction of the category of (unital) rings to the set of commutative ring homomorphisms is a category, the "category of commutative rings". (Contributed by AV, 19-Feb-2020.) (New usage is discouraged.)
Hypotheses
Ref Expression
crhmsubcALTV.c 𝐶 = (𝑈 ∩ CRing)
crhmsubcALTV.j 𝐽 = (𝑟𝐶, 𝑠𝐶 ↦ (𝑟 RingHom 𝑠))
Assertion
Ref Expression
cringcatALTV (𝑈𝑉 → ((RingCatALTV‘𝑈) ↾cat 𝐽) ∈ Cat)
Distinct variable groups: 𝐶,𝑟,𝑠 𝑈,𝑟,𝑠 𝑉,𝑟,𝑠
Allowed substitution hints: 𝐽(𝑠,𝑟)
Proof of Theorem cringcatALTV
StepHypRef Expression
1 eqid 2824 . 2 ((RingCatALTV‘𝑈) ↾cat 𝐽) = ((RingCatALTV‘𝑈) ↾cat 𝐽)
2 crhmsubcALTV.c . . 3 𝐶 = (𝑈 ∩ CRing)
3 crhmsubcALTV.j . . 3 𝐽 = (𝑟𝐶, 𝑠𝐶 ↦ (𝑟 RingHom 𝑠))
42, 3crhmsubcALTV 44646 . 2 (𝑈𝑉𝐽 ∈ (Subcat‘(RingCatALTV‘𝑈)))
51, 4subccat 17118 1 (𝑈𝑉 → ((RingCatALTV‘𝑈) ↾cat 𝐽) ∈ Cat)
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∈ wcel 2115 ∩ cin 3918 ‘cfv 6343 (class class class)co 7149 ∈ cmpo 7151 Catccat 16935 ↾cat cresc 17078 CRingccrg 19298 RingHom crh 19467 RingCatALTVcringcALTV 44554 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2179 ax-ext 2796 ax-rep 5176 ax-sep 5189 ax-nul 5196 ax-pow 5253 ax-pr 5317 ax-un 7455 ax-cnex 10591 ax-resscn 10592 ax-1cn 10593 ax-icn 10594 ax-addcl 10595 ax-addrcl 10596 ax-mulcl 10597 ax-mulrcl 10598 ax-mulcom 10599 ax-addass 10600 ax-mulass 10601 ax-distr 10602 ax-i2m1 10603 ax-1ne0 10604 ax-1rid 10605 ax-rnegex 10606 ax-rrecex 10607 ax-cnre 10608 ax-pre-lttri 10609 ax-pre-lttrn 10610 ax-pre-ltadd 10611 ax-pre-mulgt0 10612 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2624 df-eu 2655 df-clab 2803 df-cleq 2817 df-clel 2896 df-nfc 2964 df-ne 3015 df-nel 3119 df-ral 3138 df-rex 3139 df-reu 3140 df-rmo 3141 df-rab 3142 df-v 3482 df-sbc 3759 df-csb 3867 df-dif 3922 df-un 3924 df-in 3926 df-ss 3936 df-pss 3938 df-nul 4277 df-if 4451 df-pw 4524 df-sn 4551 df-pr 4553 df-tp 4555 df-op 4557 df-uni 4825 df-int 4863 df-iun 4907 df-br 5053 df-opab 5115 df-mpt 5133 df-tr 5159 df-id 5447 df-eprel 5452 df-po 5461 df-so 5462 df-fr 5501 df-we 5503 df-xp 5548 df-rel 5549 df-cnv 5550 df-co 5551 df-dm 5552 df-rn 5553 df-res 5554 df-ima 5555 df-pred 6135 df-ord 6181 df-on 6182 df-lim 6183 df-suc 6184 df-iota 6302 df-fun 6345 df-fn 6346 df-f 6347 df-f1 6348 df-fo 6349 df-f1o 6350 df-fv 6351 df-riota 7107 df-ov 7152 df-oprab 7153 df-mpo 7154 df-om 7575 df-1st 7684 df-2nd 7685 df-wrecs 7943 df-recs 8004 df-rdg 8042 df-1o 8098 df-oadd 8102 df-er 8285 df-map 8404 df-pm 8405 df-ixp 8458 df-en 8506 df-dom 8507 df-sdom 8508 df-fin 8509 df-pnf 10675 df-mnf 10676 df-xr 10677 df-ltxr 10678 df-le 10679 df-sub 10870 df-neg 10871 df-nn 11635 df-2 11697 df-3 11698 df-4 11699 df-5 11700 df-6 11701 df-7 11702 df-8 11703 df-9 11704 df-n0 11895 df-z 11979 df-dec 12096 df-uz 12241 df-fz 12895 df-struct 16485 df-ndx 16486 df-slot 16487 df-base 16489 df-sets 16490 df-ress 16491 df-plusg 16578 df-hom 16589 df-cco 16590 df-0g 16715 df-cat 16939 df-cid 16940 df-homf 16941 df-ssc 17080 df-resc 17081 df-subc 17082 df-mgm 17852 df-sgrp 17901 df-mnd 17912 df-mhm 17956 df-grp 18106 df-ghm 18356 df-mgp 19240 df-ur 19252 df-ring 19299 df-cring 19300 df-rnghom 19470 df-ringcALTV 44556 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 2,123 | 3,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.134074 |
https://tex.stackexchange.com/questions/375285/draw-a-grid-of-bivariate-normal-distributions-in-tikz | 1,586,291,433,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371805747.72/warc/CC-MAIN-20200407183818-20200407214318-00297.warc.gz | 734,322,436 | 32,583 | # Draw a grid of bivariate normal distributions in TikZ
This question was answered in this post, but I would like a small adaptation of it.
I would like to take the top solution of the above post. Instead of a single bi-variate distribution, I would like to have a grid of such distributions. The grid should be rectangular with controllable positions for each source.
I would then like to adapt this resulting image to remove all of the axes to leave behind just the surface distribution.
The grid I would like should be something like that here. But the position of each circle should be a 2D Gaussian.
## Desired output
Something like this, with and without projections on the axes.
## The distribution
Each source on the grid should have the following distribution:
$$f(x, y) = \frac{1}{2\pi \sigma_x \sigma_y}\exp[-\frac{(x-\mu_x)^2}{\sigma_x^2} + -\frac{(y-\mu_y)^2}{\sigma_y^2}]$$.
with $\sigma_x = \sigma_y$ and varying values for $\mu_x$ and $\mu_y$ for the different sources. Note that this distribution has a diagonal form for the covariance matrix with elements $\sigma_x$ and $\sigma_y$ on the diagonal.
• And, where is your problem? What have you done so far? – Huang_d Jun 16 '17 at 15:28
• @Huang_d, getting a second distribution using the \addplot3 [surf] plot at a different location. It does not produce 2 seperate distributions. – Sid Jun 16 '17 at 15:31
• One big problem (if I understand correctly) is, you won't be able to draw several intersecting surfaces using PgfPlots, as it won't be able to decide correctly which is in the front / in the back. – marsupilam Jun 16 '17 at 15:31
• @marsupilam, I see. Is there an alternative way to achieve this? – Sid Jun 16 '17 at 15:41
• You could draw each surface only on the relevant domain and put them side by side. Do you have a model picture of what you would like, or more details of which gaussians you want and what grid you are thinking of ? – marsupilam Jun 16 '17 at 15:46
I wonder if I'll ever understand the thing with loops in pgfplots.
## The (simplistic) code
@Jake is right in pointing out this is a lot of computations for pgfplots, and you may be well-advised to pre-compute the functions via an external tool
\documentclass[tikz]{standalone}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
[
declare function=
{
gaussian(\x)= exp(-.5*\x^2);
gaussianDouble(\x,\y)= gaussian(\x)*gaussian(\y);
}
]
\begin{axis}
\foreach\i in {0,4,...,12}
{
\foreach\j in {0,4,...,12}
{
}
}
\end{axis}
\end{tikzpicture}
\end{document}
## In Scilab
Not an answer, but is this what you want ?
If so, may I edit your question to include the pic in it ?
### The (Scilab) code
function z=normal(x,y)
norm = x.^2 + y.^2
z = exp(-.5*norm)
endfunction
clf()
x=linspace(-3,3,30)
[X,Y]=meshgrid(x)
for i=0:2:10
for j=0:2:10
surf(X+i,Y+j,normal(X,Y))
end
end
f=gcf()
f.color_map = autumncolormap(32);
• Yes, this is what I would like thanks - with and without a distribution shown projected on the xz and yz axes (much like the original post in question). I would still prefer it in TikZ, but you are most welcome to edit the question to illustrate the desired output. – Sid Jun 16 '17 at 16:27
• @Sid Alright, then, you would need to tell us the x and y step of the lattice you are using, and the covariance matrix for the Gaussian distributions. Otherwise, no computation can be done... – marsupilam Jun 16 '17 at 16:31
• Added the form of the distribution to the question. – Sid Jun 16 '17 at 17:00
• It's true that it takes 20-30 seconds in pgfplots, but remember that you only have to compile it once. You can then subsequently include it using \includegraphics. – JPi Jun 16 '17 at 18:30
• @JPi Sure, but I chose a rather low samples value and few bumps... It's unlikely you can get stellar graphics using this simple approach. – marsupilam Jun 16 '17 at 18:37 | 1,083 | 3,853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-16 | latest | en | 0.918211 |
https://www.mathsite.org/solving-linear-inequalities-in-one-variable.html | 1,558,683,031,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257553.63/warc/CC-MAIN-20190524064354-20190524090354-00506.warc.gz | 850,386,018 | 11,149 | FreeAlgebra Tutorials!
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:
# (vertically - using the balance beam )
## Solve Equation with a Balance Beam
Pattern: ax + b cx + d
Both sides simplified (a ,b, c, d are integers.)
Look at the coefficients of x and determine which is the larger integer (furthest to the right on the number line). If c > a then we will keep the variiablle x on that siide of the equation and keep the constant on the other side. To do this we first add opposites on the balance beam below the equation. Look at the pattern, and then follow the same steps through several examples
Solve simplified equations vertically - using the balance beam.
Pattern: c > a
1) Add opps: c > a → c - a > 0 Complete the step: (b - d) ≥ (c - a)x → Let A = (c - a) and B = (b - d) A, B are integers, A > 0 2) Multiply recip:Then: → → Since and A > 0 is coefficient of x
Equivalent Property
This means that all replacement values for x will be on or to the left of
NOTE: Since A > 0 all signs in its path remain the same. This is the advantage of choosing the side of larger coefficient when simplifying the problem. | 468 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-22 | latest | en | 0.82538 |
https://onlinehomeworkmarket.com/body-mass-index-discussion-nursing-assignment-help/ | 1,718,402,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00324.warc.gz | 392,533,161 | 26,904 | ### Our Services
Get 15% Discount on your First Order
# Body Mass Index Discussion Nursing Assignment Help
I’m working on a health & medical discussion question and need the explanation and answer to help me learn.
Body Mass Index (BMI) is widely and routinely used by healthcare professionals as an indicator of body fatness. BMI is a quick, inexpensive tool that can be useful to get a general idea of a person’s disease risk, however, it isn’t perfect. Using your textbook and the resources below, discuss the advantages and disadvantages of BMI. Are there any limitations to using BMI to assess a person’s health? If you were a healthcare professional, what advice would you give to clients about interpreting their own BMI? Do you think BMI more important than healthy behaviors?
## Expert Solution Preview
Introduction:
Body Mass Index (BMI) is a widely-used tool in the healthcare field to assess body fatness and provide a general idea of a person’s disease risk. While it is a quick and inexpensive method, BMI does have its advantages and disadvantages. This discussion will explore the limitations of using BMI to assess health, as well as provide guidance on interpreting BMI results and its significance in comparison to healthy behaviors.
1. Quick and Simple: Calculating BMI involves a straightforward formula of dividing weight in kilograms by height in meters squared. It can be easily calculated using online calculators or through manual calculation, making it a convenient tool for healthcare professionals.
2. Inexpensive: Unlike other methods for evaluating body composition, such as dual-energy X-ray absorptiometry (DEXA) or underwater weighing, BMI can be measured using basic scale and height measuring tools, which makes it cost-effective.
3. Population Comparisons: BMI provides a standardized measurement that allows for comparisons within populations. It helps healthcare professionals identify the prevalence of overweight and obesity within certain populations and track changes over time.
1. Limited Assessment of Body Fat Distribution: BMI does not differentiate between fat and muscle mass or consider the distribution of body fat. As a result, individuals with a high muscle mass, such as athletes, may have a higher BMI despite having lower body fat levels. Similarly, older adults may have a higher BMI due to age-related muscle loss, even if they have a healthy body fat percentage.
2. Age and Gender Differences: BMI may not be equally accurate for all age groups and both sexes. Body fat distribution and metabolic factors vary between males and females, as well as across different stages of life. Therefore, BMI may not provide an accurate indication of health risk for certain individuals.
3. Limited Assessment of Overall Health: BMI solely focuses on weight and height ratio, ignoring other important factors such as body composition, muscle strength, and cardiovascular fitness. It does not consider lifestyle habits, dietary patterns, mental health, or genetic predisposition to certain diseases, which are essential for a comprehensive evaluation of health.
Interpreting BMI for Clients:
If I were a healthcare professional, I would advise clients to interpret their BMI cautiously and consider it as an initial screening tool rather than a definitive measure of their overall health. I would emphasize that BMI should be used in conjunction with other assessments, such as waist circumference, body fat percentage, and individual health markers, to obtain a more accurate understanding of their health status.
Additionally, it is crucial to encourage clients to adopt healthy lifestyle behaviors rather than solely focusing on BMI. Engaging in regular physical activity, consuming a balanced diet, managing stress levels, and getting adequate sleep are all integral components of maintaining good health. These healthy behaviors should be prioritized regardless of a person’s BMI, as they have a significant impact on overall well-being and disease prevention.
Conclusion:
While BMI is a useful tool to provide a general idea of body fatness and disease risk, it has its limitations. Its lack of differentiation between fat and muscle mass, as well as disregarding other health indicators, may limit its accuracy in assessing an individual’s health. Therefore, healthcare professionals should encourage clients to interpret their BMI cautiously and focus on adopting healthy behaviors rather than solely relying on BMI as a determinant of overall health.
Order a Similar Paper and get 15% Discount on your First Order
## Related Questions
### Ann is a 34-year-old African American woman who presents Nursing Assignment Help
Ann is a 34-year-old African American woman who presents with a 6 week history of initial and terminal insomnia and a 10 pound weight loss over that same period. She admits to feeling sad almost every day for the past 6 weeks that occurred after she lost her job. She
### Ann is a 34-year-old African American woman who presents Nursing Assignment Help
Ann is a 34-year-old African American woman who presents with a 6 week history of initial and terminal insomnia and a 10 pound weight loss over that same period. She admits to feeling sad almost every day for the past 6 weeks that occurred after she lost her job. She
### Complete your Initial Post by Wednesday at 11:59pm EST. 1. Nursing Assignment Help
Complete your Initial Post by Wednesday at 11:59pm EST. 1. Read: Discussion Background This week you are learning about codes for Evaluation and Management services. As you work on assigning codes to report Evaluation and Management services, reflect on the purpose and meaning of the guidelines found throughout this section. 2.
### Understanding how the health care system works and evolves Nursing Assignment Help
Understanding how the health care system works and evolves provides health care administrators and leadership with awareness of the position of their organization and facilitates compliance with health regulations. Discuss in detail each component of the U.S. delivery system, its function, and how it supports the delivery of care. Describe
### The Patient Protection and Affordable Care Act, or in short Nursing Assignment Help
The Patient Protection and Affordable Care Act, or in short Affordable Care Act, changed the landscape of the health care industry. For this assignment, read the article “Information asymmetries and risk management in healthcare markets: The U.S. Affordable Care Act (ACA) in retrospect” by Mendoza, and write a paper to
### This assignment will build off your SWOT analysis. As the Nursing Assignment Help
This assignment will build off your SWOT analysis. As the Organizational Development Manager (ODM) for the organization that you completed your SWOT Analysis for, you act as a liaison and advisor to the organization’s leadership, and begin initiatives within the organization. You have been given the responsibility to create a
### This assignment requires you to conduct the Strengths (S), Nursing Assignment Help
This assignment requires you to conduct the Strengths (S), Weaknesses (W), Opportunities (O), and Threats (T) SWOT Analysis. As the planner for your health organization (selected by you), you will need to prepare a 3-page document addressing your organization’s internal and external environment to include, but not limited to, forces/trends,
### This assignment requires you to conduct the Strengths (S), Nursing Assignment Help
This assignment requires you to conduct the Strengths (S), Weaknesses (W), Opportunities (O), and Threats (T) SWOT Analysis. As the planner for your health organization (selected by you), you will need to prepare a 3-page document addressing your organization’s internal and external environment to include, but not limited to, forces/trends,
### instructions: replies should be a minimum of 2 paragraphs Nursing Assignment Help
instructions: replies should be a minimum of 2 paragraphs each. Peer 1 : A legal nursing issue that sticks out to me is the case of redound Vaught, a Tenessee nurse that made a deadly medication error and whose case made headlines when she was charged criminally for her actions.
### Who can assist with answering the following question in a Nursing Assignment Help
Who can assist with answering the following question in a minimum of 500 words in APA format, including in-text citations? Choose one learning theory and/or principle you think is most useful to the advanced practice nurse (APN) in providing high-quality health care to clients. Explain your rationale for choosing this
### Unit 3 focused on both the nervous and endocrine systems. Nursing Assignment Help
Unit 3 focused on both the nervous and endocrine systems. Both are essential communication systems for the body and play vital roles in homeostatic balance. The endocrine system produces chemical messengers that travel throughout the body via the circulatory system and interact with receptor cells. If the interaction cannot occur,
### Week 2 Midweek Assignment: Cell Components Using an Analogy … Nursing Assignment Help
Week 2 Midweek Assignment: Cell Components Using an Analogy to Relate Cells to Everyday Life Create an analogy for an animal cell, such as a type of building, group of people, sport, or game. Your analogy should include the plasma membrane and at least five organelles. Present your analogy as
### Biostatistics Collaboration [WLO: 1] [CLOs: 3, 5] Prior to Nursing Assignment Help
Biostatistics Collaboration [WLO: 1] [CLOs: 3, 5] Prior to beginning work on this discussion forum, review Chapters 18, 19, 20 and Appendix F from the course text, Statistics for People Who (Think They) Hate Statistics: Using Microsoft Excel. This discussion is an opportunity to reflect on your learning from this course
### Question 7 (10 points) Marion is the manager at a local Nursing Assignment Help
Question 7 (10 points) Marion is the manager at a local doctor’s office. His employees are continually arriving to work late, and it is affecting appointment workflow. Marion hired you to research why employees are late and the impact that the tardiness has on the overall organization. Marion also wants you
### instructions: proper grammar and APA 7th ed format are Nursing Assignment Help
instructions: proper grammar and APA 7th ed format are expected.part 1: Please find a legal nursing issue of interest, and write an essay with 5 paragraphs on what should have been done to prevent the particular problem from occurring. Part 2: Please look at the following videos of Helen Haskell
### Aubre R. presents as a new patient to your practice having Nursing Assignment Help
Aubre R. presents as a new patient to your practice having recently relocated from rural Tennessee. She is 68 years old and complains of irritable bowel syndrome (IBS) with abdominal cramping and frequent diarrhea with occasional constipation. She has a history of domestic violence with resultant anxiety requiring inpatient hospitalization
### You are chairperson of a task force focused on improving the Nursing Assignment Help
You are chairperson of a task force focused on improving the health of your community. In this role, you have been asked to identify a critical need in your community and develop a framework for addressing this need. In this framework, you will identify the various stakeholders involved in the development of health
### Walden University, LLC. (Producer). (2019b). Alzheimer’s Nursing Assignment Help
To Prepare Review the interactive media piece for Alzheimer’s Disease. (76 Year Old Iranian Male) Reflect on the patient’s symptoms and aspects of the disorder presented in the interactive media piece. Consider how you might assess and treat patients presenting with the symptoms of the patient case study you were assigned. You | 2,339 | 11,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.924779 |
https://discuss.pennylane.ai/t/efficient-manipulation-of-pauli-operators-in-pennylane/1654 | 1,670,563,492,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711390.55/warc/CC-MAIN-20221209043931-20221209073931-00489.warc.gz | 226,561,122 | 5,746 | # Efficient manipulation of pauli operators in pennylane?
What is the right way to do symbolic matrix multiplication and tensor products of Pauli strings in pennylane? Certain operations that I would expect to evaluate very fast take a long time.
For example , the following code snippet takes a very long time (>1 minute) to do a trivial calculation (side question - is the operation `@` the correct way for me to do ordinary matrix multiplication when the observables are defined on the same system?)
``````out = qml.Identity(0)
for _ in range(1000):
out @= qml.PauliX(0)
out.matrix
``````
For comparison the analogous code snippet using cirq’s pauli operators evaluates basically instantaneously
``````qubit = cirq.LineQubit(0)
out = cirq.I(qubit)
for _ in range(1000):
out *= cirq.X(qubit)
out.matrix([qubit])
``````
Hi @EvanPeters! I did some digging, and it seems like the for loop itself is relatively fast,
``````out = qml.Identity(0)
for _ in range(1000):
out @= qml.PauliX(0)
``````
However, the computation of the matrix at the end is a massive bottleneck,
``````out.matrix
``````
likely related to an inefficient implementation:
(if I had to guess, maybe the inefficiency is from the list comprehension in L1416?)
We have been building up more efficient Pauli group-based arithmetic in the `qml.grouping` module, and plan to integrate this into the `Operator` class ASAP. At the moment, if you are restricted to Pauli words, here is a quick function that should be much more efficient:
``````>>> def matrix(word):
... phase = 1
... prod = word[0]
... for o2 in word:
... prod, ph = qml.grouping.pauli_mult_with_phase(prod, o2)
... phase *= ph
... return ph * prod.matrix
>>> matrix([qml.PauliX(0) for i in range(1000)])
array([[0, 1],
[1, 0]])
>>> %timeit matrix([qml.PauliX(0) for i in range(1000)])
65.1 ms ± 4.61 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
``````
This seems to match the behaviour of Cirq:
``````>>> %%timeit
... qubit = cirq.LineQubit(0)
... out = cirq.I(qubit)
... for _ in range(1000):
... out *= cirq.X(qubit)
... out.matrix([qubit])
79 ms ± 5.11 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
``````
Depending on how many qubits you are using, a direct matrix approach could also work?
``````>>> %%timeit
... out = qml.Identity(0).matrix
... for _ in range(1000):
... out = out @ qml.PauliX(0).matrix
9.72 ms ± 794 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
``````
(although, my above solutions only make sense if you are looking for Pauli multiplication only; if you need a mixture of multiplication and/or tensor products, that makes it more complicated) | 757 | 2,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-49 | latest | en | 0.860613 |
http://encyclopedia2.thefreedictionary.com/equilibrium | 1,560,930,426,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998923.96/warc/CC-MAIN-20190619063711-20190619085711-00558.warc.gz | 58,250,528 | 13,991 | # equilibrium
Also found in: Dictionary, Thesaurus, Medical, Legal, Financial, Acronyms, Wikipedia.
Related to equilibrium: Equilibrium of forces
## equilibrium,
state of balance. When a body or a system is in equilibrium, there is no net tendency to change. In mechanics, equilibrium has to do with the forces acting on a body. When no force is acting to make a body move in a line, the body is in translational equilibrium; when no force is acting to make the body turn, the body is in rotational equilibrium. A body in equilibrium at rest is said to be in static equilibrium. However, a state of equilibrium does not mean that no forces act on the body, but only that the forces are balanced. For example, when a leverlever,
simple machine consisting of a bar supported at some stationary point along its length and used to overcome resistance at a second point by application of force at a third point. The stationary point of a lever is known as its fulcrum.
is being used to hold up a raised object, forces are being exerted downward on each end of the lever and upward on its fulcrum, but the upward and downward forces balance to maintain translational equilibrium, and the clockwise and counterclockwise moments of the forces on either end balance to maintain rotational equilibrium. The stability of a body is a measure of its ability to return to a position of equilibrium after being disturbed. It depends on the shape of the body and the location of its center of gravity (see center of masscenter of mass,
the point at which all the mass of a body may be considered to be concentrated in analyzing its motion. The center of mass of a sphere of uniform density coincides with the center of the sphere.
). A body with a large flat base and a low center of gravity will be very stable, returning quickly to its position of equilibrium after being tipped. However, a body with a small base and high center of gravity will tend to topple if tipped and is thus less stable than the first body. A body balanced precariously on a point is in unstable equilibrium. Some bodies, such as a ball or a cone lying on its side, do not return to their original position of equilibrium when pushed, assuming instead a new position of equilibrium; these are said to be in neutral equilibrium. In thermodynamicsthermodynamics,
branch of science concerned with the nature of heat and its conversion to mechanical, electric, and chemical energy. Historically, it grew out of efforts to construct more efficient heat engines—devices for extracting useful work from expanding hot gases.
, two bodies placed in contact with each other are said to be in thermal equilibrium when, after a sufficient length of time, their temperatures are equal. Chemical equilibriumchemical equilibrium,
state of balance in which two opposing reversible chemical reactions proceed at constant equal rates with no net change in the system. For example, when hydrogen gas, H2, and iodine gas, I2
refers to reversible chemical reactions in which the reactions involved are occurring in opposite directions at equal rates, so that no net change is observed.
## Equilibrium
The state of a body in which the forces acting on it are equally balanced.
## equilibrium
see SOCIAL EQUILIBRIUM.
## Equilibrium
in thermodynamics, such a slow transition of a thermodynamic system from one equilibrium state to another that all intermediate states may be regarded as equilibrium states. It is characterized by a very slow (infinitely slow at the limit) variation of the thermodynamic parameters of state. Any equilibrium process is a reversible process, and, conversely, any reversible process is an equilibrium process.
## equilibrium
[‚ē·kwə′lib·rē·əm]
(chemistry)
(mechanics)
Condition in which a particle, or all the constituent particles of a body, are at rest or in unaccelerated motion in an inertial reference frame. Also known as static equilibrium.
(physics)
Condition in which no change occurs in the state of a system as long as its surroundings are unaltered.
(statistical mechanics)
Condition in which the distribution function of a system is time-independent.
## equilibrium
The state of being equally balanced; a state of a body in which the forces acting on it are equally balanced.
## equilibrium
1. any unchanging condition or state of a body, system, etc., resulting from the balance or cancelling out of the influences or processes to which it is subjected
2. Physics a state of rest or uniform motion in which there is no resultant force on a body
3. Chem the condition existing when a chemical reaction and its reverse reaction take place at equal rates
4. Physics the condition of a system that has its total energy distributed among its component parts in the statistically most probable manner
5. Physiol a state of bodily balance, maintained primarily by special receptors in the inner ear
6. the economic condition in which there is neither excess demand nor excess supply in a market
References in periodicals archive ?
In order to move the concept of market equilibrium from a theoretical condition to one with practical applications, consider the following alternative definitions of real estate market equilibrium:
These alternative definitions link vacancy to equilibrium and equilibrium to rents.
For the equilibrium analysis, we use figures to analyze equilibria for both scenarios.
Rothschild and Stiglitz (1976) define the equilibrium set of contracts as the set of contracts such that each contract offered in equilibrium earns non-negative expected profits and such that there does not exist a contract outside the equilibrium set of contracts that earns, if added, non-negative expected profits.
Thus, the state of relatively stable or metastable equilibrium is defined as the state in which a system remains for a long period of time, and any slight disturbance causing the system to deviate from the metastable state does not result in the system passing into another state.
The mover will perceive his state of motion to be stable while his state of equilibrium is in fact metastable.
These conditions jointly determine the equilibrium point.
By solving the conditions for both firms simultaneously, Nash equilibrium in price strategies can be obtained.
Chapter Six introduces a decentralized version of the economy presented in Chapter Five, and Chapter Seven defines competitive equilibrium and the equilibrium price system.
Tieben reviews how the founders of and contributors to preclassical, classical, neoclassical, and heterodox economics, especially Austrian economics, employed the equilibrium concept.
The content of Tieben's historical investigation into the usage of the equilibrium is outlined and assessed in Section 2.
A distribution is a correlated equilibrium if for any player i and any pair [s*.
Site: Follow: Share:
Open / Close | 1,366 | 6,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-26 | latest | en | 0.942932 |
http://www.conversion-website.com/mass/atomic-mass-unit-to-gram.html | 1,511,500,588,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807089.35/warc/CC-MAIN-20171124051000-20171124071000-00023.warc.gz | 372,385,122 | 4,604 | # Atomic mass units to grams (amu to g)
## Convert atomic mass units to grams
Atomic mass units to grams converter above calculates how many grams are in 'X' atomic mass units (where 'X' is the number of atomic mass units to convert to grams). In order to convert a value from atomic mass units to grams (from amu to g) simply type the number of amu to be converted to g and then click on the 'convert' button.
## Atomic mass units to grams conversion factor
1 atomic mass unit is equal to 1.66053892173E-24 grams
## Atomic mass units to grams conversion formula
Mass(g) = Mass (amu) × 1.66053892173E-24
Example: Calculate how many grams are in 205 atomic mass units.
Mass(g) = 205 ( amu ) × 1.66053892173E-24 ( g / amu )
Mass(g) = 3.4041047895465E-22 g or
205 amu = 3.4041047895465E-22 g
205 atomic mass units equals 3.4041047895465E-22 grams
## Atomic mass units to grams conversion table
atomic mass units (amu)grams (g)
121.992646706076E-23
142.324754490422E-23
162.656862274768E-23
182.988970059114E-23
203.32107784346E-23
223.653185627806E-23
243.985293412152E-23
264.317401196498E-23
284.649508980844E-23
304.98161676519E-23
325.313724549536E-23
345.645832333882E-23
365.977940118228E-23
386.310047902574E-23
406.64215568692E-23
426.974263471266E-23
447.306371255612E-23
467.638479039958E-23
487.970586824304E-23
508.30269460865E-23
atomic mass units (amu)grams (g)
3004.98161676519E-22
4006.64215568692E-22
5008.30269460865E-22
6009.96323353038E-22
7001.162377245211E-21
8001.328431137384E-21
9001.494485029557E-21
10001.66053892173E-21
11001.826592813903E-21
12001.992646706076E-21
13002.158700598249E-21
14002.324754490422E-21
15002.490808382595E-21
16002.656862274768E-21
17002.822916166941E-21
18002.988970059114E-21
19003.155023951287E-21
20003.32107784346E-21
21003.487131735633E-21
22003.653185627806E-21
Versions of the atomic mass units to grams conversion table. To create a atomic mass units to grams conversion table for different values, click on the "Create a customized mass conversion table" button.
## Related mass conversions
Back to atomic mass units to grams conversion
TableFormulaFactorConverterTop | 754 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-47 | latest | en | 0.493286 |
https://math.stackexchange.com/questions/80141/given-a-surjective-linear-mapping-of-free-modules-how-do-you-show-the-correspond | 1,566,550,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00269.warc.gz | 548,058,235 | 29,545 | # Given a surjective linear mapping of free modules how do you show the corresponding matrix has an invertible minor?
The following post can is related to part c) of this problem http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=124137 and boils down to some issues I am having with use of wedge product.
Let $V,W$ be two free $R$ modules. Suppose that $f: V \rightarrow W$ is a linear mapping and let $A$ be the matrix corresponding to the linear map $f$.
How do you show if $f$ is onto and if the ideal $\subset R$ generated by the non-invertible elements is distinct from $R$ there exists a minor of $A$ which is invertible and whose order is the same dimension as $W$ and $dim(V) \geq dim(W)$?
The reason I metioned wedge product is because we are supposed to consider the power $f\wedge f \wedge \ldots \wedge f$ where we have taken $\dim(W)$ wedge products of f. | 229 | 885 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-35 | latest | en | 0.89487 |
http://www.leadtelecom.com/gb14gg/6c9455-cosine-similarity-between-two-documents | 1,685,549,854,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00607.warc.gz | 66,148,274 | 6,608 | Plagiarism Checker Vs Plagiarism Comparison. Calculate the cosine document similarities of the word count matrix using the cosineSimilarity function. ), -1 (opposite directions). When we talk about checking similarity we only compare two files, webpages or articles between them.Comparing them with each other does not mean that your content is 100% plagiarism-free, it means that text is not matched or matched with other specific document or website. Here's how to do it. Some of the most common and effective ways of calculating similarities are, Cosine Distance/Similarity - It is the cosine of the angle between two vectors, which gives us the angular distance between the vectors. COSINE SIMILARITY. I often use cosine similarity at my job to find peers. While there are libraries in Python and R that will calculate it sometimes I'm doing a small scale project and so I use Excel. As documents are composed of words, the similarity between words can be used to create a similarity measure between documents. Calculating the cosine similarity between documents/vectors. From Wikipedia: âCosine similarity is a measure of similarity between two non-zero vectors of an inner product space that âmeasures the cosine of the angle between themâ C osine Similarity tends to determine how similar two words or sentence are, It can be used for Sentiment Analysis, Text Comparison and being used by lot of popular packages out there like word2vec. Cosine Similarity will generate a metric that says how related are two documents by looking at the angle instead of magnitude, like in the examples below: The Cosine Similarity values for different documents, 1 (same direction), 0 (90 deg. In the blog, I show a solution which uses a Word2Vec built on a much larger corpus for implementing a document similarity. One of such algorithms is a cosine similarity - a vector based similarity measure. Jaccard similarity. This reminds us that cosine similarity is a simple mathematical formula which looks only at the numerical vectors to find the similarity between them. But in the ⦠So in order to measure the similarity we want to calculate the cosine of the angle between the two vectors. You have to use tokenisation and stop word removal . A similarity measure between real valued vectors (like cosine or euclidean distance) can thus be used to measure how words are semantically related. In general,there are two ways for finding document-document similarity . If you want, you can also solve the Cosine Similarity for the angle between vectors: The cosine similarity, as explained already, is the dot product of the two non-zero vectors divided by the product of their magnitudes. And then apply this function to the tuple of every cell of those columns of your dataframe. Cosine similarity between two folders (1 and 2) with documents, and find the most relevant set of documents (in folder 2) for each doc (in folder 2) Ask Question Asked 2 years, 5 months ago For more details on cosine similarity refer this link. The solution is based SoftCosineSimilarity, which is a soft cosine or (âsoftâ similarity) between two vectors, proposed in this paper, considers similarities between For simplicity, you can use Cosine distance between the documents. The cosine similarity between the two documents is 0.5. This metric can be used to measure the similarity between two objects. This script calculates the cosine similarity between several text documents. Mathematically, it measures the cosine of the angle between two vectors projected in a multi-dimensional⦠Note that the first value of the array is 1.0 because it is the Cosine Similarity between the first document with itself. It is calculated as the angle between these vectors (which is also the same as their inner product). We might wonder why the cosine similarity does not provide -1 (dissimilar) as the two documents are exactly opposite. go package that provides similarity between two string documents using cosine similarity and tf-idf along with various other useful things. Document 2: Deep Learning can be simple In the scenario described above, the cosine similarity of 1 implies that the two documents are exactly alike and a cosine similarity of 0 would point to the conclusion that there are no similarities between the two documents. Compute cosine similarity against a corpus of documents by storing the index matrix in memory. The matrix is internally stored as a scipy.sparse.csr_matrix matrix. Unless the entire matrix fits into main memory, use Similarity instead. Cosine Similarity (Overview) Cosine similarity is a measure of similarity between two non-zero vectors. The cosine of 0° is 1, and it is less than 1 for any angle in the interval (0, Ï] radians. So we can take a text document as example. TF-IDF approach. Step 3: Cosine Similarity-Finally, Once we have vectors, We can call cosine_similarity() by passing both vectors. At scale, this method can be used to identify similar documents within a larger corpus. You can use simple vector space model and use the above cosine distance. It will calculate the cosine similarity between these two. Convert the documents into tf-idf vectors . When to use cosine similarity over Euclidean similarity? The two vectors are the count of each word in the two documents. Cosine similarity is a measure of similarity between two non-zero vectors of an inner product space.It is defined to equal the cosine of the angle between them, which is also the same as the inner product of the same vectors normalized to both have length 1. Document similarity2 a bag of words or more precise a bag of word document similarity2 technical information that be... That may be new or difficult to the learner document-document similarity are.... New or difficult to the tuple of every cell of those columns of your dataframe similarity for angle! Not provide -1 ( dissimilar ) as the angle between the two.... Columns of your dataframe finding document-document similarity formula which looks only at the numerical vectors to find the similarity words. A text document as example words or more precise a bag of.. Using cosine similarity between two vectors projected in a similar direction the between. Reminds us that cosine similarity, I show a solution which uses Word2Vec! Deep Learning can be represented by the vectors are the count of each word in two... On cosine similarity of 0 consider the cosine similarities between pairs of the angle between the documents. Of your dataframe for simplicity, you can use cosine distance two non-zero divided... It is the cosine document similarities of the array is 1.0 because is. Search to construct a corpus of documents by storing the index matrix in memory dissimilar ) the. Will calculate the cosine similarity does not provide -1 ( dissimilar ) as the angle between the two documents input... Use tokenisation and stop word removal word count matrix using the cosineSimilarity function this function to tuple... Is at the numerical vectors to find the similarity between two sets more details on cosine similarity then a... Exactly opposite into RAM explained already, is the dot product of the angle between these documents... Determine the similarity between two vectors duplicates detection resulting three-dimensional vectors more details on cosine and... A bag of terms 1.0 because it is 0 then both vectors are the count of word..., in general, there are two ways for finding document-document similarity into main memory, use instead... Is a simple mathematical formula which looks only at the numerical vectors to find the between... The cooridate system ( 0,0 ) common words a cosine similarity is used to create a similarity for! Method can be used to identify similar documents within a larger corpus for implementing a document similarity using similarity. To be in terms of their subject matter first value of the vector is at the center of the documents. Both vectors are similarâ implementing a document similarity using cosine similarity between words can be by... Of 1, two documents to create a similarity measure between documents or vectors also... Is calculated as the angle between the two vectors are similarâ Yes cosine... Also the same as their inner product ) mathematical formula which looks only at the center of the is! Document similarity2 between [ 0,1 ], B ) = cosine of the vector is the! New or difficult to the learner you can use simple vector space model use! Explained already, is the cosine similarity between documents use Amazonâs book search to construct a corpus of documents storing. Algorithms is a simple mathematical formula which looks only at the numerical to. A text corpus containing all words of documents by storing the index matrix in memory the vector at... Vectors a and B is, in general, there are two cosine similarity between two documents for finding similarity... The numerical vectors to find the similarity between documents or vectors scale, this method can used. The cosine similarity measure for document clustering, there are two ways for finding similarity! Index matrix in memory contains sparse vectors ( such as tf-idf documents and! Information that may be new or difficult to the learner containing all words of by... ( Overview ) cosine similarity against a corpus of documents will calculate the cosine similarity and tf-idf along with other! Product of the angle between two vectors are similarâ in the blog I... Word frequency distribution of a document similarity using cosine similarity does not provide -1 ( dissimilar ) the... And use the above cosine distance contains sparse vectors ( such as tf-idf documents ) fits!, as explained already, is the dot product of the vector at... So in order to measure the orientation between two vectors a and B is in! Use simple vector space model and use the above cosine distance the two documents are of! The cosineSimilarity function document is a simple mathematical formula which looks only at the numerical vectors to the... 1. bag of words or more precise a bag of terms tokenisation and stop word.. Similar direction the angle between these cosine similarity between two documents ( which is also the as! Orientation between two string documents using cosine similarity and tf-idf along with various other useful things document. Between two sets likely to be in terms of their magnitudes if the two documents are.. Show a solution which uses a Word2Vec built on a much larger corpus for implementing a document similarity using cosine similarity between two documents... Clustering, there are different similarity measures available as their inner product ) center! Such as tf-idf documents ) and fits into RAM their magnitudes, two.... Metric can be used to measure the similarity we want to calculate cosine similarity - a vector based measure. Are similarâ between [ 0,1 ] are composed of words, the similarity between two string using! I guess, you can define a function to the learner each word the! Be used to identify similar documents within a larger corpus for implementing a document similarity using similarity! Be represented by the vectors are the count of each word in the blog, I will Amazonâs! The numerical vectors to find the similarity between two non-zero vectors document be... This metric can be represented by a bag of word document similarity2 the center of the two documents 0.5! The learner can be used to measure the similarity between them are complete different internally stored as scipy.sparse.csr_matrix! Because it is calculated as the angle between the documents the cooridate system ( 0,0 ) of the angle the... Be hard Yes, cosine similarity between two vectors projected in a multi-dimensional⦠bag... Of technical information that may be new or difficult to the learner document. This reminds us that cosine similarity refer this link guess, you define... This reminds us that cosine similarity for the angle between the first with! Direction the angle between vectors: Yes, cosine similarity, I will use Amazonâs search... Is 1.0 because it is calculated as the two documents are exactly.... Formula to calculate cosine similarity is a measure of how similar two documents documents or vectors is. Provides similarity between documents or vectors difficult to the tuple of every cell of those of. 1: Deep Learning can be particularly useful for duplicates detection of words or more a! Lot of technical information that may be new or difficult to the tuple of cell! Similarity of 1, two documents represented by a bag of words, the similarity them. Similarities between pairs of the angle between these vectors ( such as tf-idf documents ) and fits into.. Documents ) and fits into main memory, use similarity instead value of the resulting three-dimensional vectors into RAM be... Similar documents within a larger corpus for implementing a document similarity using cosine similarity is a but! Algorithms is a metric Duration: 6:43 documents or vectors of 1, two documents represented by the of... That these two like a lot of technical information that may be new or to... Theory I will⦠with cosine similarity is a measure of similarity between them likely! Information that may be new or difficult to the tuple of every of! Between them with itself to create a similarity measure similarity measures available is the... Use simple vector space model and use the above cosine distance apply this function to calculate the cosine the! By a bag of words, the similarity we want to calculate the cosine angle... Why the cosine of angle between vectors: Yes, cosine similarity between two projected. Corpus for implementing a document similarity using cosine similarity is a cosine similarity of 1, two documents a. - a vector based similarity measure between documents words, the similarity between documents vectors. The cosineSimilarity function why the cosine similarities between pairs of the two documents exactly! Use cosine similarity between two documents above cosine distance between two vectors are complete different the similarity between documents or vectors Word2Vec! A vector based similarity measure for document clustering, there are different similarity measures available corpus for implementing document. Text/Term/Document similarity, you can use cosine distance between the documents with cosine similarity for the between... Similarity using cosine similarity the word count matrix using the cosineSimilarity function cosine similarity between two documents simple but intuitive measure how. ) cosine similarity does not provide -1 ( dissimilar ) as the two documents represented by the product their... = cosine of angle between two objects similarity âTwo documents are composed of words, the similarity between can! The center of the cooridate system ( 0,0 ) similarity is a of! To construct a corpus of documents reminds us that cosine similarity refer this link frequency count between... Can now measure the similarity between words can be hard to find the similarity between two is. Divided by the product of their subject matter by a bag of terms of distance between text... Similar if their vectors are the count of each word in the blog, I show solution. Document can be used to identify similar documents within a larger corpus the concept text/term/document! Two sets index matrix in memory to identify similar documents within a larger corpus implementing...
Advanced Dermatology Lincolnshire, Korean Butcher Online, Fiat Stilo Reliability, Value Of Old Silver Dollars 1887, Dr Who -- Digital Spy Forum, How To Bonsai Portulacaria Afra, Healthcare Regulatory Compliance Salary, Farm House For Sale Near Kalyan, Wiki Cheese Mite, | 3,088 | 15,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.914434 |
https://www.opengl.org/discussion_boards/archive/index.php/t-144704.html | 1,531,806,903,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589573.27/warc/CC-MAIN-20180717051134-20180717071134-00168.warc.gz | 961,712,464 | 4,609 | PDA
View Full Version : View Frustum Question
azcoder
01-19-2003, 12:26 PM
Thanks in advance for any help -
What is the best way to get the coordinates of each corner of the view frustum? I would like to get the projection of the view frustum onto the xz plane after the projection and camera transformations have been applied.
Thanks again for any hints or tips
azcoder
01-19-2003, 02:12 PM
The best source of info I have found so far is at: http://www.markmorley.com/opengl/frustumculling.html
But still I am unclear - How can I extract the corners of the frustum from the combined Projection/View Matrix? The tutorial extracts plane definitions of the Normal vector / distance from origin format. I am looking for only the projection of the view frustum onto the xz plane.
Thanks again for any help
rgpc
01-19-2003, 02:34 PM
For what purpose do you require these coords?
Are you trying to draw a representation of the Frustum onto an ortho viewport? (ie. Drawing the "Camera Cone")
If so you can easily work out the Frustum as a set of vertices and then use GL transformations when you draw it...
Eg.
When using gluPerspective you have..
near
far
fovy
aspect
so for the near plane
p1.z = p2.z = p3.z = p4.z = near;
p1.x = p2.x = -near * tan(0.5f * fovy * aspect);
p3.x = p4.x = -p1.x;
p1.y = p3.y = near * tan(0.5f * fovy);
p2.y = p4.y = -p1.y
For the far plane you do the same with far instead of near in you equations.
And from there you can (a) multiply the verts by your modelview matrix or (b) use gl transforms etc to draw your shape in an Ortho/Persp view...
azcoder
01-19-2003, 06:02 PM
I want to do view frustum culling for some heightmap terrain tiles. Rather than store and test a bounding box, I want to just test a bounding square against the quadrilateral formed by projecting the frustum onto the xz plane. Perhaps I can just send each vertex of the frustum throught the view matrix. I was using gluPerspective rather than glFrustum. I'm going to swith and try that...
Thanks again for any help... | 549 | 2,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-30 | latest | en | 0.88549 |
http://www.enotes.com/homework-help/how-integrate-y-x-x-4-1-442804 | 1,462,280,470,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121534.33/warc/CC-MAIN-20160428161521-00203-ip-10-239-7-51.ec2.internal.warc.gz | 503,978,028 | 11,366 | # How integrate y = x/(x^4+1)?
Posted on
You need to use substitution technique, hence, you should come up with the following substitution, such that:
`x^2 = t => 2xdx = dt => xdx = (dt)/2`
Replacing the variable, yields:
`int x/(x^4+1)dx = int ((dt)/2)/(t^2 + 1)`
`int ((dt)/2)/(t^2 + 1) = (1/2)tan^(-1) t + c`
Replacing back `x^2` for `t` yields:
`int x/(x^4+1)dx = (1/2)tan^(-1) x^2 + c`
Hence, evaluating the indefinite integral of the given function, using substitution technique, yields `int x/(x^4+1)dx = (1/2)tan^(-1) x^2 + c.` | 207 | 544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2016-18 | longest | en | 0.593173 |
https://docsbay.net/calculate-the-gravity-at-the-surface-and-at-5-km-altitude-at-the-pole | 1,638,814,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00117.warc.gz | 299,506,015 | 4,552 | Calculate the Gravity at the Surface and at 5 Km Altitude at the Pole
Tags
ATMO 551aHomework 3Due October 22, 2010
1. Calculate the dry adiabatic temperature lapse rate of the atmosphere of Titan in K/km. Assume the atmosphere is made of 97% N2 and 3% methane.
1. Gravity calculations
2. calculate the gravity at the surface and at 5 km altitude at the pole
3. calculate the gravity at the surface and at 5 km altitude at 45o latitude
4. calculate the gravity at the surface and at 5 km altitude at the equator
5. What is the percentage change in the gravity between the surface and 5 km in each of the cases?
Consider 2 atmospheric cases. In both cases, assume the surface pressure is 1000 mb and the latitude is 45o. To make things simple, assume the air is completely dry and contains no water.
• In Case 1, the surface temperature is 280.5 K and temperature decreases with altitude at 5 K/km.
• In Case 2, the surface temperature is 288K and temperature decreases with altitude is 6.5 K/km.
1. Calculate the pressure at 5 km altitude
Use the equation below from page 10 of the notes entitled “Physical Properties of the Atmosphere”
and the average gravity to calculate the pressure at 5 km altitude.
1. sum the values of gravity at the surface and 5 km and divide by 2 to get the approximate average gravity between the surface and 5 km altitude to use in the equation
2. Calculate the temperature and the pressure at 5 km altitude for Case 1
3. Calculate the temperature and the pressure at 5 km altitude for Case 2
4. Which pressure is higher? Explain why.
(hint: Think in terms of the pressure scale height)
1. Determine the potential temperature, .
2. Calculate the potential temperature at 5 km altitude for Case 1
3. Calculate the potential temperature at 5 km altitude for Case 2
4. What would the temperature of the air parcel be if it were lowered to the surface in each Case?
5. Which value is higher? Explain why
1. Calculate the stability (d/dz) at 5 km altitude for each Case.
1. At what frequency would a parcel oscillate if it were displaced at 5 km altitude in each Case?
1. a. What is the restoring acceleration for a vertical displacement of 100 m in each Case?
b. How large is this compared to g?
1. Venus Surface Temperature: Along the lines of the argument given in class, estimate the surface temperature of Venus assuming the atmospheric temperature equals the radiative equilibrium temperature at 300 mb and the surface pressure is 92 bars. State any assumptions you make. Account for the fact that the heat capacity of CO2 changes with temperature
SHOW ALL WORK | 618 | 2,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-49 | latest | en | 0.876897 |
www.iapnrpfgm.org | 1,555,981,679,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578583000.29/warc/CC-MAIN-20190422235159-20190423021159-00481.warc.gz | 244,544,043 | 8,749 | ## FAQ
Q. How do we calculate the training load of District Instructors for ToT program of NRP FGM BNCRP?
1. First we should find out the number of Districts in a State in which we wish to start program.
2. We propose to make 4 District Instructors in each District. Hence, to calculate the number of District Instructors in a state we have to multiply the No of Districts with 4, it will give us the number of District Instructors needed in that state. It is quite possible that we may not get 4 prospective instructors from small districts. Similarly we will need more number of instructors from bigger cities.
3. To calculate the Training of Trainers (ToT) workshop number we have to divide the Number of district Instructors in a particular state by 40. As in one Training of Trainers - 40 District Instructors will be trained. (These will be labeled as District Instructor’s Training of Trainers Workshop).
• Total number of districts in MP: 50
• Total number of District Instructors needed 4 @ District : 200
• Total number of Training of Trainers = 40 @ Training of Trainers = 5
Q. What is selection procedure of the participants? (Or) What are the various steps involved in selecting the participants for the DI TOT?
Selection of Participants
- Participants for Training of Trainers – i.e. Potential District Instructors should be selected by State academic coordinator in consultation with the state IAP President & Secretary – they should preferably be Paediatricians and IAP members.
- The selection should be made only after considering the merits of candidates (teaching skills, academic understanding and willingness to propagate the skills of Basic New born Care)
- The Participants will have to give an undertaking that they will promote and propagate the mission of NRP FGM.
Q. How do we calculate the training load for Provider courses of NRP FGM BNCRP?
Under NRP FGM Project we have funding to train 200,000 people in BNCRP. Calculate number of pediatricians, obstetricians, Nursing Home doctors who are conducting deliveries and minimum two paramedical staff from each maternity homes in that district in private sector. More paramedical workers can be taken from bigger hospitals/nursing homes. Ensure every delivery will be attended by at least one person trained in NRP.
Total No of Pediatricians + Total No of Gynecologists + Total no of Paramedics (2 or more from each maternity hospital) = Total number of participants in a particular District.
The courses can be conducted in batches of 16, 24 or 32. There should be one instructor for 8 participants
Calculation of training load and planning the number of courses in a district would be done by district coordinator.
FOR EXAMPLE IN MP
- Paediatricians – 1500
- Gynaecologists – 2500
- Paramedics (Nursing Homes + Pvt Hospitals = 500) ×2 =1000
- Total Number of participants - 5000
Q. What is the source of funds for the TOT course?
Release of Finances for the TOT Course
1. The finances for these courses will be released through IAP Central office, Mumbai.
2. Finances for the Training of Trainers (ToT) workshop will be rendered by IAP NNF NRP FGM office to the organizing branch as per recommendations of State Academic Coordinators and National Coordinator.
Q. What is the source of funds for the Provider course?
1. The finances for these courses will be released through IAP Central office, Mumbai.
2. For provider courses the fund would be sent to local IAP branch account. 70% of the fund would be sent soon after registration of course and the remaining would be sent after successful completion of course and submission of course roster
Common Questions about Pre-course Planning and conducting the course
Q. How do I justify the need for a Basic Newborn Care and Resuscitation Program course and identify potential beneficiaries?
There is a need for a continuous Basic Newborn Care and Resuscitation Program in every city / district and convincing health professionals about the need for Basic Newborn Care and Resuscitation Program may prove to be a challenge.
Q. How do I identify potential beneficiaries?
To identify potential beneficiaries of the course –
• Prepare a list of maternity and general facilities (where maternity facilities are offered) in a city / district. This can be done with the help of the local government and municipality.
• A database can be created from the information gathered, which should include names of health professionals, their medical qualifications, their operational facility, contact details, etc.
• Request them to fill out an ‘experience survey form’ (download) and analyze the results. Based on the result candidates can be chosen for training.
Q. What information I can get from Internet about the course?
The website of IAP NNF NRP FGM is www.iapnrpfgm.org where you can get all information related to the present course and the future courses.
Preparations before a Course
Q. When should one start preparing and what are the preparations required to conduct a training course?
Preparations should start 1 month before a course
These include –
1. Fixing a date for the course
2. Fixing a venue
3. Registering course online
4. Registering participants for the course
5. online evaluation of participant
6. Procuring course material for participants
7. Applying for course grant from IAP NNF NRP FGM office
8. coordinating with lead instructor and other instructors , sending out invitations and confirming their participation
9. Arranging for faculty stay (if necessary)
Q. What should be the ideal date for scheduling an NRP course?
The date chosen for the course(s) plays a major part in its success or failure. AVOID scheduling your course -
- The day before or after a holiday weekend
- In close proximity to a major religious holiday
- During a major school holiday, such as Summer Vacation
Q. What should be the ideal venue for Basic Newborn Care and Resuscitation Program course?
While deciding location one should keep the following facts in mind -
- It is advisable to keep these courses in Hospitals where an ideal venue can be created. However, in some instances, an ideal venue may not exist for the Basic Newborn Care and Resuscitation Program course. As a Trainer of Basic Newborn Care and Resuscitation Program, one may need to be flexible and creative under challenging conditions.
- There are no budgetary provisions to keep courses in hotels and this practice is highly discouraged. In case hospitals are not available an alternative venue like IMA Hall, District Training Centre should be chosen.
- The venue should comprise of a hall or two or more adjoining rooms as per the number of workstations required. Make sure that if the venue is in a hall then there should be adequate distance between two workstation to prevent voice over between stations
- A model menu for working lunch will be given to course coordinator.
Q. How should you register your course on the website?
Registration of course would be done by course coordinator. For this log on to website www.iapnrpfgm.org, click on the link ‘Register your course’ given on the lower left side of the home page and fill in the details required and submit the form. After approval from iapnrpfgm office you would receive a mail confirming that your course has been approved and a course ID would be provided to you. Please save the course ID as this would be required in all future communication.
Q. How should the participants register for BNCRP Course?
Course coordinator would send a letter to all prospective candidates depicting the course ID, type of course, aim of course, venue, location, timing of the course and tell them to register themselves by logging on to website www.iapnrpfgm.org , then click the link ‘participant registration’ fill the details required and submit. If the participant doesn’t have access to internet, then get the participant to fill the registration form physically and hand it over to the course coordinator. The course coordinator will then upload the information on website.
Q. How to perform online evaluation test?
Once the participant is registered he would get a login ID and Password. Log on to the website www.iapnrpfgm.org, click the link ‘Online evaluation’ fill the login ID and Password and then give the online exam as per your course i.e. for BNCRP Provider course part 1- NSSK part 1 only, for BNCRP Provider full course – NSSK Part 1 and 2
Q. How do I procure course manual, other NRP training material and ensure availability of Manikins for training?
The course material differs for different NRP courses
• NRP Advanced Provider course – course manual is AAP Text Book of Neonatal Resuscitation, 5th Edition.
• BNCRP Provider course – course manual is NSSK manual you can get copies through your District Coordinator and send to participants one month prior to the course.
• Please get in touch with your state academic coordinator/Zonal coordinator for supplies and availability of manuals, manikins for training.
Q. When should the course material reach the Participants?
The course material should be made available to each learner at least one month prior to the date of the course to enable them to read it well before appearing for online evaluation.
Q. What arrangements should be ensured on the day of the course?
The following administrative and infrastructural arrangements should be made on the day of the course
- Mark attendance of all course attendees
- Give folders to participants and instructor
- Workstations
Q. What should be the contents of participant folder?
The participant folder should consist of
- Ball pen
- Pencil
- Sharpener
- Eraser
- Plan of the day
- Exercise Checklists
- Posters (for BNCRP full course only)
- Evaluation forms (Feedback Forms) to be filled by participants
Q. What should be the contents of instructor folder?
All the items in the participants folder plus
- Performance evaluation pre and post test papers (10-12 copies each)
- A copy of written evaluation paper
Q. How to prepare a workstation for training?
The number of workstation required for the training is one per eight participants. The workstation should be made by joining four tables (aprox. 6*3 feet) that can be detached during the practice session and again can be realigned. There should be sitting arrangement for 8 participants and the instructor. If there are more then one workstation in a hall then there should be adequate distance between them to prevent voice over. Each work station should have
- Attendance sheet
- Manikin kits: Fill the manikin with water before training
- Scissors (one per manikin kit)
- Cord tie or clamp (one per manikin kit)
- Gauze pieces (two per manikin kit)
- Gloves (two per manikin kit)
- Identification tag (one per manikin kit)
- Thermometer (one per manikin kit) only for full course
Q. Could an IAP branch accept outside Sponsorship (Academic grant) to meet the expenses of NRP Course?
For BNCRP courses, funds will be provided by IAP NNF NRP FGM office. Johnson and Johnson India has agreed to provide academic grant to train 200,000 health professionals and workers for the next 5 years in Basic NRP and Essential Newborn Care. Additional funding will not be required for these courses.
Q. Could I organize an Advanced Provider course in my city and accept Sponsorship (Academic grant) to meet the expenses of the Course?
• Advanced NRP courses in 2011 (other than those allotted in agreement with sponsors if any) and in subsequent years will be self sponsored and will be conducted on lines of PALS course of IAP.
• If funding is required it can be raised under banner of IAP local branch in the form of sponsorship for stalls, scientific session, Lunch or Dinner. However, IAP Guidelines for sponsorship should be strictly followed. Course content and duration should never be compromised. | 2,523 | 11,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.918071 |
https://10secondracing.com/products/eibach%C2%AE-07-13-bmw-m3-e90-sedan-0-8-x-0-6-pro-kit-lowering-coil-springs | 1,726,427,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00146.warc.gz | 55,943,031 | 149,653 | 20100.140
# Eibach® (07-13) BMW M3 E90/E92 - 0.8" x 0.6" Pro-Kit Lowering Coil Springs
Regular price\$350.00
/
Shipping calculated at checkout.
• In stock, ready to ship
• Question? Call us at (888) 257-1077 🏁
• Free Shipping USA (Exc. HI, AK, PR)
Compatible: (4.0L)
2007 - 2013 BMW M3 E90 Sedan
2007 - 2013 BMW M3 E92 Coupe
Note:
Tested with factory dampers
Includes vehicles with EDC (Electronic Damping Control)
STEP 1: MOTION RATIO
In developing a basic spring setup, you first step is determining your Motion Ratio. A different formula is needed for the type of suspension your race vehicle utilizes: A-arm or Beam axle. Please take into consideration the Angle Correction Factor in your computation.
### A-arm Suspension - (See Diagram 1)
MR
Motion Ratio
d1
Distance from spring centerlines to control arm inner pivot center (in) or (mm)
d2
Distance from outer ball joint to control arm inner pivot center (in) or (mm)
### Beam Axle Suspension - (See Diagram 2)
MR
Motion Ratio
d3
Distence between spring centerlines (in) or (mm)
d4
Distance between tire centerlines
### Angle Correction Factor
ACF
Angle Correction Factor
d3
Spring angle From Vertical (see diagram 1)
d4
Distance between tire centerlines
#### DIAGRAM 1 (A-ARM SUSPENSION)
The motion ratio is a lever arm effect of the control arm acting on the spring. If the spring is mounted at an angle, the reduced motion of the spring must also be taken in account.
#### DIAGRAM 2 (BEAM AXLE SUSPENSION)
The motion ratio of a live axle setup is shown here. Over two-wheel bumps, the motion ratio is 1:1. Over single wheel bumps and during body roll, the motion ratio as shown in step 1. The motion ratio is only used for calculating roll resistance, not for suspension frequencies.
## STEP 2: WHEEL RATE
Wheel Rate is the actual rate of a spring acting at the tire contact patch. This value is measured in lbs/inch or N/mm, just as spring rate is. The wheel rate can be determined by using the formula below.
### Wheel Rate (non beam)
WR
Wheel Rate (lbs/in) or (N/mm)
C
Spring Rate (lbs/in) or (N/mm)
MR
Motion Ratio
ACF
Angle Correction Factor
## STEP 3: SUSPENSION FREQUENCY
Suspension Frequency refers to the number of oscillations or "cycles" of the suspension over a fixed time period when a load is applied to the vehicle
### Wheel Rate (non beam)
SF
Suspension Frequency (cpm)*
WR
Wheel Rate (lbs/in) or (N/mm)
Sprung Weight
Vehicle corner weight less unsprung weight
### Tip 1: Calculation of Wheel Rate for a given frequency
WR
Wheel Rate (lbs/in) or (N/mm) (see step 2)
SF
SF Suspension Frequency (cpm) (see step 3)
Sprung Weight
Vehicle corner weight less unsprung weight
### Tip 2: Calculation of Spring Rate needed for a given Wheel Rate
C
Spring Rate (lbs/in) or (N/mm)
WR
Wheel Rate (lbs/in) or (N/mm) (see step 2)
MR
Motion Ratio
ACF
Angle Correction Factor
## Determining Spring Rate
All Eibach motorsport springs are tested between 20% and 70% of the spring’s total travel. This spring rate can be measured easily using the following steps:
Example Spring—Standard: 1200.250.0500 (12”Free Length, 2.5”ID, 500lb/in)
Metric: 0300.060.0100 (300mm Free Length, 60mm ID, 100N/mm)
### STEP 1: DETERMINE TRAVEL
For the spring to be rated, please refer to the specifications listed for your spring part number in this catalog and record the travel measurement.This number represents the total available travel from free height to coil bind. Our example spring travel measurements for standard: 6.25" metric: 146mm
### STEP 2: DETERMINE TEST RANGE
Calculate the first test point by taking 20% of 6.25" (which equals 1.25") or 20% of 146mm (which equals 29.20mm) and the second test point by taking 70% of 6.25" (which equals 4.375") or 70% of 146mm (which equals 102.20mm). The actual travel between these two points (3.125") or (73mm) is where we determine the spring rate.
### STEP 3: SPRING RATE TEST
Preload the spring 1.25" or 29.20mm and record the force measurement. Continue to compress the spring an additional 3.125" (total compression of 4.375") or 73mm (total compression of 102.20mm) and record the force measurement. Calculate and record the difference in force between the two points (1.25">< 4.375") or (29.20mm>< 102.20mm). In our example the difference would be approximately 1565lbs or 7300N.
### STEP 4: SPRING RATE CALCULATION
With Eibach4s precise spring rate tolerance of +/- 2% (500 x 2% = 10 lbs) the spring rate should fall between 490 and 510 lbs/in (1565 / 3.125 = 500lb) or 95N and 105N (7300N / 73mm = 100N/mm).
Install Guides:
Warranty:
## Customer Reviews
Be the first to write a review
0%
(0)
0%
(0)
0%
(0)
0%
(0)
0%
(0) | 1,328 | 4,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.831818 |
https://www.airmilescalculator.com/distance/gbe-to-orp/ | 1,721,323,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00410.warc.gz | 568,656,268 | 21,748 | # How far is Orapa from Gaborone?
The distance between Gaborone (Gaborone Sir Seretse Khama International Airport) and Orapa (Orapa Airport) is 230 miles / 369 kilometers / 199 nautical miles.
The driving distance from Gaborone (GBE) to Orapa (ORP) is 334 miles / 537 kilometers, and travel time by car is about 6 hours 20 minutes.
230
Miles
369
Kilometers
199
Nautical miles
## Distance from Gaborone to Orapa
There are several ways to calculate the distance from Gaborone to Orapa. Here are two standard methods:
Vincenty's formula (applied above)
• 229.514 miles
• 369.367 kilometers
• 199.442 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 230.415 miles
• 370.816 kilometers
• 200.225 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Gaborone to Orapa?
The estimated flight time from Gaborone Sir Seretse Khama International Airport to Orapa Airport is 56 minutes.
## Flight carbon footprint between Gaborone Sir Seretse Khama International Airport (GBE) and Orapa Airport (ORP)
On average, flying from Gaborone to Orapa generates about 59 kg of CO2 per passenger, and 59 kilograms equals 130 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Gaborone to Orapa
See the map of the shortest flight path between Gaborone Sir Seretse Khama International Airport (GBE) and Orapa Airport (ORP).
## Airport information
Origin Gaborone Sir Seretse Khama International Airport
City: Gaborone
Country: Botswana
IATA Code: GBE
ICAO Code: FBSK
Coordinates: 24°33′18″S, 25°55′5″E
Destination Orapa Airport
City: Orapa
Country: Botswana
IATA Code: ORP
ICAO Code: FBOR
Coordinates: 21°16′0″S, 25°19′0″E | 537 | 2,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.813003 |
https://math.stackexchange.com/questions/359723/law-of-divisibility-on-37/359756 | 1,618,273,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00366.warc.gz | 501,384,799 | 39,636 | # law of divisibility on $37$
how to find and prove law of divisibility on $37$?
Added:---- how to prove for$37$ that: Split off the last digit, multiply by 11, and subtract the product from the number that is left. if the result is divisible by 37 then so is the original number.divisibility criteria for 37
• define law of divisibility – clark Apr 12 '13 at 18:39
• @clark for example law of divisibility on 2 is the last digits of number must be even. – agustin Apr 12 '13 at 18:41
• so it should be an easier or faster process than doing the actual division ? – clark Apr 12 '13 at 18:44
• I guess you mean divisibility criterion, cut-the-knot.org/blue/divisibility.shtml – leonbloy Apr 12 '13 at 18:45
• Hint: $\rm 37\mid 10^3 - 1,\$ or note $\rm\: mod\ 37\!:\ 100\equiv -11,\:$ and $\rm\:{-11}^{-1}\equiv 10.$ – Math Gems Apr 12 '13 at 18:46
As $$999=37\cdot27$$
So, $10^3\equiv1\pmod{37}\implies 10^{3k}\equiv1\pmod{37}$
So, $\sum_{0\le r\le n}a_i10^i$
$$=(a_0+10a_1+100a_2)+10^3(a_3+10a_4+100a_5)+10^6(a_6+10a_7+100a_8)\cdots$$
$$\equiv (a_0+10a_1+100a_2)+(a_3+10a_4+100a_5)+(a_6+10a_7+100a_8)\cdots\pmod {37}$$
i.e., we can group by $3$ digits and add to test the divisibility by $37$
EDIT: the explanation of the link in the Question
As $11\cdot(10a+b)-1\cdot(11b-a)=111\cdot a$
$(10a+b)$ will be divisible by $111\iff (11b-a)$ is divisible by $111$
A more general idea can be found here.
Try to find how $11,1$ are identified as multiplier from the link.
HINT: $37$ divides $999=37\cdot27$, so $37\equiv-1\pmod{1000}$. Now think about the classic divisibility test for $9\equiv-1\pmod{10}$. Imagine writing the number to be tested in base $1000$.
$\rm mod\ 37\!:\ 100^{-1}\!\equiv 10\:$ so e.g. $\rm\:37\mid n = 100^2 c + 100 b + a \iff 37\mid 10^2 n \equiv c + 10 b + 10^2 a$
$37\mid \color{#0a0}{54}\color{blue}{39}\!:\quad \color{#0a0}{54}\!+\!10(\color{blue}{39}) \equiv \color{0a0}{17}\!+\!10(\color{blue}2)\equiv 0$
$\rm\!\begin{eqnarray} 37\mid \color{#0a0}{44}\color{blue}{47}\color{#c00}{40}\!: &&\!\!\color{#0a0}{44}\!+\!10(\color{blue}{47}\!+\!10(\color{#c00}{40}))\\ \equiv && \color{#0a0}7\!+\!10(\color{blue}{10}\!+\!10(\color{#c00}3))\\ \equiv && \color{#0a0}7\!+\!10(3)\equiv 0\end{eqnarray}$
$\rm\!\begin{eqnarray} 37\mid \color{#0a0}{54}\color{blue}{66}\color{#c00}{65}38\!: &&\color{#0a0}{54}\!+\!10(\color{blue}{66}\!+\!10(\color{#c00}{65}\!+\!10(38))) \\ \equiv && \color{#0a0}{17}\!+\!10(\color{blue}{29}\!+\!10(\color{#c00}{28}\!+\!10)) \\ \equiv && \color{#0a0}{17}\!+\!10(\color{blue}{29}\!+\!10)\\ \equiv &&\color{#0a0}{17}\!+\!10(2)\equiv 0 \end{eqnarray}$
Given a number in base 10, say $d_nd_{n-1}\cdots d_0$, calculate $d_0 + 10d_1+26d_2+d_3+10d4+26d_5+\cdots$ (alternating $1$, $10$, and $26$). The resulting number is divisible by 37 if and only if $d_nd_{n-1}\cdots d_0$ is divisible by $37$. | 1,190 | 2,857 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-17 | latest | en | 0.636649 |
http://stackoverflow.com/questions/3767270/my-simple-turing-machine?answertab=votes | 1,427,907,620,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131305143.64/warc/CC-MAIN-20150323172145-00063-ip-10-168-14-71.ec2.internal.warc.gz | 248,882,054 | 17,501 | # My simple turing machine
I'm trying to understand and implement the simplest turing machine and would like feedback if I'm making sense.
We have a infinite tape (lets say an array called T with pointer at 0 at the start) and instruction table:
``````( S , R , W , D , N )
S->STEP (Start at step 1)
W->WRITE (0 or 1)
D->DIRECTION (0=LEFT 1=RIGHT)
N->NEXTSTEP (Non existing step is HALT)
``````
My understanding is that a 3-state 2-symbol is the simplest machine. 3-state i don't understand. 2-symbol because we use 0 and 1 for READ/WRITE.
For example:
``````(1,0,1,1,2)
(1,1,0,1,2)
``````
Starting at step 1, if Read is 0 then { Write 1, Move Right) else {Write 0, Move Right) and then go to step 2 - which does not exist which halts program.
What does 3-state mean? Does this machine pass as turing machine? Can we simplify more?
-
I think the confusion might come from your use of "steps" instead of "states". You can think of a machine's state as the value it has in its memory (although as a previous poster noted, some people also take a machine's state to include the contents of the tape -- however, I don't think that definition is relevant to your question). It's possible that this change in terminology might be at the heart of your confusion. Let me explain what I think it is you're thinking. :)
You gave lists of five numbers -- for example, (1,0,1,1,2). As you correctly state, this should be interpreted (reading from left to right) as "If the machine is in state 1 AND the current square contains a 0, print a 1, move right, and change to state 2." However, your use of the word "step" seems to suggest that that "step 2" must be followed by "step 3", etc., when in reality a Turing machine can go back and forth between states (and of course, there can only be finitely many possible states).
• Turing machines keep track of "states" not "steps";
• What you've described is a legitimate Turing machine;
• A simpler (albeit otherwise uninteresting) Turing machine would be one that starts in the HALT state.
Edits: Grammar, Formatting, and removed a needless description of Turing machines.
Response to comment: Correct me if I'm misinterpreting your comment, but I did not mean to suggest a Turing machine could be in more than one state at a time, only that the number of possible states can be any finite number. For example, for a 3-state machine, you might label the possible states A, B, and C. (In the example you provided, you labeled the two possible states as '1' and '2') At any given time, exactly one of those values (states) would be in the machine's memory. We would say, "the machine is in state A" or "the machine is in state B", etc. (Your machine starts in state '1' and terminates after it enters state '2').
Also, it's no longer clear to me what you mean by a "simpler/est" machine. The smallest known Universal Turing machine (i.e., a Turing machine that can simulate another Turing machine, given an appropriate tape) requires 2 states and 5 symbols (see the relevant Wikipedia article).
On the other hand, if you're looking for something simpler than a Turing machine with the same computation power, Post-Turing machines might be of interest.
-
NEXTSTEP does not necessarily mean STEP+1. Using the word 'STATE' instead of 'STEP' here would not make more sense since there can be 2 defined which goes against the meaning of the word 'STATE'. Maybe there is a better word than both STEP and STATE. And yes, when I say simpler I mean a simpler definition that still can be programmed to do any computation. – Yehonatan Sep 23 '10 at 5:07
@Yehonatan: I've updated my answer in response to your comment. I'm not entirely sure I understand you correctly, though; if I'm still not adequately addressing your questions, I would appreciate clarification. – Seth Sep 23 '10 at 5:42
I believe that the concept of state is basically the same as in Finite State Machines. If I recall, you need a separate termination state, to which the turing machine can transition after it has finished running the program. As for why 3 states I'd guess that the other two states are for intialisation and execution respectively.
Unfortunately none of that is guaranteed to be correct, but I thought I'd post my thoughts anyway since the question was unanswered for 5 hours. I suspect if you were to re-ask this question on cstheory.stackexchange.com you might get a better/more definative answer.
-
So whats the difference between FSM and TM? – Yehonatan Sep 23 '10 at 5:14
@Yehonatan: An FSM has no tape. You feed an FSM input (a stream of ones and zeroes, for example), but this stream is not like a tape, since the FSM cannot rewind or modify it. – Seth Sep 23 '10 at 6:07
@Seth understood. – Yehonatan Sep 23 '10 at 7:51
"State" in the context of Turing machines should be clarified as to which is being described: (i) the current instruction, or (ii) the list of symbols on the tape together with the current instruction, or (iii) the list of symbols on the tape together with the current instruction placed to the left of the scanned symbol or to the right of the scanned symbol. Reference
- | 1,271 | 5,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2015-14 | latest | en | 0.94685 |
https://ishratamin.com/what-is-the-sharpe-ratio-in-investment-analysis/ | 1,701,894,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00771.warc.gz | 365,676,300 | 24,767 | # Understanding The Sharpe Ratio: A Key Investment Analysis Metric
Do you ever find yourself confused when it comes to analyzing investments? If so, you’re not alone. One popular metric that many investors use is the Sharpe ratio. But what exactly is the Sharpe ratio in investment analysis? Well, in short, it’s a measure that helps evaluate the risk-adjusted return of an investment. By taking into account both the return and the volatility of an investment, the Sharpe ratio offers insight into its performance. In this blog article, we’ll delve into the details of the Sharpe ratio, how it’s calculated, and why it’s an important tool for investors. So, let’s dive in and unravel the secrets of the Sharpe ratio!
## What is the Sharpe Ratio in Investment Analysis?
Investing in the financial markets can be a daunting task, especially for those who are not familiar with the intricacies of the industry. There are numerous factors to consider, such as risk and return, that can greatly impact investment decisions. One commonly used tool in investment analysis is the Sharpe ratio. The Sharpe ratio is a measure of risk-adjusted performance that helps investors evaluate the attractiveness and potential profitability of investment opportunities. In this article, we will delve into the details of what the Sharpe ratio is and how it is calculated.
### Understanding the Sharpe Ratio
The Sharpe ratio, named after its creator William F. Sharpe, is a financial metric that measures the excess return of an investment compared to the risk-free rate of return, adjusted for the standard deviation of the investment’s returns. It provides investors with a way to assess the risk-reward tradeoff of a particular investment or investment portfolio.
The Sharpe ratio takes into account both the return generated by an investment and the level of risk associated with it. It allows investors to compare different investments and determine which one offers the best risk-adjusted return. By incorporating risk, the Sharpe ratio provides a more comprehensive analysis than simply looking at returns alone.
### Calculating the Sharpe Ratio
The Sharpe ratio formula is relatively straightforward:
Sharpe Ratio = (Rp – Rf) / σp
Where:
– Rp is the average return of the investment.
– Rf is the risk-free rate of return.
– σp is the standard deviation of the investment’s returns.
Let’s break down each component of the formula:
1. Average Return (Rp): This is the average return generated by the investment over a specific time period. It can be calculated by summing up all the returns and dividing them by the number of periods.
2. Risk-Free Rate of Return (Rf): The risk-free rate is the hypothetical return an investor would earn on a risk-free investment, such as a U.S. Treasury bond. It represents the minimum return an investor expects to receive without taking on any risk.
3. Standard Deviation (σp): The standard deviation is a statistical measure of the dispersion of returns around the average return. It provides insights into the volatility or variability of the investment’s returns. A higher standard deviation indicates greater price volatility and, therefore, higher risk.
By subtracting the risk-free rate from the investment’s average return and dividing it by the standard deviation, the Sharpe ratio quantifies the excess return per unit of risk.
### Interpreting the Sharpe Ratio
The Sharpe ratio serves as a valuable tool for investors as it allows them to assess the risk-reward profile of an investment or investment portfolio. Here’s how to interpret the Sharpe ratio:
1. Positive Sharpe Ratio: A positive Sharpe ratio indicates that the investment or portfolio has generated excess returns above the risk-free rate for each unit of risk taken. The higher the Sharpe ratio, the better the risk-adjusted performance.
2. Negative Sharpe Ratio: A negative Sharpe ratio suggests that the investment or portfolio has underperformed the risk-free rate, considering the level of risk taken. This indicates poor risk-adjusted performance.
3. Sharpe Ratio of 0: A Sharpe ratio of 0 implies that the investment or portfolio has generated returns equal to the risk-free rate, with no excess return per unit of risk. It indicates a neutral risk-reward tradeoff.
### Benefits and Limitations of the Sharpe Ratio
The Sharpe ratio offers several benefits in investment analysis:
1. Risk-adjusted Comparison: The Sharpe ratio enables investors to compare different investments or portfolios on a risk-adjusted basis. This allows for a more meaningful evaluation of investment opportunities.
2. Objective Measure: By incorporating both return and risk, the Sharpe ratio provides an objective measure of an investment’s performance. It helps investors make informed decisions based on a comprehensive analysis.
3. Portfolio Optimization: Investors can use the Sharpe ratio to optimize their investment portfolios by allocating their funds to investments with higher risk-adjusted returns.
However, it is important to be aware of the limitations of the Sharpe ratio:
1. Reliance on Historical Data: The Sharpe ratio relies on historical data to calculate returns and standard deviation. Past performance may not accurately predict future performance, making the ratio subject to potential inaccuracies.
2. Risk Assumption: The Sharpe ratio assumes that the investment’s returns follow a normal distribution, which might not always be the case. In reality, financial markets can exhibit extreme or non-normal behaviors.
3. Single-Factor Analysis: The Sharpe ratio is a single-factor analysis and does not take into account other factors that may influence an investment’s performance, such as market trends or company fundamentals.
In conclusion, the Sharpe ratio is a valuable tool in investment analysis that allows investors to assess the risk-reward profile of an investment or investment portfolio. By considering both returns and risk, the Sharpe ratio provides a more comprehensive perspective than evaluating returns alone. It aids in comparing investments on a risk-adjusted basis and helps investors make informed decisions. However, it is essential to understand the limitations of the Sharpe ratio and use it in conjunction with other analysis techniques to gain a more complete understanding of investment opportunities.
### What is the Sharpe ratio in investment analysis?
The Sharpe ratio is a measure used in investment analysis to assess the risk-adjusted return of an investment. It helps investors understand whether the returns they are receiving are worth the level of risk they are taking. Calculated by dividing the excess return of an investment over the risk-free rate by its standard deviation, the Sharpe ratio provides a quantitative way to evaluate investment performance.
### How is the Sharpe ratio calculated?
To calculate the Sharpe ratio, you need the annualized excess return of the investment (the investment return minus the risk-free rate) and the standard deviation of the investment returns. The formula is as follows: Sharpe Ratio = (Average Return of Investment – Risk-Free Rate) / Standard Deviation of Investment.
### What does a higher Sharpe ratio indicate?
A higher Sharpe ratio implies a better risk-adjusted return. It indicates that an investment generates higher returns relative to the amount of risk taken. Investors often prefer investments with higher Sharpe ratios as they offer more favorable risk-return trade-offs.
### How should the Sharpe ratio be interpreted?
The Sharpe ratio should be interpreted as a measure of risk-adjusted return. A value of 1 or higher is considered good, while a negative value implies the investment underperforms the risk-free rate. However, it is important to compare the Sharpe ratios of different investments within the same context and industry for accurate assessment.
### What are the limitations of the Sharpe ratio?
The Sharpe ratio has certain limitations. It assumes that investment returns follow a normal distribution, which may not be the case in reality. Additionally, it relies on historical data and past performance, which may not be indicative of future results. Furthermore, the Sharpe ratio does not consider non-financial risks or qualitative factors that can affect investment decisions.
### Can the Sharpe ratio be negative?
Yes, the Sharpe ratio can be negative. A negative Sharpe ratio indicates that the investment’s returns are not sufficient to compensate for the risk taken. It suggests the investment has performed worse than the risk-free rate or has a higher level of risk compared to the expected returns.
### What is the risk-free rate used in the Sharpe ratio calculation?
The risk-free rate used in the Sharpe ratio calculation is typically the rate of return on a risk-free investment, such as a government bond or a Treasury bill. It represents the return that could be obtained without taking on any investment risk.
### Can the Sharpe ratio be used to compare different types of investments?
Yes, the Sharpe ratio can be used to compare different types of investments, as long as the risk-free rate and the investment returns are calculated consistently. However, it is important to consider the nature of the investments being compared and any specific factors that may impact their risk and return characteristics.
### Is the Sharpe ratio the only measure of investment performance?
No, the Sharpe ratio is not the only measure of investment performance. There are other indicators, such as the Sortino ratio, Treynor ratio, and Jensen’s alpha, which also provide insights into risk-adjusted returns. Each measure has its own strengths and limitations, and investors often consider multiple metrics to evaluate investment performance comprehensively.
## Final Thoughts
The Sharpe ratio is a widely used tool in investment analysis that helps investors evaluate the risk-adjusted performance of a particular investment or portfolio. By measuring the excess return of an investment per unit of risk, the Sharpe ratio provides valuable insights into both the return potential and the volatility associated with an investment. A higher Sharpe ratio indicates better risk-adjusted returns, while a lower ratio suggests higher volatility relative to returns. Understanding the Sharpe ratio can assist investors in making informed decisions by quantifying the risk-reward tradeoff. Incorporating this ratio into investment analysis can enhance portfolio management strategies and improve overall investment performance. | 2,016 | 10,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-50 | longest | en | 0.927814 |
http://dyerlab.github.io/applied_population_genetics/hardy-weinberg-equilibrium.html | 1,579,390,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00481.warc.gz | 53,125,933 | 16,464 | # 7 Hardy-Weinberg Equilibrium
At the heart of population genetics is the expectation that genotypes will occur at predictable frequencies given a set of assumptions about the underlying population. This is formulated in the enigmatic Hardy-Weinberg Equation, $$p^2 + 2pq + q^2 = 1$$.
In this chapter, we delve in to what that means, where it comes from, and how we can gain biological inferences from data that do not follow.
The underlying idea of Hardy-Weinberg Equilibrium (HWE for brevity) is that genotypes should occur at frequencies as predicted by probability theory alone IF (and this is the big part), the population we are looking at is mating in particular ways. Lets spend a little time talking about what that means and what these assumptions actually are.
## 7.1 Genetic Assumptions
Implicit in the description Above is a characterization of what constitutes a genotype. For simplicity, as this is the way it was originally defined, we will assume that the species we are looking at is diploid, carrying homologous chromosomes from both parents. Diploidy means that in the adult life stage, each individual has two copies of each allele, one from each parental individual. If diploidy is true, we can denote the genotype as $$AA$$ for the diploid homozygote, $$AB$$ for the heterozygote, and $$BB$$ for the other diploid homozygote. If it is not true and the species (at least at that life stage) is haploid, then will will represent the genotype as $$A$$ or $$B$$. Higher levels of ploidy (triploid, tetraploid, hexaploid, etc.) are also possible in many taxa. While the math that follows is essentially the same for these ploidy levels, the actual algebra is a bit more messy. If you are working on taxa with higher ploidy levels, you can do many of the same operations we will focus on in gstudio, consult the documentation for more information.
In addition to ploidy, genetic constraints on the formulation of the classic HWE assume that the species has two separate sexes, both of which contribute to the next generation. In general, this rules out uniparentally inherited markers such as mtDNA or cpDNA, which are transmitted in most cases by only one of the sexes during reproduction. Examining mating events between diploid individuals is something we should all remember from basic biology and genetics through the use of the Punnet Square. That was a tool designed to teach us about sexual mating, however, it is also a great example of the probabilities of mating we see in sexual reproduction. These probabilities play direct roles in the derivation of HWE.
Two additional constraints arise directly from diploid sexual reproduction. First, we assume that the likelihood of a particular allele is independent of the sex of the individual that contains it. For example, male individuals should have as many alleles (e.g., no sex chromosomes that mix ploidy of genotypes) and alleles at that can occur with equal frequency as the other sex (no sex biased alleles probabilities as may arise from say sex-biased lethals).
Finally, we will invoke both of Mendel’s Laws of Inheritance. His first law states that the alleles at a particular locus are statistically independent from each other. If there is an A allele provided by one parent, the probability of the second allele is either $$A$$ or $$B$$ is entirely independent of that first alleles state. The second law, though not necessarily relevant for many HWE applications, deals with the probability of genotypes at two loci. Here we assume that the alleles present at one locus are statistically independent of those at the other locus. We will return to this when we get to linkage but for now lets assume it is a valid assumption.
### 7.1.1 Punnet Squares
A Punnet square is a simple tool used to teach transition probabilities in basic genetics and biology courses. Here is an interactive example using the 2-allele locus (though the tool can handle other ploidy levels) that you can play with to see the kinds of offspring genotypes produced by individual matings.
The population we are examining need to have some specific demographic parameters for HWE to be valid. The most basal assumption that is invoked is that the population size must be very large. In a fact, most of the work of R.A. Fisher and J.B.S. Haldane in developing the modern synthesis relied upon this assumption as well. If population size, $$N$$, is too small, then the stochastic change (what we call genetic drift) can have serious influences on allele frequencies.
Next, and perhaps this is more for our laziness than any other reason, we assume that the generations do not overlap. This means that the during a single generation, all individuals participate in the mating events, each of which is equally likely to be selected as one participating in a mating event (e.g., random mating). For simplicity, lets denote the frequencies of these genotypes as $$f(AA) = P$$, $$f(AB) = Q$$, and $$f(BB) = R$$. If these individuals are really mating randomly with respect to their genotypes, then the probability of union between any pair is simply the product of their population frequencies.
AA AB BB
AA $$P^2$$ $$PQ$$ $$PR$$
AB $$PQ$$ $$Q^2$$ $$QR$$
BB $$PR$$ $$QR$$ $$R^2$$
More importantly for our simplicity in terminology, at the end of the mating episode, all of the next generations offspring are produced and all adults die. If there were overlapping generations or multiple mating events, we would have to integrate these processes along a continuous timescale rather than treating a single mating generation as a discrete unit. Totally possible, and even fun, but not part of the original formulation of HWE.
## 7.2 Evolutionary Assumptions
Finally, we need to invoke two evolutionary assumptions to meet the requirements of HWE. First, we must assume that there is $$AB$$solutely no mutation. A non-zero mutation rate, $$\mu > 0$$, means that the state of a particular allele, say the $$A$$ allele, has a likelihood of spontaneously becoming something other than the $$A$$ allele (e.g., the $$B$$ in this formulation). If HWE is to help us determine the genotype frequencies having one allele spontaneously mutate to another would require that we integrate the mutation rate into HWE directly. Perhaps not that surprisingly, that wasn’t integrated back in the day.
The next assumption has a very similar consequence. Namely, we must assume that the set of individuals that are participating in the mating event in this population is static. Individuals from other populations are not immigrating into this population and individuals within the population are not emigrating out of this population. Just like mutation, and when we get to migration we will see directly how similar these processes are, we will take the easy way out and assume that they do not occur.
The final assumption is based upon selection and is a very easy one to deal with. We simply assume it doesn’t happen. If selection were operating, say an extreme form such as increased lethality of the $$AA$$ genotype prior to reproduction, then we would have to both specify the way in which selection is operating as well as the its magnitude. We will come back to this topic later but for the simplicity of HWE, we make the assumption that it has no effect.
## 7.3 The Mechanics
In a sampled population, we estimate the frequencies of the genotypes as:
$$P = \frac{N_{AA}}{N}$$,
$$Q = \frac{N_{AB}}{N}$$,
and
$$R = \frac{N_{BB}}{N}$$
where $$N_{XX}$$ is the number of $$XX$$ genotypes and $$N$$ is the total number of individuals in the sample. From these genotype frequencies, we can directly estimate the frequency of each allele ($$A$$ and $$B$$) denoted as p and q as:
$$f(A) = P + \frac{Q}{2} = p$$
and
$$f(B) = R + \frac{Q}{2} = q$$
using the lower case versions of $$p$$ and $$q$$. Be careful about switching these up. The population geneticist John Nason, makes the connection that this is what it means to “mind our p’s and q’s” though I suspect the etiology of this statement has more to do with liquid volume measurements than population genetics…
Intuitively the formulation denoted above for $$p$$ and $$q$$ makes sense as the heterozygote genotype ($$AB$$) frequency, $$Q$$, is split evenly between the homozygote genotype frequencies, $$P$$ (for $$AA$$) and $$R$$ (for $$BB$$). If these are the only two alleles in the population then there is the additional restriction that $$p + q = 1$$. It is easy to expand these methods to more than two alleles, but again the original approximation was specifically set for 2-allele systems.
Here is an interactive widget that shows how the expected frequencies of the genotypes change, even under Hardy Weinberg Equilibrium, due to changes in the allele frequencies.
In R, we can approximate these formulas using the gstudio library. After loading in the library, we can create data objects representing diploid loci as follows:
library(gstudio)
hom.AA <- locus( c("A","A") )
hom.BB <- locus( c("B","B") )
het.AB <- locus( c("A","B") )
These objects have have properties associated with them appropriate for representing genetic loci.
ploidy(het.AB)
## [1] 2
is_heterozygote( het.AB )
## [1] TRUE
And can be used in a vector or data.frame or other R container just like any other data type facilitating easy integration into existing analytical workflows.
genotypes <- c( hom.AA, het.AB, hom.BB)
genotypes
## [1] "A:A" "A:B" "B:B"
To examine genotype and allele frequencies for a set of loci, I’ll make a random collection and then use those loci as an example.
loci <- sample(genotypes, 20, replace=TRUE)
loci
## [1] "B:B" "A:B" "B:B" "A:B" "A:A" "A:B" "A:A" "A:B" "B:B" "A:B" "A:A"
## [12] "A:A" "B:B" "A:B" "A:A" "B:B" "A:A" "A:B" "B:B" "A:B"
Here the function sample() makes a random draw of the arguments given in the first position (the genotypes), and selects 20 (the second argument), and since there are fewer potential values than observed ones, you need to tell it that you can sample the same genotypes more than once (the replace=TRUE part). In analytical population genetics, the use of randomization and permutation is critical and we will see this repeatedly throughout this text. Technically, this is a Monte Carlo simulation that was done. Fancy, no?
Next, we can take these data and estimate genotype frequencies. Here the expectation is also given by the code using the methodologies described in the next section.
genotype_frequencies(loci)
## Genotype Observed Expected
## 1 A:A 6 5
## 2 A:B 8 10
## 3 B:B 6 5
and allele frequencies from these loci are:
frequencies( loci )
## Allele Frequency
## 1 A 0.5
## 2 B 0.5
It should be noted that for these function, and for almost all the functions that return data in gstudio, the return object is a native data.frame. Using these tools, it is easy to take our data and extract from it the components necessary for estimating the degree to which they conform to the expectations of HWE.
## 7.4 Trans-Generational Transition Probabilities
If the assumptions in the next generation hold, then it is possible to iterate through all possible combinations of matings, noting the probability of observing each type, and estimate the frequency of offspring genotypes. This is a bit messy and is probably better displayed in tabular format.
Parents $$\mathbf{P(Parents)}$$ $$\mathbf{P(AA)}$$ $$\mathbf{P(AB)}$$ $$\mathbf{P(BB)}$$
$$AA\;x\;AA$$ $$P^2$$ $$P^2$$
$$AA\;x\;AB$$ $$PQ$$ $$PQ/2$$ $$PQ/2$$
$$AA\;x\;BB$$ $$PR$$ $$PR$$
$$AB\;x\;AB$$ $$Q^2$$ $$Q^2/4$$ $$Q^2/2$$ $$Q^2/4$$
$$AB\;x\;BB$$ $$QR$$ $$QR/2$$ $$QR/2$$
$$BB\;x\;BB$$ $$R^2$$ $$R^2$$
During the parents generation, the frequencies of $$AA$$, $$AB$$, and $$BB$$ were P, Q, and R, defining allele frequencies as p and q. If we invoke all those assumptions about genetic, demographic, and evolutionary processes being $$AB$$sent from mating and production of offspring genotypes (e.g., notice how we do not have any terms in there for mutation, being sex biased, etc.) then the frequency of genotypes at the next generation are defined in the table. For example, at the next generation, say $$t+1$$, the frequency of the $$AA$$ offspring in the population is the sum of probabilities of all matins that produced $$AA$$ genotypes.
\begin{aligned} P_{t+1} &= P_{t}^2 + \frac{P_tQ_t}{2} + \frac{Q^2_t}{4} \\\\ &= \left( P_t + \frac{Q_t}{2} \right)^2 \\\\ &= p^2 \end{aligned}
The other homozygote is the same producing an expectation of the frequency of the $$BB$$ genotype at $$t+1$$ of $$q^2$$. Similarly for the heterozygote
\begin{aligned} Q_{t+1} &= \frac{P_tQ_t}{2} + 2P_tR_t + \frac{Q^2_t}{2} + Q_tR_t\\ &= 2\left(P_t + \frac{Q_t}{2} \right)\left(R_t + \frac{Q_t}{2}\right)\\ &= 2pq \end{aligned}
Putting these together, we have an expectation that the genotypes $$AA$$, $$AB$$, and $$BB$$ will occur at frequencies of $$p^2$$, $$2pq$$, and $$q^2$$, which is exactly what HWE is. It takes only one generation of mating under the assumptions of HWE to return all genotype frequencies to HWE.
## 7.5 Consequences of HWE
The consequences of this are important for our understanding of what processes influence population genetic structure. If none of these forces are operating in the population, then the expectation is that the allele frequencies should predict genotype frequencies. However, if there are some forces that are at work, genetic, demographic, or evolutionary, then the frequencies at which we see these genotypes will deviate from those expectations. Now, biologically speaking, is there any population that conforms exactly to these expectations—definitely not! However, in many cases populations are large enough to not be influenced by small N, mutation is rare enough to not cause problems, etc.
Using HWE, we can gain some insights into the which sets of forces may be influencing the observed genotype frequencies. In large part, we examine these changes as deviations from expectations with respect to loss or gain of heterozygotes. These expectations, as depicted above, define an idealized situation against which we measure our data. This metric was specifically designed to ignore almost every process and feature that population geneticists would be interested in looking at. If our data are consistent with these expectations then we have no support for the operation of any of these processes. Boring result, no? In a larger sense, the entire field of Population Genetics is devoted to understanding how violations of these assumptions influence population genetic structure. If everything was in HWE and none of these processes were operating, there would be very little for us to do besides DNA fingerprinting and we would all be forensic examiners…
If some of these features are operating, they do have specific expectations on how they would influence the frequency of genotypes (specifically the homozygotes). Specific examples include:
Decrease Heterozygosity Increase Heterozygosity Inbreeding Coding Errors Null Alleles Gametic Gene Flow Positive Assortative Mating Negative Assortative Mating Selection against Heterozygotes Outbreeding Wahlund Effects Selection for Heterozygotes
## 7.6 Zygotic Gene Flow
In closing, it should be noted that HWE presented here is easily expandable to loci with more than two alleles as well as for loci with different ploidy levels. The same approaches apply, though it is not as clean to estimate the expectations. In the next section, we look at how we specifically test for HWE in our data.
At a more meta level, one could see the entirety of this topic and population genetics as a discipline as understanding how deviations from HWE manifest and testing for the strength of these deviations. It is becoming less common for people to actually test for HWE as there are many ways that you can have a deficiency (or excess) of heterozygotes and it is probably much better to focus on the specific processes that may causing these deviations rather than the magnitude of the deviation alone. That said, it is a nice organizing principle to use HWE as a straw man argument on which to frame the rest of this topic. | 3,870 | 16,253 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-05 | longest | en | 0.9295 |
https://www.overthinkingit.com/2017/04/11/rick-est-rick-multiverse-p-hacking/ | 1,716,781,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00246.warc.gz | 801,225,138 | 28,905 | # The Rick-est Rick, the Multiverse, and p-Hacking
## Is Rick C-137 really the Rick-est Rick in the multiverse? Let’s find out using science!
Please enjoy this guest post by Sean Nixon!
The heart of science is measurement. Seconds came out of the Middle Ages as a measure of time, meters were created in 1793 to standardize the measure of distance, amperes were developed in 1880’s to handle the study of electrical currents, and in the early 1900’s, scientists (or rather statisticians), drunk on their previous success, invented p-values to measure truth. More precisely, p-values give a measurement of how likely it is that the results from an experiment are complete rubbish. A p-value 0 means the scientist’s hypothesis is absolute truth and a p-value of 1 means the experiment gave no support for the hypothesis (all p-values lie between 0 and 1). Ronald Fisher popularized the use of p-values in the 1920s and 30s and set the standard of p<0.05 as the arbitrary cutoff for truth (usually referred to as statistical significance).This had the predictable corollary of making p-values less than 0.05 the litmus test for publishable research. Searching through data for publishable p-values is called p-hacking.
(If you are unfamiliar with p-hacking, the article “Science Isn’t Broken” has an absolutely fantastic interactive illustration that lets you try your hand at p-hacking in real time.)
Briefly, P-hacking occurs when researchers attempt to bend the numbers to fit their narrative. In an idealized situation, statistics aims to divine truth from the tea leaves of chaos, but translating this analysis into something compressible for mere mortals requires constructing a narrative around the results. It’s the difference between saying “subjects who added twenty grams of dark chocolate to their weekly diet over the course of twelve months showed an average decrease in blood pressure of 5+/- 2.43 mmHg,” and saying “dark chocolate is good for combating the effects of stress.” Now imagine that eating the dark chocolate also drove up cholesterol. Or imagine that the effect was only found in people with low sugar diets. Or imagine fifteen different diets were tried and this was the only one that produced a change. Even the most fastidious scientist must make judgement calls about which data is relevant.
Of course, there’s another side to science. In the deep, dark recesses of theoretical physics lie theories and conjectures that utterly defy measurement. The many-worlds interpretation of quantum mechanics makes sense of the underlying mathematics without producing any predictions. Or rather, the predictions literally exist in another dimension. With no observable difference between living in a single lone universe and living in one of an infinite collection of universe, the theory of the multiverse is more philosophy than science.
To understand the branching of the multiverse, imagine two dice: one Newtonian die and one Quantum die. When you roll the Newtonian die it seems like the results are random, but this is an illusion. With enough information about the initial state of the die (the density of the die, the trajectory of the throw, air resistance, etc.) there’s only one solution to Newton’s governing equations of motion. The results of the die roll is uniquely determined. In fact, Stanford mathematician Persi Diaconis has made a career out of showing when things like this aren’t really random. When you roll the Quantum die, however, something strange happens. The governing equations admit six possible solutions. And, when the underlying math has multiple solutions, scientists expect them all to exist somewhere out there. For example, the equations Dirac used to define electrons also predicted the existence of positrons, as a second solution, years before their discovery by Carl Anderson. So, when the the Quantum die appears to come up three, where do the other five possible solutions exist? The branching theory suggests that five other universes must come into being to house the other possible solutions: one universe for each solution. (If you’re a Rick and Morty fan, you might remember this as the premise of the Community episode “Remedial Chaos Theory”).
Science fiction writers have been enamored of alternate realities since long before they were a trendy scientific theory. There are parallel worlds which take the form of an otherworldly dimensions such as in Alice in Wonderland or the Narnia series. Alternate histories where the consequences of a single changed moment in time are teased out like in The Man in the High Castle or it’s spiritual predecessor Bring the Jubilee. Countless TV shows have visited a second earth where the characters’ relationships have been scrambled. And finally, there are franchises that canonically include a full multiverse like Sliders, the DC Comics continuity, or more recently Rick and Morty, Dan Harmon’s wildly popular, animated brainchild, now entering its third season. It is in this last case that the immeasurable strangeness of quantum branching and parallel realities circles back around to the practical calculations of statistics.
Selection Bias of the Rick Kind
In the penultimate episode of Rick and Morty’s first season, the show goes full multiverse. The concept had previously been introduced briefly when, after turning everyone on earth into a Cronenberg-esque horror, the titular duo literally abandon planet and set up shop in someone else’s universe. In the episode “Close Rick-counters of the Rick Kind,” our heroes visit the “Citadel of Ricks,” an entire city populated by versions of Rick Sanchez and his grandson Morty Smith from across the multiverse. Cowboy Rick, insurance salesman Rick, and an entire governing council of Ricks with avant garde haircuts. For purposes of identification, the primary Rick is designated C-137 (the serial number for his home universe).
The Council of Ricks considers Rick C-137 a trouble maker, stating: “…of all the Ricks in the central finite curve, you’re the malcontent. The rogue.”
A condemnation that Rick C-137 wears as a badge of honor, since part of being a Rick means being anti-authority. Rick C-137 believes himself to be the Rick-est Rick, and he later cheers up his grandson by musing that the Rick-est Rick would naturally have the Morty-est Morty. Like an SAT question, Ricks are to the general population what Rick C-137 is to other Ricks. Now, a reasonable person might accept this at face value, smile at the touching moment the relentlessly acerbic Rick (C-137) manages to share with Morty and move onto more important things.
However, that’s not why you read Overthinking It.
Imagine that each Rick is a datum in the enormous dataset of all possible universes. Furthermore, imagine a person’s “Rick-ness” can be distilled into a single constant, R, with large negative R values representing nice, law abiding dimwits, and large positive R values representing brilliant, anarchist asshats. This gives a mathematical framework to transform the existential assertion of being the Rick-est Rickinto a statistical question about the distribution of R values. For instance, looking at the distribution of “Rick-ness” among the general population, we would expect a big lump of average folks with a Rick-ness value near zero and any Rick (even doofus Rick, J19-zeta-7) would appear in the right tail of the distribution for large R values. See Figure 1. The threshold between regular R values and Rick level R values is indicated by a color change from yellow to blue in both distributions.
Figure 1
Similarly, when looking at just the population of Ricks there would be a lump of average Ricks centered around some large R value and then Rick C-137 would fall somewhere in the right hand tail of this already right skewed distribution. See Figure 2. For the moment, we’ll set aside the argument that the Rickest-Rick might instead be the Rick whose R value is exactly the mean (average) among Ricks.
Figure 2
Interdimensional data collection probably presents a great many challenges. Even mere terrestrial endeavors are far from perfect. For example, the United States census tends to undercount minorities; in 2010 it undercounted African-Americans by 2.1% and Hispanics by 1.5% (for a total of about 1.5 million people). Also, that Twitter poll you saw a while ago? Not super accurate.
The problem is selection bias. The impossibility of collecting exactly the data you require means that scientists, statisticians, and pollsters rely on samples (or subsets) of the total population. For example, out of a city of a few thousand Ricks, only about a hundred Ricks appear – maybe ten with actual speaking roles. The audience is left to infer that the rest of the Ricks are similar to the ones that we’re shown. You notice that one-in-twenty Rick/Morty pairs exhibit some visually striking difference (cyborg, cyclops, sea creature, Cronenbergian horror), and you presume that this holds true of the whole population. Ideally, individuals are chosen randomly to get a representative sample. And, when this ultimately fails, other techniques are deployed to try to fake how a completely random sample should look. With a few notable exceptions (like the scientist formerly know as Rick), all of the Ricks that we see in this episode have voluntarily decided to join the Rick Collective. In other words, Ricks who live and work at the Citadel are oversampled.
In a scientific study, this sort of selection bias can taint your results, and there are numerous methods to adjust for oversampling. A notably extreme example of selection bias comes from World War II: the British Airforce was deciding how to most efficiently reinforce their bombers in order to reduce losses without making the planes too heavy or costly. They had been collecting data on all the returning bombers and made plans to reinforce the spots on the planes that most commonly returned with bullet holes. Luckily, before this plan could be implemented, they consulted with the Austrian mathematician Abraham Wald for a second opinion. Wald rightly realized that the data had been tainted by selection bias; the only planes in the data set were those that had successfully returned from their mission. If enemy fire were to bring the plane down, those bullet holes would be conspicuously missing from the data. Thus, the plan was reconsidered. Eventually, the Airforce decided to reinforce the airplane parts displaying the least amount of damage in the data, figuring that the damaged areas of returned planes were of least concern.
In Rick and Morty, the selection bias serves a narrative purpose. Whether or not the original Rick truly represents the epitome of Rick-ness is immaterial; the ocean of Ricks (all geniuses, all sarcastic jerks) creates a background against which Rick C-137’s other characteristics come into stark relief. Artificially skewing the R distribution by only showing Ricks who choose to play by the council’s rules (see Figure 3) seems perfectly reasonable with this goal in mind. As a statistical outlier, Rick C-137 naturally appears to be non-conformist. But, beyond just the anti-council stance, Rick C-137 also shows a greater degree of compassion for Morty, relative to your average Rick.
Figure 3
Throughout the series, Rick C-137 belittles, endangers, neglects, traumatizes, and alienates Morty, so it may seem strange to talk of his compassion. Which is exactly why it takes an entire city of soulless bastards for Rick C-137 to look warm and fuzzy by comparison. The Ricks of the Citadel treat Mortys like pets (at best) or bits of fancy tech, complete with accessories, insurance plans, and coupons for replacement; the plot of the episode where Rick and Morty visit the Citadel involves an enormous engineering project fueled by the suffering of stolen Mortys. Rick C-137, by contrast, spends the episode doing an admittedly terrible job of apologizing to Morty and wraps things up by not quite complimenting Morty for a job well done. Baby steps, as it were. While this doesn’t make Rick C-137’s usual treatment of Morty any less comically horrifying, it does illustrate how Rick C-137 is reaching out (even if he’s failing at it) and establishes a theme that will be expanded on going forward. It’s Rick nature to be a dick, but our Rick is trying to improve.
Narrative p-hacking
It is not entirely surprising that statistics and the multiverses would find themselves so intertwined as both derive from the underlying study of probability. Where statisticians seek to study the distribution of heights or the likelihood of disease across a population of similar people, the multiverse can extend this idea ad absurdum to a population consisting of duplicates of a single person. Not just how likely you are to go bald based on your age, race, sex, etc., but how likely you are to go bald based on the fact that you are Jerry Fling living in Nashville with the stressful job of grizzly bear psychiatrist. Where statisticians seek to assess risk and predict the future, the multiverse confers reality upon all possible futures. A scenario where counterfactuals can be investigated not just hypothetically, but concretely. The only caveat? Our inability to access this data.
The many-world interpretation of quantum mechanics is like having data points from parallel universes. Picking and choosing data to build a narrative constitutes p-hacking. Thus, constructing a narrative set across the multiverse represents the fiction author’s version of p-hacking – or, as I like to think of it, narrative p-hacking. A far less pernicious analog, since tailoring a world (or many worlds) that conveys the writer’s ideas to the audience is, if not the whole point of fiction, at least strongly encouraged.
As a final thought, I’ve been, up to this point, fairly positive about the application of “narrative p-hacking.” However, there’s also a downside to using these techniques. Once you know that when people say “That’s a good question,” they are just stalling for time, once you know advertisers put “Fat Free” on things that never had fat to begin with, once you’re told how the magic is performed, the trick loses its potency. Similarly, once you start seeing the storytelling analogs of p-hacking at work, the curated multiverses lose authenticity. Rick’s unique standing among the other Ricks is really just a failure to take a random sampling. In a half-hour late-night comedy like Rick and Morty, perhaps it doesn’t matter. But what about Sliders? What about Groundhog Day, or Edge of Tomorrow? The Long Earth series? His Dark MaterialsAt some point, does narrative p-hacking become lazy storytelling?
Got another example of narrative p-hacking? Think the writer’s a nutcase? Leave it in the comments, Morty.
### 2 Comments on “The Rick-est Rick, the Multiverse, and p-Hacking”
1. asdf #
“More precisely, p-values give a measurement of how likely it is that the results from an experiment are complete rubbish. A p-value 0 means the scientist’s hypothesis is absolute truth and a p-value of 1 means the experiment gave no support for the hypothesis”
This is wrong
2. johny #
Hasn’t Rick has been proven to be the rick-est rick out of all the ricks in the multiverse, he’s more rick than the citadel and ANY other rick. This is well known, because he IS the measurement for being a rick;. He is the rick-st rick, but why does this even matter, it’s just a cool show with a cool lore behind it. | 3,268 | 15,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | longest | en | 0.940281 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.