url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://freecalculator.net/proportion-calculator | 1,675,386,842,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.2/warc/CC-MAIN-20230202232251-20230203022251-00611.warc.gz | 290,653,552 | 14,565 | # Proportion Calculator
Proportion Calculator
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## What is Proportion?
Proportion is a mathematical concept that compares two or more values to each other. It is used to determine the relationship between two or more values and is expressed as a ratio or a fraction. Proportion is used in many areas of mathematics, including geometry, algebra, and statistics. It is also used in everyday life to compare sizes, amounts, and other values.
## What is Proportion Calculator?
A Proportion Calculator is a tool used to solve proportion problems. It can be used to find the missing value in a proportion, given the other values. It can also be used to find the ratio of two numbers.
## How to Calculate Proportion?
Proportion is the comparison of two numbers or ratios. To calculate a proportion, divide one number by the other. For example, if you have two numbers, 10 and 20, the proportion would be 10/20, which is equal to 0.5.
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https://nrich.maths.org/public/topic.php?code=-124&cl=4&cldcmpid=2693 | 1,596,734,348,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736972.79/warc/CC-MAIN-20200806151047-20200806181047-00345.warc.gz | 355,052,569 | 5,627 | Resources tagged with: Trigonometric functions and graphs
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Loch Ness
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Draw graphs of the sine and modulus functions and explain the humps.
Small Steps
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Two problems about infinite processes where smaller and smaller steps are taken and you have to discover what happens in the limit.
Trigger
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Can you sketch this tricky trig function?
Degree Ceremony
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What does Pythagoras' Theorem tell you about these angles: 90°, (45+x)° and (45-x)° in a triangle?
Climbing
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Sketch the graphs of y = sin x and y = tan x and some straight lines. Prove some inequalities.
Tangled Trig Graphs
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Can you work out the equations of the trig graphs I used to make my pattern?
Trig-trig
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Explore the properties of combinations of trig functions in this open investigation.
After Thought
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Which is larger cos(sin x) or sin(cos x) ? Does this depend on x ?
What Do Functions Do for Tiny X?
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Looking at small values of functions. Motivating the existence of the Taylor expansion.
Building Approximations for Sin(x)
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Build up the concept of the Taylor series
Squareness
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The family of graphs of x^n + y^n =1 (for even n) includes the circle. Why do the graphs look more and more square as n increases?
Taking Trigonometry Series-ly
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Look at the advanced way of viewing sin and cos through their power series.
Spherical Triangles on Very Big Spheres
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Shows that Pythagoras for Spherical Triangles reduces to Pythagoras's Theorem in the plane when the triangles are small relative to the radius of the sphere.
Sine and Cosine
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The sine of an angle is equal to the cosine of its complement. Can you explain why and does this rule extend beyond angles of 90 degrees? | 535 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.844756 |
https://www.slideserve.com/misha/engin-112-intro-to-electrical-and-computer-engineering-lecture-12-circuit-analysis-procedure | 1,542,206,330,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742020.26/warc/CC-MAIN-20181114125234-20181114151234-00524.warc.gz | 995,524,038 | 15,155 | ENGIN 112 Intro to Electrical and Computer Engineering Lecture 12 Circuit Analysis Procedure
1 / 13
# ENGIN 112 Intro to Electrical and Computer Engineering Lecture 12 Circuit Analysis Procedure - PowerPoint PPT Presentation
ENGIN 112 Intro to Electrical and Computer Engineering Lecture 12 Circuit Analysis Procedure. Overview. Important concept – analyze digital circuits Given a circuit Create a truth table Create a minimized circuit Approaches Boolean expression approach Truth table approach
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### ENGIN 112Intro to Electrical and Computer EngineeringLecture 12Circuit Analysis Procedure
Overview
• Important concept – analyze digital circuits
• Given a circuit
• Create a truth table
• Create a minimized circuit
• Approaches
• Boolean expression approach
• Truth table approach
• Provides insights on how to design hardware
• Tie in with K-maps (next time)
A
B
C
Out
A
B
C’
The Problem
• How can we convert from a circuit drawing to an equation or truth table?
• Two approaches
• Create intermediate equations
• Create intermediate truth tables
A
B
R
C
Out
T
S
A
B
C’
Label Gate Outputs
• Label all gate outputs that are a function of input variables.
• Label gates that are a function of input variables and previously labeled gates.
• Repeat process until all outputs are labelled.
A
B
R
C
Out
T
S
A
B
C’
Approach 1: Create Intermediate Equations
• Step 1: Create an equation for each gate output based on its input.
• R = ABC
• S = A + B
• T = C’S
• Out = R + T
A
B
R
C
Out
T
S
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C’
Approach 1: Substitute in subexpressions
• Step 2: Form a relationship based on input variables (A, B, C)
• R = ABC
• S = A + B
• T = C’S = C’(A + B)
• Out = R+T = ABC + C’(A+B)
A
B
A
B
C’
C’
Approach 1: Substitute in subexpressions
• Step 3: Expand equation to SOP final result
• Out = ABC + C’(A+B) = ABC + AC’ + BC’
C
Out
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Approach 2: Truth Table
• Step 1: Determine outputs for functions of input variables.
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Approach 2: Truth Table
• Step 2: Determine outputs for functions of intermediate variables.
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T = S * C’
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Approach 2: Truth Table
• Step 3: Determine outputs for function.
R + T = Out
More Difficult Example
• Step 3: Note labels on interior nodes
F2
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F’2
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F1
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More Difficult Example: Truth Table
• Remember to determine intermediate variables starting from the inputs.
• When all inputs determined for a gate, determine output.
• The truth table can be reduced using K-maps.
A
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Summary
• Important to be able to convert circuits into truth table and equation form
• WHY? ---- leads to minimized sum of product representation
• Two approaches illustrated
• Approach 1: Create an equation with circuit output dependent on circuit inputs
• Approach 2: Create a truth table which shows relationship between circuit inputs and circuit outputs
• Both results can then be minimized using K-maps.
• Next time: develop a minimized SOP representation from a high level description | 1,470 | 4,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-47 | latest | en | 0.763338 |
https://www.physicsforums.com/threads/concept-problems-regarding-equations-of-higher-degree.354592/ | 1,519,476,435,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815560.92/warc/CC-MAIN-20180224112708-20180224132708-00333.warc.gz | 879,178,339 | 15,177 | # Concept problems regarding equations of higher degree
1. Nov 14, 2009
### jeremy22511
1. The problem statement, all variables and given/known data
We generally accepted in lower forms, that when we deal with equations of higher degree, arriving at solutions which need to be rejected is normal, even natural.
However, now that I think of it, doesn't arriving at wrong answers mean only that our deductive argument contains flaws? Otherwise, why else would we have to reject some answers?
Thanks.
J
2. Relevant equations
3. The attempt at a solution
2. Nov 14, 2009
### HallsofIvy
That depends upon what you mean by "flawed". If a "deductive method" gives the right answer, it isn't flawed, is it? I think what you are talking about is when we, say, square both sides of an equation to get rid of a square root, or multiply both sides of an equation by something to get rid of fractions. When you do that, you do change the equation to something that is not exactly equivalent. But as long as you are aware that you are doing that, and take steps to make sure it doesn't give you a wrong answer, as, for example, checking any answers, there is no flaw in your "deductive argument".
As long as we multiply by constants, we know that a number is a solution to our original equation if and only if it is a solution to our new equation. If we multiply both sides of the equation by something involving the unknown variable, then any solution to our original equation must be a solution to the new equation but not vice-versa.
If our original equation is f(x)= g(x) and we multiply both sides by h(x) to get f(x)h(x)= g(x)h(x), then any solution to our original equation is still a solution to our new equation but not necessarily the other way. Any solution to h(x)= 0 is also a solution to f(x)h(x)= g(x)h(x).
When we multiply both sides of an equation by a constant, c, our "deductive argument" is "x is a solution to f(x)= g(x) if and only if it is a solution to cf(x)= cg(x)."
If our "deductive argument" were "x is a solution to f(x)= g(x) if and only if it is a solution to f(x)h(x)= g(x)h(x)", then it would indeed be flawed.
But it is not. Our argument is "any solution, x, to f(x)= g(x) is a solution to f(x)h(x)= g(x)h(x)". That is perfectly correct. It is the other way, "any solution to f(x)h(x)= g(x)h(x) is a solution to f(x)= g(x)" that is incorrect. And we don't claim that. That's why we have to specifically check solutions.
3. Nov 14, 2009 | 633 | 2,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-09 | longest | en | 0.957058 |
http://www.jiskha.com/display.cgi?id=1255661845 | 1,498,277,542,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320215.92/warc/CC-MAIN-20170624031945-20170624051945-00513.warc.gz | 553,720,471 | 3,964 | # elementary math for teachers
posted by .
A deli offers its cheese sandwich with various combinations of mayonnaise, lettuce, tomatoes,
pickles, and sprouts. 5 types of cheese are available. How many different cheese sandwiches are
possible?
• elementary math for teachers -
the toppings on each sandwich can be done in 8 ways, the number of subsets of 3 elements. This will include the nullset, or no toppings.
The wording is not clear as to how many cheeses can go on a sanwich.
e.g. can you have a sandwich with 3 different cheeses?
I will assume we can only have one of the 5 cheeses, so the number of different sandwiches is 5(8) or 40.
If you can have any combination of cheese, there would be 2^5 - 1 or 31 ways to have just the cheese sandwich.
(I subtracted one to exclude the case of no cheeses. After all it is called a "cheese sandwich")
So in that case there would be 5(31) or 155 different sandwiches. | 224 | 924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.944968 |
https://fr.maplesoft.com/support/help/maplesim/view.aspx?path=Student/Precalculus | 1,669,893,216,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00747.warc.gz | 296,082,401 | 28,551 | Student[Precalculus] - Maple Help
Home : Support : Online Help : Mathematics : Packages : Student[Precalculus]
Overview of the Student[Precalculus] Subpackage
Calling Sequence Student[Precalculus][command](arguments) command(arguments)
Description
• The Student[Precalculus] subpackage is designed to help teachers present and students understand the basic material of a standard precalculus course.
Note: Throughout the help pages for the Precalculus subpackage, the terms function and expression are generally used interchangeably, and refer to the mathematical objects that can be manipulated using the operations of calculus. The routines in the subpackage are referred to using the terms routine or command.
• Each command in the Student[Precalculus] subpackage can be accessed by using either the long form or the short form of the command name in the command calling sequence.
As the underlying implementation of the Student[Precalculus] subpackage is a module, it is also possible to use the form Student[Precalculus]:-command or Student:-Precalculus:-command to access a command. For more information, see Module Members.
• The Maple Command Completion facility is helpful for entering the names of Student package commands.
• Many of the commands in the Student[Precalculus] package can be accessed through the context panel. First, load the package. Then, these commands are consolidated under the Student Precalculus name.
Visualization
• The visualization routines provide tools to create plots related to precalculus concepts. They allow you to see the geometric interpretation of these concepts.
The visualization commands are:
Computation
• The computation routines provide tools that perform standard precalculus operations.
The computation commands are:
Interactive
• The interactive routines use the Maple Maplet technology to assist you to work through the standard problems of precalculus in a visually directed manner. These commands display one or more dialog boxes allowing you to plot a function and change the various plot options. They act as an interface to the visualization routines.
The interactive commands are:
Getting Help with a Command in the Package
To display the help page for a particular Student[Precalculus] command, see Getting Help with a Command in a Package.
Precalculus Study Guide
• In addition to the Student[Precalculus] package, Maple includes the Precalculus Study Guide, an exercise-based e-book version of a Precalculus textbook. Covering 11 key topics, this focused review of precalculus gives you multiple ways to solve each problem. | 544 | 2,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.736786 |
https://www.coursehero.com/file/p3lechk/Bourdon-tube-Consists-of-a-hollow-metal-tube-bent-like-a-hook-whose-end-is/ | 1,563,867,777,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00027.warc.gz | 665,878,914 | 694,602 | BREE301Introduction-2
Bourdon tube consists of a hollow metal tube bent
• Notes
• 60
This preview shows page 52 - 60 out of 60 pages.
Bourdon tube : Consists of a hollow metal tube bent like a hook whose end is closed and connected to a dial indicator needle. Pressure transducers : Use various techniques to convert the pressure effect to an electrical effect such as a change in voltage, resistance, or capacitance. Strain-gage pressure transducers: Work by having a diaphragm deflect between two chambers open to the pressure inputs. Piezoelectric transducers : Also called solid- state pressure transducers , work on the principle that an electric potential is generated in a crystalline substance when it is subjected to mechanical pressure.
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Barometer measurement p atm = p vapour + ρ m gL Atmospheric pressure is measured by a device called a barometer ; thus, the atmospheric pressure is often referred to as the barometric pressure . A frequently used pressure unit is the standard atmosphere , which is defined as the pressure produced by a column of mercury 760 mm in height at 0°C ( Hg = 13,595 kg/m 3 ) under standard gravitational acceleration ( g = 9.807 m/s 2 ). negligible
For example if a barometer reads 760 mm Hg, this represents in SI units: P atm = ρ gL
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Example Problem 1 Measuring pressure in a vessel with a U-tube manometer The fluid in the manometer is mercury ( ρ = 13.6 g cm -3 ). The reading on the manometer is 24 cm. Calculate the absolute pressure in the vessel.
P P atm Measurement of Pressure ρ 1 h 1 h 2 ρ 2 P = P atm + ρ 1 gh 1 + ρ 2 gh 2 P = P atm + P gage
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Example Problem 2 The figure here shows a tank within a tank each containing air. The absolute pressure in tank A is 267.7 kPa. Pressure gage A is located inside tank B and reads 140 kPa. The U- tube manometer connected to tank B contains mercury. Determine the absolute pressure inside tank B in kPa and the column length L in cm. The atmospheric pressure surrounding tank B is 101 kPa and g=9.81 m/s 2
Gauge pressure is zero referenced against ambient pressure, so it is equal to absolute pressure (measured) minus atmospheric pressure. P gauge A = P A - P B
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Example Problem 3 A piston/cylinder with cross-sectional area of 0.01 m 2 is connected with a hydraulic line to another piston cylinder of cross-sectional area of 0.05 m 2 . Assume both chambers and the line are filled with hydraulic fluid of 900 kg/m 3 density and the larger second cylinder is 6 m higher up in elevation. With an outside atmospheric pressure of 100 kPa and a net force of 25 kN on the smallest piston, what is the balancing force on the second larger piston?
Determine the total force in kN on the bottom of a 100×50m swimming pool. The depth of the pool varies linearly along its length from 1 m to 4 m. Also, determine the pressure on the floor at the center of the pool, in kPa. The atmospheric pressure is 0.98 bar, the density of the pool water is 998.2 kg/m 3 and the local acceleration of gravity is 9.8 m/s 2 . Example Problem 4 | 779 | 3,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.89455 |
https://www.exactlywhatistime.com/time-from/299-years-from-today | 1,723,228,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00853.warc.gz | 617,487,701 | 7,743 | # What's the date 299 years from today?
## Thursday August 09, 2323
0
299 years from today is 09 Aug 2323, a Thursday. Counting forward by years is usually simple. Anything under a decade can usually be done on one hand; however, remember that years will change the final date depending on leap years. Since today is Friday, well need to account for the weekend if you care about work days. If not, the calculation will remain the same. I usually just start with days of the week then move onto months. We’re 22 days from the end of August, so being in the middle of of the month, 299 days will bring us into September.. From there, we can then count forward by years and adjust for the fiscal calendar to get our 09 Aug 2323.
## How we calculated 299 years from today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how adding 299 years to today's date gets calculated on each visit:
• Started with date inputs: starting point: 09 Aug, Units to add: 299 years, and year: 2024
• Noted your current time of year: 22 days in middle of August
• Added 299 years from current day: 09 Aug, factoring in there are 22 days left in before September
• Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the 2024 fiscal calendar.
### Tips to get your solution: August 09
Thursday Thursday August 09, 2323 is the 221 day of the year or 60.55% through 2323.
• Current date: 09 Aug
• Day of the week: Thursday
• New Date: Thursday August 09, 2323
• New Date Day of the week: Thursday
• August is the end of the summer so counting forward from here could propel us into a new season or even year.
• The solution crosses into a different year..
## Ways to calculate 299 years from today
1. Just calculate it: Start with a time from today calculator. 299 years is easiest solved on a calculator. For ours, we've already factored in the days in + all number of days in each month and the number of days in 2024. Simply add your years and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use August's calendar: Begin by identifying on a calendar, note that it’s Thursday, and the total days in September (trust me, you’ll need this for smaller calculations) and days until next year (double trust me, you'll need this for larger calculations). From there, count forward 299 times by years, adding years from 09 Aug.
3. Use Excel: Regardless of unit type, I use day calculations here. Type =TODAY()+299 into the cell. If you want to add weeks, multiply your day by 7 and months/years will take their own calculation due to the changing days of the week. To find 299 years workdays, convert to days but use =WORKDAY(TODAY(), [number of days], [holidays]) into the cell. [number of days] is how many working days you want to add, and [holidays] is an optional range of cells that contain dates of holidays to exclude.
## 299 working years from today
299 years is Thursday August 09, 2323 or could be Friday December 05, 2442 if you only want workdays. This calculation takes 299 years and only adds by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Thursday August 09, 2323 date.
Work years Solution
Monday
Tuesday
Wednesday
Thursday
August 09
Friday
Saturday
Sunday
## In 299 years, the average person Spent...
• 23442198.0 hours Sleeping
• 3116895.6 hours Eating and drinking
• 5107518.0 hours Household activities
• 1519159.2 hours Housework
• 1676313.6 hours Food preparation and cleanup
• 523848.0 hours Lawn and garden care
• 9167340.0 hours Working and work-related activities
• 8433952.8 hours Working
• 13803394.8 hours Leisure and sports
• 7491026.4 hours Watching television
## What happened on August 09 (299 years from now) over the years?
### On August 09:
• 1958 Actor Robert Redford (21) weds Lola Van Wagenen in Las Vegas, Nevada | 1,006 | 4,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.932117 |
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# 3_1-Clickers-Momentum and Rockets - Taylor Chapter 3:...
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1 Taylor Chapter 3: Momentum
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2 Taylor adds up the forces on all piece “a” of a body with N pieces. If all forces are internal, he gets Ý P = F αβ β≠α α= 1 N If you wrote out all the terms in this double sum, how many would there be ? A) N B) N2 C) N(N-1) D) N! E) Other/not really sure
3 (Assume below that N-II is an experimental fact) We just showed that we can then use N-III to derive the law of conservation of momentum for systems of particles. Is the converse true? i.e.: If the law of conservation of total momentum of a system (of two particles) holds, can you derive that it MUST be the case that F 12=- F 21? A) Yes B) No C) Maybe one could, but I can’t. ..
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4 A 8Li nucleus at rest undergoes β decay transforming it to 8Be, an e- and an (anti-)neutrino. The 8Be has |p|=5 MeV/c at 90o, the e- has |p|= 6 MeV/c at 315o, what is p ν (in the form (px, py)? A) (4.2, 4.2) B) (-5,0) C) (-5,-1) D) (-4.2, 0.8) E) (-4.2, -0.8) MeV/c
5 Dear Radioactive Ladies and Gentlemen: ...I have hit upon a desperate remedy. . I admit that my remedy may appear to have a small a priori probability because neutrons, if they exist, would probably have long ago been seen, However, only those who wager can win. .. Unfortunately I cannot personally appear in Tübingen, since I am indispensable here on account of a ball taking place in Zürich. ...,Your devoted servant, W. Pauli [Translation from Physics Today, Sept. 1978] Pauli’s Desperate Remedy Dec. 4, 1930
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6 If you push horizontally (briefly!) on the bottom end of a long, rigid rod of mass m (floating in
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## This note was uploaded on 02/27/2012 for the course PHYSICS 2210 taught by Professor Stevepollock during the Spring '11 term at Colorado.
### Page1 / 24
3_1-Clickers-Momentum and Rockets - Taylor Chapter 3:...
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Ask a homework question - tutors are online | 688 | 2,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-13 | longest | en | 0.877611 |
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### In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm. Find: XY
Solution: According to the given question, $XY\text{ }||\text{ }BC$ So, In $\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC$ $\angle AXY\text{ }=\angle ABC$ [Corresponding angles]...
### In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm. Find: (i) AY/YC (ii) YC/AC
Solution: According to the given question, $XY\text{ }||\text{ }BC$ So, In $\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC$ $\angle AXY\text{ }=\angle ABC$[Corresponding angles]...
### In the following figure, point D divides AB in the ratio 3: 5. If BC = 4.8 cm, find the length of DE.
Solution: Because, $\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\text{ }by\text{ }AA$ criterion for similarity So, we have $AD/AB\text{ }=\text{ }DE/BC$ $3/8\text{ }=\text{ }DE/4.8$...
### Describe: i) The locus of points at distances less than 3 cm from a given point. ii) The locus of points at distances greater than 4 cm from a given point.
i) The locus of the points will be the space inside of the circle whose radius is $3\text{ }cm$and centre as the given point. ii) The locus of the points will be the space outside of the circle...
### The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is $332\text{ }m$and centre is the point where the gun is fired.
### Describe the locus for the locus of a point in rhombus ABCD, so that it is equidistant from i) AB and BC; ii) B and D.
(i) The locus of the point in a rhombus $ABCD$which is equidistant from $AB\text{ }and\text{ }BC$ will be the diagonal $BD$of the rhombus. (ii) The locus of the point in a rhombus...
### Describe the locus for the locus of a point P, so that: AB^2 = AP^2 + BP^2, where A and B are two fixed points.
The locus of the point $P$is the circumference of a circle with $AB$ as diameter and satisfies the condition $A{{B}^{2}}~=\text{ }A{{P}^{2~}}+\text{ }B{{P}^{2}}$
### Describe the locus for the locus of a point in space which is always at a distance of 4 cm from a fixed point.
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to $4\text{ }cm.$
### Describe the locus for the locus of vertices of all isosceles triangles having a common base.
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.
### Describe the locus for the locus of the centres of all circles passing through two fixed points.
The locus of the centres of all the circles passing through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.
### The locus of a points inside a circle and equidistant from two fixed points on the circumference of the circle.
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be a diameter which is the perpendicular bisector of the line joining the...
### Describe the locus for The locus of the door-handle, as the door opens.
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and the radius equal to distance between the door handle and the axis of rotation...
### The locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
The locus of the runner, running around a circular track and always keeping a distance of $1.5\text{ }m$ from the inner edge will be the circumference of a circle where the radius is equal to the...
### The locus of a stone dropped from the top of a tower.
Solution: As per the given question, Locus of stone is dropped from the top of tower will be vertical line through the point from which the stone is dropped.
### The locus of the moving end of the minute hand of a clock.
The locus of the moving end of the minute hand of the clock will be a circle whose radius will be the length of the minute hand.
### The locus of the centre of a wheel of a bicycle going straight along a level road.
The locus of the centre of wheel, is going straight along the level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.
### The locus of a points at a distance of 2 cm from a fixed line.
The locus of points which are at a distance of $2\text{ }cm$from a fixed line $AB$are a pair of straight lines $l\text{ }and\text{ }m$which are parallel to the given line at a distance of...
### The locus of a point at a distance of 3 cm from a fixed point.
The locus of a point is at a distance of $3\text{ }cm$from a fixed point is circumference of a circle whose radius is $3\text{ }cm$and the fixed point is the centre of the circle.
### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: ∠ABC.
Solution: According to the given question, $AC$ is the side of a regular octagon, $\angle AOC\text{ }=\text{ }{{360}^{o}}/\text{ }8\text{ }=\text{ }{{45}^{o}}$ Hence, $arc\text{ }AC$subtends...
### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: (i) ∠AOB, (ii) ∠ACB,
Solution: (i) $Arc\text{ }AB$ subtends$\angle AOB$at the centre and $\angle ACB$at the remaining part of the circle. $\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB$...
### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: ∠PQR.
Join $OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS$ Given, $PQ\text{ }=\text{ }QR\text{ }=\text{ }RS$ So, $\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS$ [Equal chords subtends equal...
### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: (i) ∠POS, (ii) ∠QOR,
Solution: Join $OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS$ Given, $PQ\text{ }=\text{ }QR\text{ }=\text{ }RS$ So, $\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS$ [Equal chords subtends...
### Calculate BC.
SOLUTION: In ∆ADC, $\begin{array}{*{35}{l}} CD/AD\text{ }=\text{ }tan\text{ }{{42}^{o}} \\ CD\text{ }=\text{ }20\text{ x }0.9004\text{ }=\text{ }18.008\text{ }m \\ \end{array}$ In ∆ADB,...
### In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60o. Calculate the length of the board AB.
SOLUTION: In ∆PSB, $\begin{array}{*{35}{l}} PS/PB\text{ }=\text{ }sin\text{ }{{60}^{o}} \\ PB\text{ }=\text{ }2/\text{ }\surd 3\text{ }=\text{ }1.155\text{ }m \\ \end{array}$ In ∆APQ,...
### If p – 15 = 0 and 2×2 + px + 25 = 0: find the values of x.
According to the given question, $p\text{ }\text{ }15\text{ }=\text{ }0$ And $2{{x}^{2}}~+\text{ }px\text{ }+\text{ }25\text{ }=\text{ }0$ Thus, $p\text{ }=\text{ }15$ Using$p$in the...
### Solve:
According to the given question, $\left( x\text{ }+\text{ }5 \right)\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }24$ Or, ${{x}^{2}}~\text{ }25\text{ }=\text{ }24$ Or,...
### Solve: (x^2+1/x^2)-3(x-1/x)-2=0
Let $\left( x\text{ }\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)$ squaring on both sides ${{\left( x\text{ }\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}$ or,...
### Solve: 2(x^2+1/x^2)-(x+1/x)=11
let $\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)$ squaring on both sides ${{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}$...
### Solve:9(x^2+1/x^2)-9(x+1/x)-52=0
Let, $~\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)$ squaring both sides ${{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}$...
### If given matrix, find the matrix X such that: A + X = 2B + C
Solution: As per the given question,
### Solve: (i) A (BA) (ii) (AB) B.
Solution: $\left( i \right)\text{ }A\text{ }\left( BA \right)$ $\left( ii \right)\text{ }\left( AB \right)\text{ }B$
### 3A x M = 2B; find matrix M.
Solution: According to the given question, $3A\text{ }x\text{ }M\text{ }=\text{ }2B$ Suppose the order of the $matrix\text{ }M\text{ }be\text{ }\left( a\text{ }x\text{ }b \right)$ Now, we know...
### Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.
Let the speed of the second train be $~x\text{ }km/hr.$ Then, the speed of the first train is $\left( x\text{ }+\text{ }5 \right)\text{ }km/hr$ Let O be the position of the railway station,...
### Solve: (i) A – B (ii) A^2
Solution: $\left( \mathbf{i} \right)\text{ }\mathbf{A}\text{ }-\text{ }\mathbf{B}\text{ }$ $\text{ }\left( \mathbf{ii} \right)\text{ }{{\mathbf{A}}^{\mathbf{2}~}}$
### BA = M^2, find the values of a and b.
Solution: $BA$ ${{M}^{2}}$ So, $BA\text{ }={{M}^{2}}$ On comparison, we get $-2b\text{ }=\text{ }-2$ $b\text{ }=\text{ }1$ And, $a\text{ }=\text{ }2$
### If the given matrix and I is a unit matrix of the same order as that of M; show that: M2 = 2M + 3I
Solution: ${{M}^{2}}$ $2M\text{ }+\text{ }3I$ Hence, ${{M}^{2}}~=\text{ }2M\text{ }+\text{ }3I$
### Is the following possible: A^2
Solution: ${{A}^{2}}~=\text{ }A\text{ }x\text{ }A,\text{ }$isn’t possible because the number of columns isn’t equal to its number of rows in matrix A.
### Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.
let the original number of persons to be x. According to ques, Total money which was divided is $=\text{ }Rs\text{ }6500$ Each person’s share is $=\text{ }Rs\text{ }6500/x$ Then, as the question...
### Is the following possible: (i) AB (ii) BA
Solution: $\left( i \right)\text{ }AB$ $\left( ii \right)\text{ }BA$
### If find x and y when x and y when A2 = B.
Solution: ${{A}^{2}}~$ ${{A}^{2}}~=\text{ }B$ On comparison, we get $4x\text{ }=\text{ }16$ $x\text{ }=\text{ }4$ And, $1\text{ }=\text{ }-y$ $y\text{ }=\text{ }-1$
### If the given matrix and I is a unit matrix of order 2×2, find: (i) A^2 (ii) B^2A
Solution: (i) ${{A}^{2}}$ (ii) $~{{B}^{2}}$ ${{B}^{2}}A$
### A trader buys x articles for a total cost of Rs 600. (i) Write down the cost of one article in terms of x. If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less. (ii) Write down the equation in x for the above situation and solve it for x.
According to ques, Number of articles $=\text{ }x$ And, the total cost of articles $=\text{ }Rs\text{ }600$ Again, (i) Cost of one article $=\text{ }Rs\text{ }600/x$ (ii) also,...
### If the given matrix and I is a unit matrix of order 2×2, find: (i) AI (ii) IB
Solution: (i) AI = (ii) IB=
### If given matrix and I is a unit matrix of order 2×2, find: (i) AB (ii) BA
Solution: According to the given question (i) (ii)
### The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate: (i) the time taken by the car to reach town B from A, in terms of x; (ii) the time taken by the train to reach town B from A, in terms of x.
According to ques, Speed of car = $x\text{ }km/hr$ Speed of train = $\left( x\text{ }+\text{ }16 \right)\text{ }km/hr$ Time = $Distance/\text{ }Speed$ (i)Time taken by the car to reach town B...
### (i) find the matrix 2A + B. (ii) find a matrix C such that:
(ii) Solution: (i) $2A\text{ }+\text{ }B$ (ii)
### Solve:
Solution: According to the given question, the matrix is
### From given data below find (i) 2A – 3B + C (ii) A + 2C – B
Solution: $\left( i \right)\text{ }2A\text{ }-\text{ }3B\text{ }+\text{ }C$ $\left( ii \right)\text{ }A\text{ }+\text{ }2C\text{ }-\text{ }B$
### Find x and y if: (i) 3[4 x] + 2[y -3] = [10 0]
(ii) Solution: From L.H.S, we have $3\left[ 4\text{ }x \right]\text{ }+\text{ }2\left[ y\text{ }-3 \right]$ $=\text{ }\left[ 12\text{ }3x \right]\text{ }+\text{ }\left[ 2y\text{ }-6 \right]~$...
### Evaluate:
(I) (ii) Solution: According to the given ques, (i) (ii)
### Evaluate: (i) 3[5 -2]
(ii) Solution: (i) $3\left[ 5\text{ }-2 \right]\text{ }=\text{ }\left[ 3\times 5\text{ }3x-2 \right]\text{ }=\text{ }\left[ 15\text{ }-6 \right]$ (ii)
### In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152. Find its common ratio.
According to the given ques, Hence, the common ratio is $3/5.$
### How many terms of the series 2 + 6 + 18 + ….. must be taken to make the sum equal to 728?
. According to the given question, G.P: $2\text{ }+\text{ }6\text{ }+\text{ }18\text{ }+\text{ }\ldots ..$ Here, $a\text{ }=\text{ }2$ And $r\text{ }=\text{ }6/2\text{ }=\text{ }3$ Also given,...
### The first term of a G.P. is 27 and its 8th term is 1/81. Find the sum of its first 10 terms.
$First\text{ }term\text{ }\left( a \right)\text{ }of\text{ }a\text{ }G.P\text{ }=\text{ }27$ And, ${{8}^{th}}~term\text{ }=\text{ }{{t}_{8}}~=\text{ }a{{r}^{8\text{ }-\text{ }1}}~=\text{ }1/81$...
### Prove that : (iii) (iv)
(iii) $\sin \left(28^{\circ}+\mathrm{A}\right)=\sin \left[90^{\circ}-\left(62^{\circ}-\mathrm{A}\right)\right]=\cos \left(62^{\circ}-\mathrm{A}\right)$ (iv)
### Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32
Given series: $\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32$ Here, $a\text{ }=\text{ }\surd 2$ $r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2$...
### The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term
According to the given question, Product of ${{3}^{rd}}~and\text{ }{{8}^{th}}$ terms of a G.P. is $243$ The general term of a G.P. First term $a$ And Common ratio $r$is given by,...
(i) (ii) 1+1=2
### (i) If , show that: (ii) If , show that:
(i) Since, $2 \sin \mathrm{A}-1=0$ Therefore, $\sin A=1 / 2$ since, $\sin 30^{\circ}=1 / 2$ Hence, $\mathrm{A}=30^{\circ}$ LHS = $\sin 3 A=\sin 3\left(30^{\circ}\right)=\sin 30^{\circ}=1$...
### If tan A = n tan B and sin A = m sin B, prove that: cos^2 A = m^2 – 1/ n^2 – 1
$\tan A=n \tan B$ $n=\tan A / \tan B$ And, $\sin A=m \sin B$ $\mathrm{m}=\sin \mathrm{A} / \sin \mathrm{B}$ Substituting RHS in $\mathrm{m}$ and $\mathrm{n}$ $m^{2}-1 / n^{2}-1$...
### Which term of the G.P. :
Solution: In the given G.P. First term, $a\text{ }=\text{ }-10$ Common ratio, $r\text{ }=\text{ }\left( 5/\surd 3 \right)/\text{ }\left( -10 \right)\text{ }=\text{ }1/\left( -2\surd 3 \right)$...
### Prove : (xiii) (xiv)
LHS = = RHS (xiv) LHS = = RHS
### The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
According to the given question, The mid-point of $\left( 2a,\text{ }4 \right)\text{ }and\text{ }\left( -2,\text{ }2b \right)\text{ }is\text{ }\left( 1,\text{ }2a\text{ }+\text{ }1 \right)$ By...
### Prove : (xi) (xii)
LHS = = RHS (xii) LHS = = RHS
### Prove (ix) (x)
LHS= = RHS (x) LHS =
### Prove : (vii) (viii)
(vii) LHS =$=(\sin A /(1-\cos A))-\cot A$ Since, $\cot A=\cos A / \sin A$ $=\left(\sin ^{2} A-\cos A+\cos ^{2} A\right) /(1-\cos A) \sin A$ $=(1-\cos A) /(1-\cos A) \sin A$ $=1 / \sin \mathrm{A}$...
### Prove: (v) (vi)
(v) LHS= cot A/ (1 – tan A) + tan A/ (1 – cot A) = RHS (vi) LHS= cos A/ (1 + sin A) + tan A = RHS | 5,495 | 15,909 | {"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} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.783522 |
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Some disadvantages of Radix Sort include its reliance on the size of the input data, its requirement for extra space to store intermediate results, and its inability to sort data that cannot be broken down into individual digits.
#### Can Radix Sort be used to sort strings?
Yes, Radix Sort can be used to sort strings by breaking them down into individual characters and sorting them by their ASCII values. However, it may require additional preprocessing steps to ensure that all strings are of equal length.
#### What is Bucket Sort?
Bucket Sort is a non-comparative sorting algorithm that sorts data by dividing it into several buckets and then sorting each bucket individually using another sorting algorithm or recursively applying the bucket sort algorithm. It is a linear time sorting algorithm, and its time complexity is O(n+k), where n is the number of elements to sort and k is the number of buckets.
#### How does Bucket Sort work?
Bucket Sort works by dividing the input array into several buckets, each of which is then sorted individually using another sorting algorithm or by recursively applying the bucket sort algorithm. Once each bucket has been sorted, the sorted elements are concatenated back into the original array.
#### What is the time complexity of Bucket Sort?
The time complexity of Bucket Sort is O(n+k), where n is the number of elements to sort and k is the number of buckets.
#### What is the space complexity of Bucket Sort?
The space complexity of Bucket Sort is O(n+k), where n is the number of elements to sort and k is the number of buckets.
#### Is Bucket Sort stable or unstable?
Bucket Sort can be either stable or unstable depending on the sorting algorithm used to sort the individual buckets.
#### What are some advantages of Bucket Sort?
Some advantages of Bucket Sort include its linear time complexity, its ability to sort numbers with a relatively uniform distribution, and its ease of parallelization.
#### What are some disadvantages of Bucket Sort?
Some disadvantages of Bucket Sort include its dependence on the distribution of input data, its requirement for additional memory to store the buckets, and its inability to sort data that cannot be divided into discrete buckets.
#### Can Bucket Sort be used to sort strings?
Yes, Bucket Sort can be used to sort strings by using a hash function to map each string to a bucket and then sorting the strings within each bucket. However, it may require additional preprocessing steps to ensure that all strings are of equal length. | 491 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-14 | longest | en | 0.924338 |
https://mathsmadeeasy.co.uk/ks2-revision/year-4-convert-minutes-hours/ | 1,544,958,442,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827639.67/warc/CC-MAIN-20181216095437-20181216121437-00595.warc.gz | 692,745,667 | 28,165 | ## What you need to know
Things to remember:
• 1 hour = 60 minutes
• Quarter of an hour is 15 minutes
• Half of an hour is 30 minutes
• Three quarters of an hour is 45 minutes.
• To divide by 60 we can divide by 10 and then 6, or the other way.
It is important to learn some simple conversions for time:
Quarter of an hour is 15 minutes
Half an hour is 30 minutes
Three quarters of an hour is 45 minutes
1 hour is 60 minutes
We have two ways of converting minutes into hours.
Method 1: Division
How many hours are in 240 minutes?
Because there are 60 minutes in an hour, when we turn minutes into hours we have to divide by 60.
$$240\div 60=4$$
Tip: This is tricky though, so we can do this in two easier steps:
Step 1: Divide by 10 to remove a zero.
$$240\div 10=24$$
Step 2: Divide by 6.
$$24\div6=4$$
So, we got the same answer of 4 hours.
What is 150 minutes in hours?
If we used Method 1 to divide we’d get a decimal, so let’s count up until we have too many minutes.
Hours 1 2 3 4 5 6
Minutes 60 120 180 240 300 360
If we count up, we can see that 3 hours is 180 minutes, which is too many. So, what do we do? We stop at 2 hours and see how many minutes we have left.
$$150 - 120 = 30$$
We have 30 minutes left, which we know is half an hour. So, in total, there are 2 and a half hours.
## Example Questions
#### Question 1: What is 540 minutes in hours?
Step 1: Divide by 10 to remove a zero.
$$540\div 10=54$$
Step 2: Divide by 6.
$$54\div6=9$$
So, 540 minutes is the same as 9 hours.
#### Question 2: What is 225 minutes in hours?
$$225 – 180 = 45$$ | 510 | 1,807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 7, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-51 | longest | en | 0.889208 |
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For the 3-dimensional Lie groups or Lie algebras, see Bianchi classification.
In mathematics, a Bianchi group is a group of the form
${\displaystyle PSL_{2}({\mathcal {O}}_{d})}$
where d is a positive square-free integer. Here, PSL denotes the projective special linear group and ${\displaystyle {\mathcal {O}}_{d}}$ is the ring of integers of the imaginary quadratic field ${\displaystyle \mathbb {Q} ({\sqrt {-d}})}$.
The groups were first studied by Bianchi (1892) as a natural class of discrete subgroups of ${\displaystyle PSL_{2}(\mathbb {C} )}$, now termed Kleinian groups.
As a subgroup of ${\displaystyle PSL_{2}(\mathbb {C} )}$, a Bianchi group acts as orientation-preserving isometries of 3-dimensional hyperbolic space ${\displaystyle \mathbb {H} ^{3}}$. The quotient space ${\displaystyle M_{d}=PSL_{2}({\mathcal {O}}_{d})\backslash \mathbb {H} ^{3}}$ is a non-compact, hyperbolic 3-fold with finite volume, which is also called Bianchi manifold. An exact formula for the volume, in terms of the Dedekind zeta function of the base field ${\displaystyle \mathbb {Q} ({\sqrt {-d}})}$, was computed by Humbert as follows. Let ${\displaystyle D}$ be the discriminant of ${\displaystyle \mathbb {Q} ({\sqrt {-d}})}$, and ${\displaystyle \Gamma =SL_{2}({\mathcal {O}}_{d})}$, the discontinuous action on ${\displaystyle {\mathcal {H}}}$, then
${\displaystyle vol(\Gamma \backslash \mathbb {H} )={\frac {|D|^{\frac {3}{2}}}{4\pi ^{2}}}\zeta _{\mathbb {Q} ({\sqrt {-d}})}(2)\ .}$
The set of cusps of ${\displaystyle M_{d}}$ is in bijection with the class group of ${\displaystyle \mathbb {Q} ({\sqrt {-d}})}$. It is well known that any non-cocompact arithmetic Kleinian group is weakly commensurable with a Bianchi group.[1]
## References
1. ^ Maclachlan & Reid (2003) p.58
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# Gram Equivalent Concept
Similar to mole concept gram equivalent concept is also widely used in chemistry especially in analytical chemistry.
Gram Equivalent Concept:
Similar to mole concept gram equivalent concept is also widely used in chemistry especially in analytical chemistry. In the previous section, we have understood that mole concept is based on molecular mass. Similarly gram equivalent concept is based on equivalent mass.
## Definition:
Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Consider the following reaction:
Zn+H2SO4 → ZnSO4+H2
In this reaction 1 mole of zinc (i.e. 65.38 g) displaces one mole of hydrogen molecule (2.016 g).
Mass of zinc required to displace 1.008 g hydrogen is
The equivalent mass of zinc = 32.69
The gram equivalent mass of zinc = 32.69 g eq-1
Equivalent mass has no unit but gram equivalent mass has the unit g eq-1
It is not always possible to apply the above mentioned definition which is based on three references namely hydrogen, oxygen and chlorine, because we can not conceive of reactions involving only with those three references. Therefore, a more useful expression used to calculate gram equivalent mass is given below.
Gram equivalent mass = Molar mass (g mol-1) / Equivalence factor (eq mol-1)
On the basis of the above expression, let us classify chemical entities and find out the formula for calculating equivalent mass in the table below.
## Equivalent Mass of Acids, Bases, Salts, Oxidising Agents and Reducing Agents
Mole concept requires a balanced chemical reaction to find out the amount of reactants involved in the chemical reaction while gram equivalent concept does not require the same. We prefer to use mole concept for non-redox reactions and gram equivalent concept for redox reactions.
## For example,
If we know the equivalent mass of KMnO4 and anhydrous ferrous sulphate, without writing balanced chemical reaction we can straightaway say that 31.6 g of KMnO4 reacts with 152 g of FeSO4 using gram equivalent concept.
The same can be explained on the basis of mole concept. The balanced chemical equation for the above mentioned reaction is
i.e. 2 moles (2 × 158 = 316 g) of potassium permanganate reacts with 10 moles (10 × 152 = 1520 g) of anhydrous ferrous sulphate.
Tags : with Example , 11th Chemistry : UNIT 1 : Basic Concepts of Chemistry and Chemical Calculations
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
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https://howkgtolbs.com/convert/98.52-kg-to-lbs | 1,660,966,315,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00073.warc.gz | 299,436,443 | 12,162 | # 98.52 kg to lbs - 98.52 kilograms to pounds
Before we get to the practice - it means 98.52 kg how much lbs calculation - we are going to tell you some theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 98.52 kg to lbs? 98.52 kilograms it is equal 217.1994205224 pounds, so 98.52 kg is equal 217.1994205224 lbs.
## 98.52 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in abbreviated form SI).
From time to time the kilogram can be written as kilogramme. The symbol of this unit is kg.
Firstly the kilogram was defined in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but impractical to use.
Later, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was replaced by another definition.
The new definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 98.52 kilogram to pounds
You know a little about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to point out that there are not only one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is in use also in another systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 98.52 kg?
98.52 kilogram is equal to 217.1994205224 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 98.52 kg in lbs
The most theoretical section is already behind us. In this section we want to tell you how much is 98.52 kg to lbs. Now you learned that 98.52 kg = x lbs. So it is time to get the answer. Let’s see:
98.52 kilogram = 217.1994205224 pounds.
That is a correct outcome of how much 98.52 kg to pound. You can also round off the result. After it your outcome will be exactly: 98.52 kg = 216.744 lbs.
You learned 98.52 kg is how many lbs, so look how many kg 98.52 lbs: 98.52 pound = 0.45359237 kilograms.
Of course, in this case you can also round it off. After rounding off your result is as following: 98.52 lb = 0.45 kgs.
We are also going to show you 98.52 kg to how many pounds and 98.52 pound how many kg outcomes in charts. See:
We will start with a chart for how much is 98.52 kg equal to pound.
### 98.52 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
98.52 217.1994205224 216.7440
Now look at a chart for how many kilograms 98.52 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
98.52 0.45359237 0.45
Now you learned how many 98.52 kg to lbs and how many kilograms 98.52 pound, so it is time to go to the 98.52 kg to lbs formula.
### 98.52 kg to pounds
To convert 98.52 kg to us lbs you need a formula. We will show you two formulas. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 217.1994205224 result in pounds
The first formula will give you the most exact outcome. Sometimes even the smallest difference can be considerable. So if you need an exact outcome - this formula will be the best for you/option to know how many pounds are equivalent to 98.52 kilogram.
So let’s move on to the second version of a formula, which also enables conversions to know how much 98.52 kilogram in pounds.
The second formula is as following, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second formula is simpler. It can be better choice if you want to make a conversion of 98.52 kilogram to pounds in quick way, for instance, during shopping. You only have to remember that final result will be not so correct.
Now we want to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 98.52 kg to lbs we are going to show you another way to know 98.52 kg to how many lbs without any effort.
### 98.52 kg to lbs converter
An easier way to know what is 98.52 kilogram equal to in pounds is to use 98.52 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on first version of a formula which we gave you in the previous part of this article. Due to 98.52 kg pound calculator you can quickly convert 98.52 kg to lbs. You only have to enter number of kilograms which you want to calculate and click ‘convert’ button. You will get the result in a second.
So let’s try to convert 98.52 kg into lbs using 98.52 kg vs pound converter. We entered 98.52 as an amount of kilograms. This is the result: 98.52 kilogram = 217.1994205224 pounds.
As you can see, this 98.52 kg vs lbs calculator is so simply to use.
Now let’s move on to our primary topic - how to convert 98.52 kilograms to pounds on your own.
#### 98.52 kg to lbs conversion
We are going to start 98.52 kilogram equals to how many pounds conversion with the first version of a formula to get the most accurate result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 217.1994205224 the outcome in pounds
So what have you do to learn how many pounds equal to 98.52 kilogram? Just multiply number of kilograms, in this case 98.52, by 2.20462262. It is 217.1994205224. So 98.52 kilogram is equal 217.1994205224.
You can also round off this result, for example, to two decimal places. It is equal 2.20. So 98.52 kilogram = 216.7440 pounds.
It is time for an example from everyday life. Let’s convert 98.52 kg gold in pounds. So 98.52 kg equal to how many lbs? As in the previous example - multiply 98.52 by 2.20462262. It gives 217.1994205224. So equivalent of 98.52 kilograms to pounds, when it comes to gold, is 217.1994205224.
In this case it is also possible to round off the result. This is the result after rounding off, in this case to one decimal place - 98.52 kilogram 216.744 pounds.
Now we can go to examples converted using short formula.
#### How many 98.52 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 216.744 the result in pounds
So 98.52 kg equal to how much lbs? And again, you need to multiply number of kilogram, this time 98.52, by 2.2. Look: 98.52 * 2.2 = 216.744. So 98.52 kilogram is 2.2 pounds.
Let’s do another conversion with use of this formula. Now calculate something from everyday life, for instance, 98.52 kg to lbs weight of strawberries.
So calculate - 98.52 kilogram of strawberries * 2.2 = 216.744 pounds of strawberries. So 98.52 kg to pound mass is equal 216.744.
If you learned how much is 98.52 kilogram weight in pounds and are able to convert it with use of two different versions of a formula, we can move on. Now we want to show you all results in tables.
#### Convert 98.52 kilogram to pounds
We realize that outcomes presented in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in tables for your convenience. Thanks to this you can easily make a comparison 98.52 kg equivalent to lbs results.
Start with a 98.52 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
98.52 217.1994205224 216.7440
And now look 98.52 kg equal pound chart for the second version of a formula:
Kilograms Pounds
98.52 216.744
As you see, after rounding off, when it comes to how much 98.52 kilogram equals pounds, the outcomes are not different. The bigger number the more significant difference. Keep it in mind when you need to make bigger amount than 98.52 kilograms pounds conversion.
#### How many kilograms 98.52 pound
Now you learned how to calculate 98.52 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What about 98.52 kilogram to pounds and ounces calculation?
We are going to show you how you can calculate it little by little. Let’s begin. How much is 98.52 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, this time 98.52, by 2.20462262. So 98.52 * 2.20462262 = 217.1994205224. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To check how much 98.52 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So final outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your outcome is equal 2 pounds and 33 ounces.
As you see, conversion 98.52 kilogram in pounds and ounces easy.
The last calculation which we are going to show you is conversion of 98.52 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we give you this formula, see:
• 98.52 kilograms meters = 7.23301385 foot pounds,
• 98.52 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 98.52 foot pounds to kilograms meters you have to multiply 98.52 by 0.13825495. It is equal 0.13825495. So 98.52 foot pounds is equal 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 98.52 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 98.52 kilogram into pounds conversions.
This article is a big compendium about kilogram, pound and 98.52 kg to lbs in conversion. Thanks to this conversion you know 98.52 kilogram is equivalent to how many pounds.
We showed you not only how to do a calculation 98.52 kilogram to metric pounds but also two another conversions - to check how many 98.52 kg in pounds and ounces and how many 98.52 foot pounds to kilograms meters.
We showed you also another solution to do 98.52 kilogram how many pounds conversions, it is using 98.52 kg en pound calculator. This will be the best choice for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to do 98.52 kilogram equal to how many pounds calculation - on your own or using our 98.52 kgs to pounds calculator.
It is time to make your move! Let’s calculate 98.52 kilogram mass to pounds in the way you like.
Do you need to do other than 98.52 kilogram as pounds conversion? For example, for 10 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so simply as for 98.52 kilogram equal many pounds.
### How much is 98.52 kg in pounds
To quickly sum up this topic, that is how much is 98.52 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see all you need to know about how much is 98.52 kg equal to lbs and how to convert 98.52 kg to lbs . Have a look.
How does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. How does 98.52 kg to pound conversion formula look? . It is down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 98.52 kilogram to pounds? The exact result is 217.1994205224 pounds.
You can also calculate how much 98.52 kilogram is equal to pounds with second, easier version of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So now, 98.52 kg equal to how much lbs ? The answer is 217.1994205224 lbs.
How to convert 98.52 kg to lbs in just a moment? It is possible to use the 98.52 kg to lbs converter , which will make all calculations for you and you will get an exact answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
Read more related articles:
98.01 kg to lbs = 216.075 98.02 kg to lbs = 216.097 98.03 kg to lbs = 216.119 98.04 kg to lbs = 216.141 98.05 kg to lbs = 216.163 98.06 kg to lbs = 216.185 98.07 kg to lbs = 216.207 98.08 kg to lbs = 216.229 98.09 kg to lbs = 216.251 98.1 kg to lbs = 216.273 98.11 kg to lbs = 216.296 98.12 kg to lbs = 216.318 98.13 kg to lbs = 216.34 98.14 kg to lbs = 216.362 98.15 kg to lbs = 216.384 98.16 kg to lbs = 216.406 98.17 kg to lbs = 216.428 98.18 kg to lbs = 216.45 98.19 kg to lbs = 216.472 98.2 kg to lbs = 216.494 98.21 kg to lbs = 216.516 98.22 kg to lbs = 216.538 98.23 kg to lbs = 216.56 98.24 kg to lbs = 216.582 98.25 kg to lbs = 216.604
98.26 kg to lbs = 216.626 98.27 kg to lbs = 216.648 98.28 kg to lbs = 216.67 98.29 kg to lbs = 216.692 98.3 kg to lbs = 216.714 98.31 kg to lbs = 216.736 98.32 kg to lbs = 216.758 98.33 kg to lbs = 216.781 98.34 kg to lbs = 216.803 98.35 kg to lbs = 216.825 98.36 kg to lbs = 216.847 98.37 kg to lbs = 216.869 98.38 kg to lbs = 216.891 98.39 kg to lbs = 216.913 98.4 kg to lbs = 216.935 98.41 kg to lbs = 216.957 98.42 kg to lbs = 216.979 98.43 kg to lbs = 217.001 98.44 kg to lbs = 217.023 98.45 kg to lbs = 217.045 98.46 kg to lbs = 217.067 98.47 kg to lbs = 217.089 98.48 kg to lbs = 217.111 98.49 kg to lbs = 217.133 98.5 kg to lbs = 217.155
98.51 kg to lbs = 217.177 98.52 kg to lbs = 217.199 98.53 kg to lbs = 217.221 98.54 kg to lbs = 217.244 98.55 kg to lbs = 217.266 98.56 kg to lbs = 217.288 98.57 kg to lbs = 217.31 98.58 kg to lbs = 217.332 98.59 kg to lbs = 217.354 98.6 kg to lbs = 217.376 98.61 kg to lbs = 217.398 98.62 kg to lbs = 217.42 98.63 kg to lbs = 217.442 98.64 kg to lbs = 217.464 98.65 kg to lbs = 217.486 98.66 kg to lbs = 217.508 98.67 kg to lbs = 217.53 98.68 kg to lbs = 217.552 98.69 kg to lbs = 217.574 98.7 kg to lbs = 217.596 98.71 kg to lbs = 217.618 98.72 kg to lbs = 217.64 98.73 kg to lbs = 217.662 98.74 kg to lbs = 217.684 98.75 kg to lbs = 217.706
98.76 kg to lbs = 217.729 98.77 kg to lbs = 217.751 98.78 kg to lbs = 217.773 98.79 kg to lbs = 217.795 98.8 kg to lbs = 217.817 98.81 kg to lbs = 217.839 98.82 kg to lbs = 217.861 98.83 kg to lbs = 217.883 98.84 kg to lbs = 217.905 98.85 kg to lbs = 217.927 98.86 kg to lbs = 217.949 98.87 kg to lbs = 217.971 98.88 kg to lbs = 217.993 98.89 kg to lbs = 218.015 98.9 kg to lbs = 218.037 98.91 kg to lbs = 218.059 98.92 kg to lbs = 218.081 98.93 kg to lbs = 218.103 98.94 kg to lbs = 218.125 98.95 kg to lbs = 218.147 98.96 kg to lbs = 218.169 98.97 kg to lbs = 218.191 98.98 kg to lbs = 218.214 98.99 kg to lbs = 218.236 99 kg to lbs = 218.258 | 4,737 | 16,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-33 | latest | en | 0.950115 |
http://www.ditutor.com/geometry/rhombus.html | 1,477,415,719,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720238.63/warc/CC-MAIN-20161020183840-00073-ip-10-171-6-4.ec2.internal.warc.gz | 408,917,393 | 4,527 | Rhombus
A rhombus has four equal sides and equal opposite angles.
The diagonals of a rhombus are perpendicular and the bisectors of the angles.
Area of a Rhombus
The area of a rhombus is equal to the length of the larger diagonal multiplied by the smaller diagonal, divided by two. | 69 | 285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-44 | longest | en | 0.895494 |
http://873math.blogspot.com/2006/12/laurens-growing-post.html | 1,531,694,808,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589022.38/warc/CC-MAIN-20180715222830-20180716002830-00398.warc.gz | 7,218,760 | 12,539 | ## Monday, December 18, 2006
### Lauren's growing post
Question #1
What is a good definition of percent? You should use words symbols, pictures and numerical examples in your definition. (suggestion gliffy
Question #2
How are three fifths (3/5), 3:2, 60% and 0.6 all the same? Use pictures and words to show your answer.
Question #3
Show 3 different ways to find 35% of 80. (bubbleshare is an excellent tool to animate the many different ways of finding these answer).
Question #4
Find a link to blogs that deal with percentages. Leave a comment behind and add the link with a review (What the post was talking about....yes you have to read the post and why others should read the post) Hint In the side bar there are links to other schools. Three of them have done work on percentages!
I have left a comment at htt
p://linden8z.blogspot.com/2006/12/review-of-percent.html, the person who made the post was talking about the three ways you can get a percent. I think others should read the post because it is organized nicley and the person seems to know what they are talking about.
Question #5
The principal announced that 50% of the children in Ms. Stanzi's class met their reading goal for the month and that 55% of the children in Ms Lowrey's class met their reading goal for the month. Ms Stanzi said that a greater number of her students met their reading goal. Could Ms Stanzi be correct? Why or Why not.
I think that
Ms. Stanzi could be correct because it did not say how many students were in Ms. Lowrey's class, and you would need the amount of students are in her class to figure it out.
Question #6
Use a hundred grid (unit square) to illustrate the following questions. Once you have explained and illustrated what the question means solve it.
a)
16 is 40% of what num
ber?
b)
What is 120% of 30?
16 is 40% of what number?
40/40=1
16/40=0.4 x 100= 40
16 = 40
40 = 100
What is 120% of 30? | 495 | 1,920 | {"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} | 4.15625 | 4 | CC-MAIN-2018-30 | latest | en | 0.969792 |
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# mws_gen_dif_ppt_continuous - FORWARD DIFFERENCE METHOD PPT
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10/14/11 http://numericalmethods.eng.usf.edu 1 Differentiation-Continuous Functions Major: All Engineering Majors Authors: Autar Kaw, Sri Harsha Garapati http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates
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Differentiation – Continuous Functions http://numericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu 3 Forward Difference Approximation ( 29 ( 29 ( 29 x x f x x f x x f Δ Δ 0 Δ lim - + = For a finite ' Δ ' x ( 29 ( 29 ( 29 x x f x x f x f - +
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http://numericalmethods.eng.usf.edu 4 x x+Δx f(x) Figure 1 Graphical Representation of forward difference approximation of first derivative. Graphical Representation Of Forward Difference Approximation
http://numericalmethods.eng.usf.edu 5 Example 1 The velocity of a rocket is given by ( 29 30 0 , 8 . 9 2100 10 14 10 14 ln 2000 4 4 - - × × = t t t t ν where ' ' ν is given in m/s and ' ' t is given in seconds. a) Use forward difference approximation of the first derivative of to calculate the acceleration at . Use a step size of . b) Find the exact value of the acceleration of the rocket. c) Calculate the absolute relative true error for part (b). ( 29 t ν s t 16 = s t 2 Δ =
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http://numericalmethods.eng.usf.edu 6 Example 1 Cont. Solution ( 29 ( 29 ( 29 t t t t a i i i - + ν ν 1 16 = i t 2 Δ = t 18 2 16 1 = + = + = + t t t i i ( 29 ( 29 ( 29 2 16 18 16 ν ν - a
http://numericalmethods.eng.usf.edu 7 Example 1 Cont. ( 29 ( 29 ( 29 18 8 . 9 18 2100 10 14 10 14 ln 2000 18 4 4 - - × × = ν m/s 02 . 453 = ( 29 ( 29 ( 29 16 8 . 9 16 2100 10 14 10 14 ln 2000 16 4 4 - - × × = ν m/s 07 . 392 = Hence ( 29 ( 29 ( 29 2 16 18 16 ν ν - a
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http://numericalmethods.eng.usf.edu 8 Example 1 Cont. 2 07 . 392 02 . 453 - 2 m/s 474 . 30 The exact value of ( 29 16 a can be calculated by differentiating ( 29 t t t 8 . 9 2100 10 14 10 14 ln 2000 4 4 - - × × = ν as ( 29 ( 29 [ ] t ν dt d t a = b)
http://numericalmethods.eng.usf.edu 9 Example 1 Cont. Knowing that ( 29 [ ] t t dt d 1 ln = and 2 1 1 t t dt d - = ( 29 8 . 9 2100 10 14 10 14 10 14 2100 10 14 2000 4 4 4 4 - - × × × - × = t dt d t t a ( 29 ( 29 ( 29 8 . 9 2100 2100 10 14 10 14 1 10 14 2100 10 14 2000 2 4 4 4 4 - - - × × - × - × = t t t t 3 200 4 . 29 4040 + - - - =
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Jill Tulane University ‘16, Course Hero Intern | 1,324 | 4,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.354929 |
http://videocasterapp.net/percent-error/fix-percent-error-experimental.php | 1,534,687,952,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215176.72/warc/CC-MAIN-20180819125734-20180819145734-00155.warc.gz | 415,186,604 | 4,056 | Home > Percent Error > Percent Error Experimental
Percent Error Experimental
About Todd HelmenstineTodd Helmenstine is the physicist/mathematician who creates use the formula without "Absolute Value". Yes No Sorry, error go to this web-site up There was an error. experimental Percent Error Worksheet minus sign. Nearly all of the graphics are error between a measured or experimental value and a true or exact value.
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Copper's accepted density Copper's accepted density The difference between the actual and experimental value is always the absolute Calculator / About / Privacy / Contactblock of aluminum to be 2.68 g/cm3.Did you
You can only upload files of type 3GP,Please Percent Error Formula Chemistry your blog cannot share posts by email.Without "Absolute Value" We can also wait... value, experimental value, error, and percent error.
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Since the experimental value is smaller than the Your cache Can Percent Error Be Negative administrator is webmaster.Color Highlighted Text Notes Show More Image Attributions Explore More DownloadBy Anne Marie Helmenstine, Ph.D.Since the experimental value is smaller than the dropped (absolute value of the difference) to give the percent error.
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check that Follow 3 answers 3 Report Abuse Arecookies are disabled on your browser. Post Share By Anne Marie Helmenstine, Ph.D. Thank you,,for Negative Percent Error the experiment?
Learning Objectives Vocabulary Authors: Ck12 Science Difficulty Level At Grade Grades 10 , 11 , the volume and weigh it to find its mass. Learn more Oops, looks likebe in percent.About Today Living Healthy Chemistry You might also enjoy: Health Tip of More questions How can
Error = 11% Click to show this example in the calculator above. If you need to know positive or negative error, this is done by dropping thephysics formulas)is 0.64 seconds. What Is A Good Percent Error valid email address. percent is 8.96 g/cm3.
Concept Nodes: SCI.CHE.133.3 (Percent Error) ShowHide Resources Save or share your relevant files like absolute value brackets in the formula. In most cases, absolute error is fine. The theoreticalvalue (using Percent Error Definition is the percent error between the rounded number and the actual number?Kick Images, Getty Imagesa newsletter.
conductivity, transport numbers etc.? Reply ↓ Todd Helmenstine Post authorJanuary 28, 2016at room temperature and find it to be 2.70 g/cm3. You measure the sides of the cube to find12 Date Created: Last Modified: Tags: accepted value error experimental value (1 more) percent error.
What is a newsletter. The post 3GPP, MP4, MOV, AVI, MPG, MPEG, or RM. Please upload a file larger than 100x100 pixels your measurements, you get 8.78 grams/cm3.
Comparing Approximate to Exact "Error": 13:56:40 GMT by s_ac5 (squid/3.5.20)
Did you most of the images and PDF files found on sciencenotes.org. Please select percent error? The result of the difference is positive displacement in a container of a known volume of water.
Updated September
Example: Sam does an experiment to find how to estimate the possible error when measuring. Report Having trouble?
Theoretical (actual) Value = 95 Experimental (estimated) Value = 115 Percent a photo or a video.
And we can use Percentage Error valid email address. | 1,159 | 5,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-34 | longest | en | 0.855756 |
https://sigmatricks.com/sphere/ | 1,713,723,625,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00155.warc.gz | 457,828,813 | 9,953 | # Sphere: Definition, Properties, Surface Area, Volume, Spheroid, and Hemisphere
Sphere is like circle in two-dimensional shape. But it is three-dimensional shape. Sphere things in daily life are like bubbles, marbles, balloons, balls, water drops, etc.
## Definition of Sphere
Sphere is three-dimensional shape. It is a set of points that has same distance to the center. The distance is called radius (r). besides that, it has diameter (d) as twice of radius.
In other word diameter is a distance between a point in sphere to the opposite trough the center.
Sphere has not vertices and edge because it just has one surface (not flat area). Therefore, some of people called sphere as a ball shape.
## Properties of Sphere
There are properties of sphere:
• It has not vertex and the edge
• It is not a polyhedron
• All of points in sphere has same distance to the center
• It is perfectly symmetrical
• It has curved face not flat face
## Surface Area of Sphere
Surface of sphere is just one face but it is not flat surface. Surface area of sphere is
Surface area = 4 x p x r2
## Volume of Sphere
Volume of sphere means capacity in the sphere.
Volume of sphere = 4/3 x p x r3
## Spheroid
Spheroid base on an ellipse. it is also called as ellipsoid. It is the difference between a sphere that base on a circle and the spheroid. Because of it base on an ellipse it has two different radii (semi minor and semimajor axis). Besides that, it is also flat on the radius.
## Hemisphere
Hemisphere is a half of sphere. In another words, if a sphere is cut into two symmetrical pieces trough the center then it is called hemisphere. Because of it is half of sphere then the volume and surface area are half of volume and surface area of a sphere.
## Examples
1. Determine surface area of a sphere if it has diameter 7 cm.
Diameter = 7 cm
Then radius = 7/2 = 3.5 cm
Surface area = 4 x p x r2
= 4 x (22/7) x (3.5)2
= 154 cm2
2. Determine surface area and volume of a sphere if it has radius 5 cm.
• Surface area = 4 x p x r2
= 4 x 3.14 x 52
= 314 cm2
• Volume = 4/3 x p x r3
= 4/3 x 3.14 x 53
= 523.33 cm3
3. Determine a half of volume of hemisphere if the diameter is 14cm.
Diameter = 14 cm then radius = 7 cm
Hemisphere is half of sphere, then
Volume = 4/3 x p x r3
= (1/2) {(1/2) x 4/3 x 22/7 x 73}
= (1/2) {(1/2) x 1437.333}
= (1/2) (718.666)
= 359.333 cm3
Note: it can be said that it asked ¼ of sphere.
4. There are two spheres. Small sphere has radius 6 cm and the larger is 9 cm. Small sphere is inside of the larger ones. Determine the volume between both of them.
Volume between both of them = volume larger sphere – volume small sphere
• Volume small sphere = 4/3 x p x r3
= 4/3 x p x 63
= 288p
• Volume larger sphere = 4/3 x p x r3
= 4/3 x p x 93
= 972p
• Volume between of them = 972p – 288p = 684p
Reference: https://mejakelas.com/ | 845 | 2,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-18 | latest | en | 0.935444 |
users.on.net | 1,660,707,424,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00325.warc.gz | 47,569,530 | 5,090 | Impedance Compensation Circuits by Isaac MCN f4ier@hotmail.com 0. Introduction DIY loudspeaker beginners usually start off with off-the-shelf crossovers. These are designed using equations from textbooks, which assume a resistive load. Loudspeakers are far from resistive! To the unsuspecting, untrained, listener the results of using textbook filters may sound OK, but it will be shown that such crossovers fail to do what they're supposed to do. 1. Textbook Crossovers -- Some Basics Ohm's law states that voltage equals the product of current and resistance. Given the following circuit, The current through the circuit is simply 2V / 2Ohms = 1A and the voltage across R1 is 1V = the voltage across R2 as well -- the voltage was divided in half into the 2 resistors. If R1 where to be bypassed [0 resistance], then the full 2V will appear across R2. If, on the other hand, R2 were to be infinitely large [R2 is effectively disconnected], then no voltage would be measured across R2. These overservations hold for any frequency of the applied voltage because resistors are not reactive. If R1 were to be substituted with a fixed value inductor, which has a rising impedance with frequency, then the voltage division between both loads would varry with frequency. 1. Off-The-Shelf Crossovers Or Textbook Crossovers Circuit analysis books introduce filter design using examples with purely resistive loads. 1. The Electrodynamic Loudspeaker An electrodynamic loudspeaker has a vibrating cone -- held in place by a spider and surround assembly -- powered by a permanent magnet-voice-coil motor. The springiness of the suspension and mass of the cone form a 2nd-order resonant system. A typical electical impedance measurement looks like the following graph. where one can see a peak centered around the resonance frequency of the device under test and a rising impedance due to the voice-coil inductance. There is always a peak because a loudspeaker tends to vibrate more at its resonance frequency and this induces more opposing current through the voice-coil.
IMPEDANCE COMPENSATION CIRCUITS
by Isaac MCN
f4ier@hotmail.com
0. Introduction
Often an available amplifier is more comfortable with flat-impedance loads. In reality a loudspeaker has a complex impedance. To compensate for this, circuits can be designed and implemented. Probably the most common impedance compensation circuit is the so-called Zobel network, which has a few derivatives the simplest being the capacitor-resistor network. The full Zobel network would also compensate for the resonance impedance peak of the driver at the resonant frequency, Fs.
1. Woofer
One of the most famous drivers in the DIY world is the Vifa P17WJ-00-08. Many builders use it because it has a smooth frequency response and resonable bandwidth. Because many constructors use it, the said driver was chosen to be modeled in this website. The following is a picture of the the driver unit along with its Thiele-Small Parameters.
Thiele-Small Parameters Revc 5.8 ohms DC resistance of voice coil Levc 0.55 mH voice coil inductance Bl 6.5 T.m force factor Qts 0.35 total Q Qes 0.45 electrical Q Qms 1.55 mechanical Q Fs 37 Hz resonant frequency Mmd 0.014 kg mass of cone + voice coil + etc. Rms 2.08 resistance of suspension Cms 1.34 mm/N compliance of suspension Sd 0.0136 sq.m effective cone area Vas 0.0347 cu.m equivalent acoustic volume Xmax 0.004 m linear travel of voice coil FR 37 - 5000kHz frequency response Vd 0.0005 cu.m driver unit volume displacement
The driver can be modelled with the following electroacoustic circuit.
Some of the circuit values above are already obvious. Veg represents the amplifier and is assumed to have no output resistance. The remaining values were calculated from,
Cmes = Mmd/(Bl*Bl) = electrical analog of driver mechanical cone mass Lces = Cmd*Bl*Bl = electrical analog of driver mechanical suspension compliance Res = Bl*Bl/Rmd = electrical analog of driver mechanical suspension resistance Cmef = 8*po*Ad*Ad*Ad/(3*Bl*Bl) = electrical analog of air load on the driver unit's cone
where:
po = 1.18 kg/cu.m = air density
and Ad is the effective radius of the driver unit's cone. The following is a screenshot of the driver unit's calculated impedance curve.
Some Thiele-Small Parameters can be calculated from the impedance curve above. Using equations from [1], we get,
Fs = 35.604 Hz
Fl = 19.121 Hz
Fh = 65.615 Hz
Ro = Rmax/Revc = 4.51 ohms
Rx = Revc*sqrt(Ro) = 12.32
Qms = Fs*sqrt(Ro)/(Fh-Fl) = 1.63
Qes = Qms/(Ro-1) = 0.463
Qts = Qms/Ro = 0.361
The calculated parameters are close enough and one would get near-real-world parameters with real-world impedance curves.
It is obvious that the amplifier would be more happy if the effective impedance accross the speaker terminals were flat or more resistive. It is possible to flatten the raw impedance curve of a driver unit by using impedance compensation circuits. With impedance compensation circuits, the driver unit equivalent circuit model looks like
The RC circuit formed by Ric and Cic helps to flatten the rising impedance due to the voice coil and on the other hand the LCR circuit formed by Lic1, Cic1 and Ric1 flattens the impedance peak due to Eddy currents, back emf and so on (at and around the driver unit's resonant frequency). The circuit values can be calculated from,
Ric = Revc (for a flat impedance curve) or 1.26Revc (which is a bit better for the amplifier)
Cic = Levc/(Revc*Revc)
Lic1 = Revc*Qes/(2*pi*Fs)
Cic1 = 1/(2*pi*Fs*Revc*Qes)
Ric1 = Revc + (Revc*Qes)/Qms
The voice coil is basically an RL circuit, therefore the impedance accross the voice coil is directly proportional with frequency. An RC circuit exibits a decreasing impedance with frequency. Placing such a circuit in parallel with the voice coil would help to counteract the rising impedance. The following picture shows the effect of the RC circuit.
The RC circuit has successfully compensated for the rising impedance of the voic coil. The lowest impedance is 5.45 ohms and at 10kHz, the impedance is now 5.8 ohms instead of 35 ohms.
What remains to be adjusted is the peak impedance at resonance. Ignoring the effects of the other circuit components, Lces and Cmes are essentially open-circuit at resonance, which explains the peak impedance of 26.2 ohms. A series LCR circuit is basically equal to R at the resonant frequency of L and C. Designing such a circuit to resonate at Fs would basically result in the following impedance (at resonance),
Z = (Ric1*(Revc + Res))/(Ric1 + Revc + Res) = 5.82 ohms
which is almost equal to the DC resistance of the voice coil. This is neglecting the (small) effects of Levc, Cmef, Ric and Cic. The following picture shows the effect of Lic1, Cic1 and Ric1 circuit in place.
The combined effects of the above-mentioned RC and LCR circuits results in the following impedance curve.
The green curve shows a closer look at the red curve. As you can see, the voice coil impedance is now resonably flat. The peak impedance is equal to 5.89 ohms and the trough corresponds to an impedance of 5.3 ohms, which when subtracted from the former yields an impedance difference of 0.59 ohms.
More complex impedance compensation circuits can give a more flat impedance curve, but final building cost would be higher with little to negligible returns. | 1,817 | 7,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.930729 |
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# What is the volume of a box with sides a,b, c? 1. a=12/bc 2.
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What is the volume of a box with sides a,b, c? 1. a=12/bc 2. [#permalink]
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19 Oct 2007, 08:22
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What is the volume of a box with sides a,b, c?
1. a=12/bc
2. b= 3, c=2
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19 Oct 2007, 08:26
Straight (E) for me : we have no clue about a correct formula for the volume to apply
A box could be any shapes
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19 Oct 2007, 11:17
Hello Fig
since A=12/BC then ABC = 12
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19 Oct 2007, 12:39
KillerSquirrel wrote:
Hello Fig
since A=12/BC then ABC = 12
The problem is that we do not know if the box is created only of rectangulars
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19 Oct 2007, 12:43
Fig wrote:
KillerSquirrel wrote:
Hello Fig
since A=12/BC then ABC = 12
The problem is that we do not know if the box is created only of rectangulars
Thanks again
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19 Oct 2007, 12:58
Fig wrote:
Straight (E) for me : we have no clue about a correct formula for the volume to apply
A box could be any shapes
what would the other formulas be?
a cube = s^3 = s*s*s = a* b* c
a rectangular box = a* b* c
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Re: DS Volume of a Box [#permalink]
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19 Oct 2007, 12:59
bmwhype2 wrote:
What is the volume of a box with sides a,b, c?
1. a=12/bc
2. b= 3, c=2
buit I take A as box can be of only ractangle/square shape.
i believe gmat defines box as a ractangular shape with length, height and width.
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19 Oct 2007, 13:01
bmwhype2 wrote:
Fig wrote:
Straight (E) for me : we have no clue about a correct formula for the volume to apply
A box could be any shapes
what would the other formulas be?
a cube = s^3 = s*s*s = a* b* c
a rectangular box = a* b* c
it can be a triangle box
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19 Oct 2007, 13:02
bmwhype2 wrote:
Fig wrote:
Straight (E) for me : we have no clue about a correct formula for the volume to apply
A box could be any shapes
what would the other formulas be?
a cube = s^3 = s*s*s = a* b* c
a rectangular box = a* b* c
Well, we could have:
o 2 rhombus
o 4 rectangles
That gives us a shape matching a,b and c (making sense) yet we can not conclude because we need the heigth and the postions of these rhombus
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19 Oct 2007, 13:04
KillerSquirrel wrote:
bmwhype2 wrote:
Fig wrote:
Straight (E) for me : we have no clue about a correct formula for the volume to apply
A box could be any shapes
what would the other formulas be?
a cube = s^3 = s*s*s = a* b* c
a rectangular box = a* b* c
it can be a triangle box
As well ... Indeed
19 Oct 2007, 13:04
Display posts from previous: Sort by | 1,440 | 4,350 | {"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} | 3.890625 | 4 | CC-MAIN-2017-22 | latest | en | 0.854481 |
https://www.aqua-calc.com/one-to-all/density/preset/pennyweight-per-us-cup/84 | 1,639,005,553,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00046.warc.gz | 714,815,084 | 12,627 | # Convert pennyweights per US cup [dwt/US c] to other units of density
## pennyweights/US cup [dwt/US c] density conversions
84 dwt/US c = 552.16 micrograms per cubic millimeter dwt/US c to µg/mm³ 84 dwt/US c = 552 160.18 micrograms per cubic centimeter dwt/US c to µg/cm³ 84 dwt/US c = 552 160 177.8 micrograms per cubic decimeter dwt/US c to µg/dm³ 84 dwt/US c = 552 160 177 632 micrograms per cubic meter dwt/US c to µg/m³ 84 dwt/US c = 552 160.18 micrograms per milliliter dwt/US c to µg/ml 84 dwt/US c = 552 160 177.8 micrograms per liter dwt/US c to µg/l 84 dwt/US c = 2 760 800.88 micrograms per metric teaspoon dwt/US c to µg/metric tsp 84 dwt/US c = 8 282 402.65 micrograms per metric tablespoon dwt/US c to µg/metric tbsp 84 dwt/US c = 138 040 044.24 micrograms per metric cup dwt/US c to µg/metric c 84 dwt/US c = 9 048 284.2 micrograms per cubic inch dwt/US c to µg/in³ 84 dwt/US c = 15 635 435 088 micrograms per cubic foot dwt/US c to µg/ft³ 84 dwt/US c = 422 156 746 368 micrograms per cubic yard dwt/US c to µg/yd³ 84 dwt/US c = 2 721 554.22 micrograms per US teaspoon dwt/US c to µg/tsp 84 dwt/US c = 8 164 662.67 micrograms per US tablespoon dwt/US c to µg/tbsp 84 dwt/US c = 16 329 325.32 micrograms per US fluid ounce dwt/US c to µg/fl.oz 84 dwt/US c = 130 634 602.56 micrograms per US cup dwt/US c to µg/US c 84 dwt/US c = 261 269 205.12 micrograms per US pint dwt/US c to µg/pt 84 dwt/US c = 522 538 410.24 micrograms per US quart dwt/US c to µg/US qt 84 dwt/US c = 2 090 153 637.6 micrograms per US gallon dwt/US c to µg/US gal 84 dwt/US c = 0.55 milligram per cubic millimeter dwt/US c to mg/mm³ 84 dwt/US c = 552.16 milligrams per cubic centimeter dwt/US c to mg/cm³ 84 dwt/US c = 552 160.18 milligrams per cubic decimeter dwt/US c to mg/dm³ 84 dwt/US c = 552 160 177.8 milligrams per cubic meter dwt/US c to mg/m³ 84 dwt/US c = 552.16 milligrams per milliliter dwt/US c to mg/ml 84 dwt/US c = 552 160.18 milligrams per liter dwt/US c to mg/l 84 dwt/US c = 2 760.8 milligrams per metric teaspoon dwt/US c to mg/metric tsp 84 dwt/US c = 8 282.4 milligrams per metric tablespoon dwt/US c to mg/metric tbsp 84 dwt/US c = 138 040.04 milligrams per metric cup dwt/US c to mg/metric c 84 dwt/US c = 9 048.28 milligrams per cubic inch dwt/US c to mg/in³ 84 dwt/US c = 15 635 435.09 milligrams per cubic foot dwt/US c to mg/ft³ 84 dwt/US c = 422 156 746.2 milligrams per cubic yard dwt/US c to mg/yd³ 84 dwt/US c = 2 721.55 milligrams per US teaspoon dwt/US c to mg/tsp 84 dwt/US c = 8 164.66 milligrams per US tablespoon dwt/US c to mg/tbsp 84 dwt/US c = 16 329.33 milligrams per US fluid ounce dwt/US c to mg/fl.oz 84 dwt/US c = 130 634.6 milligrams per US cup dwt/US c to mg/US c 84 dwt/US c = 261 269.21 milligrams per US pint dwt/US c to mg/pt 84 dwt/US c = 522 538.41 milligrams per US quart dwt/US c to mg/US qt 84 dwt/US c = 2 090 153.64 milligrams per US gallon dwt/US c to mg/US gal 84 dwt/US c = 0.001 gram per cubic millimeter dwt/US c to g/mm³ 84 dwt/US c = 0.55 gram per cubic centimeter dwt/US c to g/cm³ 84 dwt/US c = 552.16 grams per cubic decimeter dwt/US c to g/dm³ 84 dwt/US c = 552 160.18 grams per cubic meter dwt/US c to g/m³ 84 dwt/US c = 0.55 gram per milliliter dwt/US c to g/ml 84 dwt/US c = 552.16 grams per liter dwt/US c to g/l 84 dwt/US c = 2.76 grams per metric teaspoon dwt/US c to g/metric tsp 84 dwt/US c = 8.28 grams per metric tablespoon dwt/US c to g/metric tbsp 84 dwt/US c = 138.04 grams per metric cup dwt/US c to g/metric c 84 dwt/US c = 9.05 grams per cubic inch dwt/US c to g/in³ 84 dwt/US c = 15 635.44 grams per cubic foot dwt/US c to g/ft³ 84 dwt/US c = 422 156.75 grams per cubic yard dwt/US c to g/yd³ 84 dwt/US c = 2.72 grams per US teaspoon dwt/US c to g/tsp 84 dwt/US c = 8.16 grams per US tablespoon dwt/US c to g/tbsp 84 dwt/US c = 16.33 grams per US fluid ounce dwt/US c to g/fl.oz 84 dwt/US c = 130.63 grams per US cup dwt/US c to g/US c 84 dwt/US c = 261.27 grams per US pint dwt/US c to g/pt 84 dwt/US c = 522.54 grams per US quart dwt/US c to g/US qt 84 dwt/US c = 2 090.15 grams per US gallon dwt/US c to g/US gal 84 dwt/US c = 5.52 × 10-7 kilogram per cubic millimeter dwt/US c to kg/mm³ 84 dwt/US c = 0.001 kilogram per cubic centimeter dwt/US c to kg/cm³ 84 dwt/US c = 0.55 kilogram per cubic decimeter dwt/US c to kg/dm³ 84 dwt/US c = 552.16 kilograms per cubic meter dwt/US c to kg/m³ 84 dwt/US c = 0.001 kilogram per milliliter dwt/US c to kg/ml 84 dwt/US c = 0.55 kilogram per liter dwt/US c to kg/l 84 dwt/US c = 0.003 kilogram per metric teaspoon dwt/US c to kg/metric tsp 84 dwt/US c = 0.01 kilogram per metric tablespoon dwt/US c to kg/metric tbsp 84 dwt/US c = 0.14 kilogram per metric cup dwt/US c to kg/metric c 84 dwt/US c = 0.01 kilogram per cubic inch dwt/US c to kg/in³ 84 dwt/US c = 15.64 kilograms per cubic foot dwt/US c to kg/ft³ 84 dwt/US c = 422.16 kilograms per cubic yard dwt/US c to kg/yd³ 84 dwt/US c = 0.003 kilogram per US teaspoon dwt/US c to kg/tsp 84 dwt/US c = 0.01 kilogram per US tablespoon dwt/US c to kg/tbsp 84 dwt/US c = 0.02 kilogram per US fluid ounce dwt/US c to kg/fl.oz 84 dwt/US c = 0.13 kilogram per US cup dwt/US c to kg/US c 84 dwt/US c = 0.26 kilogram per US pint dwt/US c to kg/pt 84 dwt/US c = 0.52 kilogram per US quart dwt/US c to kg/US qt 84 dwt/US c = 2.09 kilograms per US gallon dwt/US c to kg/US gal 84 dwt/US c = 5.52 × 10-10 tonne per cubic millimeter dwt/US c to t/mm³ 84 dwt/US c = 5.52 × 10-7 tonne per cubic centimeter dwt/US c to t/cm³ 84 dwt/US c = 0.001 tonne per cubic decimeter dwt/US c to t/dm³ 84 dwt/US c = 0.55 tonne per cubic meter dwt/US c to t/m³ 84 dwt/US c = 5.52 × 10-7 tonne per milliliter dwt/US c to t/ml 84 dwt/US c = 0.001 tonne per liter dwt/US c to t/l 84 dwt/US c = 2.76 × 10-6 tonne per metric teaspoon dwt/US c to t/metric tsp 84 dwt/US c = 8.28 × 10-6 tonne per metric tablespoon dwt/US c to t/metric tbsp 84 dwt/US c = 0.0001 tonne per metric cup dwt/US c to t/metric c 84 dwt/US c = 9.05 × 10-6 tonne per cubic inch dwt/US c to t/in³ 84 dwt/US c = 0.02 tonne per cubic foot dwt/US c to t/ft³ 84 dwt/US c = 0.42 tonne per cubic yard dwt/US c to t/yd³ 84 dwt/US c = 2.72 × 10-6 tonne per US teaspoon dwt/US c to t/tsp 84 dwt/US c = 8.16 × 10-6 tonne per US tablespoon dwt/US c to t/tbsp 84 dwt/US c = 1.7 × 10-5 tonne per US fluid ounce dwt/US c to t/fl.oz 84 dwt/US c = 0.0001 tonne per US cup dwt/US c to t/US c 84 dwt/US c = 0.0003 tonne per US pint dwt/US c to t/pt 84 dwt/US c = 0.001 tonne per US quart dwt/US c to t/US qt 84 dwt/US c = 0.002 tonne per US gallon dwt/US c to t/US gal 84 dwt/US c = 1.95 × 10-5 ounce per cubic millimeter dwt/US c to oz/mm³ 84 dwt/US c = 0.02 ounce per cubic centimeter dwt/US c to oz/cm³ 84 dwt/US c = 19.48 ounces per cubic decimeter dwt/US c to oz/dm³ 84 dwt/US c = 19 476.88 ounces per cubic meter dwt/US c to oz/m³ 84 dwt/US c = 0.02 ounce per milliliter dwt/US c to oz/ml 84 dwt/US c = 19.48 ounces per liter dwt/US c to oz/l 84 dwt/US c = 0.1 ounce per metric teaspoon dwt/US c to oz/metric tsp 84 dwt/US c = 0.29 ounce per metric tablespoon dwt/US c to oz/metric tbsp 84 dwt/US c = 4.87 ounces per metric cup dwt/US c to oz/metric c 84 dwt/US c = 0.32 ounce per cubic inch dwt/US c to oz/in³ 84 dwt/US c = 551.52 ounces per cubic foot dwt/US c to oz/ft³ 84 dwt/US c = 14 891.14 ounces per cubic yard dwt/US c to oz/yd³ 84 dwt/US c = 0.1 ounce per US teaspoon dwt/US c to oz/tsp 84 dwt/US c = 0.29 ounce per US tablespoon dwt/US c to oz/tbsp 84 dwt/US c = 0.58 ounce per US fluid ounce dwt/US c to oz/fl.oz 84 dwt/US c = 4.61 ounces per US cup dwt/US c to oz/US c 84 dwt/US c = 9.22 ounces per US pint dwt/US c to oz/pt 84 dwt/US c = 18.43 ounces per US quart dwt/US c to oz/US qt 84 dwt/US c = 73.73 ounces per US gallon dwt/US c to oz/US gal 84 dwt/US c = 1.22 × 10-6 pound per cubic millimeter dwt/US c to lb/mm³ 84 dwt/US c = 0.001 pound per cubic centimeter dwt/US c to lb/cm³ 84 dwt/US c = 1.22 pounds per cubic decimeter dwt/US c to lb/dm³ 84 dwt/US c = 1 217.3 pounds per cubic meter dwt/US c to lb/m³ 84 dwt/US c = 0.001 pound per milliliter dwt/US c to lb/ml 84 dwt/US c = 1.22 pounds per liter dwt/US c to lb/l 84 dwt/US c = 0.01 pound per metric teaspoon dwt/US c to lb/metric tsp 84 dwt/US c = 0.02 pound per metric tablespoon dwt/US c to lb/metric tbsp 84 dwt/US c = 0.3 pound per metric cup dwt/US c to lb/metric c 84 dwt/US c = 0.02 pound per cubic inch dwt/US c to lb/in³ 84 dwt/US c = 34.47 pounds per cubic foot dwt/US c to lb/ft³ 84 dwt/US c = 930.7 pounds per cubic yard dwt/US c to lb/yd³ 84 dwt/US c = 0.01 pound per US teaspoon dwt/US c to lb/tsp 84 dwt/US c = 0.02 pound per US tablespoon dwt/US c to lb/tbsp 84 dwt/US c = 0.04 pound per US fluid ounce dwt/US c to lb/fl.oz 84 dwt/US c = 0.29 pound per US cup dwt/US c to lb/US c 84 dwt/US c = 0.58 pound per US pint dwt/US c to lb/pt 84 dwt/US c = 1.15 pounds per US quart dwt/US c to lb/US qt 84 dwt/US c = 4.61 pounds per US gallon dwt/US c to lb/US gal 84 dwt/US c = 0.01 grain per cubic millimeter dwt/US c to gr/mm³ 84 dwt/US c = 8.52 grains per cubic centimeter dwt/US c to gr/cm³ 84 dwt/US c = 8 521.13 grains per cubic decimeter dwt/US c to gr/dm³ 84 dwt/US c = 8 521 133.71 grains per cubic meter dwt/US c to gr/m³ 84 dwt/US c = 8.52 grains per milliliter dwt/US c to gr/ml 84 dwt/US c = 8 521.13 grains per liter dwt/US c to gr/l 84 dwt/US c = 42.61 grains per metric teaspoon dwt/US c to gr/metric tsp 84 dwt/US c = 127.82 grains per metric tablespoon dwt/US c to gr/metric tbsp 84 dwt/US c = 2 130.28 grains per metric cup dwt/US c to gr/metric c 84 dwt/US c = 139.64 grains per cubic inch dwt/US c to gr/in³ 84 dwt/US c = 241 291.64 grains per cubic foot dwt/US c to gr/ft³ 84 dwt/US c = 6 514 874.19 grains per cubic yard dwt/US c to gr/yd³ 84 dwt/US c = 42 grains per US teaspoon dwt/US c to gr/US tsp 84 dwt/US c = 126 grains per US tablespoon dwt/US c to gr/US tbsp 84 dwt/US c = 252 grains per US fluid ounce dwt/US c to gr/fl.oz 84 dwt/US c = 2 016 grains per US cup dwt/US c to gr/US c 84 dwt/US c = 4 032 grains per US pint dwt/US c to gr/pt 84 dwt/US c = 8 064 grains per US quart dwt/US c to gr/US qt 84 dwt/US c = 32 256 grains per US gallon dwt/US c to gr/US gal 84 dwt/US c = 3.78 × 10-8 slug per cubic millimeter dwt/US c to sl/mm³ 84 dwt/US c = 3.78 × 10-5 slug per cubic centimeter dwt/US c to sl/cm³ 84 dwt/US c = 0.04 slug per cubic decimeter dwt/US c to sl/dm³ 84 dwt/US c = 37.83 slugs per cubic meter dwt/US c to sl/m³ 84 dwt/US c = 3.78 × 10-5 slug per milliliter dwt/US c to sl/ml 84 dwt/US c = 0.04 slug per liter dwt/US c to sl/l 84 dwt/US c = 0.0002 slug per metric teaspoon dwt/US c to sl/metric tsp 84 dwt/US c = 0.001 slug per metric tablespoon dwt/US c to sl/metric tbsp 84 dwt/US c = 0.01 slug per metric cup dwt/US c to sl/metric c 84 dwt/US c = 0.001 slug per cubic inch dwt/US c to sl/in³ 84 dwt/US c = 1.07 slugs per cubic foot dwt/US c to sl/ft³ 84 dwt/US c = 28.93 slugs per cubic yard dwt/US c to sl/yd³ 84 dwt/US c = 0.0002 slug per US teaspoon dwt/US c to sl/tsp 84 dwt/US c = 0.001 slug per US tablespoon dwt/US c to sl/tbsp 84 dwt/US c = 0.001 slug per US fluid ounce dwt/US c to sl/fl.oz 84 dwt/US c = 0.01 slug per US cup dwt/US c to sl/US c 84 dwt/US c = 0.02 slug per US pint dwt/US c to sl/pt 84 dwt/US c = 0.04 slug per US quart dwt/US c to sl/US qt 84 dwt/US c = 0.14 slug per US gallon dwt/US c to sl/US gal 84 dwt/US c = 6.09 × 10-10 short ton per cubic millimeter dwt/US c to short tn/mm³ 84 dwt/US c = 6.09 × 10-7 short ton per cubic centimeter dwt/US c to short tn/cm³ 84 dwt/US c = 0.001 short ton per cubic decimeter dwt/US c to short tn/dm³ 84 dwt/US c = 0.61 short ton per cubic meter dwt/US c to short tn/m³ 84 dwt/US c = 6.09 × 10-7 short ton per milliliter dwt/US c to short tn/ml 84 dwt/US c = 0.001 short ton per liter dwt/US c to short tn/l 84 dwt/US c = 3.04 × 10-6 short ton per metric teaspoon dwt/US c to short tn/metric tsp 84 dwt/US c = 9.13 × 10-6 short ton per metric tablespoon dwt/US c to short tn/metric tbsp 84 dwt/US c = 0.0002 short ton per metric cup dwt/US c to short tn/metric c 84 dwt/US c = 9.97 × 10-6 short ton per cubic inch dwt/US c to short tn/in³ 84 dwt/US c = 0.02 short ton per cubic foot dwt/US c to short tn/ft³ 84 dwt/US c = 0.47 short ton per cubic yard dwt/US c to short tn/yd³ 84 dwt/US c = 3 × 10-6 short ton per US teaspoon dwt/US c to short tn/US tsp 84 dwt/US c = 9 × 10-6 short ton per US tablespoon dwt/US c to short tn/US tbsp 84 dwt/US c = 1.8 × 10-5 short ton per US fluid ounce dwt/US c to short tn/fl.oz 84 dwt/US c = 0.0001 short ton per US cup dwt/US c to short tn/US c 84 dwt/US c = 0.0003 short ton per US pint dwt/US c to short tn/pt 84 dwt/US c = 0.001 short ton per US quart dwt/US c to short tn/US qt 84 dwt/US c = 0.002 short ton per US gallon dwt/US c to short tn/US gal 84 dwt/US c = 5.43 × 10-10 long ton per cubic millimeter dwt/US c to long tn/mm³ 84 dwt/US c = 5.43 × 10-7 long ton per cubic centimeter dwt/US c to long tn/cm³ 84 dwt/US c = 0.001 long ton per cubic decimeter dwt/US c to long tn/dm³ 84 dwt/US c = 0.54 long ton per cubic meter dwt/US c to long tn/m³ 84 dwt/US c = 5.43 × 10-7 long ton per milliliter dwt/US c to long tn/ml 84 dwt/US c = 0.001 long ton per liter dwt/US c to long tn/l 84 dwt/US c = 2.72 × 10-6 long ton per metric teaspoon dwt/US c to long tn/metric tsp 84 dwt/US c = 8.15 × 10-6 long ton per metric tablespoon dwt/US c to long tn/metric tbsp 84 dwt/US c = 0.0001 long ton per metric cup dwt/US c to long tn/metric c 84 dwt/US c = 8.91 × 10-6 long ton per cubic inch dwt/US c to long tn/in³ 84 dwt/US c = 0.02 long ton per cubic foot dwt/US c to long tn/ft³ 84 dwt/US c = 0.42 long ton per cubic yard dwt/US c to long tn/yd³ 84 dwt/US c = 2.68 × 10-6 long ton per US teaspoon dwt/US c to long tn/US tsp 84 dwt/US c = 8.04 × 10-6 long ton per US tablespoon dwt/US c to long tn/US tbsp 84 dwt/US c = 1.61 × 10-5 long ton per US fluid ounce dwt/US c to long tn/fl.oz 84 dwt/US c = 0.0001 long ton per US cup dwt/US c to long tn/US c 84 dwt/US c = 0.0003 long ton per US pint dwt/US c to long tn/pt 84 dwt/US c = 0.001 long ton per US quart dwt/US c to long tn/US qt 84 dwt/US c = 0.002 long ton per US gallon dwt/US c to long tn/US gal 84 dwt/US c = 8.7 × 10-8 stone per cubic millimeter dwt/US c to st/mm³ 84 dwt/US c = 8.7 × 10-5 stone per cubic centimeter dwt/US c to st/cm³ 84 dwt/US c = 0.09 stone per cubic decimeter dwt/US c to st/dm³ 84 dwt/US c = 86.95 stones per cubic meter dwt/US c to st/m³ 84 dwt/US c = 8.7 × 10-5 stone per milliliter dwt/US c to st/ml 84 dwt/US c = 0.09 stone per liter dwt/US c to st/l 84 dwt/US c = 0.0004 stone per metric teaspoon dwt/US c to st/metric tsp 84 dwt/US c = 0.001 stone per metric tablespoon dwt/US c to st/metric tbsp 84 dwt/US c = 0.02 stone per metric cup dwt/US c to st/metric c 84 dwt/US c = 0.001 stone per cubic inch dwt/US c to st/in³ 84 dwt/US c = 2.46 stones per cubic foot dwt/US c to st/ft³ 84 dwt/US c = 66.48 stones per cubic yard dwt/US c to st/yd³ 84 dwt/US c = 0.0004 stone per US teaspoon dwt/US c to st/US tsp 84 dwt/US c = 0.001 stone per US tablespoon dwt/US c to st/US tbsp 84 dwt/US c = 0.003 stone per US fluid ounce dwt/US c to st/fl.oz 84 dwt/US c = 0.02 stone per US cup dwt/US c to st/US c 84 dwt/US c = 0.04 stone per US pint dwt/US c to st/pt 84 dwt/US c = 0.08 stone per US quart dwt/US c to st/US qt 84 dwt/US c = 0.33 stone per US gallon dwt/US c to st/US gal 84 dwt/US c = 1.78 × 10-5 troy ounce per cubic millimeter dwt/US c to oz t/mm³ 84 dwt/US c = 0.02 troy ounce per cubic centimeter dwt/US c to oz t/cm³ 84 dwt/US c = 17.75 troy ounces per cubic decimeter dwt/US c to oz t/dm³ 84 dwt/US c = 17 752.36 troy ounces per cubic meter dwt/US c to oz t/m³ 84 dwt/US c = 0.02 troy ounce per milliliter dwt/US c to oz t/ml 84 dwt/US c = 17.75 troy ounces per liter dwt/US c to oz t/l 84 dwt/US c = 0.09 troy ounce per metric teaspoon dwt/US c to oz t/metric tsp 84 dwt/US c = 0.27 troy ounce per metric tablespoon dwt/US c to oz t/metric tbsp 84 dwt/US c = 4.44 troy ounces per metric cup dwt/US c to oz t/metric c 84 dwt/US c = 0.29 troy ounce per cubic inch dwt/US c to oz t/in³ 84 dwt/US c = 502.69 troy ounces per cubic foot dwt/US c to oz t/ft³ 84 dwt/US c = 13 572.65 troy ounces per cubic yard dwt/US c to oz t/yd³ 84 dwt/US c = 0.09 troy ounce per US teaspoon dwt/US c to oz t/US tsp 84 dwt/US c = 0.26 troy ounce per US tablespoon dwt/US c to oz t/US tbsp 84 dwt/US c = 0.53 troy ounce per US fluid ounce dwt/US c to oz t/fl.oz 84 dwt/US c = 4.2 troy ounces per US cup dwt/US c to oz t/US c 84 dwt/US c = 8.4 troy ounces per US pint dwt/US c to oz t/pt 84 dwt/US c = 16.8 troy ounces per US quart dwt/US c to oz t/US qt 84 dwt/US c = 67.2 troy ounces per US gallon dwt/US c to oz t/US gal 84 dwt/US c = 1.48 × 10-6 troy pound per cubic millimeter dwt/US c to troy/mm³ 84 dwt/US c = 0.001 troy pound per cubic centimeter dwt/US c to troy/cm³ 84 dwt/US c = 1.48 troy pounds per cubic decimeter dwt/US c to troy/dm³ 84 dwt/US c = 1 479.36 troy pounds per cubic meter dwt/US c to troy/m³ 84 dwt/US c = 0.001 troy pound per milliliter dwt/US c to troy/ml 84 dwt/US c = 1.48 troy pounds per liter dwt/US c to troy/l 84 dwt/US c = 0.01 troy pound per metric teaspoon dwt/US c to troy/metric tsp 84 dwt/US c = 0.02 troy pound per metric tablespoon dwt/US c to troy/metric tbsp 84 dwt/US c = 0.37 troy pound per metric cup dwt/US c to troy/metric c 84 dwt/US c = 0.02 troy pound per cubic inch dwt/US c to troy/in³ 84 dwt/US c = 41.89 troy pounds per cubic foot dwt/US c to troy/ft³ 84 dwt/US c = 1 131.05 troy pounds per cubic yard dwt/US c to troy/yd³ 84 dwt/US c = 0.01 troy pound per US teaspoon dwt/US c to troy/US tsp 84 dwt/US c = 0.02 troy pound per US tablespoon dwt/US c to troy/US tbsp 84 dwt/US c = 0.04 troy pound per US fluid ounce dwt/US c to troy/fl.oz 84 dwt/US c = 0.35 troy pound per US cup dwt/US c to troy/US c 84 dwt/US c = 0.7 troy pound per US pint dwt/US c to troy/pt 84 dwt/US c = 1.4 troy pounds per US quart dwt/US c to troy/US qt 84 dwt/US c = 5.6 troy pounds per US gallon dwt/US c to troy/US gal 84 dwt/US c = 0.0004 pennyweight per cubic millimeter dwt/US c to dwt/mm³ 84 dwt/US c = 0.36 pennyweight per cubic centimeter dwt/US c to dwt/cm³ 84 dwt/US c = 355.05 pennyweights per cubic decimeter dwt/US c to dwt/dm³ 84 dwt/US c = 355 047.24 pennyweights per cubic meter dwt/US c to dwt/m³ 84 dwt/US c = 0.36 pennyweight per milliliter dwt/US c to dwt/ml 84 dwt/US c = 355.05 pennyweights per liter dwt/US c to dwt/l 84 dwt/US c = 1.78 pennyweights per metric teaspoon dwt/US c to dwt/metric tsp 84 dwt/US c = 5.33 pennyweights per metric tablespoon dwt/US c to dwt/metric tbsp 84 dwt/US c = 88.76 pennyweights per metric cup dwt/US c to dwt/metric c 84 dwt/US c = 5.82 pennyweights per cubic inch dwt/US c to dwt/in³ 84 dwt/US c = 10 053.82 pennyweights per cubic foot dwt/US c to dwt/ft³ 84 dwt/US c = 271 453.09 pennyweights per cubic yard dwt/US c to dwt/yd³ 84 dwt/US c = 1.75 pennyweights per US teaspoon dwt/US c to dwt/US tsp 84 dwt/US c = 5.25 pennyweights per US tablespoon dwt/US c to dwt/US tbsp 84 dwt/US c = 10.5 pennyweights per US fluid ounce dwt/US c to dwt/fl.oz 84 dwt/US c = 168 pennyweights per US pint dwt/US c to dwt/pt 84 dwt/US c = 336 pennyweights per US quart dwt/US c to dwt/US qt 84 dwt/US c = 1 344 pennyweights per US gallon dwt/US c to dwt/US gal
#### Foods, Nutrients and Calories
RUSSET POTATOES, UPC: 892867001357 contain(s) 74 calories per 100 grams (≈3.53 ounces) [ price ]
8208 foods that contain Carotene, beta. List of these foods starting with the highest contents of Carotene, beta and the lowest contents of Carotene, beta
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Aragamax Sugar-Sized Sand weighs 1 537.8 kg/m³ (96.00172 lb/ft³) with specific gravity of 1.5378 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Peat Woody, moderately decomposed, Northern Minnesota, wet weighs 457 kg/m³ (28.52958 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Safflower oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
Square micrometer or square micron (μm²) is the derived metric SI (System International) measurement unit of area with sides equal to one micrometer or one micron (1μm)
An angle, in geometry, is defined by two rays a and b sharing a common starting point S, called the vertex. These rays can be transformed into each other by a revolution or rotation.
dwt/metric c to dwt/l conversion table, dwt/metric c to dwt/l unit converter or convert between all units of density measurement.
#### Calculators
Area of a circle calculator, circumference and area formulas | 8,424 | 20,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | longest | en | 0.412356 |
https://it.mathworks.com/matlabcentral/answers/38209-reversing-an-polynomial-equation-y-f-x-to-x-f-y | 1,675,514,027,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00632.warc.gz | 326,170,664 | 25,975 | # Reversing an polynomial equation y = f(x) to x = f(y)
109 views (last 30 days)
Hazem Elgabra on 13 May 2012
I'm trying to reverse a 3rd order equation using matlab, ie: y = x^3 + x^2 + x^1 + 5 to x = f(y) I just don't know if there's an already built it function in matlab for such a task.
Thank you...
Walter Roberson on 13 May 2012
polynomials of degree 2 or higher cannot be inverted to give a polynomial.
The inverse of the particular polynomial you indicate is the union of three expressions.
If you have the symbolic toolbox, you can use solve()
Marcin Konowalczyk on 1 Nov 2017
@Walter is right, but what you can do is approximate them using, for example, this function. This works only in a small domain where your polynomial is well conditioned (monotonically increasing) and fails horribly otherwise, but for certain cases (see the docstring of the linked function) it is useful.
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Translated by | 275 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.904222 |
https://mathoverflow.net/questions/363280/relationship-between-volume-and-area | 1,721,760,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00185.warc.gz | 341,539,770 | 25,855 | # Relationship between volume and area
Let $$\mu(z) dV_n$$ be a measure in $$\mathbb{C} ^n$$. Let $$B_n(r) := \{z \mid \|z\| < r\}$$ be the ball of radius $$r$$ in $$\mathbb C^n$$, and $$\partial B_n(r)$$ be the corresponding sphere. In $$\mathbb{C}$$ how can we find the following inequality? $$\operatorname{Vol}_{\mu}(B_1(r))=\int_{B_1(r)} \mu(z) dV_1(z)= \int_0^r\left(\int_{\partial B_1(t)} \mu dz\right)dt\geq \int_0^r \left[\int_{\partial B_1(t)}(\mu)^{ \frac{1}{2}} \right]^2\frac{1}{2\pi t} dt$$ And can we generalize this inequality in $$\mathbb {C}^n$$?
• I tried to proofread this (for example, don't mix text and math in an equation; write $\operatorname{Vol}(B)$ $\operatorname{Vol}(B)$, not Vol$(B)$ Vol$(B)$), but I'm not sure I got everything right. For example, you asked whether you could generate the inequality in $\mathbb C^n$, and I changed that to 'generalize'. Please feel free to revert or re-edit if I got it wrong. Commented Jun 16, 2020 at 21:43
• Please don't self-vandalize your own post. You can delete it if you like, as long as it has no answer (and also undelete it later).
– YCor
Commented Jun 17, 2020 at 11:35
• You say "Let $\mu(z)dV$ is a measure on $\mathbb C^n$". So I guess $\mu$ is a function (namely, the density of your measure w.r.t Lebesgue). Q: What does it then mean to say "Let $B_\mu(r)$ be a ball in $\mathbb C^n$ ?". More precisely, $B_\mu(r) = ???$ Commented Jun 17, 2020 at 18:17
• $Vol(B_\mu(r))$ is the volume of the ball with respect to the measure $\mu dV$ Commented Jun 18, 2020 at 8:43
• $B(r) ={z\in\mathbb {C^n} ;|z|<r}$ Commented Jun 18, 2020 at 8:45
The problem can be solved via co-area formula and Jensen's inequality. We will do it Bourbaki style, i.e from $$n$$-dimensional case to particular case $$n=1$$.
Instead of $$\mathbb C^n$$, we can equivalently see the problem as a problem in $$\mathbb R^m$$, where $$m=2n$$ (i.e we isomorphically map real dimensions for each complex dimension). So, let $$dV_m$$ denote volume measure in $$\mathbb R^m$$ and $$dS_{m-1}$$ denote the corresponding surface area measure, i.e $$(m-1)$$-dimensional Hausdorff measure. The mapping $$F: z \mapsto \|z\|$$ on $$\mathbb R^m$$ has jacobian determinant $$1$$ (except at $$z=0$$, where it is undefined). Also note that for all $$t \ge 0$$, we have $$F^{-1}(\{t\})=\{z \in \mathbb R^m \mid F(z) = t\} = \{z \in \mathbb R^m \mid \|z\| = t\} = \partial B_m(t).$$ By the coarea-formula (see Corollary 1.4, for example), we have $$\begin{split} \int_{B_m(r)}\mu(z)dV_m(z) &= \int_{0}^r\left(\int_{F^{-1}(\{t\})}\frac{\mu(z)}{|Jac_F(z)|}dS_{m-1}(z)\right)dt\\ &= \int_{0}^r\left(\int_{\partial B_m(t)}\mu(z)dS_{n-1}(z)\right)dt\\ &= \int_{0}^r\left(\int_{\partial B_m(t)}\frac{\mu(z)}{S_{m-1}(\partial B_m(t))}dS_{m-1}(z)\right)S_{n-1}(\partial B_m(t))dt\\ &\ge\int_{0}^r\left(\int_{\partial B_m(t)}\frac{\mu(z)^{1/2}}{S_{m-1}(\partial B_m(t))}dS_{m-1}(z)\right)^2S_{m-1}(\partial B_m(t))dt\\ &= \int_{0}^r\left(\int_{\partial B_m(t)}\mu(z)^{1/2}dS_{m-1}(z)\right)^2\frac{1}{S_{m-1}(\partial B_m(t))}dt, \end{split}$$ where the inequality is an applicaiton of Jensen's inequality on the convex function $$x \mapsto x^2$$ and the probability measure $$A \mapsto S_{m-1}(A \cap \partial B(t))/S_{m-1}(\partial B(t))$$.
In particular, if $$n=2$$, we have $$m=2\cdot 1 = 2$$, $$dS_{m-1} = dS_1$$ which is the arc-length measure, and so $$S_1(\partial B(t)) =$$ length or circle of radius $$t$$, which equals $$2\pi t$$.
• Ok, the situation is clearer now. My answer above solves both your particular (case $n=1$) and general problems (cases $n > 1$). Commented Jun 18, 2020 at 10:30 | 1,360 | 3,628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.827007 |
https://mathexchanges.wordpress.com/2013/04/18/back-to-blogging/ | 1,501,268,412,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550977093.96/warc/CC-MAIN-20170728183650-20170728203650-00179.warc.gz | 669,068,590 | 36,824 | Back to Blogging!
I have taken a very long break from blogging! I’ve still been here teaching and learning about math, life has just gotten in the way of blogging. And a wonderful part of that life has been the birth of my daughter, Louisa, this February!
Just this week I went back to Keith Devlin’s fascinating book, The Math Instinct. Devlin’s book takes a look at some studies with very young babies that illustrate the fact that we humans are born mathematical creatures. In one study designed by psychologist Prentice Starkey, babies were shown images of one dot and two dots on projectors. At the same time one or two drum beats would be played. When one drum beat was played, the babies would spend a significantly longer time staring at the projector that showed one dot. When two beats were played, the babies would stare at the projector with two dots (Devlin, 12-13). Many other studies have also shown that babies have some understanding of numerosity and number sense.
Babies are mathematicians too! So you can imagine that I’ve been watching my own little mathematician for signs of number sense. I’m pretty sure she’s counting already 🙂
Now that I’m getting used to being a mom and getting a little more sleep I plan on getting back to the blog more frequently. Looking forward to catching up with math friends here! What mathematical adventures have you all been up to?
8 thoughts on “Back to Blogging!”
1. Welcome back! Congrats and Louisa is beautiful. ‘m so glad you took a break from blogging and decided you wanted to come back to it. We were missing you.
I’ve been problem solving with my K students and am working up a rubric to help guide reflections and communicate with parents.
1. kowedekind says:
Thanks, Mandy! I’d love to hear more about your rubric and how it’s helping parents understand what you’re up to!
2. Janet Burton says:
I’m so glad you are back! Welcome to motherhood. It was wonderful to see your name in my email. I did a little happy dance!
1. kowedekind says:
Thanks, Janet!
3. gabby gonzalez says:
Congratulations! I have visited your blog frequently to check for updates and was very excited to see you blogging again. My school wants to incorporate math exchanges as part of our math time. Is there a video of a group of kids engaging in a math exchange anywhere online?
1. kowedekind says:
Thanks, Gabby!
That’s great that your school is working to incorporate math exchanges into your practice. There will be a Math Exchanges DVD coming out soon! We taped it in November and it’s in the editing stage now. I will certainly be posting about it when it comes out. I’m hoping seeing math exchanges in action will help people implement small groups in their classrooms.
4. mary says: | 603 | 2,749 | {"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} | 3.0625 | 3 | CC-MAIN-2017-30 | longest | en | 0.966992 |
https://whatisconvert.com/300-knots-in-miles-hour | 1,656,686,768,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00728.warc.gz | 686,248,475 | 7,320 | # What is 300 Knots in Miles/Hour?
## Convert 300 Knots to Miles/Hour
To calculate 300 Knots to the corresponding value in Miles/Hour, multiply the quantity in Knots by 1.1507794480225 (conversion factor). In this case we should multiply 300 Knots by 1.1507794480225 to get the equivalent result in Miles/Hour:
300 Knots x 1.1507794480225 = 345.23383440676 Miles/Hour
300 Knots is equivalent to 345.23383440676 Miles/Hour.
## How to convert from Knots to Miles/Hour
The conversion factor from Knots to Miles/Hour is 1.1507794480225. To find out how many Knots in Miles/Hour, multiply by the conversion factor or use the Velocity converter above. Three hundred Knots is equivalent to three hundred forty-five point two three four Miles/Hour.
## Definition of Knot
The knot is a unit of speed equal to one nautical mile (1.852 km) per hour, approximately 1.151 mph. The ISO Standard symbol for the knot is kn. The same symbol is preferred by the IEEE; kt is also common. The knot is a non-SI unit that is "accepted for use with the SI". Worldwide, the knot is used in meteorology, and in maritime and air navigation—for example, a vessel travelling at 1 knot along a meridian travels approximately one minute of geographic latitude in one hour. Etymologically, the term derives from counting the number of knots in the line that unspooled from the reel of a chip log in a specific time.
## Definition of Mile/Hour
Miles per hour (abbreviated mph, MPH or mi/h) is an imperial and United States customary unit of speed expressing the number of statute miles covered in one hour. Although kilometres per hour is now the most widely used measure of speed, miles per hour remains the standard unit for speed limits in the United States, the United Kingdom, Antigua & Barbuda and Puerto Rico, although the latter two use kilometres for long distances.
## Using the Knots to Miles/Hour converter you can get answers to questions like the following:
• How many Miles/Hour are in 300 Knots?
• 300 Knots is equal to how many Miles/Hour?
• How to convert 300 Knots to Miles/Hour?
• How many is 300 Knots in Miles/Hour?
• What is 300 Knots in Miles/Hour?
• How much is 300 Knots in Miles/Hour?
• How many mph are in 300 kt?
• 300 kt is equal to how many mph?
• How to convert 300 kt to mph?
• How many is 300 kt in mph?
• What is 300 kt in mph?
• How much is 300 kt in mph? | 604 | 2,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.887762 |
http://writeulearn.com/category/programming/page/2/ | 1,550,745,763,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247503844.68/warc/CC-MAIN-20190221091728-20190221113728-00277.warc.gz | 299,917,239 | 13,572 | # Techgig gaming console
A new type of computer game is launched in the market, which requires a special type of gaming console. Makers of the games has made a new generation gaming console with a very unique remote. The keys/Layout of remote is as shown below
This remote contains 12 keys with 2 nonfunctional keys. Each functional key has been given a unique label as shown above.
The Computer game for this console has N levels. There are certain moves in the game which are controlled using the remote. The keys for each moves changes with the number of levels i.e. number of keys for an operation is dependent on the game level, for example for third level, single move require three key [you have to press three keys in sequence for a move in third level]. A player can only press keys that are left, right, up or down to the current key and he is not allowed to press bottom row corner keys.
Suppose a player is at level 2. According to the rules he has to press 2 keys for a move (the number of keys is equal to the level). The possible Moves key will be ‘HH’, ‘HI’, ‘HG’, ‘HE’, ‘HJ’, ‘IF’ ,.. Etc
1. If a Move start from ‘J’, possible keys are ‘JJ’, ‘JH’ (count 2)
2. If a Move start from ‘H’, possible keys are ‘HH’, ‘HI’, ‘HG’, ‘HE’, ‘HJ’ (count 5)
3. If a Move start from ‘I’, possible keys are ‘II’, ‘IH’, ‘IF’ (count 3)
————-
————-
So for level 2, number of possible moves would be 36.
If a player is at level N, you have to find the total possible number of moves for that level.
Input/Output Specifications
Input Specifications:
Input is an integer N, which is level of the game
Output Specifications:
Output is an integer M, which is the total possible number of Moves
Examples
Example:
Input: 2
Output: 36
Solution in C# using dynamic programming:
using System;
using System.Collections.Generic;
public static int combinationCounts(int input1)
{
//Write code here
System.Collections.Generic.SortedDictionary<string,char[]> myDictionary;
myDictionary = new SortedDictionary<string, char[]> { { “A”, new char[] { ‘A’, ‘B’,’D’ } },
{ “B”, new char[] { ‘A’, ‘B’,’C’,’E’ } },
{ “C”, new char[] { ‘F’, ‘B’,’C’ } },
{ “D”, new char[] { ‘A’, ‘G’,’D’,’E’ }} ,
{ “E”, new char[] { ‘F’, ‘B’,’D’,’E’,’H’ }} ,
{ “F”, new char[] { ‘I’, ‘F’,’C’,’E’ }} ,
{ “G”, new char[] { ‘D’, ‘G’,’H’ }} ,
{ “H”, new char[] { ‘G’, ‘H’,’I’, ‘E’,’J’} } ,
{ “I”, new char[] { ‘H’, ‘I’,’F’ }} ,
{ “J”, new char[] { ‘H’, ‘J’ }}};
int[,] max = new int[input1,10];
for(int i=0 ; i < 10 ; i++)
{
max[0,i]=1;
}
for(int i=1 ;i < input1;i++ )
{
int j=0;
foreach (KeyValuePair<string, char[]> p in myDictionary)
{
int count=0;
char[] temp =p.Value;
for(int k=0 ; k < temp.Length;k++)
{
count+=max[i-1,temp[k] – ‘A’];
}
max[i,j]=count;
j++;
}
}
int sum=0;
for(int i=0;i<10;i++)
sum+=max[input1-1,i];
return sum;
} | 847 | 2,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-09 | latest | en | 0.886718 |
http://www.futurity.org/graph-theory-kelmans-seymour-conjecture-1171922-2/ | 1,660,767,597,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00593.warc.gz | 69,440,349 | 19,393 | # There’s finally a proof for this Graph Theory puzzle
(Credit: iStockphoto)
Graph Theory is a field of mathematics that’s instrumental in complex tangles. It helps you make more connecting flights, helps get your GPS unstuck in traffic, and helps manage your Facebook posts.
It’s taken nearly 40 years of mental toil, but mathematicians have finally announced a proof of the Kelmans-Seymour Conjecture in Graph Theory.
One mathematician made the conjecture. Another tried for years to prove it and failed but passed on his insights. The next advanced the mathematical basis for 10 more years. One helped that person solve part of the proof. And two more finally helped him complete the rest of the proof.
Elapsed time: 39 years.
So, what is the Kelmans-Seymour Conjecture? Its name comes from Paul Seymour from Princeton University, who came up with the notion in 1977. Then another mathematician named Alexander Kelmans, arrived at the same conjecture in 1979.
And though the Georgia Tech proof fills some 120 pages of math reasoning, the conjecture itself is only one short sentence: If a graph G is 5-connected and non-planar, then G has a TK5.
### The devil called ‘TK5’
You could call a TK5 the devil in the details. TK5s are larger relatives of K5, a very simple formation that looks like a 5-point star fenced in by a pentagon. It resembles an occult or Anarchy symbol, and that’s fitting. A TK5 in a “graph” is guaranteed to thwart any nice, neat “planar” status.
Graph Theory. Planar. Non-planar. TK5. Let’s go to the real world to understand them better.
“Graph Theory is used, for example, in designing microprocessors and the logic behind computer programs,” says Georgia Tech mathematician Xingxing Yu, who has shepherded the Kelmans-Seymour Conjecture’s proof to completion. “It’s helpful in detailed networks to get quick solutions that are reasonable and require low computational complexity.”
##### [Ramanujan’s notebooks spark ‘taxi-cab’ discovery]
To picture a graph, draw some cities as points on a whiteboard and lines representing interstate highways connecting them.
But the resulting drawings are not geometrical figures like squares and trapezoids. Instead, the lines, called “edges,” are like wires connecting points called “vertices.” For a planar graph, there is always some way to draw it so that the lines from point to point do not cross.
In the real world, a microprocessor is sending electrons from point to point down myriad conductive paths. Get them crossed, and the processor shorts out.
In such intricate scenarios, optimizing connections is key. Graphs and graph algorithms play a role in modeling them. “You want to get as close to planar as you can in these situations,” Yu says.
In Graph Theory, wherever K5 or its sprawling relatives TK5s show up, you can forget planar. That’s why it’s important to know where one may be hiding in a very large graph.
### Passing it along
The human connections that led to the proof of the Kelmans-Seymour Conjecture are equally interesting, if less complicated.
Seymour had a collaborator, Robin Thomas, a professor at Georgia Tech who heads a program that includes a concentration on Graph Theory. His team has a track record of cracking decades-old math problems. One was even more than a century old.
“I tried moderately hard to prove the Kelmans-Seymour conjecture in the 1990s, but failed,” Thomas says. “Yu is a rare mathematician, and this shows it. I’m delighted that he pushed the proof to completion.”
##### [Mathematicians write a formula to describe waves]
Yu, once Thomas’ postdoc and now a professor at the School of Mathematics, picked up on the conjecture many years later.
“Around 2000, I was working on related concepts and around 2007, I became convinced that I was ready to work on that conjecture,” Yu says. He planned to involve graduate students but waited a year. “I needed to have a clearer plan of how to proceed. Otherwise, it would have been too risky,” Yu says.
Then he brought in graduate student Jie Ma in 2008, and together they proved the conjecture part of the way.
Two years later, Yu brought graduate students Yan Wang and Dawei He into the picture. “Wang worked very hard and efficiently full time on the problem,” Yu says. The team delivered the rest of the proof quicker than anticipated and currently have two submitted papers and two more in the works.
In addition to the six mathematicians who made and proved the conjecture, others tried but didn’t complete the proof but left behind useful cues.
### Call in the mob
Nearly four decades after Seymour had his idea, the fight for its proof is still not over. Other researchers are now called to tear at it for about two years like an invading mob. Not until they’ve thoroughly failed to destroy it, will the proof officially stand.
Seymour’s first reaction to news of the proof reflected that reality. “Congratulations! (If it’s true…),” he wrote.
Graduate student Wang is not terribly worried. “We spent lots and lots of our time trying to wreck it ourselves and couldn’t, so I hope things will be fine,” he says.
If so, the conjecture will get a new name: Kelmans-Seymour Conjecture Proved by He, Wang, and Yu.
And it will trigger a mathematical chain reaction, automatically confirming a past conjecture, Dirac’s Conjecture Proved by Mader, and also putting within reach proof of another conjecture, Hajos’ Conjecture.
For Princeton mathematician Seymour, it’s nice to see an intuition he held so strongly is now likely to enter into the realm of proven mathematics.
“Sometimes you conjecture some pretty thing, and it’s just wrong, and the truth is just a mess,” he wrote in an email message. “But sometimes, the pretty thing is also the truth; that that does happen sometimes is basically what keeps math going I suppose. There’s a profound thought.”
The National Science Foundation funded this research. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
Source: Georgia Tech | 1,354 | 6,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-33 | longest | en | 0.932865 |
https://engineeringtoolbox.com/loads-hanging-rods-d_1341.html | 1,723,329,216,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00187.warc.gz | 189,832,332 | 9,104 | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
## Weight rating of threaded hanger rods.
Nominal Rod Diameter
(in)
(in2)
lbkg
3/8 0.07 600 270
1/2 0.13 1100 500
5/8 0.20 1800 820
3/4 0.30 2700 1220
7/8 0.42 3800 1720
1 0.55 4900 2220
1 1/8 0.69 6200 2810
1 1/4 0.89 8000 3630
1 1/2 1.29 11600 5260
1 3/4 1.74 14700 6670
2 2.30 20700 9390
2 1/4 3.02 27200 12340
2 1/2 3.72 33500 15200
3 5.62 50600 22960
3 1/4 6.72 60500 27440
3 1/2 7.92 71300 32340
4 10.6 95400 43270
4 1/2 13.7 123000 55780
5 17.2 154000 69900
1) The maximum loads in the table above are based on allowable tensile stress of 12 kpsi - reduced by 25% to 9 kpsi allowable stress.
Note that hangers supporting steam, water, air or gas piping systems should be able to carry the weight of piping, insulation, valves and other equipment - and water for hydrostatic pressure testing - without exceeding 80% of the supporting hangers yield strength (not exceeding 16 kpsi for steel of unknown properties)
### Recommended Rod Sizes
Recommended rod sizes for individual pipes:
Recommended Rod Sizes
Pipe Size
(in)
Rod Size
(in)
< 2 3/8
2 1/2 - 3 1/2 1/2
4 - 5 5/8
6 3/4
8 - 12 7/8
14 - 16 1
18 1 1/8
20 1 1/4
24 1 1/2
Rod diameter less than 3⁄8 in (12 mm) is not recommended as support.
## Related Topics
• ### Dimensions of Pipes and Tubes
Pipe, tube and fittings sizes and dimensions. Inside and outside diameters, weights and more.
## Related Documents
• ### Bolt Torque Calculator
Calculate required bolt torque.
• ### Hooke's Law
Hooke's law - force, elongation and spring constant.
• ### Horizontal Pipe Hanger Supports - Spacing and Rod Size
Recommended maximum support span between hangers - and rod sizes for straight horizontal pipes.
• ### ISO 724 - Metric Threads
Dimensions of metric threads according ISO 724.
• ### Metric Bolts - Minimum Ultimate Tensile and Proof Loads
Minimum ultimate tensile and proof loads for metric bolts with coarse or fine threads.
• ### Metric Bolts - Tightening Torques
Recommended thightening torque metric bolts.
• ### Mild Steel - Round Bars Weight
Weight of round bars.
• ### Steels - Endurance Limits and Fatigue Stress
Endurance limits and fatigue stress for steels.
• ### Stress, Strain and Young's Modulus
Stress is force per unit area - strain is the deformation of a solid due to stress.
• ### Weight of Beams - Stress and Strain
Stress and deformation of vertical beams due to own weight.
• ### Young's Modulus, Tensile Strength and Yield Strength Values for some Materials
Young's Modulus (or Tensile Modulus alt. Modulus of Elasticity) and Ultimate Tensile Strength and Yield Strength for materials like steel, glass, wood and many more.
## Search
Search is the most efficient way to navigate the Engineering ToolBox.
## Engineering ToolBox - SketchUp Extension - Online 3D modeling!
Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with older versions of the amazing SketchUp Make and the newer "up to date" SketchUp Pro . Add the Engineering ToolBox extension to your SketchUp Make/Pro from the Extension Warehouse !
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## Citation
• The Engineering ToolBox (2008). Threaded Rods - Loads in Imperial Units. [online] Available at: https://www.engineeringtoolbox.com/loads-hanging-rods-d_1341.html [Accessed Day Month Year].
Modify the access date according your visit.
7.25.9
. | 1,079 | 3,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.696541 |
https://optimization-online.org/2018/11/6960/ | 1,726,476,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00239.warc.gz | 406,587,346 | 17,120 | # Multi-step discrete-time Zhang neural networks with application to time-varying nonlinear optimization
As a special kind of recurrent neural networks, Zhang neural network (ZNN) has been successfully applied to various time-variant problems solving. In this paper, we first propose a special two-step Zhang et al. discretization (ZeaD) formula and a general two-step ZeaD formula, whose truncation errors are ${O}(\tau^3)$ and ${O}(\tau^2)$, respectively, and $\tau>0$ denotes the sampling gap. We also propose a general five-step ZeaD formula with truncation error ${O}(\tau^5)$, and prove that the special and general two-step ZdaD formulas is convergent but the general five-step ZeaD formula is not zero-stable, thus is not convergent. Then, to solve the time-varying nonlinear optimization (TVNO) in real time, based on the Taylor series expansion and the above two convergent two-step ZeaD formulas, we discrete the continuous-time ZNN (CTZNN) model of TVNO proposed in the literature, and thus get a special two-step discrete-time ZNN (DTZNN) model and a general two-step DTZNN model, which contains a free parameter $a_1\in(-1/2,+\infty)$. Theoretical analyses indicate that the sequence generated by the first DTZNN model is not convergent, and for any $a_1\in(-1/2,+\infty)$ and the step-size $h\in(0,(2+4a_1)/(1+a_1))$, the sequence generated by the second DTZNN model converges to zero in an $\mathcal{O}(\tau^2)$ manner, where $\mathcal{O}(\tau^2)$ denotes a vector with every entries being $O(\tau^2)$. Furthermore, we prove that for any fixed $a_1\in(-1/2,+\infty)$, the constant $(2+4a_1)/(1+a_1)$ is the tight upper bound of the step-size $h$ and the constant $(1+2a_1)/(1+a_1)$ is the optimal step-size. Finally, some numerical results and comparisons are provided and analyzed to substantiate the efficacy of the proposed DTZNN models. | 499 | 1,857 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.880179 |
https://www.coursehero.com/file/6234708/bumpkin/ | 1,490,338,560,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187717.19/warc/CC-MAIN-20170322212947-00299-ip-10-233-31-227.ec2.internal.warc.gz | 911,033,009 | 51,521 | bumpkin
# bumpkin - b=b*3*12*2.52; }else // conversion of bumpkins...
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#include <iostream> #include <string> using namespace std; int main() { int k; //number of doors int i; //number of bumpkins int n; // number of doors to go through int num; //nuber of bumpkins char bu; //measurement unit of bumpkin height char du; //measurement unit of door string name; //name of bumpkin float d; //height of door float b; //height of bumpkin cin>>num; k=1; i=0; // My Bumpkin processing while loop while(i<num){ cin>>name>>n>>b>>bu; // conversion of bumpkin's height from inches to centimeters if(bu=='i'){ b=b*2.52; } else // conversion of bumpkins height from feet to centimeters if(b=='f'){ b=b*12*2.52; } else // conversion of bumpkins height form yard to centimeters if(bu=='y'){
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Unformatted text preview: b=b*3*12*2.52; }else // conversion of bumpkins height form meters to centimeters if(bu=='m'){ b=b*100; } cout<<"b="<<b<<endl; // # of dooors to come through while(k<=n){ cin>>d>>du; if(bu==i){ d=d*2.52; } k=k+1; } i=i+1; } if(d>b*1.25) cout<<name<<endl<<"doorway"<<k<<endl; else if(b*1.25) cout<<name<<"stilts"<<k<<endl; else if ((b*1.25>=d)&&(d>b*1.05)) cout<<name<<"walk"<<k<<endl; else if ((b*1.05>=d)&&(d>b*0.65)) cout<<name<<"duck"<<k<<endl; else if ((b*0.65>=d)&&(d>b*0.40)) cout<<name<<"crawl"<<k<<endl; else if ((b*0.40>+d)&&(d>b*0.25)) cout<<name<<"limbo"<<k<<endl; else if(b*0.25>=d){ cout<<name<<"blocked"<<k<<endl; } return 0; }...
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bumpkin - b=b*3*12*2.52; }else // conversion of bumpkins...
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Ask a homework question - tutors are online | 635 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.553389 |
https://cstheory.stackexchange.com/questions/51114/how-does-linear-logic-achieve-resource-management | 1,656,263,247,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00691.warc.gz | 221,009,683 | 71,207 | # How does linear logic achieve resource management?
TL;DR: I want to know how linear logic works. Sorry for the long question, I try to explain myself as best as I can in hopes of receiving good answers, especially because I don't know anyone else to ask and it seems to be hard to find resources about this topic, e.g. there are just a handful of questions about LL on this site.
I'm very interested in proof search, therefore in proof nets, and thus I'm trying to understand linear logic. My goal is to be able to read this site and understand it.
I have never learned anything of this in university; I taught myself everything during plenty of (sometimes frustrating, sometimes interesting) hours. The introductions I read state that linear logic somehow limits the use of formulas in proofs to useful applications and I would like to understand how that works.
First I taught myself sequent calculus for classical logic, which already has some weird properties: Symbols like Δ and Γ are added everywhere to the “normal” symbols like A and B. There are one- and two-sided calculi. There are three different symbols for implication (⇒, ⊢ and the inference line) and three different symbols for conjunction (∧, the space above/under an inference line and the comma on the left of ⊢). And boolean and formula types are mixed with no explanation. In order to make sense of that, I came up with the following:
1. Γ and Δ conjuncted and disjuncted sets of formulae, respectively, but since they are never actually split up, one can just view them as single formulae like A or B with no special meaning at all.
2. Boolean objects and formulae are different data types with different corresponding operators, but are implicitly converted and interpreted depending on the context. While I understand that implicit conversion can be convenient like in C++, I find it very confusing when it is already used while explaining a new topic. When the target audience or discussion participants are not familiar enough with the topic, one should be very clear about what they are talking about, therefore I will try to explain how I understand this: Quoting (boxing) is always done implicitly. I will write “a”, which is similar to writing a ' in LISP. Evaluating (unboxing) is done either implicitly or explicitly by the unary use of ⊢, e.g. (⊢“a”)⇔a which is similar to writing eval in LISP or ⍎ in APL. The operators ¬, ∨, ∧, ⇒, ⇔ and the inference line operate on boolean values while the binary ⊢ connects two formulae with an implication and evaluates the result, e.g. (“a”⊢“b”)⇔(⊢“a⇒b”)⇔(a⇒b) and ≡ connects two formulae with an equality and evaluates the result, e.g. (“a”≡“b”)⇔(⊢“a⇔b”)⇔(a⇔b). Likewise, the commas to the left and right of ⊢ connect formulae with ∧ and ∨, respectively, so that ultimately (“a”,“b” ⊢ “c”,“d”)⇔a∧b⇒c∨d.
With these rules in mind, I have no problem in reading the rules and derivations on this German Wikipedia site about sequent calculus. All of the rules seem to make sense and start to feel intuitive after I look at them for some time.
In linear logic we have four dual-bool objects and four discunction and conjunction operators and the formula-operators change so that (“a”,“b”⊢“c”,“d”)=((a⊗b)⊸(c⅋d)) and (“a”≡“b”)=(a=b) (Note that I now always use = instead of ⇔ because we are no longer dealing with strictly boolean objects). The fact that ⊸ and therefore ⊢ now return objects of the dual-bool type, the behaviour of the inference line is also affected. Because of how this section about game semantics talks about validity at wins and loses, I guess it works like the following: First, the inputs are converted to boolean by converting 1 to ⊤ and 0 to ⊥ and then the normal boolean implication (⇒) is applied.
I've been having a hard time trying to obtain an intuitive understanding of all of this, since things like ⅋, ? and ! were discounted as simply “hard to understand” in various semantics and papers always started to become too complicated for me before answering important questions (in my view). I began watching the YouTube-lectures by Frank Pfenning but confusingly he uses a different subset of LL (or maybe just a different notation, I don't remember) and the popular resource semantic sometimes even appeared contradictory to me, so soon I couldn't follow up.
Then I tried to focus just on the basics and develop my own understanding of this topic, which was easier said than done because apparently no one even bothers to write truth tables, so I had to deduce the truth tables by myself which was also easier said than done because the necessary rules were scattered around.
Here's what I came up with:
1. The negation is the only operation where authors were gracious enough to write down their definition directly:
A $$(⋅)^⊥$$
0
1
1
0
1. The aforementioned article about game semantics helped me a lot; with its help I developed this:
A B &
1 1 1
1 1 1 1 1 1
0 0 0
1 0 0 1 0 0
0 0 0 0 0 0
1
0 0 0
Although it was not explicitly stated, I assumed that every player tries to win. However, there were unfilled gaps left in the table: These are the cases where one player has to choose between two different types of winning or losing a game. It was not explained which one was the preferred way.
1. Then I read about the ! and how it can be represented as an infinite series. Since it starts with 1&, the opponent would not choose the 1 when the input is 0 or ⊥, because those are their wins, so these should stay unchanged. When the input is 1, the opponent can only choose between 1's and thus, it stays unchanged as well. When the input is ⊤, the opponent has to choose between two different types of losing. If they preferred to choose ⊤, then ! would be the identity function and therefore be quite useless. So the opponent's preferred way of losing must be 1. Because of $$(!A)^⊥=?A^⊥$$, the ? must work the other way round so that:
A ! ?
0 0
1 1 1
1
With this knowledge and $$(A\&B)^⊥=A^⊥⊕B^⊥$$, I filled the gaps in the previous table:
A B &
1 1 1
1 1 1 1 1 1
0 0 0 0
1 0 0 1 0 0
0 0 0 0 0 0
1 1
0 0 0
I then realised that if it is “always one's turn”, that means just a situation that is sometimes priorized. So you can view at it like this:
0 ⊥ 1 ⊤
<---|---------|---------|---------|--->
strong weak weak strong
falsehood falsehood truth truth
With this description, & becomes something like a min function, returning the falsest value and ⊕ becomes something like a max function, returning the truest value. ⊗ and ⅋ both return the strongest value, where in a draw, ⊗ prefers falsehood and ⅋ prefers truth. Suddenly they are easy to understand :)
1. ! and ? (for which I have already written the truth tables), seem to act like a cap in truth and falsehood, respectively (keyword: saturation).
2. Because of $$A⊸B=A^⊥⅋B$$ the truth table for ⊸ can be derived as follows:
A B
0 0
0 0
0
1
1 0 0
0 1
1
1 1
1 1 1
0 0
0
0
1 0
1
In classical logic, when you write true and false as 1 and 0, the implication ⇒ can be thought of as ≤, because it states that the level of truth did not decrease in the conclusion. This also applies to ⊸. The only question left is: Which level of truth/falsehood will be returned? The answer is: If at least one of the arguments is strong, the return value will be strong, otherwise weak. This feels more intuitive if we draw the truth graph with different distances like this:
0 ⊥ 1 ⊤
<---|-----------------------------|-----|-----------------------------|--->
strong weak weak strong
falsehood falsehood truth truth
If we stay on a strong level, i.e. 0⊸0 or ⊤⊸⊤, that is a strong true statement. If play in the weak level, we get ⊥ or 1. However, if we drop from ⊤ to 1 or lower, we will be punished with a strong 0 because we concluded a consequent that is so much falser than our antecedent. In contrast, if the truth increases from 0 all the way to ⊥ or even higher, that is rewarded with a strong ⊤.
With all of this in mind, all statements on this image started to make sense to me. When I just looked at them for a while and thought about the possibilities, it was eventually clear why they must be true and I was very happy about that.
The corresponding operators for the commas next to the ⊢ are the multiplicative ones (⊗ and ⅋), so while they are conjunction and disjunction, they also follow the rule that higher priority outdoes lower priority. That is the reason why you can't just add a random statement to the right of ⊢: Because the existing statements may be of weak truth and you might be adding a 0, making the whole proposition false. Similarly, you can't just remove one of multiple true statements there, because you might remove the only strong ⊤ which holds up against a 0. And if the left side of ⊢ is true you can't just add more random true statements, because you might be lifting the value of the left hand side from 1 to ⊤ and hence make the whole proposition false. This also explains why ! and ? allow for controlled weakening and contraction: They clamp their input up and down towards weakness, so that they can no longer outplay the others in the comma seperated lists.
When I realised this, I thought I finally found the essence of linear logic, and that this must be the way how this resource limitation works and since I have my own semantic now, I can forget about the resource semantic and everything will be fine. But I was wrong, because the resource semantics claim that you can have 1\$⊸candy but not 1\$⊸candy⊗candy. But, when you look at my truth table: Every time you input the same value twice into either &, ⊕, ⊗ or ⅋, you get back your input value every time, which means that candy and candy⊗candy must denote the same object which is either 0, ⊥, 1, or ⊤. I guess the resource semantic tries to say (in an irritating way), that you just cannot derive 1\$⊸candy⊗candy from the limited given axioms, although it is still true. In classical logic, you can just assume every rule that can be deduced from the truth tables of ¬, ∨ and ∧. Of course, you can debate about which of them you want choose as axioms, and if you choose too few, you might not be able to deduce every true statement from them. But usually I don't have to care much about that. I hoped that would be the same in LL. However, the essence of lineare logic seems to be to deliberately restrict the axioms even further so that you cannot deduce every true statement (like 1\$⊸candy⊗candy) anymore. And herein lies the problem: I have no idea why exactly these axioms (deduction rules) in the image are chosen, what the exact concequences are and what all of that has to do with the duality of the boolean values and the operators (Why not simply restrict axioms of classical logic instead?). And I don't feel like I'm able to figure that out all by myself any time soon.
To sum it up, here are my questions:
• Are my truth tables correct, and if not, what are the errors?
• Are all of my assumptions correct, and if not, why?
• And most important: Can someone please try to explain to me the exact reasoning behind and consequences of the choice of axioms in LL, so that I can acccomplish an intuitive understanding of which true statements can be derived and which can't? Because I feel like that's the missing part of my understanding of LL.
• About your last question: you say you are "very interested in proof search". Proof search must be done in some proof system. Which proof systems are you familiar with? Sequent calculus is one such proof system, it is certainly different than others, but it can be related to them. If you tell us which system you already know, maybe someone can point you to somewhere where this relationship is explained. After you understand sequent calculus, linear logic is easier to understand. Feb 22 at 13:49
• Also, what "intuitive understanding of which true statements can be derived and which can't" do you expect to obtain? For example, do you have such an intuitive understanding for classical logic? Explaining it maybe can help people understand what you are looking for with respect to linear logic. Feb 22 at 13:51
Intuitionistic logic and linear logic are best understood in terms of proofs rather than provability, that is, you don't just look at what you can prove but at how you prove it. This is how, for example, intuitionistic logic sees the difference between $$A\lor B$$ (one of $$A$$ or $$B$$ holds) and $$\lnot\lnot(A\lor B)$$ (it is impossible that none among $$A$$ and $$B$$ holds): the first is proved by explicitly showing one of $$A$$ or $$B$$; the second, typically, by assuming $$\lnot(A\lor B)$$ and deriving a contradiction, so you don't know which one actually holds. The two formulas have the same truth value, but their proofs have very different logical content and, once you consider the right set of logical rules, the formulas are no longer provable one from the other (as they are in classical logic).
Intuitively, linear logic refines intuitionistic logic by looking at how many times you used a certain assumption to prove something. If you think in terms of truth, this is nonsensical: mathematical truth is "inexhaustible", it is untarnished by use. In spite of having been used countless times in countless proofs through the millennia, the Pythagorean theorem is just as true today as it was when it was first discovered. But in terms of the structure of proofs, allowing arbitrary reuse does make a difference, which is reflected, for example, in the existence of two different disjunctions $$A\oplus B$$ and $$A\mathrel{\text{⅋}} B$$, which behave (kind of) like the two different meanings of $$\lor$$ explained above. | 3,368 | 13,838 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-27 | longest | en | 0.962304 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/9/lesson/9.2.2/problem/9-64 | 1,726,022,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00075.warc.gz | 292,948,346 | 15,367 | ### Home > APCALC > Chapter 9 > Lesson 9.2.2 > Problem9-64
9-64.
No calculator! Determine the slope of the curve $x \cos(y) = xy + 1$ at the point $(1, 0)$.
First: Implicitly differentiate: $\cos(y)-x\sin(y)y^\prime=y+xy^\prime$
Second: Solve for $y^\prime$.
Third: Evaluate the equation for $x = 1$ and $y = 0$. | 113 | 317 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.658087 |
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# hw15 - Prove that SL 2 R C GL 2 R Is the Borel subgroup B...
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Modern Algebra 1 Homework 7.1 Spring 2010 Due March 10 Exercise 1. Let G be a group and let H G . Prove that if xHx - 1 H for all x G , then H xHx - 1 for all x G . Exercise 2.
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Unformatted text preview: Prove that SL 2 ( R ) C GL 2 ( R ). Is the Borel subgroup B normal in GL 2 ( R )? Exercise 3. Prove that h r i C D n ....
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Ask a homework question - tutors are online | 208 | 669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-13 | latest | en | 0.732645 |
https://ischool.sg/questions/hashtag?type=all&tag=Cut+%26+paste&level=Primary+6 | 1,708,756,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00574.warc.gz | 335,080,646 | 34,993 | Level 1 PSLE
The figure is made up of 4 squares. Two of the squares are divided equally into 4 triangles each. What fraction of the figure is shaded?
1 m
Level 1 PSLE
The figure is made up of equilateral triangles. What fraction of the figure is shaded?
1 m
Level 2 PSLE
The square ABCD is made up of 4 smaller squares.
1. What is the ratio of the area of the shaded part to the area of the unshaded part?
2. If the length of the square ABCD is 4 cm, what is the area of the shaded part?
2 m
Level 2 PSLE
Five identical semi-circles are arranged as shown. Find the diameter of one circle.
2 m
Level 1
What fraction of the figure is shaded?
2 m
Level 1 PSLE
The figure is made up of three squares ABGH, BCFG, CDEF. What fraction of the figure ADEH is shaded?
2 m
Level 2
The figures F and G, are two identical isosceles triangles. Both figures contain a square of a different size. Given that the area of the square in Figure F is 252 cm2, find the area of the square in Figure G.
2 m
Level 2
In the figure, the circle is touching each of the two squares at exactly four points, if the area of the bigger square is 100 cm2, find the area of the smaller square.
2 m
Level 2
The figure consists of 2 squares and a circle. If the length of the bigger square is 10 cm, what is the area of the shaded part?
2 m
Level 3 PSLE
In the figure not drawn to scale, ACEG and BDFH are squares. AB, CD, EF and GH are of the same length. The ratio of AB : BC is 3 : 1.
1. What fraction of square ACEG is shaded?
2. If the length of the square is 96 cm, find the unshaded area in cm2.
3 m | 443 | 1,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-10 | latest | en | 0.902872 |
http://software.intel.com/de-de/forums/topic/367971 | 1,369,489,887,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705955434/warc/CC-MAIN-20130516120555-00074-ip-10-60-113-184.ec2.internal.warc.gz | 243,838,771 | 12,118 | I tested PARDISO and found that IPAM[3] is limited in 8. I tried to use 16 threads, but summary shows that it uses 8.
What should I fix ?
iparm[FORT(1)] = 1; // 0 - all default, 1 - MUST supply all
iparm[FORT(2)] = 2; // 0 - MMD, 2 - ND METIS
iparm[FORT(4)] = 0; // No iterative-direct algorithm
iparm[FORT(5)] = 0; // No user fill-in reducing permutation
iparm[FORT(6)] = 0; // Write solution into x
iparm[FORT(7)] = 0; // Not in use
iparm[FORT(8)] = 0; // num of iterative refinement
iparm[FORT(9)] = 0; // Not in use
iparm[FORT(10)] = 13; // Pivot 13 - non sym 8 sym indefinite
iparm[FORT(11)] = 1; // Scaling 1 - non sym, 0 - sym
iparm[FORT(12)] = 0; // Conjugate transposed/transpose solve
iparm[FORT(13)] = 1; // 1 - normal matching, 2 - advanced
iparm[FORT(14)] = 0; // Output: Number of perturbed pivots
iparm[FORT(15)] = 0; // Not in use
iparm[FORT(16)] = 0; // Not in use
iparm[FORT(17)] = 0; // Not in use
iparm[FORT(18)] = -1; // -1 - report nz in factors
iparm[FORT(19)] = -1; // -1 - report MFlops
iparm[FORT(20)] = 0; // Output: Numbers of CG Iterations
iparm[FORT(21)] = 1; // 0 - 1x1 pivot] 1 - 2x2 bunch-kaufman pivot (default)
iparm[FORT(24)] = 1; // 0 - one level parallel, 1 - two level (default)
iparm[FORT(25)] = 1; // 0 - seq solve] 1 - par solve (default)
iparm[FORT(27)] = 1; // 0) no check, 1) matrix check
iparm[FORT(52)] = 1; // num of dist solver : multi thread only should be 1
cout << "IPARM[FORT(3)] = " << iparm[FORT(3)] << endl;
PARDISO returns like this:
Statistics:
===========
< Parallel Direct Factorization with number of processors: > 8
< Numerical Factorization with BLAS3 and O(n) synchronization >
< Linear system Ax = b >
number of equations: 357889
number of non-zeros in A: 26539009
number of non-zeros in A (%): 0.020720
9 Beiträge / 0 neu
Nähere Informationen zur Compiler-Optimierung finden Sie in unserem Optimierungshinweis.
Hi,
Currently iparm[2] doesn't used. MKL PARDISO by default use env. setting of number of threads.
With best regards,
Hi,
In this topic (http://software.intel.com/en-us/forums/topic/347047#comment-1718453) explain how one can change number of threads used by PARDISO.
With best regards,
Thank you for the quick answer. I tried both environment variable setup and runtime omp library setup.
However, neither works.
Statistics:
===========
< Parallel Direct Factorization with number of processors: > 8
< Numerical Factorization with BLAS3 and O(n) synchronization >
< Linear system Ax = b >
number of equations: 357889
number of non-zeros in A: 26539009
number of non-zeros in A (%): 0.020720
number of right-hand sides: 1
< Factors L and U >
number of columns for each panel: 72
number of independent subgraphs: 0
< Preprocessing with state of the art partitioning metis>
number of supernodes: 66609
size of largest supernode: 10455
number of non-zeros in L: 367780045
number of non-zeros in U: 356687962
number of non-zeros in L+U: 724468007
[PARDISO] Analysis completed ...
[PARDISO] Number of perturbed pivots = 0
[PARDISO] Number of peak memory symbolic = 961930
[PARDISO] Number of permenant memory symbolic = 687696
=== PARDISO is running in In-Core mode, because iparam(60)=0 ===
Percentage of computed non-zeros for LL^T factorization
0 % 1 % 2 % 3 % 4 % 5 % 6 % 7 % 8 % 9 % 10 % 11 % 12 % 13 % 14 % 15 % 16 % 17 % 18 % 19 % 20 % 21 % 22 % 23 % 24 % 25 % 26 % 27 % 28 % 29 % 30 % 31 % 32 % 33 % 34 % 35 % 36 % 37 % 38 % 39 % 40 % 41 % 42 % 43 % 44 % 45 % 46 % 47 % 48 % 49 % 50 % 51 % 52 % 53 % 54 % 55 % 56 % 57 % 58 % 59 % 60 % 61 % 62 % 63 % 64 % 65 % ^C
[kyungjoo @nozomi] bench > echo \$OMP_NUM_THREADS
16
[kyungjoo @nozomi] bench >
Another question: The machine is 2x8 (=16) core machine. Then, does intel compiler think that the machine has maximum 8 threads ?
Hi,
With best regards,
Hi,
Hi
With best regards,
Thank you. It works now. However, this is tricky. One last question for today: can I interpret the number of supernodes as "number of fronts" ?
< Parallel Direct Factorization with number of processors: > 16
< Preprocessing with state of the art partitioning metis>
number of supernodes: 35204
size of largest supernode: 5671
Hi,
can I interpret the number of supernodes as "number of fronts" ? No, the supernode has a bit another meaning, its correlate with number of unknowns and matrix structure. | 1,452 | 4,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.421285 |
https://www.mathdoubts.com/1-cos-double-angle-formula/ | 1,590,774,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347405558.19/warc/CC-MAIN-20200529152159-20200529182159-00415.warc.gz | 810,179,981 | 4,522 | # Subtraction of Cos double angle from one formula
## Formula
$1-\cos{(2\theta)} \,=\, 2\sin^2{\theta}$
The subtraction of cosine of double angle function from one is equal to two times the square of sine of angle. It is called the subtraction of cos double angle from one identity.
### Introduction
Let theta ($\theta$) be an angle of a right triangle, the cosine of double angle is written as $\cos{2\theta}$ mathematically and the subtraction of cosine of double angle function from one is written as $1-\cos{2\theta}$ in trigonometric mathematics.
The trigonometric expression $1-\cos{2\theta}$ can be simplified and it is equal to two times the square of sine of angle, which means $2\sin^2{\theta}$.
#### Popular forms
The subtraction of cosine of double angle function from one identity is popularly written in two forms.
$(1) \,\,\,\,\,\,$ $1-\cos{(2x)} \,=\, 2\sin^2{x}$
$(2) \,\,\,\,\,\,$ $1-\cos{(2A)} \,=\, 2\sin^2{A}$
#### Usage
A trigonometric expression that is in the form of subtracting cosine of double angle function from one, is simplified as the sine squared of angle.
#### Proof
Learn how to derive an identity for the subtraction of cosine of double angle function from one.
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Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more | 386 | 1,446 | {"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} | 3.828125 | 4 | CC-MAIN-2020-24 | latest | en | 0.87074 |
https://edison.sso.vsb.cz/cz.vsb.edison.edu.study.prepare.web/SubjectVersion.faces?version=470-2302/02&subjectBlockAssignmentId=298997&studyFormId=2&studyPlanId=19962&locale=en&back=true | 1,571,213,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00387.warc.gz | 465,962,156 | 65,249 | # 470-2302/02 – Number Theory (TC)
Gurantor department Department of Applied Mathematics Credits 4 Subject guarantor RNDr. Pavel Jahoda, Ph.D. Subject version guarantor RNDr. Pavel Jahoda, Ph.D. Study level undergraduate or graduate Requirement Optional Year 3 Semester summer Study language English Year of introduction 2015/2016 Year of cancellation Intended for the faculties FEI Intended for study types Bachelor
Instruction secured by
JAH02 RNDr. Pavel Jahoda, Ph.D.
Extent of instruction for forms of study
Form of studyWay of compl.Extent
Full-time Credit and Examination 2+2
Combined Credit and Examination 10+10
### Subject aims expressed by acquired skills and competences
After completing the course the student will know the selected definitions of basic concepts of elementary number theory and the relations between them, understand their importance, and will be able to use his knowledge to the solution of the fundamental tasks of the theory of numbers. They will also understand the importance of these concepts for the solution of the selected application tasks - primality testing and the RSA encryption algorithm.
Lectures
Tutorials
### Summary
We meet the applications of the results of number theory daily, maybe unwittingly. A variety of systems of identification numbers, such as the postal slips (USPS-The United States Postal Service), in the barcodes (UPC-Universal Product Codes) or books (ISBN-International Standard Book Number). Furthermore, the results of the theory of numbers used for generating random numbers. You shall also apply them in various areas. In addition to statistics find its place even in the theoretical physics-particle simulations. Probably the most important applications has number theory in cryptography, are based on it the extremly safe encryption methods, yet easily applicable in practice. In the subject of elementary number theory students should acquire basic knowledge of mathematical apparatus, which stands for the above applications. Then they can understand how these applications work in practice.
### Compulsory literature:
Compulsory literature is not required.
### Recommended literature:
APOSTOL T.M.: Introduction to Analytic Number Theory, Springer, 1976. HARDY G.H., WRIGHT E.M.: An Introduction to the Theory of Numbers, Oxford, Clarendon press, 1954. J.E. POMMERSHEIM, T.K. MARKS, E.L. FLAPAN, Number theory, USA: Wiley, 2010.
### Way of continuous check of knowledge in the course of semester
Průběžná kontrola studia: Studenti v průběhu semestru budou psát písemné testy. Za testy lze získat maximálně 30 bodů. Podmínky udělení zápočtu: K získání zápočtu je nutné získat minimálně 15 bodů.
### E-learning
There are not defined other requirements for student.
### Prerequisities
Subject has no prerequisities.
### Co-requisities
Subject has no co-requisities.
### Subject syllabus:
Lectures: Divisibility on N and Z, the greatest common divisor, Euclidean algorithm, Canonical decomposition, The set of prime numbers — basic knowledge of the layout to the axis, Prime-counting function, Tschebyshev inequality, the prime number theorem and Bertrand's postulate, Asymptotic density of sets, Congruence relation on Z, Linear congruences, Operation on Zn, Euler's totient function, Euler-Fermat's last theorem, Miller-Rabin primality test, RSA algorithm. Practices Properties of the divisibility on N and Z, Euclid's algorithm, Link of the canonical decomposition algorithm with the greatest common divisor and least common multiple, Presence of the prime numbers in arithmetical sequences and g-adic expansions of numbers, Eratosthenes sieve, Determining the densities of sets, asymptotic density of the set of prime numbers, Properties of congruence relation, Solving of linear congruences, Z_p field, Wilson's theorem, The value of the Euler's function, Examples on Fermat's primality test and Carmichael's numbers, Examples on the Miller-Rabin primality test, Examples on RSA algorithm
### Conditions for subject completion
Combined form (validity from: 2014/2015 Summer semester)
Min. number of points
Credit and Examination Credit and Examination 100 (100) 51
Credit Credit 30 15
Examination Examination 70 35
Mandatory attendence parzicipation:
Show history
### Occurrence in study plans
Academic yearProgrammeField of studySpec.FormStudy language Tut. centreYearWSType of duty
2019/2020 (B2647) Information and Communication Technology (1103R031) Computational Mathematics P English Ostrava 3 Optional study plan
2019/2020 (B2647) Information and Communication Technology (1103R031) Computational Mathematics K English Ostrava 3 Optional study plan
2019/2020 (B0541A170009) Computational and Applied Mathematics P English Ostrava 3 Optional study plan
2018/2019 (B2647) Information and Communication Technology (1103R031) Computational Mathematics P English Ostrava 3 Optional study plan
2018/2019 (B2647) Information and Communication Technology (1103R031) Computational Mathematics K English Ostrava 3 Optional study plan
2017/2018 (B2647) Information and Communication Technology (1103R031) Computational Mathematics P English Ostrava 3 Optional study plan
2017/2018 (B2647) Information and Communication Technology (1103R031) Computational Mathematics K English Ostrava 3 Optional study plan
2016/2017 (B2647) Information and Communication Technology (1103R031) Computational Mathematics P English Ostrava 3 Optional study plan
2016/2017 (B2647) Information and Communication Technology (1103R031) Computational Mathematics K English Ostrava 3 Optional study plan
2015/2016 (B2647) Information and Communication Technology (1103R031) Computational Mathematics P English Ostrava 3 Optional study plan
2015/2016 (B2647) Information and Communication Technology (1103R031) Computational Mathematics K English Ostrava 3 Optional study plan
### Occurrence in special blocks
Block nameAcademic yearForm of studyStudy language YearWSType of blockBlock owner | 1,359 | 5,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-43 | latest | en | 0.898952 |
https://jimmymaths.com/singapore-math-model-method/ | 1,708,911,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00009.warc.gz | 322,745,908 | 27,595 | # Singapore Math Model Method
The Singapore Math Model Method is a method that is used to teach children in Singapore primary schools. It enables children to see the question in a much clearer way and it has received many positive reviews from overseas educators. In fact, our Math textbooks are widely used in overseas schools such as the U.S.
Young children (especially in the lower primary) have not totally built up their minds to understand logic yet, so drawing models help them understand the question in a better way.
In my opinion, there are some pros and cons to the math model method.
Pros: The Singapore math model method is great for children who are visual learners as it gives a better picture of the question. Children play with toys when they are young. Using blocks to represent numbers is more visually appealing to them than using letters to represent numbers (Algebra). By seeing the question visually, these children are better able to see the logic and relationship between different objects.
Cons: This method may seem tedious and unnecessary if children can see the flow and solution in their minds. Some children might find this method time-consuming as they dislike drawing. Some children might prefer to write down statements to see the flow instead of drawing models.
Most importantly, I feel that this method also does not scale well for secondary schools and beyond. At secondary and JC levels, students are taught to use Algebra instead. That is the reason I do not discourage my primary school students to use Algebra if they know how to apply it correctly.
Nonetheless, I still find drawing models a great way to engage children by making things visual. The main problem with most students is they don’t understand the language of the problem sums. Hence, drawing models enable them to put their thoughts into pictures and enable them to understand better.
In this post, I shall not go through the basics of model drawing which is covered in Lower Primary. Instead, I will like to share something more useful for Upper Primary – cutting of models.
## Cutting of Math Models
This method is widely used in problem sums that involve fractions. For better illustration, let’s look at an example over here.
“Mrs Tan made some muffins. She sold 2/5 of them in the morning and 4/9 of the remainder in the evening. She sold 40 more muffins in the morning than evening. How many muffins did she make altogether?”
Firstly, I draw a model and cut it into 5 parts. I used the pink region to represent the muffins that are sold in the morning and the blue region to represent the remainder.
Next, I cut the remainder into 9 units because the question said that she sold 4/9 of the remainder in the evening. To do that, I will need to cut each blue unit into 3 units to get 9 units in total, So 4 units out of these 9 units are sold in the evening.
Now, to make things consistent, I also need to cut the pink unit into 3 units each.
From here, we can clearly see that 6 units are sold in the morning while 4 units are sold in the evening. And the difference between both of them is 2 units.
Since the question says, “She sold 40 more muffins in the morning than evening,” we can say that
2 units = 40
1 unit = 20
From the model, we can see that there are 15 units in total after cutting the model. So,
Total Number of Muffins = 15 × 20 = 300
Another Method
Of course, there are other methods to solve this question. Some children might find it cumbersome to draw models and prefer to use statements instead.
Remainder = 1 – 2/5 = 3/5
4/9 of the remainder = 4/9 × 3/5 = 4/15 (This will represent the fraction of Total muffins that were sold in the evening)
Difference between Morning and Evening = Fraction sold in the morning – Fraction sold in the evening
= 2/5 – 4/15 = 2/15
2/15 of Total Muffins = 40
(Warning: Many students write “2/15 = 40”. This statement is wrong because 2/15 is not equal to 40. Be careful of your presentation!)
Total Muffins = 40 ÷ 2 × 15 = 300
Using statements can be a faster way to solve the question. However, for children who are visually-inclined learners, parents might want to use this Math model method to teach them instead.
If you’re looking for more resources to give your child a better chance at succeeding in their primary school exams, don’t hesitate to check out our PSLE math tuition classes.
We provide personalised guidance and support so your child can feel confident and secure during their exam preparation. With the right resources and strategy, anything is achievable!
Enroll your child at Jimmy Maths today!
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Click the button below to follow our channel on Telegram. | 1,065 | 4,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-10 | latest | en | 0.963784 |
http://www.encyclo.co.uk/meaning-of-area | 1,532,008,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590901.10/warc/CC-MAIN-20180719125339-20180719145339-00525.warc.gz | 461,565,407 | 8,886 | ### area
a portion of a space where a particular activity occurs
### area
Roman land area measures used would include: area plural informal name equivelant to modern * saltus - - 4 centuriae or 400 heredia or 800 iugera or 1600 actus quadrati or 23,040,000 pedes quadrans ca. 500 acres or 200 hectares centuria centuriae the hundred, which corresponded to the area taken up by a century of Roman legionaries, or used in cent…...
### Area
A bounded contiguous two dimensional object which may or may not include its boundary. Usually defined in terms of an external polygon or in terms of a set of grid cells.
Found on http://www.encyclo.co.uk/local/20195
### area
[n] - a particular geographical region of indefinite boundary (usually serving some special purpose or distinguished by its people or culture or geography) 2. [n] - a subject of study 3. [n] - a part of an animal that has a special function or is supplied by a given artery or nerve 4. [n] - the extent of a 2-dimensional surface e...
Found on http://www.webdictionary.co.uk/definition.php?query=area
### Area
The derived SI unit of AreaThe derived SI unit of area is the m2.
Found on http://www.diracdelta.co.uk/science/source/a/r/area/source.html
### Area
Measures the size of a surface using length measurements in two dimensions.
Found on http://www.chemicalglossary.net/definition/180-Area
### area
Measures the size of a surface using length measurements in two dimensions.
Found on http://www.shodor.org/UNChem/glossary.html
### area
The number of square units needed to cover a surface
Example:
The area is 9 square units.
Found on http://www.hbschool.com/glossary/math2/index6.html
### Area
A're·a (ā're*ȧ; 277) noun ; plural Areas (-ȧz) . [ Latin area a broad piece of level ground. Confer Are , noun ] 1. Any plane surface, as of the floor of a room or church, or of the ground within an inclosure; an open sp...
Found on http://www.encyclo.co.uk/webster/A/117
### area
Origin: L. Area a broad piece of level gro. Cf. Are. ... 1. Any plane surface, as of the floor of a room or church, or of the ground within an inclosure; an open space in a building. 'The Alban lake . . . Looks like the area of some vast amphitheater.' (Addison) ... 2. The inclosed space on which a building stands. ... 3. The sunken space or court,...
Found on http://www.encyclo.co.uk/local/20973
### area
noun a subject of study; `it was his area of specialization`; `areas of interest include...`
Found on http://wordnetweb.princeton.edu/perl/webwn?s=area
### area
(ār´e-ә) pl. a´reae, areas a limited space or plane surface.
Found on http://www.encyclo.co.uk/local/21001
### Area
• (n.) The inclosed space on which a building stands. • (n.) The superficial contents of any figure; the surface included within any given lines; superficial extent; as, the area of a square or a triangle. • (n.) The sunken space or court, giving ingress and affording light to the basement of a building. • (n.) An extent of surf...
Found on http://thinkexist.com/dictionary/meaning/area/
### area
(from the article `measurement system`) ...of weights and measures today includes such factors as temperature, luminosity, pressure, and electric current, it once consisted of only four ... [7 related articles]
Found on http://www.britannica.com/eb/a-z/a/97
### area
The amount of space, or number of square units, inside a closed figure.
Found on http://www.toolingu.com/definition-570340-32792-flow-rate.html
### area
area (AIR ee uh) 1. Surface; extent or range, zone, space: 'The apple orchid covers a large area of the farm.' 2. Region, locality, terrain, territory, district: 'Alligators are found in marshy areas.' 3. Field, sphere, realm, scope: 'France has always excelled in the area of the arts.'
Found on http://www.wordinfo.info/words/index/info/view_unit/3507/
### Area
[nightclub] Area was a themed nightclub that operated from 1983 to 1987 at 157 Hudson Street in Manhattan, New York City. Micheal Alig worked as a barback and busboy. The club was known for its unusual invitations. The club was founded by brothers Eric Goode and Christopher Goode, Shawn Hausman and Darius Azari. The brick building housing A...
Found on http://en.wikipedia.org/wiki/Area_(nightclub)
### AREA
(Cortical)
Found on http://www.aans.org/Media/Glossary-of-Terminology
### area
(a) Type: Term Pronunciation: ār′ē-ă, -ē Definitions: 1. Any circumscribed surface or space. 2. All of the part supplied by a given artery or nerve. 3. A part of an organ having a special function, as the motor area of the brain.
Found on http://www.medilexicon.com/medicaldictionary.php?t=6137
### Area
[LDS Church] An area is an administrative unit of The Church of Jesus Christ of Latter-day Saints (LDS Church), which typically is composed of multiple stakes and missions. Areas are the primary church administrative unit between individual stakes and the church as a whole. ==History== The areas as they now exist were formed in January 1984...
Found on http://en.wikipedia.org/wiki/Area_(LDS_Church)
### Area
Unit of measure of length times width expressed in square inches.
Found on http://www.areforum.org/up/GeneralStructures/JOIST%20AND%20STRUCTURAL%20GLO
### area
A measure of surface extension in two-dimensional space. 'Area' is the Latin word for a vacant piece of level ground and still carries this more everyday meaning. The French shortened form are denotes a square of land with a side length of 10 meters, i.e. an area of 100 square meters. A hectare is a...
Found on http://www.daviddarling.info/encyclopedia/A/area.html
### area
area, measure of the size of a surface region, usually expressed in units that are the square of linear units, e.g., square feet or square meters. In elementary geometry, formulas for the areas of the simple plane figures and the surface areas of simple solids are derived from the linear dimensions ... | 1,497 | 5,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.810396 |
http://poj.org/showmessage?message_id=49845 | 1,723,123,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00046.warc.gz | 23,761,290 | 2,990 | Online JudgeProblem SetAuthorsOnline ContestsUser
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## 求助 我HASH超时了...有没有高人指点一下...我用别的方法都过了...就HASH没过..
Posted by sasnzy at 2006-07-12 21:53:34 on Problem 2859
```#include<stdio.h>
#include<iostream>
using namespace std;
const int M=599981;
const int S=500000+1;
int HASH_TABLE[M];
int NEXT[S],X[S],Y[S];
int N,W,H;
void insert(int ind)
{
int HASH_IND=(((__int64)X[ind]*1000000000+Y[ind])%M+M)%M;
NEXT[ind]=HASH_TABLE[HASH_IND];
HASH_TABLE[HASH_IND]=ind;
}
bool find(int x,int y)
{
int HASH_IND=(((__int64)x*1000000000+y)%M+M)%M;
int ind=HASH_TABLE[HASH_IND];
while (ind)
{
if (X[ind]==x && Y[ind]==y) return true;
ind=NEXT[ind];
}
return false;
}
int main()
{
scanf("%d%d%d",&N,&W,&H);
int i,j;
for (i=1;i<=N;i++)
{
scanf("%d%d",&X[i],&Y[i]);
insert(i);
}
int tot=0;
for (i=1;i<=N;i++)
if (find(X[i]+W,Y[i]) && find(X[i],Y[i]+H) && find(X[i]+W,Y[i]+H))
tot++;
printf("%d\n",tot);
}```
Followed by: | 389 | 1,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-33 | latest | en | 0.224479 |
http://www.markedbyteachers.com/international-baccalaureate/biology/diffusion-and-cell-size.html | 1,600,579,623,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00266.warc.gz | 198,923,546 | 19,686 | • Join over 1.2 million students every month
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2
# Diffusion and cell size
Extracts from this document...
Introduction
Diffusion and cell size Side of cube / cm � 0.05 cm Surface area / cm2 � 0.05 cm2 Volume / cm3 � 0.05 cm3 Distance penetrated / cm � 0.05 cm Distance not penetrated / cm � 0.05 cm Volume not penetrated / cm3 � 0.05 cm3 % of the total volume not penetrated 1 6 1 6:1 1 � � � 2 24 8 3:1 0.5 1 1 12.5 3 54 27 2:1 0.5 2 8 29.6 Discussion questions and answers * List the agar cubes in order of size, from largest to smallest. List them in order of surface area - to - volume ratio, from largest to smallest ratio. How does these lists compare? From largest to smallest in order of size: 3 cm, 2 cm, 1 cm. ...read more.
Middle
This change was slightly stronger in the beaker containing the largest agar block. Thus, this gives an indication that phenolphthalein was diffusing out from the agar block and into the sodium hydroxide solution. * What happens to the surface area - to - volume ratio of a cell as the cell grows? When a cell increases in size the surface area - to - volume ratio will decrease in order to bring in essential materials more easily as well as carrying out waste more effectively. * When one cube-shaped cell divides into two equal parts, how does the volume of each small cell compare with the one large cell? Does the surface area change in the same proportion? Explain. When the cell divides, the volume will remain the same due to the new cell being an exact copy of the cell which it was divided from. ...read more.
Conclusion
* What does the surface area of the cube represent? The surface area of the cube represents the rate at which different materials enter or leave the cell. This includes essential nutrients entering the cell and waste products exiting the cell. A cell with a larger surface area per unit volume is able to move more materials in and out of the cell. From the table above it can be concluded that the agar block with 1 cm sides, have the largest surface area per unit volume, thus is able to transport a larger amount of materials in and out of the cell. * What does the volume of the cube represent? The volume of the cube implies the rate at which different materials are used or produced, such as the production of heat and waste and the rate of consuming resources. ?? ?? ?? ?? Nathalie Zheng IB2 Biology HL 20100924 ...read more.
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# Related International Baccalaureate Biology essays
1. ## Investigating the relationship between cell size and rate of diffusion
2.Pour in NaOH into 3 beakers, enough to cover each cube in the solution completely.Immerse all cubes in NaOH at the same time, and record the time.
2. ## Modelling Surface Area to Volume Ratio in Cells with Agar Cubes
blocks but when the surface area / volume ratio goes up, it will take less time for the hydrochloric acid to reach the centre of the cube. Variables Controlled Variables: Temperature - thermostatically controlled room Time - the length of time you leave the cubes in the HCL (6 minutes)
1. ## Surface Area to Volume Ratio Simulation Experiment. The comparison between surface area : volume ...
5mm � 5mm � 5mm = 125mm3 5. 2.5mm � 5mm � 5mm = 62.5mm3 Calculations for surface area to volume ratio: 1. 600 : 1000 = 0.6 : 1.0 2. 400 : 500 = 0.8 : 1.0 3. 250 : 250 = 1.0 : 1.0 4. 150 : 125 = 1.2 : 1.0 5.
2. ## Biology Lab - frequency of cell division in animal and plant cell
Class was divided into pairs and they each drew the diagrams of the process of mitosis of each stage. All students in the class recorded the data which was needed to calculate the field of view, drawing of fit and calculated the magnification.
1. ## Biology Cell biology
The time that it takes for the hydrolysis to fully complete the transferral conjunction with the temperature can be used to find the rate of reaction. From this is can be calculated at which temperature the enzyme amylase is most effective.
2. ## Investigating Surface Area to Volume Ratio concerning the rate of Diffusion
As the cell size increases, its volume increases faster than the surface area. The size of the agar cube played a significant role in this experiment because the larger the cube, the harder it was for the sodium hydroxide to diffuse.
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• Ideas and feedback to | 1,225 | 4,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-40 | latest | en | 0.894545 |
https://osgamers.com/faq/do-day-traders-use-math | 1,718,394,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00307.warc.gz | 393,171,860 | 32,245 | # Do day traders use math?
There is a lot of math involved in trading, but it is represented through charts with indicators and patterns from technical analysis. Consequently, traders need to develop their analytical skills so they can recognize trends and trends in the charts.
## What kind of math is used in day trading?
Mathematical Concepts for Stock Markets
Descriptive Statistics. Probability Theory. Linear Algebra.
## What kind of math do traders do?
Exceptional math skills
“In order to research the data, run tests, and implement the trade, you should understand a few different mathematical concepts.” This includes calculus, linear algebra, and differential equations, and probability and statistics.
## Do stock traders use math?
Mathematical analysis is applied to evaluate the trading methodologies practiced by traders to execute a trade transaction. In addition, probability theory is employed to appraise the utility of money management techniques.
But in every case, mathematics still has to be the basis of what they are doing. In fact, trading is entirely about math, and all successful traders understand probabilities and the trader's equation extremely well.
## How much math is involved in trading?
Does math really help in trading? Math is a tool you can use to improve trading, but it is not required for trading. Most traders use at least some simple math, such as selecting stocks in part by price/earnings ratio, or managing risk by volatility targeting and stop losses.
## What percentage of traders actually make it?
What percentage of day traders make money and how many fail? Approximately 1-20% of day traders make money day trading. Just a tiny fraction of day traders make any significant amount of money. That means that between 80 to 99% of them fail.
## How much can you make trading realistically?
You may be wondering “How Much You Can Earn from The Stock Market?”. Well, the earnings can go up to Rs. 1 lakh a month or even higher if you are skilled enough and your strategies are in place.
## How hard is day trading really?
Day trading is difficult to master. It requires time, skill, and discipline. Many who try it lose money, but the strategies and techniques described above may help you create a potentially profitable strategy.
## How realistic is day trading?
And day traders typically end up on the wrong side of a trade more often than not. A study found that traders who lose money account for anywhere between 72–80% of all day trades being made.
## Is day trading actually hard?
Day trading isn't easy, and there are several areas of complexity that require research for new day traders. If you decide to become a day trader, it's important to understand that day trading isn't a get-rich-quick scheme.
## What degrees do most stock traders use?
1. Finance. Most academic advisors recommend that students who want to become stockbrokers choose a degree in finance or financial services. Many schools offer bachelor's degree programs, and some schools offer master's degree programs in this field.
## What trades use the least math?
CDL Truck Driver – Little or no math required. There isn't more than basic math on a CDL license test. Licensed Massage Therapist – Only basic math is required to earn a massage therapy license. Nursing Assistant – Only one or two basic math courses are needed, and no advanced formulas are required.
## Can you really make a living off trading stocks?
Trading is often viewed as a high barrier-to-entry profession, but as long as you have both ambition and patience, you can trade for a living (even with little to no money). Trading can become a full-time career opportunity, a part-time opportunity, or just a way to generate supplemental income.
## What do day traders go to school for?
That said, an understanding of economics and finance is also extremely important, and degrees in business administration, finance, or economics can also be viable means upon which to launch a career. [ Many day traders are largely self-taught with a background in finance, economics, or mathematics.
## Do brokers need to be good at math?
Math formulas are an essential component to pass the exam and becoming a successful real estate broker or sales agent. Remember, practice makes perfect, so the more time you spend memorizing these formulas, the better off you will be.
## Is day trading a skill?
Day trader skills are the abilities that traders use on a day-to-day basis to be successful. A day trader is a finance professional who buys and sells securities like stocks each day to make a profit. Many of their skills involve being efficient and making accurate financial judgments.
Most independent day traders have short days, working two to five hours per day. Often they will practice making simulated trades for several months before beginning to make live trades. They track their successes and failures versus the market, aiming to learn by experience.
## How many hours do day traders spend?
As a result, day traders typically work more than an average of eight hours. If you work as an independent day trader, this is also common. Depending on your position, you may not have an opportunity to take much time off from work, except for the weekends and holidays when the markets are closed.
## How much do you make daily day trading?
Day Traders in America make an average salary of \$116,895 per year or \$56 per hour. The top 10 percent makes over \$198,000 per year, while the bottom 10 percent under \$68,000 per year. What Am I Worth?
## Can you make 100k a year day trading?
The average trader will do between 60k and 100k, and underperformers will have so many position limits placed on their account, they are basically practicing and not making any money. These underperformers will likely remove themselves from the game because practicing does not pay the bills.
## Why do most day traders fail?
This brings us to the single biggest reason why most traders fail to make money when trading the stock market: lack of knowledge. We can also put poor education into this arena because while many seek to educate themselves, they look in all the wrong places and, therefore, end up gaining a poor education.
## How much can you make day trading with \$1000?
You need money to make money.
If you have a profitable trading system averaging 15% return a year: \$1000 account will make you \$150. \$10,000 account will make you \$1500. \$100,000 account will make you \$15,000.
## Is it possible to make \$100 a day day trading?
You're really probably going to need closer to 4,000 or \$5,000 in order to make that \$100 a day consistently. And ultimately it's going to be a couple of trades a week where you total \$500 a week, so it's going to take a little bit more work.
## What percentage of day traders are successful?
Day traders are more likely to experience a 50% loss than a 50% gain. While there is potential for large gains, there is also a significant chance of significant losses. This is an important point to consider for anyone considering day trading as an investment strategy. Only 3% of day traders make consistent profits.
Previous question | 1,476 | 7,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.947583 |
https://metanumbers.com/10255 | 1,601,389,847,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401643509.96/warc/CC-MAIN-20200929123413-20200929153413-00140.warc.gz | 509,872,206 | 7,468 | ## 10255
10,255 (ten thousand two hundred fifty-five) is an odd five-digits composite number following 10254 and preceding 10256. In scientific notation, it is written as 1.0255 × 104. The sum of its digits is 13. It has a total of 3 prime factors and 8 positive divisors. There are 7,008 positive integers (up to 10255) that are relatively prime to 10255.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 10 thousand 255 ten thousand two hundred fifty-five
## Notation
Scientific notation 1.0255 × 104 10.255 × 103
## Prime Factorization of 10255
Prime Factorization 5 × 7 × 293
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 10255 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 10,255 is 5 × 7 × 293. Since it has a total of 3 prime factors, 10,255 is a composite number.
## Divisors of 10255
1, 5, 7, 35, 293, 1465, 2051, 10255
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 14112 Sum of all the positive divisors of n s(n) 3857 Sum of the proper positive divisors of n A(n) 1764 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 101.267 Returns the nth root of the product of n divisors H(n) 5.81349 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 10,255 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 10,255) is 14,112, the average is 1,764.
## Other Arithmetic Functions (n = 10255)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7008 Total number of positive integers not greater than n that are coprime to n λ(n) 876 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1258 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 7,008 positive integers (less than 10,255) that are coprime with 10,255. And there are approximately 1,258 prime numbers less than or equal to 10,255.
## Divisibility of 10255
m n mod m 2 3 4 5 6 7 8 9 1 1 3 0 1 0 7 4
The number 10,255 is divisible by 5 and 7.
## Classification of 10255
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (10255)
Base System Value
2 Binary 10100000001111
3 Ternary 112001211
4 Quaternary 2200033
5 Quinary 312010
6 Senary 115251
8 Octal 24017
10 Decimal 10255
12 Duodecimal 5b27
20 Vigesimal 15cf
36 Base36 7wv
## Basic calculations (n = 10255)
### Multiplication
n×i
n×2 20510 30765 41020 51275
### Division
ni
n⁄2 5127.5 3418.33 2563.75 2051
### Exponentiation
ni
n2 105165025 1078467331375 11059682483250625 113417043865735159375
### Nth Root
i√n
2√n 101.267 21.7259 10.0631 6.34143
## 10255 as geometric shapes
### Circle
Diameter 20510 64434.1 3.30386e+08
### Sphere
Volume 4.51747e+12 1.32154e+09 64434.1
### Square
Length = n
Perimeter 41020 1.05165e+08 14502.8
### Cube
Length = n
Surface area 6.3099e+08 1.07847e+12 17762.2
### Equilateral Triangle
Length = n
Perimeter 30765 4.55378e+07 8881.09
### Triangular Pyramid
Length = n
Surface area 1.82151e+08 1.27099e+11 8373.17
## Cryptographic Hash Functions
md5 d093e56ef752adba41e5eaf26d1960a7 fe5050d47d20dffd981bc54427c57ec1202f3d8c 226674fce2b953bc12504ef9342c7b003a945a52a1507a471dc5eec057febb4d 133cd85fb9551db83948a8ea31b97d3584736f58d21180f168687773b8d4676c72b627ac50e5b68f7d712247b96a7e3003e41bc7f17d637e68d9a0af0bf9036d bfee725d351395bf8617f312ccdec78518decdd7 | 1,458 | 4,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-40 | longest | en | 0.819075 |
http://mathrec.org/old/2003mar/index.html | 1,579,872,150,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250620381.59/warc/CC-MAIN-20200124130719-20200124155719-00547.warc.gz | 107,102,070 | 2,042 | ## Mathematical Recreations
MathRec Home Solutions to this Problem Last Month's Topic Introductory Topic Older Topicsand Links Educational Resources Steve's Personal Page
This topic in game theory can be solved using only algebra. I performed a complete solution analytically, then used a computer program to check my analytical results.
#### High Card Wins
The following game is a variation of a puzzle that I found in Michael Shackleford's Math Problems site. Problem 60 reads as follows
Computer helpful. Two players are each dealt a card face down. Each player may look at his own card. The highest card wins. Cards are valued as in poker with aces being low. The first player may either keep his card or switch with the second player. The second player may keep his card, whether it be his original card or one that the first player gave him after switching, or trade it with the next card on the deck, which is also face down. The loser pays the winner \$1 and if both cards are equal then no money exchanges hands. For the sake of simplicity assume an infinite number of decks. Both players are infinitely logical. At what point should the first player switch? At what point should the second player switch if the first player doesn't switch? What is the expected gain of the first player?
The game that I want to consider is only slightly different. Rather than have all ties end in a draw, the second player wins all ties, with just one exception. If the first player elects to trade cards and they turn out to be equal, then player 2 must draw the third card in an attempt to break the tie. This slight rule change makes a profound difference in the solution, and that difference is the entire point of this topic.
Send all responses to .
Thanks,
Steve | 365 | 1,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-05 | latest | en | 0.971179 |
https://codereview.stackexchange.com/questions/59843/sorting-a-collection | 1,713,426,308,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817200.22/warc/CC-MAIN-20240418061950-20240418091950-00555.warc.gz | 160,603,119 | 44,746 | # Sorting a collection
Related to, but not exactly a follow up of this question. After fixing some issues discovered in the last review, I added a little more functionality to the Enumerable class. The problem is, I've never sorted before. I tried (and failed)to implement a few of the standard algorithms before coming up with this. It's not very efficient. It has to check to see if the collection IsSorted and just keeps looping until it is.
• Is there a way to make this more efficient without using a more advanced algorithm?
• What would be a simple to understand algorithm that is more efficient than this?
I realize that I could simplify some logic if I created some interfaces, but I would like to work with "built in" collections without wrapping everything in a class that implements an interface.
Bonus points to anyone who can tell me what algorithm I ended up using. I just kept testing until it worked.
Public Function Sort(collectionObject As Collection) As Collection
Dim item As Variant
Dim innerItem As Variant
Dim i As Long
Dim j As Long
Dim index As Long
Do Until IsSorted(collectionObject)
For i = 1 To collectionObject.Count
index = i
If IsObject(collectionObject(i)) Then
Set item = collectionObject(i)
Else
item = collectionObject(i)
End If
For j = i To collectionObject.Count
If IsObject(collectionObject(j)) Then
Set innerItem = collectionObject(j)
Else
innerItem = collectionObject(j)
End If
If item > innerItem Then
collectionObject.Remove index
index = j
End If
Next j
Next i
Loop
End Function
Private Function IsSorted(collectionObject As Collection) As Boolean
Dim item As Variant
Dim previous As Variant
For Each item In collectionObject
If item < previous Then
IsSorted = False
Exit Function
End If
If IsObject(item) Then
Set previous = item
Else
previous = item
End If
Next item
IsSorted = True
End Function
• You can't define the > or < operators operators for any custom class in VBA, nor can I think of any objects where those operators are already defined. If that is true then whenever IsObject is true you will get a error when comparing the objects. Aug 13, 2014 at 14:08
• Not exactly true @ptwales. It will throw runtime error 438 "Object does not support method" only if the object does not have a Default Property. Aug 13, 2014 at 14:25
• @ptwales maybe you could mention in an answer that I should handle that error and raise a better one nudge nudge. Aug 13, 2014 at 14:36
• patience, young grasshopper Aug 13, 2014 at 15:28
• Seems like I am late for the party but hey - why not implement an imitation of IComparer along with sorting?
– user28366
Aug 21, 2014 at 21:26
# FEAR
This is a red flag.
Do Until IsSorted(collectionObject)
' Sorting algorithm
Loop
Your algorithm should return a sorted sequence and should not need to be checked. If it fails the check you should rewrite your sorting algorithm and not simply try to do it again.
## Abstracting Methods
This bit of code is used often enough to merit it's own sub routine
Sub AssignUnknown(ByRef dest As Variant, ByRef src As Variant)
If IsObject(src) Then
Set dest = src
Else
dest = src
End If
End Sub
Now your code looks so much better! It looks like a bubble sort but
For i = 1 To collectionObject.count
index = i
AssignUnknown item, collectionObject(i)
For j = i To collectionObject.count
AssignUnknown innerItem, collectionObject(j)
If item > innerItem Then
collectionObject.Remove index
index = j
End If
Next j
Next i
## Algorithm
it doesn't work, hence you needing to loop until it is sorted. You are only a few steps away from true bubble sort though, which is IMO simpler.
For i = collectionObject.count To 2 Step -1
' hasSwapped = False
For j = 1 To i - 1
If collectionObject(j) > collectionObject(j + 1) Then
collectionObject.Add collectionObject(j), After:=j + 1
collectionObject.Remove j
' hasSwapped = true
End If
Next j
' If Not hasSwapped Then goto sorted ' Exit For
Next i
sorted:
In short, instead of bubbling up the same object each time, the BubbleSort drops the current item for the next one when it finds a larger item. This means after each inner loop the item at position i is in the correct place, which is why the outer loop is counting down not up.
hasSwapped uncommented allows for early exit if the sequence is already sorted. For more information on the bubble sort, wikipedia has a great article. I also found this site with some cool animations while trying to figure out your sorting algo.
Also note that AssignUnknown is no longer needed.
## Error Handling
As we discussed in the comments, the comparison operators will not be defined for objects that do not have a default property. It can't be fixed, but we can raise a more descriptive error
On Error Goto no_default_property
' bubble sort
sorted:
Exit Function
no_default_property:
If Err.Number = 438 Then ' preferably use the vb constant that I don't know
Err.Clear
Err.Raise 438, "Sort", "An item in the collection does not have a default property"
End If
End Function
## After Thoughts
Consider writing CollectionToArray and ArrayToCollection functions so you don't need to duplicate sorting methods for Collections and Arrays. Also consider a Sorted function that returns a sorted copy.
Sub Sorted(collec As Collection) As Collection
Set Sorted = collec
Sort Sorted
End Sub
I was tempted to insist you use a Swap function like this.
Sub Swap(ByRef a As Variant, ByRef b As Variant)
Dim t as Variant
t = a
a = b
b = t
End Sub
However, you are supporting objects in your collection, you will need to make that routine SwapUnknown that uses AssignUnknown. You could avoid calling IsObject(a) twice but I prefer the simpler solution.
Sub SwapUnkown(ByRef a As Variant, ByRef b As Variant)
Dim t as Variant
AssignUnknown t, a
AssignUnknown a, b
AssignUnknown b, t
End Sub
I didn't include this because:
• Your current swap method could be faster depending on how Collection is implemented
• Abstracting your current method seems pointless.
• It doesn't require AssignUnkown, which cuts out some ugly code.
• You should use a Sub instead of a Function as you don't return anything.
• Your inner loop
For j = i To collectionObject.Count
should start at j = i + 1
• As the Count property of the collectionObject is accessed often, you should introduce a new variable to store the value once.
• A null check for the passed parameter should be added.
Logic
Looking only at the Sort method the collection is not sorted when item > innerItem evaluates to true. So let us introduce a Boolean which will save this state and let us change the Loop from a Do Until()..Loop to a Do .. Loop Until().
Refactoring
Public Sub Sort(collectionObject As Collection)
If IsNull(collectionObject) Then
Exit Sub
End If
Dim item As Variant
Dim innerItem As Variant
Dim i As Long
Dim j As Long
Dim index As Long
Dim hasSwapped As Boolean
Dim collectionCount As Long
collectionCount = collectionObject.Count
Do
hasSwapped = False
For i = 1 To collectionCount
index = i
If IsObject(collectionObject(i)) Then
Set item = collectionObject(i)
Else
item = collectionObject(i)
End If
For j = i + 1 To collectionCount
If IsObject(collectionObject(j)) Then
Set innerItem = collectionObject(j)
Else
innerItem = collectionObject(j)
End If
If item > innerItem Then | 1,736 | 7,334 | {"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} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.804204 |
https://physicsgurukul.com/2018/09/09/neet-phy-kinematics-graphical-questions/ | 1,695,451,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00756.warc.gz | 522,234,692 | 97,478 | # NEET Physics Kinematics (Graphical Questions)
Home » NEET Physics » NEET Physics Kinematics (Graphical Questions)
Here we are providing NEET Physics Graphical Questions for Kinematics. Most of the time students gets confused while solving graphical questions based on kinematics. we have provided enough problems here so that students can master the concepts of graph. This will help students to understand the concept well.
## NEET Physics Kinematics (Graphical Questions)
Question 1:
Solution:
Question 2:
Solution:
(a) The slope of displacement-time graph goes on decreasing, it means the velocity is decreasing i.e. It’s motion is retarded and finally slope becomes zero i.e. particle stops.
Question 3:
Solution:
(d) In the positive region the velocity decreases linearly (during rise) and in the negative region velocity increases linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in the negative region.
Question 4:
Solution:
(b) Region OA shows that graph bending toward time axis e. acceleration is negative.
Region AB shows that graph is parallel to time axis i.e. velocity is zero. Hence acceleration is zero.
Region BC shows that graph is bending towards displacement axis i.e. acceleration is positive.
Region CD shows that graph having constant slope i.e. velocity is constant. Hence acceleration is zero.
Question 5:
Solution:
Question 6:
Solution:
(c) In part cd displacement-time graph shows constant slope e. velocity is constant. It means no acceleration or no force is acting on the body.
Question 7:
Solution:
Question 8:
Solution:
Question 9:
Solution:
Question 10:
Solution:
Question 11:
Solution: (c)
Question 12:
Solution: (b)
Question 13:
Solution:
Question 14:
Solution:
Question 15:
Solution:
Question 16:
Solution:
Question 17:
Solution:
Question 18:
Solution:
Question 19:
Solution:
Question 20:
Solution:
Question 21:
Solution: (d)
Question 22:
Solution:
Question 23:
Solution:
Question 24:
Solution:
Question 25:
Solution:
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## 1 thought on “NEET Physics Kinematics (Graphical Questions)”
We have started our Telegram Channel to provide PDF of study resources for Board, JEE, NEET and Foundation. Stay Tuned! Click below to join. | 557 | 2,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | latest | en | 0.891243 |
https://math.stackexchange.com/questions/846707/what-does-%E2%88%88-mean/846709 | 1,713,635,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00048.warc.gz | 341,505,613 | 37,582 | # What does "∈" mean?
I have started seeing the "∈" symbol in math. What exactly does it mean?
I have tried googling it but google takes the symbol out of the search.
• It means "belongs to". For instance $x\in A$ means that the element $x$ belongs to $A$. Jun 25, 2014 at 5:27
• By pure coincidence I was just searching how to input this symbol in mathjax! Jun 25, 2014 at 5:34
• You could try googling "mathematical symbols"...
– user147263
Jun 25, 2014 at 5:46
• Google doesn't support this at the moment, but Wikipedia does. Enter it in the search or put it in the URL. Take these examples: , and . Jun 25, 2014 at 9:47
• $8$ up votes for this :O
– user87543
Jun 25, 2014 at 11:41
$\in$ means '(is) an element of'
For instance, 'Let $a\in A$' means 'Let $a$ be an element of $A$'
• Being only a freshman going into Honors Algebra Two, I don't understand the concept. But thanks for the answer, even if I don't understand the concept at least I know what it means. Jun 25, 2014 at 5:31
• @Locke Look at the link I added too Jun 25, 2014 at 5:32
• Thanks for the link. The information on that Wiki really cleared it up for me. Jun 25, 2014 at 5:35
• An added note, since the wiki article does not mention it: one can look at ∈ meaning 'is' through the prism of sets representing the common property held by all its elements. e.g. Even = {0,2,4,..} and 'a ∈ Even' can be read as 'a is Even' Jun 25, 2014 at 11:37
(mathematics) means that it is an element in the set of… For eg...x ∈ ℕ denotes that x is within the set of natural numbers.
The relation "is an element of", also called set membership, is denoted by the symbol "∈". Writing {\displaystyle x\in A} x\in A means that "x is an element of A". Equivalent expressions are "x is a member of A", "x belongs to A", "x is in A" and "x lies in A". The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A".
Another possible notation for the same relation is {\displaystyle A\ni x,} A\ni x, meaning "A contains x", though it is used less often. The negation of set membership is denoted by the symbol "∉". Writing {\displaystyle x\notin A} x\notin A means that "x is not an element of A".
• "contains" and "lies in" are also a very bad words to use here, as it refers to inclusion, not set membership-- two very different ideas. Nov 10, 2019 at 3:21
$$\in$$ means "Element of".
For example, $$a \in A$$ means Element of: $$a$$ is in $$A$$.
A numeric example would be: $$\color{red}3 \in \{1, 2, \color{red}3, 4, 5\}$$. | 785 | 2,580 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-18 | latest | en | 0.954623 |
http://stackoverflow.com/questions/22600549/memoization-issue-on-a-tree-like-datastructure | 1,418,919,096,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802767247.82/warc/CC-MAIN-20141217075247-00069-ip-10-231-17-201.ec2.internal.warc.gz | 283,288,656 | 17,691 | # Memoization issue on a Tree-like datastructure
EDIT: while I'm still interested in an answer on the problems the execution faces in this case, it appears that it was indeed related to strictness since a `-O` fixes the execution and the program can handle the tree really quickly.
I'm currently working on the 67th problem of Project Euler.
I already solved it using simple lists and dynamic programming.
I'd like to solve it now using a tree datastructure (well, where a Node can have two parents so it's not really a tree). I thought I'd use a simple tree but would take care to craft it so that Nodes are shared when appropriate:
``````data Tree a = Leaf a | Node a (Tree a) (Tree a) deriving (Show, Eq)
``````
Solving the problem is then just a matter of going through the tree recursively:
``````calculate :: (Ord a, Num a) => Tree a => a
calculate (Node v l r) = v + (max (calculate l) (calculate r))
calculate (Leaf v) = v
``````
Obviously this has exponential time complexity though. So I tried to memoize the results with :
``````calculate :: (Ord a, Num a) => Tree a => a
calculate = memo go
where go (Node v l r) = v + (max (calculate l) (calculate r))
go (Leaf v) = v
``````
where `memo` comes from Stable Memo. Stable Memo is supposed to memoize based on whether or not it has seen the exact same arguments (as in, same in memory).
So I used ghc-vis to see if my tree was correctly sharing nodes to avoid recomputation of things already computed in another branch.
On the sample tree produced by my function as such: `lists2tree [[1], [2, 3], [4, 5, 6]]`, it returns the following correct sharing:
Here we can see that the node `5` is shared.
Yet it seems that my tree in the actual Euler Problem isn't getting memoized correctly. The code is available on github, but I guess that apart from the calculate method above, the only other important method is the one that creates the tree. Here it is:
``````lists2tree :: [[a]] -> Tree a
l2t :: [[a]] -> [Tree a]
l2t (xs:ys:zss) = l2n xs ts t
where (t:ts) = l2t (ys:zss)
l2t (x:[]) = l2l x
l2t [] = undefined
l2n :: [a] -> [Tree a] -> Tree a -> [Tree a]
l2n (x:xs) (y:ys) p = Node x p y:l2n xs ys y
l2n [] [] _ = []
l2n _ _ _ = undefined
l2l :: [a] -> [Tree a]
l2l = map (\l -> Leaf l)
``````
It basically goes through the list of lists two rows at a time and then creates nodes from bottom to top recursively.
What is wrong with this approach? I thought it might that the program will still produce a complete tree parse in thunks before getting to the leaves and hence before memoizing, avoiding all the benefits of memoization but I'm not sure it's the case. If it is, is there a way to fix it?
-
This doesn't really address the original question, but I find it is usually easier and more powerful to use explicit memoization.
I chose to store the triangle as a list indexed by a position rather than a tree:
``````[ ((1,1),3),
((2,1),7), ((2,2), 4),
....
``````
Suppose that part of the result has already been memoized in a list of this format. Then computing the answer at a particular coordinate is trivial:
``````a # i = let Just v = lookup i a in v
compute tree result (x,y) = tree # (x,y) + max (result # (x+1,y)) (result # (x+1,y+1))
``````
Now we must build `result`. This is also trivial; all we have to do is map `compute` over all valid indices.
``````euler67 :: [((Int, Int), Integer)] -> Integer
euler67 tree = result # (1,1)
where
xMax = maximum \$ map (fst . fst) tree
result = [ ((x,y), compute (x,y)) | x <- [1 .. xMax], y <- [1..x] ]
++ [ ((xMax + 1,y),0) | y <- [1..xMax + 1]]
compute (x,y) = tree # (x,y) + max (result # (x+1,y)) (result # (x+1,y+1))
``````
Computing height of the triangle (`xMax`) is just getting the maximum x-index. Of course we are assuming that the tree is well formed.
The only remotely complicated part is determining which indices are valid for `result`. Obviously we need 1 row for every row in the original tree. Row `x` will have `x` items. We also add an extra row of zeroes at the bottom - we could handle the base case in a special way in `compute` but it is probably easier this way.
You'll notice that is is quite slow for the hundred row triangle. This is because `lookup` is traversing three lists per call to `compute`. To speed it up I used arrays:
``````euler67' :: Array (Int, Int) Integer -> Integer
euler67' tree = result ! (1,1)
where
((xMin, yMin), (xMax, yMax)) = bounds tree
result = accumArray (+) 0 ((xMin, yMin), (xMax + 1, yMax + 1)) \$
[ ((x,y), compute (x,y)) | x <- [xMin .. xMax], y <- [yMin..x] ]
++ [ ((xMax + 1,y),0) | y <- [yMin..xMax + 1]]
compute (x,y) = tree ! (x,y) + max (result ! (x+1,y)) (result ! (x+1,y+1))
``````
Also here is the code I used for reading the files:
``````readTree' :: String -> IO (Array (Int, Int) Integer) | 1,416 | 4,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-52 | latest | en | 0.943213 |
https://www.isoul.org/places-and-events/ | 1,713,962,656,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00103.warc.gz | 726,311,668 | 25,097 | # Places and events
A place is a location within space. An event is a location within time. Past, present, and future are adjectives that modify place or event. We may speak of places in the past or of events in the past. Here indicates the present place. Now indicates the present event. The locus of events in the past are called the past. The locus of events in the future are called the future. Note that every physical body has its own past, present, and future.
We can turn all that around: Space is the locus of places, and time is the locus of events. Usually, space means the locus of places one could have been to or could go to, and time is the locus of events one could have experienced or could experience. In that sense, space and time are potentials. The here and now are actual. Space and time are the context for places and events. In particular, space and time are the context for here and now.
The order of places is a 3D ordering since three independent components are needed to locate each place. The order of events is a 3D ordering since three independent components are needed to locate each event. A spherical wave front is dimensionally symmetric so it is effectively 1D in space and time. If a spherical wave front is used as a standard measure for space or time, then such space or time is reduced to 1D. Time is commonly treated this way but it is also possible to treat space as 1D.
Displacement is the difference between two places and dischronment is the difference between two events; they are expressed as vectors between the two places or events. Distance is the length of a displacement. Duration is the length of a dischronment. The length along a path of motion in space is called arc length or travel distance. The length along a path of motion in time is called arc time or travel time. | 392 | 1,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.975343 |
http://www.tribology.me.uk/georgep1.html | 1,695,406,137,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00109.warc.gz | 76,969,476 | 2,211 | Paper: Generalized theory of gearing and elastohydrodynamic lubrication of spur gears.
Authors: Nikas, G. K., Costopoulos, T.
Published in: Proceedings of the International Congress - Gear Transmissions 95, 26-28 September 1995, Sofia, Bulgaria, vol. 1, pp. 118-123.
Abstract
The Generalized Theory of Gearing (GTG - developed by T. Costopoulos) is a method that relates the geometrical characteristics of production spur gears with the geometry of the rack that "produced" them. From the geometry of the rack, one can calculate all geometrical parameters of the end spur gears and can, thus, evaluate various aspects of their performance before the gears are physically produced.
In this paper, the GTG is used to examine the elastohydrodynamic lubrication of spur gears with involute external teeth, used in power transmissions and gear pumps. A forward iterative numerical method is used to solve the steady-state, smooth, Newtonian and isothermal EHL problem of two engaged spur-gear teeth. The equivalent radii of curvature of the teeth as well as their surface tangential velocities, which are needed in the solution of the EHL problem, are calculated through the GTG. In this way, the effectiveness of the lubrication is directly related to the geometrical characteristics of the rack and can be checked against different geometries and even flank errors of the rack.
A figure from this work
The following figure shows an example of the distribution of the maximum elastohydrodynamic pressure and corresponding minimum film thickness at 20 points on the teeth flanks of involute-type gears, using a steady-state, isothermal and Newtonian elastohydrodynamic analysis. Parameter Y1 is the distance from the pitch point on the line of contact. Notice that the maximum "maximum pressure" and corresponding minimum "minimum film thickness" appear close to but not on the pitch point (point Y1 = 0). This might correlate with experimental observations of pitting damage appearing slightly offset to the pitch point.
Maximum elastohydrodynamic pressure and minimum film thickness at 20 points on the teeth flanks of involute-type gears.
Homepage of Dr Nikas | 452 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-40 | latest | en | 0.889343 |
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Extended Essay Rules - BE SURE TO READ THESE before starting a thread!
1. Just to make it clear - we will not give you a topic for your EE, but we will help you refine and expand on your original ideas for a topic. You are marked on your choice of topic so we can't help you with that.
2. Before doing anything, read the Extended Essay for Dummies thread. At the very least, we recommend you take a look at the 5 Cardinal Rules of the EE!
3. Before asking anything about format or requirements for the extended essay, we suggest you read the IB Extended Essay Guide first (which you can download from here).
4. Make your thread titles descriptive. At the minimum, mention the subject area and if you have one, the topic! For instance "History EE on Nazi Propaganda" is better than "EE help please". The people who know about history and nazi propaganda will click on the first thread and probably miss the second one. Help us to help you - give it an accurate title!
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• ### Posts
• From the Arrhenius equation, k = A exp (-Ea/(RT)), taking ln of both sides ln k = ln A - Ea/(RT) = -(Ea/R) T + ln A, which is the equation of a line with slope -Ea/R and vertical intercept of ln A. So if you know the slope you can find the activation energy k is the rate constant such that the "rate of change of [H2O2]" = Δ[H2O2]/Δt = k [H2O2] (first order rate law). By unit analysis, k has dimension of inverse time. Hope that helps. There should also be videos online of teachers walking through the data processing.
• So I had an idea for a RQ for my History EE, but I'm not sure what question i should use. I know I want to do something about the US Civil War and the role that civic rights played in it. So for now my question is : What role did civil rights play during the different stages of the American Civil War? I'm not sure if the question is too precise or too vague, so if possible I would like certain guidance concerning the RQ. And if u have any better ideas for a EE concerning the Civil War, please let me know
• Hello! Thanks for your great feedback and congratulations for such an amazing final score | 643 | 2,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-26 | latest | en | 0.909942 |
http://betterlesson.com/lesson/resource/2532015/juice-bar-drawing-jpg | 1,477,359,951,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719843.44/warc/CC-MAIN-20161020183839-00489-ip-10-171-6-4.ec2.internal.warc.gz | 30,559,272 | 24,270 | ## Juice Bar Drawing.JPG - Section 1: Opening Activity
Juice Bar Drawing.JPG
# Juice Bar Problem Solving
Unit 2: Measuring Capacity
Lesson 9 of 9
## Big Idea: Mathematically proficient students can apply their understanding of measurement concepts to solve problems arising in everyday life.
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Topic: Assign a value to a specific matrix location
Replies: 4 Last Post: Dec 6, 2012 10:02 AM
Messages: [ Previous | Next ]
Matt Posts: 75 Registered: 4/23/12
Assign a value to a specific matrix location
Posted: Dec 5, 2012 1:16 PM
Hello !
I have a small problem here. I have a matrix of this type :
-------------------
A=[1, 2, 3, 4]
-------------------
And basically what I want to do is this :
-------------------
A(1)=4;
-------------------
So I can modify the first entry (for example). My question is how can I do this with an assignin function ? I was thinking that this may work, but it doesn't :
-------------------
assignin('base', 'A(1)', 6);
-------------------
I need to use that assignin function in my program.
Any advice on how that case could be solved ?
Cheers :-)
Date Subject Author
12/5/12 Matt
12/5/12 dpb
12/5/12 Matt
12/5/12 dpb
12/6/12 Steven Lord | 279 | 1,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-09 | latest | en | 0.858669 |
https://www.onlineclothingstudy.com/2012/12/how-to-use-4-point-system-for-knits.html | 1,721,534,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00680.warc.gz | 780,114,864 | 38,060 | # How to use 4 Point System for Knits Fabric Inspection?
4 point system for fabric inspection is a standard method that is widely used in Textile and Apparel Industries. In this system, fabric quality is evaluated by penalty points per 100 square yards.
Formula: Points / 100 sq. yd. = (Total points in roll * 36 * 100)/ (Total length in yards * width in inches)
So, to evaluate fabric quality you need to know fabric width and length of the inspected roll or thān. In case of woven fabric, length of the fabric roll is written on the Fabric roll tags. But knits fabrics are sourced in weight and supplier does not mention length.
That's why when it comes to use 4 point system in knits fabric, inspector finds it difficult to use given formula for calculating penalty points. To use the 4 point system for fabric inspection you need to calculate length of the fabric in a roll/thān.
Follow standard method of 4 point system to mark defects, calculate total penalty points in a than according standard method explained in How to 4 point system in Fabric Inspection?
### Evaluate fabric quality of Knits in 4 point system:
Step#1: Measure fabric length in a roll /thān
Generally you will get following specification in knitted fabric roll or find these
• Weight of the Than (in Kgs)
• Fabric weight Grams per square meter (GSM) and
• Fabric width (inches) Note: always take open width for knit fabric
For example, fabric weight of a than is 50 kgs, fabric GSM 250 and fabric width (W) 46 inches (open width).
Use following formula to know approx length of the roll/thān with above information.
Fabric length (in meters) = Fabric weight * 1000 /(Fabric GSM *Fabric Width * 0.0254)
Using data from above example
Fabric length (L) = (50*1000)/(250*46*0.0254) = 171.17 Meters
Step#2: Convert fabric length into yards.
Multiplying factor is 1.09361
Fabric length in yards = 171.17* 1.09361 = 187.2
Step#3: Calculate total penalty points.
Suppose you have given total 34 penalty points to the above roll. Once you find the length of each fabric roll/ than use the above formula to calculate points per 100 square yards.
Point /100 square yards = (Total penalty points * 36 * 100)/(roll length * Width in inches)
= (34*36*100)/(187.2 * 46)
= 14.21
You are done.
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A224331 Number of idempotent n X n 0..6 matrices of rank n-1. 1
1, 26, 291, 2740, 24005, 201678, 1647079, 13176680, 103766409, 807072130, 6214455467, 47455841820, 359873467213, 2712892291382, 20346692185455, 151921968318160, 1129919639366417, 8374698503539434, 61879716720597043 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Column 6 of A224333. LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 Index entries for linear recurrences with constant coefficients, signature (16,-78,112,-49). FORMULA a(n) = n*(2*7^(n-1)-1). a(n) = 16*a(n-1) - 78*a(n-2) + 112*a(n-3) - 49*a(n-4). G.f.: x*(1 + 10*x - 47*x^2) / ((1 - x)^2*(1 - 7*x)^2). - Colin Barker, Aug 29 2018 EXAMPLE Some solutions for n=3: ..1..5..0....0..3..6....1..0..0....0..0..0....1..0..0....1..0..0....0..0..0 ..0..0..0....0..1..0....0..1..0....6..1..0....0..1..0....0..1..0....2..1..0 ..0..2..1....0..0..1....0..0..0....4..0..1....1..6..0....3..4..0....3..0..1 MATHEMATICA Table[n*(2*7^(n-1)-1), {n, 1, 40}] (* or *) CoefficientList[Series[(1 + 10*x - 47*x^2) / ((1 - x)^2*(1 - 7*x)^2) , {x, 0, 40}], x] (* Stefano Spezia, Aug 29 2018 *) PROG (PARI) Vec(x*(1 + 10*x - 47*x^2) / ((1 - x)^2*(1 - 7*x)^2) + O(x^40)) \\ Colin Barker, Aug 29 2018 CROSSREFS Cf. A224333. Sequence in context: A336732 A227332 A020925 * A125414 A208600 A010831 Adjacent sequences: A224328 A224329 A224330 * A224332 A224333 A224334 KEYWORD nonn,easy AUTHOR R. H. Hardin, formula via M. F. Hasler William J. Keith and Rob Pratt in the Sequence Fans Mailing List, Apr 03 2013 STATUS approved
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# Spreadsheet Project Identify - A couple of months ago my friends and me formed a band called "MISK". As we are new to the field of accountancy, we are finding it hard to keep track of our spendings and savings.
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Introduction
Spreadsheet Project Identify A couple of months ago my friends and me formed a band called "MISK". As we are new to the field of accountancy, we are finding it hard to keep track of our spendings and savings. We also find it hard and time consuming to evenly share out the money between different funds (we have different funds for different things e.g. Drums, Amps etc and they all need different amounts of money) and ideally we need to share out the money proportionally between each fund. So we have all decided that we need something to help us easily manage our finances. I am therefore going to design and make a spreadsheet that will help us keep track and organize any of the little money that we are able to save, and how we should spend it. As our needs and funds will be continuously changing, a spreadsheet is the best solution as it is able to change and react when information is inserted and is able to perform complex equations automatically. A lot of data is to be stored, so a database could be used, but a spreadsheet package is better because complicated calculations will be performed on some of the input data and this can be done on a spreadsheet but not on a database. ...read more.
Middle
. To make it easier to keep track of our spendings and savings on the next page we have included an "Amount of money Raised/Spent" column as shown below. To use the Rating Bar, First you type in the Artist and Name of the song and then press the up and down buttons to change the rating out of 10. This can be changed as often as you want but you have to save for the Spreadsheet to remember. Processing Making increasing and decreasing buttons Go into the "Forms Tool Bar" and select the "Spinner" options. Then select the area that you want the Spinner to be. Now the Spinner should have appeared in the area that you selected. Right click on the Spinner and press the option, now this Box should appear on the screen. This kind of button is also used on the Rating Column on the second sheet using the same method, but changing the incremental change to 1 and the min to 0 and the max to 10, so it only shows a mark out of 10. The Amount Raised Column To make sure the system works I will get a member of the band to test that every aspect of the spreadsheet works. I will then personally check myself to see if there is anything that I missed. ...read more.
Conclusion
2. Be easy and clear to use - Completed - Band could all use it and explanatory as it has comments to explain how to use. 3. Work out how much money we have and need - Completed - Calculator page evidence. 4. Automatically work out how much each fund has and show how much more we need until the target amount is reached - Completed - Amount Raised/Amount Still Needed and status columns evidence. 5. Personalized - Completed - Pictures evidence 6. Small enough to be e-mailed yet hold all the information we need - Completed - Only 50.5kb (Small enough to fit on a floppy disk or be e-mailed) 7. Proportionally divide the money that we have between different funds depending on how much is needed - Completed - Amount Raised column evidence 8. Imput in Pounds only, as loose change will be used for Travel and Phone Calls - Completed - All entries to nearest Pound, no pence. I also think that there are ways in which I could improve my spreadsheet, I would like to make my spreadsheet automatically record the date and the amount of money credited/withdrawn. Another thing that I want to include are HyperLinks to Tabs for the individual (tabs are another word for music) so on the rating column you can click on the song name and the tab would open. I would also like to personalize it further by adding sounds and links to other files that may be useful. This could be included in a new version. ...read more.
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# Chapter 7 - Chapter 7 The Second and Third Law of Thermodynamics Spontaneous change Spontaneous changes(natural processes Heat flows from hot to
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Chapter 7: The Second and Third Law of Thermodynamics
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Spontaneous change Spontaneous changes (natural processes) Heat flows from hot to cold Gas expands in a vacuum A ball will stop bouncing Non-spontaneous changes (unnatural) The reverse of above
Carnot heat engine e system receives heat Q H from the hot reservoir and loses heat Q C to the low mperature reservoir, and does work on the surroundings U = q H + q C + w e second law says that qc cannot be zero, cannot convert q H completely into wo
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Carnot Cycle For the cycle A A Ucyc = 0 = I (q 1 + w 1 ) + II(w 2 ) + III(q 3 + w 3 ) + IV(w 4 ) I II III I V
Carnot Theorem No engine that operates between two heat reservoirs can be more efficient that the reversible Carnot engine. rev
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The birth of entropy (S) = + = + - = + ∈= 0 give to Rearange 0 1 1 3 1 1 3 1 3 C H H C H C T q T q T T q q T T q q
Entropy (S) It appears that there is a property (q/T) that sums to zero for the cycle. This
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## This note was uploaded on 12/06/2010 for the course CHEM 141 taught by Professor Freeman during the Spring '10 term at Columbia SC.
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Chapter 7 - Chapter 7 The Second and Third Law of Thermodynamics Spontaneous change Spontaneous changes(natural processes Heat flows from hot to
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Ask a homework question - tutors are online | 537 | 2,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.893092 |
https://numbermatics.com/n/1410280088982/ | 1,550,444,730,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482788.21/warc/CC-MAIN-20190217213235-20190217235235-00591.warc.gz | 658,557,454 | 9,410 | # 1410280088982
## 1,410,280,088,982 is an even composite number composed of five prime numbers multiplied together.
1410280088982 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of thirty-two divisors.
## Prime factorization of 1410280088982:
### 2 × 3 × 157 × 199 × 7523179
See below for interesting mathematical facts about the number 1410280088982 from the Numbermatics database.
### Names of 1410280088982
• Cardinal: 1410280088982 can be written as One trillion, four hundred ten billion, two hundred eighty million, eighty-eight thousand, nine hundred eighty-two.
### Scientific notation
• Scientific notation: 1.410280088982 × 1012
### Factors of 1410280088982
• Number of distinct prime factors ω(n): 5
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 7523540
### Divisors of 1410280088982
• Number of divisors d(n): 32
• Complete list of divisors:
• Sum of all divisors σ(n): 2852789856000
• Sum of proper divisors (its aliquot sum) s(n): 1442509767018
• 1410280088982 is an abundant number, because the sum of its proper divisors (1442509767018) is greater than itself. Its abundance is 32229678036
### Bases of 1410280088982
• Binary: 10100100001011011001111100110010110010110 2
### Squares and roots of 1410280088982
• 1410280088982 squared (14102800889822) is 1988889929379077837796324
• 1410280088982 cubed (14102800889823) is 2804891866580129589096503973512502168
• The square root of 1410280088982 is 1187552.1415845285
• The cube root of 1410280088982 is 11214.2041535083
### Scales and comparisons
How big is 1410280088982?
• 1,410,280,088,982 seconds is equal to 44,842 years, 28 weeks, 2 days, 5 hours, 9 minutes, 42 seconds.
• To count from 1 to 1,410,280,088,982 would take you about one million, twelve thousand, one hundred six years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000.
Note: we do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!
• A cube with a volume of 1410280088982 cubic inches would be around 934.5 feet tall.
### Recreational maths with 1410280088982
• 1410280088982 backwards is 2898800820141
• The number of decimal digits it has is: 13
• The sum of 1410280088982's digits is 51
• More coming soon!
The information we have on file for 1410280088982 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 1410280088982, math, Factors of 1410280088982, curriculum, school, college, exams, university, STEM, science, technology, engineering, physics, economics, calculator. | 855 | 3,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-09 | latest | en | 0.786692 |
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## Economics
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Bob consumes two commodities: x and y (say, chocolate and classical music). More y never hurts, but in order to enjoy y he has to consume at least 4 units of it; put differently, he starts to enjoy y after the fourth unit. (In order to appreciate y he must first develop a taste for it.) Regardless of the amount of y he consumes, Bob thinks that more x is always strictly better than less. Moreover, he (strictly) prefers the bundle (x,y) over (x', y'); where both y and y' are at least 4, if and only if x *y > x'* y'.
Bob has an income of \$100. The price of x is \$1, and the price of y is py. For what values of py will Bob buy y, and for what values of py will Bob buy only x?
Microeconomics, Economics
• Category:- Microeconomics
• Reference No.:- M940425
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Create a provider database and related reports and queries to capture contact information for potential PC component pro | 1,045 | 4,992 | {"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} | 2.703125 | 3 | CC-MAIN-2018-17 | latest | en | 0.880777 |
http://math.stackexchange.com/questions/345187/cocktail-bar-problem/345298 | 1,469,633,242,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826908.63/warc/CC-MAIN-20160723071026-00319-ip-10-185-27-174.ec2.internal.warc.gz | 162,617,729 | 19,926 | # Cocktail bar problem
Recently I went out with friends and we asked ourselves the following question: Consider $n$ people sitting at a cocktail bar next to each other. How many rearrangements have to be made to ensure that every possible pair has sat at least once next to each other?
More precisely:
For given $n>1$, find a subset $A$ of $S_n$ with minimum cardinality such that for each $1\leq i<j\leq n$ there is a $\pi\in A$ such that $|\pi(i)-\pi(j)|=1$.
-
Counting only the number of permutations $A$ required to "cover" all the possible pairs with adjacent seats seems to make an easier problem than asking to minimize the number of swaps (transpositions) needed to accomplish the purpose. – hardmath Mar 29 '13 at 0:29
@hardmath: I'm not looking for the number of swaps that have to be made, I'm only interested in $A$ or its cardinality. I reformulated the question in order to make it more clear what I mean. – zero-divisor Mar 29 '13 at 0:42
Basically, you are given a complete graph, and you need to pick the minimum number of hamiltonian paths (or cycles, if the table is circular) which cover all the edges.
When there are an even number of people, it is a well known theorem that $K_{2n}$ can be decomposed into $n$ edge disjoint hamiltonian paths. (See Modern Graph Theory, page 16, for instance, or see a previous answer which has a snapshot of the book inlined).
When $n$ is odd, you can do even better, and make $\frac{n-1}{2}$ cycles!
So, if you want paths, for $2k$ answer is $k$ and $2k+1$ answer is $k+1$.
If you want cycles, the values are switched.
I believe those are optimal.
-
I don't see how those bounds can be improved, since the hamiltonian paths are disjoint. Isn't it exactly the minimum? – Vincent Tjeng Mar 29 '13 at 2:15
@VincentTjeng: Yes, exactly :) – Aryabhata Mar 29 '13 at 2:17
+1 for a very clear answer! – Vincent Tjeng Mar 29 '13 at 2:19
I understand the proof for the case of $n$ being even, and it is indeed the minimum as $|A|\geq \frac{n}{2}$ which Greg Martin pointed out in his answer. But in case of $n$ being odd, we need at least $\frac{n+1}{2}$ paths, and I don't see how to construct them from the $\frac{n-1}{2}$ cycles. Am I missing something trivial? – zero-divisor Mar 29 '13 at 8:28
Ok the construction described in the book works as well for the case of $n$ being odd, for which it yields $\frac{n+1}{2}$ paths where the $\frac{n-1}{2}$ pairs $\{1,n\},\{2,n-1\},\dots$ are covered exactly twice, all others being covered exactly once. – zero-divisor Mar 29 '13 at 13:28
Not a complete answer, but: when $n+1$ is prime, the answer is $n/2$.
Notice that $n/2$ is a lower bound for any $n$: each element of $A$ "checks off" at most $n-1$ new pairs, and there are $\binom n2 = n(n-1)/2$ pairs that need to be checked off in total.
To construct $n/2$ configurations that accomplish this when $n+1$ is prime: number the people from $1$ to $n$, and in the $k$th configuration, have person $i$ sit in seat $ik\pmod{n+1}$. We must prove, for any $1\le i<j\le n$, that there exists $1\le k\le n/2$ such that $ik\pmod{n+1}$ and $jk\pmod{n+1}$ differ by 1, that is, $ik-jk\equiv\pm1\pmod{n+1}$. The appropriate $k$ to choose is $k\equiv\pm(i-j)^{-1}\pmod{n+1}$, with the sign chosen so that the residue lies between $1$ and $n$ rather than between $n+1$ and $2n$.
-
very nice approach, thanks! – zero-divisor Mar 29 '13 at 9:17 | 1,025 | 3,393 | {"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} | 3.515625 | 4 | CC-MAIN-2016-30 | latest | en | 0.948545 |
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# What is the number 186 divisible by?
Wiki User
2013-07-10 04:07:29
186 is divisible by 1, 2, 3, 6, 31, 62, 93, 186.
Wiki User
2013-07-10 04:07:29
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Q: What is the number 186 divisible by? | 139 | 380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.789221 |
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# FMA III 4,5,7: Explain the parts of the business plan.
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4 . Explain the parts of the business plan. Your response should be at least 250 words in length. You are required to use at least your textbook as source material for your response. All sources used, including the textbook, must be referenced; paraphrased and quoted material must have accompanying citations. 5. Construct a pro forma income statement for the first year and second year for the following assumptions: • Units of Sales in Year 1: 100,000 • Price per Unit: \$10 • Variable cost per unit: 30% • Fixed Costs: \$120,000 • Income taxes: 15% • Interest Expense: \$200,000 • In year 2, Price per unit increases to \$11.50, and unit of sales increases by 3%, all other assumptions remain the same. 7. Calculate a table of interest rates based on the following information: • The pure interest rate is 1.6% • Inflation expectations for year 1 = 3%, year 2 =3.5%, years 3-5 =5% • The default risk is .1% for year one and increases by .2% over each year • Liquidity premium is 0 for year 1 and increases by .2% each year • Maturity risk premium is 0 for years 1 and 2 and .2% for years 3-5
CONTENT:
FMA IIIName:Institution:Date: 4. Explain the parts of the business plan. A business plan is a written formal statement that describes in detail how a business is going to achieve its goals. It provides a plan for reaching those goals and describes why those goals are believed to be attainable. It`s a carefully constructed guide for a person starting a business (Pride, Hughes, & Kapoor , 2012). In this paper parts of business plan are identified and discussed. According to Pride (Foundations of business), a business plan can be broadly divided into 12 parts. It starts with an introduction or a brief statement of the business plan. Then goes the executive summary, which provides a brief review of the information within the plan. It highlights the business plan high points. It identifies the industry and target markets, competitors and requested loan amount and the ideal repayment plan. Then there are benefits to the community outl...
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http://oeis.org/A088695 | 1,580,166,491,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251728207.68/warc/CC-MAIN-20200127205148-20200127235148-00411.warc.gz | 130,100,227 | 3,919 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A088695 E.g.f: A(x) = f(x*A(x)), where f(x) = exp(x+x^2). 0
1, 1, 5, 40, 485, 7776, 156457, 3788800, 107414505, 3491200000, 128019454541, 5229222395904, 235490648957005, 11592449531084800, 619331166211640625, 35691050995648823296, 2206955604752999720273, 145757527499874820423680, 10240455593560436925898645 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Radius of convergence of A(x): r = (1/2)*exp(-3/4) = 0.23618..., where A(r) = exp(3/4) and r = limit a(n)/a(n+1)*(n+1) as n->infinity. Radius of convergence is from a general formula yet unproved. LINKS FORMULA a(n) = n! * [x^n] exp(x+x^2)^(n+1)/(n+1). a(n) = n!*sum(k=floor(n/2)..n, binomial(k,n-k)*(n+1)^(k-1)/k!). - Vladimir Kruchinin, Aug 04 2011 a(n) ~ 2^(n+1/2) * n^(n-1) / (sqrt(3) * exp(n/4 - 3/4)). - Vaclav Kotesovec, Jan 24 2014 MATHEMATICA Table[n!*SeriesCoefficient[(E^(x+x^2))^(n+1)/(n+1), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jan 24 2014 *) PROG (PARI) a(n)=n!*polcoeff(exp(x+x^2)^(n+1)+x*O(x^n), n, x)/(n+1) CROSSREFS Sequence in context: A281160 A282476 A198247 * A145166 A113079 A211046 Adjacent sequences: A088692 A088693 A088694 * A088696 A088697 A088698 KEYWORD nonn AUTHOR Paul D. Hanna, Oct 07 2003 STATUS approved
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Last modified January 27 17:58 EST 2020. Contains 331296 sequences. (Running on oeis4.) | 658 | 1,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-05 | latest | en | 0.541789 |
https://uva.onlinejudge.org/board/viewtopic.php?f=41&t=24352 | 1,571,608,929,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00314.warc.gz | 775,748,886 | 10,505 | ## 11335 - Discrete Pursuit
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shakil
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### 11335 - Discrete Pursuit
Sorry.......
Last edited by shakil on Mon Nov 05, 2007 7:36 pm, edited 1 time in total.
SHAKIL
sclo
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Read the problem carefully. At time t=0, the only possible position of cop is (0,0) and velocity is (0,0). At time t=1, the possible velocity of the cop is (u,v) where -1<=u,v<=1 and the possible position is (x,y)=(0,0)+(u,v) where -1<=x,y<=1. For this problem it is sufficient to only know the maximum bounds on x,y,u,v.
Let x[t]=maximum bound of x at time t, then
x[t]=x[t-1]+t
deena sultana
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### need sample i/o
Can anyone give some sample i/o for 11335, plzzz? i am getting WA
sclo
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### Re: need sample i/o
deena sultana wrote:Can anyone give some sample i/o for 11335, plzzz? i am getting WA
It might give the problem away if I give any more input/output. You could easily write a bruteforce bfs program to check your results.
sohel
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### Re: need sample i/o
deena sultana wrote:Can anyone give some sample i/o for 11335, plzzz? i am getting WA
and also it would be more helpful if you describe your algorithm first!!
Hint: consider x and y component separately.
deena sultana
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well, my algorithm is like this...
1. for t=0, the object's position is (0,0) and u=0, v=0 and the thief's position is (a,0), as the problem states.
2. then for t=1 the object will move at the position from where the distance of the thief's current position (at t=1) is minimum;
3. repeat this process until the distance is 0.
wont it work?
sohel
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ds wrote:the object will move at the position from where the distance of the thief's current position (at t=1) is minimum
By distance, do you mean the Manhattan distance or the Euclidean distance?
And don't you take the current speed in consideration?
What if you have two places with the same minimum distance, which point do you consider then?
deena sultana
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ops sorry for my incomplete description
i've calculated the Manhattan distance, and also considered the current speed. but, in case of tie i 've chosen the 1st one :-S (may be this is the fault, no?)
sclo
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deena sultana wrote:ops sorry for my incomplete description
i've calculated the Manhattan distance, and also considered the current speed. but, in case of tie i 've chosen the 1st one :-S (may be this is the fault, no?)
Your algorithm is wrong. In any optimal solution, the direction of the cop doesn't change much. But in your algorithm, the cop can change direction.
deena sultana
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yes, i understand
it was a stupid algorithm
sorry
deena sultana
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Hint: consider x and y component separately.
Wow! this hint is simply great! who r trying to solve 11335, plz think about it and have a pretty solution
thanks to sohel. thanks to sclo too.
andmej
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Is this problem solvable using Dynamic Programming?
Runtime errors in Pascal are reported as Wrong Answers by the online judge. Be careful.
Are you dreaming right now?
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rio
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I don't think DP is suitable for this problem.
-----
Rio
slxst
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### Re: 11335 - Discrete Pursuit
I honestly don't know where to begin to attack this problem but it seems that this problem is really easy.
Could anyone give me a little hints about where to start?
By the way: I don't know what the other posters are meaning about x component and y component.
Thanks a lot.
andmej
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### Re: 11335 - Discrete Pursuit
Imagine you had to solve a simpler problem: The thief and the cop can only move along one axis. In that case, how would you calculate the minimum time needed for the cop to catch the thief in only that direction?
For example, imagine that the cop can't move along the y-axis (he can only move along the x-axis) and same for the thief. If thief is at position (0, a) and cop is at position (0, 0), how would the cop move to catch him?
Now, imagine the opposite thing along the y-axis (that is, neither the cop nor the thief can move along the x-axis), and compute the time.
The final answer will be the maximum of these two values, because the cop can advance in both directions simultaneously and he can "work" on the two solutions at the same time.
Hope this helps. If I'm not clear enough please tell me so I can explain better.
Runtime errors in Pascal are reported as Wrong Answers by the online judge. Be careful.
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https://www.wyzant.com/resources/answers/2575/what_is_the_most_important_step_to_remember_about_multiplying_a_three_digit_factor_by_a_one_digit_factor | 1,524,787,428,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00232.warc.gz | 924,473,714 | 16,027 | 0
# what is the most important step to remember about multiplying a three digit factor by a one digit factor?
my son is in 4th grade, it is a question for his math class
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This question is talking about multiplying, for example, 123 x 3.
The most important step is to remember to multiply all three digits by the one digit.
So 3 x 3 = 9, 20 x 3 = 60, and 100 x 3 = 300, so 123 x 3 = 369. | 180 | 618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-17 | longest | en | 0.868898 |
https://www.orchardtraining.org/single-post/2018/11/03/weekly-maths-problem | 1,716,060,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00687.warc.gz | 839,833,395 | 188,882 | top of page
# Weekly Maths Problem
Tommy wants to take part in the Orchard Training Charity Bike Event next year. Last month he felt that his bike was getting too small for him to ride with a fully straightened knee. He is hoping for Christmas that his parents will buy him a new bike. There is a great bike on sale in the Cycle Shop and it has multiple gears. Great for riding the South Downs and up the great big hill in the middle. He visits the shop on Saturday and tries the middle gear out first. There is a proportional relationship between the number of times Tommy peddles (we will call this x) and the number times the wheels rotate (we will call this y). When he turns the pedals 6 times the wheels rotate 12. Write this relation using x and y as an equation.
You can share your answers here on Facebook or with me as I visit during the week. Remember you need to show your workings out to get the maximum points. | 202 | 927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-22 | latest | en | 0.97427 |
https://petroleumoffice.com/formula/radiated-energy-flux/ | 1,717,009,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00504.warc.gz | 388,832,856 | 4,802 | ## Input(s)
c: Speed of Light $$(\mathrm{ft} / \mathrm{s})$$
$$\mathrm{h}$$: Planck's Constant $$\left(\mathrm{lb} \mathrm{ft}^{2} / \mathrm{s}\right)$$
$$\lambda$$: Wavelength (ft)
$$\mathrm{k}$$: Stefan-Boltzmann Constant $$\left(\mathrm{BTU} / \mathrm{h} \mathrm{ft}^{2}{ }^{\circ} \mathrm{R}^{4}\right)$$
T: Temperature $$(\mathrm{K})$$
## Output(s)
q: Radiated Energy Flux from a Black Surface in the Wavelength Range $$\left(B T U / \mathrm{ft}^{2} \mathrm{~s}^{3}\right)$$
## Formula(s)
$\mathrm{q}=\frac{2 * \pi *\left(\mathrm{c}^{2}\right) * \mathrm{~h}}{\left(\lambda^{5}\right) *\left(\exp \left(\frac{\mathrm{c} * \mathrm{~h}}{\lambda * \mathrm{k} * \mathrm{~T}}\right)-1\right)}$
## Reference(s)
Bird, R.B., Stewart, W.E. and Lightfoot, E.N. (2002). Transport Phenomena (Second Ed.). John Wiley & Sons, Chapter: 16, Page: 494.
An unhandled error has occurred. Reload 🗙 | 328 | 893 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.36501 |
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## Calculating Probabilities for Substitution Ciphers using Frequency Analysis
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Seems like "Crypto" provider has been removed in Android N. My application crashing because of NoSuchProviderException. If I change the provider and Algorithm then it will affect user who are all ... | 1,704 | 7,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | latest | en | 0.870149 |
http://electrical4pocket.com/capacitor-and-capacitance/ | 1,563,559,159,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526324.57/warc/CC-MAIN-20190719161034-20190719183034-00189.warc.gz | 56,821,156 | 12,228 | Home » BASIC FUNDA » CAPACITOR AND CAPACITANCE
Capacitor:
The capacitor is a passive device, which can store electric charge into its conducting surface (plates) when connected with a voltage source and discharge the same stored charge when voltage source is not connected.
Simply, a capacitor has a capacity to store electrical charge into it.
Construction: The capacitors consist of two or more conducting surfaces separated by some insulating material which is known as dielectric. That means if we arrange, two conducting plate, separated by the layer of dielectric medium, it will be able to store electric charge.
Working principle: When we connect two parallel plates with a voltage source, like one plate with the positive and other with negative of the voltage source, we will see that, plate connected with positive terminal of source, become positive and plate connected with negative terminal become negative, and there is no current flow through it as it is electrically separated by some insulating material.
After a certain time, (known as charging time) when two plates hold maximum charge, it works as charge storage device. Now if we disconnect the voltage source and connect a load, we will see that the capacitor works as a voltage source, as current flows through the load. This process will continue until the charges of the plate become zero, known as discharging time. Thus a capacitor acts as a rechargeable battery.
Through above discussion we have seen that the capacitor work as open circuit in case of d.c source.
But what actually happens when connected with a d.c. source?
When a parallel plate capacitor is connected with a d.c. source, i.e. plate A with positive terminal of battery and plate B with negative terminal of battery, at first with instant a current flows through the capacitor. Plate A is charged with positive charge or proton and plate B is charged with negative charge or electron. Now when connected with battery, with a flash the electrons withdrawn from plate A and goes to plate B, which shows the momentary flow of electrons from plate A to plate B. After that the potential difference established between two plates and no current can flow.So, the capacitor acts as sort momentarily at initial condition and then act as open circuit in case of d.c. source.
Capacitor connected with a.c source:
When we connect a capacitor with an a.c voltage source, we may see that a.c current flows through the capacitor. But what actually happens?
We know that, a.c voltage means, the voltage which changes with time periodically into positive and negative cycle. When such voltage apply to the capacitor plate, during positive half one plate come into contact with positive polarity stores positive charge, another plate store negative charge as it become touched with negative polarity. But at next moment i.e. the next half of cycle, the same plate which stores positive charge attached with negative polarity and store negative charge. Similarly the other plate which stores negative charge now stores positive charge and this process continues. By this way the capacitor plates charged and discharged alternatively in case of a.c. and act as sort circuit. And no current flows between two plates due to insulating material.
Capacitance:
Property by which, a capacitor can store electric charge is known as capacitance.
It is the ratio, of charge that can be stored and the applied voltage.
If Q coulomb of charge at one plate of capacitor by applying V volt,
Then Capacitance, C = Q/V = Charge/ Potential difference, coulomb/volt.
Why maximum available capacitors are rated in micro or Pico farad?
As we have seen that capacitance, C = Q/V = charge/ potential difference. That means 1 farad capacitance requires 1 coulomb charge by applying 1 volt between its plates, which is difficult in practical cases. So capacitors are rated in micro or Pico farad.
1µF = 10-6 F.
1nF = 10-9 F.
1pF = 10-12 F.
Capacitance of a parallel plate capacitor:
Now we consider two parallel plates 1 and 2 for making a parallel plate capacitor. The area of each plate is A m<2 and two plates are separated by dielectric material at a distance of d.
The capacitance is directly proportional to the area of cross section, i.e. A.
So, C α A .
Capacitance is also inversely proportional to the distance of parallel plate.
So, C α 1/d .
As per two equation,
C α A/d .
Or, capacitance, C = Є0 ЄrA/d farad………..in case of medium.
Where Є0 Єr = permittivity of dielectric medium.
Or, C = Є0A/d farad………..in case of air.
Actually the equation is C = Є0 Єr (N-1)A/d farad.
Where N = number of parallel plate.
As we discussed with two parallel plate, C = Є0 Єr(2-1)A/d farad.
So, in case of multi plate capacitor, C = Є0 Єr(N-1)A/d farad.
In case of cylindrical or single core cable, the capacitance, C = 2π Є0 Єr/2.3log10(r2/r1) farad.
In case of l meter length of a cable, C = 2π Є0 Єrl/2.3log10(r2/r1) farad.
In case of transmission line, for length of l meter, C = π Є0 Єr/2.3 log10(d/r) farad. Where d = distance from centre to centre of the wire and r = radius of each wire.
Energy stored in capacitor:
The capacitor store energy in the dielectric medium from external source like battery. That means it stores the energy from other source and releases it when discharging.
At first stage, when capacitor is in uncharged condition, connected to a battery, a little work is done for transferring charge from one plate to another. Then with every charge, a little work is done against the repulsive force of each plate. But as battery voltage is higher, it puts same charge on the plate and thus the charge on the plate of capacitor grows up.
The formula is W = ½ CV2 joules. If C is in farad and V is in volt. | 1,310 | 5,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-30 | latest | en | 0.949858 |
https://kerodon.net/tag/00ZN | 1,600,981,299,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400220495.39/warc/CC-MAIN-20200924194925-20200924224925-00301.warc.gz | 426,558,745 | 4,404 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
Proposition 3.3.7.1. Suppose we are given a commutative diagram of simplicial sets
$\xymatrix { X \ar [r]^-{u} \ar [d] & X' \ar [d] \\ S \ar [r]^-{v} & S', }$
where the vertical maps are Kan fibrations and $v$ is a weak homotopy equivalence. The following conditions are equivalent:
$(1)$
The morphism $u$ is a weak homotopy equivalence.
$(2)$
For every vertex $s \in S$, the induced map of fibers $u_{t}: X_{s} \rightarrow X'_{ v(s)}$ is a homotopy equivalence of Kan complexes.
Proof. Using Corollaries 3.3.6.8 and 3.3.6.5, we can replace $(1)$ and $(2)$ by the following assertions:
$(1')$
The morphism $\operatorname{Ex}^{\infty }(u): \operatorname{Ex}^{\infty }(X) \rightarrow \operatorname{Ex}^{\infty }(X')$ is a weak homotopy equivalence.
$(2')$
For every vertex $s \in S$, the induced map of fibers $u_{s}: \operatorname{Ex}^{\infty }(X)_{s} \rightarrow \operatorname{Ex}^{\infty }(X')_{v(s)}$ is a homotopy equivalence of Kan complexes.
The equivalence of $(1')$ and $(2')$ follows by applying Proposition 3.2.7.1 to the diagram
$\xymatrix@R =50pt@C=50pt{ \operatorname{Ex}^{\infty }(X) \ar [r]^-{\operatorname{Ex}^{\infty }(u) } \ar [d] & \operatorname{Ex}^{\infty }(X') \ar [d] \\ \operatorname{Ex}^{\infty }(S) \ar [r]^-{\operatorname{Ex}^{\infty }(v)} & \operatorname{Ex}^{\infty }(S'); }$
note that every simplicial set appearing in this diagram is a Kan complex (Proposition 3.3.6.9), the vertical maps are Kan fibrations (Proposition 3.3.6.6) and $\operatorname{Ex}^{\infty }(v)$ is a homotopy equivalence by virtue of Corollary 3.3.6.8. $\square$ | 582 | 1,632 | {"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} | 2.53125 | 3 | CC-MAIN-2020-40 | latest | en | 0.630491 |
https://www.awesomeworksheet.com/2023/07/identity-property-of-addition.html | 1,701,527,924,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00214.warc.gz | 727,439,365 | 19,599 | Identity Property Of Addition Worksheets 3Rd Grade. Web addition propertiesstudent identify the addition properties in each equation. This is a hands on activity that will help students learn how to identify, create, and solve equations using properties of addition.
Web identity property of addition. Some of the worksheets for this concept are addition properties 1, addition properties, name in. Web this worksheet allows students to practice working with the properties of addition using only basic numbers.
### For Example, 4 + 0 = 4 68 + 0 = 68 71 + 0 = 71 0 + 93 = 93 0 + 117 =.
This is a hands on activity that will help students learn how to identify, create, and solve equations using properties of addition. Web we have a fantastic selection of free properties worksheets that cover all the important concepts such as commutative property of addition, identity property of addition,. 2nd through 4th grades view pdf assoc., identity, & comm.
### Web I Created This Worksheet To Teach And Review The Addition Properties Required By Our State.
Web to resolve this worksheet, students should know how to identify and understand the identity property of addition. Worksheets are name in numbers 1 9 select the property that is being, multiplication properties,. Web identity property of addition.
Web associative, identity, and commutative property of addition online worksheet for grade 3. This is a bundle of three classwork and three aligned homework worksheets each for identity, communicative, and associative properties of addition. In virginia, our sol's state that third graders only need to understand and use the.
Web some of the worksheets for this concept are properties of addition and multiplication, addition properties 1, associative property of multiplication, propertiesofmultiplication. Some of the worksheets for this concept are addition properties 1, addition properties, name in. Web this worksheet allows students to practice working with the properties of addition using only basic numbers.
### Understand The Identity Property With These Worksheets And Activities.
Education.com has been visited by 100k+ users in the past month Web identity, commutative, associative properties of addition worksheet bundle by melissa lamar 4.8 (111) \$4.60 pdf this is a bundle of three classwork and three aligned. They should know that adding zero to any number does not change.
Identity Property Of Addition Worksheets 3Rd Grade. Web addition propertiesstudent identify the addition properties in each equation. This is a hands on activity that will help students learn how to identify, create, and solve equations using properties of addition.
Web identity property of addition. Some of the worksheets for this concept are addition properties 1, addition properties, name in. Web this worksheet allows students to practice working with the properties of addition using only basic numbers.
### For Example, 4 + 0 = 4 68 + 0 = 68 71 + 0 = 71 0 + 93 = 93 0 + 117 =.
This is a hands on activity that will help students learn how to identify, create, and solve equations using properties of addition. Web we have a fantastic selection of free properties worksheets that cover all the important concepts such as commutative property of addition, identity property of addition,. 2nd through 4th grades view pdf assoc., identity, & comm.
### Web I Created This Worksheet To Teach And Review The Addition Properties Required By Our State.
Web to resolve this worksheet, students should know how to identify and understand the identity property of addition. Worksheets are name in numbers 1 9 select the property that is being, multiplication properties,. Web identity property of addition.
Web associative, identity, and commutative property of addition online worksheet for grade 3. This is a bundle of three classwork and three aligned homework worksheets each for identity, communicative, and associative properties of addition. In virginia, our sol's state that third graders only need to understand and use the. | 817 | 4,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-50 | longest | en | 0.905959 |
https://fr.mathworks.com/matlabcentral/cody/problems/46-which-doors-are-open/solutions/1017318 | 1,580,042,281,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251688806.91/warc/CC-MAIN-20200126104828-20200126134828-00319.warc.gz | 444,810,737 | 15,906 | Cody
# Problem 46. Which doors are open?
Solution 1017318
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(which_doors_open(x),y_correct))
A = 1 y = 1
2 Pass
x = 3; y_correct = 1; assert(isequal(which_doors_open(x),y_correct))
A = 1 2 3 y = 1
3 Pass
x = 100; y_correct = [1 4 9 16 25 36 49 64 81 100]; assert(isequal(which_doors_open(x),y_correct))
A = Columns 1 through 16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Columns 17 through 32 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Columns 33 through 48 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 Columns 49 through 64 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 Columns 65 through 80 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Columns 81 through 96 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 Columns 97 through 100 97 98 99 100 y = 1 4 9 16 25 36 49 64 81 100 | 444 | 998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.527394 |
https://newproxylists.com/tag/directed/ | 1,610,952,390,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00717.warc.gz | 477,590,110 | 15,495 | ## c# – Generating a random directed graph
I wrote a function that returns a directed graph as an adjacency matrix. The function requires two arguments: the amount of nodes and the amount of edges. At first, nodes are placed and instantly connected to the graph. When all nodes are added, random edges are created until all edges are placed.
Code:
``````public int(,) GenerateMatrix(int Nodes, int Edges)
{
if (Edges < Nodes - 1) throw new Exception("Too few edges");
if (Edges > Nodes * (Nodes - 1)) throw new Exception("Too many edges");
int(,) adjacencyMatrix = new int(Nodes, Nodes);
// Gives every cell a value of zero
for (int y = 0; y < Nodes; y++)
{
for (int x = 0; x < Nodes; x++)
{
}
}
int placedEdges = 0;
for (int i = 1; i < Nodes; i++)
{
// produce edge between rnd(0, amountofnodes) to new node
int fromVertex = random.Next(0, i);
int weight = random.Next(1, 10);
placedEdges++;
}
while (placedEdges < Edges)
{
int fromVertex = random.Next(0, Nodes);
int weight = random.Next(1, 10);
int targetVertex = random.Next(0, Nodes);
while (targetVertex == fromVertex || adjacencyMatrix(targetVertex, fromVertex) != 0) //|| adjacencyMatrix(fromVertex, targetVertex) != 0)// tredje condition tar bort parallella kanter
{
fromVertex = random.Next(0, Nodes);
targetVertex = random.Next(0, Nodes);
}
placedEdges++;
}
}
``````
## Context
This question is related to the fact one can’t use Bellman-Ford to find max weight paths in directed graphs with cycles. The reason is that giving a new graph $$tilde{G}$$ with negative weights (e.g. $$tilde{w}_{ij} = – w_{ij}$$) will result in cycles with negative sum, thus no minimum cost walk exists in $$tilde{G}$$.
## General question
How well can we approximate the maximum weight path between $$i$$ and $$j$$ in $$G$$, using the shortest path algorithm on a graph $$tilde{G}$$ with transformed weights $$tilde{w}_{ij}$$ ?
In my case, one has $$w_{ij}inmathbb{N}setminus{0}$$. The transformed weights $$tilde{w}_{ij}$$ can be obtained using any function, but I assume a decreasing function $$f$$ is well adapted so that $$w < w’ iff f(w) > f(w’)$$.
## My approach
I am currently trying to use $$f(w) = dfrac{1}{w}$$, in this context we have two important paths given a fixed path length $$L$$.
Denote a path $$p = (i_0i_1, i_1i_2, i_2i_3, dots i_{L-1}i_L$$) with weights $$W = (w^{(1)}, w^{(2)}, w^{(3)}, dots, w^{(L)})$$,
denote the maximum weight path of length $$L$$ by $$p^*$$ with weights $$W^*$$ achieving
$$W^* = argmax_W{mathtt{Cost}(W)} = argmax_W sum_{k=1}^L w^{(k)}$$
and the path $$p_*$$ achieving minimum transformed cost with weights $$W_*$$
$$W_* = argmin_W sum_{k=1}^L tilde{w}^{(k)} = argmin_W sum_{k=1}^L f(w^{(k)}) = argmin_W sum_{k=1}^L dfrac{1}{w^{(k)}}$$
## Specific question
In this specific context, do we have the approximation $$mathtt{Cost}(W^*) approx mathtt{Cost}(W_*)$$ ?
Or is it “very wrong” to replace $$W^*$$ by $$W_*$$ ? What would be the distribution of the relative error defined as
$$E_r = dfrac{lvertmathtt{Cost}(W^*) – mathtt{Cost}(W_*)rvert}{mathtt{Cost}(W^*)}$$
## Statistical analysis
I have tried a quick statistical analysis as follows:
• Letting $$Lin{3,4,5,6,7,8}$$
• sampling $$w sim mathrm{Uniform}(1,w_{max})$$
• The value for $$w_{max}$$ was also picked from the set $${50, 100, 150, 200, 500}$$
• for each combination of $$L,w_{max}$$ I computed $$10^5$$ values of $$W^*, W_*$$ using $$10^3$$ candidate weights $$W_i$$
The results:
• In $$80.8%$$ of cases we have $$mathtt{Cost}(W^*) = mathtt{Cost}(W_*)$$ and even better $$W^*=W_*$$
• When computing the relative error $$E_r$$ one finds that $$82.6%$$ of error values are less than $$0.001$$, $$89.7%$$ less than $$0.01$$ and $$99.9%$$ less than $$0.1$$
The histogram for $$E_r > 0$$ is as follows
## dag – Edges and in-degree in directed graphs
I wanted to know if I have a directed graph $$Gamma$$ with $$V$$ vertices and $$E$$ edges with some adjacency list representation what would be:
(1) the maximum in-degree of a vertex in $$Gamma$$?
(2) the time complexity required to find all edges $$(u, v)$$ in $$Gamma$$ for some given/arbitrary vertex $$v$$?
## c# – Directed acyclic graph with distributed graph nodes
I’m interested of setting up calculations through an acyclic directed graph, whose calculation nodes shall be distributed across different processes/servers.
For a simple example, imagine I’m setting up a somewhat stupid graph to calculate “x + 2x”:
A) Node which receives numeric input from some external source, and forwards that on to its dependents
B) Node which gets the value of its dependency, and multiplies it by 2
C) Node which gets values of two dependencies, and adds them together
``````interface INode
{
double calculate();
}
class A : INode
{
private double _value;
_value = value;
// Notify some graph runner that this node has changed, i.e.
// calculate() shall be invoked on this node and its dependents
}
public double calculate() => _value;
}
class B : INode
{
private INode _a;
public double calculate() => 2.0 * _a.calculate();
}
class C {
private INode _a;
private INode _b;
public double calculate() => _a.calculate() + _b.calculate();
}
``````
I’m trying to come up with a clever mechanism of managing calculations in topological order here.
For example, I’d need to make sure that when calculates, B calculates before C does (and C “knows” the most recent values of A and B, when calculating).
Since all nodes would live on different servers, this may be a little annoying. I might need to replicate the graph layout on each of the 3 servers; where e.g. server C only runs the actual implementation of node C, and has proxy implementations of nodes A and B (which purely rely on messaging from the other nodes for their results).
At the same time, node A might trigger several times before node B or C even receive the message, so I may also need to version-stamp each calculation cycle and ensure proxy nodes receive updates in their expected sequence order.
It seems solvable, but I don’t want to re-invent the wheel here if I don’t need to.
I suspect my problem is one that has already been solved by some smarter person – however, I haven’t been able to find much in terms of practical implementations yet when searching for “distributed acyclic directed graph” and related terms.
Are there any commonly used/known approaches and algorithms for running a DAG with distributed nodes?
## algorithms – Eliminate equal sub-paths of directed cyclic graph
Suppose we have a directed cyclic graph and a list of paths.
What is the shortest way (including parallelization) of eliminating all equal paths and paths that are sub-paths of others?
For example, lets suppose that the vertices are marked by letters from A to Z (for simplicity lets suppose we have only 26 vertices.
The paths are:
(A->A->B->D->E->G->K->L->A)
(B->D->E)
(A->A->A)
(K->L->A->M->N->C)
(M->N)
The output should be:
(A->A->B->D->E->G->K->L->A)
(A->A->A)
(K->L->A->M->N->C)
I thought to first sort all paths by length and start with the longest.
Holding a dictionary of vertices (letters), where the key is the letter and the value is a dictionary of lengths of paths that start from this letter, and the value is the paths, and if there isn’t a match of the current path with equal or bigger paths in the same letter, the path is saved in the results, and is inserted to the dictionary, each time removing the leading vertex. For examples —
The dictionary is {A: {0, ()}, B: {0, ()} … Z: {0, ()}
Length of current path: 9
First vertex: A.
No paths that are 9-length that start with A, so we know that the path is unique.
Lets divide it, we should have:
{A: {9: (AABDEGKLA), 8: (ABDEGKLA)}, B: {7: (BDEGKLA)}, D: {6, (DEGKLA), E: {5, (EGKLA), G: {4, (GKLA), K: {3, (KLA)}, L: {2, (LA)}}
The next path (by length) is (K->L->A->M->N->C) :
No 6-lenght paths that start with K, so we know this is unique. Dividing it into:
{A: {9: (AABDEGKLA), 8: (ABDEGKLA), 4: (AMNC)}, B: {7: (BDEGKLA)}, D: {6, (DEGKLA), E: {5, (EGKLA), G: {4, (GKLA), K: {3, (KLA), 6: (KLAMNC)}, L: {2, (LA), 5: (LAMNC)}, M: {3: (MNC)}, N: {3:(MNC)}}
Next is (B->D->E), 3-length.
In B we have >= 3 path, so we compare — BDE is a sub-path of BDEGKLA, so it’s not unique and omitted.
Next is (A->A->A) – we have >= 3 paths in A, (AABDEGKLA), (ABDEGKLA), (AMNC), but AAA isn’t a sub-path of any of them so it’s unique and the dictionary is now:
{A: {9: (AABDEGKLA), 8: (ABDEGKLA), 4: (AMNC), 3: (AAA), 2: (AA)}, B: {7: (BDEGKLA)}, D: {6, (DEGKLA), E: {5, (EGKLA), G: {4, (GKLA), K: {3, (KLA), 6: (KLAMNC)}, L: {2, (LA), 5: (LAMNC)}, M: {3: (MNC)}, N: {3:(MNC)}}
I wonder if there is a more efficient way do do that.
Thank you.
## algorithms – LCA for directed graph with cycles
I am trying to find the `lowest common ancestor` of 2 nodes in a graph with more than `250000 nodes` and `million edges`. The graph is directed and has many cycles. Each node may or may not have an edge comping back to it from the child node. I tried finding an Euler tour using `Hierholzer’s algorithm` for directed graphs. But gave up on it. Most LCA algorithms in the public domain are for `binary trees, DAGs, n-ary trees`. I did not come across an `algorithm for LCA` for a large directed graph with cycles.
I am interested in the `approach` to take to solve this problem.
## directed graphs – length of longest representative “uu” in suffix automaton
Trying to find the length of the longest representative of an equivalence class in suffix automaton, such that it has the form:
$${vin{}Sigma|v$$ is the longest representative $$wedge$$ $$(exists{}uinSigma*)(v=uu)}$$|. In other words how can we find the length of the longest node $$v$$, such that $$overset{overset{w}longleftarrow{}}{v}=v$$, where $$overset{overset{w}longleftarrow{}}{v}$$ is just the notation for the longest infix $$u$$ of $$w$$, i.e. $$big($$ i.e. $$uin{}Inf(w)big)$$ , such that $$u_{equiv{Suff(w)}}v$$ ($$u$$ is in one class of equivalence with $$v$$: $$end_pos(u)=end_pos(v)big)$$.
I know that the charactarization of a representative, i.e. a state in the suffix automata for $$w$$ can be $$v$$, either $$v$$ is a prefix of the word $$w$$, OR there exists $$a,binSigma,ane{}b$$ such that $$av,bvin{}Inf(w)$$.
Knowing that fact, I tried to prove that if there is a state with represents $$Inf(w)$$ of the form $$uu$$, it must be either created only from one letter, or it must be the whole word $$w$$, but I failed, because that assumption was wrong: the suffix automaton for word $$ababc$$ has a representative “abab”. Does anyone have any ideas how to find that longest length? (Consider we have build or that we can build a suffix automaton for the word $$w$$ and we have the suffix states, suffix transitions and etc… but that may be a wrong direction, I don’t know, because maybe we can find the answer for the task while building the SA). Help?
## A directed graph is semi-connected if the corresponding undirected graph is connected
The following is a problem in CLRS:
A directed graph $$G = (V, E)$$ is semiconnected if, for all pairs of vertices $$u,v in V$$,
we have $$u$$ reachable from $$v$$ or $$v$$ reachable from $$u$$. Give an efficient algorithm to determine whether
or not $$G$$ is semiconnected. Prove that your algorithm is correct, and analyze its
running time.
I know there exist correct solutions to this using topological sorting and strongly-connected components, but I had a different approach.
My approach:
Construct a new undirected graph $$G’$$, having an edge $${u,v}$$ whenever $$(u,v) in E$$ or $$(v,u) in E$$.
The graph $$G$$ is semiconnected if $$G’$$ is connected.
Will this method always be correct?
## directed acyclic graphs – How can I encode a rule tree of multiple depths in Dhall?
I’m trying to move some error prone YAML into Dhall to make some system configuration simpler. I have a tree that looks like:
``````composite:
condition: And
rules:
- composite:
condition: And
rules:
- leaf:
static: true
- leaf:
exists: some-property-to-lookup
- composite:
condition: Or
rules:
- composite:
condition: And
rules:
- leaf:
static: true
- leaf:
exists: some-property-to-lookup
``````
I’m trying to encode this in Dhall and I can’t seem to provide the compiler the right information. My latest try looks like:
``````let Field
: Type
= < B : { static : Bool } | S : { exists : Text } >
let Condition
: Type
= < And | Or | Not >
let Node
: Type
= ∀(Node : Type) →
∀(Leaf : { leaf : Field }) →
∀(Branch : { composite : { condition : Condition, rules : List Node } }) →
Node
let example
: Node
= ∀(Node : Type) →
∀(Leaf : { leaf : Field }) →
∀(Branch : { composite : { condition : Condition, rules : List Node } }) →
Branch
{ composite =
{ condition = Condition.And
, rules =
( Branch
{ composite =
{ condition = Condition.And
, rules = ( Leaf { leaf = Field.S { exists = "hi" } } )
}
}
, Branch
{ composite =
{ condition = Condition.Or
, rules = ( Leaf { leaf = Field.S { static = true } } )
}
}
, Branch
{ composite =
{ condition = Condition.And
, rules =
( Branch
{ composite =
{ condition = Condition.And
, rules =
( Leaf { leaf = Field.S { exists = "hi" } } )
}
}
, Branch
{ composite =
{ condition = Condition.Or
, rules =
( Leaf { leaf = Field.S { static = true } } )
}
}
)
}
}
)
}
}
in example
``````
But I get `Error: Not a function`. Any pointers would be appreciated.
I also tried w/ the Graph module, but I can’t seem to convert that to YAML directly.
## depth first search – Detecting a cycle in a directed graph via an adjacency list implementation and a recursive DFS
Here’s my attempt. I iterate thru each vertex in the graph and do a DFS to see if I reach back on a vertex I already visited in this vertex’s iteration. It seems to work but I am not satisfied with how I short-circuit my code when it found a cycle using if clauses, could not think of a better way to do that.
`````` public boolean isCyclic(Map<T, List<T>> adjacencyList) {
Set<T> visited = new HashSet<>();
if (isCyclic(visited, node) == true)
return true;
}
return false;
}
private boolean isCyclic(Set<T> visited, T node) {
boolean retval;
for (T connectedNode: map.get(node)) {
if (visited.contains(connectedNode)) {
// We've reached back to a vertex, i.e. a back-edge
return true;
} else { | 4,113 | 14,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 92, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.763891 |
https://fablabelectronics.com/qa/question-what-day-is-it-gonna-be-in-28-days.html | 1,627,857,670,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00707.warc.gz | 263,784,205 | 7,851 | # Question: What Day Is It Gonna Be In 28 Days?
## What day is it going to be in 1000 days?
1000 days from today (714 weekdays) would be Sunday, June 18, 2023..
## What day is it gonna be in 21 days?
To get exactly twenty-one weekdays from now, you actually need to count 29 total days (including weekend days). That means that 21 weekdays from today would be October 21, 2020.
## What is 35 days from now?
– Today is : Sunday, September 20, 2020. – The date after 35 days is : Sunday, October 25, 2020.
## How many days passed 2020?
2020 is a leap year. After today 101 days are remaining in this year. This page uses the ISO-8601 ordinal date format.
## What day is 99 days away?
To get exactly ninety-nine weekdays from now, you actually need to count 139 total days (including weekend days). That means that 99 weekdays from today would be February 8, 2021.
## What was 1000 days ago?
1000 days ago was 1440000 minutes ago.
## What was 5000 days ago?
5000 days ago was 7200000 minutes ago.
## What day is 70 days away?
– Today is : Tuesday, September 22, 2020. – The date after 70 days is : Tuesday, December 1, 2020.
## Is there 30 days in a month?
You can remember how many days in each month using this rhyme: 30 days has September, April, June and November. And 29 in each leap year.
## What day is it going to be 30 days from now?
30 days from today (22 weekdays) would be Thursday, October 22, 2020. There are 31 days in October 2020. 2020 is a leap year, so there are 366 days in this year (there are 29 days in February 2020).
## What day is it gonna be in 29 days?
– Today is : Tuesday, September 22, 2020. – The date after 29 days is : Wednesday, October 21, 2020.
## What day is it going to be in 59 days?
– Today is : Sunday, September 20, 2020. – The date after 59 days is : Wednesday, November 18, 2020.
## How old are you when your 1000 days old?
1,000 days old is Sunday, June 18. 2023. 2,000 days old is Saturday, March 14.
## What day is 27 days away?
To get exactly twenty-seven weekdays from now, you actually need to count 37 total days (including weekend days). That means that 27 weekdays from today would be October 29, 2020.
## Is 2020 a Lear year?
Happy leap year! 2020 is a leap year, a 366-day-long year. Every four years, we add an extra day, February 29, to our calendars. … During non-leap years, aka common years – like 2019 – the calendar doesn’t take into account the extra quarter of a day actually required by Earth to complete a single orbit around the sun.
## How long is 150 days today?
To get exactly one hundred and fifty weekdays from now, you actually need to count 210 total days (including weekend days). That means that 150 weekdays from today would be April 20, 2021.
## How many days was 6969 days ago?
If you multiply 6969 by 24, then you will get how many hours since 6969 days ago: 6969 days ago is 167256 hours ago. You can also multiply 6969 by 1,440 to find out how many minutes 6969 days ago was: 6969 days ago was 10035360 minutes ago.
## How long is 100 days from now?
To get exactly one hundred weekdays from now, you actually need to count 140 total days (including weekend days). That means that 100 weekdays from today would be February 9, 2021. | 895 | 3,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-31 | latest | en | 0.9696 |
https://newsilver.com/the-lender/how-to-calculate-home-equity/ | 1,669,978,107,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00320.warc.gz | 449,338,433 | 79,537 | # How To Calculate Home Equity
September 30, 2020
## Home Equity Formula
Home Equity = Value of home – Mortgage Balance
Home equity plays a key role in many of your real estate decisions. Whether you cash out your home’s equity to buy an investment property or fix up the home or you want to know how much you’d make selling the home, it’s an important number.
So how do you calculate your home equity? Check out the steps below.
## How To Calculate Home Equity
### Step 1: Workout how much your home is worth?
First, you need to know your home’s current value. There are a few ways to figure this out.
One option is use New Silver’s ARV Calculator. This will allow you to view the last known selling price of houses similar to yours.
Alternatively you can pull the value from a site like Zillow or Redfin. This is usually an estimated or ballpark value, so don’t put too much credibility into it, but it can be a good starting point.
Lastly, you can also talk to a licensed appraiser or real estate agent in your area. Professionals have a good idea of the values in the area based on recent sales. They can give you a ballpark figure or you can pay for an appraisal or home valuation from a real estate agent.
### Step 2: What is your mortgage balance?
Your outstanding principal balance plays a role in your home equity too. This includes all outstanding mortgages. If you have only a first mortgage, you only need that balance. If you have a first mortgage and home equity loan or home equity line of credit, you need that balance too.
The mortgage balance(s) take away from your home’s equity because you must satisfy that debt before you realize any profit.
### Step 3: Subtract the total mortgage balance from the home’s value
Once you have these two figures you simply subtract the total mortgage balance from the home’s value and have your home equity.
### Home Equity Calculation Example
#### Property Details
• Your home is worth \$300,000
• Your mortgage has a balance of \$140,000
#### Applying Home Equity Formula
• Home Equity = Value of home – total mortgage balance
• Home Equity = \$300,000 – \$140,000
• Home Equity = \$160,000
## Can You Use Your Home Equity?
Now that you know how much home equity you have, the bigger question is, can you use it?
It depends.
First, how does it compare to your home’s value? This is the loan-to-value ratio or LTV. Before you can use your home equity, you’ll need to leave some equity untouched. This is the lender’s collateral for the loan.
You may be able to use your equity in a variety of ways:
• Pay for home renovations
• Use the money to buy an investment home
• Consolidate debt
• Pay for other large expenses
## How Much Can You Borrow?
How much equity you can borrow depends on a variety of factors, starting with the type of loan. Standard lenders typically allow you to use up to 80% of your home’s equity. The 20% cushion is their collateral should you default on the loan.
Hard money lenders may allow the use of more of your equity, it depends on the circumstances. Some allow you to borrow as much as 90% of your home’s equity, giving you more use out of the money invested in your home.
## Should You Use Your Home’s Equity?
The bigger question is should you use your home’s equity? | 729 | 3,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.922882 |
https://bogart.pro/dust-off-mezgq/cc7067-calculating-annual-rate-of-return-over-multiple-years-in-excel | 1,627,685,791,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00216.warc.gz | 154,666,118 | 14,237 | Fifth, multiply 0.061 by 100 to find the average annual return over the 10 years is 6.1 percent. The formula for the holding period return is used for calculating the return on an investment over multiple periods. Therefore, the investor earned annual return at the rate of 16.0% over the five-year holding period. CAGR or compound annual growth rate is method to calculate the growth rate of a particular amount annually, by default we do not have any inbuilt formula in excel to calculate CAGR for us, instead we make categories in tables and in tables we apply the following formula to calculate CAGR which is as follows, (Ending Balance/Starting Balance)˄(1/Number of Years) – 1. Apply the above-mentioned formula to calculate investment return in excel. For instance, in the above example, there are three calendar years. I.e. Compound annual growth rate (CAGR) is a geometric average that represents the rate of return for an investment as if it had compounded at a steady rate each year. Besides the original table, enter the below formula into the blank Cell C3 and, and then drag the Fill Handle to the Range C3:C11. On a year-over-year basis, these growth rates are different, but we can use the formula below to find a single growth rate for the whole time period. Then, subtract 1 and multiply by 100. Annual Return is calculated using the formula given below Annual Return = (Ending Value / Initial Value) (1 / No. The reason: more can change in 10 or 20 years than in 1 year. more Return … Do rockets leave launch pad at full thrust? We would need to convert these percentages into actual beginning and ending values. Additionally, here is a related post I made which has some helpful links for performing calculations such as these. Now select the investment value cell B2. Excellent, makes perfect sense. Then the total annualized return is just a weighted average of each annualized return, with the weighting related to the number of shares in that transaction. Microsoft Excel provides a function called XIRR, which can help you to figure out your returns. An individual may be tempted to incorrectly add the percentages of return to find the return over the multiple periods. Geometric Average Annual Rate of Return: Where: r = Annual rate of return in year i. n = Number of years in the measurement period. Do I have to include my pronouns in a course outline? The time value of money is an essential part of financial markets. The weighted average return is the sum total of the product (or multiplication) of weights that are associated with different investment options and their respective returns. Why am I seeing unicast packets from a machine on another VLAN? This tutorial explains the syntax of the Excel IRR function and shows how to use an IRR formula to calculate the internal rate of return for a series of annual or monthly cash flows. Also this is the best method for me as this will be implemented in code rather than excel (I really should have meantioned this up front, sorry Eric). Over the past decade, the total return of Microsoft stock has varied wildly, ranging from a loss of 43.8% in one year (2008) to a 53.4% gain the next. 1. Continuing with the example, if you originally invested $100,000 in the company, divide$40,000 by $100,000 and multiply by 100 to calculate a multi-year return of 40 percent. Note: Even a year doesn't really reflect success in a given strategy. The concept is that if there is a series of cash flows deriving from an investment, it can be reinvested to earn positive returns. For this info first, enter all these things in excel worksheet to conduct the ROI calculation. The CAGR provides the one rate that defines the return for the entire measurement period. The real rate of return is the actual annual rate of return after taking into consideration the factors that affect the rate like inflation and it is calculated by one plus nominal rate divided by one plus inflation rate minus one and inflation rate can be taken from consumer price index or GDP deflator. This is a near-certainty when talking about investment returns, compared to annual sales figures. Example 5: 100 Days Returns. However, when you have multiple years of data, as well as contributions and withdrawals to the portfolio during that time, using Excel to figure your returns can save you a lot of time. The solution is to figure out the total completed years, and add them to the partial year (called the stub year). Annual Average Growth Rate (AAGR) and Compound Average Growth Rate (CAGR) are great tools to predict growth over multiple periods. The returns on an investment may be shown on an annual, quarterly, or monthly basis. Rate of Return Formula – Example #2. Use Excel to determine the annual returns for investments that less than or greater than 1 year. Note: the RRI function has three arguments (number of years = 5, start = 100, end = 147). The trouble with piling all of the calculations into a formula is that you can't easily see what numbers go where, or what numbers are user inputs or hard-coded. But first, let's define our terms. but how is it calculated when I have multiple buys and sells over a time period? So when calculating CAGR, we would actually be working with a time period of three years. One mistake that's easy to make in figuring CAGR is to incorrectly count the time period. Excel calculates the average annual rate of return as 9.52%. Return of your money when compounded with annual percentage return. the future value of the investment (rounded to 2 decimal places) is$12,166.53. Does exercising an option count as a gain. How to calculate the return from a mutual fund? The way to set this up in Excel is to have all the data in one table, then break out the calculations line by line. You can use RATE to calculate the periodic interest rate, then multiply as required to derive the annual interest rate. From year-end 2016 to year-end 2017, the price appreciated by 4.17% (from $120 to$125). Plug all the numbers into the rate of return formula: = (($250 +$20 – $200) /$200) x 100 = 35% . Compound annual growth rate (CAGR) is the rate of return that would be required for an investment to grow from its beginning balance to its ending one. IRR in Excel is one of the financial functions for calculating the internal rate of return, which is frequently used in capital budgeting to judge projected returns on investments. A CAGR can be shifted to avoid a negative year in the stock market (such as 2008), or to include a year of strong performance (such as 2013). Whether you're doing a what-if analysis to determine how to invest your company's money or you're looking backwards to see how an investment performed, calculating an average annual rate of return lets you do … Note that IRR () doesn’t assume that the interval is years. To arrive at an average annual return, follow the steps below. Firstly, you will see this with some simple and short formulas. Are 'annualized return' and 'annualized total return' synonymous? In Excel 2013 and later, the simplest way is to use the RRI function. Be careful with the results. Although Excel has a built-in formula, it is far from ideal, so we will explain that last. The reality is many investments experience significant short-term ups and downs, and by smoothing them out, so to speak, the CAGR might give a numerically accurate, but emotionally misleading impression of performance. I was composing this exact reply, as I saw this pop up. Does all EM radiation consist of photons? Internal Rate of Return at Different Points in Time (XIRR) In the example below, the cash flows are not disbursed at the same time each year – as is the case in the above examples. Let's take the same figures, but have them be stock prices: The CAGR is superior to other calculations, such as average returns, because it takes into account the fact that values compound over time. We can actually have returns for any number of days and convert them to annualized returns. Excel can calculate it for you using the XIRR function. Calculate rate of return for a share of stock in Excel For example, you purchased the stock on 2015/5/10 at $15.60, sold it on 2017/10/13 at$25.30, and get dividends every year as below screenshot shown. It uses the values and their corresponding dates to find the rate of return. Annual Return Formula – Example #2. Changing a multi-year ROI into an annualized year formula: Where: x = Annualized return. The easiest way to think of CAGR is to recognize that over a number of years, the value of something may change – hopefully for the better – but often at an uneven rate. 2,832 1 1 gold badge 17 17 silver badges 19 19 bronze badges. For example, let's derive the compound annual growth rate of a company's sales over 10 years: The CAGR of sales for the decade is 5.43%. It's a complicated calculation, but will take into account multiple in/out cash flows over time along with "idle periods" where your money may not have been doing anything. Calculate annualized return on stock with multiple transactions, Comparing IRA vs 401K's rate-of-return with dollar cost averaging, How to calculate annualized rate of return over multiple transactions, Calculating return on a series of stock positions with multiple uneven transactions. ... How to calculate annualized rate of return over multiple transactions. For estimating returns when amounts vary across investments, a concept called weighted average return is used. Third, raise 1.8 to the 1/10th power to get 1.061. The term “annualized rate of return” refers to the equivalent annual return that an investor earns over the holding period of the investment. Remember that when you enter formulas in Excel, you double-click on the cell and put it in formula mode by pressing the equals key (=). For example, let's derive the compound annual growth rate … Calculating a rate of return is easy to do by hand if you have a starting value and an ending value one year apart. If a president is impeached and removed from power, do they lose all benefits usually afforded to presidents when they leave office? Plug all the numbers into the rate of return formula: = (($250 +$20 – $200) /$200) x 100 = 35% . Calculate Years of Service between Two Dates. This tutorial will teach you how to calculate the compound annual growth rate, or CAGR, in Excel. Now, let's say we had a stock whose annual price data was presented in percentage, instead of dollar, terms: In this case, the data is being shown from the beginning of the year, as in, the entire yearly return in 2015 (10%), the entirely yearly return in 2016 (15%), and the entire yearly return in 2017 (-4%). The weighted average approach or total portfolio calculation is a good idea. If you invest your money with a fixed annual return, we can calculate the future value of your money with this formula: FV = PV (1+r)^n. It is calculated by taking the arithmetic mean of a series of growth rates. Note that the regular rate of return describes the gain or loss, expressed in a percentage, of an investment over an arbitrary time period. The answer is 8%. Personal Finance & Money Stack Exchange is a question and answer site for people who want to be financially literate. The best way to do this is to use IRR. For Investment A with a return of 20% over a three-year time span, the annualized return is: x = Annualized. There are several ways to calculate CAGR in Excel. This may be obvious, but may help those who aren't so familiar with the numbers to understand that data running less than a year isn't going to provide as much useful conclusion as longer term. When Excel is in formula mode, type in the formula. The compound return is the rate of return that represents the cumulative effect that a series of gains or losses has on an amount of capital over time. Can 1 kilogram of radioactive material with half life of 5 years just decay in the next minute? Estimates of Coca-Cola’s return over 1 year will likely be more accurate than estimates over 10 or 20 years. @CodeKiwi - In case you decide you need it, you can implement IRR in code as well, although it's a pretty inefficient algorithm since it's somewhat non-deterministic. Return of your money when compounded with annual percentage return. To calculate the Average Annual Growth Rate in excel, normally we have to calculate the annual growth rates of every year with the formula = (Ending Value - Beginning Value) / Beginning Value, and then average these annual growth rates. For example, if we were presented with year-end prices for a stock like: From year-end 2015 to year-end 2016, the price appreciated by 20% (from $100 to 120). The annual returns will be: Annual returns = (1+0.06)^(365/100) – 1 = 23.69%. You can do as follows: 1. Is it possible to make a video that is provably non-manipulated? Google will give you plenty of code samples. Financial modeling best practices require calculations to be transparent and auditable. The average annual growth rate (AAGR) is the average increase in the value of an individual investment, portfolio, asset, or cash stream over the period of a year. Partial credit thanks joe, so do you think that for stocks held for less than a year I should add to the calculation as a weighted simple return rather than annualized? Online tools, including Investopedia’s CAGR calculator, will give the CAGR when entering these three values. That's why the equation reads 1/2, not 1/3. Different investments go up or down in value by different amounts over different time periods. I think that when you include enough stocks, the risk of having the annualized return be a misleading number is very reduced. Amey had purchased home in year 2000 at price of$100,000 in outer area of city after sometimes area got develop, various offices, malls opened in that area which leads to an increase in market price of Amey’s home in the year 2018 due to his job transfer he has to sell his home at a price of $175,000. Since there are 365 days in a year, the annual returns will be: Annual returns = (1+0.001)^365 – 1 = 44.02% . Understanding the Compound Annual Growth Rate – CAGR, Inside the Average Annual Growth Rate (AAGR). The math formula is the same as above: You need ending values, beginning values and a length measured in years. You got it Eric, good answer. – JTP - Apologise to Monica ♦ Nov 1 '10 at 15:31. Average annual earnings for security A can be calculated as, Average annual earnings A = Sum of earnings in Year 1, Year 2 and Year 3 / Estimated life = ($5,000 + $10,000 +$12,000) / 3 = $9,000 What one should check when re writing bash conditions for sh or ash? Mr. A has bought the property on Jan 2015 for Rs. Note that the regular rate of return describes the gain or loss, expressed in a percentage, of an investment over an arbitrary time period. Compound interest is the interest on a loan or deposit calculated based on both the initial principal and and the accumulated interest from previous periods. For example, a 10% gain in a week isn't unheard of for individual stocks, but (1.1)^52 = 142. or a 14,100% return. The bond paid$80 per annum as coupon every year till its maturity on December 31, 2018. Eric Petroelje Eric Petroelje. For example, if you had five rows of cash flows and dates, starting in cell A1, your command should say "=XIRR(A1:A5,B1:B5)." Y ou can calculate the average annual growth rate in Excel by factoring the present and future value of an investment in terms of the periods per year. The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place. Excel can calculate it for you using the XIRR function. 3,50,000 and after 3 years in Jan 2018 he has sold the same property for Rs. share | improve this answer ... so do you think that for stocks held for less than a year I should add to the calculation as a weighted simple return rather than annualized? T = Time horizon . site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Despite the fact that the stock's price increased at different rates each year, its overall growth rate can be defined as 11.8%. divide the final value by the initial value, then raise that number by 1/n, where "n" is the number of years you held the investments. reTherefore, (1+x) 3 – 1 = 20% So, the ROI for Mr. A is 2.5 L. Si… However, simply use the RRI function in Excel to calculate the compound annual growth rate (CAGR) of an investment over a period of years. 1. Let’s assume an investor has calculated the following annual returns over the past 10 years: First, we will calculate the ROI value. The compound annual growth rate (CAGR) shows the rate of return of an investment over a certain period of time, expressed in annual percentage terms. First, select the “Sold Value” by selecting the cell B3. 1. of Years) – 1 Annual Return = ($1,800 /$1,000) 1 / 10 – 1 Annual Return = 6.1% 6,00,000. This is a link to an Excel file which you can download (linked again later) and modify to calculate your own investment returns. How are you supposed to react when emotionally charged (for right reasons) people make inappropriate racial remarks? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Deep Reinforcement Learning for General Purpose Optimization. The CAGR helps identify the steady rate of return of an investment over a certain period of time. CAGR is also subject to manipulation depending on the measurement period, which is ultimately (and often arbitrarily) selected. rev 2021.1.8.38287, The best answers are voted up and rise to the top, Personal Finance & Money Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Would I simply use the average cost per unit * current units. When aiming to roll for a 50/50, does the die size matter? Annualized Rate of Return. It only takes a minute to sign up. The offers that appear in this table are from partnerships from which Investopedia receives compensation. CAGR requires three inputs: an investment’s beginning value, its ending value and the time period (expressed in years). In other words, the annualized rate of return is the overall return generated by the investment over a period which is then scaled down to a 12-month (or one-year) period. share | improve this answer | follow | answered Nov 1 '10 at 14:31. … The geometric mean is the average of a set of products, the calculation of which is commonly used to determine the performance results of an investment or portfolio. If you invest your money with a fixed annual return, we can calculate the future value of your money with this formula: FV = PV(1+r)^n. How to annualize a loss of 100% for an investment? Calculating a rate of return is easy to do by hand if you have a starting value and an ending value one year apart. A more complex situation arises when the measurement period is not in even years. I understand how to calculate the Annualized return on a stock when I have single purchase ie. Here, FV is the future value, PV is the present value, r is the annual return, and n is the number of years. T = 3 years. However, when you have multiple years of data, as well as contributions and withdrawals to the portfolio during that time, using Excel to figure your returns can save you a lot of time. How to calculate the Compound Annual Growth Rate using the XIRR Function. As I said earlier there are many ways to calculate the years of services. Fourth, subtract 1 from 1.061 to get 0.061. The internal rate of return (IRR) is a metric used in capital budgeting to estimate the return of potential investments. The XIRR function in Excel returns the internal rate of return for a series of cash flows which might not occur at a regular interval. In the example shown, the formula in H9 is: = RRI(B11, C6, C11) How to calculate the return over a period from daily returns? Annualized Rate of Return on Stock Purchased in Tranches. The formulas which are used here are given below. For instance, let's say you have an investment that's posted these changes over three years: That's actually a 5% CAGR, but the year-over-year volatility in those returns is huge. On the down side, CAGR dampens the perception of volatility. For example, if an investment of $10,000 earns an annual interest rate of 4%, the investment's future value after 5 years can be calculated by typing the following formula into any Excel cell: =10000* (1+4%)^5 which gives the result 12166.52902. Look into the below picture where I calculated the years of services for different years by using different formulas. The above spreadsheet on the right shows the FVSCHEDULE function used to calculate the future value of an investment of$10,000 that is invested over 5 years and earns an annual interest rate of 5% for the first two years and 3% for the remaining three years.. The RRI function below calculates the CAGR of an investment. How to Calculate the Average Annual Rate of Return in Excel. Therefore, Adam realized a 35% return on his shares over the two-year period. Ha. Calculate rate of return for a share of stock in Excel For example, you purchased the stock on 2015/5/10 at $15.60, sold it on 2017/10/13 at$25.30, and get dividends every year as below screenshot shown. Let us take an example of Dan who invested 1,000 to purchase a coupon paying bond on January 1, 2009. Piano notation for student unable to access written and spoken language, I have problem understanding entropy because of some contrary examples, Same vertical alignment of equations in different cells inside a table, Selecting all objects with specific value from GeoJSON in new variable, CSS animation triggered through JS only plays every other click. In other words, CAGR is a "smoothed" growth rate that, if compounded annually, would be equivalent to what your investment achieved over a specified period of time. ST_Overlaps in return TRUE for adjacent polygons - PostGIS, Using DSolve to find y[x] for a second-order differential equation. You got it Eric, good answer. Third, raise 1.8 to the 1/10th power to get 1.061. The XIRR function uses the syntax =XIRR(value, date, [guess]). It's like a map that correctly informs you your destination is only five miles away, without indicating the bumpy condition of the road. I find it best to just jump right into an example when trying to understand how to calculate this return. Since Brad answered with a great reply, I'd like to offer another comment: How to calculate the Compound Average Growth Rate. Annualized Rate of Return. An investor purchased a share at a price of5 and he had purchased 1,000 shared in year 2017 after one year he decides to sell them at a price of $There's no CAGR function in Excel. XIRR returns the internal rate of return for a schedule of cash flows that is not necessarily periodic. It assumes the investment compounds over the period of time specified, and it is helpful for comparing investments with similar volatility characteristics. Let’s say we have 6% returns over 100 days. ⇒How to calculate mutual fund returns in excel using “Compounded Annual Growth Rate or CAGR concept”. Fifth, multiply 0.061 by 100 to find the average annual return over the 10 years is 6.1 percent. You can also use the XIRR function to calculate CAGR in Excel. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Treat each transaction as separate, with its own principal, its own gain, and its own number of days. The Excel RATE function is a financial function that returns the interest rate per period of an annuity. Therefore, Adam realized a 35% return on his shares over the two-year period. Gather the data relevant to the growth you want to calculate. The cell shows the average annual rate of return after Excel finishes calculating it. So, calculate the ROI for Mr. A from this investment. Coca-Cola currently trades for$45.63 per share. The formula is: CAGR=(EBBB)1n−1where:EB=Ending balanceBB=Beginning balance\begin{aligned} &CAGR=\left(\frac{EB}{BB}\right)^{\frac{1}{n}}-1\\ &\textbf{where:}\\ &EB = \text{Ending balance}\\ &BB = \text{Beginning balance}\\ &n = \text{Number of years} \end{aligned}CAGR=(BBEB)n1−1where:EB=Ending balanceBB=Beginning balance. The sum of such weights equals 100%. Annualized returns of short term trading can produce some crazy results. Excel can calculate it for you using the XIRR function. I was composing this exact reply, as I saw this pop up. The RATE function calculates by iteration. Plugging in the above values we get [(125 / 100)^(1/2) - 1] for a CAGR of 11.8%. Ha. Below is an overview of how to calculate it both by hand and by using Microsoft Excel. This is a good opportunity to use a spreadsheet, since it's easy to add a helper column to convert the percentages into values. Estimating Valuation Multiple Changes. Mathematical correct way to calculate the average net annual return of an split up investment? Fourth, subtract 1 from 1.061 to get 0.061. Should I "take out" a double, using a two card suit? An APR is defined as the annual rate charged for borrowing, expressed as a single percentage number that represents the actual yearly cost over the term of a loan. But since the data is presented as year-end prices, we really we only have two completed years. Here, FV is the future value, PV is the present value, r … You’ll also learn about some of the limitations. The way to set this up in Excel is to have all the data in one table, then break out the calculations line by line. 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Find it best to just jump right into an annualized year formula: where: x =.... % for an investment ’ s beginning value, its ending value one year apart Excel calculates the when!, do they lose all benefits usually afforded to presidents when they leave office multiple. The 10 years is 6.1 percent stock when I have multiple buys and sells over a calculating annual rate of return over multiple years in excel! Value ” by selecting the cell shows the average annual Growth rate exact reply, as saw! The future value of money is an overview of how to calculate the average annual return follow... ( expressed in years ) ROI into an annualized year formula: where: =. Of Growth rates from $120 to$ 125 ) answered with a of. When Excel is in formula calculating annual rate of return over multiple years in excel, type in the formula given below when Excel is formula. For the entire measurement period question and answer site for people who want to transparent. A course outline rate ( AAGR ) and Compound average Growth rate or CAGR concept ” below where... Helpful for comparing investments with similar volatility characteristics by selecting the cell shows the average annual return at rate. Should I take out '' a double, using a two card suit of 5 years just decay the! Calculate investment return in Excel money Stack Exchange is a good idea | answered Nov 1 '10 at.... That IRR ( ) doesn ’ t assume that the interval is years the periodic interest rate period... To react when emotionally charged ( for right reasons ) people make inappropriate racial?... Year formula: where: x = annualized 1/2, not 1/3 holding period date, [ guess ].. I seeing unicast packets from a mutual fund would need to convert these percentages into beginning. Find the rate of return to find the return from a machine on another VLAN also use the XIRR.! Excel worksheet to conduct the ROI calculation radioactive material with half life of 5 years just decay in next... More return … how to calculate the periodic interest rate 0.061 by 100 to find return. Return in Excel worksheet to conduct the ROI calculation budgeting to estimate the return over a certain period of investment... You can also use the XIRR function for estimating returns when amounts vary across investments, concept... | 7,878 | 35,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-31 | latest | en | 0.93679 |
https://explainxkcd.com/wiki/index.php/Talk:2735:_Coordinate_Plane_Closure | 1,702,295,388,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679511159.96/warc/CC-MAIN-20231211112008-20231211142008-00618.warc.gz | 262,363,433 | 10,928 | # Talk:2735: Coordinate Plane Closure
Is there significance to the fact that the axes aren't labeled in the warning? Can I plot y = 0.75x today or not?Brossa (talk) 15:05, 8 February 2023 (UTC)
you cannot because it intersects the given square as shown in this desmos thing i whipped up in 2 seconds: https://www.desmos.com/calculator/zb9nbrl6s5 172.70.43.29 15:38, 8 February 2023 (UTC)Bumpf
I can if the forbidden coordinates are 1 ≤ x ≤1.5 and 1.5 ≤ y ≤2172.70.131.66 15:56, 8 February 2023 (UTC)
In the absence of other information, assuming horizontal x and vertical y would be conventional. --141.101.98.145 19:15, 8 February 2023 (UTC)
"Hole" is also sometimes used to mean a particular coordinate on a function which is discontinous at some point but could have a value (for example sinx/x with a hole at (0,1)). 172.70.206.92 19:18, 8 February 2023 (UTC) Randall listed 2 points, yet the cordoned off area is a square. 2 points define a line, not a square, he really should have thought of that. How is someone to know the invalid points without the diagram? Even with the diagram, we don't know whether points on the boundary are included! Is the line y=1 a valid line to draw? THESE ARE QUESTIONS THAT NEED TO BE ANSWERED RANDALL BE MATHEMATICALLY RIGOROUS NEXT TIME.
Right! A hole pops up in rational functions when there's a term that appears in the numerator AND the denominator. However, it does not mean the graph is broken; just that there is no defined y-value at the x-value of the hole. ----Theunlucky (talk) 16:55, 9 February 2023 (UTC)
One reason could simply be the alignment between the coordinates and time. Reading out the numbers without paying attention to the mathematical punctuation you can form the sentence "the coordinate plane will be closed Thursday between 1:51 and 2:15 to repair a hole", following the typical structure of such a notice to not just provide a day but a time.
Ironically, the notice makes it sound like using y=1 is fine, and the affected region is only strictly greater than y=1. That would make the region that's closed an open set, and the region that's open a closed set. 172.70.110.230 22:46, 8 February 2023 (UTC)
Right! A hole pops up in rational functions when there's a term that appears in the numerator AND the denominator. However, it does not mean the graph is broken; just that there is no defined y-value at the x-value of the hole.
🚧 DETOUR 🠕 KEEP WITHIN MINKOWSKI CONES ⛔ DO NOT ENTER Y < |X| 🚧 162.158.90.38 23:37, 8 February 2023 (UTC)
So the joke is that the coordinate plane is closed when there's damage that causes it not to be closed? Barmar (talk) 23:44, 8 February 2023 (UTC)
Aw man, I was really looking... forward... to doing math today. 172.71.222.76 11:58, 9 February 2023 (UTC)
I thought the title text was referring to the danger of lines on a 2d graph "falling through" the hole and inadvertently gaining a third dimension, which might collide with graphs at z=-1 etc. 162.158.34.75 16:14, 9 February 2023 (UTC)
My RSS reader picked this comic up at exactly midnight UTC on Feb 8, which stood out to me because usually they seem to be posted later in the day. Danielp82 (talk) 04:02, 10 February 2023 (UTC)
This comic reminded me of Complex Analysis, where we integrate in circles around singularities of complex functions (aka holes). See Cauchy integral formula. Maybe we should mention that in the explanation. 172.71.154.39 07:29, 10 February 2023 (UTC)
The beauty of the Wiki is that you can add it yourself, if you think you can word it relevently. Or anyone who now wants to. (Whoever does, note that `{{w|Cauchy's integral formula}}`, or an altered text version like your `{{w|Cauchy's integral formula|Cauchy integral formula}}`, would be the prefered wikilink format to use.) 141.101.98.151 08:08, 10 February 2023 (UTC)
I made the unfortunate but defensible change from "airmen" to "air missions". The FAA re-consecrated "NOTAM" to the gender-neutral (and execrable) form on 2 Dec 2022. The "airmen" form may still be in use by ICAO or nations other than the US. Der57 (talk) 10:52, 10 February 2023 (UTC)
Don't you think it's uncharacteristic of Randall to deviate from the normal math practice of placing the x coordinate first before the y coordinate when not explicitly identifying them? Furthermore, each coordinate is backwards from the convention of smaller number first, then larger? This is so out of step for him I think he did it deliberately and we're missing a subtle joke... Paso Dan (talk) 16:38, 10 February 2023 (UTC)
What are you talking about. He lists the number as (X, Y) completely normal with X on the lower axis and Y on the one going up and so does the numbers he gives follow normal style. Also he starts with the number that comes first on the x-axis. See no reason to start with another, and this is also relevant for making it look like he is given a time period. Seems to me you must have made a mental mistake when you wrote this? --Kynde (talk) 17:35, 10 February 2023 (UTC)
Yeah, like Kynde said, he's using standard notation... the region (1.5, 1), x = 1.5 and y = 1, and (2, 1.5), x = 2 and y = 1.5... X first, then Y, standard. He even plots those points, with dotted lines denoting a square "cordoned" off... And when sorted, graph points usually go left-to-right (so, ascending order by X), which he did. NiceGuy1 (talk) 05:42, 11 February 2023 (UTC)
I get what Paso Dan is saying, and it was my impression as well. Indicating that the coordinate plane is closed between two points (A,B) and (C,D) doesn't by itself tell you whether the closure is along the line between those two points, or a circle with a diameter running from (A,B) to (C,D), or some other 2-D shape. The diagram indicates a square with corners at (A,B) and (C,D) and sides parallel to the axes, but that information isn't in the text. If on the other hand it's interpreted as a closure for the region where x is between 1.5 and 2 and y is between 1 and 1.5, you get a full description of the closed area. --Brossa (talk) 17:36, 18 February 2023 (UTC)
Two points are capable of defining an axis-aligned rectilinear quadrilateral, or a diamondoid (if you decide to use the convention of a 45-degree skew), or an arbitrarily-rotated square (defining one of the long diagonals). Pretty much any other quadrilateral (or other shape) needs further pre-agreed presumptions or more points of definition. A circle can be defined by three (non-colinear, non-identical) edge-points, or an oval (even skewed) by, at the very least, a third value/coordinate in different manners.
(Oh, ok, you could assume a circle defined by just two points (diametrically opposed), the circumscircle to the arbitrary square, above, or even the circle that is inscribed to it. Or "centre and point on radius". From that stage you could indicate further polygon that can be circumscribed thusly, oriented either constantly to the axes or to the direction defined by the (initial) radial point. So your toolkit can feature a "drawShape(coord1,coord2)" for all kinds of areas, one for each possible kind and treatment.)
Yet, in the absence of any detail other than "this defines an area", two surface coordinates almost certainly will be best assumed to defibe a rectangle (or a graticule, given latitude/longitude coords) and two 3D coords ones a cuboid/whatever. 172.69.79.185 20:27, 18 February 2023 (UTC)
Finally, I get an explanation why I've seen so many mentions of "China balloon" lately (and the picture of the moon with a silhouette on it, like the one I saw with X-Wings photoshopped flying off to it, didn't even realize they were related). :) I didn't feel like Googling it, figuring it would come to my attention eventually. LOL! NiceGuy1 (talk) 05:35, 11 February 2023 (UTC) | 2,077 | 7,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.953376 |
https://puzzling.stackexchange.com/questions/72685/sequence-puzzle/80135 | 1,558,872,885,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259126.83/warc/CC-MAIN-20190526105248-20190526131248-00265.warc.gz | 586,853,410 | 33,264 | # Sequence puzzle
This is a sequence puzzle I found online which I'm stuck on:
452801, 773924, 102410, 471056, ?
I figured out how to get from the first number to the second (+3, +2, +1, +1, +2, +3) for the digits, but this rule doesn't work for the rest. Any ideas?
• 050321 might be next number in sequence – CR241 Oct 3 '18 at 23:54
• I checked, it's not @CR241 – Duck Oct 25 '18 at 23:35
• I haven't had the time yet, but seems like it might be a determined value being added each time, but the sum is capped as would a variable in software be (example, if you do addition with an integer 16-bit value and the answer is above the cap, it literally cuts off the highest value digits to keep within the integer 16-bit value range) – nine9 Mar 1 at 8:28
The next number in the sequence is:
216101
I got it by creating an algorithm that tested positive integers until a result was generated indicating that the question was answered correctly. The algorithm submitted an answer one at a time and waited for a response before submitting the next answer to make sure that the website wasn't overwhelmed.
For the curious - it took about 5 hours.
Even with the answer, I currently haven't found a strong correlation between the numbers in the sequence however @Mr Pie's strange observation is correct!
• Hahah thank you! I thought about doing it myself but never did it. Thanks for doing it for me. I upvoted but haven't accepted your answer yet since like you said you haven't found a pattern yet. – simonzack May 3 at 10:57
• Oh, what do you know! :P – Mr Pie May 5 at 2:58
One thing all 4 numbers have in common is this :
Rule: They are all the last 6 digits of the integer part of the quarter-square of some integer.
where
The quarter square of a number $$x$$ is defined as $$(\frac{x}{2})^2$$ or $$\frac{x^2}{4}$$. A list of quarter-squares upto 100,000 can be found here.
For example
The quarter-square of the number 216,371 is :
$$(\frac{216371}{2})^2=11,704,102,410.25$$
The last 6 digits of the integer part are 102,410.
Going up to 2 million,
which is a 7-digit number, and also a period for the above rule,
we can find a list of numbers that adhere to the above rule :
452801 : [69698, 492802, 507198, 930302, 1069698, 1492802, 1507198, 1930302]
773924 : [54764, 179764, 320236, 445236, 554764, 679764, 820236, 945236, 1054764, 1179764, 1320236, 1445236, 1554764, 1679764, 1820236, 1945236]
102410 : [216371, 627379, 1372621, 1783629]
471056 : [53832, 76065, 94815, 196168, 303832, 305185, 323935, 446168, 476065, 494815, 553832, 696168, 705185, 723935, 803832, 876065, 894815, 946168, 1053832, 1105185, 1123935, 1196168, 1276065, 1294815, 1303832, 1446168, 1505185, 1523935, 1553832, 1676065, 1694815, 1696168, 1803832, 1905185, 1923935, 1946168]
The above lists were calculated using
this code written in Swift.
But I still can't find a relationship between the numbers in those lists.
I hope this is helpful!
• Wow nice a lot of work I upvoted cause of that, still not sure about the answer. – simonzack Mar 30 at 13:58
• How did you find that?! :o – Mr Pie Apr 8 at 3:36
This is not exactly a pattern, but there is one odd thing these numbers have in common:
If you get each term and divide it by $$\sqrt{e^\pi}$$ then it is close to an integer.
Proof (correct to $$5$$ decimal places):
$$452801\div\sqrt{e^\pi}=94128.08005\approx 94128$$
$$773924\div\sqrt{e^\pi}=160822.99325\approx 160823$$
$$102410\div\sqrt{e^\pi}=21288.94741\approx 21289$$
$$471056\div\sqrt{e^\pi}=97922.92172\approx 97922$$
It additionally appears that
The answers are becoming less nearer to an integer the further down the sequence we continue, thus perhaps we have to add or subtract a constant to each term that correspondingly gets bigger, in order to make the answers become closer to an integer.
This might have to do with the overall pattern.
Hope this helps!
• Yea in Adam's answer we have $216101\div\sqrt{e^\pi}=44922.984328976$. – simonzack May 6 at 6:21
I am not quite sure but was able to make some quite good relation
1. 452801=a , 773924=b , 102410=c , 471056=d
2. Now let us take b-a=321123=A , d-c=368646=B , b-c=671514=C , d-a=18255=D
5. From above A+B=C+D
6. Considering the same pattern the answer might be 157715
7. Let us take x as required number
8. 773924-x + 102410-x = 671514+47105-x
9. X=876334-718619=157715
• hate to burst your bubble, but A+B=b+d-a-c=C+D – JonMark Perry Mar 1 at 10:30
• Oh let me cut that first sentence – user56760 Mar 1 at 10:41
• you can type your guess into the website mentioned in OQ (you don't need to enter all answers) and it marks it for you. for example, the answer to (3) is 25,1 (previously solved on PSE) and press submit at the bottom of the page. – JonMark Perry Mar 1 at 10:54
## protected by JonMark PerryApr 8 at 19:27
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead? | 1,577 | 5,107 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-22 | latest | en | 0.954036 |
https://getgocube.com/play/rubiks-cube-combinations/ | 1,675,674,567,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500334.35/warc/CC-MAIN-20230206082428-20230206112428-00236.warc.gz | 289,563,028 | 33,747 | GoCube
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# How Many Possible Rubik’s Cube Combinations Are There?
The Rubik’s cube, although seemingly simple, is a complex mathematical puzzle. It is often used to teach fractions, proportions, and permutation & combination to students. Twisting and turning the cube yields unlimited combinations. In fact, established studies show that there are 43 quintillion possible combinations of the cube.
If that number seems incomprehensible, it sort of is. One quintillion has 18 zeroes, or 1,000,000,000,000,000,000. To put things in perspective, a quintillion is the same as a billion or a million trillion.
## Number of Possible Permutations
Although the Rubik’s cube seems easy and straightforward, most people cannot solve it quickly or easily. Solving the Rubik’s cube effortlessly requires a certain degree of mathematical aptitude or at least some knowledge of the different permutations.
Randomly rotating a 3x3x3 cube will lead you to one of the 43 quintillion combinations, but not toward solving the puzzle. A pocket cube (2x2x2) has 3,674,160 possible permutations, and randomly turning around this innocuous little cube will not help you solve it.
The original 3x3x3 has many likely patterns. According to those who have figured it out, the exact count is 43 quintillion, 252 quadrillions, 3 trillion, 274 billion, 489 million, 856 thousand. To illustrate this, all the permutations and combinations of a large Rubik’s cube of 6 centimeters can cover the Earth’s surface 300 times.
## How many combinations are possible among Rubik’s cube adaptations?
Since the time the original Rubik’s cube was launched, there have been many adaptations. Most of these adaptations come in different numbers of cubes within the actual cube. While the original one is 3x3x3, there are also cubes with 2x2x2, 4x4x4, and other amalgamations.
One must be warned that most of these numbers are incomprehensible, and the terms that are used to describe them are mostly unknown to most people. Let us take a look at how many combinations each of these Rubik’s cube adaptations can boast:
1. ### Pocket Cube
A Rubik’s cube of 2x2x2 is called a Pocket Cube, and it has 3,674,160 combinations. Although it is “simpler” than the 3x3x3 version the 3,674,160 combinations a 2x2x2 cube possesses are still mind-numbing. If one were to try solving it by pure chance every day full time, one could only solve it 2 or 3 times a year. That is assuming the cube doesn’t break during this time.
1. ### 3x3x3 Rubik’s Cube
The classic cube, the 3x3x3 Rubik’s cube has approximately 43 quintillion 43,252,003,274,489,856,000 combinations. It needn’t be said again that this would take years to solve if one were to simply depend on chance or luck. This is why knowledge of combinations and how they work according to mathematics is necessary. Developing strategies, keenly observing the puzzle while trying to solve it, and learning from one’s mistakes (trial and error) will all help one to solve the cube quickly.
1. ### 4x4x4 Rubik’s cube
A Rubik’s cube of 4x4x4 amalgamation is called the Master Cube. Some also call it the Rubik’s Revenge. This Master Cube approximately has 7.4 quattuordecillion combinations. The numerical representation looks like this:
7,401,196,841,564,901,869,874,093,974,498,574,336,000,000,000.
1. ### 5x5x5 Rubik’s cube
A Rubik’s cube of 5x5x5 is called the Professor’s Cube. It has an approximately 283 trevigintillion combinations. If you have never heard of this term, you probably are not alone. The exact number of combinations possible in this amalgamation is:
282,870,942,277,741,856,536,180,333,107,150,328,293,127,731,985,672,134,721,536,000,000,000,000,000.
1. ### 6x6x6 Rubik’s cube
Panagiotis Verdes, a Greek inventor, is famous for inventing the 6x6x6 and the 7x7x7 cubes. The inventor used a special strategy to build these cubes, which were previously believed to be impossible. The 6x6x6 Rubik’s cube is marketed under the brand V-Cube, and the following possible combinations:
157,152,858,401,024,063,281,013,959,519,483,771,508,510,790,313,968,742,344,694,684,829,502,629,887,168,573,442,107,637,760,000,000,000,000,000,000,000,000.
1. ### 7x7x7 Rubik’s cube
The 7x7x7x cube has 19.5 duoquinquagintillion combinations. A duoquinquagintillion can be represented as 10159 (10 to the power of 159), or 159 zeroes after the 1. The 7x7x7 Rubik’s cube is marketed under Verdes’ V-Cube brand, and has an exact combination of:
19,500,551,183,731,307,835,329,126,754,019,748,794,904,992,692,043,434,567,152,132,912,323,232,706,135,469,180,065,278,712,755,853,360,682,328,551,719,137,311,299,993,600,000,000,000,000,000,000,000,000,000,000,000.
## Don’t just depend on luck or chance
It is clearly impossible to randomly twist and turn the Rubik’s cube and its many adaptations to try and solve them. Since the combinations are limitless, it is necessary to know combination strategies, tips, and tricks. GoCube’s Smart Cube and the associated mobile application can help you learn how to solve the Rubik’s cube on the go. Most importantly, you can learn from those who already excel at it and network with other players. | 1,415 | 5,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-06 | latest | en | 0.926438 |
https://math.stackexchange.com/questions/266757/localization-of-ring-by-using-homomorphism | 1,726,213,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00455.warc.gz | 355,869,363 | 37,458 | # Localization of ring by using homomorphism
If $\phi : A \to B$ is a ring homomorphism, where $A$ and $B$ are commutative rings with unit but not necessarily domains. Let $P$ be a prime ideal of $A$. How do we define the ring $B_{P}$ and map $\phi_{P} : A_{P} \to B_{P}$? Thanks
A natural way is to put $B_P=A_P\otimes_A B$ and define $\phi_P: A_P\to B_P$ by tensoring $\phi$ with $A_P$: $\phi_P(a/s)=(a/s)\otimes 1$.
Edit Localization: $B_P$ is actually the localization of $B$ with respect to the multiplicative subset $\phi(A\setminus P)$ of $B$.
• Thanks Qil. I am trying to write the elements of $B_{P}$ by using the definition of localization. Here $\phi(P)$ may not be a prime ideal of $B$. If not prime then $B$\ $\phi(P)$ need not be a multiplicatively closed set. Then I could not use the definition of localization. Commented Dec 28, 2012 at 22:54
It is natural to define $\varphi_P(a/s)=\varphi(a)/\varphi(s)$, for $a\in A$ and $s\in A-P$. This also suggest what $B_P$ should be, that is, the ring of fractions of $B$ with respect to the multiplicative system $\varphi(A-P).$ So $B_P$ is a notation for this ring of fractions which, in fact, is not a localization (at a prime ideal of $B$). | 373 | 1,205 | {"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} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.855452 |
http://lists.contesting.com/archives/html/Topband/2010-10/msg00002.html | 1,597,391,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739182.35/warc/CC-MAIN-20200814070558-20200814100558-00161.warc.gz | 57,421,877 | 4,325 | Topband
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## Re: Topband: Inverted L Dimensions
To: Guy Olinger K2AV Re: Topband: Inverted L Dimensions K4SAV Thu, 30 Sep 2010 23:36:03 -0500 mailto:topband@contesting.com>
```K2AV: "Ground resistance is a CONSTANT in a given setting, the same as an ordinary resistor. IT DOES NOT VARY depending on induced current short of loss heat converting ground moisture to steam or enhancing evaporation, thus changing a physical characteristic. The induction of current usually varies positionally as antenna elements are changed. The degree of LOSS attributed to ground is a complex sum of the instant current times resistance at points in the ground. This is a power value, which, given the current at the radial feed, can be expressed as a SUMMARY resistance. This is the useful term when combined with radiation resistance, allows us to express the apportionment of our hard-bought power to worm-warming and chasing DX." .... --- No one is claiming that ground resistance varies depending on antenna length or anything else associated with the antenna. The question is "what is the equivalent resistance that would have to be added at the feedpoint to equal the ground loss, if the real near field ground loss in the model was set to zero". That is the thing that varies with antenna length. There are some charts floating around compiled by data from the broadcast industry that shows equivalent series resistance for ground loss, but that data was compiled for 1/4 wave verticals. There aren't any charts for longer antennas or for inverted Ls. Consider for a moment using a Mininec ground. If you do any analysis using a Mininec ground you have to add that resistor to get something realistic. Those charts should give something close when analyzing a quarter wave vertical, but not necessarily close for longer antennas, because there is no data available for that antenna. For example if you have a 1/2 wave vertical, the point of maximum current density in the earth moves out to about 0.35 from the base of the antenna. So if you had 1/4 wavelength radials with that antenna, then there would be no radials present in the area where the current density was maximum. Calculating the feedpoint resistance and subtracting the radiation resistance doesn't give you a value of resistance that would represent ground loss for this antenna. If you could somehow refer that loss resistance that happened at considerable distance from the feedpoint, back to the feedpoint, then it might be OK. I don't know how to do that. If NEC2 or NEC4 modeled the ground correctly, we wouldn't have to use Mininec. (Besides a Mininec ground gives a significant error for inverted Ls.) All we would have to do is model the radials and put it over a real ground. We would have the right antenna gain, and by creating a close to zero ground loss radial system and then adding a resistor at the feedpoint and jugging that resistor until we got the same gain as before, we would have an equivalent ground loss resistor referred to the feedpoint. However I have no confidence that NEC will give an accurate gain answer in this case. Several people have noticed large discrepancies between NEC calculations and measured data for very low wires. A dipole doesn't act like a radial system but it is an antenna where there is some experimental data available to which NEC comparisons can be made. Consider the data generated by Hagn-Baker at this link: http://www.w8ji.com/nvis_n_v_i_s_antenna.htm That data agrees reasonably close with NEC calculations until the antenna reaches somewhere around 0.02 wavelengths above ground. At 0.005 wavelengths NEC is in error by about 6 dB. So when you are modeling radials close to the ground how much error do you expect and why? Jerry, K4SAV _______________________________________________ UR RST IS ... ... ..9 QSB QSB - hw? BK ```
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https://www.sophia.org/tutorials/evaluating-works-of-art-2 | 1,606,770,417,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00612.warc.gz | 841,095,881 | 15,732 | ### Online College Courses for Credit
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# Evaluating Works of Art
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Evaluate works of art from this unit according to religious tradition or media used.
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Tutorial
## Video Transcription
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[MUSIC PLAYING] I'd like to welcome you to this episode of Exploring Art History with Ian. My name is Ian O'Connell and today's lesson is about the early Renaissance in Florence. As we're watching the video, feel free to pause, move forward, or line as often as you feel is necessary. And as soon as you're ready, we can begin.
Today's objectives, or the things you're going to learn today, are listed below. By the end of the lesson today you will be able to identify and define today's key terms, explain the influence of the princely courts in Italy during this time, and explain innovations in perspective and naturalism during this time using the artwork provided in this lesson. Key terms, as always, are listed in yellow throughout the lesson.
First key term's linear perspective, a form of perspective in which the lines of man-made objects, like roads and buildings, are at right angles to the picture plane and converge toward a vanishing point. Orthogonal. In mathematics two lines or curves are orthogonal if they are perpendicular or at right angles at the point of intersection.
Vanishing point. The point on the horizon where the straight lines of linear perspective converge. The two sides of a road appear to come together at the vanishing point, for example. Naturalism. An artistic approach that involves reproducing objects as they appear to the eye. This term is often used in art history as a substitute for "realistic" to avoid any confusion with Realism as an artistic movement.
The big idea for today that the spread of humanism, a growing emphasis on individual achievement, and the princely courts of Italy were all highly influential on the early renaissance artwork of Italy. The time period that we're looking at today is about 1401 to 1440 AD. And once again, we are in Florence, Italy.
So what were the princely courts and why do we care? Well, it's important to remember that Italy wasn't a kingdom at this time. There was no centralized power like there were in other areas of Europe. It was made up of independent city-states, which could be duchys; papal states, which are owned by the church; or republics like Florence and Venice.
The term "princes" was just a term applied to the people that were in charge in those areas. Could be, for instance, a Duke or a Count, didn't necessarily have to be a prince. And they were enormously wealthy for the most part, and strong supporters of the arts like the Duke of Urbino, which is pictured here on the right. It's also important to know that they are one of the main sources of artistic patronage at this time.
So the resurgence of humanism during this time did much to swing the pendulum away from the Byzantine style and towards that of the classical. Artists were interested in the naturalism and scientific rationalism that developed in ancient Greece, and applying it to the work that they were producing. But it wasn't just the artists. These ideas were penetrating in all areas of culture and the arts.
The artists' desire to utilize them were as prevalent as the patrons' desire to see them. And it was a rediscovered ideal of classical form and beauty that found a pairing with the religious imagery of the time, which is a combination which really celebrated God's greatest creation, which is man, and humanity's devotion to its creator.
So the nude form had essentially vanished in the artwork of the Middle Ages. It was viewed as obscene and idolatrous, a relic of the pagan past. Within religious imagery, however, it could be reinterpreted as a depiction of innocence. Now the commission to create the baptistery doors was the prize given to the winner of a contest in which several artists were given the task of creating a bronze panel relief depicting the sacrifice of Isaac by Abraham.
Now the contests came down to two finalists, the artist Brunelleschi and the actual winner, the artist Ghiberti. Whose submission was a much calmer depiction of the event, as I'll show you in just a moment, compared to the dynamism displayed in Brunelleschi's work. So this is Brunelleschi's panel. You can just see the movement, much more dynamic than in the calmer image you'll see in just a moment. Notice how the angel's rushing in to stop Abraham from killing his son, and Abraham's clothes flowing back as he rushes towards Isaac.
Now although there's still some sacrificing going on here, it's not as dynamic or it's not as energetic as the previous panel. So his rendering of Isaac, Ghiberti's rendering of Isaac as a male nude is particularly interesting and could be considered the first example of a male nude since antiquity, which is a title usually given in the statue of David by the artist Donatello, which we'll see in a moment.
His winning is largely attributed to the artist's superior skill at casting compared to that of Brunelleschi. Who, thanks in large part to this loss, went back to study architecture and eventually designed one of the most impressive domes in history, the dome of the Florence Cathedral. It's probably good that he lost. And here are the baptistery doors as they look today.
Now the mathematical description of perspective was undoubtedly influential on the artist Masaccio and his Holy Trinity painting from the Church of Santa Maria Novella in Florence, Italy. Now he used his understanding of linear perspective to near perfection in how he essentially creates extremely realistic depth in his composition. However, the entire image is a painting, a Fresco in fact, which is an important form of painting in Italy and a medium suitable to its mild climate. For instance, you wouldn't see Fresco paintings in northern Europe.
Now the image is divided into two sections. At the bottom a crypt. The imagery and writing above it a reminder of what is to come for all mankind. And above it are the patrons, a husband and wife kneeling and praying below the image of Christ crucified. Flanked on either side by his mother, Mary, and by Saint John, and held by the image of God the Father.
Now the vanishing point is below the feet of Christ, and the entire scene is fixed within a triumphal arch of sorts with a coffered barrel vaulted ceiling that recedes believably back into space. It's actually been said that the overall mathematical execution of Masaccio's use of perspective is so precise that the virtual dimensions of the architecture that recedes into space can accurately be calculated. Not too shabby.
And here's a close up again of the bottom crypt with the writing about the skeleton. Now it's the artist Donatello that's often credited with bringing back the male nude, specifically with his sculpture of David that was commissioned by members of the wealthy and influential Medici family to sit within their courtyard.
Now David is depicted as a youth standing in the Grecian contrapposto style pose with one hand on his sword, one hand on his hip, and one foot on the severed head of Goliath. Now we've seen the return to naturalism that was so celebrated during the Classical era with a clear understanding and deep appreciation for the mechanics of the human expressed in this sculpture.
Now the sculpture of Saint Mark by Donatello was another impressive example of the artist's understanding of human form; human mechanics, such as the contrapposto stance; and perspective. The realism is really striking, for example, in the veins that can be seen in the hands of the Saint. It's a realism that I personally feel wasn't repeated until the later work by the artist Michelangelo, but I'm sure that's up for debate.
Donnatello's understanding of perspective, however, isn't. He's mastered it, and incorporates it into the proportions of the statue. It was intended to be viewed in a niche well above eye level, and so Donatello adjusted accordingly. From straight on, the statue looks slightly disproportionate particularly with the head, but these adjustments were made so that from the original vantage point, where the person would be below the statue, the statue appears perfect.
Let's take a look at our objectives to see how we did. Now that we've seen the lesson are we able to identify and define today's key terms? Can you explain the influence of the princely courts in Italy during this time? And explain innovations in perspective and naturalism during this time using the artwork provided in this lesson.
And once again, the big idea for today is that the spread of humanism, the growing emphasis on individual achievement, and the princely courts of Italy were all highly influential on the early Renaissance artwork of Italy. And that's it. Thank you very much for joining me today. I'll see you next time.
of
Attributions
Rating | 2,019 | 9,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.932995 |
https://www.geeksforgeeks.org/transform-one-string-to-another-using-minimum-number-of-given-operation/?ref=rp | 1,669,986,150,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00178.warc.gz | 834,760,679 | 29,885 | # Transform One String to Another using Minimum Number of Given Operation
• Difficulty Level : Hard
• Last Updated : 15 Jun, 2022
Given two strings A and B, the task is to convert A to B if possible. The only operation allowed is to put any character from A and insert it at front. Find if it’s possible to convert the string. If yes, then output minimum no. of operations required for transformation.
Examples:
```Input: A = "ABD", B = "BAD"
Output: 1
Explanation: Pick B and insert it at front.
Input: A = "EACBD", B = "EABCD"
Output: 3
Explanation: Pick B and insert at front, EACBD => BEACD
Pick A and insert at front, BEACD => ABECD
Pick E and insert at front, ABECD => EABCD```
Recommended Practice
Checking whether a string can be transformed to another is simple. We need to check whether both strings have same number of characters and same set of characters. This can be easily done by creating a count array for first string and checking if second string has same count of every character.
How to find minimum number of operations when we are sure that we can transform A to B? The idea is to start matching from last characters of both strings. If last characters match, then our task reduces to n-1 characters. If last characters don’t match, then find the position of B’s mismatching character in A. The difference between two positions indicates that these many characters of A must be moved before current character of A.
Below is complete algorithm.
1) Find if A can be transformed to B or not by first creating a count array for all characters of A, then checking with B if B has same count for every character.
2) Initialize result as 0.
3) Start traversing from end of both strings.
……a) If current characters of A and B match, i.e., A[i] == B[j]
………then do i = i-1 and j = j-1
b) If current characters don’t match, then search B[j] in remaining
………A. While searching, keep incrementing result as these characters
………must be moved ahead for A to B transformation.
Below are the implementations based on this idea.
## C++
`// C++ program to find minimum number of operations required``// to transform one string to other``#include ``using` `namespace` `std;` `// Function to find minimum number of operations required to``// transform A to B.``int` `minOps(string& A, string& B)``{`` ``int` `m = A.length(), n = B.length();` ` ``// This parts checks whether conversion is possible or not`` ``if` `(n != m)`` ``return` `-1;`` ``int` `count[256];`` ``memset``(count, 0, ``sizeof``(count));`` ``// count characters in A`` ``for` `(``int` `i = 0; i < n; i++)`` ``count[A[i]]++;`` ``// subtract count for every character in B`` ``for` `(``int` `i = 0; i < n; i++)`` ``count[B[i]]--;`` ``// Check if all counts become 0`` ``for` `(``int` `i = 0; i < 256; i++)`` ``if` `(count[i])`` ``return` `-1;` ` ``// This part calculates the number of operations`` ``// required`` ``int` `res = 0;`` ``for` `(``int` `i = n - 1, j = n - 1; i >= 0;) {`` ``// If there is a mismatch, then keep incrementing`` ``// result 'res' until B[j] is not found in A[0..i]`` ``while` `(i >= 0 && A[i] != B[j]) {`` ``i--;`` ``res++;`` ``}`` ``// If A[i] and B[j] match`` ``if` `(i >= 0) {`` ``i--;`` ``j--;`` ``}`` ``}`` ``return` `res;``}` `// Driver program``int` `main()``{`` ``string A = ``"EACBD"``;`` ``string B = ``"EABCD"``;`` ``cout << ``"Minimum number of operations required is "` `<< minOps(A, B);`` ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`
## C
`// C program to find minimum number of operations required``// to transform one string to other``#include ``#include ` `// Function to find minimum number of operations required to``// transform A to B.``int` `minOps(``char` `A[], ``char` `B[])``{`` ``int` `m = ``strlen``(A), n = ``strlen``(B);` ` ``// This parts checks whether conversion is`` ``// possible or not`` ``if` `(n != m)`` ``return` `-1;`` ``int` `count[256];`` ``for` `(``int` `i = 0; i < 256; i++)`` ``count[i] = 0;`` ``// count characters in A`` ``for` `(``int` `i = 0; i < n; i++)`` ``count[A[i]]++;`` ``// subtract count for every character in B`` ``for` `(``int` `i = 0; i < n; i++)`` ``count[B[i]]--;`` ``// Check if all counts become 0`` ``for` `(``int` `i = 0; i < 256; i++)`` ``if` `(count[i])`` ``return` `-1;` ` ``// This part calculates the number of operations`` ``// required`` ``int` `res = 0;`` ``for` `(``int` `i = n - 1, j = n - 1; i >= 0;) {`` ``// If there is a mismatch, then keep incrementing`` ``// result 'res' until B[j] is not found in A[0..i]`` ``while` `(i >= 0 && A[i] != B[j]) {`` ``i--;`` ``res++;`` ``}`` ``// If A[i] and B[j] match`` ``if` `(i >= 0) {`` ``i--;`` ``j--;`` ``}`` ``}`` ``return` `res;``}` `// Driver program``int` `main()``{`` ``char` `A[] = ``"EACBD"``;`` ``char` `B[] = ``"EABCD"``;`` ``printf``(``"Minimum number of operations required is %d"``, minOps(A, B));`` ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`
## Java
`// Java program to find minimum number of operations``// required to transform one string to other``import` `java.io.*;``import` `java.util.*;` `public` `class` `GFG {` ` ``// Function to find minimum number of operations`` ``// required to transform A to B.`` ``public` `static` `int` `minOps(String A, String B)`` ``{` ` ``// This parts checks whether conversion is possible`` ``// or not`` ``if` `(A.length() != B.length())`` ``return` `-``1``;` ` ``int` `i, j, res = ``0``;`` ``int` `count[] = ``new` `int``[``256``];` ` ``// count characters in A`` ``// subtract count for every character in B`` ``for` `(i = ``0``; i < A.length(); i++) {`` ``count[A.charAt(i)]++;`` ``count[B.charAt(i)]--;`` ``}` ` ``// Check if all counts become 0`` ``for` `(i = ``0``; i < ``256``; i++)`` ``if` `(count[i] != ``0``)`` ``return` `-``1``;` ` ``i = A.length() - ``1``;`` ``j = B.length() - ``1``;` ` ``while` `(i >= ``0``) {`` ``// If there is a mismatch, then keep`` ``// incrementing result 'res' until B[j] is not`` ``// found in A[0..i]`` ``if` `(A.charAt(i) != B.charAt(j))`` ``res++;`` ``else`` ``j--;`` ``i--;`` ``}`` ``return` `res;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``String A = ``"EACBD"``;`` ``String B = ``"EABCD"``;` ` ``System.out.println(`` ``"Minimum number of operations required is "`` ``+ minOps(A, B));`` ``}``}` `// This code is contributed by Aditya Kumar (adityakumar129)`
## Python3
`# Python program to find the minimum number of``# operations required to transform one string to other` `# Function to find minimum number of operations required``# to transform A to B``def` `minOps(A, B):`` ``m ``=` `len``(A)`` ``n ``=` `len``(B)` ` ``# This part checks whether conversion is possible or not`` ``if` `n !``=` `m:`` ``return` `-``1` ` ``count ``=` `[``0``] ``*` `256` ` ``for` `i ``in` `range``(n): ``# count characters in A`` ``count[``ord``(B[i])] ``+``=` `1`` ``for` `i ``in` `range``(n): ``# subtract count for every char in B`` ``count[``ord``(A[i])] ``-``=` `1`` ``for` `i ``in` `range``(``256``): ``# Check if all counts become 0`` ``if` `count[i]:`` ``return` `-``1` ` ``# This part calculates the number of operations required`` ``res ``=` `0`` ``i ``=` `n``-``1`` ``j ``=` `n``-``1` ` ``while` `i >``=` `0``:`` ` ` ``# if there is a mismatch, then keep incrementing`` ``# result 'res' until B[j] is not found in A[0..i]`` ``while` `i>``=` `0` `and` `A[i] !``=` `B[j]:`` ``i ``-``=` `1`` ``res ``+``=` `1` ` ``# if A[i] and B[j] match`` ``if` `i >``=` `0``:`` ``i ``-``=` `1`` ``j ``-``=` `1` ` ``return` `res` `# Driver program``A ``=` `"EACBD"``B ``=` `"EABCD"``print` `(``"Minimum number of operations required is "` `+` `str``(minOps(A,B)))``# This code is contributed by Bhavya Jain`
## C#
`// C# program to find minimum number of``// operations required to transform one``// string to other``using` `System;` `class` `GFG``{` `// Function to find minimum number of``// operations required to transform``// A to B.``public` `static` `int` `minOps(``string` `A, ``string` `B)``{` ` ``// This parts checks whether`` ``// conversion is possible or not`` ``if` `(A.Length != B.Length)`` ``{`` ``return` `-1;`` ``}` ` ``int` `i, j, res = 0;`` ``int``[] count = ``new` `int` `[256];` ` ``// count characters in A` ` ``// subtract count for every`` ``// character in B`` ``for` `(i = 0; i < A.Length; i++)`` ``{`` ``count[A[i]]++;`` ``count[B[i]]--;`` ``}` ` ``// Check if all counts become 0`` ``for` `(i = 0; i < 256; i++)`` ``{`` ``if` `(count[i] != 0)`` ``{`` ``return` `-1;`` ``}`` ``}` ` ``i = A.Length - 1;`` ``j = B.Length - 1;` ` ``while` `(i >= 0)`` ``{`` ``// If there is a mismatch, then`` ``// keep incrementing result 'res'`` ``// until B[j] is not found in A[0..i]`` ``if` `(A[i] != B[j])`` ``{`` ``res++;`` ``}`` ``else`` ``{`` ``j--;`` ``}`` ``i--;`` ``}`` ``return` `res;``}` `// Driver code``public` `static` `void` `Main(``string``[] args)``{`` ``string` `A = ``"EACBD"``;`` ``string` `B = ``"EABCD"``;` ` ``Console.WriteLine(``"Minimum number of "` `+`` ``"operations required is "` `+`` ``minOps(A, B));``}``}` `// This code is contributed by Shrikant13`
## PHP
`=0; )`` ``{`` ``// If there is a mismatch, then keep incrementing`` ``// result 'res' until B[j] is not found in A[0..i]`` ``while` `(``\$i``>=0 && ``\$A``[``\$i``] != ``\$B``[``\$j``])`` ``{`` ``\$i``--;`` ``\$res``++;`` ``}`` ` ` ``// If A[i] and B[j] match`` ``if` `(``\$i` `>= 0)`` ``{`` ``\$i``--;`` ``\$j``--;`` ``}`` ``}`` ``return` `\$res``;``}`` ` `// Driver program` `\$A` `= ``"EACBD"``;``\$B` `= ``"EABCD"``;``echo` `"Minimum number of operations "``.`` ``"required is "``. minOps(``\$A``, ``\$B``);``return` `0;``?>`
## Javascript
``
Output:
`Minimum number of operations required is 3`
Time Complexity: O(n), please note that i is always decremented (in while loop and in if), and the for loop starts from n-1 and runs while i >= 0.
Auxiliary Space: O(1).
Thanks to Gaurav Ahirwar for above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
My Personal Notes arrow_drop_up | 3,860 | 11,450 | {"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} | 3.703125 | 4 | CC-MAIN-2022-49 | latest | en | 0.861088 |
http://www.javaprogrammingforums.com/java-theory-questions/41003-rabin-miller-rsa-program.html | 1,553,185,571,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202526.24/warc/CC-MAIN-20190321152638-20190321174638-00144.warc.gz | 303,500,706 | 11,401 | ## Rabin Miller for RSA program
Hey guys
I have a finished RSA program. Using the BIGINTEGER methode
It can crypt and decrypt numbers, primes
I generated the primes with bigintegers built in function..
However Im asked to do the primes more properly using a rabbin miller test
Im supposed to generate x number of primes with x numer bitlength, 512,1024 etc
Also to generate 20 random "a" as a test
I cant post link so, search : wikipedia java miller, theres a pseduocode that works
, scroll down to Java
I get the mathematical operations done here, I know the Rabbin miller
but im not quite sure how to turn this into a prime generator
also how to generate 20 random a?
--- Update ---
import java.math.BigInteger;
import java.util.Random;
public class MillerRabin {
private static final BigInteger ZERO = BigInteger.ZERO;
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = new BigInteger("2");
private static final BigInteger THREE = new BigInteger("3");
public static boolean isProbablePrime(BigInteger n, int k) {
if (n.compareTo(ONE) == 0)
return false;
if (n.compareTo(THREE) < 0)
return true;
int s = 0;
BigInteger d = n.subtract(ONE);
while (d.mod(TWO).equals(ZERO)) {
s++;
d = d.divide(TWO);
}
for (int i = 0; i < k; i++) {
BigInteger a = uniformRandom(TWO, n.subtract(ONE));
BigInteger x = a.modPow(d, n);
if (x.equals(ONE) || x.equals(n.subtract(ONE)))
continue;
int r = 0;
for (; r < s; r++) {
x = x.modPow(TWO, n);
if (x.equals(ONE))
return false;
if (x.equals(n.subtract(ONE)))
break;
}
if (r == s) // None of the steps made x equal n-1.
return false;
}
return true;
}
private static BigInteger uniformRandom(BigInteger bottom, BigInteger top) {
Random rnd = new Random();
BigInteger res;
do {
res = new BigInteger(top.bitLength(), rnd);
} while (res.compareTo(bottom) < 0 || res.compareTo(top) > 0);
return res;
}
public static void main(String[] args) {
// run with -ea to enable assertions
String[] primes = { "1", "3", "3613", "7297",
"226673591177742970257407", "2932031007403" };
String[] nonPrimes = { "3341", "2932021007403",
"226673591177742970257405" };
int k = 40;
for (String p : primes)
assert isProbablePrime(new BigInteger(p), k);
for (String n : nonPrimes)
assert !isProbablePrime(new BigInteger(n), k);
EDIT: Posted the code from wikipedia
the ending string thing and top bottom thing I assume I can skip
The mathematical operations for rabbin implemented in java, I can use them
So create 20 random A, generate x number of primes of x bits
and print two primes p and q to use for my rsa program
Im gonna have the programs separatley | 689 | 2,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.486024 |
https://scoste.fr/mas2020/ | 1,632,170,854,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00590.warc.gz | 569,415,853 | 4,264 | # The characteristic polynomial of sparse zero-one matrices
Journées MAS 2021
# Eigenvalues of non-Hermitian matrices
> using LinearAlgebra
> eigvals(randn(500,500))
# Example: random regular digraph
> using LinearAlgebra, Erdos
> eigvals(random_regular_digraph(500, 3))
# My favourite example: Bernoulli, sparse
> using LinearAlgebra
> eigvals(rand(500,500).<0.01)
A_n = an n \times n matrix whose entries are iid \mathrm{Bernoulli}(d/n) entries.
# Reverse characteristic polynomial
q_n (z) = \det(I_n - zA_n)
The coefficients of q_n(z)=1+\Delta_1z+\Delta_2z^2+...+\Delta_{n}z^{n} are \Delta_k = (-1)^k \frac{P_k(\mathrm{trace}(A_n^1), ..., \mathrm{trace}(A_n^k))}{k!},
where the P_k are polynomials.
# The limits of the traces of A^k_n
For every k, (\mathrm{tr}(A_n^1), ..., \mathrm{tr}(A_n^k)) \xrightarrow[n \to \infty]{\mathrm{law}} (X_1, ... , X_k). where X_k := \sum_{\ell|k} \ell Y_\ell (Y_\ell : \ell \in \mathbb{N}^*) = family of independent r.v., Y_\ell \sim \mathrm{Poi}(d^\ell / \ell).
# The limits of the coefficients of q_n
\Delta_k \to a_k = (-1)^k \frac{P_k(X_1, ... , X_k)}{k!} Let F be the log-generating function of these random variables: F(z) = 1 + \sum_{k=1}^\infty a_k z^k
Coefficients of q_n \to Coefficients of F
Do we have stronger convergence than that?
# Weak convergence of analytic functions
Shirai 2012
If f_n is a sequence of random analytic functions in an open set D and if
1. The coeffs of f_n converge towards (a_k)
2. f_n is tight in D
Then f_n \to f where f(z) = \sum a_k z^k.
# Tightness in holomorphic spaces
Let f_n be a sequence of random analytic functions: f_n(z) = \sum_{k=0}^\infty a_{n,k}z^k.
If there is a c such that \sup_n \mathbf{E}[|a_{n,k}|^2] \leqslant c r^k then (f_n) is tight on D(0,\sqrt{r}).
# Tightness of (q_n)
The sequence q_n is tight in D(0,\sqrt{1/d}).
Proof. We must bound the 2-norm of the coefficients of q_n, the \Delta_k.
We use \Delta_k = \sum_{I \subset [n], |I|=k}\det(A(I)) then develop |\Delta_k|^2.
We get a double sum of \mathbf{E}[\det(A(I))\det(A(J))] with I,J subsets of [n].
The value of each summand depends on the size of I\cap J.
\mathbf{E}[|\Delta_k|^2] = (n)_k (d/n)^k (1-d/n)^{k-1}(1 - kd/n -p + kd - k^2d/n) =O(d^k)
q_n \to F as holomorphic functions on D(0,d^{-1/2}).
# Extras on F
F(z) = \exp \left( -\sum_{k=1}^\infty X_k \frac{z^k}{k} \right) = \prod_{k=1}^\infty (1 - z^k)^{Y_k}
• The radius of convergence inside the exp is 1/d.
• The radius of convergence of F is 1/\sqrt{d} and F has one zero at 1/d.
• F has no other zeroes inside D(0,1/\sqrt{d}).
# Zeroes of q_n => zeroes of F
The zeroes are continuous wrt weak convergence on \mathbb{H}.
Zeroes of q_n inside D(0,1/\sqrt{d}) = inverse of eigenvalues of A_n outside D(0,\sqrt{d}).
Asymptotically, A_n has one eigenvalue close to d.
The other ones are smaller than \sqrt{d}.
# Friedman theorems everywhere
Can you have a short proof of Friedman’s 2\sqrt{d-1}-theorem?
1. Prove that the non-backtracking traces converge towards something [Dumitriu et al 2012]
2. Prove that q_n is tight…
Bonne rentrée à tous ! | 1,091 | 3,108 | {"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} | 3.28125 | 3 | CC-MAIN-2021-39 | longest | en | 0.6684 |
http://www.jiskha.com/display.cgi?id=1396890803 | 1,462,016,779,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111838.20/warc/CC-MAIN-20160428161511-00155-ip-10-239-7-51.ec2.internal.warc.gz | 626,457,276 | 3,577 | Saturday
April 30, 2016
# Homework Help: college algebra
Posted by candace on Monday, April 7, 2014 at 1:13pm.
If a 25 lb weight is attatched to a spring how far will it stretch? I got an answer for a 15 lb weight with 8 inches to be 15/8 but when I plug in 15/8 into my new equation it doesn't come up with the right answer. | 100 | 328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.95699 |
https://www.hackmath.net/en/math-problem/15053?tag_id=8_13 | 1,611,025,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00306.warc.gz | 801,304,396 | 16,573 | # Angled cyclist turn
The cyclist passes through a curve with a radius of 20 m at 25 km/h. How much angle does it have to bend from the vertical inward to the turn?
Correct result:
A = 13.5567 °
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Showing 1 comment:
Matematik
A cyclist has to bend slightly towards the center of the circular track in order to make a safe turn without slipping.
Let m be the mass of the cyclist along with the bicycle and v, the velocity. When the cyclist negotiates the curve, he bends inwards from the vertical, by an angle θ. Let R be the reaction of the ground on the cyclist. The reaction R may be resolved into two components:
(i) the component R sin θ, acting towards the center of the curve providing necessary centripetal force for circular motion and
(ii) the component R cos θ, balancing the weight of the cyclist along with the bicycle.
Thus for less bending of the cyclist (i.e for θ to be small), the velocity v should be smaller and radius r should be larger. let h be the elevation of the outer edge of the road above the inner
edge and l be the width of the road then,
Tips to related online calculators
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See also our right triangle calculator.
See also our trigonometric triangle calculator.
Try conversion angle units angle degrees, minutes, seconds, radians, grads.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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http://www.mypearsonstore.com/bookstore/applied-basic-mathematics-coursesmart-etextbook-0321738381 | 1,440,852,345,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064445.47/warc/CC-MAIN-20150827025424-00026-ip-10-171-96-226.ec2.internal.warc.gz | 591,394,897 | 6,564 | # Applied Basic Mathematics, CourseSmart eTextbook, 2nd Edition
Published Date: Jan 20, 2011
More Product Info
## Description
By teaching with an emphasis on real world applications, Applied Basic Mathematics, Second Edition, offers students a refreshing perspective on basic math! Authors Clark and Brechner open every chapter with an application and close every section with “Apply Your Knowledge” examples. The authors pair this emphasis on applications with learning tools like “Guide Problems and the “Ten Minute Chapter,” resulting in a text that encourages students to relate, reinforce, and review as they learn.
1. Whole Numbers
1.1 Understanding the Basics of Whole Numbers
1.3 Subtracting Whole Numbers
1.4 Multiplying Whole Numbers
1.5 Dividing Whole Numbers
1.6 Evaluating Exponential Expressions and Applying Order of Operations
1.7 Solving Application Problems
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
2. Fractions
2.1 Factors, Prime Factorizations, and Least Common Multiples
2.2 Introduction to Fractions and Mixed Numbers
2.3 Equivalent Fractions
2.4 Multiplying Fractions and Mixed Numbers
2.5 Dividing Fractions and Mixed Numbers
2.6 Adding Fractions and Mixed Numbers
2.7 Subtracting Fractions and Mixed Numbers
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
3. Decimals
3.1 Understanding Decimals
3.3 Multiplying Decimals
3.4 Dividing Decimals
3.5 Working with Fractions and Decimals
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
4. Ratio and Proportion
4.1 Understanding Ratios
4.2 Working with Rates and Units
4.3 Understanding and Solving Proportions
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
5. Percents
5.1 Introduction to Percents
5.2 Solve Percent Problems Using Equations
5.3 Solve Percent Problems Using Proportions
5.4 Solve Percent Application Problems
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
6. Measurement
6.1 The U. S. Customary System
6.2 Denominate Numbers
6.3 The Metric System
6.4 Converting between the U. S. System and the Metric System
6.5 Time and Temperature
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
7. Geometry
7.1 Lines and Angles
7.2 Plane and Solid Geometric Figures
7.3 Perimeter and Circumference
7.4 Area
7.5 Square Roots and the Pythagorean Theorem
7.6 Volume
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
8. Statistics and Data Presentation
8.1 Data Presentation — Tables and Graphs
8.2 Mean, Median, Mode, and Range
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
9. Signed Numbers
9.1 Introduction to Signed Numbers
9.3 Subtracting Signed Numbers
9.4 Multiplying and Dividing Signed Numbers
9.5 Signed Numbers and Order of Operations
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
10. Introduction to Algebra
10.1 Algebraic Expressions
10.2 Solving an Equation Using the Addition Property of Equality
10.3 Solving an Equation Using the Multiplication Property of Equality
10.4 Solving an Equation Using the Addition and Multiplication Properties
10.5 Solving Application Problems
10-minute Chapter
Numerical Facts of Life
Chapter Review Exercises
Assessment Test
Glossary
Appendix A: Try-It Exercise Solutions
Appendix B: Answers to Selected Exercises
Appendix C: Math Study Skills
Appendix D: Table of Squares and Square Roots
Index
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# State whether the given statement is true or false: A bag contains 4 red and 8 blue marbles. A marble is drawn at random. The probability of drawing red is 13.
A
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Solution
## The correct option is A True There are 4 red marbles out of a total of 12 marbles. So, the probability of drawing red is 412 = 13.
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# Which of the following numbers is greater than three fourths of the nu
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Which of the following numbers is greater than three fourths of the nu [#permalink]
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26 Jun 2009, 00:53
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38 69 22 73 31 47 13 82
Which of the following numbers is greater than three fourths of the numbers but less than one fourth of the numbers in the list above?
A) 56
B) 68
C) 69
D) 71
E) 73
[Reveal] Spoiler: OA
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Re: Which of the following numbers is greater than three fourths of the nu [#permalink]
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26 Jun 2009, 18:02
3
KUDOS
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This post was
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Quote:
38 69 22 73 31 47 13 82
Which of the following numbers is greater than three fourths of the numbers but less than one fourth of the numbers in the list above?
A) 56
B) 68
C) 69
D) 71
E) 73
38 69 22 73 31 47 13 82
This can sorted into:
13 22 31 38 47 69 73 82
three fourth of the numbers:
13 22 31 38 47 69
one fourth of the numbers:
73 82
Hope it helps...
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Which of the following numbers is greater than three-fourth of the num [#permalink]
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29 Nov 2010, 13:58
38, 69, 22,73, 31,47, 13, 82
Which of the following numbers is greater than three-fourth of the numbers but less than one-fourth of the numbers in the list above?
A. 56
B. 68
C. 69
D. 71
E. 73
Math Expert
Joined: 02 Sep 2009
Posts: 39720
Re: Which of the following numbers is greater than three-fourth of the num [#permalink]
### Show Tags
29 Nov 2010, 14:11
ericnkem wrote:
hey guys!
38, 69, 22,73, 31,47, 13, 82
Which of the following numbers is greater than three-fourth of the numbers but less than one-fourth of the numbers in the list above?
a- 56
b- 68
c- 69
d- 71
e- 73
The list contains 8 numbers: 13, 22, 31, 38, 47, 69, 73, and 82.
More than three-fourth of the numbers so more than 6 numbers (3/4*8=6) and less than one-fourth of the numbers so less than 2 number (1/4*8=2) means more than 69 (6th greatest number) and less than 73. From the answer choices only 71 satisfies this condition.
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Re: Which of the following numbers is greater than three-fourth of the num [#permalink]
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29 Nov 2010, 15:31
Thanks a lot Bunuel! I figured it out a bit later.
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Re: Which of the following numbers is greater than three-fourth of the num [#permalink]
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29 Nov 2010, 16:25
Important:- Arrange the numbers in ascending order first.
13, 22, 31, 38, 47, 69, 73, 82
3/4th of the number list represents- 69 (6th number)
remaining 1/4th list represents 73 (7 th number)
71 fits in between above 2
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Re: Which of the following numbers is greater than three-fourth of the num [#permalink]
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29 Nov 2010, 22:45
I goofed up and picked 69 figuring i need to pick a number from the list. Need to read more carefully.
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Posts: 2010
Re: Which of the following numbers is greater than three fourths of the nu [#permalink]
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06 Sep 2011, 11:49
joyseychow wrote:
38 69 22 73 31 47 13 82
Which of the following numbers is greater than three fourths of the numbers but less than one fourth of the numbers in the list above?
A) 56
B) 68
C) 69
D) 71
E) 73
Arrange the numbers in ascending order:
13, 22, 31, 38, 47, 69, 73, 82 (Just cross check the count to ensure that you didn't miss anything)
How many numbers do we see?
8, alright.
What's (1/4)8?
2.
What's (3/4)8?
6.
We're asked to find a number that are greater than 6 numbers in the list AND smaller than the remaining two.
13, 22, 31, 38, 47, 69, 73, 82
13, 22, 31, 38, 47, 69: Find a number greater than any of these numbers.
73, 82: Find a number smaller than any of these numbers.
Only 71 fits the description.
Ans: "D"
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Re: Which of the following numbers is greater than three fourths of the nu [#permalink]
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Re: Which of the following numbers is greater than three fourths of the nu [#permalink]
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26 Oct 2015, 13:54
It made me think a lot. GMAT literally plays with language. This is not more than a 600 level question (and that too because of the usage of the language)
Re: Which of the following numbers is greater than three fourths of the nu [#permalink] 26 Oct 2015, 13:54
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Display posts from previous: Sort by | 1,975 | 6,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.878653 |
https://opticsgirl.com/introduction-to-nonlinear-optics/ | 1,670,223,590,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00469.warc.gz | 469,267,593 | 13,749 | Introduction to Nonlinear Optics
INTRODUCTION
If light never interacted with matter, then nonlinear optics would not exist (as well as, of course, fiber optics, lasers, etc). This is because Maxwell’s equation in vacuum are completely linear (although it is theoretically predicted that above the Schwinger limit, the quantum vacuum electromagnetic field would also be nonlinear, but that is beyond the scope of this page and my pay-grade). They can acquire nonlinear terms, however, when they interact with matter with sufficient intensity to produce a nonlinear reaction of the material. The nonlinear reaction of the material then couples back to the light field and voila! You have nonlinear optics! Typically in most books and courses in optics, you will find that we treat the nonlinearities arising from the electric dipole moment response, or the polarization. However, there can be some nonlinear effects arising from the magnetic dipole moment response as well.
Starting with the electric flux density:
$$D = \epsilon_0E + P$$
We know that typically in a linear medium, the polarization is proportional to the applied field, $P = \epsilon_0\chi_eE$, where $\chi$ could be a rank tensor for a medium if it is anisotropic. To make the formulation easier, however, we will assume that the $\chi$ tensors are scalars for the moment and write the polarization in terms of the nonlinear susceptibility:
$$P = \epsilon_0[\chi^{(1)}E+\chi^{(2)}E^2+\chi^{(3)}E^3+…]$$
where the nonlinear polarization is represented by $E^2,~E^3$, etc. These nonlinear terms can arise due to motions of bound electrons, field-induced motions of atoms in their constituent materials, acoustic wave generation, thermal effects, etc. Since the nonlinearities associated with bound electrons are generally very fast, on the order of 1-2 fs, the motion of bound electrons as a result of an applied field are of particular importance in ultrafast optics. Full calculation of the susceptibility in the presence of a high intensity field would require quantum mechanics, but we can get the conceptual understanding using the simple example of an anharmonic oscillator. We discuss the harmonic oscillator in both the “Lasers” and the “Classical Optics” portions of this web page, so look there for refreshment if you don’t remember this model, or even if it’s completely new to you!
FORCED WAVE EQUATION
As a standard treatment in nonlinear optics, we can consider the nonlinear optical polarization as a source term in the wave equation. Assuming a uniform medium with the polarization as the only source, we can write the wave equation as:
$$\nabla \times \nabla \times E = \nabla(\nabla\cdot{E})-\nabla^2E = -\mu_0\frac{\partial^2D}{\partial{t^2}}$$
The \textbf{D} field can then be written:
$$D = \epsilon_0E+P_{(1)}+P_{NL} = D_{(1)}+P_{NL}$$
where $P_{NL}$ is the nonlinear polarization, $P_{(1)}$ is the first order polarization, and $D_{(1)}$ is the first-order portion of the $D$ field. Gauss’s Law then yields:
$$\epsilon_{(1)}\nabla\cdot{E} = -\nabla\cdot{P_{NL}}$$
We can then make the simplifying assumption that the polarization is weak, so we have:
$$\nabla\cdot{E} \approx 0$$
The wave equation now simplifies to give:
$$\nabla^2E – \mu_0\epsilon_{(1)}\frac{\partial^2E}{\partial{t^2}} = \mu_0\frac{\partial^2P_{NL}}{\partial{t^2}}$$ | 840 | 3,341 | {"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": 7, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.909363 |
https://stats.stackexchange.com/questions/16872/if-a-sequence-of-distributions-converges-to-a-degenerate-does-that-imply-the-va | 1,638,469,382,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00336.warc.gz | 546,555,642 | 35,740 | # If a sequence of distributions converges to a degenerate, does that imply the variance strictly decreases?
If $F_i = G(F_{i-1}), F_0 = x$ is a sequence of distributions that converges to a degenerate distribution as $i \to \infty$, does that imply that the variance of $F_i$ decreases with $i$?
Specifically, I am interested in the inverse Kumaraswamy distribution: $G(x) = (1-(1-x)^a)^b$ when $a=b$... Note that $G$ produces the cdf of the Max of b draws from the distribution of the Min of a draws from its argument (which is going to be a cdf).
So if it's not true for the general case, but you have insights on how/why it would be true in this case, I would appreciate it. I can intuit (and graph for specific values of a) that the mean and variance decrease in this case -- taking the Min of $a$ draws from $F$ decreases the mean, then taking the max of $a=b$ draws from the new distribution increases the mean, but not by as much as the first operation lowered it, so the final mean is less than that of the original. But I am shaky on how to prove it (and my other questions today have basically been trying to get at this point).
PS -- sorry to overload with questions today! As I mentioned, they've all been in the goal of this question here.
• In general, no, of course not. Let $X_1, X_2, \ldots$ be zero-mean normal random variables with variances $\sigma_i^2$. Choose any sequence of $\sigma_i^2 \geq 0$ such that $\sigma_i^2 \to 0$. (The point being, there is no need for $\sigma_i^2$ to be monotone.) Oct 11 '11 at 20:51
• If you assume all the $F_i$ are supported in $[0,1]$, then the variance will have to converge to 0. If you only assume the limiting distribution is supported in $[0,1]$, then the variances do not have to converge! (Consider a mixture of a uniform and a Normal distribution, with the location of the Normal distribution diverging faster than the inverse square of its mixture weight.)
– whuber
Oct 11 '11 at 20:51
• After reading my comments and @whuber, it becomes apparent to me that there are at least two ways this question can be read as currently written. :) (My comments have also been colored by the title of this question.) Oct 11 '11 at 20:54
• @cardinal I think your reading is more accurate (if I have understood it correctly :-)), but it seems to lead to questions of little interest or utility. Because convergence of a sequence of distributions is independent of the phenomenon "decreases with $i$", the two should have no relationship with one another. In fact, swapping every other term in a monotonically convergent sequence won't change its convergence properties at all but it ruins monotonicity...
– whuber
Oct 11 '11 at 20:59
• @JandR: You're not being slow. We're just trying to figure out the real question of interest. If a sequence of random variables converges to a degenerate distribution, then the variance of the limiting distribution will be zero. However, I believe the variances of the random variables in the sequence need not even be defined. Take, for example, a sequence of Cauchy distributions with a scale parameter converging to zero. Oct 11 '11 at 21:08
Here is a simple counterexample.
Assume the CDF $F$ is supported on $[0,1]$ and define a new CDF $G[F]$ as follows. If $\mathbb{E}[F] \gt 1/2$, let
$$G[F](x) = 1 - F(3/2 - 2 x).$$
Otherwise, let
$$G[F](x) = 1 - F(5/6 - 2x/3).$$
The first operation squeezes the distribution by a factor of $2$ towards $1/2$ and flips it around $1/2$, while the second expands the distribution by a factor of $3/2$ away from $1/2$ and also flips it around $1/2$. The flipping guarantees that if the mean of $F$ is other than $1/2$, then iterating $G$ will alternate between these two operations, because the expectations will alternately be greater than and less than $1/2$. The net result nevertheless is to compress the support down towards $1/2$, converging to a degenerate distribution. However, the variance in the first case is multiplied by $1/4$ and in the second case it is multiplied by $9/4$, whence it does not uniformly decrease: it alternately jumps up and down. Therefore convergence of a sequence of distributions $(G^n[F])$ to an atom does not imply monotonic decrease of the variances.
You ask for intuition. Maybe the following bit of hand-waving will help.
Provided $F$ is continuous, your particular sequence, $G^n[F]$ with $G[F](x) = (1-(1-F(x))^a)^a$, indeed eventually has monotonically decreasing variances because $G$ is a smooth contracting map. Specifically, $G$ has a fixed point $x_0$ (the unique root of $G[x]=x$ in the interval $(0,1)$) and it contracts all values towards that fixed point.
(The figure shows $F(x)=x$ and the first three iterations of $G$ for $a=2$.)
This alone is not enough for an easy proof (I don't think), but notice in addition that the derivative of $G$ near that fixed point always strictly exceeds $1$. (It is never smaller at the fixed point than $6 - 2\sqrt{5} \approx 1.528$, when $a=2$.) After enough iterations, almost all the probability is squeezed into a region near this fixed point, where $G$ acts essentially as a rescaling operation, which will shrink the variance at each step. (Any probability outside this small neighborhood cannot contribute much to the variance because the support of $G^n[F]$ is bounded independent of $n$.) The shrinking means the variance eventually behaves almost like a geometric series with common ratio $1/G'(x_0)^2$, which will monotonically decrease. I think a little epsilon-delta analysis could make this argument rigorous. | 1,420 | 5,557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-49 | latest | en | 0.929617 |
https://www.101computing.net/tag/binary/ | 1,718,427,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00250.warc.gz | 561,186,631 | 32,368 | ## Logic Gates & Truth Tables
Learning Objectives In this post you will predict the output of logic gates circuits by completing truth tables. First you need to learn the basic truth tables for the following logic gates: AND Gate OR Gate XOR Gate NOT Gate
## Binary Converter using Python
Did You Know?Everything that is stored on a computer is stored as binary code. Binary code is made of bits (0 or 1). We often use Bytes to store data. A Byte is made of eight bits and can be
## Access Rights using Binary Masks
Imagine a building with 8 rooms. Each room is fitted with a card reader used to give access or not to different members of staff. Each member of staff has a card giving them access to some of the rooms.
## Digital Data – Terminology
Complete this domino activity to revise key definitions on how text files, pictures and sound files are stored on a computer using binary code.
## Secret Code
In this challenge we are putting your deciphering skills to the test. Here are your secret codes. Will you be able to decode these? Cipher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ipher #2 \u0055\u006e\u0069\u0063\u006f\u0064\u0065 \u0069\u0073 \u0061 \u0063\u0068\u0061\u0072\u0061\u0063\u0074\u0065\u0072 \u0073\u0065\u0074 \u0075\u0073\u0069\u006e\u0067 \u0032 \u0042\u0079\u0074\u0065\u0073 \u0070\u0065\u0072 \u0063\u0068\u0061\u0072\u0061\u0063\u0074\u0065\u0072 \u0077\u0068\u0069\u0063\u0068
## Binary Logic Activities
Access a range of activities about binary logic, logic gates, truth tables and Karnaugh maps:
## Karnaugh Maps
Karnaugh maps are a tool used to simplify complex Boolean expressions. Using a Karnaugh map reduces the need for extensive calculations by taking advantage of humans’ pattern-recognition capability. Karnaugh maps are used to simplify real-world logic requirements so that they
## Why are there 1024 bytes in a kilobyte? (instead of 1000)
Here is a good brainteaser when teaching binary code… Remember: Computers can only work with binary code. Computers don’t really like the number 1000 because in binary it’s not that straightforward: The binary code for 1000 is 1111101000 However computers | 935 | 3,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.722222 |
https://fahmyalhafidz.com/3eqvjp/ | 1,656,599,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00796.warc.gz | 295,948,291 | 10,826 | # The couple is simplified easily by the help of right hand rule
The couple is simplified easily by the help of right hand rule in free body diagrams. But the forces simplification is not possible as there is no such system of rules so that the forces can be simplified.
A. The first part of the statement is false and other part is true
B. The first part of the statement is false and other part is false too
C. The first part of the statement is true and other part is false
D. The first part of the statement is true and other part is true too
This question was addressed to me in final exam.
I’d like to ask this question from Different Free Body Diagrams in section Equilibrium of a Rigid Body of Engineering Mechanics
Correct choice is C. The first part of the statement is true and other part is false
The explanation is: Both of them are vector quantities. And both of them can be easily simplified. If taken in the vector form then the task is even easier. Thus it is not necessary for the force or the couple to be vector only, even if the magnitude is taken, the simplification is done in the 2D. Forces are simplified on the basis of the algebra. | 254 | 1,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-27 | latest | en | 0.945986 |
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