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https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=265851 | 1,526,984,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00218.warc.gz | 718,991,960 | 3,860 | # carpentry 1-ratio and proportion mechanical advantage and percentage
The flashcards below were created by user heeres on FreezingBlue Flashcards.
1. the easiest way to compare ratios is?
to write them as a fraction first
2. how do you compare if ratios are =?
cross multiply
3. how can direct proportion ratio calculations be solved?
cross multiply then solve for the missing term.
4. always set up direct proportion ratios so the order is?
the order is the same on both sides of the equal sign
5. what is an example of an indirect proportion?
the greater the speed the less time it takes to reach the distination
6. a third class lever creates? | 144 | 649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-22 | latest | en | 0.915479 |
https://ncertbookspdf.com/chapter-12-triangles-class-6-dav-secondary-mathematics-2/ | 1,725,716,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00629.warc.gz | 389,206,370 | 17,288 | # Chapter 12 | Triangles | Class-6 DAV Secondary Mathematics
Are you looking for DAV Maths Solutions for class 6 then you are in right place, we have discussed the solution of the Secondary Mathematics book which is followed in all DAV School. Solutions are given below with proper Explanation please bookmark our website for further updates!! All the Best !!
## Unit 12 Worksheet 2 || Triangles
1. In the given triangle, name the—
(a) vertex opposite to side QR.
(b) side opposite to vertex Q.
(c) side opposite to vertex R.
(d) vertex opposite to side PQ.
2. Draw any triangle ABC and mark the points—
(a) X, Y in the interior of the ABC.
(b) R, S in the exterior of the ABC.
3. What do you mean by a triangular region?
4. Look carefully at the following figure and name—
(a) all the triangles formed.
(b) the triangles which have the point P in its exterior.
(c) the triangles which have the point P in its interior.
5. In the given figure, name the triangles which have—
(a) A as one vertex (any two)
(b) B as one vertex (any two)
(c) E as one vertex (any two)
(d) DF as one side (any two)
(e) BC as one side
(f) AE as one side | 284 | 1,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.915005 |
http://crypto.stackexchange.com/questions?page=139&sort=active | 1,469,647,009,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827077.13/warc/CC-MAIN-20160723071027-00140-ip-10-185-27-174.ec2.internal.warc.gz | 53,485,428 | 26,554 | All Questions
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ECDSA vs RSA: Performance on Android platform and surprising results
For our privacy-preserving protocol, an encrypted channel is established. In order to protect our system from man-in-the-middle attacks, signature-based approach is used. After we've implemented it ... | 788 | 3,298 | {"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} | 2.765625 | 3 | CC-MAIN-2016-30 | latest | en | 0.908033 |
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## Linear Function Examples Explanation and Application
### 3.4 Solving Real-Life Problems Big Ideas Math
Linear Equations Maths Accelerator. Linear equations are the simplest or to draw the graph of a linear equation It's easy to think of algebra as an abstract notion that has no use in real life., Linear Equations. Linear equations are those which make straight lines when graphed. Real life examples include: Calculating wages based on an hourly pay rate.
### Real-life application of a linear function brainmass.com
Real Life Application Linear Function TutorVista. Linear Algebra: Real World Uses . exercises; It finds the maximum or minimum of linear functions in many variables subject to economics application- profit, Linear Equations. Linear equations are those which make straight lines when graphed. Real life examples include: Calculating wages based on an hourly pay rate.
Linear Functions A linear function is In many applications, However, this form isn’t really as useful, since the constant b = 7 has no real meaning. Real-life examples of linear equations include distance and rate One application of linear equations When Will You Use Linear Equations in Real Life? A:
2.1 QUADRATIC FUNCTIONS AND MODELS Linear function many real-life applications—especially those involving 7.2 TWO-VARIABLE LINEAR SYSTEMS • Use systems of linear equations in two variables to model and solve real-life problems.
Question 103770: I need a real-life application of a linear function. State the application, give the equation of the linear function, and state what the x and y in Real world linear equations in action as well as free worksheet that goes hand in hand with this page's real world ,word problems.
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Practical Applications of Algebra. It's easy to think of algebra as an abstract notion that has no use in real life. Understanding the… Read more Home » Linear Equations » Applications of Linear Systems. Applications of Linear Systems. We will see many of these applications again in future lessons,
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2.1 QUADRATIC FUNCTIONS AND MODELS Linear function many real-life applications—especially those involving Real Life Application Linear Function Introduction to Linear function: Linear function is a polynomial function that has only one variablewith first degree .
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Are there any real life applications of general vector spaces? girls at a bar may be listed among "real life applications", and real-valued functions ? Linear Equations. Linear equations are those which make straight lines when graphed. Real life examples include: Calculating wages based on an hourly pay rate
### Real World Linear Graphs SlideShare
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### 3.4 Solving Real-Life Problems Big Ideas Math
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1-1 Linear Equations and Applications 3 The left and right members represent real numbers for all other replacements of x by real numbers. The solution set for an Solve various word problems that involve real world relationships that can be represented by linear equations or functions. Linear word problems. 800
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Applications of Linear Functions. Real world applications can also be modeled with multiple lines such as if two trains travel toward each Real Life Linear Model. 126 Chapter 3 Writing Linear Equations and Linear Systems Use what you learned about solving real-life problems to complete EXAMPLE 1 Real-Life Application
126 Chapter 3 Writing Linear Equations and Linear Systems Use what you learned about solving real-life problems to complete EXAMPLE 1 Real-Life Application I have provided a real life application of a linear function and shown how to obtain various quantities using this equation. I also have found the volume of a
Practical Applications of Algebra. It's easy to think of algebra as an abstract notion that has no use in real life. Understanding the… Read more Linear equations are the simplest or to draw the graph of a linear equation It's easy to think of algebra as an abstract notion that has no use in real life.
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## Linear word problems Algebra I Math Khan Academy
Linear function examples in real life StudyPug. 2/08/2015В В· Real Life Application of Linear Function Real Life Applications of Linear Equations Application Of Functions In Real Life Situations, Applications of Linear Functions. Real world applications can also be modeled with multiple lines such as if two trains travel toward each Real Life Linear Model..
### Real World Linear Graphs SlideShare
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Apply concepts of linear functions to real life problems, Algebra 1 students What are the applications of functions in real life? What is an example of a linear function's real life situation? do they have any real life applications?
In real world applications such as those Using linear equations in science. Linear equations can be used to describe many (or daily life), Linear Algebra: Real World Uses . exercises; It finds the maximum or minimum of linear functions in many variables subject to economics application- profit
1-1 Linear Equations and Applications 3 The left and right members represent real numbers for all other replacements of x by real numbers. The solution set for an Linear Algebra: Real World Uses . exercises; It finds the maximum or minimum of linear functions in many variables subject to economics application- profit
Linear Algebra: Real World Uses . exercises; It finds the maximum or minimum of linear functions in many variables subject to economics application- profit ... Linear Inequalities Systems of Linear Equations Functions and Relations Application of Function Composition. from a model of a real-life situation
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One of the most helpful ways to apply linear equations in everyday life is Linear Equations Used in Everyday Life? "How Are Linear Equations Used in Everyday APPLICATIONS OF LINEAR EQUATIONS The linear equation y mx b is a formula that determines a value of y for each given value of x. In this section you will study linear
126 Chapter 3 Writing Linear Equations and Linear Systems Use what you learned about solving real-life problems to complete EXAMPLE 1 Real-Life Application What are the real life applications of correlation and convolution? I am interested in knowing the real life application distribution functions gives the pdf of
In real world applications such as those Using linear equations in science. Linear equations can be used to describe many (or daily life), We believe that by presenting simple practical applications of algebra, students will gain a and real numbers, including before solving linear equations and
Real-life examples of linear equations include distance and rate One application of linear equations When Will You Use Linear Equations in Real Life? A: Solve various word problems that involve real world relationships that can be represented by linear equations or functions. Linear word problems. 800
Programa Bilingüe Español – Inglés. “IES Floridablanca”.- DNL: Matemáticas. Function: Concepts. Linear Functions. Functions in the Real World ... for a linear scale, but for most applications a linear of linear functions in our every day life.The following are the some example of real life linear
126 Chapter 3 Writing Linear Equations and Linear Systems 3.4 Solving Real-Life Problems How can you use a linear equation in two variables to model and solve a real Real Life Application Linear Function: Lot of real application of linear functions is always around us. We can found many examples of linear functions in our every
In real world applications such as those Using linear equations in science. Linear equations can be used to describe many (or daily life), Real Life Application Linear Function: Lot of real application of linear functions is always around us. We can found many examples of linear functions in our every
Students take a look at applications of linear functions. Explain some of the linear relationships that exist in the real about real-world applications, 2/12/2012В В· Real life applications of simultaneous equations. So I don't think that is an overused real-life application. Linear programming in D1 textbooks or
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Real world linear equations in action as well as free worksheet that goes hand in hand with this page's real world ,word problems. One of the most helpful ways to apply linear equations in everyday life is Linear Equations Used in Everyday Life? "How Are Linear Equations Used in Everyday
Real Life Application Linear Function: Lot of real application of linear functions is always around us. We can found many examples of linear functions in our every ... Linear Inequalities Systems of Linear Equations Functions and Relations Application of Function Composition. from a model of a real-life situation
In this guide, we'll go over some linear function examples to help you better understand the logic and application of linear functions. 7.2 TWO-VARIABLE LINEAR SYSTEMS • Use systems of linear equations in two variables to model and solve real-life problems.
1-1 Linear Equations and Applications 3 The left and right members represent real numbers for all other replacements of x by real numbers. The solution set for an Real World Linear Graphs Real World Linear Graphs 1. Real Life Straight Line Graphs and Why they are Important. Interesting applications of graph theory
LESSON 1: Solve systems of linear functions by graphing - TWO lines on the same graph?!? LESSON 2: Solve systems of linear equations by graphing - with some real-life REAL LIFE APPLICATION OF FUNCTIONS Some Real - Life Applications of Linear Functions Cost of the purchased item Total hours of travel and distance Transport fare
### What are the applications of functions in real life? Quora
Algebra 2 Application of Function Composition Algebra 2. Practical Applications of Algebra. It's easy to think of algebra as an abstract notion that has no use in real life. Understanding the… Read more, Application of Statistics in Daily Life Non-Linear Functions in Real Life. Using Nonlinear Functions in Real Life Situations Related Study Materials..
### Real World Linear Graphs SlideShare
Solve systems of linear equations by graphing with some. 2.1 QUADRATIC FUNCTIONS AND MODELS Linear function many real-life applications—especially those involving https://en.wikipedia.org/wiki/Linear_function In this section we will revisit some of the applications we saw in the linear application Applications of Quadratic Equations. get either two real distinct.
Are there any real life applications of general vector spaces? girls at a bar may be listed among "real life applications", and real-valued functions ? Solve various word problems that involve real world relationships that can be represented by linear equations or functions. Linear word problems. 800
Real World Linear Graphs Real World Linear Graphs 1. Real Life Straight Line Graphs and Why they are Important. Interesting applications of graph theory Apply concepts of linear functions to real life problems, Algebra 1 students
to have this math solver on your website, free of what are the difference between functions and linear equations ; expressing real life problems in algebraic APPLICATIONS OF LINEAR EQUATIONS The linear equation y mx b is a formula that determines a value of y for each given value of x. In this section you will study linear
Linear equations are the simplest or to draw the graph of a linear equation It's easy to think of algebra as an abstract notion that has no use in real life. ... Linear Inequalities Systems of Linear Equations Functions and Relations Application of Function Composition. from a model of a real-life situation
Linear Equations. Linear equations are those which make straight lines when graphed. Real life examples include: Calculating wages based on an hourly pay rate Practical Applications of Algebra. It's easy to think of algebra as an abstract notion that has no use in real life. Understanding the… Read more
Real Life Application Linear Function Introduction to Linear function: Linear function is a polynomial function that has only one variablewith first degree . LESSON 1: Solve systems of linear functions by graphing - TWO lines on the same graph?!? LESSON 2: Solve systems of linear equations by graphing - with some real-life
Linear Functions A linear function is In many applications, However, this form isn’t really as useful, since the constant b = 7 has no real meaning. 126 Chapter 3 Writing Linear Equations and Linear Systems 3.4 Solving Real-Life Problems How can you use a linear equation in two variables to model and solve a real
Applications of Linear Functions. Real world applications can also be modeled with multiple lines such as if two trains travel toward each Real Life Linear Model. 2/12/2012В В· Real life applications of simultaneous equations. So I don't think that is an overused real-life application. Linear programming in D1 textbooks or
Some years ago I used mobile phone or internet rates (for example, with basic fees and a given charge per minute or by data volume) to introduce and motivate the In this guide, we'll go over some linear function examples to help you better understand the logic and application of linear functions.
Robot Moe is assembling memory cards for computers. At 9:00 am, 52 cards had been assembled. At 11:00 am, a total of 98 had been made. Assuming the production rate is 2/08/2015В В· Real Life Application of Linear Function Real Life Applications of Linear Equations Application Of Functions In Real Life Situations
Mathematics Applied to Physics and Engineering Applications and Use of the Inverse Functions. Examples on how to apply and use inverse functions in real life Tutorial on how to apply and use inverse functions Applications and Use of the Inverse Functions. Examples on how to aplly and use inverse functions in real life
View all posts in Ballapur category | 4,492 | 22,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-05 | latest | en | 0.905492 |
https://ximpledu.com/en-us/limit-of-log-a-1-plus-x-over-x/ | 1,611,409,377,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538082.57/warc/CC-MAIN-20210123125715-20210123155715-00514.warc.gz | 1,081,901,196 | 3,973 | # Limit of (loga (1 + x))/x
How to use the limit of Limit of (loga (1 + x))/x to solve the given limit with a logarithmic function: formula, 1 example, and its solution.
## Formula
### Formula
The limit of (loga (1 + x))/x as x → 0 is
1/(ln a).
## Example
### Solution
First write the limit part and [log2 (1 + x)].
The inner part of [log2 (1 + x)] is (1 + x).
So write x
in the denominator.
To undo the denominator x,
write x in the numerator.
Write the denominator sin x.
So [log2 (1 + x)]/(sin x)
= [(log2 (1 + x))/x]⋅[x/(sin x)].
As x → 0,
(log2 (1 + x))/x → 1/(ln 2).
As x → 0,
(sin x)/x → 1.
So x/(sin x) → 1.
Limit of (sin x)/x
[1/(ln 2)]⋅1 = 1/(ln 2)
So 1/(ln 2) is the answer. | 266 | 702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-04 | latest | en | 0.78614 |
https://www.physicsforums.com/threads/force-to-open-the-gate.869748/ | 1,660,853,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00651.warc.gz | 835,135,371 | 15,979 | # Force to open the gate
foo9008
## Homework Statement
the gate is 3m wide , is hinged at point A , and rests against a smooth wall at B . Calculate the force on the gate due to seawater pressure and horizontal force exerted by wall at B . I have attached the sample answer below . I dont understand why the author use FH(6 ) instead of FH(10)
## The Attempt at a Solution
IMO , it should be FH(10) , am i right ? 10 is the distance from A to B
#### Attachments
• 167.PNG
19.8 KB · Views: 346
Homework Helper
Gold Member
In what direction does FH act?
foo9008
In what direction does FH act?
Horizontal | 162 | 610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-33 | latest | en | 0.887369 |
https://www.physicsforums.com/threads/solving-for-energy-involving-hyperbolic.278749/ | 1,510,974,328,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00312.warc.gz | 865,210,399 | 14,414 | # Solving for energy involving hyperbolic
1. Dec 10, 2008
### nick227
1. The problem statement, all variables and given/known data
What energy (in eV) should a beam of electrons have so that 0.1% of them are able to tunnel through a barrier of height 7.0eV and 1.0 nm wide? Start with the equation for T(E) and set it up with 1/T(E) on one side and let E/U=x for the unknown. Solve the equation for x and then E.
2. Relevant equations
T(E) (1+.25(U2/(E(U-E)))sinh2($$\alpha$$L))
$$\alpha$$ = ((2m(U-E))1/2)/h
3. The attempt at a solution
I get
0=x-x2-.25sinh2($$\alpha$$L)
this is after I make E/U=x. How can i solve for x and then solve for E?
2. Dec 12, 2008
### buffordboy23
Have you tried to look for a numerical solution?
EDIT: If you need to solve it analytically, you may need to rewrite the hyperbolic sine term another way:
http://en.wikipedia.org/wiki/Hyperbolic_tangent#Standard_algebraic_expressions
I actually derived this whole equation once. It required the use of an alternate expression for the hyperbolic sine term.
The numerical solution seems easiest.
Last edited: Dec 12, 2008 | 327 | 1,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-47 | longest | en | 0.902671 |
https://dgtal-team.github.io/doc-nightly/structDGtal_1_1concepts_1_1CSeparableMetric.html | 1,721,657,929,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00326.warc.gz | 182,923,103 | 6,798 | DGtal 1.5.beta
DGtal::concepts::CSeparableMetric< T > Struct Template Reference
Aim: defines the concept of separable metrics. More...
#include <DGtal/geometry/volumes/distance/CSeparableMetric.h>
Inheritance diagram for DGtal::concepts::CSeparableMetric< T >:
[legend]
## Public Types
typedef T::Point Point
Public Types inherited from DGtal::concepts::CMetricSpace< T >
typedef T::Point Point
typedef T::Space Space
typedef T::Value Value
typedef T::RawValue RawValue
## Public Member Functions
BOOST_CONCEPT_USAGE (CSeparableMetric)
void checkConstConstraints () const
Public Member Functions inherited from DGtal::concepts::CMetricSpace< T >
BOOST_CONCEPT_ASSERT ((CSpace< Space >))
BOOST_CONCEPT_ASSERT ((CQuantity< Value >))
BOOST_CONCEPT_ASSERT ((CQuantity< RawValue >))
BOOST_CONCEPT_USAGE (CMetricSpace)
void checkConstConstraints () const
## Private Attributes
myX
Point u
Point v
Point w
Point start
Point end
bool myBool
DGtal::Dimension dim
## Detailed Description
### template<typename T> struct DGtal::concepts::CSeparableMetric< T >
Aim: defines the concept of separable metrics.
Description of concept 'CSeparableMetric'
Separable metrics are metrics satsifying the monotonicity property. More formally, in dimension 2, consider two points $$p(x,y)$$, $$q(x',y')$$ with $$x<x$$. Let $$r( x'',0)$$ be a point on the x-axis such that $$d(p,r) = d(q,r)$$ and $$s(u,0)$$ be another point on the x-axis. A metric $$d$$ is monotonic if
$u < x'' \implies d(p,s) \leq d(q,s)$
and
$u > x'' \implies d(p,s) \geq d(q,s)$
# Valid expressions and semantics
Name Expression Type requirements Return type Precondition Semantics Post condition Complexity
hiddenBy predicate hiddenBy(u,v,w,startingPoint,endPoint,dim) u,v,w,startingPoint,endPoint of type Point, dim of type DGtal::Dimension startingPoint and endPoint only differ by their dim-th coordinate returns true if the intersection between the segment [startingPoint,endPoint] and the Voronoi cell associated with v is empty (hidden on the segment by u and w Voronoi cells). -
# Notes
Template Parameters
T the type that should be a model of concepts::CSeparableMetric.
Definition at line 101 of file CSeparableMetric.h.
## ◆ Point
template<typename T >
typedef T::Point DGtal::concepts::CSeparableMetric< T >::Point
Definition at line 105 of file CSeparableMetric.h.
## ◆ BOOST_CONCEPT_USAGE()
template<typename T >
DGtal::concepts::CSeparableMetric< T >::BOOST_CONCEPT_USAGE ( CSeparableMetric< T > )
inline
Definition at line 106 of file CSeparableMetric.h.
107 {
109 }
## ◆ checkConstConstraints()
template<typename T >
void DGtal::concepts::CSeparableMetric< T >::checkConstConstraints ( ) const
inline
Definition at line 110 of file CSeparableMetric.h.
111 {
112 // const method dummyConst should take parameter myA of type A and return
113 // something of type B
114 ConceptUtils::sameType( myBool, myX.hiddenBy(u,v,w,start,end,dim) );
115 }
void sameType(const T &, const T &)
Definition: ConceptUtils.h:117
## ◆ dim
template<typename T >
private
Definition at line 121 of file CSeparableMetric.h.
## ◆ end
template<typename T >
Point DGtal::concepts::CSeparableMetric< T >::end
private
Definition at line 119 of file CSeparableMetric.h.
## ◆ myBool
template<typename T >
bool DGtal::concepts::CSeparableMetric< T >::myBool
private
Definition at line 120 of file CSeparableMetric.h.
## ◆ myX
template<typename T >
T DGtal::concepts::CSeparableMetric< T >::myX
private
Definition at line 118 of file CSeparableMetric.h.
## ◆ start
template<typename T >
Point DGtal::concepts::CSeparableMetric< T >::start
private
Definition at line 119 of file CSeparableMetric.h.
## ◆ u
template<typename T >
private
Definition at line 119 of file CSeparableMetric.h.
## ◆ v
template<typename T >
private
Definition at line 119 of file CSeparableMetric.h.
## ◆ w
template<typename T >
private
Definition at line 119 of file CSeparableMetric.h.
The documentation for this struct was generated from the following file: | 1,080 | 4,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.501737 |
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### One Paragraph Summary
Always explore your data visually. Whatever specific hypothesis you have when you go out to collect data is likely to be worse than any of the hypotheses you’ll form after looking at just a few simple visualizations of that data. The most effective hypothesis testing framework in existence is the test of intraocular trauma.
### Context
This morning, I woke up to find that Neil Kodner had discovered a very convenient CSV file that contains geospatial data about every valid US zip code. I’ve been interested in the relationship between places and zip codes recently, because I spent my summer living in the 98122 zip code after having spent my entire life living in places with zip codes below 20000. Because of the huge gulf between my Seattle zip code and my zip codes on the East Coast, I’ve on-and-off wondered if the zip codes were originally assigned in terms of the seniority of states. Specifically, the original thirteen colonies seem to have some of the lowest zip codes, while the newer states had some of the highest zip codes.
While I could presumably find this information through a few web searches or could gather the right data set to test my idea formally, I decided to blindly plot the zip code data instead. I think the results help to show why a few well-chosen visualizations can be so much more valuable than regression coefficients. Below I’ve posted the code I used to explore the zip code data in the exact order of the plots I produced. I’ll let the resulting pictures tell the rest of the story.
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ``` ```zipcodes <- read.csv("zipcodes.csv") ggplot(zipcodes, aes(x = zip, y = latitude)) + geom_point() ggsave("latitude_vs_zip.png", height = 7, width = 10) ggplot(zipcodes, aes(x = zip, y = longitude)) + geom_point() ggsave("longitude_vs_zip.png", height = 7, width = 10) ggplot(zipcodes, aes(x = latitude, y = longitude, color = zip)) + geom_point() ggsave("latitude_vs_longitude_color.png", height = 7, width = 10) ggplot(zipcodes, aes(x = longitude, y = latitude, color = zip)) + geom_point() ggsave("longitude_vs_latitude_color.png", height = 7, width = 10) ggplot(subset(zipcodes, longitude < 0), aes(x = longitude, y = latitude, color = zip)) + geom_point() ggsave("usa_color.png", height = 7, width = 10)```
### Picture
#### (Longitude, Latitude) Heatmap without Non-States
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Click here to close (This popup will not appear again) | 773 | 3,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.915835 |
https://www.sciencefacts.net/standing-waves.html | 1,618,768,983,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038507477.62/warc/CC-MAIN-20210418163541-20210418193541-00042.warc.gz | 1,086,026,482 | 14,691 | Home / Physics / Standing Waves
# Standing Waves
## Definition: What is Standing Wave?
Standing wave, also called a stationary wave, is a combination of two waves moving in opposite directions, each having the same amplitude and frequency. This phenomenon is a result of interference of these two waves with their amplitudes are either adding or canceling out. Because the observed wave pattern is characterized by points that appear to be standing still, the pattern is often called a standing wave pattern.
### Standing Wave Animation
Image Courtesy: Physicsclassroom.com
## Examples of Standing Waves
These are a few situations where standing waves can be observed.
• Two people shaking either end of a jump rope
• Ocean or pool waves hitting a wall and bouncing back in the opposite direction
• Plucking the strings of a guitar. A guitarist can purposely alter the pitch of the sound by placing a finger on the string to shorten its length that is vibrating.
• Wind instruments that have air columns in them. The effective length of the tube determines the sound the pipe makes.
• In plates, rods, and diaphragms that are parts of percussion instruments and form an essential feature in many musical instruments
• In optical media, such as optical cavities and waveguides.
## How are Standing Waves Formed?
Standing waves can form under a variety of conditions. However, they are easily demonstrated in a medium that is finite or bounded. In general, standing waves can be produced by any two identical waves traveling in opposite directions that have the right wavelength. In a bounded medium, the vibrational waves emitted by a source is reflected from one end of the medium and interfere with the incident waves. The interference of these two waves produces a resultant wave that is stationary, i.e., a pattern that does not appear to move.
There is always energy associated with standing waves. This energy travels through a medium as a transverse pulse.
## Terms Associated with Standing Waves
Nodes: Points on a standing wave where the wave stays in a fixed position over time because of destructive interference.
Antinodes: Points on a standing wave where the wave vibrates with maximum amplitude due to constructive interference.
Fundamental frequency: Lowest frequency of a standing wave that has the fewest number of nodes and antinodes.
Harmonics: Higher frequencies that areintegral multiples of the fundamental frequency.
## Standing Wave Equation
Consider two sine waves of the same angular frequency (ω), wavelength (λ), wavenumber (k), and amplitude (A) that move in opposite directions. The displacements (y) of the waves as a function of position (x) and time (t) are described by
y1 = (A/2) sin (kx – ωt)
y2 = (A/2) sin (kx + ωt)
where the first wave is moving toward the right (positive x), and the second wave is moving toward the left (negative x). The factor of 1/2 is placed for convenience. Adding these two waves gives the total displacement at any point and time:
yT = y1 + y2 = (A/2) sin (kx – ωt) + (A/2) sin (kx + ωt) = (A/2)[sin (kx – ωt) + sin (kx + ωt)]
Recall that,
Sin (A+B) = sin A cos B + cos A sinB
Sin (A-B) = sin A cos B – cos A sin B
Therefore,
Sin (A-B) + Sin (A+B) = 2 sin A cos B
Hence,
yT = (A/2) 2sin (kx) cos (ωt)
Or, yT = sin (kx) cos (ωt)
In the boldface equation above, the displacement varies with distance as sin kx, but changing the time does not shift the location of peaks and troughs. It merely changes the amplitude. For example, at x=0, the displacement is always zero, no matter what the value of t. Hence, the velocity of a standing wave is zero.
## Standing Waves in a String
Standing waves are readily observed on a string, where the ends are fixed, and the string does not move. Places, where the string is not vibrating, are the nodes. The nodes limit the wavelengths that are possible, which in turn determines the possible frequencies since the speed of sound is fixed. The lowest frequency is called the fundamental frequency. Higher frequencies are all multiples of the fundamental frequency and are called harmonics. The various harmonics are also called the normal modes of vibration of the string.
## Standing Waves in Air Columns – Open and Closed Tube
Standing waves are a common phenomenon in air columns, pipes, and tubes. Waves are created at one end of the tube by something that vibrates. The waves then travel to the end of the tube and reflect. A standing wave is created by the waves traveling in each direction. Since the wave is traveling in the air, it will move at the speed of sound in air at the ambient temperature.
The length of a tube determines the frequency of a standing wave in the tube. Several complications arise depending on if one or both ends are closed or open. If an end is open, the air can move freely, and so there is maximum air movement or a displacement antinode at that end.
An open tube or pipe is defined as one that is open at both ends. For such a tube, a standing wave can form if the wavelength of sound results in antinodes at both ends.
Many musical instruments are made of a tube that is closed at one end but opened at the other. Such an instrument is known as a closed tube or pipe. Like an open tube, they can form a standing wave with the sound of an appropriate frequency. In this case, the resultant wave pattern forms an antinode at the open end and a node at the closed end.
## Applications of Standing Wave
• Detect leakage in pipeline systems
• X-ray standing wave (XSW) method is applied for investigation of the real structure of crystals
• For calibrating commercial and reference attenuators used in laboratories
Article was last reviewed on Thursday, August 6, 2020 | 1,283 | 5,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.936328 |
https://community.qlik.com/thread/176715 | 1,516,331,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00006.warc.gz | 647,012,968 | 22,100 | 3 Replies Latest reply: Aug 17, 2015 12:07 AM by jagan mohan rao appala
# Formula on both sides of the expression in the expression editor
Hi,
I have 3 columns: bdate, camnt and ramnt. I am trying to calculate the MTD sales based on these values.
sales = (camnt - ramnt)
However bdate is in a different format and requires a formula to be converted to date format:
bdate1 = date(floor(bdate/ 86400 + 25569))
Now my MTD expression looks something like this:
Sum({<\$(=date(floor(bdate/ 86400 + 25569))) = {">=\$(=MonthStart(max(date(floor(bdate/ 86400 + 25569))))) <=\$(=(max(date(floor(bdate/ 86400 + 25569)))))"}>} (camnt-ramnt))
However, all I get is the sales for that particular date which I have selected and not the cummulative sales of that month till that date. Can someone please tell me what am I doing wrong here?
Regards,
Saurav
• ###### Re: Formula on both sides of the expression in the expression editor
You can't use an expression instead of a field name on the left side of equal sign of a set analysis set modifier.
Transform your date values when you load your field in the script:
date(floor(bdate/ 86400 + 25569)) as bdate,
camnt,
ramnt
FROM ...;
Then your expression should be straight forward using this field.
• ###### Re: Formula on both sides of the expression in the expression editor
Once you have transformed your bdate using swuehl's suggestion in the script. Try this expression:
Sum({\$<bdate = {"\$(='>=' & Date(MonthStart(Max(bdate))) & '<=' & Date(Max(bdate)))"}>} (camnt-ramnt))
• ###### Re: Formula on both sides of the expression in the expression editor
Hi,
This is not possible, you have to format and arrive a new column in script as swuehl's suggested.
Regards,
jagan. | 462 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-05 | longest | en | 0.804594 |
https://infinitylearn.com/surge/question/mathematics/what-is-the-value-of/ | 1,701,657,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00145.warc.gz | 357,255,862 | 110,254 | What is the value of
What is the value of
1. A
2. B
3. C
4. D
Fill Out the Form for Expert Academic Guidance!l
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Solution:
Concept- We have to find the value of . The only way to solve the number with the power is write that number in its factor form and it will cancel the power. To solve the power apply the formula of exponentials i.e., .
Given that,
It can be written as, Hence, the correct answer is option (2) .
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Verify OTP Code (required) | 197 | 840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.805021 |
https://www.physicsforums.com/threads/how-late-did-the-man-arrive-to-catch-the-train.961096/ | 1,718,793,184,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00531.warc.gz | 830,572,012 | 35,125 | # How Late Did the Man Arrive to Catch the Train?
• Jenny Physics
In summary, a man arrives T seconds late to a train station and watches the last three cars of the train pass by. The next-to-the-next-to-last car takes time t to pass by the passenger, the next-to-last car takes time t' to pass by and the last car takes time t'' to pass by. Assuming the train's acceleration a is constant, the value of T can be determined using the equations of kinematics and the known length of the train. However, if we assume that there is no elastic deformation of the train, only two of the three time measurements are necessary to determine T. This is because the third time measurement is related to the other two through a simple substitution, and the solution for T can be
Jenny Physics
## Homework Statement
A man arrives ##T## seconds late to a train station. He watches the last three cars of the train pass by.
The next-to-the-next-to-last car takes time ##t## to pass by the passenger, the next-to-last car takes time ##t'## to pass by and the last car takes time ##t''## to pass by.
Assume the train's acceleration ##a## is constant. What is ##T##?
Kinematics.
## The Attempt at a Solution
The train departs at ##t=0## (man's time). At the time ##t_{l}=T## that the man arrives at the train station, the speed of the train is ##v_{l}=aT##. If the length of a car is ##L## then at time ##t_{2}=t_{l}+t## we have ##L=\frac{1}{2}at^{2}+v_{l}t##, at time ##t_{3}=t_{l}+t+t'## we have ##2L=\frac{1}{2}a(t+t')^{2}+v_{l}(t+t')## and at time ##t_{4}## we have ##3L=\frac{1}{2}a(t+t'+t'')^{2}+v_{l}(t+t'+t'')##. Therefore ##at^{2}+2v_{l}t=\frac{1}{2}a(t+t')^{2}+v_{l}(t+t')## or ##at^{2}+2aTt=\frac{1}{2}a(t+t')^{2}+aT(t+t')##. Solving for ##T## we get
$$T=\frac{t^{'2}+2tt'-t^{2}}{2(t-t')}$$
If we use ##L## and ##3L## we get instead ##\frac{3}{2}at^{2}+3v_{l}t=\frac{1}{2}a(t+t'+t'')^{2}+v_{l}(t+t'+t'')## or ##3at^{2}+6aTt=a(t+t'+t'')^{2}+2aT(t+t'+t'')## which leads to
$$T=\frac{(t+t'+t'')^{2}-3t^{2}}{2(2t-t'-t'')}$$
We could then equal these two expressions for ##T## and get some relation between ##t,t',t''##. This seems right to me but is it?
If acceleration "a" is contant, and each each car has length "L", then it's just enough to use t an t': the difference (t'-t) is proportional to the acceleration "a" and to "L" (and does not even depend on the initial speed of train when it starts going from the station)
We can also say (t''-t' =t'-t) if we assume a cartesian space-time (we ignore the effects of relativity contracting space "L" caused by acceleration where cars would possibly no longer have equal lengths): the last third car is not different from the second last car, we suppose the length of the train and each car is conserved (the train does not approaches the speed of light for the observer).
The two equations for T above should then be equivalent: just make the substitution (t''=2t'-t) implied by (t''-t' =t'-t) in the second formula, you should get the first one...
Jenny Physics
The unknowns are ## L ##, ## a ## , and ## v_l ##, with ## T=\frac{v_l}{a} ##. Three equations and 3 unknowns. I think what you did on your first attempt was correct, but I need to check it.
Jenny Physics
Note: we must also assume that there's no elastic deformation of cars (this elastic deformation exists for non relativist speeds, but for solid trains it should may be neglected... that's not really true in actual trains where elastic deformation really occurs at each attachment between each pair of cars, and is compensated by elastic springs, so not all cars start accelerating at the same time; the first car start moving when the last car is still steady for some measurable fraction of seconds, sometimes more for very long trains where we can really see the significant variation of total length during accelerations or brakes...)
The second equation gives T=(t'+t)/2 after the substitution.
So you can determine (a) directly from the average of (t') and (t), and from (L).
You don't need the third variable v_l, for the speed of the train when the last car exits the station, which only depends on the speed v_0 [of the front of the train when it leaves the station at time=0, but in your problem v_0=0] and from the impulsion (a), and (L) and the number of cars in the train.
Note that your solution based on the second does not depend at all on (t'').
You could also make the same substitution of variables in the 1st equation for T and get a solution that depends only on (t') and (t'') but not on (t).
Conclusion: you don't need three measurements of time, two of them are enough.
You may compare use the 3 measurements only to check that there was no elastic deformation of the train: if the two solutions gives different results, the difference is proportional to the elastic elongation of the train (not really caused by relativity, but by friction of car wheels on rails and by friction of wheels on their axis), which is more or less proportional to the acceleration (a) and to some mechnical properties of the building materials of cars used to resist this elongation.
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Jenny Physics
@Jenny Physics I believe both of your answers are correct. For this problem, ## t, t' ##, and ## t'' ## are all related, but I don't think the simple relation given in post 2 is correct. To determine that relation, simply make both of your answers equal, (assuming all of your algebra is correct), and simplify the relation as much as possible.
Jenny Physics
You can check by yourself: the substitution works on both formulas to give a solution to your problem, that does depends only on two time measurements (not three). The solution is necessarily the same if there's no elastic elongation of the train. And you've noted yourself that T=v_l/a:
We've got too many variables for a single solution of the non-elastic train: you only need to know t, t', and L to determine a and v_l, you don't even need to know how many cars were in the train.
If you need three times, you have a solution for an uniformly elastic train (assuming also that friction of wheels on rails or friction of wheels on their axis is uniform all along the train and the rails on which it runs), or possibly one of the cars from which you measured the passage duration was actually longer (not all cars are equal; but here also you can generally know the length of each car, by their known construction, or known location in the station when the train was parked there: these locations may be even marked on the waiting dock ground of the station even if the train is now gone; for trains like automatic metros, passenger cars usually are all with equal length, except the first and/or last car with engines, so that they align to the steps or protection doors that give access to the cars, and so that trains may have more or less cars depending on expected trafic of passengers)
Jenny Physics
Also for trains that transport very heavy weights, the cars cannot be safely built to be completely rigid. They would not resist to elongation forces and would crack. Instead, the structure is built with an elastic floor supported by non-rigid, non-straight horitontal bars which are slightly curved. The charge is then hold on one or several several points fixed along the bars supporting also the rotating wheel axis, and within cabins of passenger cars, the floor is slightly elastic (with several gliding plates).
The structure of trains cannot be completely rigid (including side walls to resist lateral accelerations when trains take curves: the cabin is constantly changing form; rails are also slightly elastic and can be elongated and slide on the balast, otherwise they would also crack when trains are passing on them, and they must also adapt to changes of temperature caused by weather conditions).
The inelastic trains and railways do not really exist. Any attempt to rigidify them have failed catastrophically, just like the fact that they require regular maintenance to check for the existence of cracks: this is mandatory routine inspection for planes, but needed as well for trains and railways, because the elasticity of materials is not eternal and small cracks in the structure will fataly occur; but the construction may help maintaining a good enough elasticity with minimum damages. In trains the elastic springs connecting cars, and the elastic curved bars supporting cabins are standard ; just like the elastic construction of wings in planes, or the elastic suspensions in all transportation systems with wheels, and the elastic transmission of every motor "engines" -- including for animal traction, or transmission of human force by flexible chains on a bicycle, without which bicycles would be too risky for human bones (even if they are protected by elastic muscles and membranes whose "maintenance" is ensured by the renewal of new living cells to recycle the damaged cells, replace them and reglue them with elastic fluids, a natural process which is much longer in bones than in muscles and membranes, that cannot resist high pressures for long like what bones allow).
Jenny Physics
Philippe Verdy said:
Also for trains that transport very heavy weights, the cars cannot be safely built to be completely rigid. They would not resist to elongation forces and would crack. Instead, the structure is built with an elastic floor supported by non-rigid, non-straight horitontal bars which are slightly curved. The charge is then hold on one or several several points fixed along the bars supporting also the rotating wheel axis, and within cabins of passenger cars, the floor is slightly elastic (with several gliding plates).
The structure of trains cannot be completely rigid (including side walls to resist lateral accelerations when trains take curves: the cabin is constantly changing form; rails are also slightly elastic and can be elongated and slide on the balast, otherwise they would also crack when trains are passing on them, and they must also adapt to changes of temperature caused by weather conditions).
The inelastic trains and railways do not really exist. Any attempt to rigidify them have failed catastrophically, just like the fact that they require regular maintenance to check for the existence of cracks: this is mandatory routine inspection for planes, but needed as well for trains and railways, because the elasticity of materials is not eternal and small cracks in the structure will fataly occur; but the construction may help maintaining a good enough elasticity with minimum damages. In trains the elastic springs connecting cars, and the elastic curved bars supporting cabins are standard ; just like the elastic construction of wings in planes, or the elastic suspensions in all transportation systems with wheels, and the elastic transmission of every motor "engines" -- including for animal traction, or transmission of human force by flexible chains on a bicycle, without which bicycles would be too risky for human bones (even if they are protected by elastic muscles and membranes whose "maintenance" is ensured by the renewal of new living cells to recycle the damaged cells, replace them and reglue them with elastic fluids, a natural process which is much longer in bones than in muscles and membranes, that cannot resist high pressures for long like what bones allow).
That's all very well and good, but does nothing to address the OP's puzzle. Please restrict your replies to homework help requests to matters that directly pertain the problem at hand and in the spirit in which it was framed. Relativity was not mentioned, strength of materials was not mentioned, neither was flexibility nor does manner of deformation of the train or tracks appear in the original problem statement.
If you wish to open a discussion about incidental matters that occur to you after reading a homework help request, the technical forums would be a better place to post. Members their would be more than happy to engage such topics.
DaveE, Merlin3189 and Jenny Physics
The comments was about the contest that claimed there was a need of 3 time variables in the problem.
I argue that only 2 are needed but only under some strict inelastic conditions that actually never exist in any real train.
At least 3 time variables are needed for the minimum elastic problem which is the real one but more variables may be needed if we want to be very precise (because elasticity is not uniform across actual materials, and not constant over time when materials are aging).
These are real physical contraints far from the "ideal" situation of the theoretical-only initial problem: we can use this simple solution however, if we don't seek very high precision and accept an actual margin of error.
The highest error will be most probably in the three measurements (t,t',t'') of time and in the knowledge of the exact value of (L).
You can minimize the errors if you have (t,t',t'') by solving the two equations for T using only the two largest times (t+t', t+t'+t''), if these measures were taken on the same clock with the same known absolute precision.
Jenny Physics
Philippe Verdy said:
The comments was about the contest that claimed there was a need of 3 time variables in the problem.
I argue that only 2 are needed but only under some strict inelastic conditions that actually never exist in any real train.
At least 3 time variables are needed for the minimum elastic problem which is the real one but more variables may be needed if we want to be very precise (because elasticity is not uniform across actual materials, and not constant over time when materials are aging).
These are real physical contraints far from the "ideal" situation of the theoretical-only initial problem: we can use this simple solution however, if we don't seek very high precision and accept an actual margin of error.
The highest error will be most probably in the three measurements (t,t',t'') of time and in the knowledge of the exact value of (L).
You can minimize the errors if you have (t,t',t'') by solving the two equations for T using only the two largest times (t+t', t+t'+t''), if these measures were taken on the same clock with the same known absolute precision.
That's all very well, but the assumption that the train has constant acceleration trumps all that!
Jenny Physics
"constant acceleration of the train" ? So why I was conter-argued that we needed 3 time measurements? The experiment exposed in the problem is already testing the elastic model if it really needs the 3 time measurement variables (in which case there's NO constant acceleration of the **whole** train: the experiment effectively uses 3 distinct measurements on selected **parts** of the train, assuming already that acceleration may be variable). It's not clear what is "constant aceleration of the train", it may just mean the constant acceleration of the leading engined car...
Jenny Physics
Philippe Verdy said:
We can also say (t''-t' =t'-t)
Let us make a concrete model to test this assertion. In our model the train is constructed from materials we store in the same box that we keep our physicists best stuff, like massless pulleys and light flexible strings. In other words, no flexing, stretching, or other shenanigans.
Each car has length L = 15 m. Acceleration is a constant a = 1 m/s2. The passenger arrives when the train is already moving at vo = 5 m/s.
We can calculate the speed of the train for each instant when a car has passed, and the duration of time that car took to pass:
Not the same.
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jbriggs444, haruspex, Klystron and 2 others
Philippe Verdy said:
We've got too many variables for a single solution of the non-elastic train: you only need to know t, t', and L to determine a and v_l, you don't even need to know how many cars were in the train.
We are not given L. It is an unknown constant.
Jenny Physics
Philippe Verdy said:
We can also say (t''-t' =t'-t)
Clearly not. Add more carriages. Asymptotically, the time per carriage tends to zero, so the differences will tend to zero.
Philippe Verdy said:
the difference (t'-t) is proportional to the acceleration "a"
Did you mean inversely proportional to a? Even if so, that would require some proof. It is not evidently true.
Jenny Physics
@haruspex, you're wrong on BOTH your claims.
The problem specifies "v_l" this limits the number of cars because "v_l" is already the speed reached at the station where the train started to go.
This implies a fixed number of cars which can "tend to infinity", so the time per carriage cannot tend to zero.
Suppose "v_l" is not known, then the number of cars must be known because the problem searches the time since when the train quitted the station, at the measurement location.
The difference of time between two cars really grows proportionaly to acceleration: if there was no acceleration at all (=0), this difference of times between carriages (t'-t) or (t''-t') between cars of equal length is also zero, so it can '''certainly not''' be inversely proportional to this null acceleration (otherwise this time difference would be... infinite !).
In fact such type of measurement is how we traditionally measure speeds, on naval ships when there's no other visible fixed point, by dragging a rope with regular "knots", the knots being like the points separating cars in a train, and consequently also accelerations by measuring time differences between equal number of knots (this measured speed however is relative to the speed and direction of water current and assumes this current is constant)
I agree with @haruspex on the first claim (post 14), and I mentioned this in post 5 above. And I also agree, as a couple of people have stated, that they only needed to give ## t ## and ## t' ##, and that is sufficient to solve for the time that the passenger was late, as the OP correctly computed with the first answer they got for ## T ##.
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PeroK and Jenny Physics
Philippe Verdy said:
The problem specifies "v_l" this limits the number of cars because "v_l" is already the speed reached at the station where the train started to go.
This implies a fixed number of cars which can "tend to infinity", so the time per carriage cannot tend to zero.
The idea is not to add cars to the front of the train, but to imagine cars following the "last" car in the problem. So the passenger arrives just as some car in a long string of cars is passing him. Clearly, as the number of cars passing at a constant acceleration grows, the velocity must grow and hence time to pass for each must diminish. The same must hold true for only three cars.
haruspex, Jenny Physics and Charles Link
I don't know where the formulas used by gneill to compute successive speeds come from "magically".
The problem speaks about a train where the passenger clearly knows the number of cars (he was arriving in the station where he expected to see the first car before it left the station: how does he know then he is counting the last cars if he does not know how the train is composed?
Yes there are is some implied variable not clearly stated about train composition (the fixed number N of cars in the train, or its total length NL), but also an additional "polluting" unneeded variable in the 3 measurements of time.
The measurement of times between cars and the knowledge of length L per car directly gives average speeds for the time each car passes in front of the traveler, like knots on a naval ship. It is the difference of these times that gives differences of speeds, which is proportional (not inversely proportional) to the "constant" acceleration, by definition of an acceleration.
These measurements are not *instantaneous* measurements of speeds and acceleration, but just averaged values between separate instants (the actual definition would require car lengths to tend to zero: this is no longer a train with cars of finite length, this is a single thread made with infinitely narrow cars, and the "3 last time measurements" would be all zero in that case).
Now comes the problem: the speed of train is also not specified: "He watches the last three cars of the train pass by." but we can assume the passenger was knowing how many cars there were in the train, and knew the length of each car, or that he knows the speed of the train when he arrives to the station.
Otherwise there's no solution at all to the problem if we have absolutely no knowledge at all for what is in front of the train when the passenger arrived to the station and no way to know the instantaneous speed of the train (which is why the passenger takes times measurements between the last 3 cars passing in front of him).
If the passenger assumes that the train is infinitely long and only sees the last 3 cars and knows absolutely nothing about the length of the train, this is non-sense. The passenger arrives in a station whose docking area has a known finite length were he could wait for the train before coming too late to take it. The total length of the train is necessarily not infinite, and must be known (otherwise it could not even park along the waiting dock of the station, these docks would also be need to be infinitely long, and the traveller would have no chance at all to take the train on time in the last car where he would have reserved a seat, he would always be too late to take it), otherwise this is not a real train station.
The problem states that the train contains more than 3 cars. The traveler coming to the station cannot walk indefinitely, the station is necessarily limited in total length, and this also limits the total length of the train.
So you can argue that the problem did not state all necessary variables, but still it exposes an unnecessary time variable. Something must be missing, the problem is not complete in the case we don't allow any assumption about:
- what is a traveler (a human that cannot walk at infinite speed to reach a specific car position in due time before the train starts going),
- what is a train (a non-infinite vehicle of finite length),
- what is a station (a non-infinite dock area where travelers can wait their train and walk to the right position to enter the correct car in which he will find his seat).
Let's assume that the position where the traveler arrived too late in the station was the correct position to enter the correct car in the train (so he would have not needed to walk along the dock if he was in time when the train arrived and was parked waiting for travelers to come inside), then we know that he's looking at the trail of the train and not interested at all in what was on the front side: we reduce the train only to these last 3 cars, the rest of the train does not matter (he does not need to know then how many cars have already passed in front of that position)
May be the train only had these 3 cars and there's no other cars in front, the traveler could not measure a non zero speed of the train because it has not already left the station so its speed must be zero, and the passenger is not too late to take his train, and this gives a contradiction. So the train necessarily has more than 3 cars (at least 4, but not an infinity: it is still limited by the total length of the station).
Jenny Physics
Philippe Verdy said:
The difference of time between two cars really grows proportionaly to acceleration:
Let's see...
Suppose one car takes Δt, the next takes Δt', and the speed at the boundary is v.
I get 2v(Δt-Δt')=a(Δt2+Δt'2).
Maybe you have some other formulation, using something other than v?
Edit: using L instead of v:
L(Δt-Δt')=aΔtΔt'(Δt+Δt')
So if we were to double a then the factor that must be applied to all the time deltas to get the same L is 1/√2.
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Jenny Physics
There's a way to represent the problem geometrically: if the acceleration of the train is constant, the graph of the position x of the start of the train according to time forms an vertical parabolic curve. The 3 measurements are made at regular intervals (of length L) along the vertical x axis, and indicate 4 positions on the parabolic arc made at regularly spaced vertical positions.
The traveller does not measure all points on the arc, only the 3 nodes on the parabolic arc, and its time measurements are vertical distance along the time axis between 4 instants. Basically the travelers only knows 3 chords of the parabolic arc, and wants to know the length to the origin of both axis where the parabolic arc was tangeant to the horizontal axis (speed=0 at x=0).
How many parobolic curves like this can exist that passes by the 4 points of the plane ? if the origin of x and y-axis are unknown (time measurements are indications of absolute time with an unknown origin at T horizontal distance of the instant he saw the first car passing), we know that the parabolic arc must have its focal axis vertical, this does not leave many choices: there's a single vertical parabole passing throught the four nodes which are equally spaced vertically and whose relative horizontal distances (measured time differences) is also determined by the problem.
This is a geometric problem, it's not possible to have multiple parabols. And in fact only 3 nodes are needed to define the parabole. A 4th one may give no solution at all.
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Philippe Verdy said:
There's a way to represent the problem geometrically: if the acceleration of the train is constant, the graph of the position x of the start of the train according to time forms an horizontal parabolic curve. The 3 measurements are made at regular intervals (of length L) along the x axis, and indicate 4 positions on the parabolic arc.
The traveler does not measure all points on the arc, only the 3 nodes on the parabolic arc, and its time measurements are vertical distance along the time axis between 4 instants. Basically the travelers only knows 3 chords of the parabolic arc, and wants to know the length to the point on the x-axis where the parabolic arc is vertical.
Your construction is accurate, and as I think we have all concluded, we only need two adjacent intervals, which comprise 3 points, to locate the origin of the parabola. The 3rd interval is extra info.
Jenny Physics
Your construction is accurate, and as I think we have all concluded, we only need two adjacent intervals, which comprise 3 points, to locate the origin of the parabola. The 3rd interval is extra info.
It's extra info that allows checking the solution, or verifying that there was no elastic elongation or that all cars of the train have the same length L (if the train is inelastic)
So now assume we don't know L: it's the same as saying that the x vertical axis has no known unit. We can use any arbitrary unit because this only scales that axis and scaling vertically any vertical parabol still gives a parabol... L does not matter we can set to 1.
As well if we don't know the origin of time, we can fix it arbitrarily: this just slides horizontally the vertical parabol along the horizontal time axis, and this translation is still a parabolic arc. We can set the one instant arbitrarily at position t=0.
So we can finally position the 4 nodes on our arbitrarily chosen base. But at least the horizontal time axis has a known scale (in seconds).
Two chords are fully positioned on the resulting graph. The only unknown position is the position where the parabolic arc is tangeant to the horizontal time axis. But the parabolic arc is uniquely defined by 3 of the 4 nodes whose position are known on our arbitrarily chosen coordinate system. It is enough to determine T because the time axis has a known scale.
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Jenny Physics
Philippe Verdy said:
I don't know where the formulas used by gneill to compute successive speeds come from "magically".
Basic kinematic equations:
##v_f^2 - v_o^2 = 2 a d##
##v_f = v_o + a t##
The problem speaks about a train where the passenger clearly knows the number of cars (he was arriving in the station where he expected to see the first car before it left the station: how does he know then he is counting the last cars if he does not know how the train is composed?
In the problem as given he sees the last three cars with no other cars following. We are also told explicitly in the problem that they are the last three cars.
Yes there are is some implied variable not clearly stated about train composition (the fixed number N of cars in the train, or its total length NL), but also an additional "polluting" unneeded variable in the 3 measurements of time.
It would be nice to see your two-time solution in detail. Remember, the length of the train cars is not a given.
The measurement of times between cars and the knowledge of length L per car directly gives average speeds for the time each car passes in front of the traveler, like knots on a naval ship. It is the difference of these times that gives differences of speeds, which is proportional (not inversely proportional) to the "constant" acceleration, by definition of an acceleration.
We are NOT given L. So you need another variable to replace it.
These measurements are not *instantaneous* measurements of speeds and acceleration, but just averaged values between separate instants (the actual definition would require car lengths to tend to zero: this is no longer a train with cars of finite length, this is a single thread made with infinitely narrow cars, and the "3 last time measurements" would be all zero in that case).
The definitions of the values are clearly stated in the problem statement. In pure thought experiment such as this one we can assume that they are rigorously known givens.
Now comes the problem: the speed of train is also not specified: "He watches the last three cars of the train pass by." but we can assume the passenger was knowing how many cars there were in the train, and knew the length of each car, or that he knows the speed of the train when he arrives to the station.
No. While we as analysts may be mildly interested in such musings, all we're told is that the last three cars pass by the passenger. What he otherwise knows about the train (it's color maybe?) is irrelevant to the problem as stated.
Otherwise there's no solution at all to the problem if we have absolutely no knowledge at all for what is in front of the train when the passenger arrived to the station and no way to know the instantaneous speed of the train (which is why the passenger takes times measurements between the last 3 cars passing in front of him).
We know that the train has some speed as the cars are passing. What's in front of those three cars is irrelevant to the problem as given. You're free to use your imagination or to determine some realistic number cars that lay ahead of the last three. But that will not change the problem statement.
If the passenger assumes that the train is infinitely long and only sees the last 3 cars and knows absolutely nothing about the length of the train, this is non-sense.
Agreed. So why go there?
The passenger arrives in a station whose docking area has a known finite length were he could wait for the train before coming too late to take it. The total length of the train is necessarily not infinite, and must be known (otherwise it could not even park along the waiting dock of the station, these docks would also be need to be infinitely long, and the traveller would have no chance at all to take the train on time in the last car where he would have reserved a seat, he would always be too late to take it), otherwise this is not a real train station.
The problem states that the train contains more than 3 cars. The traveler coming to the station cannot walk indefinitely, the station is necessarily limited in total length, and this also limits the total length of the train.
In fact it's not a real train station. This is a math puzzle. All of this talk about infinities and docking space and walking times are utterly irrelevant to the puzzle as it is stated. It's just obfuscation to no good purpose.
Station length is irrelevant to the specific problem as it is spelled out. For our purposes the train might be exactly three cars long being pulled by a single engine.
So you can argue that the problem did not state all necessary variables, but still it exposes an unnecessary time variable. Something must be missing, the problem is not complete in the case we don't allow any assumption about:
- what is a traveler (a human that cannot walk at infinite speed to reach a specific car position in due time before the train starts going),
- what is a train (a non-infinite vehicle of finite length),
- what is a station (a non-infinite dock area where travelers can wait their train and walk to the right position to enter the correct car in which he will find his seat).
All irrelevant musings. We are told that he watches the last three cars go by. That is the only fact about his location and what happens that you can get from the problem statement. Everything else is irrelevant speculation. And we'd still like to see your algebraic two-time variable solution.
Let's assume that the position where the traveler arrived too late in the station was the correct position to enter the correct car in the train (so he would have not needed to walk along the dock if he was in time when the train arrived and was parked waiting for travelers to come inside), then we know that he's looking at the trail of the train and not interested at all in what was on the front side: we reduce the train only to these last 3 cars, the rest of the train does not matter (he does not need to know then how many cars have already passed in front of that position)
May be the train only had these 3 cars and there's no other cars in front, the traveler could not measure a non zero speed of the train because it has not already left the station so its speed must be zero, and the passenger is not too late to take his train, and this gives a contradiction. So the train necessarily has more than 3 cars (at least 4, but not an infinity: it is still limited by the total length of the station).
Three cars plus an engine would suffice.
Jenny Physics
@gneill I do like the latest input from @Philippe Verdy in post 20. It does suggest that you should be able to compute ## a ## in terms of ## L ##, ## t ##, and ##t' ##, which I believe I just did, and also then determine how far away the head of the train is when the passenger arrived, which is ## s=\frac{1}{2}aT^2 ##, in terms of ## L ##, ##t ##, and ## t' ##. ## \\ ## Some of his inputs, I think, were going off on a tangent, but post 20 looks like a very useful one.
Jenny Physics
@gneill I do like the latest input from @Philippe Verdy in post 20. It does suggest that you should be able to compute ## a ## in terms of ## L ##, ## t ##, and ##t' ##, which I believe I just did, and also then determine how far away the head of the train is, which is ## s=\frac{1}{2}aT^2 ##, in terms of ## L ##, ##t ##, and ## t' ##. ## \\ ## Some of his inputs, I think, were going off on a tangent, but post 20 looks like a very useful one.
I agree that the geometric approach is elegant and a worthy contribution to the thread. Unfortunately for the OP, the problem statement does not give us L.
Jenny Physics
@gneill I do like the latest input from @Philippe Verdy in post 20. It does suggest that you should be able to compute ## a ## in terms of ## L ##, ## t ##, and ##t' ##, which I believe I just did, and also then determine how far away the head of the train is when the passenger arrived, which is ## s=\frac{1}{2}aT^2 ##, in terms of ## L ##, ##t ##, and ## t' ##. ## \\ ## Some of his inputs, I think, were going off on a tangent, but post 20 looks like a very useful one.
The timings depend on the ratio of ##a/L##. If you assume the engine is the same length as a carriage, then from two timings you can calculate the number of carriages. And, from three timings, you can calculate the the length of the engine compared to the length of each carriage. It would be a bit messy though.
Jenny Physics
PeroK said:
The timings depend on the ratio of ##a/L##. If you assume the engine is the same length as a carriage, then from two timings you can calculate the number of carriages. And, from three timings, you can calculate the the length of the engine compared to the length of each carriage. It would be a bit messy though.
Interesting. How?
Jenny Physics said:
Interesting. How?
Here is my solution for ## a ##: ## \\ ## ## L=v_o t+\frac{1}{2}at^2 ## ## \\ ## ## L=(v_o+at)t'+\frac{1}{2} at'^2=v_o t' +a(tt'+\frac{1}{2}t'^2) ## ## \\ ## Eliminate ## v_o ## by miltiplying the first equation by ## t' ## and the second by ## t ## and subtracting. ## \\ ## The result is ## a=\frac{(t-t')(2L)}{tt'(t+t')} ## if my algebra is correct.
Jenny Physics
There's another difficulty in this problem: it's clear that even if we restrict the conditions to use only 2 times measurements, the solution for T is a zero of a quadratic equation.
Now let's project the origin of the parabol at time -T along the x axis: we get an x0 position which is NOT necessarily an integer multiple of L ! (To simplify thing, the x-axis should be scaled by dividing it by L: it counts the number of cars.
We know that the travelers arrives just to see the last 3 cars passing "completely" in front of him. This does not necessarily means that the traveler arrived at the station exactly at the time where the first of the last 3 cars was passing there, it may have been there between the two times the last 4 cars were starting to pass and the last 3 ones. but he must have waited a bit before seeing the exact moment where there remained exactly 3 cars passing.
If the resulting coordinate on the x-axis of the origin of the parabolic arc is not an integer, this means that the traveler was not at the correct position to take its train just before the train started to leave the station (he would have needed to walk a little along the waiting dock, by a distance between 0 and L) ! So the traveler was in fact a bit more late to take his train because he also needed to walk to the correct start position (at walking speed, not specified).
Jenny Physics
@Philippe Verdy We know that. You are making it overly complicated then. We need to assume he arrived just as he observed the edge of one car right in front of him. Otherwise, we have no way of computing anything precisely.
Jenny Physics
Jenny Physics said:
Interesting. How?
The easy case, counting the engine as the first carriage (same length):
If the train starts from rest, the time ##t_n## for ##n## carriages to pass satisfies:
##nL = \frac12 a t_n^2##
The time for the n+1 st carriage to pass is then:
##T_{n+1} = t_{n+1} - t_n = \sqrt{\frac{2a}{L}}(\sqrt{n+1} - \sqrt{n})##
Hence you have:
##\frac{T_{n+2}}{T_{n+1}} = \frac{\sqrt{n+ 2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}}##
That's going to be different for every ##n##, so you just need to work out for which ##n## that equation holds.
Now, if there's an engine of length ##K## ...
Jenny Physics
PeroK said:
The easy case, counting the engine as the first carriage (same length):
If the train starts from rest, the time ##t_n## for ##n## carriages to pass satisfies:
##nL = \frac12 a t_n^2##
The time for the n+1 st carriage to pass is then:
##T_{n+1} = t_{n+1} - t_n = \sqrt{\frac{2a}{L}}(\sqrt{n+1} - \sqrt{n})##
Hence you have:
##\frac{T_{n+2}}{T_{n+1}} = \frac{\sqrt{n+ 2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}}##
That's going to be different for every ##n##, so you just need to work out for which ##n## that equation holds.
Now, if there's an engine of length ##K## ...
@PeroK If you choose an ## L ##, e.g. ##L=30 ## m, (and tell us what it is), and choose your own ## a ##, e.g. ## a=1 ## m/sec^2, but don't tell us what it is, if you give us ## t ## and ## t' ## (in seconds) from your formula, (you are free to select ## n ## but don't tell us what it is), I believe our calculations for ## T ##, ##a ##, and ## s=\frac{1}{2}aT^2 ## would allow us to readily compute both the value for ## a ## as well as the number of cars in the train that had already passed as the passenger arrived. Such a test would basically be showing that our algebra is correct. ## \\ ## Edit: I tried it using ## n=9 ## (and 10, and 11), and ## a=1 ## and ## L=30 ##, and it computed like it should.
Last edited:
Jenny Physics
Jenny Physics said:
some relation between t,t′,t′′
(t'-t-t")3=t'3-t3-t"3?
Last edited:
Jenny Physics
@Philippe Verdy We know that. You are making it overly complicated then. We need to assume he arrived just as he observed the edge of one car right in front of him. Otherwise, we have no way of computing anything precisely.
If you assume that, then there's no general solution with the rigid train !
I repeat what we know: we have a unique parabolic arc with vertical passing by three nodes in the cartesian plane, and the horizontal axis of distance is subvided at regular intervals of length L. The measured times are not constrained and can be any real value.
But then the extreme point of the unique ellipse passing by the three nodes is NOT (in the general case) located (on the distance axis of coordinates) at an exact multiple of L because that coordinate is a zero of a quadratic function. We then have a contradiction in the problem or we must seek other reasons:
- the clock is not exact but the error margins allows an imprecision for the determination of the position of the extreme point of the eliipse that includes a region of the 2D plane where it falls between two integer multiples of L.
- if the clock is precise enough and there's no integer multiple L for locating that point of the elliptic arc, the only solution is to admit elasticity. If we suppsoe that the train is rigid (when measured inside the train, then the elasticity is what the traveler observes outside the the train in the station: he must see the effect of relativity.
- now if the three mesures give 3 different locations for the extreme point of the ellipse, the elasticity is no longer an hypothese, it is the only solution and must be observed in the train as well as in the station, independantly of general relativity (whose we can predict the effect precisely).
The important point is that the solution cannot respect the condition that all cars must have the same length L, because it's impossible in the general case! Or the curve of movement is not elliptic, i.e. the acceleration "a" cannot be constant indefinitely !
And we know that acceleration cannot be constant because general relativity cannot support speeds higher than c, so the elliptic movement is certainly wrong (it is only possible as an approximation at low speeds): the two branches of the curve of positions are necessarily decelerating, because the derived curve of speeds is not a straight line, but is a sigmoid whose two branches will necessarily be converging to -c and +c. But, let's ignore the first branch which concerns the train before he arrived in the station there remains only the positive half-sigmoid branch for speeds ! The general relativity applies a negative acceleration on top of the positive acceleration ignited by the engines of the train.
And this is an interesting result, because the traveller with his own precise-enough clock can measure the effect of general relativity, just by observing the passing train (or equivalently by taking place in an elevator falling freely in front of stages of a building, and measuring the time when it passes in front of eah of them; here also we know that it cannot fall freely at constant acceleration, because the acceleration will be zero at the center of Earth and will become negative to continue "falling" to the other side of Earth; the elevator will reach its maximum speed which is about 28000 km/h in the middle of Earth after about 20 minutes and will then decelerate to come to the surface of at the antipode, where it will emerge and will not go above the initial height from where it was initially falling).
We can then observe general relativity on Earth by just looking at trains accelerating in front of us: this just requires a precise enough clock.
And on short distances in a laborary, we can also observe general relativity by measuring the time it takes for a small accelerated graduated wheel to measure interval of times where the regularly spaced graduations passes in front of a sensor: we should see significant differences depending only on the initial angular position of the wheel where it started rotating, and on the precision of the clock.
Last edited:
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3K | 10,331 | 45,229 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.817472 |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_21 | 1,709,538,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00544.warc.gz | 106,471,007 | 10,892 | # 1954 AHSME Problems/Problem 21
## Problem 21
The roots of the equation $2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5$ can be found by solving:
$\textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0$
## Solution
Make the substitution $t=\sqrt{x}$. Then $2t+\frac{2}{t}=5\implies 2t^2-5t+2=0$. Then $2x-5\sqrt{x}+2=0\implies 2x+2=5\sqrt{x}\implies (2x+2)^2=25x\implies 4x^2+8x+4=25x\implies 4x^2-17x+4=0$, $\fbox{C}$
## See Also
1954 AHSC (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 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 All AHSME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | 494 | 1,023 | {"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.341648 |
https://forums.wolfram.com/mathgroup/archive/2012/Aug/msg00166.html | 1,723,633,819,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00594.warc.gz | 205,806,703 | 7,823 | problems withn 3d models
• To: mathgroup at smc.vnet.net
• Subject: [mg127639] problems withn 3d models
• From: Roger Bagula <roger.bagula at gmail.com>
• Date: Fri, 10 Aug 2012 02:41:40 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
```The first problem is the 3ds files comes out invalid.
The second problem is that in obj files a phantom triangle
shows up.
And lastly the stl file appears to be empty.
It might just be my machine with an over worked
version 8 Mathematica,
but it might also be something wrong with my code?
Clear[x, y, z, f, g]
f[t_] = Sqrt[Abs[Sin[t]]]*Cos[t]^2;
g[t_] = Sqrt[Abs[Sin[t]]]*Sin[t]*Cos[t];
x = (1 - f[t]^2)/(1 + f[t]^2);
y = 2 f[t]/(1 + f[t]^2);
z = 2 g[t]/(1 + f[t]^2);
w = {x*(10 + Cos[p]), y*(10 + Cos[p + 2*Pi/3]),
z*(10 + Cos[p + 2*Pi/3])}
g1 = ParametricPlot3D[w, {t, 0, 2*Pi}, {p, 0, 2*Pi}, Axes -> None,
Boxed -> False, ViewPoint -> {10, 0, 0}, PlotPoints -> {64, 16},
Mesh -> False, ColorFunction -> "Rainbow"]
g2 = ParametricPlot3D[{-w[[1]], w[[2]], -w[[3]]} + {20, 0, 0}, {t, 0,
2*Pi}, {p, 0, 2*Pi}, Axes -> None, Boxed -> False,
ViewPoint -> {10, 0, 0}, PlotPoints -> {64, 16}, Mesh -> False,
ColorFunction -> "Rainbow", PlotRange -> All]
Show[{g1, g2}, PlotRange -> All]
b = Reverse[
Union[Flatten[
Table[{Blue,
Cylinder[{10*{x, y, z} /.
t -> 2*Pi*n/128, (10*{-x, y, -z} + {20, 0, 0}) /.
t -> 2*Pi*n/128}, 1/40]}, {n, 0, 127}]]]];
g3 = Show[Graphics3D[b], Boxed -> False, PlotRange -> All]
gw = Show[{g1, g2, g3}, PlotRange -> All]
Export["BB_strings.3ds", gw]
Export["BB_strings.obj", gw]
Export["BB_strings.stl", gw]
```
• Prev by Date: Re: Trying to quickly split lists at the point of
• Next by Date: Re: Trying to quickly split lists at the point of
• Previous by thread: Re: Trying to quickly split lists at the point of | 734 | 1,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.683556 |
http://www.geeksforgeeks.org/dynamic-programming-set-4-longest-common-subsequence/ | 1,498,634,178,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322873.10/warc/CC-MAIN-20170628065139-20170628085139-00187.warc.gz | 541,819,321 | 18,591 | # Dynamic Programming | Set 4 (Longest Common Subsequence)
We have discussed Overlapping Subproblems and Optimal Substructure properties in Set 1 and Set 2 respectively. We also discussed one example problem in Set 3. Let us discuss Longest Common Subsequence (LCS) problem as one more example problem that can be solved using Dynamic Programming.
LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”. So a string of length n has 2^n different possible subsequences.
It is a classic computer science problem, the basis of diff (a file comparison program that outputs the differences between two files), and has applications in bioinformatics.
Examples:
LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.
LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The naive solution for this problem is to generate all subsequences of both given sequences and find the longest matching subsequence. This solution is exponential in term of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem.
1) Optimal Substructure:
Let the input sequences be X[0..m-1] and Y[0..n-1] of lengths m and n respectively. And let L(X[0..m-1], Y[0..n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0..m-1], Y[0..n-1]).
If last characters of both sequences match (or X[m-1] == Y[n-1]) then
L(X[0..m-1], Y[0..n-1]) = 1 + L(X[0..m-2], Y[0..n-2])
If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then
L(X[0..m-1], Y[0..n-1]) = MAX ( L(X[0..m-2], Y[0..n-1]), L(X[0..m-1], Y[0..n-2])
Examples:
1) Consider the input strings “AGGTAB” and “GXTXAYB”. Last characters match for the strings. So length of LCS can be written as:
L(“AGGTAB”, “GXTXAYB”) = 1 + L(“AGGTA”, “GXTXAY”)
2) Consider the input strings “ABCDGH” and “AEDFHR. Last characters do not match for the strings. So length of LCS can be written as:
L(“ABCDGH”, “AEDFHR”) = MAX ( L(“ABCDG”, “AEDFHR”), L(“ABCDGH”, “AEDFH”) )
So the LCS problem has optimal substructure property as the main problem can be solved using solutions to subproblems.
2) Overlapping Subproblems:
Following is simple recursive implementation of the LCS problem. The implementation simply follows the recursive structure mentioned above.
## C/C++
```/* A Naive recursive implementation of LCS problem */
#include<bits/stdc++.h>
int max(int a, int b);
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
if (m == 0 || n == 0)
return 0;
if (X[m-1] == Y[n-1])
return 1 + lcs(X, Y, m-1, n-1);
else
return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Driver program to test above function */
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
int m = strlen(X);
int n = strlen(Y);
printf("Length of LCS is %d\n", lcs( X, Y, m, n ) );
return 0;
}
```
## Java
```/* A Naive recursive implementation of LCS problem in java*/
public class LongestCommonSubsequence
{
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char[] X, char[] Y, int m, int n )
{
if (m == 0 || n == 0)
return 0;
if (X[m-1] == Y[n-1])
return 1 + lcs(X, Y, m-1, n-1);
else
return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
return (a > b)? a : b;
}
public static void main(String[] args)
{
LongestCommonSubsequence lcs = new LongestCommonSubsequence();
String s1 = "AGGTAB";
String s2 = "GXTXAYB";
char[] X=s1.toCharArray();
char[] Y=s2.toCharArray();
int m = X.length;
int n = Y.length;
System.out.println("Length of LCS is" + " " +
lcs.lcs( X, Y, m, n ) );
}
}
// This Code is Contributed by Saket Kumar
```
## Python
```# A Naive recursive Python implementation of LCS problem
def lcs(X, Y, m, n):
if m == 0 or n == 0:
return 0;
elif X[m-1] == Y[n-1]:
return 1 + lcs(X, Y, m-1, n-1);
else:
return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
# Driver program to test the above function
X = "AGGTAB"
Y = "GXTXAYB"
print "Length of LCS is ", lcs(X , Y, len(X), len(Y))
```
Output:
`Length of LCS is 4`
Time complexity of the above naive recursive approach is O(2^n) in worst case and worst case happens when all characters of X and Y mismatch i.e., length of LCS is 0.
Considering the above implementation, following is a partial recursion tree for input strings “AXYT” and “AYZX”
``` lcs("AXYT", "AYZX")
/ \
lcs("AXY", "AYZX") lcs("AXYT", "AYZ")
/ \ / \
lcs("AX", "AYZX") lcs("AXY", "AYZ") lcs("AXY", "AYZ") lcs("AXYT", "AY")```
In the above partial recursion tree, lcs(“AXY”, “AYZ”) is being solved twice. If we draw the complete recursion tree, then we can see that there are many subproblems which are solved again and again. So this problem has Overlapping Substructure property and recomputation of same subproblems can be avoided by either using Memoization or Tabulation. Following is a tabulated implementation for the LCS problem.
## C/C++
```/* Dynamic Programming C/C++ implementation of LCS problem */
#include<bits/stdc++.h>
int max(int a, int b);
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
int L[m+1][n+1];
int i, j;
/* Following steps build L[m+1][n+1] in bottom up fashion. Note
that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
for (i=0; i<=m; i++)
{
for (j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i-1] == Y[j-1])
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = max(L[i-1][j], L[i][j-1]);
}
}
/* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */
return L[m][n];
}
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Driver program to test above function */
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
int m = strlen(X);
int n = strlen(Y);
printf("Length of LCS is %d\n", lcs( X, Y, m, n ) );
return 0;
}
```
## Java
```/* Dynamic Programming Java implementation of LCS problem */
public class LongestCommonSubsequence
{
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char[] X, char[] Y, int m, int n )
{
int L[][] = new int[m+1][n+1];
/* Following steps build L[m+1][n+1] in bottom up fashion. Note
that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i-1] == Y[j-1])
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = max(L[i-1][j], L[i][j-1]);
}
}
return L[m][n];
}
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
return (a > b)? a : b;
}
public static void main(String[] args)
{
LongestCommonSubsequence lcs = new LongestCommonSubsequence();
String s1 = "AGGTAB";
String s2 = "GXTXAYB";
char[] X=s1.toCharArray();
char[] Y=s2.toCharArray();
int m = X.length;
int n = Y.length;
System.out.println("Length of LCS is" + " " +
lcs.lcs( X, Y, m, n ) );
}
}
// This Code is Contributed by Saket Kumar
```
## Python
```# Dynamic Programming implementation of LCS problem
def lcs(X , Y):
# find the length of the strings
m = len(X)
n = len(Y)
# declaring the array for storing the dp values
L = [[None]*(n+1) for i in xrange(m+1)]
"""Following steps build L[m+1][n+1] in bottom up fashion
Note: L[i][j] contains length of LCS of X[0..i-1]
and Y[0..j-1]"""
for i in range(m+1):
for j in range(n+1):
if i == 0 or j == 0 :
L[i][j] = 0
elif X[i-1] == Y[j-1]:
L[i][j] = L[i-1][j-1]+1
else:
L[i][j] = max(L[i-1][j] , L[i][j-1])
# L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1]
return L[m][n]
#end of function lcs
# Driver program to test the above function
X = "AGGTAB"
Y = "GXTXAYB"
print "Length of LCS is ", lcs(X, Y)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
```
Time Complexity of the above implementation is O(mn) which is much better than the worst case time complexity of Naive Recursive implementation.
The above algorithm/code returns only length of LCS. Please see the following post for printing the LCS.
Printing Longest Common Subsequence
You can also check the space optimized version of LCS at
Space Optimized Solution of LCS | 2,816 | 8,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-26 | latest | en | 0.903358 |
http://smallbusiness.chron.com/compute-futa-suta-taxes-accounting-22260.html | 1,406,886,856,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510274967.3/warc/CC-MAIN-20140728011754-00147-ip-10-146-231-18.ec2.internal.warc.gz | 263,643,403 | 15,366 | # How to Compute FUTA & SUTA Taxes for Accounting
by Matt McGew, Demand Media
The Federal Unemployment Tax Act requires employers to pay a tax on a portion of every employee's earnings to fund state unemployment agencies. State Unemployment Tax Acts also require employers to pay a tax to fund the state unemployment agencies. Both FUTA and SUTA use established tax bases that determine the maximum amount of employee compensation subject to these taxes. Additionally, FUTA offers employers a credit for taxes paid under SUTA.
Step 1
Determine your SUTA tax rate by consulting your state's payroll tax instructions. The SUTA tax rate varies from state to state. For an example, let's assume your state's SUTA rate is 5 percent.
Step 2
Determine your state's SUTA tax base by consulting your state's payroll tax instructions. The tax base is the maximum amount of wages taxed per employee. Many states use the same tax base as FUTA -- \$7,000 as of September of 2011. However, the actual SUTA tax base may vary from state to state. For our example, let's assume your state's SUTA tax base is also \$7,000.
Step 3
Determine your employee's state taxable base. If the employee's total compensation for the year exceeds the maximum amount established by your state, you would use the state's tax base. For example, if an employee earned \$25,000 for the tax year and the state's SUTA tax base is \$7,000, you would use the \$7,000 figure. If, on the other hand, the employee earned less than \$7,000, you would use the actual compensation figure as the employee's taxable base.
Step 4
Multiply the employee's tax base by the state SUTA tax rate. Continuing the same example, \$7,000 x .05 = \$350. This figure represents the SUTA taxes the employer must pay for the employee.
Step 5
Multiply the employee's federal tax base by the current FUTA rate. As of September of 2011, the maximum federal tax base is \$7,000 and the tax rate is 6 percent. Continuing the same example, \$7,000 x .06 = \$420.
Step 6
Subtract the employee's SUTA taxes from the FUTA tax calculation. Continuing the same example, \$420 - \$350 = \$70. This figure represents the SUTA taxes the employer must pay for the employee.
#### References (1)
Since 1992 Matt McGew has provided content for on and offline businesses and publications. Previous work has appeared in the "Los Angeles Times," Travelocity and "GQ Magazine." McGew specializes in search engine optimization and has a Master of Arts in journalism from New York University.
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https://www.doubtnut.com/question-answer/let-z-be-a-complex-number-and-a-be-a-real-parameter-such-that-z2-z-a20-then-648230984 | 1,653,407,369,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573053.67/warc/CC-MAIN-20220524142617-20220524172617-00003.warc.gz | 816,355,602 | 93,764 | HomeEnglishClass 12MathsChapterComplex Numbers
Let z be a complex number and ...
# Let z be a complex number and a be a real parameter such that z^2+z+a^2=0, then
Text Solution
locus of z is a pair of straight lineslocus of z is a circlearg(z)=+-(5pi)/3|z|=-2|a| | 84 | 266 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.96875 | 3 | CC-MAIN-2022-21 | latest | en | 0.575881 |
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PHY 101 Exam 4
# PHY 101 Exam 4 - f = N F = ma The only force acting on the...
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x x 2 2 x F ma The only force acting on the crate is friction. f N (mg) ma a g (0.250)(9.8m /s ) 2.45m /s = = μ = μ = = μ = = ! ! nd A B First, we assume a direction of motion, which for this problem would be the y-direction, and let it be positive. Let's apply Newton's 2 Law to each block individually: Block A: T = m a Block B: m g T - B B B A B B A m a Because the blocks are connected, the acceleration must be the same. Notice that we have +T and -T in the 2 equations. If we add them together, T will be eliminated. m g a(m m ) m g (1.60 a (m m ) = = + = = + 2 2 kg)(9.8m / s ) 3.14m / s (3.40 1.60)kg = + f N = μ ! !
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The apparent weight is the Normal force. 2 2 2 2 F ma Everything is happening in the y-direction, so we can drop the vector notation. v N mg m Substitute 7mg for the Normal force. r v 7mg mg m Notice that the pilot's mass cancels out. r v (320m / s) r 8g 6 = ± - = - = = = ! ! 2
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PHY 101 Exam 4 - f = N F = ma The only force acting on the...
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Ask a homework question - tutors are online | 476 | 1,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-09 | latest | en | 0.84554 |
https://math.stackexchange.com/questions/1966328/lin-alg-subspace/1966330 | 1,561,516,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000044.37/warc/CC-MAIN-20190626013357-20190626035357-00371.warc.gz | 492,097,167 | 34,604 | Lin Alg. Subspace
I am brand new to learning about subspaces in my linear Algebra class. Ive tried to follow khan academy but to no avail. I encountered these practice problems in the textbook. However my textbook inconveniently provides answers for odd numbered problems only.
I want to further my understanding. Could someone please explain how to go about these problems?
Given vector space V = the set of all 2 × 2 matrices.....
Does the set of all 2 × 2 nonsingular matrices form a subspace of
V? Explain.
Does the set of all 2 × 2 singular matrices form a subspace of V?
Explain.
Thank you :)
• Always assume it is not a subspace first. Can you think of properties which might not hold for particular elements in these sets? – David Peterson Oct 13 '16 at 4:38
To get you started... Let $\operatorname{GL}_2$ denote the set of $2 \times 2$ nonsingular matrices. The identity $I$ is in $\operatorname{GL}_2$... but what about $I - I = 0$? | 237 | 952 | {"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} | 3.125 | 3 | CC-MAIN-2019-26 | latest | en | 0.87695 |
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Clear All
P Pankaj Sanodiya
The inductance of the inductor The capacitance of the capacitor Voltage supply Frequency of voltage supply . Here, we have Impedance Now, Current in the circuit will be where, The negative sign is just a matter of the direction of current.so, here But, since the value of R is zero(since our circuit have only L and C) Hence Now, RMS value of this current: .
P Pankaj Sanodiya
As we know, in the case of a parallel RLC circuit: The current will be minimum when Which is also the condition of natural frequency. Hence the total current is minimum when source frequency is equal to the natural frequency. RMS value of current in R RMS value in Inductor RMS value in capacitor Capacitor current and inductor current will cancel out each other so the current flowing in...
P Pankaj Sanodiya
Given, The capacitance of the capacitor The resistance of the circuit Voltage supply Frequency of voltage supply The maximum current in the circuit Hence maximum current in the circuit is 3.9A. b) In the case of capacitor, we have So, So the time lag between max voltage and max current is : At high frequencies, tends to zero. which indicates capacitor acts as a conductor at high...
P Pankaj Sanodiya
In the case of a capacitor, we have So, So the time lag between max voltage and the max current is :
P Pankaj Sanodiya
Given, The capacitance of the capacitor The resistance of the circuit Voltage supply Frequency of voltage supply The maximum current in the circuit Hence maximum current in the circuit is 3.24A.
P Pankaj Sanodiya
Given, The inductance of the coil the resistance of the coil Supply voltage Supply voltage frequency a) Now, as we know peak voltage = (RMS Voltage) Peak voltage Now, The impedance of the circuit : Now peak current in the circuit : Hence peak current is in the circuit. The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit...
P Pankaj Sanodiya
Let the voltage in the circuit be and Current in the circuit be Where is the phase difference between voltage and current. V is maximum At t = 0 is maximum At Hence, the time lag between voltage maximum and the current maximum is . For phase difference we have Hence time lag between the maximum voltage and the maximum current is
P Pankaj Sanodiya
Given, The inductance of the coil the resistance of the coil Supply voltage Supply voltage frequency Now, as we know peak voltage = (RMS Voltage) Peak voltage The impedance of the circuit : Now peak current in the circuit : Hence peak current is 1.82A in the circuit.
P Pankaj Sanodiya
If the resistance is added to the circuit, the whole energy will dissipate as heat eventually. energy will keep moving between the capacitor and inductor with reducing in magnitude in each cycle and eventually all energy will be dissipated.
P Pankaj Sanodiya
The energy will be shared equally when the energy in the capacitor is half of the maximum energy it can store.i.e. From Here, we got So Now, we know the charge on the capacitor, we can calculate time for which From here, Hence for these times, the total energy will be shared equally between capacitor and inductor.
P Pankaj Sanodiya
The stored energy will we purely magnetic when the pure electrical stored is zero. i.e. when the charge on the capacitor is zero, all energy will be stored in the inductor. So, t for which charge on the capacitor is zero is Hence at these times, the total energy will be purely magnetic.
P Pankaj Sanodiya
at any instant, the charge on the capacitor is: Where time period : Now, when the total energy is purely electrical, we can say that this is possible when Hence Total energy will be purely electrical(stored in a capacitor) at .
P Pankaj Sanodiya
Given, The inductance of the inductor: The capacitance of the capacitor : The initial charge on the capacitor: The natural angular frequency of the circuit: Hence the natural angular frequency of the circuit is . The natural frequency of the circuit: Hence natural frequency of the circuit is 159Hz.
P Pankaj Sanodiya
Given, The inductance of the inductor: The capacitance of the capacitor : The initial charge on the capacitor: Total energy present at the initial moment: Hence initial energy in the circuit is 1J. Since we don't have any power-consuming element like resistance in the circuit, the energy will be conserved
P Pankaj Sanodiya
Potential difference across any element = NOW, The potential difference across capacitor: Potential difference across the inductor The potential difference across Resistor Potential difference across LC combination Hence at resonating frequency potential difference across LC combination is zero.
P Pankaj Sanodiya
Given, Variable frequency supply voltage = 230V Inductance Capacitance Resistance Now, The impedance of the circuit is at Resonance Condition : Hence, Impedance at resonance is 40. Now, at resonance condition, impedance is minimum which means current is maximum which will happen when we have a peak voltage, so Current in the Resonance circuit is Given by Hence amplitude of the current...
P Pankaj Sanodiya
Given, Variable frequency supply voltage = 230V Inductance Capacitance Resistance a) Resonance angular frequency in this circuit is given by : Hence this circuit will be in resonance when supply frequency is 50 rad/sec.
P Pankaj Sanodiya
Given, Range of the frequency in which radio can be tune = The effective inductance of the Circuit = Now, As we know, where is tuning frequency. For getting the range of the value of a capacitor, let's calculate the two values of the capacitor, one maximum, and one minimum. first, let's calculate the minimum value of capacitance which is the case when tuning frequency = 800KHz. Hence... | 1,365 | 5,863 | {"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": 80, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-16 | latest | en | 0.779213 |
https://www.jiskha.com/display.cgi?id=1267646730 | 1,519,396,760,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814787.54/warc/CC-MAIN-20180223134825-20180223154825-00085.warc.gz | 867,320,654 | 3,693 | # Maths
posted by .
What roots can I use to sketch this graph of:
y = x^4 - x^2
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More Similar Questions | 635 | 1,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-09 | longest | en | 0.775178 |
https://goprep.co/ex-1.4-q3-maria-invested-rs-80-000-in-a-business-she-would-i-1nkoq8 | 1,606,552,195,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00269.warc.gz | 328,474,774 | 39,702 | # Maria invested Rs. 80,000 in a business. She would be paid interest at 5% per annum compounded annually. Find(i) the amount standing to her credit at the end of second year and (ii) the interest for the third year.
i. Principle = 80,000
Rate of interest per annum = 5%
Time = years
Compound interest for 2 years
Amount =
=
= 88,200
Compound interest = Amount – principle
= 80,000 - 88,200 = 8,200
8,200 will be credited at the end of 2 years
ii. Compound interest for 3 years
Amount =
=
= 92,610
Compound interest = Amount – principle
= 80,000 – 92,610 = 12,610
Compound interest for 3rd year
= compound interest of 3 years – compound interest of 2 years
= 12,610 - 8,200
= 4,410
Compound interest for the 3rd year is 4,410
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Conversion Result: ```British peck = 0.00909219858976 volume (volume) ``` Related Measurements: Try converting from "British peck" to barrel, beer gallon (English beer gallon), coomb, cord (of wood), cup, displacement ton, dry pint, freight ton, jigger, liter, load, noggin, oil arroba (Spanish oil arroba), peck (dry peck), strike, tablespoon, UK bushel (British bushel), UK peck (British peck), UK pint (British pint), wey, or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: British peck = .23319464 amphora (Greek amphora), .60047619 balthazar, 1.97 beer gallon (English beer gallon), 9,092.2 cc (cubic centimeters), .00917394 displacement ton, 2,459.55 dram fluid (fluid dram), 8.26 dry quart, .26687831 firkin, .0080272 freight ton, 76.86 gill, .05046018 koku (Japanese koku), .72346529 oil arroba (Spanish oil arroba), 2.29 omer (Israeli omer), 1.03 peck (dry peck), 19.22 pint (fluid pint), .01906274 pipe, 1.6 rehoboam, .87660758 tou (Chinese tou), 1 UK peck (British peck), 16 UK pint (British pint).
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# Problem 1193. Insert zeros into vector
Solution 190228
Submitted on 12 Jan 2013 by Peter Wittenberg
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 2 3 4 5]; n=2; y_correct = [1 0 0 2 0 0 3 0 0 4 0 0 5 0 0]; assert(isequal(insert_zeros(x,n),y_correct))
``` ```
2 Pass
%% x = [1 2 3 4 5]; n=1; y_correct = [1 0 2 0 3 0 4 0 5 0 ]; assert(isequal(insert_zeros(x,n),y_correct))
``` ```
3 Pass
%% x = [1 2]; n=5; y_correct = [1 0 0 0 0 0 2 0 0 0 0 0]; assert(isequal(insert_zeros(x,n),y_correct))
``` ``` | 258 | 648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.586972 |
http://mathhelpforum.com/pre-calculus/61632-co-ordinate-geometry-print.html | 1,529,586,748,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00270.warc.gz | 212,060,227 | 2,726 | # co ordinate geometry
• Nov 25th 2008, 12:30 PM
co ordinate geometry
can anyone help??
got this question,,
The points ABC have co ordinates (2, 1) (b,3) and (5,5) where b>3
and angle ABC=90degrees ??
I know the answer just have no idea how to work it out.
• Nov 25th 2008, 12:39 PM
skeeter
(slope of AB)(slope of BC) = -1
• Nov 25th 2008, 12:43 PM
Plato
In a right triangle \$\displaystyle a^2 + b^2 = c^2\$.
So \$\displaystyle (b-2)^2 +(3-1)^2 + (b-5)^2 + (3-5)^2 = (5-2)^2 + (5-1)^2\$
• Nov 25th 2008, 01:02 PM | 220 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | latest | en | 0.798973 |
http://gmatclub.com/forum/odds-and-evens-113229.html?fl=similar | 1,484,738,178,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280266.9/warc/CC-MAIN-20170116095120-00207-ip-10-171-10-70.ec2.internal.warc.gz | 122,315,742 | 57,930 | odds and evens! : GMAT Problem Solving (PS)
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# odds and evens!
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04 May 2011, 22:37
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Each digit in the 2 digit number G is halved to form a new 2 digit number H.Which of the following could be the sum of G and H?
A.153
B.150
c.137
D.129
E.89
[Reveal] Spoiler: OA
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04 May 2011, 22:56
the unit digit combinations are 6,3; 8,4; 4,2 giving 9, 2 and 6 as the units digit of the sum.
also if the digits are x and y then G = 10x + y and H = 5x + 0.5y G + H = 1.5 ( 10x + y).
thus with these two criteria. D fits the bill.
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05 May 2011, 00:16
Amit:pls explain lil briefly .Not exactly clear
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### Show Tags
05 May 2011, 06:34
02468 - Even digits
12349 - Odd digits
So the set of halves of even digits gives (odds can't be halved):
01234
86
43
---
129
With some trial and error :
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05 May 2011, 07:35
AnkitK wrote:
Each digit in the 2 digit number G is halved to form a new 2 digit number H.Which of the following could be the sum of G and H?
A.153
B.150
c.137
D.129
E.89
Say H=xy [x is the ten's place digit after being halved and y is the unit's place digit after being halved]
H=10x+y
G=(2x)(2y)
G=10*(2x)+2y
Possible values of 2x=2,4,6,8; 2y=0,2,4,6,8
Possible values of x=1,2,3,4; y=0,1,2,3,4
G+H=20x+2y+10x+y=30x+3y=3(10x+y)
Thus, the sum must be a multiple of 3; options C and E are out.
Let's try other options:
A. 3(10x+y)=153 i.e. 10x+y=51 i.e. 10*5+1=51; x=5(Not a possible value of x) and y=1
B. 3(10x+y)=150 i.e. 10x+y=50 i.e. 10*5+0=50; x=5(Not a possible value of x) and y=0
D. 3(10x+y)=129 i.e. 10x+y=43 i.e. 10*4+3=43; x=4(Possible) and y=3(Possible)
Ans: D
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### Show Tags
05 May 2011, 13:45
AnkitK wrote:
Each digit in the 2 digit number G is halved to form a new 2 digit number H.Which of the following could be the sum of G and H?
A.153
B.150
c.137
D.129
E.89
For the question to make sense, G must be even. So we are adding an even number G to G/2, and the answer will be 3*(G/2), and thus must be a multiple of 3. Further, G < 100, so 3G/2 is less than 150. The only possible answer is thus D.
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05 May 2011, 15:23
G = 10x + y
H = 5x + y/2
so G + H = 15x + 3/2y. Multiply by 2 and you get 2 ( G+H )= 3 (10 x + y) so G+H must be a multiple of 3 and G a multiple of 2 (obviously otherwise we cant divide G) and we know G less than or equal to 88 (highest 2 digits even) and so H less than or equal to 44 (half G), so G+H less than 132.
ABC out and E out because not a multiple of 3
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### Show Tags
06 May 2011, 14:49
G is even
H can be even or odd
G's unit digit possibilities are 0 2 4 6 8
H's unit digit possibilities are 0 1 2 3 4
No checking the answers we can see only D fits the criteria.
Re: odds and evens! [#permalink] 06 May 2011, 14:49
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Display posts from previous: Sort by | 1,909 | 5,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-04 | latest | en | 0.806833 |
https://www.convertunits.com/from/step/to/kyu | 1,590,416,096,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00322.warc.gz | 622,887,467 | 10,667 | ## ››Convert step to kyu
step kyu
How many step in 1 kyu? The answer is 0.00032808398950131.
We assume you are converting between step and kyu.
You can view more details on each measurement unit:
step or kyu
The SI base unit for length is the metre.
1 metre is equal to 1.3123359580052 step, or 4000 kyu.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between steps and kyu.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of step to kyu
1 step to kyu = 3048 kyu
2 step to kyu = 6096 kyu
3 step to kyu = 9144 kyu
4 step to kyu = 12192 kyu
5 step to kyu = 15240 kyu
6 step to kyu = 18288 kyu
7 step to kyu = 21336 kyu
8 step to kyu = 24384 kyu
9 step to kyu = 27432 kyu
10 step to kyu = 30480 kyu
## ››Want other units?
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ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 419 | 1,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.852206 |
https://en.wikiversity.org/wiki/Hyperbolic_functions/R/Introduction/Section | 1,718,756,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00836.warc.gz | 190,566,616 | 12,659 | # Hyperbolic functions/R/Introduction/Section
## Definition
The function defined for ${\displaystyle {}x\in \mathbb {R} }$ by
${\displaystyle {}\sinh x:={\frac {1}{2}}{\left(e^{x}-e^{-x}\right)}\,,}$
is called hyperbolic sine.
## Definition
The function defined for ${\displaystyle {}x\in \mathbb {R} }$ by
${\displaystyle {}\cosh x:={\frac {1}{2}}{\left(e^{x}+e^{-x}\right)}\,,}$
is called hyperbolic cosine.
## Lemma
The functions hyperbolic sine and hyperbolic cosine
have the following properties.
1. ${\displaystyle {}\cosh x+\sinh x=e^{x}\,.}$
2. ${\displaystyle {}\cosh x-\sinh x=e^{-x}\,.}$
3. ${\displaystyle {}(\cosh x)^{2}-(\sinh x)^{2}=1\,.}$
### Proof
${\displaystyle \Box }$
## Lemma
The function hyperbolic sine is strictly increasing, and the function hyperbolic cosine is strictly decreasing on ${\displaystyle {}\mathbb {R} _{\leq 0}}$ and strictly increasing on ${\displaystyle {}\mathbb {R} _{\geq 0}}$.
### Proof
See exercise and exercise.
${\displaystyle \Box }$
## Definition
The function
${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \tanh x={\frac {\sinh x}{\cosh x}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}},}$
is called hyperbolic tangent. | 405 | 1,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-26 | latest | en | 0.618324 |
https://www.stat.math.ethz.ch/pipermail/r-help/2007-January/123109.html | 1,695,587,583,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.30/warc/CC-MAIN-20230924191454-20230924221454-00391.warc.gz | 1,133,328,953 | 2,763 | # [R] mcmcsamp and variance ratios
Martin Henry H. Stevens hstevens at muohio.edu
Thu Jan 4 19:28:26 CET 2007
```On Jan 4, 2007, at 11:18 AM, Douglas Bates wrote:
> On 1/3/07, Martin Henry H. Stevens <hstevens at muohio.edu> wrote:
>> Hi folks,
>> I have assumed that ratios of variance components (Fst and Qst in
>> population genetics) could be estimated using the output of mcmcsamp
>> (the series on mcmc sample estimates of variance components).
>
>> What I have started to do is to use the matrix output that included
>> the log(variances), exponentiate, calculate the relevant ratio, and
>> apply either quantile or or HPDinterval to get confidence intervals.
>
>> This seems too simple but I can't think of what is wrong with it.
>
> Why bother exponentiating? I'm not sure what ratios you want but if
> they are ratios of two of the variances that are columns of the matrix
> then you just need to take the difference of the logarithms. I expect
> that the quantiles and HPDintervals would be better behaved, in the
> sense of being based on a distribution that is close to symmetric, on
> the scale of the logarithm of the ratio instead of the ratio itself.
>
> Quantiles calculated for the logarithm of the ratio will map to
> quantiles of the ratio. However, if you really do feel that you must
> report an HPDinterval on the ratio then you would need to exponentiate
> the logarithm of the ratio before calculating the interval.
> Technically the HPD interval of the ratio is not the same as
> exponentiating the end points of the HPDinterval of the logarithm of
> the ratio but I doubt that the differences would be substantial.
My collaborator (the evolutionary biologist on this project) is very
skeptical of the results I have been providing. Most of the Qst ratios,
Qst = Var[population] / ( Var[population] + Var[genotype] )
have values close to 0.5 (0.45--0.55) and wider confidence intervals
(e.g. 0.2--0.8) than they have tended to see in the literature.
I suspect that this derives from our tiny sample sizes: 24 genotypes
total, distributed among 9 populations (2-3 genotypes within each
population).
Our variances (shrinkage estimates) frequently do not differ from
zero. My model building using AIC results in the removal of most of
the variance components. We only stuck the terms back in the model in
order to get SOME number for these.
The biologist (supported by a biometrician) wants to bootstrap or
jackknife the models. I will be very skeptical if all of a sudden
they get qualitatively different estimates and intervals.
Does my perspective make sense? All comments appreciated.
-Hank
Dr. Hank Stevens, Assistant Professor
338 Pearson Hall
Botany Department
Miami University
Oxford, OH 45056
Office: (513) 529-4206
Lab: (513) 529-4262
FAX: (513) 529-4243
http://www.cas.muohio.edu/~stevenmh/
http://www.muohio.edu/ecology/
http://www.muohio.edu/botany/
"E Pluribus Unum"
``` | 763 | 2,926 | {"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} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.936863 |
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https://metanumbers.com/394 | 1,701,343,075,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00223.warc.gz | 458,672,563 | 7,330 | # 394 (number)
394 is an even three-digits composite number following 393 and preceding 395. In scientific notation, it is written as 3.94 × 102. The sum of its digits is 16. It has a total of 2 prime factors and 4 positive divisors. There are 196 positive integers (up to 394) that are relatively prime to 394.
## Basic properties
• Is Prime? no
• Number parity even
• Number length 3
• Sum of Digits 16
• Digital Root 7
## Name
Name three hundred ninety-four
## Notation
Scientific notation 3.94 × 102 394 × 100
## Prime Factorization of 394
Prime Factorization 2 × 197
Composite number
Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 2 Total number of prime factors rad 394 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 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 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 394 is 2 × 197. Since it has a total of 2 prime factors, 394 is a composite number.
## Divisors of 394
1, 2, 197, 394
4 divisors
Even divisors 2 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ 4 Total number of the positive divisors of n σ 594 Sum of all the positive divisors of n s 200 Sum of the proper positive divisors of n A 148.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 19.8494 Returns the nth root of the product of n divisors H 2.6532 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 394 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 394) is 594, the average is 148.5.
## Other Arithmetic Functions (n = 394)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 196 Total number of positive integers not greater than n that are coprime to n λ 196 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 79 Total number of primes less than or equal to n r2 8 The number of ways n can be represented as the sum of 2 squares
There are 196 positive integers (less than 394) that are coprime with 394. And there are approximately 79 prime numbers less than or equal to 394.
## Divisibility of 394
m n mod m
2 0
3 1
4 2
5 4
6 4
7 2
8 2
9 7
The number 394 is divisible by 2.
• Semiprime
• Deficient
• Polite
• Square Free
## Base conversion 394
Base System Value
2 Binary 110001010
3 Ternary 112121
4 Quaternary 12022
5 Quinary 3034
6 Senary 1454
8 Octal 612
10 Decimal 394
12 Duodecimal 28a
20 Vigesimal je
36 Base36 ay
## Basic calculations (n = 394)
### Multiplication
n×y
n×2 788 1182 1576 1970
### Division
n÷y
n÷2 197 131.333 98.5 78.8
### Exponentiation
ny
n2 155236 61162984 24098215696 9494696984224
### Nth Root
y√n
2√n 19.8494 7.33104 4.45527 3.30445
## 394 as geometric shapes
### Circle
Diameter 788 2475.58 487688
### Sphere
Volume 2.56199e+08 1.95075e+06 2475.58
### Square
Length = n
Perimeter 1576 155236 557.2
### Cube
Length = n
Surface area 931416 6.1163e+07 682.428
### Equilateral Triangle
Length = n
Perimeter 1182 67219.2 341.214
### Triangular Pyramid
Length = n
Surface area 268877 7.20813e+06 321.7
## Cryptographic Hash Functions
md5 28f0b864598a1291557bed248a998d4e bc62305b6cff49d43aed5f6550716c89890a3ccc 04d19fde0a08b17aca69491e714bea43565384d12a63626e08477662cc03780e c2ecf9109ef89a73d081161c67eb0eb1cdb1377e35af79ad2928b3cd84e86902af3f2b0007c2713463aaa44069e12536b6471cfcebb99007923e09b76de03c58 3dfae353e61ec8330bcd35b5a06c20a2485492c7 | 1,271 | 3,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.836856 |
https://pzck.wirtualnemapy.pl/word-problems-for-grade-2-multiplication-pdf.html | 1,610,783,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703505861.1/warc/CC-MAIN-20210116074510-20210116104510-00266.warc.gz | 521,125,858 | 9,068 | # Word problems for grade 2 multiplication pdf
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• Worldpercent27s dumbest drivers,Word Problem 2nd Grade Worksheets Two Step Word Problems Grade ... #174992,Grade 2 Multiplication and Division Word Problems Name: _____ Class: _____ Question 1 5 boys share 35 candies equally. How many candies does each boy get? Question 2 24 children are put into equal groups. There are 4 children in each group. How many equal groups are there? Question 3
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Word Problem With Multiplication of Two Decimals - Daniel earns \$8.80 per hour at his part-time job. On Sunday, he worked 5.5 hours. What were his total earnings for the day?
• Video volume booster apkFriday October 16th, 2020 Distance Learning / Online Learning and Server Upgrade Fun Hey all, So this has been a fun couple weeks. :P For those who missed the last couple updates, the site was completely recoded (not just for distance learning, but also for some other updates coming soon). ,Multi-Step Word Problem Unit Grade 4 In this unit you will find: ๏a complete, comprehensive unit plan ๏12, two-step word problems and solutions ๏6 word problems and solutions for students to analyze ๏student Planning Sheet ๏assessment and solutions Suggestions: ๏Make packets of student problems - copy and staple both sheets of problems and the sheet of problems to analyze.
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6 Multiplication Word Problems Grade 3 Pdf 2 - Use these free worksheets to master letters, sounds, words, reading, writing, numbers, colors, shapes and other preschool and kindergarten skills.All worksheets are pdf documents for quick printing.
• Which of the following statements about bilingual education is trueWord problems Here is a list of all of the skills that cover word problems! These skills are organized by grade, and you can move your mouse over any skill name to preview the skill. To start practising, just click on any link. IXL will track your score, and the questions will automatically increase in difficulty as you improve! ,The following collection of free 4th grade maths word problems worksheets cover topics including addition, subtraction, multiplication, division, mixed operations, fractions, and decimals. Students need to gain a strong understanding of place value in order to understand the relationship between digits and how these relationships apply to ...
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Math Word Problems Grade 9 Worksheets Number Line Multiplication Worksheet Multiplication Template Free Multiplication Worksheets Grade 4 Multiplication Worksheets Grade 7 Thanksgiving Multiplication Worksheets Math Exercise For Class 2 Free Printable For Preschool Activity Addition Of Polynomials Worksheet PDF Addition Of Polynomials Worksheet PDF Reading Paragraph About Teacher Super ...
• Nms easy frigate fuelThis is an online free math test for third grade. This test comprises of word problems which use mixed operations including addition, subtraction, multiplication, and division. The students have to use the correct operation according to each question to answer the problems.
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Nov 07, 2011 · Multiplication Word Problems - word problems to practice multiplying 3-digit numbers by 2-digit numbers Multiplication Video - [11:15] 2-digit times a 2-digit number from the Khan Academy Sum Sense (Multiplication) - Students practice single digit multiplication by dragging numbers to complete the sentence.
• Wife died childbirth redditThese 2-digit by 1-digit multiplication word problems give students practice solving word problems using a variety of multiplication strategies with a work mat and 24 task cards. Also included is a set of 24 worksheets featuring the same multiplication problems as individual printables.
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• Solve multiplication word problems with factors up to and including 10. Represent the problem using arrays, pictures, and/or equations with a symbol for the unknown number to represent the problem. • Solve division word problems with a divisor and quotient up to and including 10.
• Filezilla could not start transferTo help second-grade students learn to solve word problems, teach them to use the following steps: Survey the math problem: Read the word problem to get an idea of its general nature. Talk with your students about the problem and discuss which parts are most important. Read the math problem: Read the question again. This time, focus on the ...
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In 4th grade math, the multiplicative comparison is crucial as it helps in understanding multiplication as scaling in Grade 5 and sets the stage for proportional reasoning in Grade 6. Introducing multiplication with area and perimeter allows for new and interesting multiplication word problems as children learn to calculate with larger numbers ...
• Cpu benchmark e5 2697 v4Multiplication Word Problems . Multiplication word problem worksheets. 1, 2 and 3 digit word problems for multiplication. 3rd grade math worksheets for problem solving with multiplication.
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We have a complete K-12 math curriculum library. One of the largest available anywhere. Check grade levels below. Click on the links below to view sample pages.
• Homebridge hueUse multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. 1. Worksheet Multiplication and Division. Division Word Problems Worksheet (2 Pages)
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4.1 The student will a) identify orally and in writing the place value for each digit in a whole number expressed through millions; b) compare two whole numbers expressed through millions, using symbols (>, , or = ); and c) round whole numbers expressed through millions to the nearest thousand, ten thousand, and hundred thousand.
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2-Digit Multiplication Word Problems Worksheets. Task 4th grade and 5th grade children with solving the six story problems in each pdf here. Watch kids train themselves to solve our free word problem worksheets on multiplying 2-digit by 2-digit numbers.
• S3 to dynamodbGrade 5 multiplication word problems Worksheet. Search form. Search . To print this worksheet: click the "printer" icon in toolbar below. To save, click the "download" icon. Sign Up For Our FREE Newsletter! * By signing up, you agree to receive useful information and to our ...
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Regarding point (2) above: It can be really frustrating (and embarassing) to spend fifteen minutes solving a word problem on a test, only to realize at the end that you no longer have any idea what "x" stands for, so you have to do the whole problem over again. I did this on a calculus test — thank heavens it was a short test! — and, trust ...
• Craigslist livingston mt rentalsMultiplication Word Problems (eBook), Remedia Publications Grade Level: 3-6Interest Level: N/AReading Level: N/ASolve it with math! This 16-lesson learning unit starts with two ,By the end of first grade, children should be able to recognize basic number patterns including addition, subtraction, word problems, measurements, and more. Comparisons and recognition of very simple geometric shapes make up the remainder of their expected knowledge base for first grade, and multiplication games for 1st grade can support this ...
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Practice solving word problems by using one of the following multiplication strategies: create an array, skip counting, repeated addition, or writing a multiplication sentence. Part two of three. 2nd grade
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2-Digit Multiplication Word Problems Worksheets. Task 4th grade and 5th grade children with solving the six story problems in each pdf here. Watch kids train themselves to solve our free word problem worksheets on multiplying 2-digit by 2-digit numbers.
• Amigel supplierRewrite the number 1 as an equivalent fraction i.e. 2/2, 3/3, 4/4, etc. 5.NF.5b Solve word problems involving multiplication of fractions and mixed numbers. 5.NF.6 Represent the product of fractions in simplest form. 5.NF.6 Write equations to represent word problems involving multiplication of fractions. 5.NF.6 ,Multiplication Worksheets Times Table Worksheets Brain Teaser Worksheets Picture Analogies Cut and Paste Worksheets Pattern Worksheets Dot to Dot worksheets Preschool and Kindergarten – Mazes Size Comparison Worksheets. Top Worksheets New Worksheets Most Popular Math Worksheets . First Grade Worksheets Most Popular Worksheets New Worksheets ...
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Grade 5 Multiplication Word Problems - Displaying top 8 worksheets found for this concept.. Some of the worksheets for this concept are Multiplication word problems, Multiplication word problems, Multiplication and division word problems no problem, Fifthgradecommoncore wordproblems operationsand, Sample work from, Multiplication word problems, Single digit multiplication 1, Word problem ...
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Displaying top 8 worksheets found for - Division Word Problems. Some of the worksheets for this concept are Division word problems, Division, Division 1, Grade 2 multiplication and division word problems, Word problem with division of two decimals work 19, Division word problems, Multiplication and division word problems no problem, Dividing fractions word problems 1.
• Thompson center thunderhawk 50 cal price• Solve multiplication word problems with factors up to and including 10. Represent the problem using arrays, pictures, and/or equations with a symbol for the unknown number to represent the problem. • Solve division word problems with a divisor and quotient up to and including 10.,Multiplication Word Problems (1-step word problems) Here are some examples of multiplication word problems that can be solved in one step. We will illustrate how block models (tape diagrams) are used in the Singapore math approach to help you to visualize the multiplication word problems in terms of the information given and the answer that ...
Jan 19, 2017 · Now, my 8-year-olds begin third grade in September doing multiplication and division word problems using variables. No matter that many of them still haven't grasped addition and subtraction, and they don't know their times facts all that well, these are the expectations.
• Angka bocoran toto 4d hari iniDec 16, 2020 - Explore Suzette Moore's board "Multiplication Activities", followed by 324 people on Pinterest. See more ideas about multiplication, 3rd grade math, math classroom.
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This is an online free math test for third grade. This test comprises of word problems which use mixed operations including addition, subtraction, multiplication, and division. The students have to use the correct operation according to each question to answer the problems.
• Fiio m5 updateWord Problem Multiplication Sentence Total Partner Practice Multiplication Matters Game Directions: Cut out all cards on page 15. Lay all cards face down in 3 separate piles: "Word Problems", "Multiplication Sentence", and "Total." Mix them up. First partner chooses 1 of any card and puts it under the correct heading.,Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.2 Interpret whole-number quotients of whole numbers 4.OA.1, 4.OA.2 Interpret and solve a 3.OA.3 Use multiplication and division to solve word problems involving equal groups, arrays, and measuremen t multiplication equation as a comparison
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• Smart start breathalyzerGrade 2 Multiplication and Division Word Problems Name: _____ Class: _____ Question 1 3 girls share 24 toys equally. How many toys does each girl get? _____ girls share _____ toys equally. How many toys does each girl get? Question 2 A hamburger costs 3 dollars . Petra has 20 dollars.,Grade 5 multiplication word problems Worksheet. Search form. Search . To print this worksheet: click the "printer" icon in toolbar below. To save, click the "download" icon. Sign Up For Our FREE Newsletter! * By signing up, you agree to receive useful information and to our ...
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Foster critical thinking skills with practice problems, video hints, and full step-by-step solutions, all clearly aligned with Common Core standards.
• Minecraft bedrock mutant creatures modMultiplication Word Problems . Multiplication word problem worksheets. 1, 2 and 3 digit word problems for multiplication. 3rd grade math worksheets for problem solving with multiplication.
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Solve the following multiplication problems using the Egyptian method. 14 x 16 23 x 13 34 x 126 1 28 2 56 4 112 8 224 13 364 1 35 2 70 4 140 8 280 16 560 17 595 1 58 2 116 4 232 8 464 16 928 32 1856 45 2,610 | 2,915 | 13,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.912576 |
https://matchmaticians.com/questions/9uoznv/9uoznv-green-s-theorem-question | 1,721,898,405,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00893.warc.gz | 319,294,015 | 17,040 | # Green's Theorem
In the following exercise you must explain with detail how to use the Green's Theorem to get the solution.
Find the area bounded by the hypocycloid $r(t)=(4cos(t)+ cos(4t), 4sin(t) - sin(4t))$
• Mathe
+2
Hi, there must be an error in your question because the second component of the curve simplifies into 3sin(t)
• Mathe
+1
For the curve to be a hypocycloid, r(t) must be (4cos(t)+cos(4t),4sin(t)−sin(4t))
• You're right it was a typo mistake.
Answers can only be viewed under the following conditions:
1. The questioner was satisfied with and accepted the answer, or
2. The answer was evaluated as being 100% correct by the judge. | 188 | 658 | {"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} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.94826 |
https://ask.libreoffice.org/t/help-using-an-advanced-calc-formula/23363 | 1,674,913,029,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00617.warc.gz | 117,715,644 | 5,067 | Help using an advanced calc formula
Hello, I have attached a document I need help with:
Benefit deduction example.ods
Basically I need the DEDUCTION (\$) column to display the result I am after based on the information shown below the table.
It’s kind of hard to describe my question, considering I’m not sure what formula I should use to achieve the intended result. Hopefully the information supplied in the attached document should make everything self- explanatory.
I’m not trying to be lazy here, it’s just that the formula required to work out the result I need is a bit advanced for me as I only just started using calc.
Could anyone who is good at using calc please assess the attached document and explain the formula I need to use here? I would be extremely grateful as I’m a little out of my depth here.
See Chapter 7 in the Calc manual - all is explained there.
There are many ways. One way is to use one or more IF statements. ie IF (X) Then (Y) Else Z.
So, If salary greater than 100 and less than 200, THEN reduction equal (salary - 100) * 0.3 Else reduction = 0
If salary greater than 300, reduction equals \$30 + (salary - 300) * 0.7
Perhaps something like this it’s what you are looking for.
148554239973229.ods | 291 | 1,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.911725 |
https://kachina2012.wordpress.com/category/27-37-137/ | 1,600,840,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00555.warc.gz | 488,717,864 | 33,273 | # 2012 unv-IEL-ing = YHVH = ‘The Lord’ = I37 = 5wa5tika
### What exactly is the significance of 27, 37, to the numbers 137 and 64?
@ 4:37 of this video a much younger Charles connects MOLEEDS to DNA and 27, 37, 137, and when I am 64.
IEL mirrored on its x-axis or rotated 90 degrees twice (180 degrees) looks like 73I
i.e. they are polar opposites and essentially the 180 degrees indicates they are out-of-phase.
And 73I if read from right to the left yields I37.Who cares about the number I37 and the language of numbers, and how they translate into ‘maps’?
Well a fella called Pythagoras would be one, CF is another.
But Pythagoras is dead, and clearly an animated Charlie Fleischer is not.
Something I noticed, @ 5:58 he mentions drawing a map of Toronto, where I was born.
• @ 6:37 CF says “the point is numbers are maps”
• @ 7:00 we see Euler who died in 1783, and CF connects how his IDEAS paved the way to modern string theory.
• @ 9:40 I noted that Sir Issac Newton has a most interesting date of birth/death
4 January 1643 – 31 March 1727
[OS: 25 December 1642 – 20 March 1727]
• @ 11:17 “I think therefore I am”
• @ 11:33 “He unified algebra and geoMetry”
• @ 11:47 we see an image of Descartes who helped us with learning how to convert geometric xyz into an algebraic form.
Left Brain and Right Brain in communication…note the year on the stamp.
Published in 1637, the Discours de la méthode (Discourse on the Method) stands at a crucial point of transition in the philosophical and scientific thinking of René Descartes, and please note in the year 1637, we have the inverted 9.
• @ 11:55 In 1793, French Revolutionary forces captured Monaco, and it remained under French control until 1814.
What is the significance of MONaco?
Obviously the French Connection is a good reason why MONaco and Indonesia have the same flags…RED on top WHITE on the bottom.
That BTW is the REVERSE of what the French would have found in Egyptian History.
AHA!
• @12:55 JC FGauSS has many of my favorite numbers, (3×7) 137 and 55, and my address #43 in his birth/death cycle too.Johann Carl Friedrich Gauss (Latin: Carolus Fridericus Gauss) (30 April 1777 – 23 February 1855) was a German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.
Here is a 137 WOW involving GauSS…
Gauss also discovered that every positive integer is representable as a sum of at most three (3) triangular numbers on (1)0 July (7) and then jotted down in his diary the famous words, “Heureka! num = Δ + Δ + Δ”
IF I0= I
3 I 7
….IT gets better Gau55 died in the year I8SS and NOW the WOW MON and MOM…another 55.
Gauss’s presumed method was to realize that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050
• @12:58 the golden section and HELVETIAHelvetia is the female national personification of Switzerland, officially Confœderatio Helvetica, the “Helvetic Confederation”.
The Old Swiss Confederacy (German: Alte Eidgenossenschaft) was the precursor of modern-day Switzerland. It was a loose federation of largely independent small states called cantons that existed from the late 13th or early 14th century until 1798, when it was invaded by the French Republic, who transformed it into the short-lived Helvetic Republic.
AHA!
How does the former Helvetic Confederation connect to Hopi Prophecy?
So how do I (raphaEL) connect the Fibonacci numbers 1, 1, 2, 3, 5, 8 to the numbers 137?
• @ 13:37 Where is the going?
A summation of 27 years work by Charles Fleischer….follow the voice/character of the white rabbit.
Roger and ME.
• @ 14:09 Charles discusses ‘ATOM SMASHING’, the image he shows looks similar to this one.
• @ 14:12 he begins his discussion of MOLEEDS.
(…a 27 year white rabbit synthesis)
• @ 16:56 Charles merges MOLEEDS 1/37 with the GOLDEN RATIO based on Fibonacci numbers 1, 1, 2, 3, 5, 8.
@ 4:37 of this video a much younger Charles connects MOLEEDS to DNA.
more to ponder:
Can we extract the codes I have found 11258 and 137 from Sir Isaac Newton?
Sir IN yields INRI S and SIR Isaac Newton….gotta love the SS, the Subtle Signature we all leave behind in our ‘wakes’.
As our titanic illusions go down with the captain of the ship of fools….
Can we extract the codes I have found 11258 and 137 and MON and place them all onto ONE KEY found in researching the TAROt?
By the way something profound to consider at this junction of where the NUMBER collides with an IDEA, one of my archetypal guides has been CJ along with JC.
Carl Jung, and just by coincidence his ‘guide’ was an entity he referred to as PhILEmon.
By coincidence we can extract from the name Philemon, PH, the ILE and the MON that we find on the tarot key I recovered, with the assistance of the 20th century mystic C.C. Zain.
And after Pythagoras came Plato who suggested IDEAS give rise to shape and form.
And if we merge Pythagoras’ number + Plato’s IDEAS about IDEAS we find ourselves on Carl Jung’s couch 2000+ years later discussing the importance of archetypes, and the role of acausal synchronicity, with Wolfgang Pauli, who died in room I37 in 1958.
And I37 is a number that he and all the other physicist/mystics eventually become obeSSed with.
BUT we need to go back to the importance of the roles of these other folks play throughout the various incarnations of the scriptures over thousands of year…until modern physics penetrated the ATOM and found its secrets.
Today I37 has been described as the following:
I37 is the DNA of light.
I37 is the fingerprint of an atom, or was it adam?
But using allegory where the NUMBER/IDEA/SHAPE/FORM collide, how would this Oral Tradition evolve.
How would SOUNDS being put onto paper using writing, appear in the text we read?
How do we hear what cannot be heard using our eyes?
Shall we ask Galileo GaliLEI to explain using his heretical style?
Or how about the esoteric giant called ELIphas LEvI, an adopted name?
Important to NOTE who is in the center closest to the Tabernacle were.
The priestly LEvItes of course.
Thus LEvI blue jeans/genes take on a whole new meaning.
Funny how meSSiahs are far too often associated with the color blue.
ELIjah, GabrIEL, RazIEL, DanIEL, EzekIEL
The following are the Archangels associated with the ten sephiroth of the Tree of Life:
Metatron KetherIEL, RazIEL, ophanIEL, TzadqIEL, KhamaEL, RaphaEL MelakhIEL, HanIEL, MichaEL or MikhaEL BenIEL, GabrIEL, SandalPHON or SandalPHON MalkuthaEL is associated with the bottom sephiroth number ten.
The angelic names, a subset of angels associated with each of the ten sephiroth, reveals the following pattern.
With the exception of Kether (whose symbol is the swastika), the other nine groups of angels corresponding to each of the sephiroth ALL end withIM.
ALL of the Archangelic names associated with the Planets end in either IEL or AEL.
i.e. KaSSIEL is the archangel of Saturn and RaphaEL is the archangel of Mercury (the meSSenjah )
Each planet has an intelligence and ALL of them end in IEL except the MOON which has two intelligences attributed to it.
One ends in the suffix IM and the other EL
Each planet has a subset of spirit names.
With the exception of the Sun, the Moon, and Mercury, all other planet spirits have an EL ending.
The Archangels associated with each sign of the zodiac ALL end in IEL, with these exceptions:
Capricorn and Aries end archangels have a AEL suffix and Taurus’ archangel is AsmodEL.
##### http://www.amazon.com/Self-Initiation-I … 1567181368
All of the above info, with descriptions of each archangel, angel, spirit is found in this book from pages 157 – 166.
(this guide has 792 pages)
Seeing a pattern yet?
Don’t worry you will, we are obviously being lead to the same ONE IDEA…follow me folks to the buried treasure deep inside each of you, ewe, and magnetic U, the treasure is always marked by an X
##### DEM BONES Sir Issac Newton’s Coat of Arms next to Photo 51 The first ever x-ray of DNA – double hELIx
The pirate Sir Issac Newton’s Coat of Arms ‘beside itself’, i.e. what all SAGES see…the X marking the treasure is the same perspective aslooking down/from above on the double helix discovered by Rosalind Franklin, an X-ray diffraction expert…
Who is Rosalind Franklin?
She was in fact a ‘SuperWoman’, able to use her X-RAY talents and abilities the way a ‘SuperMan’ would.
She was the matriarch that x-rayed the DNA TEMPLE originally…that I have recovered embedded in ancient TEMPLEs…
It’s hard for us to imagine now how such a thing could have happened. It’s almost as if there were a conspiracy to credit two “nice” men, instead of a Jewish woman, with the DNA discovery that revolutionized the twentieth century.
She was the woman who planted the SEED, an IDEA into the fertile creative brains of Francis Crick and James Watson.
The Unsung Hero Who Discovered The Double Helix
There is now ample evidence from multiple sources that Franklin’s colleagues and graduate students at King’s College showed her x-ray images of DNA to Watson and Crick without her permission or knowledge.
The so-called Photo 51 (pictured above) provided proof that DNA’s structure was probably a helix. Several witnesses – including Crick and Watson themselves – say that the two researchers saw this photo before “discovering” the double helix.
http://io9.com/#!5761388/the-unsung-her … uble-helix
Do you have any IDEA what it is like unraveling history and all of the intellectual egotistical thievery that occurred in the name of science, as ignorant humans clawed their way to the top?
The TRUTH is wonderfully liberating, it feels just right….so why do the sheeple avoid it?
I can feel this TRUTH about 137 in ME bONEs folks.
Which bones?
DEM BONES of course.
2bee continued…
namaste
p.s.
I do hope you noticed all of the 27 along with 37 and 137.
If not Michael Joyce is a fine mentor;
God’s signature in Genesis1.1, Light & Hydrogen
http://www.numberscience.plus.com/Godsig.html
All connected to phi and PHI?
Which are RA-tios?
Does RAphaEL need to teach you about RA now?
Let ME know.
_________________
KEY 528=Swastika=ancient Spherical Standing Wave Theory
“A theory is more impressive the greater is the simplicity of its premise, the more different are the kinds of things it relates and the more extended its range of applicability…” | 2,774 | 10,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.899224 |
https://mtp.tools/converters/byte-to-exabit-calculator | 1,571,782,883,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987824701.89/warc/CC-MAIN-20191022205851-20191022233351-00528.warc.gz | 604,048,824 | 5,508 | # Bytes (B) to Exabits (Eb) calculator
Input the amount of bytes you want to convert to exabits in the below input field, and then click in the "Convert" button. But if you want to convert from exabits to bytes, please checkout this tool.
## Formula
Formula used to convert B to Eb:
F(x) = x / 125000000000000000
For example, if you want to convert 15 B to Eb, just replace x by 15 [B]:
15 B = 15/125000000000000000 = 1.2e-16 Eb
## Steps
1. Divide the amount of bytes by 125000000000000000.
2. The result will be expressed in exabits.
## Byte to Exabit Conversion Table
The following table will show the most common conversions for Bytes (B) to Exabits (Eb):
Bytes (B) Exabits (Eb)
0.001 B 0 Eb
0.01 B 0 Eb
0.1 B 0 Eb
1 B 0 Eb
2 B 0 Eb
3 B 0 Eb
4 B 0 Eb
5 B 0 Eb
6 B 0 Eb
7 B 0 Eb
8 B 0 Eb
9 B 0 Eb
10 B 0 Eb
20 B 0 Eb
30 B 0 Eb
40 B 0 Eb
50 B 0 Eb
60 B 0 Eb
70 B 0 Eb
80 B 0 Eb
90 B 0 Eb
100 B 0 Eb
A byte is a unit of digital information that represents eight (8) bits. It is widely used in computing and in digital communications. The symbol used to represent a byte is B (upper-case letter B). It was designated by the International Electrotechnical Commission (IEC) and Institute of Electrical and Electronics Engineers (IEEE).
The amount of bits that a byte represent changed over the years, but nowadays a byte represents 8 bits. With 1 byte you can represent numbers from 0 to 255.
A exabit is a unit of measurement for digital information and computer storage. The prefix exa (which is expressed with the letter E) is defined in the International System of Units (SI) as a multiplier of 10^18 (1 quintillion). Therefore, 1 exabit is equal to 1,000,000,000,000,000,000 bits and equal to 1,000 petabits. The symbol commonly used to represent a exabit is Eb (sometimes as Ebit).
## FAQs for Byte to Exabit calculator
### What is Byte to Exabit calculator?
Byte to Exabit is a free and online calculator that converts Bytes to Exabits.
### How do I use Byte to Exabit?
You just have to insert the amount of Bytes you want to convert and press the "Convert" button. The amount of Exabits will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Byte to Exabit work on?
Byte to Exabit calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 705 | 2,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | latest | en | 0.84241 |
https://woodstock-online.com/the-force-f-is-f-ma-acceleration/ | 1,696,064,920,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00662.warc.gz | 661,586,100 | 11,846 | # The force f is f = ma. Acceleration
TheMathematician Isaac Newton was born January 4, 1643, Woolsthorpe Manor,United Kingdom. The English physics studied the basic force of gravity and isone of the world’s greatest scientist. But, beneath his truth was a story neverto be spoken of. Newton made discoveries in optics, motion, and mathematics.Newton thought that white light had particles of even other colors as well. In1866 Newton came home from Trinity college due to the horrible disease that hadspread around the country.Duringall of the disease Newton has become fashionably astonished and started tostudy the force of gravity. While Newton was sitting under a tree an apple hitthe ground or on him.
Seeing an apple fall from the tree, made him to think whyit fell straight down instead of at an angle. This leading to Newton and hisdiscoveries about motion. Newton made the three laws of motion each will beexplained. Any object at rest will stay at rest until a force pushes the objectto move. Here is the second law.
The relationship between an object’s mass m,its acceleration a, and the applied force f is f = ma. Acceleration and forceare in this law the direction of the force vector is the same as the directionof the acceleration vector. Finally, here is the last law of motion. With everyaction there is are a same or different action upon it.
Down towards the end of his life Newton lived at apark, Cranberry Park, in Winchester, England, he lived with his niece and hisniece’s wife. By this time, Newton has become one of the greatest scientists ofall in England. His science has been unchallenged by others and probably won’tbe any other person in a century. Newton hadbecome a wealthy man, investing in charity’s and other businesses. Before his deathbed came when he was younger he was elected to Trinity college and was emittedto succeed in his life. The professor exempted him from necessity of tutoringbut imposed the duty of delivering an annual course of lectures.
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In1687, there was 18 months of nonstop writing and working, Newton publishedPrincipium of Mathematica, mostly well known as Principia. It is the to beknown as the greatest book of philosophy’s. After the publication Newton hadbecome famous and very popular from his book. Principia tells exactly all theforces of motion, with three basic laws of motion.Hereare the Three Laws of Motion . Here aresome websites that I have used to help me get my information · www.history.
com/topics/isaac-newton· https://www.grc.nasa.gov/www/k-12/airplane/newton.
html· teachertech.rice.edu/Participants/louviere/Newton/· https://www.britannica.
com/science/Newtons-laws-of-motion Citations “UP, UP, and AWAY!” Newton’s 3 Lawsof Motion, teachertech.rice.edu/Participants/louviere/Newton/law3.html. “Isaac Newton.
” Biography.com, A NetworksTelevision, 1 Aug. 2017, www.biography.com/people/isaac-newton-9422656.
x
Hi!
I'm Dora!
Would you like to get a custom essay? How about receiving a customized one? | 877 | 3,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.974142 |
https://www.gamedev.net/forums/topic/333377-freesolid-gjk/ | 1,542,744,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746639.67/warc/CC-MAIN-20181120191321-20181120213321-00281.warc.gz | 881,560,369 | 26,677 | FreeSolid GJK
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Does FreeSolid support single float precision? I'd have problems with FreeSolid 2.0, because the approximation loop for find the closest simplex points never stops! due rounding precision isues. Honestly, I don't understand GJK algorithm very well. The loop exit condition is one of the trickiest part of the algorithm, because it does somehow of absolute and relative error checking that it isn't clear to me. This is the trouble part of solid:
void closest_points(const Convex& a, const Convex& b,
const Transform& a2w, const Transform& b2w,
Point& pa, Point& pb) {
static Vector zero(0, 0, 0);
Vector v = a2w(a.support(zero)) - b2w(b.support(zero));
Scalar dist = v.length();
Vector w;
bits = 0;
all_bits = 0;
Scalar mu = 0;
#ifdef STATISTICS
num_iterations = 0;
#endif
while (bits < 15 && dist > abs_error) {
last = 0;
last_bit = 1;
while (bits & last_bit) { ++last; last_bit <<= 1; }
p[last] = a.support((-v) * a2w.getBasis());
q[last] = b.support(v * b2w.getBasis());
w = a2w(p[last]) - b2w(q[last]);
set_max(mu, dot(v, w) / dist);
if (dist - mu <= dist * rel_error) break;
if (degenerate(w)) {
#ifdef STATISTICS
++num_irregularities;
#endif
break;
}
y[last] = w;
all_bits = bits|last_bit;
#ifdef STATISTICS
++num_iterations;
if (num_iterations > 1000) catch_me();
#endif
if (!closest(v)) {
#ifdef STATISTICS
++num_irregularities;
#endif
break;
}
dist = v.length();
}
compute_points(bits, pa, pb);
}
I remind that I forces FreeSolid to use single float precision. So, Is FreeSolid 2.0 robust? It's logical that it takes 15 iterations for find the closest points of two spheres? Have you ever have a good experience with FreeSolid 2.0? Please tell me if you know another GJK collision package. Thanks. Att: A desperate game programmer.
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Well for GJK, the number 15 is significant because it's the number of possible "voronoi regions" for a tetrahedron ("3-simplex"). Basically, GJK goes around selecting tetrahedra made of points on your spheres, such that the tetrahedron converges very quickly to where some point on it to the origin represents the closest point between the two spheres.
Now, I can envision, with 2 small spheres (or spheres far from the origin) if your epsilon value is smaller than than the support function can produce, your loop might go infinite.
The code you present just looks like a precalculation for the real Voronoi computation, like "only test these certain regions" though I don't look very close. Is there an epsilon you can increase? You don't list enough to say for certain "this is unstable".
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232 views
### Plot from scratch (without explicit data)
I'm trying to draw a plot which I've been given in a scan: What is the correct way to plot a custom shaped curve --- a freehand tool counterpart. I used quick curve but it's hard to tell extremum ... | 3,334 | 12,451 | {"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} | 2.78125 | 3 | CC-MAIN-2014-41 | latest | en | 0.913535 |
https://community.qlik.com/t5/Qlik-Sense-App-Development/Visualization-Percent-of-Total/td-p/809996 | 1,542,425,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743248.7/warc/CC-MAIN-20181117020225-20181117042225-00332.warc.gz | 608,385,222 | 40,226 | # Qlik Sense App Development
Discussion board where members can learn more about Qlik Sense App Development and Usage.
Contributor
## Visualization & Percent of Total
I'm having trouble trying to get a visualization in Qlik Sense to display the percent of total. Here's what I have:
1. Dimension: Product
2. Spend: Sum({<[Mode (Trip)]={'A'}>}[Paid Fare])
This correctly displays the total spend by product in a horizontal bar chart. However, I'd like to show the percent of total spend by product. This was more obvious in QlikView but I'm having difficulty learning how to do this in Qlik Sense.
Thanks for your suggestions.
Tags (5)
1 Solution
Accepted Solutions
Employee
## Re: Visualization & Percent of Total
Hi , in QlikView there was a 'relative' checkbox to quickly convert. In qlik sense you need to write it into the expression. Try both of the following. You'll need to format as a percent if it works too...
Sum({<[Mode (Trip)]={'A'}>}[Paid Fare])
/
Sum( total {<[Mode (Trip)]={'A'}>}[Paid Fare])
OR
Sum({<[Mode (Trip)]={'A'}>}[Paid Fare])
/
Sum( total <Product> {<[Mode (Trip)]={'A'}>}[Paid Fare])
3 Replies
Employee
## Re: Visualization & Percent of Total
Hi , in QlikView there was a 'relative' checkbox to quickly convert. In qlik sense you need to write it into the expression. Try both of the following. You'll need to format as a percent if it works too...
Sum({<[Mode (Trip)]={'A'}>}[Paid Fare])
/
Sum( total {<[Mode (Trip)]={'A'}>}[Paid Fare])
OR
Sum({<[Mode (Trip)]={'A'}>}[Paid Fare])
/
Sum( total <Product> {<[Mode (Trip)]={'A'}>}[Paid Fare])
Contributor
## Re: Visualization & Percent of Total
That worked (first one). Thanks!
Employee
## Re: Visualization & Percent of Total
cool. the 2nd one is if you have a chart with multiple dimensions (Product and something else) , it will do a percent of subtotal based on the individual product totals . good luck ! | 498 | 1,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-47 | latest | en | 0.838542 |
http://metamath.tirix.org/mpeuni/ofmresval | 1,721,511,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00355.warc.gz | 18,533,624 | 1,809 | # Metamath Proof Explorer
## Theorem ofmresval
Description: Value of a restriction of the function operation map. (Contributed by NM, 20-Oct-2014)
Ref Expression
Hypotheses ofmresval.f ( 𝜑𝐹𝐴 )
ofmresval.g ( 𝜑𝐺𝐵 )
Assertion ofmresval ( 𝜑 → ( 𝐹 ( ∘f 𝑅 ↾ ( 𝐴 × 𝐵 ) ) 𝐺 ) = ( 𝐹f 𝑅 𝐺 ) )
### Proof
Step Hyp Ref Expression
1 ofmresval.f ( 𝜑𝐹𝐴 )
2 ofmresval.g ( 𝜑𝐺𝐵 )
3 ovres ( ( 𝐹𝐴𝐺𝐵 ) → ( 𝐹 ( ∘f 𝑅 ↾ ( 𝐴 × 𝐵 ) ) 𝐺 ) = ( 𝐹f 𝑅 𝐺 ) )
4 1 2 3 syl2anc ( 𝜑 → ( 𝐹 ( ∘f 𝑅 ↾ ( 𝐴 × 𝐵 ) ) 𝐺 ) = ( 𝐹f 𝑅 𝐺 ) ) | 289 | 496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.351266 |
https://socratic.org/questions/what-are-the-intercepts-of-2-x-3-2-25 | 1,575,556,550,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481076.11/warc/CC-MAIN-20191205141605-20191205165605-00205.warc.gz | 552,490,669 | 5,833 | # What are the intercepts of y=-2(x+3)^2+25?
Y-intercept is at (0,7) ; X-intercepts are at $\left(\frac{5}{\sqrt{2}} - 3 , 0\right) \mathmr{and} \left(- \frac{5}{\sqrt{2}} - 3 , 0\right)$
Y-intercept: putting x=0 in the equation we get $y = - 2 \cdot 9 + 25 = 7$ So Y-intercept is at (0,7); X-ntercept : putting y=0 in the equation we get $2 {\left(x + 3\right)}^{2} = 25 \mathmr{and} {\left(x + 3\right)}^{2} = \frac{25}{2} \mathmr{and} x + 3 = \frac{5}{\sqrt{2}} \mathmr{and} x + 3 = - \frac{5}{\sqrt{2}} \mathmr{and} x = - 3 + \frac{5}{\sqrt{2}} \mathmr{and} x = - 3 - \frac{5}{\sqrt{2}}$ graph{-2(x+3)^2+25 [-80, 80, -40, 40]}[Ans] | 297 | 636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 4.21875 | 4 | CC-MAIN-2019-51 | longest | en | 0.720732 |
https://cp3.irmp.ucl.ac.be/projects/madgraph/wiki/CernSchool2011?action=diff&version=1 | 1,539,994,453,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512499.22/warc/CC-MAIN-20181019232929-20181020014429-00236.warc.gz | 648,165,532 | 5,324 | # Changes between Initial Version and Version 1 of CernSchool2011
Ignore:
Timestamp:
Mar 20, 2012 4:15:49 PM (7 years ago)
Comment:
--
### Legend:
Unmodified
v1 == Basics of QCD for the LHC, Cern School 2011 == 5 Lectures are given on QCD and MC's developments for the LHC. The PDF of the lectures can be found here: [attachment:QCD1.pdf QCD1 : Intro and QCD fundamentals] [attachment:QCD2.pdf QCD2: QCD in the final state] [attachment:QCD3.pdf QCD3: QCD in the initial state] [attachment:QCD4.pdf QCD4: From accurate QCD to useful QCD] [attachment:QCD5.pdf QCD5: Advanced QCD with applications to the LHC] The examples and exercises are related to physics at the LHC. === Basic aspects of QCD === ===== Various Tests and exercises ===== A collection of test, exercises, and web applications on pQCD can be found in [attachment:QCD-exercises.pdf QCD-exercises.pdf]. ===== Exercise on recursive relations ===== Think about the computation of q \bar q + n \gamma in a recursive way, by trying not calculate the same quantities twice. Show that the complexity of the calculation goes from n! (factorial) to 2^{n-1} (exponential) . === Going NLO === 1. pp>H at LO (1-loop): details of the calculation (Mathematica Notebook) [attachment:HiggsGG-LO-mtfinite.nb HiggsGG-LO-mtfinite.nb] 1. pp>H at NLO: details of the calculation (Mathematica Notebook) [attachment:HiggsGG-NLO.nb higgsGG-NLO.nb] 1. pp>H at NLO: cross section evaluation for the LHC (Mathematica Notebook+PDF libraries to be compiled) [attachment:phenHiggs.tar.gz phenHiggs.tar.gz]. A summary of the results can be found in [attachment:Higgs.pdf Higgs.pdf]. === Selected exercises on QCD that can be solved with a Matrix Element generator. === * [:DeadCone:Radiation from heavy quarks]: the dead cone in %$e^+e^- \to Q \bar Q g$%. * [:GluonSpin:Spin of the gluon]: Vector vs scalar in the angular correlations of %$e^+ e^- \to$% 4 jets. * [:2Jets:Jets] : Di-jet kinematics and rates in pp collisions. * [:3Jets:3 Jets] : Energy distributions in 3-jet events in pp collisions. * [:tt:top production] : %$t \bar t$% production, Tevatron vs LHC. * [:WAsymm:Drell-Yan]: rapidity asymmetry at the Tevatron and LHC. * [:TopDecaySpinCorrelations: Spin correlations in top decay], by [http://www.hep.phy.cam.ac.uk/theory/webber/ Bryan Webber]. === Monte Carlo integration === A short introduction to the techniques of Monte Carlo integration. Exercises proposed during lecture are collected in this Mathematica Notebook: [attachment:mc101.nb mc101.nb]. === LHC Phenomenology === * [:DiscoverTheHiggs:SM Higgs discovery at the LHC]: Three important channels * [:WarmingUpChallenge:Simple Black Boxes]: New gauge bosons === How to familiarize with Software.MadGraph === All exercises proposed can be "solved" or checked with Software.MadGraph / Software.MadEvent. Here is how to familiarize with the code. * Logon to the Software.MadGraph web site and register. * Familiarize with the code by generating a few processes in QED and QCD trying to guess which diagrams appear. What is the minimum number of jets have to be asked for in %$e^+e^-$% collisions so that the triple gauge vertex appear? * Look at the new physics models and check the particle and interactions content. * Generate events for a few selected processes and look at the plots: * ttbar production with decays: pp>tt~>bb~mu+e-ve~vm * VV production: pp>VV> leptons, with V=Z,W. * Single top + Higgs: pp>tHj (QCD=0, QED=3, j=gudsc,p=gudscb). Show that there is a large negative interference between the diagrams -- Main.FabioMaltoni - 2011-09-09 | 980 | 3,564 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-43 | latest | en | 0.835342 |
http://car4car.top/two-step-equations-worksheet-with-exles/ | 1,590,694,043,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00367.warc.gz | 22,911,867 | 18,136 | Two Step Equations Worksheet With Exles
In Free Printable Worksheets171 views
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Top | 2,154 | 10,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-24 | latest | en | 0.936367 |
http://www.logicalaptitude.com/reasoning/number-series-reasoning-questions/ | 1,604,051,004,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00281.warc.gz | 144,252,802 | 8,410 | # Number Series Reasoning Questions
Number Series Exercise 2 – Find the missing number in the given Number series. Select the correct answer among Multiple choice given for each Question.
Q 1- 37, 38, 34, 43, 27, …. ?
• 49
• 54
• 52
• 48
Q 2- 24, 27, 30, 31, 36, 35, …. ?
• 42
• 40
• 39
• 45
Q 3 – 5, 7, 11, 19, …?
• 28
• 30
• 33
• 35
Q 4- 11, 3, 14, 17, 31, …?
• 39
• 48
• 42
• 38
Q 5 – Complete the missing number 32, 23, 27, 72, 94, …?
• 84
• 49
• 78
• 68
Q 6 – Find the missing number 17, 8, 19, 16, 21, .?., 23
• 11
• 25
• 24
• 26
Q 7- Complete the number series 6, 7, 13, 20, …, 53
• 29
• 31
• 33
• 35
Q 8- 96, 90, …, 60, 36
• 80
• 82
• 84
• 78
Q 9 – Find the missing number 9, 5, 21, 45, 95, …, 392
• 204
• 206
• 192
• 184
Q 10 Complete this series 3, 20, 63, 144,275, …?
• 354
• 468
• 548
• 554
For more Reasoning Question, please refer here
Also refer here for latest GRE updates. | 445 | 902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | longest | en | 0.530395 |
http://www.velocityreviews.com/forums/t823345-p2-algorithm-help.html | 1,394,761,958,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678683421/warc/CC-MAIN-20140313024443-00028-ip-10-183-142-35.ec2.internal.warc.gz | 592,781,531 | 10,102 | Velocity Reviews > Ruby > algorithm help
# algorithm help
Pit Capitain
Guest
Posts: n/a
08-02-2005
Martin DeMello schrieb:
> Ara.T.Howard <(E-Mail Removed)> wrote:
>
>>here dark is the array having values like
>>
>> [ 1, 2, 3, 6789, 6790, 6791 ]
>>
>>that i want to reduce to a list of ranges.
>>
>>in reality the ranges are huge and there will, in amost all cases except
>>crossing the poles, be only one. also note that 'dark' is a sorted list. my
>>current optimization is
>>
>> min, max = dark[0], dark[dark.size - 1]
>>
>> ranges =
>> if((max - min + 1) != dark.size)
>> slow_search dark
>> else
>> [ min .. max ]
>> end
>>
>> ranges.each do |range|
>> #
>> # munge data based on ranges which are small and fast
>> #
>> end
>>
>>i can't think of anything better than brute force for the 'slow_search' but
>>thought i'd throw it out there...
>
> This might work better: binary search for {x | (x - min) == (i_x - i_min)},
> set i_min to i_x+1 and repeat. For an added optimisation, run two loops
> in parallel, searching from each end.
Just in case you haven't implemented Martin's binary search idea yet:
def gap?( data, min, max )
data[ max ] - data[ min ] > max - min
end
def find_gaps( data, min, max, acc )
if gap?( data, min, max )
if max - min == 1
acc << max
else
mid = ( max + min ) / 2
find_gaps( data, min, mid, acc )
find_gaps( data, mid, max, acc )
end
end
end
def gaps( data )
acc = [ 0 ]
find_gaps( data, 0, data.size - 1, acc )
acc << data.size
end
def ranges( *data )
gaps = gaps( data )
( 1 ... gaps.size ).map { |idx|
data[ gaps[ idx - 1 ] ] .. data[ gaps[ idx ] - 1 ]
}
end
p ranges( 1, 2, 3, 6789, 6790, 6791 )
# => [1..3, 6789..6791]
Regards,
Pit
Ara.T.Howard
Guest
Posts: n/a
08-03-2005
On Tue, 2 Aug 2005, Pit Capitain wrote:
>> This might work better: binary search for {x | (x - min) == (i_x - i_min)},
>> set i_min to i_x+1 and repeat. For an added optimisation, run two loops
>> in parallel, searching from each end.
>
> Just in case you haven't implemented Martin's binary search idea yet:
>
> def gap?( data, min, max )
> data[ max ] - data[ min ] > max - min
> end
>
> def find_gaps( data, min, max, acc )
> if gap?( data, min, max )
> if max - min == 1
> acc << max
> else
> mid = ( max + min ) / 2
> find_gaps( data, min, mid, acc )
> find_gaps( data, mid, max, acc )
> end
> end
> end
>
> def gaps( data )
> acc = [ 0 ]
> find_gaps( data, 0, data.size - 1, acc )
> acc << data.size
> end
>
> def ranges( *data )
> gaps = gaps( data )
> ( 1 ... gaps.size ).map { |idx|
> data[ gaps[ idx - 1 ] ] .. data[ gaps[ idx ] - 1 ]
> }
> end
>
> p ranges( 1, 2, 3, 6789, 6790, 6791 )
> # => [1..3, 6789..6791]
sorry it took me so long to respond pit and martin... got sidetracked. your
binary search idea was perfect! for the actual data i'll be working with this
should be either O(1) or O(log(n)) - thought it can obviously degrade in the
general case. i used your idea in essentially the reverse pit, here's what i
ended up with:
harp:~ > cat a.rb
def sequences list
distance = lambda do |a,b|
(b - a).abs
end
continuous = lambda do |range, list|
first, last = range.first, range.last
a, b = list[first], list[last]
(list.empty? or (distance[a,b] == distance[first,last])) ? (a .. b) : nil
end
discrete = lambda do |range, list|
first, last = range.first, range.last
edge = last + 1
edge >= list.length or distance[list[last], list[edge]] > 1
end
sequence = lambda do |range, list|
first, last = range.first, range.last
list[first] .. list[last]
end
last = list.length - 1
a, b = 0, last
accum = []
while a <= b and a < list.length and b < list.length
range = a .. b
if continuous[ range, list ]
if discrete[ range, list ]
accum << sequence[ range, list ]
a, b = b + 1, last # move a and b up
else
b = b + distance[b, last] / 2 # move b up
end
else
b = a + distance[a, b] / 2 # move b down
end
end
accum
end
tests = [
[],
[0],
[0,1],
[0,1,2],
[0,1,2,3],
[0,3],
[0,3,5],
[0,3,5,7],
[0,1,3,4],
[0,1,3,4,6,7],
[0,1,3,4,6,7,9,10],
[0,1,3,4,6,7,10],
[0,3,4,6,7,9,10],
[0,1,3],
[0,3,4],
[0,-1],
[0,-1,-2],
[0,-1,-3,-4],
[0,-1,-3,-4,-6],
[0,-1,-3,-4,-6,-7],
[-1,-3,-4,-6,-7],
[-3,-4],
[-3,-4,-6, *(-42 .. -.to_a.reverse],
]
tests.each do |a|
printf "%-37.37s => %37.37s\n", a.inspect, sequences(a).inspect
a.reverse!
printf "%-37.37s => %37.37s\n", a.inspect, sequences(a).inspect
end
harp:~ > ruby a.rb
[] => []
[] => []
[0] => [0..0]
[0] => [0..0]
[0, 1] => [0..1]
[1, 0] => [1..0]
[0, 1, 2] => [0..2]
[2, 1, 0] => [2..0]
[0, 1, 2, 3] => [0..3]
[3, 2, 1, 0] => [3..0]
[0, 3] => [0..0, 3..3]
[3, 0] => [3..3, 0..0]
[0, 3, 5] => [0..0, 3..3, 5..5]
[5, 3, 0] => [5..5, 3..3, 0..0]
[0, 3, 5, 7] => [0..0, 3..3, 5..5, 7..7]
[7, 5, 3, 0] => [7..7, 5..5, 3..3, 0..0]
[0, 1, 3, 4] => [0..1, 3..4]
[4, 3, 1, 0] => [4..3, 1..0]
[0, 1, 3, 4, 6, 7] => [0..1, 3..4, 6..7]
[7, 6, 4, 3, 1, 0] => [7..6, 4..3, 1..0]
[0, 1, 3, 4, 6, 7, 9, 10] => [0..1, 3..4, 6..7, 9..10]
[10, 9, 7, 6, 4, 3, 1, 0] => [10..9, 7..6, 4..3, 1..0]
[0, 1, 3, 4, 6, 7, 10] => [0..1, 3..4, 6..7, 10..10]
[10, 7, 6, 4, 3, 1, 0] => [10..10, 7..6, 4..3, 1..0]
[0, 3, 4, 6, 7, 9, 10] => [0..0, 3..4, 6..7, 9..10]
[10, 9, 7, 6, 4, 3, 0] => [10..9, 7..6, 4..3, 0..0]
[0, 1, 3] => [0..1, 3..3]
[3, 1, 0] => [3..3, 1..0]
[0, 3, 4] => [0..0, 3..4]
[4, 3, 0] => [4..3, 0..0]
[0, -1] => [0..-1]
[-1, 0] => [-1..0]
[0, -1, -2] => [0..-2]
[-2, -1, 0] => [-2..0]
[0, -1, -3, -4] => [0..-1, -3..-4]
[-4, -3, -1, 0] => [-4..-3, -1..0]
[0, -1, -3, -4, -6] => [0..-1, -3..-4, -6..-6]
[-6, -4, -3, -1, 0] => [-6..-6, -4..-3, -1..0]
[0, -1, -3, -4, -6, -7] => [0..-1, -3..-4, -6..-7]
[-7, -6, -4, -3, -1, 0] => [-7..-6, -4..-3, -1..0]
[-1, -3, -4, -6, -7] => [-1..-1, -3..-4, -6..-7]
[-7, -6, -4, -3, -1] => [-7..-6, -4..-3, -1..-1]
[-3, -4] => [-3..-4]
[-4, -3] => [-4..-3]
[-3, -4, -6, -8, -9, -10, -11, -12, - => [-3..-4, -6..-6, -8..-42]
[-42, -41, -40, -39, -38, -37, -36, - => [-42..-8, -6..-6, -4..-3]
your ideas were essential to getting the peice of code i'm working on now
running in a reasonable amount of time - thanks to all the contributors!
-a
--
================================================== =============================
| email :: ara [dot] t [dot] howard [at] noaa [dot] gov
| phone :: 303.497.6469
| My religion is very simple. My religion is kindness.
| --Tenzin Gyatso
================================================== =============================
Daniel Berger
Guest
Posts: n/a
08-03-2005
Ara.T.Howard wrote:
<snip>
> your ideas were essential to getting the peice of code i'm working on now
> running in a reasonable amount of time - thanks to all the contributors!
>
> -a
No applause please. Just throw money (at RubyCentral).
Regards,
Dan
Ara.T.Howard
Guest
Posts: n/a
08-03-2005
On Thu, 4 Aug 2005, Daniel Berger wrote:
> Ara.T.Howard wrote:
>
> <snip>
>
>> your ideas were essential to getting the peice of code i'm working on now
>> running in a reasonable amount of time - thanks to all the contributors!
>>
>> -a
>
> No applause please. Just throw money (at RubyCentral).
how? i bought the new book...
-a
--
================================================== =============================
| email :: ara [dot] t [dot] howard [at] noaa [dot] gov
| phone :: 303.497.6469
| My religion is very simple. My religion is kindness.
| --Tenzin Gyatso
================================================== ============================= | 3,059 | 7,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2014-10 | latest | en | 0.712923 |
http://razhayesheitanparastan.com/r05780/7fq5qao0pimszgiu/ | 1,580,310,109,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00385.warc.gz | 134,449,620 | 9,593 | # Gene And Chromosome Mutation Worksheet Answer Key Genetic Techzulla Bio Worksheets Genetics Dna
Tuesday, September 3rd 2019. | Multiplication Worksheets
## Inherited Traits Worksheet
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You may pick the of rows and columns for your own arrays, in addition to description given to draw on the variety. You pick the array of columns and rows employed for arrays. The for each be separately varied to make unique sets of Multiplication difficulties. They separately varied to build different of multiplication worksheets. Proceed to the and see the way they rise by 10's. There really are a great deal of themed activities and math modalities available via a search.
If students finish early, then they need to finish Still another matter. Before they could find out about multiplication sentences, they must comprehend the of . Researching the subsequent days students are exposed to a of different types of narrative . Once the concept of multiplication through arrays, a very simple word problem used to anchor the . They time and assistance to discover for themselves the relationship of equal numbers of in each row a specific amount of rows. The students may create similar varieties of issues with and also pose the difficulties to one another. They know what they know about this issue, the things they to understand, then they learned.
## Main Idea Worksheets Middle School
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to the Football field and see the numbers rise by 10's. There are in fact main reason parents decide to their own children. You might also allow numbers don't meet any the divisibility to the difficulty for those . The numbers for each factor could possibly be separately varied to different of Multiplication difficulties. Shape number 35 be circle this of patterns develops in a standard way. There are a lot of to a memorize and practice math facts. You can the selection of decimals at the dividend to get those difficulties. | 440 | 2,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.904242 |
andersonbuohs.blogzet.com | 1,561,028,273,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999210.22/warc/CC-MAIN-20190620105329-20190620131329-00034.warc.gz | 11,736,102 | 6,022 | # The Secrets and techniques of Cash4 Method Uncovered
Cash4 can be a lottery game with 10,000 possible number combinations. Only one number combination is absorbed in every drawing therefore the likelihood of you winning within the Straight play is certainly one in \$ 10, 000. The odds is probably not as bad as is also to lottery games, though with a lot of possible combinations, punching the jackpot remains to be difficult. Even if you tend to participate in the Box option, how many possible combinations continues to be plentiful. There are 2500 combinations for 4-way, 1667 for 6-way, 833 for 12-way, and 417 for 24-way. As for the quads, you will find ten of these from 0000 to 9999.
Statistics And Probability
Choosing or picking out a number combination according to gut feeling alone is not a great idea. You need to have a superb background in statistics and probability to help keep winning in lottery games, including Cash4. If you merely choose numbers that you just believe are lucky, it could take you hundreds or 1000s of tries before you win a decent amount of cash. You might even never win by any means. You should at the very least make an effort to work with a statistical strategy that works so you're able to increase your chances inside the game.
Repeated Numbers
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The ?repeated numbers? method is quite a bit less simple because you can still find it. You need to allot significant amounts of time towards examining the effects of previous drawings. It might be just a little difficult should you be the person who not deal well with numbers. However, you'll want to keep in mind that this can be a get more info lot better than only making your bets without making use of worthwhile strategies. Not only do you waste your cash, moreover, you may squander your chances.
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Websites With Calculators
You might want to look for websites that could carry out the calculations in your case if you still find it too difficult to perform the tallying and calculations yourself. Using a virtual calculator would also reduce the probabilities of committing errors. There might be one problem though. It might be hard to discover a website or software that may perform the Cash4 calculations without cost. There might be websites that may attempt to get compensated within you. If you think the buying price of the calculator is too expensive, then you definately should just try it for yourself or have someone else practice it in your case.Article Source: how Art Robinson can guide you to win the Pick 4 Lottery on: WWW.GUARANTEED4NUMBERS.NET | 772 | 3,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-26 | latest | en | 0.960678 |
https://www.examrace.com/Bank-Clerical/Bank-Clerical-Practice-Questions/Aptitude-Questions/Aptitude-Logical-Reasoning-Simple-Interest-Part-1.html | 1,596,939,362,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738380.22/warc/CC-MAIN-20200809013812-20200809043812-00174.warc.gz | 656,422,187 | 9,377 | # Aptitude Logical Reasoning Simple Interest 2020 Bank Clerical Part 1
Get unlimited access to the best preparation resource for Bank-PO : fully solved questions with step-by-step explanation- practice your way to success.
1. Find the principle on a certain sum of money at per annum for years if the amount being Rs.1120?
A. Rs.1000
B. Rs.1115
C. Rs.1090
D. Rs.1300
2. What sum of money will produce Rs.70 as simple interest in 4 years at percent?
A. Rs.525
B. Rs.500
C. Rs.550
D. Rs.555
3. At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
A.
B.
C.
D.
4. What is the rate percent when the simple interest on Rs.800 amount to Rs.160 in 4 Years?
A.
B.
C.
D.
5. Find the simple interest on Rs.500 for 9 months at 6 paisa per month?
A. Rs.345
B. Rs.270
C. Rs.275
D. Rs.324
6. A certain sum amounts to Rs.1725 in 3 years and Rs.1875 in 5 years. Find the rate per annum?
A.
B.
C.
D.
7. At what rate percent on simple interest will a sum of money double itself in 30 years?
A.
B.
C.
D.
8. A certain sum of money at simple interest amounted Rs.840 in 10 years at per annum, find the sum?
A. Rs.500
B. Rs.512
C. Rs.520
D. Rs.646
9. In what time a sum of money double itself at per annum simple interest?
A. 29 years
B. years
C. years
D. years
10. The simple interest on a sum of money will be Rs.600 after 10 years. If the principal is trebled after 5 years what will be the total interest at the end of the tenth year?
A. Rs.700
B. Rs.800
C. Rs.1200
D. Rs.1900 | 477 | 1,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.866939 |
https://en.wikibooks.org/wiki/Chess_Opening_Theory/1._e4/1...e5/2._Qh5/2...Nc6/3._Bc4/3...g6/4._Qf3/4...Nf6 | 1,709,021,882,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00298.warc.gz | 226,213,082 | 16,325 | # Chess Opening Theory/1. e4/1...e5/2. Qh5/2...Nc6/3. Bc4/3...g6/4. Qf3/4...Nf6
< Chess Opening Theory | 1. e4 | 1...e5 | 2. Qh5 | 2...Nc6 | 3. Bc4 | 3...g6 | 4. Qf3
# Parham Attack
Parham Attack
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Position in Forsyth-Edwards Notation (FEN)
Moves: 1. e4 e5 2. Qh5 Nc6 3. Bc4 g6 4. Qf3 Nf6
### 4...Nf6
Black blocks the attack from the queen. Although the black knight is "pinned" to the pawn (If it were to move, white would play Qxf7#), Black has been handed the moves ...Bg7 and ...O-O on a platter. Another idea is if White doesn't defend d4, instead playing 5. Nc3?, (which is also a common response between low rating players) Black has 5...Nd4! 6. Qd1 Bg7.
## Theory table
For explanation of theory tables, see theory table and for notation, see algebraic notation.
4 5 6 7
Main line Qf3
Nf6
Ne2
Bg7
Nbc3
d6
d3
O-O
=/+
Variation Nc3 ...
...
Nc3? Qd1
Bg7
-/+ | 408 | 952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.837065 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1224/1/bu/b/ | 1,603,412,034,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00148.warc.gz | 788,070,158 | 47,278 | # Properties
Label 1224.1.bu.b Level $1224$ Weight $1$ Character orbit 1224.bu Analytic conductor $0.611$ Analytic rank $0$ Dimension $4$ Projective image $D_{8}$ RM discriminant 8 Inner twists $4$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1224 = 2^{3} \cdot 3^{2} \cdot 17$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 1224.bu (of order $$8$$, degree $$4$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.610855575463$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{8})$$ Defining polynomial: $$x^{4} + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image $$D_{8}$$ Projective field Galois closure of 8.0.153158089019904.1
## $q$-expansion
The $$q$$-expansion and trace form are shown below.
$$f(q)$$ $$=$$ $$q + \zeta_{8} q^{2} + \zeta_{8}^{2} q^{4} + ( 1 - \zeta_{8}^{3} ) q^{7} + \zeta_{8}^{3} q^{8} +O(q^{10})$$ $$q + \zeta_{8} q^{2} + \zeta_{8}^{2} q^{4} + ( 1 - \zeta_{8}^{3} ) q^{7} + \zeta_{8}^{3} q^{8} + ( 1 + \zeta_{8} ) q^{14} - q^{16} -\zeta_{8}^{2} q^{17} + ( -\zeta_{8} + \zeta_{8}^{2} ) q^{23} -\zeta_{8} q^{25} + ( \zeta_{8} + \zeta_{8}^{2} ) q^{28} + ( \zeta_{8}^{2} + \zeta_{8}^{3} ) q^{31} -\zeta_{8} q^{32} -\zeta_{8}^{3} q^{34} + ( \zeta_{8} + \zeta_{8}^{2} ) q^{41} + ( -\zeta_{8}^{2} + \zeta_{8}^{3} ) q^{46} + ( -\zeta_{8} + \zeta_{8}^{3} ) q^{47} + ( 1 - \zeta_{8}^{2} - \zeta_{8}^{3} ) q^{49} -\zeta_{8}^{2} q^{50} + ( \zeta_{8}^{2} + \zeta_{8}^{3} ) q^{56} + ( -1 + \zeta_{8}^{3} ) q^{62} -\zeta_{8}^{2} q^{64} + q^{68} + ( -1 - \zeta_{8} ) q^{71} + ( -\zeta_{8}^{2} + \zeta_{8}^{3} ) q^{73} + ( -1 - \zeta_{8}^{3} ) q^{79} + ( \zeta_{8}^{2} + \zeta_{8}^{3} ) q^{82} + ( \zeta_{8} - \zeta_{8}^{3} ) q^{89} + ( -1 - \zeta_{8}^{3} ) q^{92} + ( -1 - \zeta_{8}^{2} ) q^{94} + ( -\zeta_{8}^{2} + \zeta_{8}^{3} ) q^{97} + ( 1 + \zeta_{8} - \zeta_{8}^{3} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 4q^{7} + O(q^{10})$$ $$4q + 4q^{7} + 4q^{14} - 4q^{16} + 4q^{49} - 4q^{62} + 4q^{68} - 4q^{71} - 4q^{79} - 4q^{92} - 4q^{94} + 4q^{98} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1224\mathbb{Z}\right)^\times$$.
$$n$$ $$137$$ $$613$$ $$649$$ $$919$$ $$\chi(n)$$ $$-1$$ $$-1$$ $$-\zeta_{8}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
53.1
−0.707107 + 0.707107i −0.707107 − 0.707107i 0.707107 + 0.707107i 0.707107 − 0.707107i
−0.707107 + 0.707107i 0 1.00000i 0 0 0.292893 0.707107i 0.707107 + 0.707107i 0 0
485.1 −0.707107 0.707107i 0 1.00000i 0 0 0.292893 + 0.707107i 0.707107 0.707107i 0 0
773.1 0.707107 + 0.707107i 0 1.00000i 0 0 1.70711 0.707107i −0.707107 + 0.707107i 0 0
1205.1 0.707107 0.707107i 0 1.00000i 0 0 1.70711 + 0.707107i −0.707107 0.707107i 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
8.b even 2 1 RM by $$\Q(\sqrt{2})$$
51.g odd 8 1 inner
408.be odd 8 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1224.1.bu.b yes 4
3.b odd 2 1 1224.1.bu.a 4
8.b even 2 1 RM 1224.1.bu.b yes 4
17.d even 8 1 1224.1.bu.a 4
24.h odd 2 1 1224.1.bu.a 4
51.g odd 8 1 inner 1224.1.bu.b yes 4
136.o even 8 1 1224.1.bu.a 4
408.be odd 8 1 inner 1224.1.bu.b yes 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
1224.1.bu.a 4 3.b odd 2 1
1224.1.bu.a 4 17.d even 8 1
1224.1.bu.a 4 24.h odd 2 1
1224.1.bu.a 4 136.o even 8 1
1224.1.bu.b yes 4 1.a even 1 1 trivial
1224.1.bu.b yes 4 8.b even 2 1 RM
1224.1.bu.b yes 4 51.g odd 8 1 inner
1224.1.bu.b yes 4 408.be odd 8 1 inner
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{23}^{4} + 2 T_{23}^{2} + 4 T_{23} + 2$$ acting on $$S_{1}^{\mathrm{new}}(1224, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 + T^{4}$$
$3$ 1
$5$ $$1 + T^{8}$$
$7$ $$( 1 - T )^{4}( 1 + T^{4} )$$
$11$ $$1 + T^{8}$$
$13$ $$( 1 + T^{2} )^{4}$$
$17$ $$( 1 + T^{2} )^{2}$$
$19$ $$( 1 + T^{4} )^{2}$$
$23$ $$( 1 + T^{2} )^{2}( 1 + T^{4} )$$
$29$ $$1 + T^{8}$$
$31$ $$( 1 + T^{2} )^{2}( 1 + T^{4} )$$
$37$ $$1 + T^{8}$$
$41$ $$( 1 + T^{2} )^{2}( 1 + T^{4} )$$
$43$ $$( 1 + T^{4} )^{2}$$
$47$ $$( 1 + T^{4} )^{2}$$
$53$ $$( 1 + T^{4} )^{2}$$
$59$ $$( 1 + T^{4} )^{2}$$
$61$ $$1 + T^{8}$$
$67$ $$( 1 - T )^{4}( 1 + T )^{4}$$
$71$ $$( 1 + T )^{4}( 1 + T^{4} )$$
$73$ $$( 1 + T^{2} )^{2}( 1 + T^{4} )$$
$79$ $$( 1 + T )^{4}( 1 + T^{4} )$$
$83$ $$( 1 + T^{4} )^{2}$$
$89$ $$( 1 + T^{4} )^{2}$$
$97$ $$( 1 + T^{2} )^{2}( 1 + T^{4} )$$ | 2,515 | 4,968 | {"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} | 2.546875 | 3 | CC-MAIN-2020-45 | latest | en | 0.213002 |
http://stackoverflow.com/questions/9407784/can-an-arraylist-be-a-two-dimensional-array | 1,443,993,780,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736676092.10/warc/CC-MAIN-20151001215756-00146-ip-10-137-6-227.ec2.internal.warc.gz | 293,329,724 | 18,191 | # Can an ArrayList be a two-dimensional array?
I'm brand new to Java, programming, and StackOverflow. I need to use a list where I can add or remove things and don't know the initial size (like ArrayList) but I also need it to be two dimensional. I've read on Google and StackOverflow and I can't find a concrete answer. Is this a possibility? And if not can you point me in the right direction? Thanks in advance.
-
ArrayList of ArrayLists? – Nishant Feb 23 '12 at 5:41
``````ArrayList<ArrayList> arrList2D = new ArrayList<ArrayList>(2);
``````
`arrList2D` is a 2D ArrayList.
-
Yay thanks!! I've been messing around with my code for a few hours so I figured I'd just ask. I'm glad I did! High Five yo! – Punkrockie Feb 23 '12 at 5:49
Note that the `2` in the argument to the constructor has nothing to do with it being 2d. That is just the initial capacity (can grow later) of the first dimension. – Thilo Feb 23 '12 at 5:51
Thanks this place is awesome! – Punkrockie Feb 23 '12 at 5:53
well you can always try writing a simple code and see if this work. By the way you can use Array list in array list and I am pretty sure it will be a very bad idea.
-
heh, why is it a bad idea? – Gleno Feb 23 '12 at 5:49
what is so very bad about an `ArrayList<ArrayList<?>>` ? – Thilo Feb 23 '12 at 5:50
@Thilo there is no guarantee that the member lists will be the same size, so you'll constantly have to be careful about indexing. It's not bad per se, but it gets tricky. – jackrabbit Feb 23 '12 at 6:32
That is the same with a two dimensional array, and it has advantages, too. – Thilo Feb 23 '12 at 7:17
@Thilo the reason jackrabbit described is the one I was thinking . secondly the way arraylist grows is by copying the existing arrays into the next memory location. Having the double dimension array list , I think, will lead to big overhead when the number of rows are growing. – Geek Feb 23 '12 at 14:54 | 529 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | latest | en | 0.923465 |
https://education.blurtit.com/1133730/how-do-i-find-a-surface-area-of-a-room | 1,620,465,985,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988858.72/warc/CC-MAIN-20210508091446-20210508121446-00617.warc.gz | 238,767,691 | 12,328 | How Do I Find A Surface Area Of A Room?
5 Answers
Oddman answered
Multiply the length of the perimeter of the room by the wall height to get the wall area. Multiply length times width for the area of a rectangular shape such as a ceiling or floor. Usually, you are interested in square feet, so the measurements need to be expressed in terms of feet.
Example
Suppose you have a room that is 10' by 12' with a ceiling that is 7' 10" above the floor. The perimeter of the room is 10' + 12' + 10' + 12' = 2(10' + 12') = 2*22' = 44'. The height in feet is 7 10/12 ft = 7 5/6 ft. (An inch is 1/12 ft.) The total wall area is
(44')(7 5/6') = (44*7 + 44*5/6) ft2 = (308 + 36 2/3) ft2 = 344 2/3 ft2
The ceiling or floor area of the rectangular room is the product of length and width,
(10')(12') = 120 ft2 The total surface area of the room is the sum of the areas of ceiling, walls, and floors:
120 ft2 + 344 2/3 ft2 + 120 ft2 = 584 2/3 ft2.
Of course, if you are figuring for paint, you just add up the areas that will be painted.
thanked the writer.
Kerry answered
www.ehow.com Here's how to find the surface area of 3D objects.
thanked the writer.
Oddman answered
Formulas for the area of many polygon, circular, and elliptical shapes have been developed. To find the area of such a shape, you determine the appropriate dimensions of the shape, match them with the variables of the formula, and evaluate.
For many regular solids, formulas for surface area have also been developed. For solids having faces in the shape of a polygon, it is the sum of the areas of each of the polygon faces. A regular tetrahedron, for example, has 4 congruent triangular faces--so the area of the surface is the 4 times the area of one of the faces.
For solids with curved surfaces, formulas have been developed for a few: Cones and sections of cones, spheres and sections of spheres, cylinders. Certain elliptical shapes also have surface area formulas that have been worked out.
If you can write an equation for the curve that generates the surface, calculus may help you determine the equation for the surface area.
What constitutes a suitable method of measurement of surface area depends on what you are measuring and how accurate you need to be. Measuring the area of a snowflake requires different methods than measuring the area of a battleship. Finding the surface area of the leaves of a tree is a different proposition from finding the area of your living room wall (unless you live in a tree house).
thanked the writer.
Anonymous answered
I just want to know how to find surface area
thanked the writer.
Anonymous answered
How do I find surface area for a room that is 8feet 0 inches long and 7feet 5inches wide
thanked the writer. | 701 | 2,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-21 | longest | en | 0.938986 |
https://www.proprofs.com/quiz-school/story.php?title=pump-ninja | 1,709,019,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00841.warc.gz | 975,393,478 | 95,777 | # Are You A Pump Ninja?
Approved & Edited by ProProfs Editorial Team
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| By Amessamore
A
Amessamore
Community Contributor
Quizzes Created: 1 | Total Attempts: 43
Questions: 5 | Attempts: 43
Settings
.
• 1.
### Given a specific fluid to be pumped, at what viscosity range should you begin to apply viscous corrections to the hydraulic performance for a centrifugal pump?
• A.
5 to 15 centipoise
• B.
Above 300 centipoise
• C.
Above 600 centipoise
• D.
Centrifugal pumps do not require viscosity corrections.
A. 5 to 15 centipoise
Explanation
Viscous corrections should be applied to the hydraulic performance of a centrifugal pump when the viscosity of the fluid being pumped falls within the range of 5 to 15 centipoise. This means that if the fluid's viscosity is below 5 centipoise or above 15 centipoise, there is no need to make any adjustments to the pump's performance based on viscosity.
Rate this question:
• 2.
### For a given size piping system, adding a second identical pump in parallel will …
• A.
Increase the flow rate to a point determined by the system curve.
• B.
Double the flow.
• C.
• D.
None of these answers is correct.
A. Increase the flow rate to a point determined by the system curve.
Explanation
Adding a second identical pump in parallel to a piping system will increase the flow rate. However, the increase in flow rate will be limited by the system curve, which represents the relationship between flow rate and pressure drop in the system. Therefore, the flow rate will increase up to a point determined by the system curve, but it will not necessarily double.
Rate this question:
• 3.
### Water will boil at what temperature (F)?
• A.
It depends on the pressure.
• B.
212
• C.
100
• D.
33
• E.
600
A. It depends on the pressure.
Explanation
The boiling point of water is not a fixed temperature but rather depends on the pressure applied. At sea level, water boils at 212 degrees Fahrenheit, but at higher altitudes where the atmospheric pressure is lower, water boils at a lower temperature. Similarly, at higher pressures, such as in a pressure cooker, water can reach higher temperatures before boiling. Therefore, the correct answer is that the boiling temperature of water depends on the pressure.
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• 4.
### When starting up a centrifugal pump, the suction valve should always be ...
• A.
Fully open.
• B.
Partially open, then opened the rest of the way a few minutes afterward.
• C.
Fully closed.
• D.
It doesn't matter, because it is a centrifugal pump.
A. Fully open.
Explanation
When starting up a centrifugal pump, the suction valve should always be fully open. This is because a centrifugal pump relies on the flow of liquid through the impeller to generate pressure and create the necessary suction. By fully opening the suction valve, it allows for maximum flow of liquid into the pump, preventing any potential cavitation or damage to the pump. Partially opening the valve or keeping it fully closed can restrict the flow and cause issues with the pump's performance. Therefore, it is important to always start the centrifugal pump with the suction valve fully open.
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• 5.
### Which fluid, when subjected to cavitation conditions, will normally create the most damage in a pump (assuming all other conditions are the same)?
• A.
Water at 68 F
• B.
Gasoline (87 octane) at 75 F
• C.
Ethylene glycol mixed with water at 50 percent (180 F)
• D.
Ethanol at 75 F
• E.
Naval jet fuel JP5 at 68 F
• F.
Water at 300 F
A. Water at 68 F
Explanation
When subjected to cavitation conditions, the fluid that will normally create the most damage in a pump is water at 68 F. Cavitation occurs when the pressure of a fluid drops below its vapor pressure, causing the formation and collapse of vapor bubbles. These collapsing bubbles can create intense shockwaves that can erode the pump's impeller and other components, leading to damage. Water is particularly prone to cavitation due to its low vapor pressure, and at 68 F, it is in a temperature range where cavitation is more likely to occur.
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Related Topics | 1,036 | 4,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.896515 |
http://gigasquidsoftware.com/blog/2016/02/10/fairy-tale-word-vectors/ | 1,726,277,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00093.warc.gz | 12,297,163 | 10,303 | This post continues our exploration from the last blog post Why Hyperdimensional Socks Never Match. We are still working our way through Kanerva’s paper. This time, with the basics of hypervectors under our belts, we’re ready to explore how words can be expressed as context vectors. Once in a high dimensional form, you can compare two words to see how similar they are and even perform reasoning.
To kick off our word vector adventure, we need some words. Preferring whimsy over the Google news, our text will be taken from ten freely available fairy tale books on http://www.gutenberg.org/.
### Gather ye nouns
Our goal is to assemble a frequency matrix, with all the different nouns as the rows and the columns will be the counts of if the word appears or not in the document. Our matrix will be binary with just 1s and 0s. The document will be a sentence or fragment of words. A small visualization is below.
Noun Doc1 Doc2
flower 1 0
king 0 1
fairy 1 0
gold 1 1
The size of the matrix will be big enough to support hypervector behavior, but not so big as to make computation too annoyingly slow. It will be nouns x 10,000.
The first task is to get a set of nouns to fill out the rows. Although, there are numerous online sources for linguistic nouns, they unfortunately do not cover the same language spectrum as old fairy tale books. So we are going to collect our own. Using Stanford CoreNLP, we can collect a set of nouns using Grimm’s Book as a guide. There are about 2500 nouns there to give us a nice sample to play with. This makes our total matrix size ~ 2500 x 10,000.
Now that we have our nouns, let’s get down to business. We want to create an index to row to make a `noun-idx` and then create a sparse matrix for our word frequency matrix.
The next thing we need to do is to have some functions to take a book, read it in, split it into documents and then update the frequency matrix.
### Random indexing for the win
The interesting thing about the update method is that we can use random indexing. We don’t need to worry about having a column for each document. Because of the nature of hyperdimensions, we can randomly assign 10 columns for each document.
The whole book is processed by slurping in the contents and using a regex to split it up into docs to update the matrix.
We can now run the whole processing with:
On my system, it only takes about 3 seconds.
Great! Now we have hypervectors associated with word frequencies. They are now context word vectors. What can we do with them.
### How close is a king to a goat?
One of the things that we can do with them is find out a measure of how closely related the context of two words are by a measure of their cosine similarity. First, we need a handy function to turn a string word into a word vector by getting it out of our frequency matrix.
Then we can make another nice function to compare two words and give a informational map back.
Let’s take a look at the similarities of some words to king.
As expected, the royal family is closer to the king then a guard or goat is.
One of the interesting things is that now we can do addition and subtraction with these word vectors and see how it affects the relation with other words.
### Boy + Gold = King, Boy + Giant = Jack
We can take a look at how close boy and king are together by themselves.
Now we can add some gold to the boy and that new word vector will be closer to king than boy was alone.
Doing the same for boy and jack, we find that adding a giant moves the context closer.
Amusingly, a frog and a princess make a prince.
We can take this even farther by subtracting words and adding others. For example a similarity to the word queen can be obtained by subtracting man from king and adding woman.
Similarly, a contextual closeness to father can be gotten from subtracting woman from mother and adding man.
But wait, that’s not all. We can also do express facts with these word vectors and reason about them.
### Reasoning with word vector with the database as a hyperdimensional value
The curious nature of hypervectors allows the storage of multiple entity, attributes in it and allow the retrieval of the likeness of them later by simple linear math - using only xor multiplication and addition. This gives us the database as a value in the form of a high dimensional vector.
For an example, say we want to express the fact that Hansel is a brother of Gretel. We can do this by adding the xor product of brother with hansel and the product of brother with Gretel.
Also we can express that Jack is a brother of Hansel.
We can add these two facts together to make a new hypervector value.
Now we can actually reason about them and ask questions. Is Jack a brother of Hansel? With a high cosine similarity, we can assume the answer is likely.
What about someone unrelated. Is Cinderella the brother of Gretel? - No
Is Jack the brother of Gretel - Yes
We can take this further by adding more facts and inventing a relation of our own.
### Siblings in Hyperspace
Let’s invent a new word vector that is not in our nouns - siblings. We are going to create new random hypervector to represent it.
We will define it in terms of word vectors that we already have. That is, siblings will be a the sum of brother + sister. We XOR multiply it by siblings to associate it with the hypervector.
Now we can add some more facts. Gretel is a sister of Hansel.
Gretel is also a sister of Jack.
Collecting all of our facts into one hypervector (as a database).
Now we can ask some for questions.
Are Hansel and Gretel siblings? - Yes
Are John and Roland siblings - No
Are Jack and Hansel siblings? - Yes
It is interesting to think of that nothing is stopping us at this point from retracting facts by simply subtracting the fact encoded word vectors from our “database” value and making a new value from it.
### Conclusions
In this fun, but casual exploration of word vector we have seen the potential for reasoning about language in a way that uses nothing more complicated than addition and multiplication. The ability to store dense information in hypervectors, extract it with simple methods, and flexibly collect it randomly, shows its versatility and power. Hyperdimensional vectors might hold the key to unlocking a deeper understanding of cognitive computing or perhaps even true artificial intelligence.
It is interesting to note that this technique is not limited to words. Other applications can be done the same way. For example a video recommendation using a hypervector with movie titles. Or perhaps even anomaly detection using sensor readings over a regular weekly time period.
Looking over our journey with word vectors. At the beginning it seemed that word vectors were magical. Now, after an understanding of the basics, it still seems like magic.
If you are interested in exploring further, feel free to use my github hyperdimensional-playground as a starting point. | 1,512 | 6,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.92236 |
http://www.jiskha.com/members/profile/posts.cgi?name=chloe&page=18 | 1,369,070,960,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699138006/warc/CC-MAIN-20130516101218-00019-ip-10-60-113-184.ec2.internal.warc.gz | 544,483,213 | 3,196 | Monday
May 20, 2013
# Posts by chloe
Total # Posts: 301
Thank you!
You are writing to a reader who favors a different solution than the one you recommend. Which of the following is MOST likely to persuade the reader? (Points: 5) Describe your reader's solution first, explain why it won't work, and then go on to describe your own solu...
I am sorry this is my first time on here and do not know but i think it is C but is not sure
You have four reasons for saying "no." Two of them are very strong reasons, one has a tiny loophole and one is very weak. In your message you should: (Points: 5) include all four reasons. omit the very weak reason and use the other three. use only the two strong rea...
Math
I am having trouble figuring out how to solve these logarithms. Could someone please help! log2(log4x)=1 and solve for x and y: (1/2)^x+y= 16 logx-y8=-3
Math
I am having trouble figuring out how to solve these logarithms. Could someone please help! log2(log4x)=1 and solve for x and y: (1/2)^x+y= 16 logx-y8=-3
I am having trouble figuring out how to solve this logarithms. Could someone please help! log2(log4x)=1 and solve for x and y: (1/2)^x+y= 16 logx-y8=-3
Math
A woman wants to measure the height of a nearby tower. She places a pole in the shadow of the tower so that the shadow of the pole is exactly covered by the shadow of the tower. The distance between the pole and the tower is , and the pole casts a shadow that is long. How tall... | 391 | 1,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2013-20 | longest | en | 0.955185 |
https://ml-notes.akkefa.com/en/latest/algorithms/greedy_algorithms.html | 1,713,231,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00077.warc.gz | 365,195,624 | 7,387 | # Greedy Algorithms
## Divide and Conquer
Divide and conquer is a general algorithm design paradigm: divide the problem into smaller subproblems, solve the subproblems recursively, and then combine the solutions to the subproblems to solve the original problem.
### Largest pair sum in an unsorted array
Given an unsorted of distinct integers, find the largest pair sum in it. For example, the largest pair sum in {12, 34, 10, 6, 40} is 74.
Brute force solution
numbers = [2,1,0,8,15,7,-1,6]
print(numbers)
max_pair_sum = 0
for i in numbers:
for j in numbers:
if i != j:
max_pair_sum = max(max_pair_sum, i + j)
print(max_pair_sum)
# time Complexity = n^2
# space Complexity = 1
[2, 1, 0, 8, 15, 7, -1, 6]
23
Best solution
1. Initialize the first = Integer.MIN_VALUE second = Integer.MIN_VALUE
2. Loop through the elements a) If the current element is greater than the first max element, then update second max to the first max and update the first max to the current element.
3. Return (first + second)
def findLargestSumPair(arr, n):
# Initialize first and second
# largest element
if arr[0] > arr[1]:
first_big_number = arr[0]
second_big_number = arr[1]
else:
first_big_number = arr[1]
second_big_number = arr[0]
# Traverse remaining array and
# find first and second largest
# elements in overall array
for i in range(2, n):
# If current element is greater
# than first then update both
# first and second
if arr[i] > first_big_number:
second_big_number = first_big_number
first_big_number = arr[i]
# If arr[i] is in between first
# and second then update second
elif arr[i] > second_big_number and arr[i] != first_big_number:
second_big_number = arr[i]
return (first_big_number , second_big_number)
first, second = findLargestSumPair(numbers, len(numbers))
print(first, second)
print(first + second)
# time Complexity = n
# space Complexity = 1
15 8
23
### Max subarray problem
Given an array of integers, find the contiguous subarray that has the largest sum. Return the sum.
Brute force solution
Draw graph which show as stock values.
import numpy as np
import matplotlib.pyplot as plt
numbers = [2,1,0,8,15,7,-1,6]
print(numbers)
x = range(len(numbers))
y = numbers
plt.title("Line graph")
plt.plot(x, y, color="red")
plt.show()
Matplotlib is building the font cache; this may take a moment.
[2, 1, 0, 8, 15, 7, -1, 6]
max_contiguous_sum = 0
for index,v in enumerate(numbers):
for j in range(index ,len(numbers)):
# print(index, j)
# print(numbers[index:j+1])
max_contiguous_sum = max(max_contiguous_sum, sum(numbers[index:j+1]))
print(max_contiguous_sum)
38
Divide and Conquer solution
import sys
def maxSubArraySum(arr):
# Base case: when there is only one element in the array
if len(arr) == 1:
return arr[0]
# Recursive case: divide the problem into smaller sub-problems
m = len(arr) // 2
# Find the maximum subarray sum in the left half
left_max = maxSubArraySum(arr[:m])
# Find the maximum subarray sum in the right half
right_max = maxSubArraySum(arr[m:])
# Find the maximum subarray sum that crosses the middle element
left_sum = -sys.maxsize - 1
right_sum = -sys.maxsize - 1
sum = 0
# Traverse the array from the middle to the right
for i in range(m, len(arr)):
sum += arr[i]
right_sum = max(right_sum, sum)
sum = 0
# Traverse the array from the middle to the left
for i in range(m - 1, -1, -1):
sum += arr[i]
left_sum = max(left_sum, sum)
cross_max = left_sum + right_sum
# Return the maximum of the three subarray sums
return max(cross_max, max(left_max, right_max))
print(maxSubArraySum(numbers))
38
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
sns.set()
def max_subarray(arr):
if len(arr) == 1:
return arr[0]
else:
mid = len(arr) // 2
left = max_subarray(arr[:mid])
right = max_subarray(arr[mid:])
cross = max_crossing_subarray(arr, mid)
return max(left, right, cross)
def max_crossing_subarray(arr, mid):
left_sum = -np.inf
right_sum = -np.inf
sum = 0
for i in range(mid - 1, -1, -1):
sum += arr[i]
if sum > left_sum:
left_sum = sum
max_left = i
sum = 0
for j in range(mid, len(arr)):
sum += arr[j]
if sum > right_sum:
right_sum = sum
max_right = j
return left_sum + right_sum
arr = np.random.randint(-10, 10, 10)
arr = np.array(numbers)
max_subarray(arr)
38
def max_subarray(arr):
max_sum = -np.inf
for i in range(len(arr)):
sum = 0
for j in range(i, len(arr)):
sum += arr[j]
if sum > max_sum:
max_sum = sum
max_left = i
max_right = j
return max_sum
max_subarray(arr)
38 | 1,280 | 4,492 | {"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} | 3.890625 | 4 | CC-MAIN-2024-18 | latest | en | 0.765944 |
https://schooltutoring.com/blog/guessing-on-the-sat-is-it-worth-it/ | 1,534,333,161,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210058.26/warc/CC-MAIN-20180815102653-20180815122653-00326.warc.gz | 780,836,266 | 29,245 | • Sign-In
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# Guessing on the SAT, is it Worth it?
Get Help from SchoolTutoring Academy
One fact of the SAT examination that causes many students stress is the deduction of marks for incorrect answers. I hope to dispel all of the myths regarding this rule, and explain how to maximize your potential marks on the test.
To begin with, let’s discuss about the purpose of mark deduction for incorrect answers. The rule is that for multiple choice sections, each correct answer receives 1 point, each question left blank receives zero points, and each incorrect answer results in the deduction of 1/4 of a point. It’s important to remember that this does NOT apply to the long answer mathematics section, or the essay section. Let’s assume that a student guesses on a multiple choice question with five answers. Then, they have a 1/5 chance of guessing the correct answer and receiving a mark, and a 4/5 chance of guessing incorrectly and losing 1/4 of a mark. Using a bit of probability math, this results in an average of zero points over a number of questions. This prevents students from gaining points through random guessing.
At this point you may be wondering whether you should ever guess on the SAT. It turns out that you should, but only under certain, limited circumstances. One time when it makes sense to guess when you are almost certain the answer is one of two or three options. This is most likely to happen on the vocabulary section, where knowing one of the words in a pair eliminates two or three of the incorrect options. Let’s assume that you have eliminated three of the options. Then there is a 50% chance of guessing an incorrect answer, and a 50% chance of guessing the correct answer. This averages out to a gain of 3/8 marks per guess in this situation. This means it is definitely advantageous to make a guess. However, the majority of the time, it will be much harder to eliminate answers, such as in the mathematics section. This means that the only area where it really makes sense to guess is in the vocabulary section.
If you have no idea what the answer is, simply move on, and do not waste time trying to guess.
In summation, the only time when it makes sense to guess is when you can almost instantly eliminate 2 or 3 answers, usually in the vocabulary section. Otherwise, simply move on, and save your time to gain points on questions you are able to answer correctly.
This article was written for you by Tobias one of our tutors with TestPrep Academy. | 585 | 2,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-34 | latest | en | 0.92587 |
https://www.jiskha.com/questions/1227158/1-Which-solid-has-one-base-that-is-a-triangle-and-three-lateral-surfaces-that-are | 1,553,076,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202324.5/warc/CC-MAIN-20190320085116-20190320111116-00106.warc.gz | 800,613,292 | 5,515 | # pre-algebra NEED HELP!
1.Which solid has one base that is a triangle and three lateral surfaces that are triangles?
A:triangular pyramid***
B:triangular prism
C:rectangular prism
D:rectangular pyramid
2:Sara needs to cover the top and the sides of the cylinder below.
8in 12in
About how many squat inches of paper will Sara need?
A:352in²***
B:402in²
C:804in²
D:1,005in²
3.what is the surface area of a sphere with a radius of 4 meter rounded to the nearest square meter?
A:50m²
B:101m²
C:201m²
D:268m²***
4.What is the volume of asphere with a radius of 5 meters rounded to the nearest square meter?
A:560m³
B:314m³
C:340m³
D:524m³***
5.What is the volume of a sphere with a radius of 7 cm rounded to the nearest cubic centimeter?
A:4,310cm³
B:1,437cm³***
C:615cm³
D:359cm³
Correct me if I'm wrong
1. 👍 0
2. 👎 0
3. 👁 186
1. 👍 0
2. 👎 0
posted by GummyBears
2. #1, 4,5 are correct
#3
SA of a sphere = 4π r^2
so if r = 4
SA = 4π(16) = appr 201
#2, can't see the diagram, will assume r = 8, height = 12 , cover the top but not the bottom
SA = πr^2 + 2πrh
= π(64) + 2π(8)(12) = appr 804
1. 👍 0
2. 👎 0
posted by Reiny
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More Similar Questions | 1,153 | 4,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-13 | latest | en | 0.891707 |
https://csstudy.in/class-12-physics-notes-pdf-free-download/ | 1,680,406,231,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00581.warc.gz | 216,101,921 | 33,680 | Class 12 Physics Notes PDF Chapter 1 to 14 for free download. Chapter wise Physics Class Notes PDF of NCERT book. The best revision notes are very useful for get better marks in Board exams and qualify the JEE and NEET entrance exam.
### Chapter 1 : ELECTRIC CHARGES AND FIELDS
Topics:
1.1 Introduction
1.2 Electric Charge
1.3 Conductors and Insulators
1.4 Charging by Induction
1.5 Basic Properties of Electric Charge
1.6 Coulomb’s Law
1.7 Forces between Multiple Charges
1.8 Electric Field
1.9 Electric Field Lines
1.10 Electric Flux
1.11 Electric Dipole
1.12 Dipole in a Uniform External Field
1.13 Continuous Charge Distribution
1.14 Gauss’s Law
1.15 Applications of Gauss’s Law
Click Below for Chapter 1 : ELECTRIC CHARGES AND FIELDS Notes PDF
### Chapter 2 : ELECTROSTATIC POTENTIAL AND CAPACITANCE
Topics:
2.1 Introduction
2.2 Electrostatic Potential
2.3 Potential due to a Point Charge
2.4 Potential due to an Electric Dipole
2.5 Potential due to a System of Charges
2.6 Equipotential Surfaces
2.7 Potential Energy of a System of Charges
2.8 Potential Energy in an External Field
2.9 Electrostatics of Conductors
2.10 Dielectrics and Polarisation
2.11 Capacitors and Capacitance
2.12 The Parallel Plate Capacitor
2.13 Effect of Dielectric on Capacitance
2.14 Combination of Capacitors
2.15 Energy Stored in a Capacitor
Click Below for Chapter 2 : ELECTROSTATIC POTENTIAL AND CAPACITANCE Notes PDF
### Chapter 3 : CURRENT ELECTRICITY
Topics:
3.1 Introduction
3.2 Electric Current
3.3 Electric Currents in Conductors
3.4 Ohm’s law
3.5 Drift of Electrons and the Origin of Resistivity
3.6 Limitations of Ohm’s Law
3.7 Resistivity of Various Materials
3.8 Temperature Dependence of Resistivity
3.9 Electrical Energy, Power
3.10 Combination of Resistors — Series and Parallel
3.11 Cells, emf, Internal Resistance
3.12 Cells in Series and in Parallel
3.13 Kirchhoff’s Rules
3.14 Wheatstone Bridge
3.15 Meter Bridge
3.16 Potentiometer
Click Below for Chapter 3 : CURRENT ELECTRICITY Notes PDF
### Chapter 4 : MOVING CHARGES AND MAGNETISM
Topics:
4.1 Introduction
4.2 Magnetic Force
4.3 Motion in a Magnetic Field
4.4 Motion in Combined Electric and Magnetic Fields
4.5 Magnetic Field due to a Current Element, Biot-Savart Law
4.6 Magnetic Field on the Axis of a Circular Current Loop
4.7 Ampere’s Circuital Law
4.8 The Solenoid and the Toroid
4.9 Force between Two Parallel Currents, the Ampere
4.10 Torque on Current Loop, Magnetic Dipole
4.11 The Moving Coil Galvanometer
Click Below for Chapter 4 : MOVING CHARGES AND MAGNETISM Notes PDF
### Chapter 5 :MAGNETISM AND MATTER
Topics:
5.1 Introduction
5.2 The Bar Magnet
5.3 Magnetism and Gauss’s Law
5.4 The Earth’s Magnetism
5.5 Magnetisation and Magnetic Intensity
5.6 Magnetic Properties of Materials
5.7 Permanent Magnets and Electromagnets
Click Below for Chapter 5 :MAGNETISM AND MATTER Notes PDF
### Chapter 6 : ELECTROMAGNETIC INDUCTION
Topics:
6.1 Introduction
6.2 The Experiments of Faraday and Henry
6.3 Magnetic Flux
6.5 Lenz’s Law and Conservation of Energy
6.6 Motional Electromotive Force
6.7 Energy Consideration: A Quantitative Study
6.8 Eddy Currents
6.9 Inductance
6.10 AC Generator
Click Below for Chapter 6 : ELECTROMAGNETIC INDUCTION Notes PDF
### Chapter 7 : ALTERNATING CURRENT
Topics:
7.1 Introduction
7.2 AC Voltage Applied to a Resistor
7.3 Representation of AC Current and Voltage by Rotating Vectors — Phasors
7.4 AC Voltage Applied to an Inductor
7.5 AC Voltage Applied to a Capacitor
7.6 AC Voltage Applied to a Series LCR Circuit
7.7 Power in AC Circuit: The Power Factor
7.8 LC Oscillations
7.9 Transformers
Click Below for Chapter 7 : ALTERNATING CURRENT Notes PDF
Chapter 8 : ELECTROMAGNETIC WAVES
Topics:
8.1 Introduction
8.2 Displacement Current
8.3 Electromagnetic Waves
8.4 Electromagnetic Spectrum
Click Below for Chapter 8 : ELECTROMAGNETIC WAVES Notes PDF
### CHAPTER NINE : RAY OPTICS AND OPTICAL INSTRUMENTS
Topics:
9.1 Introduction
9.2 Reflection of Light by Spherical Mirrors
9.3 Refraction
9.4 Total Internal Reflection
9.5 Refraction at Spherical Surfaces and by Lenses
9.6 Refraction through a Prism
9.7 Some Natural Phenomena due to Sunlight
9.8 Optical Instruments
Click Below for CHAPTER NINE : RAY OPTICS AND OPTICAL INSTRUMENTS Notes PDF
### Chapter 10: WAVE OPTICS
Topics:
10.1 Introduction
10.2 Huygens Principle
10.3 Refraction and Reflection of Plane Waves using Huygens Principle
10.4 Coherent and Incoherent Addition of Waves
10.5 Interference of Light Waves and Young’s Experiment
10.6 Diffraction
10.7 Polarisation
Click Below for Chapter 10: WAVE OPTICS Notes PDF
### Chapter 11:DUAL NATURE OF RADIATION AND MATTER
Topics:
11.1 Introduction
11.2 Electron Emission
11.3 Photoelectric Effect
11.4 Experimental Study of Photoelectric Effect
11.5 Photoelectric Effect and Wave Theory of Light
11.6 Einstein’s Photoelectric Equation: Energy Quantum of Radiation
11.7 Particle Nature of Light: The Photon
11.8 Wave Nature of Matter
11.9 Davisson and Germer Experiment
Click Below for Chapter 11:DUAL NATURE OF RADIATION AND MATTER Notes PDF
### Chapter 12: ATOMS
Topic:
12.1 Introduction
12.2 Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom 415
12.3 Atomic Spectra
12.4 Bohr Model of the Hydrogen Atom
12.5 The Line Spectra of the Hydrogen Atom
12.6 DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation
Click Below for Chapter 12: ATOMS Notes PDF
### Chapter 13: NUCLEI
Topics:
13.1 Introduction
13.2 Atomic Masses and Composition of Nucleus
13.3 Size of the Nucleus
13.4 Mass-Energy and Nuclear Binding Energy
13.5 Nuclear Force
13.7 Nuclear Energy
Click Below for Chapter 13: NUCLEI Notes PDF
### Chapter 14 : SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS
Topics:
14.1 Introduction
14.2 Classification of Metals, Conductors and Semiconductors
14.3 Intrinsic Semiconductor
14.4 Extrinsic Semiconductor
14.5 p-n Junction
14.6 Semiconductor Diode
14.7 Application of Junction Diode as a Rectifier
14.8 Special Purpose p-n Junction Diodes
14.9 Digital Electronics and Logic Gates
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Our ncert solutions for Solutions 9. Force and Laws of Motion - Additional -Questions 3 | Class 9 Science - Toppers Study is the best material for English Medium students cbse board and other state boards students.
Solutions ⇒ Class 9th ⇒ Science ⇒ 9. Force and Laws of Motion
# Solutions 9. Force and Laws of Motion - Additional -Questions 3 | Class 9 Science - Toppers Study
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## Solutions 9. Force and Laws of Motion - Additional -Questions 3 | Class 9 Science - Toppers Study
CBSE board students who preparing for class 9 ncert solutions maths and Science solved exercise chapter 9. Force and Laws of Motion available and this helps in upcoming exams 2023-2024.
### You can Find Science solution Class 9 Chapter 9. Force and Laws of Motion
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## Solutions 9. Force and Laws of Motion - Additional -Questions 3 | Class 9 Science - Toppers Study
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##### Solutions ⇒ Class 9th ⇒ Science
1. Matter in Our Surroundings
2. Is Matter around us Pure
3. Atoms and Molecules
4. Structure of The Atom
5. The Fundamental Unit of Life
6. Tissues
7. Diversity in Living Organisms
8. Motion
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10. Gravitation
11. Work and Energy
12. Sound
13. Why Do We Fall ill
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New Books | 720 | 2,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.82134 |
https://fr.mathworks.com/matlabcentral/answers/1914440-circular-time-plots-using-polarhistogram | 1,716,505,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00569.warc.gz | 235,512,580 | 28,040 | # Circular time plots using polarhistogram?
15 vues (au cours des 30 derniers jours)
Sadegh Rahimi le 17 Fév 2023
Commenté : William Rose le 1 Mar 2023
Hi there,
I don't know how to make circular plots in Matlab. They can provide insight into patterns of activity based on time of day and are frequently used in research articles (like the attached figure). Is it possible to use polarhistogram to draw it? Thanks for your comments and suggestions!
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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### Réponse acceptée
William Rose le 1 Mar 2023
Suppose you have the wind direction in degrees at 1 minute intervals for 1 day:
windDir=360*rand(1,1440);
Make a histogram of the wind:
polarhistogram(windDir)
Maybe you don;t like that histogram because the degrees do not look like a compass. Suppose also that you want the histogram to have 15 degree segments (pi/12) instead of 30 degrees. Make those fixes as follows::
figure; polarhistogram(windDir, 'BinWidth',pi/12)
Good luck!
##### 3 commentairesAfficher 1 commentaire plus ancienMasquer 1 commentaire plus ancien
Sadegh Rahimi le 1 Mar 2023
Truely nice, thank you very much
William Rose le 1 Mar 2023
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Translated by | 384 | 1,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.608019 |
https://www.easycalculation.com/multiplication-table-17.html | 1,596,506,834,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735851.15/warc/CC-MAIN-20200804014340-20200804044340-00033.warc.gz | 651,821,791 | 5,987 | English
# Multiplication Table of Seventeen
This 17 (Seventeen) times multiplication table will help you to learn the multiplication for 17 through 20 times. The value of 17 x 1 is 17 and 17 x 20 is 340. This table is more helpful for students and children to learn multiplication table of Seventeen.
## 17 Times Table
Multiplication Table : 17
0 x 17 0
1 x 17 17
2 x 17 34
3 x 17 51
4 x 17 68
5 x 17 85
6 x 17 102
7 x 17 119
8 x 17 136
9 x 17 153
10 x 17 170
11 x 17 187
12 x 17 204
13 x 17 221
14 x 17 238
15 x 17 255
16 x 17 272
17 x 17 289
18 x 17 306
19 x 17 323
20 x 17 340 | 234 | 583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.568585 |
http://wordyenglish.com/lojban/hrefgram2/c10-s11.html | 1,371,623,397,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708142617/warc/CC-MAIN-20130516124222-00021-ip-10-60-113-184.ec2.internal.warc.gz | 286,877,899 | 4,069 | VocabularyLiteratureEnglishing WritingAccents漢語
## Chapter 10: Imaginary Journeys: The Lojban Space/Time Tense System
### 11. Space interval modifiers: FEhE
The following cmavo is discussed in this section:
``` fe'e FEhE space interval modifier flag
```
Like time intervals, space intervals can also be continuous, discontinuous, or repetitive. Rather than having a whole separate set of selma'o for space interval properties, we instead prefix the flag “fe'e” to the cmavo used for time interval properties. A space interval property would be placed just after the space interval size and/or dimensionality cmavo:
```✥11.1 ko vi'i fe'e di'i
sombo le gurni
You-imperative [1-dimensional] [space:] [regularly]
sow the grain.
Sow the grain in a line and evenly!
```
```✥11.2 mi fe'e ciroi
tervecnu lo selsalta
I [space:] [three places]
```
```✥11.3 ze'e roroi ve'e
fe'e roroi ku
li re su'i re du li vo
[whole time] [all times] [whole space]
[space:] [all places]
The-number 2 + 2 = the-number 4.
Always and everywhere, two plus two is four.
```
As shown in ✥11.3, when a tense comes first in a bridi, rather than in its normal position before the selbri (in this case “du”), it is emphasized.
The “fe'e” marker can also be used for the same purpose before members of ZAhO. (The cmavo “be'a” belongs to selma'o FAhA; it is the space direction meaning “north of”.)
```✥11.4 tu ve'abe'a
fe'e co'a rokci
that-yonder [medium space interval - north]
[space] [initiative] is-a-rock.
That is the beginning of a rock extending to my north.
That is the south face of a rock.
```
Here the notion of a “beginning point” represented by the cmavo “co'a” is transferred from “beginning in time” to “beginning in space” under the influence of the “fe'e” flag. Space is not inherently oriented, unlike time, which flows from past to future: therefore, some indication of orientation is necessary, and the “ve'abe'a” provides an orientation in which the south face is the “beginning” and the north face is the “end”, since the rock extends from south (near me) to north (away from me).
Many natural languages represent time by a space-based metaphor: in English, what is past is said to be “behind us”. In other languages, the metaphor is reversed. Here, Lojban is representing space (or space interval modifiers) by a time-based metaphor: the choice of a FAhA cmavo following a VEhA cmavo indicates which direction is mapped onto the future. (The choice of future rather than past is arbitrary, but convenient for English-speakers.)
If both a TAhE (or ROI) and a ZAhO are present as space interval modifiers, the “fe'e” flag must be prefixed to each. | 711 | 2,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.774079 |
https://www.mpi-inf.mpg.de/departments/algorithms-complexity/teaching/summer19/dist-seq-algo/ | 1,563,282,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524548.22/warc/CC-MAIN-20190716115717-20190716141717-00284.warc.gz | 771,824,415 | 7,574 | # Basic Information
Lectures: Tuesday, 10:15-12:00, E1.4 024 Saeed Amiri and Pranabendu Misra April 9, 2019 Friday 26.04 ; Monday 29.04 and then every 2 weeks on Friday Cosmina Croitoru 26.04 16-18 5 There will be oral exams at the end of the semenster. Please subscribe here: https://lists.mpi-inf.mpg.de/listinfo/algorithms Basic knowledge of algorithms, graph theory and probability will be assumed.
# Description
In this course we study distributed and sequential algorithms for several graph theory problems. These problems, which are abstractions of various real world problems, are well studied in computer science. The aim of this course is to understand the challenges that arise in solving these problems in the above settings, and the techniques and methods to solve them.
The plan(tentative) is to study sequential and distributed algorithms for the following:
• Minimum Dominating Set approximation
• Low Diameter Decomposition of graphs and applications
• Max Cut approximation
• Approximate Minimum 2-Edge Connected Subgraph
• Maximum Matching approximation
• Low stretch Spanners
• Lovasz Local Lemma
# Schedule
### Lectures
Date Topic References Homework 09.04.2019 Minimum Dominating Set Minimum Dominating Set (Sec 7.1) Sheet 1 16.04.2019 Low Diameter Decomposition I Network Decomposition (Sec 1.5) Sheet 2 23.04.2019 Low Diameter Decomposition II Sequential and Randomized Construction No Exercise Sheet 30.04.2019 Spanners I Graph Spanners Sheet 3 07.05.2019 Distributed Coloring Lecture Notes Sheet 4 21.05.2019 Spanners II Lecture Notes Sheet 5 28.05.2019 Introduction to LLL Lecture Notes (See Chapter 2) No Exercises 04.06.2019 LLL Lecture Notes (also see the references) Sheet 6 11.06.2019 Deterministic Network Decomposition Lecture Notes Sheet 8 18.06.2019 Graph Connectivity I: Sequential Algorithms Kargar's Mincut AlgorithmMinimum k-Connected Subgraph Sheet 9 (Preliminary) 25.06.2019 Graph Connectivity II: Distributed Algorithms Lecture Notes No Homework 02.07.2019 Distributed Algorithms for All Pairs Shortest Paths Distributed Routing No Homework
# Announcements
1. Please subscribe to our Mailing List: https://lists.mpi-inf.mpg.de/listinfo/algorithms | 546 | 2,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.725297 |
https://blog.jverkamp.com/2023/12/09/aoc-2023-day-9-stackinator/ | 1,721,546,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00861.warc.gz | 116,279,313 | 7,007 | # AoC 2023 Day 9: Stackinator
## Source: Day 9: Mirage Maintenance
Full solution for today (spoilers!)
## Part 1
Given a list of terms, repeatedly calculate the differences of terms until these differences are 0. So:
0 3 6 9 12 15
3 3 3 3 3
0 0 0 0
Calculate the sum of next terms for each sequence (18 for this one).
There’s nothing particularly interesting about the types and parsing here:
#[derive(Debug)]
pub struct Equation {
pub terms: Vec<i64>,
}
fn equation(s: &str) -> IResult<&str, Equation> {
let (s, terms) = separated_list1(space1, complete::i64)(s)?;
Ok((s, Equation { terms }))
}
pub fn equations(s: &str) -> IResult<&str, Vec<Equation>> {
separated_list1(newline, equation)(s)
}
Honestly, I could have just kept Vec all through.
What’s a bit more interesting is a method to build the stack for an Equation:
impl Equation {
// Generate a 'stack' of vecs, starting from the original terms
// Each subsequent vec is the vec of differences in terms (and thus 1 element shorter)
// Stop when that list is all 0s
pub fn stack(&self) -> Vec<Vec<i64>> {
let mut stack = vec![];
stack.push(self.terms.clone());
loop {
let bottom = stack.last().unwrap();
if bottom.iter().all(|t| *t == 0) {
return stack;
}
stack.push(
bottom
.iter()
.zip(bottom.iter().skip(1))
.map(|(a, b)| *b - *a)
.collect(),
);
}
}
}
I enjoy the zip(...) of iter() and iter().skip(1).
With that, we just need to fill in the next numbers from ‘bottom up’:
fn main() -> Result<()> {
let stdin = io::stdin();
let (s, equations) = parse::equations(&input).unwrap();
assert_eq!(s.trim(), "");
let result = equations
.iter()
.map(|equation| {
// Build a stack of differences until we get to 0
let mut stack = equation.stack();
// From the bottom up, add the last value to the differences beneath it
for i in (0..stack.len() - 1).rev() {
let next = stack[i].last().unwrap() + stack[i + 1].last().unwrap();
stack[i].push(next);
}
// The new last value of the top line (the original list)
*stack[0].last().unwrap()
})
.sum::<i64>();
println!("{result}");
Ok(())
}
And there we have it.
## Part 2
Instead of summing the next terms for each equation, sum what would be the term before the current list.
Almost, exactly the same code, we just have two differences:
fn main() -> Result<()> {
let stdin = io::stdin();
let (s, equations) = parse::equations(&input).unwrap();
assert_eq!(s.trim(), "");
let result = equations
.iter()
.map(|equation| {
// Build the stacks as in part 1, but reverse them all (so we can generate a new 'first' element)
// Alternatively, use a VecDeque
let mut stack = equation.stack();
stack.iter_mut().for_each(|v| v.reverse());
// Same (from the bottom up), but this time we're subtracting
for i in (0..stack.len() - 1).rev() {
let next = stack[i].last().unwrap() - stack[i + 1].last().unwrap();
stack[i].push(next);
}
// The new last value is the value 'before' the original list
*stack[0].last().unwrap()
})
.sum::<i64>();
println!("{result}");
Ok(())
}
Specifically, we for_each(|v| v.reverse()) and subtract instead of add.
That’s really all there is to it!
## Performance
Nothing much to say here:
$just time 9 1 hyperfine --warmup 3 'just run 9 1' Benchmark 1: just run 9 1 Time (mean ± σ): 80.1 ms ± 3.3 ms [User: 30.6 ms, System: 11.3 ms] Range (min … max): 77.2 ms … 89.4 ms 32 runs$ just time 9 2
hyperfine --warmup 3 'just run 9 2'
Benchmark 1: just run 9 2
Time (mean ± σ): 79.6 ms ± 3.9 ms [User: 30.3 ms, System: 11.1 ms]
Range (min … max): 77.2 ms … 95.5 ms 32 runs
Within a margin of error of each other, which is to be expected. The first reverse may have added some time, but it only depends on the length of input, not what the number of terms is.
If we had far larger input, I might try to optimize this (I expect there are direct equations for these), but other than that… we’re already sub 100ms and it’s been a long day–so away we go! | 1,140 | 3,937 | {"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} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.625213 |
https://www.arxiv-vanity.com/papers/2001.08705/ | 1,601,230,673,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00339.warc.gz | 713,852,389 | 59,940 | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
The Eternal Game Chromatic Number of Random Graphs
Vojtĕch Dvor̆ák Department of Pure Maths and Mathematical Statistics, University of Cambridge, UK [ Rebekah Herrman Department of Mathematical Sciences, The University of Memphis, Memphis, TN [ and Peter van Hintum Department of Pure Maths and Mathematical Statistics, University of Cambridge, UK [
Abstract.
The eternal graph colouring problem, recently introduced by Klostermeyer and Mendoza [10], is a version of the graph colouring game, where two players take turns properly colouring a graph. In this note, we study the eternal game chromatic number of random graphs. We show that with high probability for odd , and also for even when for some . The upper bound applies for even and any other value of as well, but we conjecture in this case this upper bound is not sharp. Finally, we answer a question posed by Klostermeyer and Mendoza.
Vojtĕch Dvor̆ák] Rebekah Herrman] Peter van Hintum]
1. Introduction
The vertex colouring game was introduced by Brams [7] in 1981; it was later rediscovered by Bodlaender [1]. In this game, two players, Alice and Bob, take turns choosing uncoloured vertices from a graph, , and assigning a colour from a predefined set , such that the resulting partial colouring of is proper. Bob wins, if at some stage, he or Alice chooses a vertex that cannot be properly coloured. Alice wins if each chosen vertex can be properly coloured. The game chromatic number is the smallest integer such that if there are colours, Alice has a winning strategy in the vertex colouring game. This number is well defined, as Alice can win if the number of colours is at least the number of vertices. The vertex colouring game has been well studied [4, 5, 8, 11]. In particular, Bohman, Frieze and Sudakov [2] studied the game chromatic number of random graphs and found that with high probability, , where all logarithms have base . Keusch and Steger [9] improved the result to with high probability, implying with high probability by a classic result of Bollobás [3]. Both of the results require lower bounds on decaying with slowly. Frieze, Haber and Lavrov [6] studied the game on sparse random graphs, finding that for , , where is at least a large constant. This vertex colouring game requires Alice and Bob to colour the vertices once, attaching no value to the colouring that is produced at the end of the round. In a variant of the game called the eternal vertex colouring game recently introduced by Klostermeyer and Mendoza [10], the focus is shifted by continuing the game after a colouring is produced. In the eternal vertex colouring game, there is a fixed set of colours . The game consists of rounds, such that in each round, every vertex is coloured exactly once. The first round proceeds precisely the same as the vertex colouring game, with Alice taking the first turn. During all further rounds, players keep choosing vertices alternately. After choosing a vertex, the player assigns a colour to the vertex which is distinct from its current colour such that the resulting colouring is proper. Each vertex retains its colour between rounds until it is recoloured. Bob wins if at any point the chosen vertex does not have a legal recolouring, while Alice wins if the game is continued indefinitely. The eternal game chromatic number is the smallest number such that Alice has a winning strategy. Note that if , there will always be a colour available for every vertex, so is well-defined. As Alice and Bob alternate their turns, the parity of the order of the graph determines whose turn it is at the beginning of the second round. For even order, Alice always has the first move, while for odd order Bob gets to play first in all even rounds. This game has not been well studied, but Klostermeyer and Mendoza [10] obtained some basic results pertaining to paths, cycles, and balanced bipartite graphs. In this paper, we determine for odd by putting together the following two results.
Theorem 1.1.
For all constant, for odd , with high probability,
χ∞g(Gn,p)=(1+o(1))pn2.
Theorem 1.2.
For all constant, for even , with high probability,
χ∞g(Gn,p)≤(1+o(1))pn2.
Moreover, when for some ,
χ∞g(Gn,p)=(1+o(1))pn2.
The difference in the even and odd cases is because when is odd, Bob moves first in the second round. Also, note that we made no efforts to optimize terms. For the unresolved case when is even and , we conjecture the following.
Conjecture 1.3.
such that with high probability,
χ∞g(Gn,p)≤(1−ϵ)pn2.
The structure of the paper is as follows. In Section 2, we prove the upper bound for . In this proof, we make no distinction between odd and even values of . In Section 3, we prove the corresponding lower bound for odd . In Section 4, we prove a generalization of the result in Section 3, which we then use in Section 5 to get the lower bound for the case for some . Along the way, we use various structural results about the random graph . As the proofs of these are usually quite easy but technical, we collect all of them in Appendix A. Finally in Section 6, we provide answer to one of the questions posed in the paper of Klostermeyer and Mendoza. Throughout the paper we will use the following conventions. We say that a result holds in with high probability (whp) if the probability that it holds tends to as . The neighbourhood of a vertex in a graph, , denoted , will be the vertices of that is connected to, as well as itself. The partition of a set will refer to a collection of disjoint non-empty subsets whose union is .
2. Upper bound
In this section, we show the following proposition.
Proposition 2.1.
For any fixed , whp .
To prove this, we formulate a deterministic strategy for Alice and prove that whp this strategy enables her to prevent Bob from winning when the game is played with colours. The biggest danger facing Alice is that at the end of some round, Bob would manage to introduce all the colours in the neighbourhood of at least one vertex. He could then win by choosing one of those vertices at the beginning of the next round. Thus, her strategy should be to ensure that at any point of each round, she has coloured roughly as many vertices in the neighbourhood of any single vertex as Bob has, and she should use few colours on them. If, at some point during a round, a vertex has many colours in its neighbourhood compared to other vertices, Alice might be forced to colour it so Bob cannot win by choosing it later that same round. Fortunately for Alice, the number of times she is forced to colour a vertex with many different coloured neighbours is so few that she can still follow her strategy. Consider the following four properties of a graph .
1. [label=()]
2. Every vertex of has degree at most
3. There exists a constant such that does not contain sets , with , , such that every is connected to at least more vertices in than in .
4. There exist constants , and such that the following holds: for any colouring of by colours, the number of vertices that have all but at most colours in their neighbourhood is at most .
5. There exist constants and such that, in any colouring of by colours, the number of vertices that have all but at most of the colours in their neighbourhood is at most .
We prove in creftypecap A that each of these holds whp in . For the remainder of this paragraph, we will assume 1 through 4 hold for the graph . Note that as we are assuming finitely many properties, each of which holds whp, then whp all of them hold simultaneously. For a particular round of the game, let and denote the sets of vertices played by Alice and Bob respectively in the first moves of that round. We shall define the vertices that threaten Alice’s chance of winning as dangerous.
Definition 2.2.
For a fixed round of play, let denote the set of dangerous vertices at moves, denoted . A vertex belongs to if for some number of moves , Bob has played at least times more in the neighbourhood of than Alice has, i.e. .
We additionally define vertices that Alice can colour to maintain some symmetry in the game as follows.
Definition 2.3.
Let be a finite subset of vertices of a graph, . For a vertex , we say that a vertex mirrors with respect to if and for any , contains an edge if and only if it contains an edge .
Let be the set of colours used in the game. We call a colour large if it is at least , and small otherwise. Alice will use the following strategy at the move of a round: from the list below, she chooses the first point that applies, and colours the corresponding vertex with smallest colour available to that vertex. If there are multiple vertices for the same point on the list, she chooses one of these arbitrarily.
1. If there is a vertex that misses less than colours in its neighborhood and such that has not yet been coloured in the current round, she chooses .
2. If Bob played a vertex for his previous move, is not dangerous, and there is a vertex which mirrors with respect to , she chooses .
3. She chooses an arbitrary vertex.
We shall prove that because 1, 2, 3, and 4 hold, selecting vertices in order of priority will ensure that Bob can never win the eternal vertex colouring game for sufficiently large . To show Bob cannot win, we prove the following lemma.
Lemma 2.4.
For sufficiently large, at the beginning of the round of play for , the following two conditions hold: During the round, there was no vertex such that number of times Bob played in neighbourhood of was more than greater than number of times Alice played in neighbourhood of . Alice used no more than colours in round. Then, by playing according to the above described strategy, Alice ensures the following: Bob does not win during round During round, there is no vertex such that number of times Bob played in neighbourhood of is more than greater than number of times Alice played in neighbourhood of . Alice uses no more than colours in the round.
Note that Lemma 2.4 implies that, if Alice plays according to the strategy described above, Bob can never win the eternal graph colouring game.
Proof.
The first step is to establish that at beginning of the round, each vertex misses more than colours in its neighbourhood, so that there is no immediate threat to Alice. In the first round, this is immediate, as no colour is used yet. When , Alice uses at most colours in the neighbourhood of any vertex . Bob played at most more moves in the neighbourhood of than Alice did, so by property 1, Bob played at most colours in the neighborhood of . Hence, at least colours are missing from the neighbourhood of . Now, if some vertex misses at most colours at any point during the round, then in particular at least one of the times , this vertex missed at most colours. By property 3, we conclude there are at most vertices that, at some point in this round, have missed at most colours. Recall that colouring vertices that miss at most colours is of the highest priority in Alice’ strategy. If Bob were to create a vertex seeing all colours that was not yet played in this round, Alice must have spend the previous moves playing in other vertices missing at most colours in their neighbourhoods. However, this contradicts the fact that there were at most such vertices. Hence, Alice can colour all such vertices in time. Next, note that Alice uses (and in fact only constantly many) moves that are arbitrary. If Alice colours an arbitrary vertex, then either Bob played a dangerous vertex for his previous move or she cannot mirror Bob on the current set of dangerous vertices. By property 2, there are at most vertices declared dangerous during the round, so Bob can play in a dangerous vertex no more than times. On the other hand, consider if Bob did not play dangerous vertex and Alice cannot mirror his move on , the set of dangerous vertices. If we partition the rest of the graph into classes according to which vertices of they are connected to, Bob must have just played last vertex from one of these classes. Hence, Alice must have played at most arbitrary moves in any particular round. Following this strategy, Alice also ensures that Bob will play in the neighbourhood of any vertex at most more than Alice does. Indeed, once Bob has played more colours in the neighbourhood of any vertex, , it is declared dangerous. She then plays in neighbourhood of whenever Bob does, except times when she plays a move of type 1 or an arbitrary vertex. Finally, note that Alice uses large colours only if the vertex she wants to colour is connected to all the small colours. If at any point during the round a vertex is connected to all the small colours, then at least one of the times , this vertex must have been missing at most colours. By property 4, there could have only been at most vertices that were connected to all small colours at some point in this round. Hence, as she can colour all other vertices with small colours, Alice uses at most colours during the round. ∎
3. Lower bound for odd n
In this section, we prove the lower bound for the eternal game chromatic number on a graph with an odd number of vertices.
Proposition 3.1.
For any fixed, whp .
For convenience, we shall let . Fix any vertex of the graph . We will show that whp, Bob can ensure that in the first round, all colours are in the closed neighbourhood of in . Bob then wins in the first move of the second round, by choosing . Bob can introduce all colours in by playing in whenever Alice does, thus ensuring he plays in at least roughly half of the vertices in while introducing a new colour every time. Some set of vertices outside might at some point be adjacent to all unplayed vertices of . If Alice were to play some colour not appearing in in all these vertices, this colour could no longer be introduced to . Fortunately for Bob, the number of such sets will be very limited, and thus Bob can take care of them in time. Consider following two properties of a random graph.
1. [label=()]
2. Whp, every vertex of has degree at least .
3. For all , and , there exist positive constants , such that in
• Whp, there exist no 3 vertices such that the number of vertices in the neighbourhood of , but not in the neighbourhood of or is at most
• Whp, for any set of size in the graph, there exist at most mutually disjoint pairs of vertices such that at most vertices of are not in
Henceforth, we assume our graph has both properties, and fix and . In creftypecap A, we show that indeed whp has these properties. Note that if at some stage there exists a colour that does not appear in and all vertices not yet played in are adjacent to a vertex of colour , then will not appear in , which is contrary to Bob’s goal. We introduce the ideas of a double block and being away from becoming a double block in order to describe a strategy Bob should take to achieve his goal of filling the neighbourhood of a vertex with several colours.
Definition 3.2.
A pair of vertices and is called a double block if at some stage in the round, all uncoloured vertices in are in the neighbourhood of either or and neither or (if coloured) is coloured with a colour appearing in .
Definition 3.3.
A pair of vertices and is said to be away from becoming a double block, if all but at most of the uncoloured vertices in are in the neighbourhood of either or and neither or (if coloured) is coloured with a colour appearing in .
Bob will play according to the following strategy. From the list below, he picks the highest point that applies.
1. If there exists a colour that appears at least twice outside of but not in , then Bob plays it in if it is a valid colouring.
2. If at least colours appear nowhere in the graph and there are at least uncoloured vertices in the neighbourhood of , then Bob does the blocking moves in the chronological order they were called for, if legally possible. Blocking moves are called for if a pair of unplayed vertices is away from becoming a double block. Blocking moves consist of the following steps. First colour vertex a colour not yet appearing in the graph, say . Second, unless Alice plays in vertex or introduces in , play in and repeat for vertex . If Alice plays in vertex with a colour not appearing in , play in , and finish by playing in on the next move.
3. If legally possible, Bob introduces colours appearing once outside of into , in the order in which they were introduced to the game.
4. If legally possible, Bob introduces new colours to .
5. Otherwise Bob does anything.
Note that (2) might involve up to four moves for any pair close to becoming a double block. If in between these four moves a situation as in (1) arises, situation (1) takes priority.
Claim 3.4.
There are no more than moves of type (2) used in the first round.
Proof.
Let denote the set of vertices that are uncoloured in when the last blocking moves were played. By the definition of type (2) moves, , so property 2 gives the result. ∎
Let be the first move after which precisely colours are missing in during the first round. We collect the following observations about : exists and at , at least vertices in are uncoloured We shall show that . After Bob’s first moves, Alice has also played moves. As , at this stage at least vertices in are uncoloured. Bob spent at most moves playing according to (2), and when he did not, he always introduced a new colour in , if he legally could. Note that if there were still colours missing from and there were uncoloured vertices in , moves of type (4) were always legal. Hence, unless all colours appear in the graph, Bob played at least colours in and hence in . At , at most colours are missing in Between two consecutive moves of Bob before T, the number of colours appearing outside of but not in can increase by at most 2. In fact it only increases if Bob makes a move of type (2). Hence, there are at most such colours at time , and result follows. At , Bob has played at most (1) moves. Let be the number of colours appearing outside and not in . Note that between two consecutive moves of Bob up to time , increases only if Bob plays a (2) move, in which case it increases by at most 2. On the other hand, note that Bob only plays (1) moves directly after Alice plays a colour already appearing outside . Hence, decreases whenever Bob plays a (1) move. By 3.4, there were at most (2) moves, so there were at most (1) moves. No pair of vertices is closer than to becoming a double-block at any point up to We know from 2 that at the beginning of the game, no pair is closer than to becoming a double block. Up to time , whenever a pair gets closer than to becoming a double block, no (3)-(5) moves are played until this pair is eliminated. However, there are at most (1) and (2) moves played until . Hence, no pair can be closer than to becoming a double-block up to . Every colour that does not appear in at appears at most once in Note that the only time before that there is a colour appearing twice outside , but not inside, is directly after Alice has played this colour. In response, Bob immediately plays that same colour in , which is possible as no pair of vertices is a double block. Hence, if Bob made the last move before , the statement follows. If the last move before was by Alice, she must have introduced a new colour into the graph by the definition of , which again implies the statement. Next we claim:
Claim 3.5.
In the moves of Bob following , he will introduce all colours in .
Proof.
Moves of type (2) are no longer played after by their definition. In the next moves, of which are made by Bob and the by Alice, no complete double-block can be created, as all are at least moves away. Since at at least vertices of are still uncoloured, during the next moves, there are ample uncoloured vertices in . Hence, Bob can and will introduce a new colour to every move until all colours appear there, as he ceases the (2) moves. ∎
Thus we see that Bob will ensure that all colours appear in during the first round and he will win in the first move of the second round by picking .
4. Generalization of the lower bound for odd n
creftypecap 3.1 doesn’t trivially extend to even , as it is not enough for Bob to let all colours appear in the neighbourhood of a fixed vertex because Alice could use her first move in the second round to remove one of the colours from this neighbourhood. If Bob can manage to play all colours in the neighbourhood of two vertices, with no colour appearing uniquely in the intersection of the neighbourhoods, then Alice can not. This limits how Bob can colour the intersection of two neighbourhoods of vertices. By increasing the number of vertices that simultaneously see all colours, the size of this intersection can be reduced. Our aim is to show that if for some , then for any fixed , Bob can choose vertices and play all of colours in the neighbourhoods of these vertices. The correction term comes from the intersection of the neighbourhoods of these vertices. As can be taken arbitrarily large, this shows for even and . In this section, we prove a generalization of creftypecap 3.1, showing that if is partitioned into constantly many parts and each of the parts is assigned a set of colours of size roughly half the size of the part, Bob can guarantee all these colours to appear in the parts by the end of the first round. This generalizes the notion that Bob could achieve this in the single set . In the next section, we will fix some set of vertices of constant size and induce partition , where . We show in creftypecap 5.2, that for the special case , there exists an appropriate way of assigning colours to the ’s such that each vertex in will see all colours after the first round.
Proposition 4.1.
and , if are disjointly chosen independent sets of vertices of the graph with and with , then whp Bob can guarantee that at the the end of round all of the colours in appear in .
To prove creftypecap 4.1 we will use the following generalization of the structural result in creftypecap 2.
Lemma 4.2.
For all , and , there exist positive constants , such that
• For any set of size at least in the graph, whp there exist no -sets of vertices such that at most vertices of are not in
• Whp, for any set of size at least in the graph, there exist at most mutually disjoint -sets of vertices such that at most vertices of are not in
In order to prove Propositon 4.1, we will define the concepts of an end stage, m-block and away from becoming an m-block.
Definition 4.3.
Let be the first move after which at most of the colours in are missing from , if this exists. After , say is in its end stage.
Definition 4.4.
Given disjoint sets , at some stage of the round we say a set is an -block if for some , is not in its end stage, every uncoloured vertex in is in the neighbourhood of some , and no is coloured in some colour also appearing in .
Definition 4.5.
Given disjoint sets , at some stage of the round we say a set is away from becoming a -block if for some , is not in its end stage, all but of the uncoloured vertices in is in the neighbourhood of some , and no is coloured in some colour also appearing in .
Proof of creftypecap 4.1.
Let and let and as in creftypecap 4.2. Bob will play the move of the highest priority that he legally can according to the following list:
1. If for some , some colour appears times in the graph, but it is missing from more than of the ’s for which , Bob plays it in any of ’s where it does not yet appear.
2. If for some , is in its end stage, Bob plays the missing colours into it, copying the colour Alice played if it was missing.
3. If there is block closer than moves away from becoming an m-block and at least vertices in the corresponding are uncoloured, Bob kills it. By killing it, we mean the following sequence of moves. Colour the first vertex of our -set by some colour that appears less than times in the graph and is missing from at most of the relevant ’s. Then make sure in the next moves that this colour also appears in all of its designated ’s. Repeat this procedure for all vertices of our -set.
4. If for some , some colour appears times in the graph, but it is still missing from more than of ’s, Bob plays it in any of ’s where it does not yet appear.
5. Bob plays any colour in not yet used in to that , if possible in the same as Alice played in the previous move.
6. Bob plays anything anywhere.
Note that (3) might involve up to moves. If in between these moves a situation as in (1) or (2) arises, those are resolved first.
Claim 4.6.
Let . There were no more than (3) moves called for.
Proof.
Let denote the set of vertices that are still uncoloured in when the last blocking moves were called for, for this . creftypecap 4.2 says called for at most (3) moves. Hence, in total at most (3) moves are called for. ∎
Claim 4.7.
There were no more than moves of types (1),(2),(3) and (4) during the first round of the game.
Proof.
Note that at most colours appear at least times in the graph. Moreover, these colours prompt a (1) or (4) move at most times. Finally, there are at most (2) moves. Hence, there are at most , given . ∎
We collect the following observations about : exists and, at , at least vertices in are still uncoloured At the end of round one there were moves in . Moreover, by creftypecap 4.7 there were at most (1)-(4) moves. After Bob’s first in , he has played at most (1)-(4) moves. He also played in after every move of Alice in that set, except the times when he played (1)-(4) moves. Thus, at least of the vertices in are uncoloured. As , this gives the result. Let . At , Bob has played at most (1) moves. For a colour , let be the number such that colour is missing from of its designated sets. Let be the number of times appears in the graph. If , then Bob is forced to play a (1) move. If , then this induces a (4) move. Let . Note that if , then Bob must play a (1),(2),(3) or (4) move. If increases between consecutive moves of Bob, he must have played a (2) or (3) move. Moreover, increases by at most 2 in that case. On the other hand, if Bob is prompted to play a (1) move, decreases by at least . Hence, there are at most as many (1) moves as there are (2) and (3) moves, i.e. at most (1) moves. No pair of vertices is closer than to becoming a block at any point up to By creftypecap 4.2, at the beginning of the game no -set of vertices is closer than to becoming a block. Whenever, up to time , a -set gets closer than to becoming a block, no (4)-(6) moves are played until this pair is eliminated. However, there are at most (1) and (3) moves played until . Hence, no -set gets closer than to becoming a block up to . Every designated colour that does not appear in at appears at most times in our graph By the definition of (1) moves, some colour can never appear times in our graph, yet not appear in some of ’s with . Next we claim:
Claim 4.8.
In the moves of Bob following , he will introduce all colours in .
Proof.
Note that while there are still colours missing from in its end stage, Bob only plays (1) and (2) moves, both of which copy the colour Alice played. Hence, the colours missing from can be played at most times before being played into . At that stage, the colour is played at most times and no -set is closer than to becoming a block, so no block will be formed in the endstage of . Hence, we can still play this colour in . As we can introduce all the missing colours and we play at most (1) moves, we need at most moves to introduce them all. ∎
Thus, since Bob can introduce all colours into during the end game, the proof of creftypecap 4.1 is complete. ∎
Having proven creftypecap 4.1, we are ready to look at even .
5. Even n
In this section, we shall prove that for particular values of , we can achieve the same lower bound for even as for odd .
Proposition 5.1.
Let for some , and . Then whp .
For convenience write . For given , fix , such that .
Lemma 5.2.
Let be a set of vertices and for some . There exists , and a function , assigning to every subset , colours, such that for every .
To prove this lemma we will use the following auxiliary lemma. Let be the set of all partitions of the set .
Lemma 5.3.
Consider any . Let and , then there exists , such that for all ;
∑T:A∈T∈B(X)g(T)=p|A|(1−p)l−|A|
Proof.
Define as
g(T)={k−l(k−1)!(|T|−1)! if |T|≤l0 else
Fix and evaluate . Consider ordered partitions of into potentially empty sets. Each of these contributes exactly to this sum. To see this, consider a particular ordered partition of into potentially empty sets, with non-empty sets. This corresponds to a partition of into parts, which has weight . Note that to find ordered partitions into potentially empty sets corresponding to , we need to pick the locations of the sets among the options. There are ways to do this. Hence, every ordered partition of into potentially empty sets contributes weight exactly to the sum. Noting that there are exactly ordered partitions of into potentially empty sets, we can evaluate the sum as
∑T:A∈T∈B(X)g(T)=(k−1)l−|A|k−l=(1k)|A|(k−1k)l−|A|=p|A|(1−p)l−|A|
Proof of creftypecap 5.2.
Let as in creftypecap 5.3 and set . Note that . Consider any linear order on and let
where is such that . Let
f:P(X)→P((p/2−pl/2−ϵ)n);X′↦⋃T⊃X′f′(T)
Hence;
⋃X′:x∈X′⊂Xf(X′) =⋃X′:x∈X′⊂X⋃T:T⊃X′f′(T) =⋃T∈B(X)f′(T)
Proof of creftypecap 5.1.
Fix with . Sample all edges incident to . For , let . Note that whp for any . Use creftypecap 5.2 to find , such that . Now sample all the other edges in the graph. By creftypecap 4.1, whp Bob can guarantee that at the end of round one the colours in appears in . By construction of , all vertices in will see all colours and, moreover, there is no single vertex whose recolouring changes that. Regardless of Alice’ first move in the second round, Bob can choose a vertex that sees all colours in his first move in the second round. Thus, the proposition follows. ∎
Note that the condition is essential. In fact, if , then whp for any three vertices and , it is impossible to assign colours to , such that and for every . Crucially, creftypecap 5.3 fails to hold. Hence, given that Alice can play in roughly half the vertices in for all , at most two vertices at the end of every round can see all colours. Some must appear uniquely in the intersection, so Alice can recolour one of these in the first move of the second round. Hence, we cannot expect Bob to win at the beginning of round two. However, it is not immediately clear whether Bob cannot reintroduce the colour successfully. We believe this cannot be done, so we conjecture that for all such that whp , as stated in creftypecap 1.3.
6. Answer to a question of Klostermeyer and Mendoza
We conclude the paper by answering a question posed by Klostermeyer and Mendoza in their original paper. They define other variants of the eternal chromatic game on graph. One of them is greedy colouring game, where Bob must colour whatever vertex he chooses with the smallest colour possible. Let be the smallest number such that when this game is played with colours on , Alice is guaranteed to win. Further, they consider the variant of game when not only Bob, but also Alice, must use the smallest colour available for each vertex she chooses, and define to be eternal number of the game played with these rules. Note that clearly since Alice can, if she wishes so, choose the smallest colour for each vertex she chooses in any variant of the game and analogously . Klostermeyer and Mendoza pose the following question about these new variants of the game.
Question 6.1.
Let be a graph with subgraph or induced subgraph . Is it necessarily true that ? Is it necessarily true that ?
Indeed, it is not true. Consider the following example.
For , and .
Proof.
For note that Alice starts every round as the number of vertices is even. Every round she will first play in the central vertex which will become the unique element from not yet appearing in the graph. All the other vertices will now become the former colour of the central vertex. For . assume for a contradiction 3 colours suffice and note that Bob begins the second round. Let for the central vertex. Then is either adjacent to two different colours or is monochromatic. In the former case, Bob plays in and finds that there is no colour available, a contradiction.
In the latter case, Bob plays in , bringing the number of colours in to two. Hence, Alice cannot play in . She can also not bring down the number of colours in as it contains at least three vertices. Thence, when Bob gets to play his second move in the second round, and plays , he finds no colours available, again a contradiction. ∎
Note that is an (induced) subgraph of , and
χ∞2g(G)≤χ∞3g(G)≤χ∞2g(H)≤χ∞3g(H)
This answers all the subquestions in the negative. Finally, note that while there is no clear relationship between and for general graphs , in our definition of strategy of Alice in section 2, we let her always play the smallest colour available, and so in particular we show whp.
Acknowledgements
The authors are grateful to their PhD supervisor Béla Bollobás for his useful comments and support.
Appendix A Proofs of structural results in random graph
In this appendix, we provide proofs of various structural results about that were used in earlier proofs. Some of them will be shown in a more general form. One of our main tools will be the following well-known form of Hoeffding’s Inequality.
Lemma A.1.
For any and , and .
Note that Hoeffding’s inequality implies 1 from Section 2 and 1 from Section 3. To prove 3 and 4 from Section 2, we first prove the following result.
Lemma A.2.
For all , , there exist constants such that whp the following holds. For any colouring of with colours, number of vertices that have all but at most colours in their neighbourhood is at most .
Proof.
Let , , and . Note that in any colouring of by colours, we have colours appearing at most times each. Assume there exists a set of vertices missing at most colours each. For satisfying ), there exists a set of colours appearing at most times each such that no vertex in has any colour from . In particular, there must be mutually disjoint sets of vertices , such that for each , and each vertex in is joined to at least sets in our graph. Now, we consider the probability that such structure exists in . For sufficiently large, we find there are ways of choosing each of the sets . So for such large , we have at most
(nKlogn)(2(n2/α))αn4 ≤nKlogn2αn4(eαn2)αn4 =exp(K(logn)2+αn4logn+α4nlog(eα))
ways to choose sets . Now for any such fixed choice, the probability that these sets satisfy the conditions is at most
P(Bin(αn4,q)≥(q+1−q2)αn4)Klogn≤exp⎛⎝−2((1−q)2)2αn4Klogn⎞⎠
The union bound then gives that the probability of finding appropriate
exp(K(logn)2+
The result follows. ∎
To conclude 3, simply plug in , and note that if some value of works, then any smaller one does too, so we can insist on being not too large. To conclude 4, plug in and note that presence of other colours only helps us, as the result would still hold even if all the other vertices were also coloured in small colours. The following implies 2 from section 2.
Lemma A.3.
Fix any and greater than . Assume is fixed, such that . Whp, if are disjoint subsets with , then there are less than vertices connected to at least more vertices in than in .
Proof.
Assume for a contradiction and there exist as stated in the lemma such that there are at least vertices connected to at least more vertices in than in . Let be a collection of such vertices. Let and . Note that and , so that
e(B′,S)−e(A′,S)≥e(B,S)−e(A,S)−K2≥Kδn−K2≥Kδn/2.
Hence, either or . The probability of the former (the latter follows analogously) is given by;
P(Bin(|B′|⋅|S|,p)≥(|B′|Kp)+δKn/8) ≤exp⎛⎝−2(δn8|B′|)2|B′|K⎞⎠ ≤exp(−δ2nK2ϵ)
We can choose sets and in at most
(n|A|)2(nK)≤23n
ways. Thus, the probability that any such sets and exist is at most
2exp(−δ22ϵnK+3nlog2)→0
provided . ∎
The proof of 2 from section 3 follows from the fact that for sufficiently small and positive, whp there exists no three vertices such that the number of vertices in the neighbourhood of , but not in the neighbourhood of or is at most by Hoeffding’s Inequality. 2 follows directly from the following setting . creftypecap 4.2 follows in the same manner, this time using the particular we need.
Lemma A.4.
Fix , , . Then for any , whp does not contain any collection of sets such that are all mutually disjoint, , and for every , all but at most vertices of are connected to at least one vertex in | 8,633 | 36,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.920807 |
http://www.physicsforums.com/showthread.php?t=270928 | 1,386,948,516,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164954485/warc/CC-MAIN-20131204134914-00040-ip-10-33-133-15.ec2.internal.warc.gz | 479,659,941 | 8,709 | # Tension Problem (two ropes + weight)
by DarkNightwing
Tags: ropes, tension, weight
P: 3 Oh I knew I'd have to come crawling onto the internet in search of help, help me physics friends !! 1. Determine the tension in each of the ropes holding the object. Rope T1 is at an angle of 60 degrees, rope T2 is at an angle of 55 degrees, the suspended block has a mass of 300kg. 2. w = mg 3. 300kg * 9.8 m/s = 2940 N, or the weight of the block, but since this I've just divided the weight by the sin of each rope's given angle. Completely baffled and very tired, looking for a push in the right direction so my brain can worky again. The answers are: T1 = 1862 N , T2 = 1620 N but I have no clue how to arrive at them
P: 154
Quote by DarkNightwing Oh I knew I'd have to come crawling onto the internet in search of help, help me physics friends !! 1. Determine the tension in each of the ropes holding the object. Rope T1 is at an angle of 60 degrees, rope T2 is at an angle of 55 degrees, the suspended block has a mass of 300kg. 2. w = mg 3. 300kg * 9.8 m/s = 2940 N, or the weight of the block, but since this I've just divided the weight by the sin of each rope's given angle. Completely baffled and very tired, looking for a push in the right direction so my brain can worky again
Sum your forces to put the system in equilibrium.
P: 3 Ok so in other words the tension in T1 and T2 adds up to the weight of the block? T1 + T2 = 2940 N My problem is determining the tension in each rope, I thought maybe you took: 2940 / sin(60) = but this comes out to over 3300 some and that's more force than would be required to hold the block
HW Helper
P: 5,346
## Tension Problem (two ropes + weight)
Without knowing where the ropes are attached - hence the angles with respect to the walls or the ceiling ... it doesn't really matter.
Treat the tensions as vectors. They each have vertical and horizontal components.
The vertical components of the tensions must equal the weight. (It's in equilibrium.)
And the horizontal components add to 0. (It's in equilibrium.)
P: 74 Since the object is in equilibrium, the sum of the forces in the y-components must be balanced, and same with your x-components. This means (if angles are above the horizontal) T1sin60 + T2sin55 = the weight of the object Since you have 2 unknowns you need 2 equation, so you must use the forces in the x-directions to get another equation -T1cos60 + T2cos55 = 0 ---> One of these must be negative cause they should be in opposite directions Rearrange to solve for either T1 or T2 and plug into the other equation to solve for the other
P: 3 Doh I forgot to state that I have the answers (it's a practice test, reviewing for Friday): T1 = 1862 N , T2 = 1620 N Also putting these in the topical post
P: 2,422 does it matter which one we make negative
P: 1 To solve this: We know that the y component of the 2 tension must add together to equal the downward force of the block (This makes 0 net force = block not moving). Therefore: h1sin55 + h2sin60 + (-9.8)(300) = 0 We also know that the x components of the 2 tension must cancel each other out because the block doesnt move left or right. h1cos55 + (- h2cos60) = 0 Use substitution to solve the problem and your done. Hope this helps!
Related Discussions Introductory Physics Homework 10 Introductory Physics Homework 3 Introductory Physics Homework 1 Introductory Physics Homework 5 Introductory Physics Homework 1 | 896 | 3,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2013-48 | longest | en | 0.943093 |
https://howkgtolbs.com/convert/42.54-kg-to-lbs | 1,685,418,783,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00652.warc.gz | 334,811,209 | 12,170 | # 42.54 kg to lbs - 42.54 kilograms to pounds
Do you need to know how much is 42.54 kg equal to lbs and how to convert 42.54 kg to lbs? Here it is. This whole article is dedicated to kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to emphasize that all this article is dedicated to one number of kilograms - that is one kilogram. So if you need to learn more about 42.54 kg to pound conversion - keep reading.
Before we get to the practice - it means 42.54 kg how much lbs conversion - we are going to tell you few theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 42.54 kg to lbs? 42.54 kilograms it is equal 93.7846462548 pounds, so 42.54 kg is equal 93.7846462548 lbs.
## 42.54 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in abbreviated form SI).
Sometimes the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but totally impractical to use.
Then, in 1889 the kilogram was described by the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “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 can be also divided to 100 decagrams and 1000 grams.
## 42.54 kilogram to pounds
You know some information about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want to highlight that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is just equal 0.45359237 kilograms. One avoirdupois pound can be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was given 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 42.54 kg?
42.54 kilogram is equal to 93.7846462548 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 42.54 kg in lbs
Theoretical part is already behind us. In next part we will tell you how much is 42.54 kg to lbs. Now you know that 42.54 kg = x lbs. So it is high time to know the answer. Let’s see:
42.54 kilogram = 93.7846462548 pounds.
That is an exact result of how much 42.54 kg to pound. It is possible to also round off the result. After rounding off your outcome is exactly: 42.54 kg = 93.588 lbs.
You learned 42.54 kg is how many lbs, so look how many kg 42.54 lbs: 42.54 pound = 0.45359237 kilograms.
Naturally, this time you can also round off this result. After it your outcome will be as following: 42.54 lb = 0.45 kgs.
We also want to show you 42.54 kg to how many pounds and 42.54 pound how many kg results in tables. See:
We want to start with a chart for how much is 42.54 kg equal to pound.
### 42.54 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
42.54 93.7846462548 93.5880
Now see a table for how many kilograms 42.54 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
42.54 0.45359237 0.45
Now you learned how many 42.54 kg to lbs and how many kilograms 42.54 pound, so we can move on to the 42.54 kg to lbs formula.
### 42.54 kg to pounds
To convert 42.54 kg to us lbs you need a formula. We are going to show you a formula in two different versions. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 93.7846462548 result in pounds
The first formula will give you the most accurate outcome. Sometimes even the smallest difference can be considerable. So if you need a correct outcome - this version of a formula will be the best solution to convert how many pounds are equivalent to 42.54 kilogram.
So let’s move on to the shorer formula, which also enables conversions to learn how much 42.54 kilogram in pounds.
The shorter formula is as following, see:
Amount of kilograms * 2.2 = the outcome in pounds
As you can see, this version is simpler. It can be the best choice if you need to make a conversion of 42.54 kilogram to pounds in easy way, for example, during shopping. Just remember that final result will be not so exact.
Now we are going to show you these two versions of a formula in practice. But before we are going to make a conversion of 42.54 kg to lbs we want to show you easier way to know 42.54 kg to how many lbs without any effort.
### 42.54 kg to lbs converter
An easier way to check what is 42.54 kilogram equal to in pounds is to use 42.54 kg lbs calculator. What is a kg to lb converter?
Converter is an application. It is based on first version of a formula which we showed you above. Thanks to 42.54 kg pound calculator you can easily convert 42.54 kg to lbs. Just enter number of kilograms which you want to convert and click ‘convert’ button. The result will be shown in a flash.
So try to convert 42.54 kg into lbs using 42.54 kg vs pound calculator. We entered 42.54 as an amount of kilograms. It is the result: 42.54 kilogram = 93.7846462548 pounds.
As you can see, this 42.54 kg vs lbs converter is easy to use.
Now we are going to our primary issue - how to convert 42.54 kilograms to pounds on your own.
#### 42.54 kg to lbs conversion
We are going to begin 42.54 kilogram equals to how many pounds conversion with the first version of a formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 93.7846462548 the outcome in pounds
So what need you do to know how many pounds equal to 42.54 kilogram? Just multiply amount of kilograms, this time 42.54, by 2.20462262. It is 93.7846462548. So 42.54 kilogram is equal 93.7846462548.
You can also round it off, for instance, to two decimal places. It gives 2.20. So 42.54 kilogram = 93.5880 pounds.
It is high time for an example from everyday life. Let’s convert 42.54 kg gold in pounds. So 42.54 kg equal to how many lbs? As in the previous example - multiply 42.54 by 2.20462262. It gives 93.7846462548. So equivalent of 42.54 kilograms to pounds, if it comes to gold, is equal 93.7846462548.
In this case you can also round off the result. Here is the result after rounding off, this time to one decimal place - 42.54 kilogram 93.588 pounds.
Now we are going to examples converted with a short version of a formula.
#### How many 42.54 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 93.588 the result in pounds
So 42.54 kg equal to how much lbs? And again, you have to multiply amount of kilogram, in this case 42.54, by 2.2. Have a look: 42.54 * 2.2 = 93.588. So 42.54 kilogram is equal 2.2 pounds.
Do another conversion using this version of a formula. Now convert something from everyday life, for instance, 42.54 kg to lbs weight of strawberries.
So convert - 42.54 kilogram of strawberries * 2.2 = 93.588 pounds of strawberries. So 42.54 kg to pound mass is 93.588.
If you know how much is 42.54 kilogram weight in pounds and are able to calculate it using two different versions of a formula, we can move on. Now we want to show you all results in charts.
#### Convert 42.54 kilogram to pounds
We realize that results shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Thanks to this you can quickly compare 42.54 kg equivalent to lbs outcomes.
Let’s start with a 42.54 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
42.54 93.7846462548 93.5880
And now see 42.54 kg equal pound chart for the second formula:
Kilograms Pounds
42.54 93.588
As you see, after rounding off, when it comes to how much 42.54 kilogram equals pounds, the results are not different. The bigger amount the more considerable difference. Keep it in mind when you need to make bigger number than 42.54 kilograms pounds conversion.
#### How many kilograms 42.54 pound
Now you learned how to calculate 42.54 kilograms how much pounds but we will show you something more. Are you curious what it is? What about 42.54 kilogram to pounds and ounces conversion?
We are going to show you how you can convert it little by little. Let’s begin. How much is 42.54 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 42.54, by 2.20462262. So 42.54 * 2.20462262 = 93.7846462548. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 42.54 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result is 2 pounds and 33 ounces.
As you can see, conversion 42.54 kilogram in pounds and ounces quite simply.
The last conversion which we are going to show you is calculation of 42.54 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate it you need another formula. Before we give you it, let’s see:
• 42.54 kilograms meters = 7.23301385 foot pounds,
• 42.54 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 42.54 foot pounds to kilograms meters you have to multiply 42.54 by 0.13825495. It is exactly 0.13825495. So 42.54 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 42.54 foot pounds is 0.14 kilogram meters.
We hope that this calculation was as easy as 42.54 kilogram into pounds conversions.
This article was a big compendium about kilogram, pound and 42.54 kg to lbs in calculation. Due to this calculation you learned 42.54 kilogram is equivalent to how many pounds.
We showed you not only how to make a conversion 42.54 kilogram to metric pounds but also two other conversions - to check how many 42.54 kg in pounds and ounces and how many 42.54 foot pounds to kilograms meters.
We showed you also another way to do 42.54 kilogram how many pounds calculations, that is with use of 42.54 kg en pound calculator. This will be the best solution for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you are able to do 42.54 kilogram equal to how many pounds calculation - on your own or using our 42.54 kgs to pounds converter.
So what are you waiting for? Convert 42.54 kilogram mass to pounds in the best way for you.
Do you need to do other than 42.54 kilogram as pounds calculation? For example, for 10 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 42.54 kilogram equal many pounds.
### How much is 42.54 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 42.54 kg in pounds , we prepared for you an additional section. Here we have for you all you need to remember about how much is 42.54 kg equal to lbs and how to convert 42.54 kg to lbs . It is down below.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 42.54 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 42.54 kilogram to pounds. The exact result is 93.7846462548 pounds.
There is also another way to calculate how much 42.54 kilogram is equal to pounds with another, easier type of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 42.54 kg equal to how much lbs ? The result is 93.7846462548 lbs.
How to convert 42.54 kg to lbs in an easier way? It is possible to use the 42.54 kg to lbs converter , which will make all calculations for you and you will get a correct 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. | 3,625 | 13,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.942083 |
https://pkg.go.dev/github.com/goulash/stat@v1.0.0/dist | 1,685,817,996,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00429.warc.gz | 517,398,697 | 14,961 | # dist
package
Version: v1.0.0 Latest Latest
Go to latest
Published: Aug 4, 2020 License: MIT
## Documentation ¶
### Overview ¶
Package dist provides statistical probability distributions for random variables.
### Constants ¶
View Source
```const (
NaN = 0x7FF8000000000001
Inf = 0x7FF0000000000000
NegInf = 0xFFF0000000000000
)```
Nabbed these from math...
### Variables ¶
This section is empty.
### Functions ¶
#### func PDF ¶
`func PDF(d DistP, a, b float64) (p float64)`
PDF returns the probability that a value lands between a and b, where a <= b.
### Types ¶
#### type Binomial ¶
```type Binomial struct {
// contains filtered or unexported fields
}```
Binomial distribution with parameter n and p.
TODO.
#### type Continuous ¶
```type Continuous interface {
Float64() float64
}```
#### func Highpass ¶
`func Highpass(c Continuous, low float64) Continuous`
#### func Lowpass ¶
`func Lowpass(c Continuous, high float64) Continuous`
#### func Midpass ¶
`func Midpass(c Continuous, low, high float64) Continuous`
#### type Discrete ¶
```type Discrete interface {
Int63() int64
}```
#### func Ceil ¶
`func Ceil(c Continuous) Discrete`
#### func Floor ¶
`func Floor(c Continuous) Discrete`
#### func Round ¶
`func Round(c Continuous) Discrete`
#### type Dist ¶
```type Dist interface {
DistP
// Q returns the p-quantile of the distribution, this is the inverse CDF function.
Q(p float64) (x float64)
}```
Dist is implemented by all distributions that give the inverse CDF function. Because this is not possible for all distributions, it remains optional.
#### type DistP ¶
```type DistP interface {
// Name of the distribution
String() string
// P returns the probability that a value from the distribution is less than x.
// That is, P represents the cumulative probability function of the distribution.
P(x float64) (p float64)
}```
#### type Exponential ¶
```type Exponential struct {
// contains filtered or unexported fields
}```
Exponential distribution with rate of arrival.
#### func NewExponential ¶
`func NewExponential(s rand.Source, lambda float64) *Exponential`
#### func (*Exponential) Float64 ¶
`func (e *Exponential) Float64() float64`
#### func (*Exponential) Mean ¶
`func (e *Exponential) Mean() float64`
#### func (*Exponential) P ¶
`func (e *Exponential) P(x float64) float64`
#### func (*Exponential) Q ¶
`func (e *Exponential) Q(p float64) float64`
#### func (*Exponential) String ¶
`func (e *Exponential) String() string`
#### type HyperExponential ¶
```type HyperExponential struct {
// contains filtered or unexported fields
}```
HyperExponential distribution with k rates of arrival.
#### func NewHyperExponential ¶
`func NewHyperExponential(s rand.Source, probs, lambdas []float64) *HyperExponential`
#### func (*HyperExponential) Float64 ¶
`func (e *HyperExponential) Float64() float64`
#### func (*HyperExponential) Mean ¶
`func (e *HyperExponential) Mean() float64`
#### func (*HyperExponential) SecondMoment ¶
`func (e *HyperExponential) SecondMoment() float64`
#### func (*HyperExponential) String ¶
`func (e *HyperExponential) String() string`
#### func (*HyperExponential) Var ¶
`func (e *HyperExponential) Var() float64`
#### type LogNormal ¶
```type LogNormal struct {
// contains filtered or unexported fields
}```
LogNormal distribution with mean and standard deviation.
See:
```http://stackoverflow.com/questions/23699738
http://blogs.sas.com/content/iml/2014/06/04/simulate-lognormal-data-with-specified-mean-and-variance.html
```
#### func NewLogNormal ¶
`func NewLogNormal(rs rand.Source, m, s float64) *LogNormal`
#### func (*LogNormal) Float64 ¶
`func (n *LogNormal) Float64() float64`
#### func (*LogNormal) Mean ¶
`func (n *LogNormal) Mean() float64`
#### func (*LogNormal) Std ¶
`func (n *LogNormal) Std() float64`
#### func (*LogNormal) String ¶
`func (n *LogNormal) String() string`
#### func (*LogNormal) Var ¶
`func (n *LogNormal) Var() float64`
#### func (*LogNormal) Z ¶
`func (n *LogNormal) Z(x float64) float64`
#### type Normal ¶
```type Normal struct {
// contains filtered or unexported fields
}```
Normal distribution with mean and standard deviation.
#### func NewNormal ¶
`func NewNormal(s rand.Source, mean, std float64) *Normal`
#### func (*Normal) Float64 ¶
`func (n *Normal) Float64() float64`
#### func (*Normal) Mean ¶
`func (n *Normal) Mean() float64`
#### func (*Normal) Std ¶
`func (n *Normal) Std() float64`
#### func (*Normal) String ¶
`func (n *Normal) String() string`
#### func (*Normal) Var ¶
`func (n *Normal) Var() float64`
#### func (*Normal) Z ¶
`func (n *Normal) Z(x float64) float64`
#### type Null ¶
`type Null struct{}`
#### func NewNull ¶
`func NewNull() *Null`
#### func (Null) Float64 ¶
`func (n Null) Float64() float64`
#### func (Null) Int63 ¶
`func (n Null) Int63() int64`
#### func (Null) Mean ¶
`func (n Null) Mean() float64`
#### func (Null) Std ¶
`func (n Null) Std() float64`
#### func (Null) String ¶
`func (n Null) String() string`
#### func (Null) Var ¶
`func (n Null) Var() float64`
#### type Poisson ¶
```type Poisson struct {
// contains filtered or unexported fields
}```
Poisson returns random numbers according to a poisson distribution.
The poisson distribution models the number of arrivals in a set time interval when the inter-arrival times are exponentially distributed. This is why
#### func NewPoisson ¶
`func NewPoisson(s rand.Source, lambda float64) *Poisson`
#### func (Poisson) Int63 ¶
`func (p Poisson) Int63() int64`
#### func (Poisson) Mean ¶
`func (p Poisson) Mean() float64`
#### func (Poisson) String ¶
`func (p Poisson) String() string`
#### func (Poisson) Var ¶
`func (p Poisson) Var() float64`
#### type Stairs ¶
```type Stairs struct {
// contains filtered or unexported fields
}```
Stairs returns the index of the first probability value that exceeds the random value between 0.0 and 1.0.
For example, given the following list:
```[0.0, 0.3, 0.6, 0.6, 0.9]
```
We can imagine the following stairs, with indices from 0 to 4.
``` .
.____.____|
.____|
.____|
0 0.3 0.6 0.6 0.9
0 1 2 3 4
```
The probability for index 0 and 3 is 0.0. The probability for the rest is is the value minus the previous divided by the last value in the list, in this case 0.9. The indices 1, 2, and 4 all have the same probability then. Therefore, we get the same result if we pass in the list:
```[0, 3, 6, 6, 9]
```
The only requirement on the numbers in the list are that they are monotonically increasing. Failing this requirement will cause NewStairs to panic.
#### func NewStairs ¶
`func NewStairs(s rand.Source, p ...float64) *Stairs`
#### func (*Stairs) Int63 ¶
`func (s *Stairs) Int63() int64`
#### func (*Stairs) Mean ¶
`func (s *Stairs) Mean() float64`
#### func (*Stairs) P ¶
`func (s *Stairs) P(x int64) (p float64)`
#### func (*Stairs) Q ¶
`func (s *Stairs) Q(p float64) (x float64)`
#### func (*Stairs) String ¶
`func (s *Stairs) String() string`
#### type Uniform ¶
```type Uniform struct {
// contains filtered or unexported fields
}```
Uniform gives a uniform distribution between a and b.
#### func NewUniform ¶
`func NewUniform(s rand.Source, a, b float64) *Uniform`
#### func (*Uniform) Float64 ¶
`func (u *Uniform) Float64() float64`
#### func (*Uniform) Mean ¶
`func (u *Uniform) Mean() float64`
#### func (*Uniform) P ¶
`func (u *Uniform) P(x float64) (p float64)`
#### func (*Uniform) Q ¶
`func (u *Uniform) Q(p float64) (x float64)`
#### func (*Uniform) String ¶
`func (u *Uniform) String() string`
#### type UniformDiscrete ¶
```type UniformDiscrete struct {
// contains filtered or unexported fields
}```
UniformDiscrete gives a discrete uniform distribution between a and b.
#### func NewUniformDiscrete ¶
`func NewUniformDiscrete(s rand.Source, a, b int64) *UniformDiscrete`
NewUniformDiscrete returns a discrete uniform distribution in [a, b). Note: this will not work if (b-a) is greater than 2**63.
#### func (*UniformDiscrete) Int63 ¶
`func (u *UniformDiscrete) Int63() int64`
#### func (*UniformDiscrete) Mean ¶
`func (u *UniformDiscrete) Mean() float64`
#### func (*UniformDiscrete) P ¶
`func (u *UniformDiscrete) P(x int64) (p float64)`
#### func (*UniformDiscrete) Q ¶
`func (u *UniformDiscrete) Q(p float64) (x float64)`
#### func (*UniformDiscrete) String ¶
`func (u *UniformDiscrete) String() string` | 2,421 | 8,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.593676 |
https://aviation.stackexchange.com/questions/56222/what-is-compressibility-drag/57430?r=SearchResults&s=2%7C10.9590 | 1,618,242,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00176.warc.gz | 209,688,566 | 42,625 | # What is compressibility drag?
I don't mean wave drag here. What is compressibility drag, which is understood to be a form of miscellaneous drag?
• Are you asking about when the air "stacks up" as the aircraft tries to push it, the air getting "thicker" and causing more drag? Or are you asking about something else? – Ron Beyer Oct 20 '18 at 16:23
• This description of yours about a drag increase due to air compression brings light to the matter. That may be it. I found compressibility drag as a subcategory of miscellaneous drag and was curious as to what that was...since wave drag was classified separately in zero lift drag as well as in drag due to lift; both these concepts are clear to me. Just compressibility drag was a doubt. Could you explain further if there is anything else interesting about it. – Guha.Gubin Oct 20 '18 at 17:46
• A far as I could find, compressibilty drag is wave drag. It makes sense that it is a part of both the zero lift drag and induced drag. – Orbit Oct 22 '18 at 20:03
• @Orbit, yes, that is what I thought as well. Turns out, it is categorized separately & hence, it means something else. I get a feeling that Ron Beyer is right, but would appreciate it if someone could confirm this. – Guha.Gubin Oct 23 '18 at 12:22
Let's do a Gedankenexperiment:
Think of air streaming around a body as flowing inside a stack of flexible tubes. The walls of the tubes are impenetrable, infinitesimally thin and follow the local streamlines faithfully. When the body approaches at subsonic speed, the air in the tubes near that body makes way for it by speeding up: This reduces the needed cross section and lowers static pressure, so the total pressure will stay constant. On the back side of the body the air slows down again and the tubes regain their old cross section and static pressure. Bernoulli in action.
When the speed nears the speed of sound, however, the speeding up is joined by a drop in density. Still, the air near the body speeds up but that will not change the cross section as much as before, because now this speed increase is coupled with a loss of density. The cross section still drops, but not by as much as before. More tubes have to bend away from the body and need the air in them to speed up so the body can squeeze through. More general: A change in body thickness (more precisely: The second derivative of its cross section) will work on more tubes, so its effects do not die down as quickly as in subsonic speed.
At the speed of sound the cross section decrease due to speed changes is exactly balanced by the drop in density, so the same mass of air needs more volume and eats up all the gain from increased speed. Now there is a wall of air which cannot yield facing the approaching body. That is the sound barrier. In reality, the speed around that body does not reach the speed of sound at the same station in all tubes, so there are mildly sub- and supersonic sections which will allow it to squeeze through. Still, drag is much increased and depends heavily on details in the body contour.
At supersonic speed density changes more than speed, so in order to reduce its cross section, the air in the tubes will slow down in order to make way for the body. Since it has no advance warning of the approaching body, it does so in a shock. As a consequence, the cross section of the stream tube can now be reduced because density increases in that slower air past the shock. Static pressure increases also so total pressure can stay constant again. The drag coefficient drops with further increasing Mach number because the density change becomes dominant, allowing the body to squeeze through the air more easily.
This thought experiment was explained in 1951 to researchers at NACA Langley by Adolf Busemann. One person in the audience, a young fellow named Richard Whitcomb, used the insight he gained to formulate the area rule a few weeks later.
• So would you call this wave or compressibility drag? – Daniel Nov 27 '18 at 21:39
• @Daniel: Wave drag is not dying down as speed increases above Mach 1, being caused by the local inclination of the structure. The drag described here is proportional to the second derivative of cross section over the direction of flow, and it peaks around Mach 1. But the boundaries are fuzzy, I admit. It comes down to how you define wave drag precisely and what fraction of the total drag is then called wave drag. – Peter Kämpf Nov 28 '18 at 5:40
• Yeah, right. Some people might say that this is called both wave and compressibility drag. But as I showed in my answer, some might say that this is only wave drag, and compressibility drag is something different. – Daniel Nov 28 '18 at 11:14
• @PeterKämpf: Wave drag could happen anywhere where a shock results. It need not be where a local inclination occurs. It could be on the smooth surface of an airfoil, where the accelearation is so high that Mach 1 is reached, causing a shock. The adverse pressure gradients and the flow separation that may result is wave drag. Area rule supposedly lowers wave drag (as I have learnt it; however, how it does so isn't very clear to me). Your explanation sheds light on this and implies that area rule reduces compressibility drag and not wave drag. Is this what you mean? Appreciate your feedback. – Guha.Gubin Nov 28 '18 at 11:33
• @Guha.Gubin: No. The shocks in subsonic flow cause mostly pressure drag from separation. This is definitely no wave drag, because here a compression shock results (indirectly) in suction. Wave drag is also pressure drag, but caused by overpressure on forward-facing rsp. suction at rear facing surfaces due to compression shocks rsp. expansion fans in supersonic flow. And yes, the area rule is for reducing compressibility drag. – Peter Kämpf Nov 28 '18 at 11:44
Compressibility drag is a type of parasite drag caused by the compression of air ahead of an aircraft traveling at high speed. An aircraft not designed for supersonic flight will experience it as it approaches Mach 1. The effects are noticeable once the aircraft reaches a Mach number of 0.6 to 0.7 and the coefficient of drag rises by 0.005. In subsonic aircraft design it is also considered the limit of normal economic operation of the aircraft.
• however, this sounds like wave drag. – Guha.Gubin Oct 27 '18 at 13:20
• The way I understand it, wave drag is not compressibility drag, it is a component of drag that presents itself due to compressibility drag. – Juan Jimenez Oct 27 '18 at 22:00
Unfortunately, the definition of these two terms is not consistent across literature. Often, they are both used to describe the same effect: the increase in drag due to the presence of shockwaves.
However, sometimes a differentiation is made between the terms, depending on the way that the total drag is decomposed. You may find that compressibility drag is used to describe the increase in drag due to an increase in mach number at constant lift (thus a composition into zero-lift, lift-dependent and compressibility drag), whereas wave drag is used for the drag which is "physically" caused by the presence of shock waves.
In that case, the values can differ. Take for instance a certain flight condition at transonic speed and increase your angle of attack while keeping your mach number constant. Compressibility drag (according to this definition) then stays constant, whereas wave drag will increase. Check out the following document for some more clarification: http://mail.tku.edu.tw/095980/drag.pdf | 1,691 | 7,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-17 | longest | en | 0.962998 |
https://www.tutorialspoint.com/how-to-count-the-number-of-occurrences-per-year-quarter-month-week-in-excel | 1,702,284,198,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00428.warc.gz | 1,161,016,772 | 21,725 | # How to Count the Number of Occurrences Per Year/Quarter/Month/Week in Excel?
In many industries, including finance, sales, and project management, tracking and evaluating data based on time intervals is a typical activity. Excel offers a productive way to handle this difficulty with its robust features and functionalities.
This article will take you step-by-step through the process of setting up your data, using the proper formulas, and producing precise count results, regardless of your level of Excel proficiency. By the end of this tutorial, you will have a firm grasp on how to use Excel's features to effectively count occurrences throughout various time periods. So let's get started and discover how to use Excel to calculate the quantity of events that occur each year, quarter, month, and week!
## Count the Number of Occurrences Per Year/Quarter/Month/Week
Here we will first get the result using the formula, then use the autofill handle to complete the task. So let us see a simple process to know how you can count the number of occurrences per year, quarter, month, or week in Excel.
### Step 1
Consider an Excel sheet where you have a list of dates similar to the below image.
First, to count the occurrences, click on an empty cell and enter the formula as =SUMPRODUCT((MONTH($A$2:$A$24)=F2)*(YEAR($A$2:$A$24)=$E$2)) and click enter to get the first value.
Empty cell > Formula > Enter.
In the formula, A2:A24 is the range of dates, and E2 is a year.
### Step 2
Then, to get all the other values, drag down using the autofill handle from the first value. The final result will be similar to the image below.
This is how you can count the number of occurrences per year, quarter, month, or week in Excel.
## Conclusion
In this tutorial, we have used a simple example to demonstrate how you can count the number of occurrences per year, quarter, month, or week in Excel to highlight a particular set of data.
Updated on: 21-Aug-2023
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https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks/42677465 | 1,606,460,268,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00088.warc.gz | 505,276,268 | 51,648 | # How do you split a list into evenly sized chunks?
I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.
I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.
I was looking for something useful in `itertools` but I couldn't find anything obviously useful. Might've missed it, though.
Related question: What is the most “pythonic” way to iterate over a list in chunks?
• For users that want to avoid an arbitrarily small final chunk, look over at Splitting a list into N parts of approximately equal length – wim Feb 20 at 21:15
• @wim, that solution has been marked as not working since 2017. There was a rounding error causing certain requests to fail. – sanderdatema Jul 24 at 7:32
• @sanderdatema Not the accepted answer. Review the other answers. – wim Jul 24 at 18:25
• @wim Fair enough, but then you might want to add a comment here with a link to the correct answer, because it's not clear from your comment that you didn't mean the accepted answer and the link just points at the main question. I suppose you mean your own comment there? – sanderdatema Jul 24 at 19:42
As per this answer, the top-voted answer leaves a 'runt' at the end. Here's my solution to really get about as evenly-sized chunks as you can, with no runts. It basically tries to pick exactly the fractional spot where it should split the list, but just rounds it off to the nearest integer:
``````from __future__ import division # not needed in Python 3
def n_even_chunks(l, n):
"""Yield n as even chunks as possible from l."""
last = 0
for i in range(1, n+1):
cur = int(round(i * (len(l) / n)))
yield l[last:cur]
last = cur
``````
Demonstration:
``````>>> pprint.pprint(list(n_even_chunks(list(range(100)), 9)))
[[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, 66],
[67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77],
[78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88],
[89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]]
>>> pprint.pprint(list(n_even_chunks(list(range(100)), 11)))
[[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, 66, 67, 68, 69, 70, 71, 72],
[73, 74, 75, 76, 77, 78, 79, 80, 81],
[82, 83, 84, 85, 86, 87, 88, 89, 90],
[91, 92, 93, 94, 95, 96, 97, 98, 99]]
``````
Compare to the top-voted `chunks` answer:
``````>>> pprint.pprint(list(chunks(list(range(100)), 100//9)))
[[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],
[66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76],
[77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87],
[88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98],
[99]]
>>> pprint.pprint(list(chunks(list(range(100)), 100//11)))
[[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, 66, 67, 68, 69, 70, 71],
[72, 73, 74, 75, 76, 77, 78, 79, 80],
[81, 82, 83, 84, 85, 86, 87, 88, 89],
[90, 91, 92, 93, 94, 95, 96, 97, 98],
[99]]
``````
• This solution seems to fail in some situations: - when n > len(l) - for l = [0,1,2,3,4] and n=3 it returns [[0], [1], [2]] instead of [[0,1], [2,3], [4]] – DragonTux Sep 5 '16 at 10:45
• @DragonTux: Ah I wrote the function for Python 3 - it gives `[[0, 1], [2], [3, 4]]`. I added the future import so it works in Python 2 as well – Claudiu Sep 5 '16 at 17:24
• Thanks a lot. I keep forgetting the subtle differences between Python 2 and 3. – DragonTux Sep 9 '16 at 15:18
One more solution
``````def make_chunks(data, chunk_size):
while data:
chunk, data = data[:chunk_size], data[chunk_size:]
yield chunk
>>> for chunk in make_chunks([1, 2, 3, 4, 5, 6, 7], 2):
... print chunk
...
[1, 2]
[3, 4]
[5, 6]
[7]
>>>
``````
I realise this question is old (stumbled over it on Google), but surely something like the following is far simpler and clearer than any of the huge complex suggestions and only uses slicing:
``````def chunker(iterable, chunksize):
for i,c in enumerate(iterable[::chunksize]):
yield iterable[i*chunksize:(i+1)*chunksize]
>>> for chunk in chunker(range(0,100), 10):
... print list(chunk)
...
[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]
... etc ...
``````
letting r be the chunk size and L be the initial list, you can do.
``````chunkL = [ [i for i in L[r*k:r*(k+1)] ] for k in range(len(L)/r)]
``````
Use list comprehensions:
``````l = [1,2,3,4,5,6,7,8,9,10,11,12]
k = 5 #chunk size
print [tuple(l[x:y]) for (x, y) in [(x, x+k) for x in range(0, len(l), k)]]
``````
You could use numpy's array_split function e.g., `np.array_split(np.array(data), 20)` to split into 20 nearly equal size chunks.
To make sure chunks are exactly equal in size use `np.split`.
``````def chunk(lst):
out = []
for x in xrange(2, len(lst) + 1):
if not len(lst) % x:
factor = len(lst) / x
break
while lst:
out.append([lst.pop(0) for x in xrange(factor)])
return out
``````
I wrote a small library expressly for this purpose, available here. The library's `chunked` function is particularly efficient because it's implemented as a generator, so a substantial amount of memory can be saved in certain situations. It also doesn't rely on the slice notation, so any arbitrary iterator can be used.
``````import iterlib
print list(iterlib.chunked(xrange(1, 1000), 10))
# prints [(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), (11, 12, 13, 14, 15, 16, 17, 18, 19, 20), ...]
``````
The answer above (by koffein) has a little problem: the list is always split into an equal number of splits, not equal number of items per partition. This is my version. The "// chs + 1" takes into account that the number of items may not be divideable exactly by the partition size, so the last partition will only be partially filled.
``````# Given 'l' is your list
chs = 12 # Your chunksize
partitioned = [ l[i*chs:(i*chs)+chs] for i in range((len(l) // chs)+1) ]
``````
• But if the chunk size does exactly divide the number of elements then this includes a zero-length list at the end. – Arthur Tacca Dec 3 '18 at 15:34
At this point, I think we need the obligatory anonymous-recursive function.
``````Y = lambda f: (lambda x: x(x))(lambda y: f(lambda *args: y(y)(*args)))
chunks = Y(lambda f: lambda n: [n[0][:n[1]]] + f((n[0][n[1]:], n[1])) if len(n[0]) > 0 else [])
``````
• lambda function are slow. list comprehension would be faster – Sanjay Poongunran Jul 24 '18 at 19:47
• @SanjayPoongunran thanks for you feedback, but this is Python, we're not here for performance (we would write in C), but for readability. – Julien Palard Jul 24 '18 at 22:29
• @JulienPalard Oh yes, readability is what this reply is all about. – Ibolit Oct 16 '18 at 7:44
Here's an idea using itertools.groupby:
``````def chunks(l, n):
c = itertools.count()
return (it for _, it in itertools.groupby(l, lambda x: next(c)//n))
``````
This returns a generator of generators. If you want a list of lists, just replace the last line with
`````` return [list(it) for _, it in itertools.groupby(l, lambda x: next(c)//n)]
``````
Example returning list of lists:
``````>>> chunks('abcdefghij', 4)
[['a', 'b', 'c', 'd'], ['e', 'f', 'g', 'h'], ['i', 'j']]
``````
(So yes, this suffers form the "runt problem", which may or may not be a problem in a given situation.)
• Again this fails if the sub-iterators are not evaluated in order in the generator case. Let c = chunks('abcdefghij', 4) (as generator). Then set i0 = next(c); i1 = next(c); list(i1) //FINE; list(i0) //UHHOH – Peter Gerdes Dec 19 '17 at 10:19
• @PeterGerdes, thank you for noting that omission; I forgot because I always used the groupby generators in order. The documentation does mention this limitation: "Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible." – itub Dec 19 '17 at 16:12
• @PeterGerdes I think this can be solved using enumerate instead, like so: `[[x for _, x in it] for _, it in itertools.groupby(enumerate(l), lambda x: x[0]//n)]` (list(it) is a list of (index, element) pairs due to enumerate) – Yuri Feldman May 13 '19 at 18:18
I don't think I saw this option, so just to add another one :)) :
``````def chunks(iterable, chunk_size):
i = 0;
while i < len(iterable):
yield iterable[i:i+chunk_size]
i += chunk_size
``````
I have one solution below which does work but more important than that solution is a few comments on other approaches. First, a good solution shouldn't require that one loop through the sub-iterators in order. If I run
``````g = paged_iter(list(range(50)), 11))
i0 = next(g)
i1 = next(g)
list(i1)
list(i0)
``````
The appropriate output for the last command is
`````` [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
``````
not
`````` []
``````
As most of the itertools based solutions here return. This isn't just the usual boring restriction about accessing iterators in order. Imagine a consumer trying to clean up poorly entered data which reversed the appropriate order of blocks of 5, i.e., the data looks like [B5, A5, D5, C5] and should look like [A5, B5, C5, D5] (where A5 is just five elements not a sublist). This consumer would look at the claimed behavior of the grouping function and not hesitate to write a loop like
``````i = 0
out = []
for it in paged_iter(data,5)
if (i % 2 == 0):
swapped = it
else:
out += list(it)
out += list(swapped)
i = i + 1
``````
This will produce mysteriously wrong results if you sneakily assume that sub-iterators are always fully used in order. It gets even worse if you want to interleave elements from the chunks.
Second, a decent number of the suggested solutions implicitly rely on the fact that iterators have a deterministic order (they don't e.g. set) and while some of the solutions using islice may be ok it worries me.
Third, the itertools grouper approach works but the recipe relies on internal behavior of the zip_longest (or zip) functions that isn't part of their published behavior. In particular, the grouper function only works because in zip_longest(i0...in) the next function is always called in order next(i0), next(i1), ... next(in) before starting over. As grouper passes n copies of the same iterator object it relies on this behavior.
Finally, while the solution below can be improved if you make the assumption criticized above that sub-iterators are accessed in order and fully perused without this assumption one MUST implicitly (via call chain) or explicitly (via deques or other data structure) store elements for each subiterator somewhere. So don't bother wasting time (as I did) assuming one could get around this with some clever trick.
``````def paged_iter(iterat, n):
itr = iter(iterat)
deq = None
try:
while(True):
deq = collections.deque(maxlen=n)
for q in range(n):
deq.append(next(itr))
yield (i for i in deq)
except StopIteration:
yield (i for i in deq)
``````
python `pydash` package could be a good choice.
``````from pydash.arrays import chunk
ids = ['22', '89', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '1']
chunk_ids = chunk(ids,5)
print(chunk_ids)
# output: [['22', '89', '2', '3', '4'], ['5', '6', '7', '8', '9'], ['10', '11', '1']]
``````
for more checkout pydash chunk list
• neat! and this is what actualy sits under the hood of pydash.arrays.chunk: chunks = int(ceil(len(array) / float(size))) return [array[i * size:(i + 1) * size] for i in range(chunks)] – darkman Mar 27 at 14:47
``````>>> def f(x, n, acc=[]): return f(x[n:], n, acc+[(x[:n])]) if x else acc
>>> f("Hallo Welt", 3)
['Hal', 'lo ', 'Wel', 't']
>>>
``````
If you are into brackets - I picked up a book on Erlang :)
• Works with any iterable
• Inner data is generator object (not a list)
• One liner
```In [259]: get_in_chunks = lambda itr,n: ( (v for _,v in g) for _,g in itertools.groupby(enumerate(itr),lambda (ind,_): ind/n))
In [260]: list(list(x) for x in get_in_chunks(range(30),7))
Out[260]:
[[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]]
```
• g = get_in_chunks(range(30),7); i0=next(g);i1=next(g);list(i1);list(i0); Last evaluation is empty. Hidden requirement about accessing all the sublists in order seems really bad here to me because the goal with these kind of utils is often to shuffle data around in various ways. – Peter Gerdes Dec 19 '17 at 10:30
Like @AaronHall I got here looking for roughly evenly sized chunks. There are different interpretations of that. In my case, if the desired size is N, I would like each group to be of size>=N. Thus, the orphans which are created in most of the above should be redistributed to other groups.
This can be done using:
``````def nChunks(l, n):
""" Yield n successive chunks from l.
Works for lists, pandas dataframes, etc
"""
newn = int(1.0 * len(l) / n + 0.5)
for i in xrange(0, n-1):
yield l[i*newn:i*newn+newn]
yield l[n*newn-newn:]
``````
(from Splitting a list of into N parts of approximately equal length) by simply calling it as nChunks(l,l/n) or nChunks(l,floor(l/n))
• seems to yield some empty chunks (len=26, 10) , or a final very unbalanced chunk (len=26, 11). – idij Nov 27 '14 at 13:11
I have come up to following solution without creation temorary list object, which should work with any iterable object. Please note that this version for Python 2.x:
``````def chunked(iterable, size):
stop = []
it = iter(iterable)
def _next_chunk():
try:
for _ in xrange(size):
yield next(it)
except StopIteration:
stop.append(True)
return
while not stop:
yield _next_chunk()
for it in chunked(xrange(16), 4):
print list(it)
``````
Output:
``````[0, 1, 2, 3]
[4, 5, 6, 7]
[8, 9, 10, 11]
[12, 13, 14, 15]
[]
``````
As you can see if len(iterable) % size == 0 then we have additional empty iterator object. But I do not think that it is big problem.
• What do you think the following code should produce? i=0 – Peter Gerdes Dec 19 '17 at 10:03
• Try only executing list(it) on every other iteration through the loop, i.e. add a counter and check if it 0 mod 2. The expected behavior is to only print every other line of your output. The actual behavior is to print every line. – Peter Gerdes Dec 19 '17 at 10:10
No magic, but simple and correct:
``````def chunks(iterable, n):
"""Yield successive n-sized chunks from iterable."""
values = []
for i, item in enumerate(iterable, 1):
values.append(item)
if i % n == 0:
yield values
values = []
if values:
yield values
``````
Since I had to do something like this, here's my solution given a generator and a batch size:
``````def pop_n_elems_from_generator(g, n):
elems = []
try:
for idx in xrange(0, n):
elems.append(g.next())
return elems
except StopIteration:
return elems
``````
This works in v2/v3, is inlineable, generator-based and uses only the standard library:
``````import itertools
def split_groups(iter_in, group_size):
return ((x for _, x in item) for _, item in itertools.groupby(enumerate(iter_in), key=lambda x: x[0] // group_size))
``````
• Just do a `(list(x) for x in split_groups('abcdefghij', 4))`, then iterate through them: as opposed to many examples here this would work with groups of any size. – Andrey Cizov Feb 24 '18 at 21:55
An abstraction would be
``````l = [1,2,3,4,5,6,7,8,9]
n = 3
outList = []
for i in range(n, len(l) + n, n):
outList.append(l[i-n:i])
print(outList)
``````
This will print:
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
I dislike idea of splitting elements by chunk size, e.g. script can devide 101 to 3 chunks as [50, 50, 1]. For my needs I needed spliting proportionly, and keeping order same. First I wrote my own script, which works fine, and it's very simple. But I've seen later this answer, where script is better than mine, I reccomend it. Here's my script:
``````def proportional_dividing(N, n):
"""
N - length of array (bigger number)
n - number of chunks (smaller number)
output - arr, containing N numbers, diveded roundly to n chunks
"""
arr = []
if N == 0:
return arr
elif n == 0:
arr.append(N)
return arr
r = N // n
for i in range(n-1):
arr.append(r)
arr.append(N-r*(n-1))
last_n = arr[-1]
# last number always will be r <= last_n < 2*r
# when last_n == r it's ok, but when last_n > r ...
if last_n > r:
# ... and if difference too big (bigger than 1), then
if abs(r-last_n) > 1:
#[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 7] # N=29, n=12
# we need to give unnecessary numbers to first elements back
diff = last_n - r
for k in range(diff):
arr[k] += 1
arr[-1] = r
# and we receive [3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2]
return arr
def split_items(items, chunks):
arr = proportional_dividing(len(items), chunks)
splitted = []
for chunk_size in arr:
splitted.append(items[:chunk_size])
items = items[chunk_size:]
print(splitted)
return splitted
items = [1,2,3,4,5,6,7,8,9,10,11]
chunks = 3
split_items(items, chunks)
split_items(['a','b','c','d','e','f','g','h','i','g','k','l', 'm'], 3)
split_items(['a','b','c','d','e','f','g','h','i','g','k','l', 'm', 'n'], 3)
split_items(range(100), 4)
split_items(range(99), 4)
split_items(range(101), 4)
``````
and output:
``````[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]
[['a', 'b', 'c', 'd'], ['e', 'f', 'g', 'h'], ['i', 'g', 'k', 'l', 'm']]
[['a', 'b', 'c', 'd', 'e'], ['f', 'g', 'h', 'i', 'g'], ['k', 'l', 'm', 'n']]
[range(0, 25), range(25, 50), range(50, 75), range(75, 100)]
[range(0, 25), range(25, 50), range(50, 75), range(75, 99)]
[range(0, 25), range(25, 50), range(50, 75), range(75, 101)]
``````
If you don't care about the order:
``````> from itertools import groupby
> batch_no = 3
> data = 'abcdefgh'
> [
[x[1] for x in x[1]]
for x in
groupby(
sorted(
(x[0] % batch_no, x[1])
for x in
enumerate(data)
),
key=lambda x: x[0]
)
]
[['a', 'd', 'g'], ['b', 'e', 'h'], ['c', 'f']]
``````
This solution doesn't generates sets of same size, but distributes values so batches are as big as possible while keeping the number of generated batches.
A generic chunker for any iterable, which gives the user a choice of how to handle a partial chunk at the end.
Tested on Python 3.
`chunker.py`
``````from enum import Enum
class PartialChunkOptions(Enum):
INCLUDE = 0
EXCLUDE = 1
ERROR = 3
class PartialChunkException(Exception):
pass
"""
A chunker yielding n-element lists from an iterable, with various options
on_partial=PartialChunkOptions.INCLUDE (the default):
include the partial chunk as a short (<n) element list
on_partial=PartialChunkOptions.EXCLUDE
do not include the partial chunk
on_partial=PartialChunkOptions.ERROR
raise a RuntimeError if a partial chunk is encountered
"""
on_partial = PartialChunkOptions(on_partial)
iterator = iter(iterable)
while True:
vals = []
for i in range(n):
try:
vals.append(next(iterator))
except StopIteration:
if vals:
if on_partial == PartialChunkOptions.INCLUDE:
yield vals
elif on_partial == PartialChunkOptions.EXCLUDE:
pass
yield vals + [pad] * (n - len(vals))
elif on_partial == PartialChunkOptions.ERROR:
raise PartialChunkException
return
return
yield vals
``````
`test.py`
``````import chunker
chunk_size = 3
for it in (range(100, 107),
range(100, 109)):
print("\nITERABLE TO CHUNK: {}".format(it))
print("CHUNK SIZE: {}".format(chunk_size))
for option in chunker.PartialChunkOptions.__members__.values():
print("\noption {} used".format(option))
try:
for chunk in chunker.chunker(it, chunk_size, on_partial=option):
print(chunk)
except chunker.PartialChunkException:
print("PartialChunkException was raised")
print("")
``````
output of `test.py`
``````
ITERABLE TO CHUNK: range(100, 107)
CHUNK SIZE: 3
option PartialChunkOptions.INCLUDE used
[100, 101, 102]
[103, 104, 105]
[106]
option PartialChunkOptions.EXCLUDE used
[100, 101, 102]
[103, 104, 105]
[100, 101, 102]
[103, 104, 105]
[106, None, None]
option PartialChunkOptions.ERROR used
[100, 101, 102]
[103, 104, 105]
PartialChunkException was raised
ITERABLE TO CHUNK: range(100, 109)
CHUNK SIZE: 3
option PartialChunkOptions.INCLUDE used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
option PartialChunkOptions.EXCLUDE used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
option PartialChunkOptions.ERROR used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
``````
This question reminds me of the Raku (formerly Perl 6) `.comb(n)` method. It breaks up strings into `n`-sized chunks. (There's more to it than that, but I'll leave out the details.)
It's easy enough to implement a similar function in Python3 as a lambda expression:
``````comb = lambda s,n: (s[i:i+n] for i in range(0,len(s),n))
``````
Then you can call it like this:
``````some_list = list(range(0, 20)) # creates a list of 20 elements
generator = comb(some_list, 4) # creates a generator that will generate lists of 4 elements
for sublist in generator:
print(sublist) # prints a sublist of four elements, as it's generated
``````
Of course, you don't have to assign the generator to a variable; you can just loop over it directly like this:
``````for sublist in comb(some_list, 4):
print(sublist) # prints a sublist of four elements, as it's generated
``````
As a bonus, this `comb()` function also operates on strings:
``````list( comb('catdogant', 3) ) # returns ['cat', 'dog', 'ant']
``````
I've created these two fancy one-liners which are efficient and lazy, both input and output are iterables, also they doen't depend on any module:
First one-liner is totally lazy meaning that it returns iterator producing iterators (i.e. each chunk produced is iterator iterating over chunk's elements), this version is good for the case if chunks are very large or elements are produced slowly one by one and should become available immediately as they are produced:
Try it online!
``````chunk_iters = lambda it, n: ((e for i, g in enumerate(((f,), cit)) for j, e in zip(range((1, n - 1)[i]), g)) for cit in (iter(it),) for f in cit)
``````
Second one-liner returns iterator that produces lists. Each list is produced as soon as elements of whole chunk become available through input iterator or if very last element of last chunk is reached. This version should be used if input elements are produced fast or all available immediately. Other wise first more-lazy one-liner version should be used.
Try it online!
``````chunk_lists = lambda it, n: (l for l in ([],) for i, g in enumerate((it, ((),))) for e in g for l in (l[:len(l) % n] + [e][:1 - i],) if (len(l) % n == 0) != i)
``````
Also I provide multi-line version of first `chunk_iters` one-liner, which returns iterator producing another iterators (going through each chunk's elements):
Try it online!
``````def chunk_iters(it, n):
cit = iter(it)
def one_chunk(f):
yield f
for i, e in zip(range(n - 1), cit):
yield e
for f in cit:
yield one_chunk(f)
``````
You can use Dask to split a list into evenly sized chunks. Dask has the added benefit of memory conservation which is best for very large data. For best results you should load your list directly into a dask dataframe to conserve memory if your list is very large. Depending on what exactly you want to do with the lists, Dask has an entire API of functions you can use: http://docs.dask.org/en/latest/dataframe-api.html
``````import pandas as pd
split = 4
my_list = range(100)
df = dd.from_pandas(pd.DataFrame(my_list), npartitions = split)
my_list = [ df.get_partition(n).compute().iloc[:,0].tolist() for n in range(split) ]
# [[1,2,3,..],[26,27,28...],[51,52,53...],[76,77,78...]]
``````
``````import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[range(10, 20),
range(20, 30),
range(30, 40),
range(40, 50),
range(50, 60),
range(60, 70),
range(70, 75)]
``````
Confer this implementation's result with the example usage result of the accepted answer.
Many of the above functions assume that the length of the whole iterable are known up front, or at least are cheap to calculate.
For some streamed objects that would mean loading the full data into memory first (e.g. to download the whole file) to get the length information.
If you however don't know the the full size yet, you can use this code instead:
``````def chunks(iterable, size):
"""
Yield successive chunks from iterable, being `size` long.
https://stackoverflow.com/a/55776536/3423324
:param iterable: The object you want to split into pieces.
:param size: The size each of the resulting pieces should have.
"""
i = 0
while True:
sliced = iterable[i:i + size]
if len(sliced) == 0:
# to suppress stuff like `range(max, max)`.
break
# end if
yield sliced
if len(sliced) < size:
# our slice is not the full length, so we must have passed the end of the iterator
break
# end if
i += size # so we start the next chunk at the right place.
# end while
# end def
``````
This works because the slice command will return less/no elements if you passed the end of an iterable:
``````"abc"[0:2] == 'ab'
"abc"[2:4] == 'c'
"abc"[4:6] == ''
``````
We now use that result of the slice, and calculate the length of that generated chunk. If it is less than what we expect, we know we can end the iteration.
That way the iterator will not be executed unless access.
An old school approach that does not require itertools but still works with arbitrary generators:
``````def chunks(g, n):
"""divide a generator 'g' into small chunks
Yields:
a chunk that has 'n' or less items
"""
n = max(1, n)
buff = []
for item in g:
buff.append(item)
if len(buff) == n:
yield buff
buff = []
if buff:
yield buff
`````` | 8,602 | 26,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-50 | latest | en | 0.94923 |
http://cathyduffyreviews.com/homeschool-reviews-core-curricula/math-supplements/math-supplements-4th-grade/horse-lovers-math/ | 1,508,358,207,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823114.39/warc/CC-MAIN-20171018195607-20171018215607-00390.warc.gz | 64,454,883 | 10,936 | # Horse Lover's Math
Horse Lover’s Math: Understanding Math through Horses might be the perfect way to help a math-averse horse-loving girl grasp the importance and usefulness of math. This 175-page book should be ideal for girls in about fourth or fifth grade. The book teaches about horses as much or more than it teaches math applications, and it’s not just fluff. This is the real deal if a child wants to learn more about horses and their care, competitions, and equipment.
The book is presented in seven sections by horse-related topics rather than by math topics. Students practice various math skills having to do with measurement, addition, subtraction, multiplication, division, fractions, money (purchasing and budgeting), area, perimeter, time, and temperature. However, math skills are applied as they fit into the topics rather than in a progression of difficulty.
Some of the main topics covered in the book are participation in a schooling show, attending horse camp, grooming horses, riding clothes, a horse’s skeleton and conformation, calculating a horse’s height and weight, stable management and record keeping, show jumping, barrel racing, thoroughbred racing, and western trail class. There are many more facets covered, including a few make-and-do projects.
Horse Lover’s Math is designed for students to write their answers directly in the book for most responses, so you will need a copy for each student. The amount of math isn’t overwhelming. If a girl has not yet learned a particular math skill, it might be a good time to introduce it even if she doesn’t try to master it at that point. All of the math is applied through word problems and completion of charts and graphs.
Plentiful black-and-white photos and lovely, hand drawn illustrations are an important feature of the book since they show what is being discussed or are used for math applications.
Parents will appreciate the answer key at the back of the book as well as the summaries of key objectives taught in each section. The summaries might help parents easily document what has been learned.
Author Deborah Stacey lives in Canada, so she has produced two versions of the book: Canadian Metric or Imperial Version. The Imperial Version uses measurement and money systems familiar to homeschoolers in the U.S. The Amazon link below takes you to the Imperial Version even though there is nothing on the page to indicate this.
This book is labeled Level 1, so there might additional volumes in future.
### Pricing Information
All prices are provided for comparison only and are subject to change. Click on prices to verify their accuracy.
\$25
#### Horse Lover's Math Level 1
Get BIG SAVINGS on homeschool curriculum at the Co-op! Enter CATHY as the Referral Code on the sign up form and get 500 FREE SmartPoints
Get a FREE subscription to Cathy's E-Newsletter
### Instant Key
• Suitable For: one-on-one or small group | 592 | 2,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-43 | latest | en | 0.953827 |
https://askit.ro/author/ionut-ciuta/ | 1,723,608,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00457.warc.gz | 80,435,722 | 12,441 | ## Write C++ program to find the largest number among three numbers
A number will be largest if number is greater than both the other numbers.
``````#include <iostream>
using namespace std;
int main() {
float a, b, c;
cin >> a >> b >> c;
if(a >= b && a >= c)
cout << "Largest number: " << a;
if(b >= a && b >= c)
cout << "Largest number: " << b;
if(c >= a && c >= b)
cout << "Largest number: " << c;
return 0;
}
Input: 1 2 3
Largest number: 3``````
[mai mult...]
## Write a program to swap two numbers in C++
We will use a temporary variable to store one of the numbers to perform the swap operation.
[mai mult...]
## Program to Check if a number is even or odd in C++
To Check if a given number is even or odd we simply divide the given number by 2, if the remainder is 0 then it is even otherwise odd.
[mai mult...]
## Write a program for Adding two numbers in C++
To Add two numbers in C++ we will read two numbers a and b from the user then perform add operation to add a and b together to print the addition of two numbers in C++.
[mai mult...]
## C++ Program to Add Two Distances System Using Structures
In this program, a structure `Distance` containing two data members (inch and feet) is declared to store the distance in the inch-feet system.Here, two structure variables d1 and d2 are created to store the distance entered by the user. And, the sum variable stores the sum of the distances.
[mai mult...]
## C++ Program to Add Complex Numbers by Passing Structure to a Function
In this program, two complex numbers entered by the user are stored in the structures num1 and num2.
• These two structures are passed to `addComplexNumbers()` function which calculates the sum and returns the result to the `main()` function.
• This result is stored in the structure complexSum.
• Then, the sign of the imaginary part of the sum is determined and stored in the `char` variable signOfImag.
• If the imaginary part of complexSum is positive, then signofimag is assigned the value `'+'`. Else, it is assigned the value `'-'`
• We then adjust the value of complexSum.imag.
• This code changes complexSum.imag to positive if it is found to be of negative value.
• This is because if it is negative, then printing it along with signofimag will give us two negative signs in the output.
• So, we change the value to positive to avoid sign repetition.
• After this we finally display the sum.
[mai mult...]
## C++ Program to Calculate Difference Between Two Time Period
In this program, user is asked to enter two time periods and these two periods are stored in structure variables t1 and t2 respectively. Then, the `computeTimeDifference()` function calculates the difference between the time periods and the result is displayed on the screen from the `main()` function without returning it (call by reference).
[mai mult...]
## C++ Program to Store and Display Information Using Structure
In this program, a structure, student is created.
This structure has three members: name (string), roll (integer) and marks (float). Then, we created a structure array of size 10 to store information of 10 students. Using for loop, the program takes the information of 10 students from the user and displays it on the screen.
[mai mult...]
## Write a C++ program to find the largest number among three numbers
A number will be largest if number is greater than both the other numbers.
[mai mult...]
## Write a program to swap two numbers in C++
We will use a temporary variable to store one of the numbers to perform the swap operation.
[mai mult...] | 864 | 3,570 | {"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} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.759195 |
https://brainmass.com/business/capital-asset-pricing-model/measuring-risk-and-rates-of-return-273834 | 1,545,060,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828507.84/warc/CC-MAIN-20181217135323-20181217161323-00242.warc.gz | 539,857,664 | 22,112 | Explore BrainMass
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# Measuring risk and rates of return
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
6-10. (Measuring risk and rates of return):
a. Given the holding-period returns shown here, compute the average returns and the standard deviations for the Zemin Corporation and for the market.
MONTH ZEMIN CORP. MARKET
1 6% 4%
2 3 2
3 -1 1
4 -3 -2
5 5 2
6 0 2
b. If Zemin's beta is 1.54 and the risk-free rate is 8 percent, what would be an appropriate required return for an investor owning Zemin? (Note: Because the returns of Zemin Corporation are based on monthly data, you will need to annualize the returns to make them compatible with the risk-free rate. For simplicity, you can convert from monthly to yearly returns by multiplying the average monthly returns by 12.)
c. How does Zemin's historical average return compare with the return you believe to be a fair return, given the firm's systematic risk?
#### Solution Preview
6-10. (Measuring risk and rates of return)
a. Given the holding-period returns shown here, compute the average returns and the standard deviations for the Zemin Corporation and for the market.
MONTH ZEMIN CORP. MARKET
1 6% 4%
2 3% 2%
3 -1% 1%
4 -3% -2%
5 5% 2%
6 0% 2%
Zemin Corp
Month X= X 2 =
1 6% 0.0036
2 3% 0.0009
3 -1% 0.0001
4 -3% 0.0009
5 5% 0.0025
6 0% 0
Total= 10.00% 0.0080
n=no of observations= 6
Average return= 1.67% =10.%/6
variance={ΣX 2 - n(Mean) 2 }/(n-1)= 0.001265 =(0.008-6*0.0167^2)/(6-1)
standard deviation of return (SD)=√Variance= 3.56% =√0.001265
These are monthly returns and standard ...
#### Solution Summary
Compute the average returns and the standard deviations for the Zemin Corporation and for the market. Calculates appropriate required return for an investor owning Zemin using Capital Asset Pricing Model (CAPM).
\$2.19
## Portfolio Return and Standard Deviation, SML and Required Rate of Return, Futures Contracts, Future Margin, S&P 500 index futures contracts, Futures Speculation,Futures Hedging, Call Options, Put Options, Portfolio Performance Measures (Sharpe Measure, Jensen Measure, Treynor Measure)
Portfolio Return and Standard Deviation:
1. You are trading in a market that has only two securities available. Security A has an expected return of 8 percent and a standard deviation of 40 percent. Security B has an expected return of 20 percent and a standard deviation of 120 percent.
a) If you place half of your money in each stock, what is your expected return?
b) If you place 40 percent of your money in A and the remaining 60 percent in B, what is your expected return?
c) If the correlation between the returns of Securities a. and b. above is 0.8
what are the variance and the standard deviation of the returns of each of the two portfolios
you found in parts a. and 1 b. above?
2. You are trading in a market that has only two securities. Security C has an expected return of 6% and a standard deviation of 2.5% while security D has an expected return of 15% and a standard deviation of 8%. The correlation of returns between the two securities is -1.
a) If you place half of the money in each stock, then what is the expected return of
the portfolio?
b) If you place 40% of the money in stock C and the remaining in stock D, then what is the expected return of the portfolio?
c) What is the portfolio standard deviation in b above?
SML and Required Rate of Return:
3. a) A security has a beta of 1.5 when the risk-free rate is 4 percent and the expected return on the market is 12 percent. Calculate the expected return on the security.
b) If the beta on the security in a) increases to 2, what is the new expected return? Why does the expected return increase as the beta increases?
SML and Security Selection:
4. Use the following information to answer the questions below:
STOCK
BETA
CURRENT PRICE
EXPECTED PRICE
EXPECTED DIVIDEND
A -1.50
\$ 40
\$ 32
\$ 0.00
B
0.75
\$ 40
\$ 46
\$ 1.50
C
1.75
\$ 80
\$92
\$ 1.45
Assume that the Return on the Market is 12% and the Risk-free rate is 2%.
a) What are the required rates of return for the three stocks?
b) What are the estimated rates of return for the three stocks?
c) What is your investment strategy concerning the three stocks?
Futures Margin:
5. Zack Wheat has just bought four September 5,000-bushel corn futures contracts at \$1.75 per bushel. The initial margin requirement is 3%. The maintenance margin requirement is 80% of the initial margin requirement.
a) How many dollars in initial margin must Zack put up?
b) If the September price of corn rises to \$1.85, how much equity is in Zack's commodity account? Compute the profit/loss
c) If the September price of corn falls to \$1.70, how much equity is in Zack's commodity account? Compute percentage profit/loss
Futures Speculation:
6. Sleeper Sullivan bought 10 December S&P 500 index futures contracts at 310. If the index rises to 318, what is Sleeper's dollar profit? The multiplier for the S&P 500 index futures contract is 250.
7. The margin requirements on the S&P 500 futures contract are 10%, and the stock index is currently at 1,200. Each contract has a multiplier of 250. How much margin must be put up for each contract sold? If the futures price falls by 1% to 1,188 what will happen to margin account of the investor who holds one contract? What will be the investor's percentage return based on the amount put up as margin?
Futures Hedging:
8. The S&P index is currently is at 1,400. You manage a \$7 million indexed equity portfolio. The S&P 500 futures contract has a multiplier of 250.
a) If you are bearish on the stock market, how many contracts should you sell to fully eliminate your risk over the next 6 months?
b) How would your hedging strategy change if instead of holding an indexed portfolio, you hold a portfolio of socks with a beta of 0.60? How many contracts would now choose to sell? Would your hedged position be riskless?
Stock Options:
Call Options:
9. When the underlying stock trades at \$40, a call with an exercise price of \$40 is priced at \$5. Assume that you buy this call. Calculate your profits/losses for stock prices at expiration of \$20, \$30, \$40, \$50, and \$60.
Put Options:
10. Joe Six-pack bought a put option on Cisco for the month of January 2001 with a strike price of \$50. He paid a premium of \$2.25 per option. If Cisco closes at \$46.50 on the exercise date, then compute the percent of profit/loss.
Portfolio Performance Measures:
11. Given the following:
Portfolio Return Standard Deviation Beta
1 9% 5% 0.80
2 13% 7% 1.10
3 22% 18% 1.25
Market 14% 12% 1.00
Risk-Free Security 10%
Compute:
1. Sharpe Measure
2. Jensen Measure
3. Treynor Measure
And Rank the portfolios by each measure. Are the rankings consistent? Why? Or Why not?
View Full Posting Details | 1,816 | 6,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-51 | longest | en | 0.775093 |
https://www.impetus.no/support/manual/?command=CURVE | 1,709,477,530,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00465.warc.gz | 801,159,728 | 25,520 | #### CURVE
###### Parameters and functions
*CURVE
"Optional title"
cid, $sf_{\mathrm{x}}$, $sf_{\mathrm{y}}$, type${}_{\mathrm{x}}$, type${}_{\mathrm{y}}$
$x_1$, $y_1$
.
$x_{\mathrm{n}}$, $y_{\mathrm{n}}$
##### Parameter definition
VariableDescription
cid Curve identification number
$sf_{\mathrm{x}}$ Scale factor for abscissa values
default: 1
$sf_{\mathrm{y}}$ Scale factor for ordinate values
default: 1
type${}_{\mathrm{x}}$ Abscissa data type
options: time, length, disp, velo, acc, force, stress, strain, pressure, temp, energy, none
type${}_{\mathrm{y}}$ Ordinate data type
options: time, length, disp, velo, acc, force, stress, strain, pressure, temp, energy, none
$x_1$, $y_1$ First abscissa ordinate pair
$x_{\mathrm{n}}$, $y_{\mathrm{n}}$ Last abscissa ordinate pair
##### Description
Definition of a piecewise linear curve. Note that a curve and a function (see FUNCTION) can not have the same ID. Abscissa and ordinate data types only need to be specified if the curve is to be subjected to an automatic unit conversion (see example below).
##### Example
*CURVE
1, 1.0, 1.0
0.0, 0.0
1.0, 2.0
2.0, 8.0
###### Unit system conversion
Conversion of material parameters from MM/TON/S to SI units:
*UNIT_SYSTEM
SI
~convert_from_MMTONS
*MAT_METAL
1, 2.7e-9, 70.0e3, 0.3
1
*CURVE
1, 0, 0, strain, stress
0.0, 200.0
1.0, 280.0
~end_convert
The input above is equivalent to:
*UNIT_SYSTEM
SI
*MAT_METAL
1, 2700.0, 70.0e9, 0.3
1
*CURVE
1
0.0, 200.0e6
1.0, 280.0e6 | 525 | 1,472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-10 | latest | en | 0.355962 |
https://web2.0calc.com/questions/math_4973 | 1,620,839,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989766.27/warc/CC-MAIN-20210512162538-20210512192538-00383.warc.gz | 641,239,397 | 5,355 | +0
# Math
0
31
1
Wilfred has a twenty-cent stamp, a fifty-cent stamp and a one-dollar stamp. How many different postage amounts can he make with these stamps?
Apr 16, 2021
#1
+31646
+1
25
50
100
100 + 25 + 50
25 + 50
25 + 100
50 + 100 seven amounts
Apr 16, 2021 | 101 | 294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | longest | en | 0.872889 |
https://www.physicsforums.com/threads/thevenin-and-norton-equivalent.694279/ | 1,510,981,754,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00051.warc.gz | 836,116,636 | 19,109 | # Thevenin and Norton Equivalent
1. May 29, 2013
### xlu2
1. The problem statement, all variables and given/known data
Find Vt and In
2. Relevant equations
KCL
3. The attempt at a solution
I found Rt to be 5.94 ohms (1/(1/7+1/9)+2)
Then I am stuck.
I know to find Vt (which is equal to Vopen circuit), I have to do a KCL in an open circuit, so my KCL equation is
2+Voc/3=Voc/4
So Voc=-12V=Vt?
If the Voc is correct, then In is just Voc/Rt.
Would you anyone help me with finding Voc?
2. May 29, 2013
### milesyoung
3. May 29, 2013
### xlu2
4. May 29, 2013
### Staff: Mentor
Your KCL to find Voc doesn't look right. If you want to use nodal analysis and make a node where three resistors meet, then you'll need another node, too, at the top of the 4Ω resistor.
There are several different ways to find Voc. For example, you could use the current divider rule to find the current through the path with the 9Ω resistor (and hence the potential across it). Or perhaps first convert the 2A source and 4Ω resistor into their Thevenin equivalent and then use the voltage divider rule to find the potential at the top of the 9Ω resistor...
5. May 29, 2013
### milesyoung
Sorry that was a complete brainfart on my part. In my mind there was a connection between the upper nodes of the 3 and 2 ohm resistors.
6. May 29, 2013
### xlu2
KCL for where the three resistor meet: Voc/2=Voc/9+Voc/3?
KCL for 4 ohms node: 2+Voc/3=Voc/4?
Is the current through 9 ohms 2*9/(1/(1/(4+3)+1/9))? (4 and 3 are in series which are then in parallel with 9)
How do you convert 2A and 4 ohms resitor into their Thevenin equivalent?
7. May 29, 2013
### Staff: Mentor
No, those aren't right. In your equations you've neglected the potentials that lie at the remote ends of the branches. You are summing currents, and the current through a given branch depends upon the potential across that branch.
That means, if a branch (even a single resistor) lies between two nodes, the potential across that branch is equal to the difference in potential of the two nodes. So if there are two nodes A and B with potentials VA and VB, and they have a resistor R between them, then the current through R going from node A to node B is (VA - VB)/R. Note that the potentials of both nodes where R connect are referenced.
No, that doesn't look right. The 4Ω resistor is in parallel with the (3+9)Ω branch.
The same way you would find the Thevenin equivalent for any circuit. Consider the current source and resistor in isolation. What's the open-circuit voltage? What's the Thevenin resistance?
8. May 29, 2013
### xlu2
So (Voc-V1)/2+Voc/9+(Voc-V2)/3+Voc/4=2? But how would I find V1 and V2 if I label V1 to be the node joining the 3 resistors and V2 joining 3 and 4?
For Thevenin equivalent, Voc=2*4=8V and R=4 ohms?
9. May 29, 2013
### Staff: Mentor
No, there's still something wrong with your equations.
Let's redraw the circuit to remove the "Y" artistry, label the nodes, and establish the ground reference:
Note that Voc means Open Circuit voltage. That means NO CURRENT can flow through the 2Ω resistor, and it can be ignored. Voc will be the same potential as V2. Now can you write the two node equations?
You might note that with the 2Ω resistor being open circuited that the second node (V2) is not really required to "solve" the circuit; it's added in the middle of the branch ( combined (3+9)Ω branch) in order to facilitate finding Voc. Otherwise you could solve for the one node voltage V1 and then use it to apply the voltage divider rule to the branch.
Yes, that's right. If you do that you should be able to find the potential across the 9Ω resistor (which is the same as V2 in the diagram above) using the voltage divider rule.
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122 | 1,064 | 3,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-47 | longest | en | 0.940393 |
https://cloud.r-project.org/web/packages/glmtlp/readme/README.html | 1,660,351,813,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00134.warc.gz | 183,910,977 | 5,940 | glmtlp
The goal of glmtlp is to fit generalized linear models with l0, l1 and truncated lasso penalty with a fast procedure.
Installation
You can install the released version of glmtlp from CRAN with:
``install.packages("glmtlp")``
Examples for Gaussian Regression Models
The following are three examples which show you how to use `glmtlp` to fit gaussian regression models:
``````library(glmtlp)
data("gau_data")
colnames(gau_data\$X)[gau_data\$beta != 0]
#> [1] "V1" "V6" "V10" "V15" "V20"``````
``````# Cross-Validation using TLP penalty
cv.fit <- cv.glmtlp(gau_data\$X, gau_data\$y, family = "gaussian", penalty = "tlp", ncores=2)
coef(cv.fit)[abs(coef(cv.fit)) > 0]
#> intercept V1 V6 V10 V15 V20
#> -0.009692127 1.240201054 0.883169910 0.725706771 1.125980744 0.981390567
plot(cv.fit)``````
``````# Single Model Fit using TLP penalty
fit <- glmtlp(gau_data\$X, gau_data\$y, family = "gaussian", penalty = "tlp")
coef(fit, lambda = cv.fit\$lambda.min)
#> intercept V1 V2 V3 V4 V5
#> -0.009692127 1.240201054 0.000000000 0.000000000 0.000000000 0.000000000
#> V6 V7 V8 V9 V10 V11
#> 0.883169910 0.000000000 0.000000000 0.000000000 0.725706771 0.000000000
#> V12 V13 V14 V15 V16 V17
#> 0.000000000 0.000000000 0.000000000 1.125980744 0.000000000 0.000000000
#> V18 V19 V20
#> 0.000000000 0.000000000 0.981390567
predict(fit, X = gau_data\$X[1:5, ], lambda = cv.fit\$lambda.min)
#> [1] 0.1906169 2.2279315 -1.4255560 0.9313560 -2.8151758
plot(fit, xvar = "log_lambda", label = TRUE)``````
``````# Cross-Validation using L0 penalty
cv.fit <- cv.glmtlp(gau_data\$X, gau_data\$y, family = "gaussian", penalty = "l0", ncores=2)
coef(cv.fit)[abs(coef(cv.fit)) > 0]
#> intercept V1 V6 V10 V15 V20
#> -0.009687042 1.240319880 0.883378583 0.725607300 1.125958218 0.981544178
plot(cv.fit)``````
``````# Single Model Fit using L0 penalty
fit <- glmtlp(gau_data\$X, gau_data\$y, family = "gaussian", penalty = "l0")
coef(fit, kappa = cv.fit\$kappa.min)
#> intercept V1 V2 V3 V4 V5
#> -0.009687042 1.240319880 0.000000000 0.000000000 0.000000000 0.000000000
#> V6 V7 V8 V9 V10 V11
#> 0.883378583 0.000000000 0.000000000 0.000000000 0.725607300 0.000000000
#> V12 V13 V14 V15 V16 V17
#> 0.000000000 0.000000000 0.000000000 1.125958218 0.000000000 0.000000000
#> V18 V19 V20
#> 0.000000000 0.000000000 0.981544178
predict(fit, X = gau_data\$X[1:5, ], kappa = cv.fit\$kappa.min)
#> [1] 0.190596 2.228306 -1.425994 0.931749 -2.815322
plot(fit, xvar = "kappa", label = TRUE)``````
``````# Cross-Validation using L1 penalty
cv.fit <- cv.glmtlp(gau_data\$X, gau_data\$y, family = "gaussian", penalty = "l1", ncores=2)
coef(cv.fit)[abs(coef(cv.fit)) > 0]
#> intercept V1 V4 V6 V10 V11
#> -0.03589837 1.09063811 -0.03020625 0.70163818 0.56874685 0.04658280
#> V14 V15 V19 V20
#> 0.01969490 0.96583003 -0.06633302 0.81279715
plot(cv.fit)``````
``````# Single Model Fit using L1 penalty
fit <- glmtlp(gau_data\$X, gau_data\$y, family = "gaussian", penalty = "l1")
coef(fit, lambda = cv.fit\$lambda.min)
#> intercept V1 V2 V3 V4 V5
#> -0.03589837 1.09063811 0.00000000 0.00000000 -0.03020625 0.00000000
#> V6 V7 V8 V9 V10 V11
#> 0.70163818 0.00000000 0.00000000 0.00000000 0.56874685 0.04658280
#> V12 V13 V14 V15 V16 V17
#> 0.00000000 0.00000000 0.01969490 0.96583003 0.00000000 0.00000000
#> V18 V19 V20
#> 0.00000000 -0.06633302 0.81279715
predict(fit, X = gau_data\$X[1:5, ], lambda = cv.fit\$lambda.min)
#> [1] 0.03978789 1.83217872 -0.97812631 0.54083363 -2.01470751
plot(fit, xvar = "lambda", label = TRUE)``````
Examples for Logistic Regression Models
The following are three examples which show you how to use `glmtlp` to fit logistic regression models:
``````data("bin_data")
colnames(bin_data\$X)[bin_data\$beta != 0]
#> [1] "V1" "V6" "V10" "V15" "V20"``````
``````# Cross-Validation using L1 penalty
cv.fit <- cv.glmtlp(bin_data\$X, bin_data\$y, family = "binomial", penalty = "tlp", ncores=2)
coef(cv.fit)[abs(coef(cv.fit)) > 0]
#> intercept V6 V20
#> -0.1347141 0.8256183 0.9940325
plot(cv.fit)
#> Warning: Removed 98 row(s) containing missing values (geom_path).
#> Warning: Removed 98 rows containing missing values (geom_point).``````
``````fit <- glmtlp(bin_data\$X, bin_data\$y, family = "binomial", penalty = "tlp")
coef(fit, lambda = cv.fit\$lambda.min)
#> intercept V1 V2 V3 V4 V5 V6
#> -0.1347141 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.8256183
#> V7 V8 V9 V10 V11 V12 V13
#> 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#> V14 V15 V16 V17 V18 V19 V20
#> 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.9940325
predict(fit, X = bin_data\$X[1:5, ], type = "response", lambda = cv.fit\$lambda.min)
#> [1] 0.42562483 0.89838483 0.09767039 0.90898462 0.20822294
plot(fit, xvar = "log_lambda", label = TRUE)``````
``````# Cross-Validation using L1 penalty
cv.fit <- cv.glmtlp(bin_data\$X, bin_data\$y, family = "binomial", penalty = "l0", ncores=2)
coef(cv.fit)[abs(coef(cv.fit)) > 0]
#> intercept V1 V6 V20
#> -0.07161141 0.82529133 1.00648111 1.10064640
plot(cv.fit)``````
``````fit <- glmtlp(bin_data\$X, bin_data\$y, family = "binomial", penalty = "l0")
coef(fit, kappa = cv.fit\$kappa.min)
#> intercept V1 V2 V3 V4 V5
#> -0.07161141 0.82529133 0.00000000 0.00000000 0.00000000 0.00000000
#> V6 V7 V8 V9 V10 V11
#> 1.00648111 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000
#> V12 V13 V14 V15 V16 V17
#> 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000
#> V18 V19 V20
#> 0.00000000 0.00000000 1.10064640
predict(fit, X = bin_data\$X[1:5, ], kappa = cv.fit\$kappa.min)
#> [1] -0.2806496 2.9558219 -2.3128480 2.9849274 -0.9936015
plot(fit, xvar = "kappa", label = TRUE)``````
``````# Cross-Validation using L1 penalty
cv.fit <- cv.glmtlp(bin_data\$X, bin_data\$y, family = "binomial", penalty = "l1", ncores=2)
coef(cv.fit)[abs(coef(cv.fit)) > 0]
#> intercept V1 V3 V4 V5 V6
#> -0.04593027 0.79539914 0.06262547 0.19102595 0.06793078 1.05865751
#> V8 V9 V10 V11 V13 V15
#> -0.06306159 -0.09101621 0.71050104 0.01838134 0.38267285 0.74771918
#> V19 V20
#> 0.26189535 1.08151085
plot(cv.fit)``````
``````fit <- glmtlp(bin_data\$X, bin_data\$y, family = "binomial", penalty = "l1")
coef(fit, lambda = cv.fit\$lambda.min)
#> intercept V1 V2 V3 V4 V5
#> -0.04593027 0.79539914 0.00000000 0.06262547 0.19102595 0.06793078
#> V6 V7 V8 V9 V10 V11
#> 1.05865751 0.00000000 -0.06306159 -0.09101621 0.71050104 0.01838134
#> V12 V13 V14 V15 V16 V17
#> 0.00000000 0.38267285 0.00000000 0.74771918 0.00000000 0.00000000
#> V18 V19 V20
#> 0.00000000 0.26189535 1.08151085
predict(fit, X = bin_data\$X[1:5, ], type = "response", lambda = cv.fit\$lambda.min)
#> [1] 0.32934063 0.91919949 0.02726264 0.94908173 0.02600642
plot(fit, xvar = "lambda", label = TRUE)``````
References
Shen, X., Pan, W., & Zhu, Y. (2012). Likelihood-based selection and sharp parameter estimation. Journal of the American Statistical Association, 107(497), 223-232. https://doi.org/10.1080/01621459.2011.645783.
Shen, X., Pan, W., Zhu, Y., & Zhou, H. (2013). On constrained and regularized high-dimensional regression. Annals of the Institute of Statistical Mathematics, 65(5), 807-832. https://doi.org/10.1007/s10463-012-0396-3
Li, C., Shen, X., & Pan, W. (2021). Inference for a Large Directed Graphical Model with Interventions. arXiv preprint arXiv:2110.03805. https://arxiv.org/abs/2110.03805
Tibshirani, R., Bien, J., Friedman, J., Hastie, T., Simon, N., Taylor, J., & Tibshirani, R. J. (2012). Strong rules for discarding predictors in lasso‐type problems. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 74(2), 245-266. https://doi.org/10.1111/j.1467-9868.2011.01004.x.
Yang, Yi, and Hui Zou. “A coordinate majorization descent algorithm for l1 penalized learning.” Journal of Statistical Computation and Simulation 84.1 (2014): 84-95. https://doi.org/10.1080/00949655.2012.695374. | 3,672 | 9,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-33 | latest | en | 0.551108 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3856/2/a/n/ | 1,611,002,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00576.warc.gz | 884,571,405 | 59,092 | # Properties
Label 3856.2.a.n Level $3856$ Weight $2$ Character orbit 3856.a Self dual yes Analytic conductor $30.790$ Analytic rank $1$ Dimension $12$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$3856 = 2^{4} \cdot 241$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3856.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$30.7903150194$$ Analytic rank: $$1$$ Dimension: $$12$$ Coefficient field: $$\mathbb{Q}[x]/(x^{12} - \cdots)$$ Defining polynomial: $$x^{12} - 3 x^{11} - 14 x^{10} + 44 x^{9} + 65 x^{8} - 219 x^{7} - 123 x^{6} + 444 x^{5} + 105 x^{4} - 328 x^{3} - 45 x^{2} + 18 x - 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 241) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{11}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -\beta_{8} q^{3} + ( 1 - \beta_{9} ) q^{5} + ( -\beta_{4} + \beta_{5} ) q^{7} + ( 1 - \beta_{5} + \beta_{6} - \beta_{7} + \beta_{9} + \beta_{11} ) q^{9} +O(q^{10})$$ $$q -\beta_{8} q^{3} + ( 1 - \beta_{9} ) q^{5} + ( -\beta_{4} + \beta_{5} ) q^{7} + ( 1 - \beta_{5} + \beta_{6} - \beta_{7} + \beta_{9} + \beta_{11} ) q^{9} + ( -2 + \beta_{1} - \beta_{3} + \beta_{8} ) q^{11} + ( -\beta_{2} + \beta_{5} + \beta_{7} + \beta_{10} - \beta_{11} ) q^{13} + ( -1 + \beta_{2} + \beta_{4} + \beta_{10} - \beta_{11} ) q^{15} + ( -1 - \beta_{3} + \beta_{4} - \beta_{5} + \beta_{8} - \beta_{11} ) q^{17} + ( 1 - \beta_{1} + \beta_{4} + \beta_{6} + \beta_{7} + \beta_{9} + \beta_{10} ) q^{19} + ( -2 - 2 \beta_{1} + 2 \beta_{3} - \beta_{4} - 2 \beta_{6} + \beta_{7} - 2 \beta_{10} + \beta_{11} ) q^{21} + ( -5 + \beta_{1} - \beta_{2} + 3 \beta_{3} - \beta_{4} - 2 \beta_{5} - \beta_{6} - \beta_{9} - 3 \beta_{10} + \beta_{11} ) q^{23} + ( 2 - \beta_{1} + \beta_{2} - \beta_{3} + 2 \beta_{5} - \beta_{6} + \beta_{11} ) q^{25} + ( -1 + 2 \beta_{1} - \beta_{5} - \beta_{6} - 2 \beta_{7} + \beta_{9} - \beta_{10} + \beta_{11} ) q^{27} + ( -1 - \beta_{1} + \beta_{3} - \beta_{6} + \beta_{8} - 2 \beta_{10} - \beta_{11} ) q^{29} + ( -\beta_{1} + \beta_{2} + 3 \beta_{4} - \beta_{6} + \beta_{9} + \beta_{10} ) q^{31} + ( -3 - \beta_{2} + \beta_{3} + \beta_{4} - \beta_{5} + 2 \beta_{8} - \beta_{11} ) q^{33} + ( -2 + \beta_{1} + \beta_{2} - \beta_{5} - 2 \beta_{7} + \beta_{8} ) q^{35} + ( -1 + 2 \beta_{2} + \beta_{4} - \beta_{5} + \beta_{6} - \beta_{7} + \beta_{8} + \beta_{9} + \beta_{10} ) q^{37} + ( -3 - 2 \beta_{1} - \beta_{6} + 3 \beta_{7} + 2 \beta_{8} - \beta_{9} - 2 \beta_{10} ) q^{39} + ( -2 + \beta_{1} + \beta_{2} - 2 \beta_{4} - \beta_{5} - \beta_{6} - \beta_{7} + 2 \beta_{9} + \beta_{11} ) q^{41} + ( -\beta_{1} - \beta_{3} + \beta_{6} + 2 \beta_{7} + \beta_{8} + \beta_{9} ) q^{43} + ( -2 + \beta_{2} - \beta_{5} + 2 \beta_{8} + \beta_{9} + \beta_{11} ) q^{45} + ( -2 - \beta_{2} - \beta_{4} + \beta_{5} - \beta_{8} - 2 \beta_{9} + \beta_{10} - 2 \beta_{11} ) q^{47} + ( 2 - 2 \beta_{2} - 2 \beta_{3} + 2 \beta_{4} + 2 \beta_{5} + 3 \beta_{6} + 2 \beta_{7} - 2 \beta_{8} + 3 \beta_{10} - 2 \beta_{11} ) q^{49} + ( 2 + 2 \beta_{1} - \beta_{2} - 3 \beta_{3} + 2 \beta_{4} + \beta_{5} + 2 \beta_{6} - 2 \beta_{7} + \beta_{8} + 3 \beta_{10} - 2 \beta_{11} ) q^{51} + ( 1 - \beta_{1} + \beta_{4} + \beta_{5} - \beta_{7} + \beta_{9} ) q^{53} + ( -1 + \beta_{1} - 3 \beta_{2} + \beta_{3} - 2 \beta_{4} - \beta_{6} + \beta_{7} + \beta_{9} - 2 \beta_{10} + 2 \beta_{11} ) q^{55} + ( -1 - \beta_{2} + \beta_{3} - \beta_{4} + 2 \beta_{5} - 2 \beta_{6} + 3 \beta_{7} - 2 \beta_{8} - 2 \beta_{9} - \beta_{10} ) q^{57} + ( -4 - 2 \beta_{1} + 3 \beta_{3} - 3 \beta_{5} + \beta_{8} - 3 \beta_{10} + 2 \beta_{11} ) q^{59} + ( -1 + \beta_{3} + \beta_{6} - \beta_{7} - 2 \beta_{8} - 2 \beta_{9} + \beta_{10} - \beta_{11} ) q^{61} + ( 3 - \beta_{2} - \beta_{3} - \beta_{4} + 4 \beta_{5} + \beta_{6} + 3 \beta_{7} - \beta_{9} + 3 \beta_{10} - 2 \beta_{11} ) q^{63} + ( -3 + 2 \beta_{1} - \beta_{2} - \beta_{3} + \beta_{4} - \beta_{5} + \beta_{6} + \beta_{9} + \beta_{10} - \beta_{11} ) q^{65} + ( -3 + \beta_{2} + 2 \beta_{3} - \beta_{4} - 3 \beta_{5} - 2 \beta_{6} - \beta_{7} + \beta_{8} + 2 \beta_{9} + 2 \beta_{11} ) q^{67} + ( 4 \beta_{1} + \beta_{2} - 2 \beta_{3} - 2 \beta_{4} + \beta_{5} + \beta_{6} - \beta_{7} + 2 \beta_{8} + \beta_{9} + 2 \beta_{10} + \beta_{11} ) q^{69} + ( -9 + 2 \beta_{1} - \beta_{2} - \beta_{3} - 2 \beta_{4} + \beta_{5} - \beta_{6} + \beta_{7} + 2 \beta_{8} - \beta_{9} - 2 \beta_{11} ) q^{71} + ( -1 + 2 \beta_{2} - \beta_{3} - \beta_{6} + \beta_{9} + 3 \beta_{11} ) q^{73} + ( 1 - 4 \beta_{1} - \beta_{2} + \beta_{3} + 2 \beta_{5} + 2 \beta_{6} + 2 \beta_{7} + \beta_{10} - 3 \beta_{11} ) q^{75} + ( -2 + \beta_{1} + \beta_{3} + 2 \beta_{4} - 3 \beta_{5} - \beta_{7} + \beta_{8} ) q^{77} + ( 2 + \beta_{2} + 2 \beta_{5} - 2 \beta_{7} + 4 \beta_{10} ) q^{79} + ( -1 + 2 \beta_{1} - 3 \beta_{3} + \beta_{4} - 2 \beta_{5} + 2 \beta_{6} - 2 \beta_{7} + \beta_{8} - 2 \beta_{11} ) q^{81} + ( 1 - 2 \beta_{1} + 3 \beta_{2} - 2 \beta_{3} + 2 \beta_{4} - \beta_{5} + \beta_{6} - 2 \beta_{7} + 2 \beta_{8} + 3 \beta_{9} - \beta_{10} + 3 \beta_{11} ) q^{83} + ( -2 \beta_{1} - 2 \beta_{2} + \beta_{3} - \beta_{4} + \beta_{5} + 3 \beta_{7} - \beta_{8} - \beta_{10} - \beta_{11} ) q^{85} + ( 4 - \beta_{2} - 5 \beta_{3} - \beta_{4} + 5 \beta_{5} + 2 \beta_{6} - \beta_{7} + 3 \beta_{10} - \beta_{11} ) q^{87} + ( -1 - 2 \beta_{2} + 2 \beta_{3} - \beta_{4} - \beta_{5} + \beta_{6} + 3 \beta_{7} - 3 \beta_{9} - 3 \beta_{10} ) q^{89} + ( 2 + 2 \beta_{1} - 2 \beta_{3} - 2 \beta_{4} + \beta_{5} + 2 \beta_{6} - 3 \beta_{7} + 2 \beta_{9} + \beta_{10} + \beta_{11} ) q^{91} + ( 4 - 5 \beta_{3} + \beta_{4} + \beta_{5} + 6 \beta_{6} + \beta_{7} + 3 \beta_{8} - 2 \beta_{9} + 3 \beta_{10} - 3 \beta_{11} ) q^{93} + ( -3 - \beta_{3} + 3 \beta_{4} - 3 \beta_{5} + 3 \beta_{6} + \beta_{9} + 2 \beta_{10} - \beta_{11} ) q^{95} + ( -4 - \beta_{1} + 2 \beta_{2} - \beta_{3} + \beta_{4} - \beta_{5} - \beta_{7} + 3 \beta_{8} + 3 \beta_{9} - \beta_{10} + \beta_{11} ) q^{97} + ( -1 - \beta_{1} + \beta_{2} - 2 \beta_{3} + \beta_{4} + 2 \beta_{5} + \beta_{7} + \beta_{8} - 2 \beta_{9} + \beta_{10} - 2 \beta_{11} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$12q - q^{3} + 6q^{5} - 3q^{7} + 15q^{9} + O(q^{10})$$ $$12q - q^{3} + 6q^{5} - 3q^{7} + 15q^{9} - 22q^{11} - 5q^{13} - 13q^{15} - 4q^{17} + 6q^{19} - 14q^{21} - 32q^{23} + 4q^{25} + 5q^{27} + 6q^{29} - 8q^{31} - 24q^{33} - 15q^{35} - 8q^{37} - 31q^{39} - q^{41} + 2q^{43} - 15q^{45} - 34q^{47} - 9q^{49} + 3q^{51} + 5q^{53} + 3q^{55} - 22q^{57} - 26q^{59} - 26q^{61} + 4q^{63} - 25q^{65} - 6q^{67} - 2q^{69} - 94q^{71} - 22q^{73} - 7q^{77} - 9q^{79} + 4q^{81} + 8q^{83} + 4q^{85} - 4q^{87} - 3q^{89} + 20q^{91} + 12q^{93} - 33q^{95} - 29q^{97} - 36q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{12} - 3 x^{11} - 14 x^{10} + 44 x^{9} + 65 x^{8} - 219 x^{7} - 123 x^{6} + 444 x^{5} + 105 x^{4} - 328 x^{3} - 45 x^{2} + 18 x - 1$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$\nu^{2} - 3$$ $$\beta_{3}$$ $$=$$ $$($$$$\nu^{11} + \nu^{10} - 23 \nu^{9} - 13 \nu^{8} + 184 \nu^{7} + 54 \nu^{6} - 611 \nu^{5} - 94 \nu^{4} + 768 \nu^{3} + 94 \nu^{2} - 213 \nu - 8$$$$)/8$$ $$\beta_{4}$$ $$=$$ $$($$$$-4 \nu^{11} + 9 \nu^{10} + 57 \nu^{9} - 119 \nu^{8} - 273 \nu^{7} + 488 \nu^{6} + 538 \nu^{5} - 663 \nu^{4} - 422 \nu^{3} + 168 \nu^{2} + 34 \nu + 7$$$$)/8$$ $$\beta_{5}$$ $$=$$ $$($$$$7 \nu^{11} - 10 \nu^{10} - 130 \nu^{9} + 172 \nu^{8} + 869 \nu^{7} - 1046 \nu^{6} - 2491 \nu^{5} + 2625 \nu^{4} + 2774 \nu^{3} - 2230 \nu^{2} - 745 \nu + 85$$$$)/16$$ $$\beta_{6}$$ $$=$$ $$($$$$-13 \nu^{11} + 48 \nu^{10} + 160 \nu^{9} - 690 \nu^{8} - 569 \nu^{7} + 3330 \nu^{6} + 597 \nu^{5} - 6481 \nu^{4} - 470 \nu^{3} + 4634 \nu^{2} + 951 \nu - 169$$$$)/16$$ $$\beta_{7}$$ $$=$$ $$($$$$13 \nu^{11} - 44 \nu^{10} - 180 \nu^{9} + 662 \nu^{8} + 829 \nu^{7} - 3426 \nu^{6} - 1629 \nu^{5} + 7333 \nu^{4} + 1798 \nu^{3} - 5698 \nu^{2} - 1335 \nu + 141$$$$)/16$$ $$\beta_{8}$$ $$=$$ $$($$$$11 \nu^{11} - 30 \nu^{10} - 158 \nu^{9} + 432 \nu^{8} + 773 \nu^{7} - 2086 \nu^{6} - 1631 \nu^{5} + 4025 \nu^{4} + 1654 \nu^{3} - 2750 \nu^{2} - 741 \nu + 93$$$$)/8$$ $$\beta_{9}$$ $$=$$ $$($$$$-11 \nu^{11} + 34 \nu^{10} + 150 \nu^{9} - 492 \nu^{8} - 669 \nu^{7} + 2398 \nu^{6} + 1223 \nu^{5} - 4717 \nu^{4} - 1178 \nu^{3} + 3354 \nu^{2} + 737 \nu - 109$$$$)/8$$ $$\beta_{10}$$ $$=$$ $$($$$$25 \nu^{11} - 76 \nu^{10} - 340 \nu^{9} + 1086 \nu^{8} + 1513 \nu^{7} - 5178 \nu^{6} - 2777 \nu^{5} + 9809 \nu^{4} + 2718 \nu^{3} - 6602 \nu^{2} - 1643 \nu + 201$$$$)/16$$ $$\beta_{11}$$ $$=$$ $$($$$$31 \nu^{11} - 90 \nu^{10} - 450 \nu^{9} + 1340 \nu^{8} + 2237 \nu^{7} - 6822 \nu^{6} - 4819 \nu^{5} + 14249 \nu^{4} + 5014 \nu^{3} - 10758 \nu^{2} - 2497 \nu + 381$$$$)/16$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$\beta_{2} + 3$$ $$\nu^{3}$$ $$=$$ $$\beta_{11} - \beta_{10} + \beta_{9} - \beta_{6} - \beta_{5} - \beta_{4} + \beta_{3} + 5 \beta_{1} - 1$$ $$\nu^{4}$$ $$=$$ $$2 \beta_{11} + \beta_{9} - \beta_{8} - 2 \beta_{7} - \beta_{6} - \beta_{5} + \beta_{3} + 8 \beta_{2} + 15$$ $$\nu^{5}$$ $$=$$ $$11 \beta_{11} - 9 \beta_{10} + 10 \beta_{9} - \beta_{7} - 8 \beta_{6} - 11 \beta_{5} - 9 \beta_{4} + 9 \beta_{3} + \beta_{2} + 29 \beta_{1} - 10$$ $$\nu^{6}$$ $$=$$ $$20 \beta_{11} + 2 \beta_{10} + 10 \beta_{9} - 14 \beta_{8} - 19 \beta_{7} - 10 \beta_{6} - 9 \beta_{5} - 2 \beta_{4} + 11 \beta_{3} + 56 \beta_{2} + 88$$ $$\nu^{7}$$ $$=$$ $$93 \beta_{11} - 67 \beta_{10} + 78 \beta_{9} - 3 \beta_{8} - 12 \beta_{7} - 54 \beta_{6} - 94 \beta_{5} - 68 \beta_{4} + 71 \beta_{3} + 11 \beta_{2} + 181 \beta_{1} - 77$$ $$\nu^{8}$$ $$=$$ $$163 \beta_{11} + 24 \beta_{10} + 76 \beta_{9} - 138 \beta_{8} - 146 \beta_{7} - 78 \beta_{6} - 71 \beta_{5} - 30 \beta_{4} + 98 \beta_{3} + 379 \beta_{2} + 4 \beta_{1} + 558$$ $$\nu^{9}$$ $$=$$ $$720 \beta_{11} - 478 \beta_{10} + 557 \beta_{9} - 59 \beta_{8} - 106 \beta_{7} - 355 \beta_{6} - 733 \beta_{5} - 499 \beta_{4} + 542 \beta_{3} + 88 \beta_{2} + 1179 \beta_{1} - 544$$ $$\nu^{10}$$ $$=$$ $$1256 \beta_{11} + 191 \beta_{10} + 522 \beta_{9} - 1189 \beta_{8} - 1056 \beta_{7} - 566 \beta_{6} - 561 \beta_{5} - 323 \beta_{4} + 822 \beta_{3} + 2542 \beta_{2} + 76 \beta_{1} + 3665$$ $$\nu^{11}$$ $$=$$ $$5372 \beta_{11} - 3384 \beta_{10} + 3821 \beta_{9} - 748 \beta_{8} - 845 \beta_{7} - 2351 \beta_{6} - 5486 \beta_{5} - 3655 \beta_{4} + 4093 \beta_{3} + 630 \beta_{2} + 7881 \beta_{1} - 3713$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−2.02418 1.54879 −0.342147 0.0822506 2.49073 2.01020 −1.28632 1.63125 −2.59703 −1.32986 2.70063 0.115670
0 −2.93498 0 1.44091 0 −0.381245 0 5.61411 0
1.2 0 −2.81087 0 0.334961 0 4.24623 0 4.90098 0
1.3 0 −2.18519 0 −0.548903 0 −1.82459 0 1.77508 0
1.4 0 −1.81824 0 4.31963 0 −0.690569 0 0.306010 0
1.5 0 −1.22208 0 −3.14843 0 −0.136122 0 −1.50653 0
1.6 0 −0.500591 0 1.92585 0 0.852319 0 −2.74941 0
1.7 0 0.126224 0 0.612768 0 −1.03110 0 −2.98407 0
1.8 0 1.16790 0 1.75438 0 −5.06139 0 −1.63601 0
1.9 0 1.20534 0 3.49051 0 0.744578 0 −1.54716 0
1.10 0 2.18147 0 −3.40432 0 3.83334 0 1.75880 0
1.11 0 2.50808 0 0.533570 0 −0.354992 0 3.29045 0
1.12 0 3.28295 0 −1.31091 0 −3.19647 0 7.77775 0
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1.12 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$-1$$
$$241$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3856.2.a.n 12
4.b odd 2 1 241.2.a.b 12
12.b even 2 1 2169.2.a.h 12
20.d odd 2 1 6025.2.a.h 12
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
241.2.a.b 12 4.b odd 2 1
2169.2.a.h 12 12.b even 2 1
3856.2.a.n 12 1.a even 1 1 trivial
6025.2.a.h 12 20.d odd 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(3856))$$:
$$T_{3}^{12} + \cdots$$ $$T_{5}^{12} - \cdots$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{12}$$
$3$ $$64 - 400 T - 992 T^{2} + 960 T^{3} + 1540 T^{4} - 725 T^{5} - 888 T^{6} + 210 T^{7} + 224 T^{8} - 25 T^{9} - 25 T^{10} + T^{11} + T^{12}$$
$5$ $$62 - 347 T + 339 T^{2} + 1071 T^{3} - 2193 T^{4} + 497 T^{5} + 1301 T^{6} - 797 T^{7} - 68 T^{8} + 134 T^{9} - 14 T^{10} - 6 T^{11} + T^{12}$$
$7$ $$4 + 53 T + 200 T^{2} + 131 T^{3} - 588 T^{4} - 855 T^{5} + 263 T^{6} + 854 T^{7} + 245 T^{8} - 96 T^{9} - 33 T^{10} + 3 T^{11} + T^{12}$$
$11$ $$128 + 1460 T + 5900 T^{2} + 9672 T^{3} + 3100 T^{4} - 9811 T^{5} - 12739 T^{6} - 5545 T^{7} - 215 T^{8} + 553 T^{9} + 177 T^{10} + 22 T^{11} + T^{12}$$
$13$ $$-52672 - 441248 T - 90512 T^{2} + 288472 T^{3} + 69802 T^{4} - 64645 T^{5} - 15049 T^{6} + 6470 T^{7} + 1425 T^{8} - 296 T^{9} - 62 T^{10} + 5 T^{11} + T^{12}$$
$17$ $$154144 - 302576 T - 261264 T^{2} + 295792 T^{3} + 159474 T^{4} - 87027 T^{5} - 35221 T^{6} + 9972 T^{7} + 2997 T^{8} - 370 T^{9} - 97 T^{10} + 4 T^{11} + T^{12}$$
$19$ $$-3556280 + 3617301 T + 903711 T^{2} - 1614089 T^{3} + 58969 T^{4} + 247081 T^{5} - 28947 T^{6} - 16891 T^{7} + 2538 T^{8} + 524 T^{9} - 86 T^{10} - 6 T^{11} + T^{12}$$
$23$ $$-116949436 - 276824423 T - 182523158 T^{2} - 31479187 T^{3} + 11455889 T^{4} + 5191210 T^{5} + 372702 T^{6} - 138011 T^{7} - 28306 T^{8} - 627 T^{9} + 304 T^{10} + 32 T^{11} + T^{12}$$
$29$ $$58109390 - 179077009 T + 164527164 T^{2} - 50865568 T^{3} - 3169769 T^{4} + 4236578 T^{5} - 355685 T^{6} - 116722 T^{7} + 15216 T^{8} + 1375 T^{9} - 213 T^{10} - 6 T^{11} + T^{12}$$
$31$ $$-318193616 - 468437780 T + 77270368 T^{2} + 108211396 T^{3} + 841016 T^{4} - 8192365 T^{5} - 688338 T^{6} + 208450 T^{7} + 22930 T^{8} - 2167 T^{9} - 262 T^{10} + 8 T^{11} + T^{12}$$
$37$ $$50796928 - 18033984 T - 29177888 T^{2} + 5067040 T^{3} + 4698614 T^{4} - 619307 T^{5} - 323567 T^{6} + 36249 T^{7} + 10466 T^{8} - 928 T^{9} - 159 T^{10} + 8 T^{11} + T^{12}$$
$41$ $$-63338 - 4034251 T - 930334 T^{2} + 9341574 T^{3} - 976432 T^{4} - 2920817 T^{5} - 471938 T^{6} + 92737 T^{7} + 21111 T^{8} - 708 T^{9} - 262 T^{10} + T^{11} + T^{12}$$
$43$ $$12503272 + 1488169 T - 25360675 T^{2} + 4519982 T^{3} + 6500883 T^{4} - 657869 T^{5} - 569920 T^{6} + 18272 T^{7} + 18808 T^{8} + 26 T^{9} - 237 T^{10} - 2 T^{11} + T^{12}$$
$47$ $$53297792 - 37689968 T - 69239488 T^{2} - 11362328 T^{3} + 11628792 T^{4} + 4772881 T^{5} + 233524 T^{6} - 179952 T^{7} - 34508 T^{8} - 851 T^{9} + 332 T^{10} + 34 T^{11} + T^{12}$$
$53$ $$-3014 + 298853 T + 6874728 T^{2} - 11787807 T^{3} + 1527793 T^{4} + 1538756 T^{5} - 270515 T^{6} - 65448 T^{7} + 12170 T^{8} + 1019 T^{9} - 195 T^{10} - 5 T^{11} + T^{12}$$
$59$ $$-25476160 - 96501648 T - 59605584 T^{2} + 36571744 T^{3} + 47289076 T^{4} + 17302787 T^{5} + 2412075 T^{6} - 67258 T^{7} - 57444 T^{8} - 5338 T^{9} + 22 T^{10} + 26 T^{11} + T^{12}$$
$61$ $$10893274 - 29191311 T + 16419914 T^{2} + 10617016 T^{3} - 8091392 T^{4} - 1560634 T^{5} + 801476 T^{6} + 117122 T^{7} - 22505 T^{8} - 3955 T^{9} + 20 T^{10} + 26 T^{11} + T^{12}$$
$67$ $$4538509504 - 4714883120 T - 678403520 T^{2} + 669419464 T^{3} + 75258180 T^{4} - 29459511 T^{5} - 3555172 T^{6} + 474596 T^{7} + 62307 T^{8} - 2947 T^{9} - 429 T^{10} + 6 T^{11} + T^{12}$$
$71$ $$-12017198348 - 39886445545 T - 42976926619 T^{2} - 21246049133 T^{3} - 5606812300 T^{4} - 801669175 T^{5} - 46063490 T^{6} + 3647013 T^{7} + 918997 T^{8} + 79770 T^{9} + 3737 T^{10} + 94 T^{11} + T^{12}$$
$73$ $$2219968 + 25741920 T + 64034288 T^{2} - 76860088 T^{3} - 83908922 T^{4} - 7168229 T^{5} + 3607447 T^{6} + 533877 T^{7} - 28097 T^{8} - 7860 T^{9} - 208 T^{10} + 22 T^{11} + T^{12}$$
$79$ $$-1277319040 - 3418562576 T + 1017318496 T^{2} + 2331826216 T^{3} + 163831840 T^{4} - 90986869 T^{5} - 7904508 T^{6} + 1163461 T^{7} + 109307 T^{8} - 5783 T^{9} - 581 T^{10} + 9 T^{11} + T^{12}$$
$83$ $$98860915136 + 10813065520 T - 10895951696 T^{2} - 1151271176 T^{3} + 452061900 T^{4} + 42544271 T^{5} - 9197429 T^{6} - 669374 T^{7} + 100342 T^{8} + 4386 T^{9} - 548 T^{10} - 8 T^{11} + T^{12}$$
$89$ $$-1500609440 - 3427797584 T - 829552176 T^{2} + 490376448 T^{3} + 124536598 T^{4} - 21479633 T^{5} - 5233031 T^{6} + 305681 T^{7} + 79002 T^{8} - 1663 T^{9} - 479 T^{10} + 3 T^{11} + T^{12}$$
$97$ $$107861318 - 194920563 T - 162786822 T^{2} + 64432913 T^{3} + 85302025 T^{4} + 27963948 T^{5} + 3390903 T^{6} - 118070 T^{7} - 70766 T^{8} - 5577 T^{9} + 85 T^{10} + 29 T^{11} + T^{12}$$ | 8,963 | 17,088 | {"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} | 2.5625 | 3 | CC-MAIN-2021-04 | latest | en | 0.396743 |
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# Show the steps to solve each equation. Then use your calculator to verify your solution
ISBN: 9781559537636 383
## Solution for problem 12 Chapter 4.2
Discovering Algebra: An Investigative Approach | 2nd Edition
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Problem 12
Show the steps to solve each equation. Then use your calculator to verify your solution. a. 8 12m = 17 b. 2r + 7 = 24 c. 6 3w = 42
Step-by-Step Solution:
Step 1 of 3
Math 113 August 29th, 2018 In class notes We start with a true or false statement: The sum of any two odd counting numbers will be divisible by four. After this statement, we can test out the statement: 5+7=12 Twelve is divisible by four, so that’s true. Let’s try some others. 9+3=12 Once again twelve is divisible by four. 13+3=16 Sixteen divided by four is four so that works. 21+11=32 Thirty-two divided by four is eight. It seems to be that this statement would be true, however if you provide your own examples you may discover some of these outcomes: 3+3=6 Six is not divisible by four so this makes the statement false. 49+1=50 Fifty is not divisible by four, so once again the statement is proven wro
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Step 3 of 3
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# ncverilog simulation overflow
Last post Mon, Jun 25 2012 9:19 AM by Andrew Beckett. 2 replies.
Started by dzkxybx 24 Jun 2012 09:15 PM. Topic has 2 replies and 1504 views
• #### Sun, Jun 24 2012 9:15 PM
• dzkxybx
• Joined on Mon, Jun 25 2012
• Posts 1
• Points 20
ncverilog simulation overflow
hi, i am new here. I have a problem as below when i do my design simulation: 4038444601 OverflowSimulation stopped via \$stop(1) at time 40384446008 PS + 2how could this happens? thanks!and my timescale is 1ns/10ps
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• Post Points: 20
• #### Mon, Jun 25 2012 5:18 AM
• Quek
• Joined on Wed, Oct 14 2009
• Singapore, 00-SG
• Posts 1,063
• Points 16,155
Re: ncverilog simulation overflow
Hi dzkxybxFrom the limited info which you had provided, it is very hard to guess the cause of the problem. If you are not using the latest version of ncverilog, would you please re-try using the latest version and see if the same problem occurs?Best regardsQuek
• Post Points: 20
• #### Mon, Jun 25 2012 9:19 AM
Re: ncverilog simulation overflow
The limit for some considerable time has been about 9223 seconds - and you're hitting a limit much less than that. Are you using some prehistoric version of ncverilog? What does "ncverilog -version" say?e.g. simulating "ncverilog testmax.v" where testmax.v is: `timescale 1ps/1fsmodule top;reg a;initial a=1;always #100000000000 begin a=~a; \$strobe(\$time,,"a=%b",a);endendmoduleGives:... 9223368439474176 a=0 9223369655226368 a=1ncsim: *E,TRTOVF: simulated time overflow. File: ./testmax.v, line = 9, pos = 7 Scope: top Time: 9223370870978560 PS + 0./testmax.v:9 always #100000000000 begin
• Post Points: 5 | 637 | 2,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-35 | latest | en | 0.748586 |
http://stackoverflow.com/questions/12695315/algorithm-for-custom-datepicker/12696067 | 1,394,279,740,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999654396/warc/CC-MAIN-20140305060734-00096-ip-10-183-142-35.ec2.internal.warc.gz | 175,956,107 | 15,960 | # algorithm for custom DatePicker
I'm working on a custom `UIPickerView` that will show times. I need them to match what the website shows below.
The time will start at `x` but for now that is midnight (it can be anytime). It will then increment intervals of `i`. For the below image `i = 7`.
The problem I am running into is when I change the hour, for example 1am, the minutes will need to change as well so that the user can only pick a minute for that hour, that is in increments of `i`, but I'm not sure the correct algorithm to run.
For now I am using `60 % i == 0` then use the `UIDatePicker` and setting `minuteInterval` to `i`, but if `60 % i != 0` that is where I need to use this customer picker.
Can anyone help me with an algorithm that would provide me the number of rows for the hours, minutes and as well populate the hours and minutes based on the interval?
-
I can't tell you anything about objective-c, or ios, or uipickerview... For the algorithm. Here is some pseudo code in no language in particular:
``````i = 7 // this is i in your example
offset=420 // This is where you could say you want
// to start at 7:00 (420 minutes after midnight)
// you could use any number here obviously.
counter = offset
while (counter<1440+offset){
counter = counter+i
minutes = counter%60
hours = counter/60 // Use integer division
}
``````
The modulous gives you your minutes every time by giving you the remainder.. So in your example minutes would go from 56 => 63 => 3 which is what you want.
if you just want to know what the minutes and hours will be for a given interval (suppose the intervals go from 0 to 206 (there are 206 7 minute intervals in a day) then you would do the following:
``````minutes = (i*intervalNum) % 60
hours = (i*intervalNum) / 60 // Integer division!
``````
If you need the number of intervals in a time period you could do the following:
``````keep a hashmap (again I don't know how objective-c handles these) that maps an hour to a list of the minutes that mark intervals. In your example you could have a hashmap that looks like:
{0 => (0,7,14,21,28,35,42,49,56),
1 => (3,10,17,24,31,38,45,52,59),
2 => (6,13,20,27,34,41,48,55),
...
}
``````
You would build this map during the first round through. To get the number of intervals in the 2-3 timeslot just return map[2].length (or whatever its equivalent).
Obviously this is one of MANY ways to do this and one would need much more context to go into any further detail.
Good Luck
-
this definitly gets me pointed in the right direction. One thing though is how would I get the number of minute intervals for a given period? In my screenshot they both have 8 - 7 minute intervals, except the last one has 7 - 7 minute intervals. – Bot Oct 2 '12 at 19:33
See my edit above. – ajon Oct 2 '12 at 19:51
Thanks to ajon I came up with this code
``````NSInteger counter = startTime;
while (counter + minuteInterval < endTime) {
counter = counter + minuteInterval;
NSLog(@"Hours: %d Minutes: %d", counter / 60, counter % 60);
}
``````
I had to do a little bit of modification but am posting my code incase anyone else runs into this issue. I am marking his as the answer since it provided the base for this code.
- | 855 | 3,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.928201 |
https://www.physicsforums.com/threads/time-distance-speed-conversion-problem.384116/ | 1,591,434,380,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00472.warc.gz | 830,435,841 | 16,143 | # Time = distance/speed conversion problem
## Homework Statement
The velocity of light is 1.86 x 108 mi/hr. The distance of Mercury from the Sun is approximately 5.78 x 107 km. How many minutes will it take for light from the Sun to travel to Mercury?
## Homework Equations
1 mi = 1.609 km
time = distance/speed
60 min/1 hr
## The Attempt at a Solution
1.609 km/1 mi * 1.86 x 108 mi/hr = 2.99 x 108 km/hr
time = 5.78 x 107 km / 2.99 x 108 km/hr = 1.933 x 10-1 hr
60 min/1 hr * 1.933 x 10-1 hr = 115.98 x 10-1 min
That answer does not make sense to me. What did I do wrong here?
Related Precalculus Mathematics Homework Help News on Phys.org
rock.freak667
Homework Helper
I think you converted km to mi incorrectly.
1 kilometer = 0.621371192 miles
HallsofIvy
Homework Helper
I think you converted km to mi incorrectly.
1 kilometer = 0.621371192 miles
Yes, so one mile is 1/.621371192 km= 1.60934400 miles which is what he used. And he was converting from miles to kilometers.
However, Mathnomolous, the speed of light is 1.86 x 108 miles per second, not miles per hour!
You should have divided by 60 to convert second to minutes, not multiplied to convert hours to minutes.
Last edited by a moderator:
Your speed of light is incorrect. It's not 2.99 x 108 km/hr, it's 2.99 x 108 m/s, which is something like 6.71 x 108 mi/hr or 186,000 miles per second.
Last edited:
I typed the problem as it appears on my chemistry textbook. I also asked the professor to do the problem in class, with the same result. I never imagined the textbook had that kind of error. Thanks for the correction. I should be able to obtain the right answer now. | 477 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-24 | longest | en | 0.917732 |
http://www.askiitians.com/mock-questions-papers/set-a-paper1/mathematics/ | 1,513,580,100,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948609934.85/warc/CC-MAIN-20171218063927-20171218085927-00491.warc.gz | 319,774,103 | 32,395 | Click to Chat
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This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
1. The order and degree of the differential equations of the family of hyperbolas centred at origin; principal axis along co-ordinate axis and of length √(a2+λ) and √(b2+λ) where a, b are fixed real numbers and λ is a real parameter is
(A) 1, 1 (B) 2, 1
(C) 1, 2 (D) 3, 2
8. Let f: [k, k + 1, …, 2007] ® [1, 2, 3, …, n] be defined by f(x) = [ 2007 / x ] (where [.] denotes the greatest integer function). The maximum value of k such that it is not possible to make f an onto function for any vale of n is
(A) 50 (B) 40
(C) 51 (D) none of these
9. Let f and g be continuous and differentiable functions. If f(0) = f(2) = f(4); f(1) + f(3) = 0; g(0) = g(2) = g(4) = 0 and if f(x) = 0 and g’(x) = 0 do not have a common root, then the minimum number of zeros of f’(x)g’(x) + f(x) g”(x) in [0, 4] is
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USE CODE: CHEM20 | 607 | 1,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-51 | latest | en | 0.519834 |
https://www.ibpsguide.com/ssc-quantitative-aptitude-questions-day-27/ | 1,627,967,408,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00283.warc.gz | 811,685,292 | 73,576 | # SSC Quantitative Aptitude Questions (Day-27)
Dear Aspirants, you can find the Quantitative Aptitude questions with detailed explanations for the SSC exams. Nowadays the competitive level of the exam has been increasing consistently. Due to the great demand for the government job, the level of the toughness reached greater. Candidates have to enhance the preparation process in order to drive in the right path. It doesn’t need to clear the prescribed cutoff. You must have to score good marks more than the cut off marks to get into the final provisional list. Here we have updating the Quantitative Aptitude questions with detailed explanations on a daily basis. You can practice with us and measure your level of preparation. According to that you can sculpt yourself in a proper way. SSC aspirants kindly make use of it and grab your success in your career.
Start Quiz
1) If sec θ= 13/12 and  is acute, then what is the value of (√(cotθ+tanθ))?
(a) 13/(2√15)
(b) 12/(2√13)
(c) 13/(2√5)
(d) 2/13
2) If (1+tan2 θ) = 625/49  and θ is acute, then what is the value of (√(sinθ+cosθ))?
(a) 1
(b) 5/4 √(31/42)
(c)
(d) 5/7
3) If sin θ + sin2 θ = 1, then what is the value of (cos12 θ + 3 cos10 θ + 3 cos8 θ + cos6 θ – 1)?
(a) -1
(b) 0
(c) 1
(d) 2
4) If cot θ = √11 and θ is acute, then what is the value of (cosec2 θ + sec2 θ) /(cosec2 θ – sec2 θ)
(a) 2/3
(b) -6/5
(c) 3/4
(d) 7/6
5) If (cos θ/(1+sin  θ)) + (cos θ/(1- sin  θ))=2 and  is acute, then what is the value (in degrees) of θ?
(a) 30
(b) 45
(c) 60
(d) 90
Directions (6-10): The given pie chart shows the distribution (in degrees) of cars sold of different models by a company in 2015-16.
6) If the number of cars sold of model D is 40500, then how many more cars of model E are sold than that of A?
(a) 8100
(b) 16200
(c) 24300
(d) 13500
7) If the number of cars sold of model D is 40500, what is the ratio between the number of cars sold of model D and E?
(a) 9 : 5
(b) 6 : 5
(c) 11 : 9
(d) 9 : 7
8) If the number of cars sold of model D is 72900, then what is the total number of cars sold of all the models together by the company?
(a) 291600
(b) 208100
(c) 162000
(d) 214160
9) If number of cars sold of model C are 22000, then what will be the difference in the number of cars sold of model A and B?
(a) 800
(b) 1200
(c) 1000
(d) 1500
10) If 5% of the total number of cars sold of model E is 750, then what is the average number of cars sold of all the models?
(a) 15000
(b) 14400
(c) 16800
(d) 14000
secθ=13/12 [Triplet, 5, 12, 13]
cotθ=(12/ 5) & tanθ= (5/12)
So, √(cotθ+tanθ)=√(12/5+5/12)=13/√60
=13/(2√15)
(1+tan 2 θ) = sec2 θ = 625/49
sec θ = 25/7
sin θ = 24/25
cos θ = 7/25
√sin θ + cos θ
= √31/5
sin θ = (1 – sin2 θ)
sin2 θ = cos4 θ
1 – cos2 θ = cos4 θ
1 = cos4 θ + cos2 θ
1 = (cos4 θ + cos2 θ )3
1 = cos12 θ + 3 cos10 θ + 3 cos8 θ + cos6 θ
cos12 θ + 3 cos10 θ + 3 cos8 θ + cos6 θ – 1 = 0
cot θ = √11 ⇒ cot2 θ = 11
(cosec2 θ + sec2 θ) /(cosec2 θ – sec2 θ) = 1/(cos2 θ – sin2 θ) = 1/ (cos 2θ) = 1/(2 cos2θ – 1)
cos2 θ = 1/2 ⇒ (1/(2×(1/2)-1)) = -(6/5)
put θ = 45°, (cos θ/(1+sin  θ)) + (cos θ/(1- sin  θ)) = 2 √2
No. of cars = (40600/90)Â 30=13500
Required ratio = 40500/33750 = 162/135 = 6/5
Required No. of models cars = 810 × 360 = 291600
Difference in the no. of ears = 200 × 5 = 1000 | 1,486 | 3,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-31 | latest | en | 0.834143 |
https://studysoup.com/tsg/calculus/468/precalculus-with-trigonometry-concepts-and-applications/chapter/23044/5-2 | 1,638,229,886,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00386.warc.gz | 623,024,390 | 9,212 | ×
×
# Solutions for Chapter 5-2: Composite Argument and Linear Combination Properties
## Full solutions for Precalculus with Trigonometry: Concepts and Applications | 1st Edition
ISBN: 9781559533911
Solutions for Chapter 5-2: Composite Argument and Linear Combination Properties
Solutions for Chapter 5-2
4 5 0 258 Reviews
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##### ISBN: 9781559533911
Precalculus with Trigonometry: Concepts and Applications was written by and is associated to the ISBN: 9781559533911. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 5-2: Composite Argument and Linear Combination Properties includes 33 full step-by-step solutions. This textbook survival guide was created for the textbook: Precalculus with Trigonometry: Concepts and Applications, edition: 1. Since 33 problems in chapter 5-2: Composite Argument and Linear Combination Properties have been answered, more than 51067 students have viewed full step-by-step solutions from this chapter.
Key Calculus Terms and definitions covered in this textbook
• Absolute value of a real number
Denoted by |a|, represents the number a or the positive number -a if a < 0.
P(A or B) = P(A) + P(B) - P(A and B). If A and B are mutually exclusive events, then P(A or B) = P(A) + P(B)
• Annual percentage yield (APY)
The rate that would give the same return if interest were computed just once a year
• Arrow
The notation PQ denoting the directed line segment with initial point P and terminal point Q.
• Composition of functions
(f ? g) (x) = f (g(x))
• Compound interest
Interest that becomes part of the investment
• Convergence of a sequence
A sequence {an} converges to a if limn: q an = a
• Directed line segment
See Arrow.
• Equilibrium price
See Equilibrium point.
• Frequency
Reciprocal of the period of a sinusoid.
• Grapher or graphing utility
Graphing calculator or a computer with graphing software.
• Independent events
Events A and B such that P(A and B) = P(A)P(B)
• Line of travel
The path along which an object travels
• Mapping
A function viewed as a mapping of the elements of the domain onto the elements of the range
• Orthogonal vectors
Two vectors u and v with u x v = 0.
• Outliers
Data items more than 1.5 times the IQR below the first quartile or above the third quartile.
• Permutations of n objects taken r at a time
There are nPr = n!1n - r2! such permutations
• Principle of mathematical induction
A principle related to mathematical induction.
• Pythagorean identities
sin2 u + cos2 u = 1, 1 + tan2 u = sec2 u, and 1 + cot2 u = csc2 u
• Row operations
See Elementary row operations. | 654 | 2,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.865939 |
https://simpleinternettools.com/fibonacci-sequence-generator/ | 1,652,706,844,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00317.warc.gz | 613,760,356 | 3,079 | # Online Fibonacci Sequence Generator
This is an online Fibonacci sequence generator.
Input a limit, below, and the generator will calculate the Fibonacci sequence until it reaches that limit.
### What is the Fibonacci Sequence?
The Fibonacci sequence is a series of numbers in which each number is the sum of the previous two. For example, 0, 1, 1, 2, 3, 5, 8, …
### What are the Fibonacci Numbers?
The Fibonacci numbers are a series of numbers in which each number is the sum of the previous two consecutive preceding numbers. The sequence goes: 1, 1, 2, 3, 5, 8, 13, 21, and so on.
The Fibonacci sequence was introduced to Europe by Leonardo of Pisa (Leonardo Bonacci) in 1202 and it’s named after him. It has been popularly used in mathematics and art because it exhibits many properties that are aesthetically pleasing.
The Fibonacci sequence can be found everywhere from nature to architecture to urban planning.
### Fibs in Modern Mathematics: Their Powers and Applications to Nature
Fibonacci numbers are a sequence of numbers in which each term is the sum of the two previous terms.
Fibonacci numbers are used in nature to generate patterns and create patterns in nature's hierarchical systems like ecosystems and plant growth. They are also used for generating sequences in market-based simulations like financial markets or the stock market.
### The History of Fibs in Mathematics: From Ancient Greeks to Contemporary Mathematicians
Fibs are an important mathematical tool that was first used by the Ancient Greeks. They were very good at using this tool to quickly calculate the areas of figures. This is because they had no use for multiplication tables back then due to their limited technology and mathematical knowledge.
In addition, Fibs came in handy when they needed to figure out how much a ship would weigh if it carried a certain amount of cargo.
In recent years, fibs have been used in fields such as statistics and economics where data collection is difficult because of its large quantity. In these fields, people use fib to generate large sample sizes of data that is representative of the entire population being studied. | 445 | 2,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-21 | longest | en | 0.953137 |
https://tutorstips.com/question-96-chapter-5-of-2-a-t-s-grewal-12-class-part-a-vol-1/ | 1,696,354,147,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00166.warc.gz | 629,341,946 | 69,564 | # Question 96 Chapter 5 of +2-A – T.S. Grewal 12 Class Part – A Vol. 1
Question No.96 Chapter No.5 - T.S. Grewal +2 Book 2019-Solution
Question 96 Chapter 5 of +2-A
96. L, M and N were partners in a firm sharing profits in the ratio of 3 : 2 : 1. Their Balance Sheet on 31st March, 2015 was as follows:
Liabilities Assets Creditors 1,68,000 Bank 34,000 General Reserve 42,000 Debtors 46,000 Capital’s A/c Stock 2,20,000 L 1,20,000 Investment 60,000 M 80,000 Furniture 20,000 N 40,000 2,40,00 Machinery 70,000 4,50,000 4,50,000
On the above date, O was admitted as a new partner and it was decided that:
(i) The new profit-sharing ratio between L, M, N and O will be 2 : 2 : 1 : 1.
(ii) Goodwill of the firm was valued at 1,80,000 and O brought his share of goodwill premium in cash.
(iii) The market value of investments was 36,000.
(iv) Machinery will be reduced to 58,000.
(v) A creditor of 6,000 was not likely to claim the amount and hence was to be written off.
(vi) O will bring proportionate capital so as to give him 1/6th share in the profits of the firm.
Prepare Revaluation Account, Partners’ Capital Accounts and the Balance Sheet of the new firm.
## The solution of Question 96 Chapter 5 of +2-A: –
Revaluation Account Particular Amount Particular Amount To Investments A/c 24,000 By Creditors A/c 6,000 To Machinery A/c 12,000 Loss on Revaluation L’s Capital 15,000 M’s Capital 10,000 N’s Capital 5,000 30,000 36,000 36,000
Partners’ Capital Account Particulars L M N O To Loss on Revaluation A/c 15,000 10,000 5,000 To Balance c/d 1,56,000 84,000 42,000 56,400 1,71,000 94,000 47,000 56,400
Particulars L M N O By Balance B/d 1,20,000 80,000 40,000 By Bank A/c (WN2) – – – 56,400 By Premium for Goodwill A/c 30,000 – – – By General Reserve A/c 21,000 14,000 7,000 – 1,71,000 94,000 47,000 56,400
Balance Sheet Liabilities Amount Assets Amount Creditors 1,62,000 Bank (34,000+56,400+30,000) 1,20,400 Debtors 46,000 Stock 2,20,000 Capital A/cs: Investments 36,000 L 1,56,000 Furniture 20,000 M 84,000 Machinery 58,000 N 42,000 O 56,400 3,38,400 5,00,400 5,00,400
Working Note:-
Calculation of Sacrificing Ratio
Sacrificing Ratio = Old Ratio – New Ratio
L’s Sacrificing Ratio = 3 – 2 6 6
= 3 – 2 6
= 1 6
M’s Sacrificing Ratio = 2 – 2 6 6
= 2 – 2 6
= 0 6
N’s Sacrificing Ratio = 1 – 1 6 6
= 1 – 1 6
= 0 6
O’s Share of Goodwill = 1,80,000 X 1 6 = 30,000
30,000 will be credited to L’s Capital A/c, as he is the only sacrificing Partner
Calculation of O’s Proportionate Capital
Adjustment of Old Capital of L = 1,20,000 + 21,000 + 30,000 – 15,000 = 1,56,000 Adjustment of Old Capital of M = 80,000 + 14,000 – 10,000 = 84,000 Adjustment of Old Capital of N = 40,000 + 7,000 – 5,000 = 72,000 Total Adjustment Capital = 1,56,000 + 84,000 + 72,000 = 2,82,000
O’s Proportionate Capital = Total Adjusted Capital X O’s Profit Share X Reciprocal of combined new Share of Old Partner
O’s Share of Goodwill = 2,82,000 x 1 x 5 6 5 = 56,400
T.S. Grewal’s Double Entry Book Keeping +2 (Vol. I: Accounting for Not-for-Profit Organizations and Partnership Firms)
• Chapter No. 1 – Financial Statement of Not-For-Profit Organisations
• Chapter No. 2 – Accounting for Partnership Firms – Fundamentals
• Chapter No. 3 – Goodwill: Nature and Valuation
• Chapter No. 4 – Change in Profit-Sharing Ratio Among the Existing Partners
• Chapter No. 5 – Admission of a Partner
• Chapter No. 6 – Retirement/Death of a Partner
• Chapter No. 7 – Dissolution of a Partnership Firm
### T.S. Grewal’s Double Entry Book Keeping (Vol. II: Accounting for Companies)
• Chapter No. 1 – Financial Statements of a Company
• Chapter No. 2 – Financial Statement Analysis
• Chapter No. 3 – Tools of Financial Statement Analysis – Comparative Statements and Common- Size Statements
• Chapter No. 4 – Accounting Ratios
• Chapter No. 5 – Cash Flow Statement
error: Content is protected !! | 1,305 | 3,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | longest | en | 0.873797 |
https://forum.wilmott.com/viewtopic.php?f=34&t=67488&view=print | 1,590,450,587,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390437.8/warc/CC-MAIN-20200525223929-20200526013929-00111.warc.gz | 361,391,959 | 4,150 | Page 1 of 11
### About solving a transport equation
Posted: December 15th, 2008, 9:49 am
### About solving a transport equation
Posted: December 15th, 2008, 4:52 pm
Using the Fichera approach I conclude no BCs at x = 0, 1. If u(x,0) = f(x) then we get 'compatibility' conditions u(0,t) = f(0) and u(1,t) = f(1). I did a quick energy analysis and it looks conservative (decreasing in time). So some one-step 2nd order methods (e.g. Box) should be OK.Maybe there is an exact solution or MOC... I leave that to others.
### About solving a transport equation
Posted: December 15th, 2008, 8:03 pm
Since this is probably homework, I will just hint that the characteristics are easilyfound and this will also answer the OP's questions.
### About solving a transport equation
Posted: December 16th, 2008, 7:48 am
It's not a homeworkIt's kind of a project which has to be done.Need more detailed explanation.
### About solving a transport equation
Posted: December 16th, 2008, 1:58 pm
QuoteOriginally posted by: macrovueIt's not a homeworkIt's kind of a project which has to be done.Need more detailed explanation.Did you do research on this topic? What's wrong what the tips you already have received?
### About solving a transport equation
Posted: December 16th, 2008, 3:56 pm
Yeah -- we need more proof that this is not homework Discuss exactly where this problem and the questions come from, and what you have tried so far.
### About solving a transport equation
Posted: December 16th, 2008, 6:30 pm
Well, we know where we're going but we don't where we've been ..Let me try to reconstruct this unusual 1st order hyperbolic PDE (disclaimer: only a guess!)u_t + u_x = 0 on real linedefine y = coth(x)then u_t + (1+y)(1-y)u_y = 0 on [-1, 1]]define z = 1+ythen u_t + z(2-z)u_z = 0 on [0,2]Exercise: in the future we define y = coth(cx) where c is a hot spot parameter.etc.??
### About solving a transport equation
Posted: December 16th, 2008, 11:11 pm
Daniel,I didn't mean for you to discuss, but macrovue should explain the origin of his problem -- if he wants more detailed hints.
### About solving a transport equation
Posted: December 17th, 2008, 9:09 am
QuoteOriginally posted by: AlanDaniel,I didn't mean for you to discuss, but macrovue should explain the origin of his problem -- if he wants more detailed hints.I meant it for mv, indeed.
### About solving a transport equation
Posted: December 18th, 2008, 12:27 pm
Looks as if this transport thread is not going anywhere. The original post has been removed.
### About solving a transport equation
Posted: May 24th, 2009, 6:11 pm
Original post was:QuoteI have to define initial and boundary condition for a transport PDE: u_t+x(1-x)u_x=0with x and t is between [0,1], to solve this equation, what kind of numerical methodand boundary condition do you recommend and why?What kind of numerical error do you expect?Detailed explanation will be appreciated in advance.I have had a look at this, let me just take a similar PDE (but containing the essential difficulties) take the IBVP on (0, infinity) with BC at x = 0 and nothing at infinity of course. I transform to [0,1) to get a PDE similar to above.In general, numerical BCs need to be defined at x = 1 and it is very tricky (spurious reflection) when we take centred 2nd order approximation to du/dx at x = 1. This was a big pain in the past with ADI for Asian options.Now there is no BC at x = 1 and we just get du/dt = 0 there. It is possible to get a nice energy inequality in L2 space that say that the solution is bounded by IC, BC and RHS terns (use Gronwall's lemma).Numerically, we have in essence numerical BC at both x = 0 and x = 1, so we can then solve an ODE using Crank Nicolson in time, for example.In general, numerical BCs need to be defined at x = 1 and it is very tricky (spurious reflection) when we take centred 2nd order approximation to du/dx at x = 1. This was a big pain in the past with ADI for Asian options and Cheyette.Now there is no BC at x = 1 and we just get du/dt = 0 there. It is possible to get a nice energy inequality in L2 space that say that the solution is bounded by IC, BC and RHS terns (use Gronwall's lemma).Numerically, we have in essence numerical BC at both x = 0 and x = 1, so we can then solve an ODE using Crank Nicolson in time, for example. I have done this in the case of a skew symmetric matrix as well as my predictor corrector_type scheme (second order) and it works very well and no spurious reflections in the upstream direction. A spin off is that the BCs are not an issue anymore.Conclusion: No BC needed for PDE, but we need BC for the FDM scheme (and choose clever ones).BTW nice problem. Pity the thread had such a short ½ life.
### About solving a transport equation
Posted: December 18th, 2009, 7:07 am
This model pde is a good one to look at since in is defined on the real line instead of the half line. It is already transformed to [-1,1]. No BC is needed but rather a compatibility condition with the initial condition at the end points.Numerically, I used Backward Time Centred Space (BTCS) using LU decomposition and ADE and the results look good. Accuracy is first and second order, respectively.The model was U_t + aU_x = 0 U(x,0) = f(x)Solution is U(x,t) = f(x - at)The results apply to other cases, for example half-line, Asian-type pde.So, original pde is transfomed by using y = tanh(x), for example.
### Re: About solving a transport equation
Posted: January 23rd, 2020, 8:35 pm
The OP post (deleted) is similar to the rather innocuous-looking degenerate pde that crops up here and there
[$]\frac{\partial u}{\partial t} - (1-y)^2\frac{\partial u}{\partial y} = 0 [$] on the interval [$](0,1)[$] (1)
with initial condition
[$]u(y,0) = f(y)[$] on the interval [$](0,1)[$] (2)
Questions on (1), (2):
1. Does it have a solution?
2. Is there a closed form for the solution?
3. What 'happens' at the boundaries [$]y = 0[$] and [$]y = 1[$]. Describe the 'physics' flow.
4. What are the numerical boundary conditions.
/// BTW OP's pde was
[$]\frac{\partial u}{\partial t} + (1+y)(1-y)\frac{\partial u}{\partial y} = 0 [$] on the interval [$](-1,1)[$] AFAIR
### Re: About solving a transport equation
Posted: January 24th, 2020, 5:06 am
I'll play a little. For 2, I get
[$] u(y,t) = f \left( \xi(y,t) \right)[$] where the characteristic [$]\xi(y,t) = \frac{y + t - t y}{1 + t - t y}[$].
### Re: About solving a transport equation
Posted: January 26th, 2020, 8:33 pm
I'll play a little. For 2, I get
[$] u(y,t) = f \left( \xi(y,t) \right)[$] where the characteristic [$]\xi(y,t) = \frac{y + t - t y}{1 + t - t y}[$].
Some questions/remarks:
1. How do you get the characteristics? I tried and got [$]\xi(y,t) = \frac{-1 + t - t y}{1 - y}[$]. I must be doing something wrong.
2. The solution does not need any boundary conditions.
3. Plugging in, [$]u(0,t) = f(t/(1+t))[$]. Does this mean anything?
4. [$]u(1,t) = f(1))[$],
5. What happens in the empty quarter [$]y > 1[$]? for mean, it is beyond infinity.
4 is what I use for numerical BC but [$]y=0[$] is an outflow boundary so it does not matter what happens there. In [$]0 < y < 1[$] I have a 2nd order explicit scheme whereas for our Anchor article I used 1st order upwinding. | 2,088 | 7,246 | {"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} | 3.5625 | 4 | CC-MAIN-2020-24 | latest | en | 0.944337 |
https://www.mathworks.com/matlabcentral/cody/problems/5-triangle-numbers/solutions/960676 | 1,597,351,497,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00188.warc.gz | 638,344,779 | 17,327 | Cody
# Problem 5. Triangle Numbers
Solution 960676
Submitted on 6 Sep 2016 by John
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
n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
n = 30; t = 465; assert(isequal(triangle(n),t)) | 158 | 467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.677054 |
http://forums.wolfram.com/mathgroup/archive/1996/Jan/msg00072.html | 1,527,421,355,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00493.warc.gz | 116,884,270 | 7,176 | Sqrt Tip
• To: mathgroup at smc.vnet.net
• Subject: [mg3032] Sqrt Tip
• From: BobHanlon at aol.com
• Date: Wed, 24 Jan 1996 03:34:42 -0500
```Mathematica version 2.2 on a Macintosh does not simplify the product
of square roots:
In[1]:=
Sqrt[x] Sqrt[y] // Simplify
Out[1]=
Sqrt[x] Sqrt[y]
As a result, it overlooks some straightforward simplifications.
For example,
In[2]:=
Sqrt[1-x] Sqrt[1+x]/Sqrt[1 - x^2] // Simplify
Out[2]=
Sqrt[1 - x] Sqrt[1 + x]
-----------------------
2
Sqrt[1 - x ]
One would like to modify the definition of Sqrt to correct this as
follows:
In[3]:=
Unprotect[Sqrt];
Sqrt/: Sqrt[a_] Sqrt[b_] := Sqrt[a b];
Protect[Sqrt];
This generates the error message
TagSetDelayed::tagnf:
The reason that this failed is apparent if the FullForm of Sqrt is
inspected
In[6]:=
Sqrt[x] // FullForm
Out[6]//FullForm=
Power[x, Rational[1, 2]]
Sqrt is represented internally using Power. Consequently, the
modification must be made to Power rather than Sqrt.
In[7]:=
Unprotect[Power];
Power/: Sqrt[a_] Sqrt[b_] := Sqrt[a b];
Protect[Power];
After, modifying Power we obtain the desired result:
In[10]:=
Sqrt[1-x] Sqrt[1+x]/Sqrt[1 - x^2] // Simplify
Out[10]=
1
_______________
Bob Hanlon
bobhanlon at aol.com
==== [MESSAGE SEPARATOR] ====
```
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• Next by thread: Sqrt Tip | 463 | 1,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-22 | latest | en | 0.713377 |
https://convertoctopus.com/300-millimeters-to-yards | 1,653,689,834,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00471.warc.gz | 237,751,370 | 7,350 | Conversion formula
The conversion factor from millimeters to yards is 0.0010936132983377, which means that 1 millimeter is equal to 0.0010936132983377 yards:
1 mm = 0.0010936132983377 yd
To convert 300 millimeters into yards we have to multiply 300 by the conversion factor in order to get the length amount from millimeters to yards. We can also form a simple proportion to calculate the result:
1 mm → 0.0010936132983377 yd
300 mm → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 300 mm × 0.0010936132983377 yd
L(yd) = 0.32808398950131 yd
The final result is:
300 mm → 0.32808398950131 yd
We conclude that 300 millimeters is equivalent to 0.32808398950131 yards:
300 millimeters = 0.32808398950131 yards
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 3.048 × 300 millimeters.
Another way is saying that 300 millimeters is equal to 1 ÷ 3.048 yards.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred millimeters is approximately zero point three two eight yards:
300 mm ≅ 0.328 yd
An alternative is also that one yard is approximately three point zero four eight times three hundred millimeters.
Conversion table
millimeters to yards chart
For quick reference purposes, below is the conversion table you can use to convert from millimeters to yards
millimeters (mm) yards (yd)
301 millimeters 0.329 yards
302 millimeters 0.33 yards
303 millimeters 0.331 yards
304 millimeters 0.332 yards
305 millimeters 0.334 yards
306 millimeters 0.335 yards
307 millimeters 0.336 yards
308 millimeters 0.337 yards
309 millimeters 0.338 yards
310 millimeters 0.339 yards | 480 | 1,777 | {"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} | 4.15625 | 4 | CC-MAIN-2022-21 | latest | en | 0.865793 |
https://www.archivemore.com/is-7-a-coefficient/ | 1,716,188,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00459.warc.gz | 587,048,782 | 8,174 | ## Is 7 a coefficient?
The coefficients are the numbers that multiply the variables or letters. Thus in 5x + y – 7, 5 is a coefficient. In 5x + y – 7 the terms are 5x, y and -7 which all have different variables (or no variables) so there are no like terms. Constants are terms without variables so -7 is a constant.
## How do you identify terms coefficients and constants?
Each term in an algebraic expression is separated by a + sign or J sign. In , the terms are: 5x, 3y, and 8. When a term is made up of a constant multiplied by a variable or variables, that constant is called a coefficient. In the term 5x, the coefficient is 5.
## Is a constant considered a coefficient?
I would like to see this taught as: after combining like terms, all constant factors, including the constant term, are coefficients.
## What is a constant or coefficient?
Parts of an Equation A number on its own is called a Constant. A Coefficient is a number used to multiply a variable (4x means 4 times x, so 4 is a coefficient)
## How do you find the coefficient of Class 7?
A coefficient is a numerical value that is multiplied with a variable. For example, the coefficient of 7x is 7.
## What is a coefficient number?
A coefficient is a number multiplied by a variable. Examples of coefficients: In the term 14 c 14c 14c , the coefficient is 14. In the term g, the coefficient is 1.
## Can coefficient of variance be more than 100?
For the pizza delivery example, the coefficient of variation is 0.25. This value tells you the relative size of the standard deviation compared to the mean. Analysts often report the coefficient of variation as a percentage. If the value equals one or 100%, the standard deviation equals the mean.
## Can Mean be greater than 1?
There’s no problem with the expectation being bigger than 1. However, since the expectation is a weighted average of the values of the random variable, it always lies between the minimal value and the maximal value.
## Is a higher or lower coefficient of variation better?
The coefficient of variation (CV) is the ratio of the standard deviation to the mean. The higher the coefficient of variation, the greater the level of dispersion around the mean. It is generally expressed as a percentage. The lower the value of the coefficient of variation, the more precise the estimate.
## What does the coefficient of determination tell us?
The coefficient of determination is a statistical measurement that examines how differences in one variable can be explained by the difference in a second variable, when predicting the outcome of a given event.
## What is the use of coefficient of variation?
The coefficient of variation shows the extent of variability of data in a sample in relation to the mean of the population. In finance, the coefficient of variation allows investors to determine how much volatility, or risk, is assumed in comparison to the amount of return expected from investments.
## What is the difference between standard deviation and coefficient of variance?
The coefficient of variation (CV) is a measure of relative variability. It is the ratio of the standard deviation to the mean (average). For example, the expression “The standard deviation is 15% of the mean” is a CV….Coefficient of Variation Example. | 707 | 3,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-22 | latest | en | 0.916179 |
https://www.studypool.com/discuss/1090056/solve-the-system-of-equations-by-the-addition-method-3?free | 1,511,101,784,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805649.7/warc/CC-MAIN-20171119134146-20171119154146-00668.warc.gz | 841,927,307 | 14,048 | Solve the system of equations by the addition method
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
X+y=5
X-y=7
Jul 24th, 2015
x + y = 5
x - y = 7
2x = 12
2x/2 = 12/2
x = 6
Now for y, put the value of x in any equation(say first), then
6 + y = 5
6 - 6 + y = 5 - 6
y = - 1
Hence x = 6 y = - 1
Solution : {6, - 1}
Jul 24th, 2015
Hope you got it.
Jul 24th, 2015
...
Jul 24th, 2015
...
Jul 24th, 2015
Nov 19th, 2017
check_circle | 206 | 479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-47 | latest | en | 0.816899 |
https://sellfy.com/p/kTdg/ | 1,521,928,581,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651007.67/warc/CC-MAIN-20180324210433-20180324230433-00541.warc.gz | 702,994,454 | 6,941 | 1) Once a Woman won \$1 Million in scratch off game from a lottery. Some years later, she won \$1 million in another scratch off game. In the first game, she beat odds of 1 in 6.2 million to win. In the second, she beat odds of 1 in 805,600.
a) What is the probability that an individual would win 1 million in both games if they bought one scratch off ticket from each game?
b) What is the probability that an individual would win \$1 million twice in the second scratch of
2) A diamond can be classified as either gem-quality or industrial grade. 87% of diamonds are classified as industrial-grade.
(a) 2 diamonds are chosen at random. What is the probability that both diamonds are industrial-grade?
(B) 9 diamonds are chosen at random. what is the probability that all 9 diamonds are industrial-grade?
(c) what is the probability that at least 1 of 9 randomly selected diamonds is gem-quality?
(d) would it be unusual that at least 1 of 9 diamonds is gem-quality? | 228 | 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} | 3.09375 | 3 | CC-MAIN-2018-13 | latest | en | 0.963507 |
https://convertoctopus.com/1468-inches-to-meters | 1,620,296,145,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00207.warc.gz | 215,837,695 | 7,709 | Conversion formula
The conversion factor from inches to meters is 0.0254, which means that 1 inch is equal to 0.0254 meters:
1 in = 0.0254 m
To convert 1468 inches into meters we have to multiply 1468 by the conversion factor in order to get the length amount from inches to meters. We can also form a simple proportion to calculate the result:
1 in → 0.0254 m
1468 in → L(m)
Solve the above proportion to obtain the length L in meters:
L(m) = 1468 in × 0.0254 m
L(m) = 37.2872 m
The final result is:
1468 in → 37.2872 m
We conclude that 1468 inches is equivalent to 37.2872 meters:
1468 inches = 37.2872 meters
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter is equal to 0.026818854727628 × 1468 inches.
Another way is saying that 1468 inches is equal to 1 ÷ 0.026818854727628 meters.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand four hundred sixty-eight inches is approximately thirty-seven point two eight seven meters:
1468 in ≅ 37.287 m
An alternative is also that one meter is approximately zero point zero two seven times one thousand four hundred sixty-eight inches.
Conversion table
inches to meters chart
For quick reference purposes, below is the conversion table you can use to convert from inches to meters
inches (in) meters (m)
1469 inches 37.313 meters
1470 inches 37.338 meters
1471 inches 37.363 meters
1472 inches 37.389 meters
1473 inches 37.414 meters
1474 inches 37.44 meters
1475 inches 37.465 meters
1476 inches 37.49 meters
1477 inches 37.516 meters
1478 inches 37.541 meters | 455 | 1,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-21 | latest | en | 0.868812 |
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