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https://atcoder.jp/contests/abc136/tasks/abc136_b | 1,721,135,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00208.warc.gz | 98,715,807 | 5,820 | Contest Duration: - (local time) (100 minutes) Back to Home
B - Uneven Numbers /
Time Limit: 2 sec / Memory Limit: 1024 MB
### 制約
• 1 \leq N \leq 10^5
### 入力
N
### 出力
N 以下の正の整数のうち、桁数が奇数であるようなものの個数を出力せよ。
### 入力例 1
11
### 出力例 1
9
11 以下の正の整数のうち、桁数が奇数であるようなものは 1, 2, \ldots, 99 個です。
### 入力例 2
136
### 出力例 2
46
1, 2, \ldots, 9 に加えて、100, 101, \ldots, 13637 個の数も桁数が奇数です。
### 入力例 3
100000
### 出力例 3
90909
Score : 200 points
### Problem Statement
Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).
### Constraints
• 1 \leq N \leq 10^5
### Input
Input is given from Standard Input in the following format:
N
### Output
Print the number of positive integers less than or equal to N that have an odd number of digits.
### Sample Input 1
11
### Sample Output 1
9
Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.
### Sample Input 2
136
### Sample Output 2
46
In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.
### Sample Input 3
100000
### Sample Output 3
90909 | 452 | 1,234 | {"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} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.312456 |
http://oalevelsolutions.com/past-papers-solutions/cambridge-international-examinations/as-a-level-mathematics-9709/pure-mathematics-p1-9709-01/year-2019-may-june-p1-9709-12/cie_19_mj_9709_12_q_11/ | 1,619,012,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039544239.84/warc/CC-MAIN-20210421130234-20210421160234-00521.warc.gz | 69,978,169 | 13,781 | # Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | May-Jun | (P1-9709/12) | Q#11
Hits: 467
Question
The diagram shows part of the curve and the minimum point M.
i.Find the expressions for and
ii.Find the coordinates of M.
The shaded region is bounded by the curve, the y-axis and the line through M parallel to the x-axis.
iii.Find, showing all necessary working, the area of the shaded region.
Solution
i.
We are given;
First, we find .
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Hence;
Rule for differentiation of is:
Rule for differentiation of is:
Next, we find .
Rule for integration of is:
Rule for integration of is:
ii.
We are required to find the coordinates of point M which is minimum point on the curve.
Coordinates of stationary point on the curve can be found from the derivative of equation of the curve by equating it with ZERO. This results in value of x-coordinate of the stationary point on the curve.
We have found derivative of equation of the curve in (i);
We equate it with ZERO;
One possible value of implies that there is only one stationary point on the curve at this value of .
To find y-coordinate of the stationary point on the curve, we substitute value of x-coordinate of the stationary point on the curve (found by equating derivative of equation of the curve with ZERO) in the equation of the curve.
We have equation of the curve;
Substitute ;
However, it is evident from the diagram that point M lies on positive x and y axes.
Therefore;
It is also evident from the diagram that point M has not y-coordinate as ZERO.
Therefore;
Hence, coordinates of the minimum point are .
iii.
We are required to find the area of the shaded region.
It is evident from the diagram that;
To find the area of region under the curve , we need to integrate the curve from point to
along x-axis.
We are given equation of the curve as;
We need equation of the line through point M and parallel to x-axis.
To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
We already have coordinates of a point on the line .
We need slope of the line.
If two lines are parallel to each other, then their slopes and are equal;
Since line is parallel to the x-axis and slope of x-axis is ZERO, therefore;
Hence, now we can write equation of the line.
Point-Slope form of the equation of the line is;
Hence;
First, we find area under the curve.
It is evident from the diagram that for area under the curve the varies from to .
Hence;
We have found in (i) that;
Therefore;
Next, we find area under line.
It is evident from the diagram that for area under the curve the varies from to .
Hence;
Rule for integration of is:
Finally, | 704 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-17 | latest | en | 0.913835 |
https://www.studypool.com/discuss/1198181/positive-integers-in-college-algebra?free | 1,495,536,936,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00350.warc.gz | 965,758,540 | 14,122 | ##### positive integers in college algebra
Algebra Tutor: None Selected Time limit: 1 Day
The sum of the two numbers is 37, and the greater number is 8 less than twice the lesser number
Oct 5th, 2015
x+y=37
x=2y-8
2y-8+y=37
3y=45
y=15
x=2*15-8=22
Oct 5th, 2015
Find two positive integers that satisfy the given requirements.
Oct 5th, 2015
The answer is 22 and 15.
Oct 5th, 2015
Thank you, that's what I thought, just wanted to make sure. Thank you again :)
Oct 5th, 2015
...
Oct 5th, 2015
...
Oct 5th, 2015
May 23rd, 2017
check_circle | 197 | 552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-22 | longest | en | 0.920198 |
http://cosi.org/zoo/item/eli5-angular-momentum | 1,490,449,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188926.39/warc/CC-MAIN-20170322212948-00211-ip-10-233-31-227.ec2.internal.warc.gz | 79,890,039 | 10,728 | Today
10am - 5pm
# "ELI5: Angular Momentum"
Written by Paul Sutter on Monday, 13 March 2017. Posted in From The Desk of...The Chief Scientist
Jarod Smith, a member of the COSI on Wheels team, asked me for help in crafting a quick-and-easy way to explain conservation of angular momentum to young kids, and more importantly sometimes to their parents.
The topic comes up in a fun and simple demo. Sit on a chair that can rotate, and hold a spinning bicycle wheel in your hands. Flip the wheel over and presto-chango you start spinning in your chair. Magic! I mean, science!
I like to think of momentum as the amount of "oomph" an object has - how much it can pack a punch if it were to hit you. A small object (like a bullet) traveling fast enough can hurt, and a big object (like a truck) can be a pain pretty much no matter how fast it's going.
Angular momentum is then oomph going in a circle. It's conserved, which means the total amount of oomph must be the same.
For very young kids, I just refer to it as spin. The bicycle wheel in your hands is spinning really fast in one direction. That's the total amount of spin that you+wheel can have. When you flip the wheel over, you're taking away the spin in that direction, so some has to be added: you yourself start spinning to compensate.
Ultimately, while we can explain what's going on as best we can, I think in some cases it's sufficient to let the demo do the talking. Kids are developing an intuitive sense of angular momentum conservation, and that's something we can build on when they're older.
### About the Author
#### Paul Sutter
Paul Sutter is COSI's Chief Scientist. He is an astrophysicist and offers a wealth of knowledge about our universe. In addition to his COSI position, Paul Sutter is a Cosmological Researcher and Community Outreach Coordinator at The Ohio State University's Center for Cosmology and AstroParticle Physics (CCAPP). | 447 | 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} | 2.734375 | 3 | CC-MAIN-2017-13 | latest | en | 0.959941 |
https://support.sas.com/documentation/cdl/en/ormplpug/66106/HTML/default/ormplpug_intpoint_sect067.htm | 1,575,593,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482954.0/warc/CC-MAIN-20191206000309-20191206024309-00408.warc.gz | 560,381,664 | 4,733 | ### Balancing Total Supply and Total Demand
#### When Total Supply Exceeds Total Demand
When total supply of a network problem exceeds total demand, PROC INTPOINT adds an extra node (called the excess node) to the problem and sets the demand at that node equal to the difference between total supply and total demand. There are three ways that this excess node can be joined to the network. All three ways entail PROC INTPOINT generating a set of arcs (henceforth referred to as the generated arcs) that are directed toward the excess node. The total amount of flow in generated arcs equals the demand of the excess node. The generated arcs originate from one of three sets of nodes.
When you specify the THRUNET option, the set of nodes that generated arcs originate from are all demand nodes, even those demand nodes with unspecified demand capability. You indicate that a node has unspecified demand capability by using a missing D value instead of an actual value for demand data (discussed in the section Missing S Supply and Missing D Demand Values). The value specified as the demand of a demand node is in effect a lower bound of the number of flow units that node can actually demand. For missing D demand nodes, this lower bound is zero.
If you do not specify the THRUNET option, the way in which the excess node is joined to the network depends on whether there are demand nodes with unspecified demand capability (nodes with missing D demand) or not.
If there are missing D demand nodes, these nodes are the set of nodes that generated arcs originate from. The value specified as the demand of a demand node, if not missing D, is the number of flow units that node can actually demand. For a missing D demand node, the actual demand of that node may be zero or greater.
If there are no missing D demand nodes, the set of nodes that generated arcs originate from are the set of supply nodes. The value specified as the supply of a supply node is in effect an upper bound of the number of flow units that node can actually supply. For missing S supply nodes (discussed in the section Missing S Supply and Missing D Demand Values), this upper bound is zero, so missing S nodes when total supply exceeds total demand are transshipment nodes, that is, nodes that neither supply nor demand flow.
#### When Total Supply Is Less Than Total Demand
When total supply of a network problem is less than total demand, PROC INTPOINT adds an extra node (called the excess node) to the problem and sets the supply at that node equal to the difference between total demand and total supply. There are three ways that this excess node can be joined to the network. All three ways entail PROC INTPOINT generating a set of arcs (henceforth referred to as the generated arcs) that originate from the excess node. The total amount of flow in generated arcs equals the supply of the excess node. The generated arcs are directed toward one of three sets of nodes.
When you specify the THRUNET option, the set of nodes that generated arcs are directed toward are all supply nodes, even those supply nodes with unspecified supply capability. You indicate that a node has unspecified supply capability by using a missing S value instead of an actual value for supply data (discussed in the section Missing S Supply and Missing D Demand Values). The value specified as the supply of a supply node is in effect a lower bound of the number of flow units that the node can actually supply. For missing S supply nodes, this lower bound is zero.
If you do not specify the THRUNET option, the way in which the excess node is joined to the network depends on whether there are supply nodes with unspecified supply capability (nodes with missing S supply) or not.
If there are missing S supply nodes, these nodes are the set of nodes that generated arcs are directed toward. The value specified as the supply of a supply node, if not missing S, is the number of flow units that the node can actually supply. For a missing S supply node, the actual supply of that node may be zero or greater.
If there are no missing S supply nodes, the set of nodes that generated arcs are directed toward are the set of demand nodes. The value specified as the demand of a demand node is in effect an upper bound of the number of flow units that node can actually demand. For missing D demand nodes (discussed in the section Missing S Supply and Missing D Demand Values), this upper bound is zero, so missing D nodes when total supply is less than total demand are transshipment nodes, that is, nodes that neither supply nor demand flow. | 931 | 4,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.87967 |
https://aarushinema.com/2021/06/11/classes-and-objects-point2d-class/ | 1,653,173,230,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00000.warc.gz | 129,524,922 | 56,626 | # Classes and Objects: Point2D Class
Question: Create two Point2D object for a point with the specified x- and y- coordinates, use the distance method to compute the distance between the two points, use the midpoint method to find the midpoint of the line joining the two points, and use the toString() method to return a string representation of the
Sample Input and Output
Enter coordinates of point 1
1 2
Enter coordinates of point 2
3 4
Point 1 is Point2D [x = 1.0, y = 2.0]
Point 2 is Point2D [x = 3.0, y = 4.0]
The distance between Point 1 and Point 2 is 2.8284271247461903
The midpoint of the line joining Point 1 and Point 2 is Point2D [x = 2.0, y = 3.0]
```package Ch9;
import javafx.geometry.Point2D;
import java.util.Scanner;
/**
* Question: Create two Point2D object for a point with the specified x- and y- coordinates,
* use the distance method to compute the distance between the two points, use the midpoint method to find the midpoint
* of the line joining the two points, and use the toString() method to return a string representation of the point.
*
* Created by aarushi on 10/6/21.
*/
public class ChEx03 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter coordinates of point 1");
double x1 = sc.nextDouble();
double y1 = sc.nextDouble();
System.out.println("Enter coordinates of point 2");
double x2 = sc.nextDouble();
double y2 = sc.nextDouble();
Point2D p1 = new Point2D(x1, y1);
Point2D p2= new Point2D(x2, y2);
System.out.println("Point 1 is " + p1.toString());
System.out.println("Point 2 is " + p2.toString());
System.out.println("The distance between Point 1 and Point 2 is " + p1.distance(p2));
System.out.println("The midpoint of the line joining Point 1 and Point 2 is " + p1.midpoint(p2));
}
}
/* Sample Input and Output
Enter coordinates of point 1
1 2
Enter coordinates of point 2
3 4
Point 1 is Point2D [x = 1.0, y = 2.0]
Point 2 is Point2D [x = 3.0, y = 4.0]
The distance between Point 1 and Point 2 is 2.8284271247461903
The midpoint of the line joining Point 1 and Point 2 is Point2D [x = 2.0, y = 3.0]
*/```
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 632 | 2,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.7783 |
https://www.merchantnavydecoded.com/isothermal-and-isentropic-air-compression-process-safeties-of-air-compressor-on-ships/ | 1,721,110,666,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514737.55/warc/CC-MAIN-20240716050314-20240716080314-00181.warc.gz | 771,295,479 | 69,166 | # Isothermal and isentropic air compression process | Safeties of air compressor on ships
There are two ways of air compression: isothermal and isentropic
## Isothermal air compression:
We know that the isothermal process is a constant temperature process, which is not optimal for compressing air since it will be a very slow process. In an isothermal process, the work done is less and the area of the curve is less. The area below the dotted line represents an isothermal process in the figure below.
          Isothermal compression can be a useful process in certain applications, such as in refrigeration and air conditioning systems. In these applications, compressing a gas while keeping its temperature constant can help to prevent the gas from overheating and potentially damaging the equipment.
## Isentropic air compression:
Adiabatic air compression is a thermodynamic process in which air is compressed without any heat transfer between the air and its surroundings. In other words, no heat is added or removed from the air during the compression process.
 During adiabatic compression, the temperature of the air increases due to the work done on the gas by the external force compressing it. This increase in temperature can be calculated using the adiabatic compression equation, which is given by:
P2/P1 = (V1/V2)^γ
where P1 and V1 are the initial pressure and volume of the air, P2 and V2 are the final pressure and volume of the air, and γ is the ratio of specific heats of the air (which is approximately 1.4 for air).
The work done during adiabatic compression is also greater than in isothermal compression, as the air is being compressed against its own pressure due to the increase in temperature. This results in a higher compression ratio, which can be useful in some applications where a high-pressure output is required. Â
Adiabatic compression is commonly used in compressors for industrial applications such as air compressors, gas turbines, and engines. However, the increase in temperature during adiabatic compression can also be a disadvantage, as it can cause the air to reach high temperatures that may cause damage to the compressor or other components if not properly managed. As a result, adiabatic compression is often combined with other cooling methods to prevent the air from overheating.  Â
         In an isothermal process the air will be compressed by a compressor in a single stage so work done will be more to produce high-pressure compressed air. But in the multistage process we see in the curve work done is the less blue area in the pic. This green area will save the work done because the isentropic line approaches the isothermal line due to which the air temperature drops to ambient temperature and workdone is saved. For this reason, multistage is used over a single stage compressor.
## Safeties of air compressor in ship:
### For Lube oil low-pressure trip:
In ships, compressors use a splash lubrication system. Many times this lube oil low pressure can damage the compressor as the lube oil pressure will drop, due to which we need to shut the compressor in such case.
### Discharge valve temperature:
After the discharge of air the temperature of the air should be within a certain limit, when this temperature rises above a certain limit we should take action because it is a fire hazard. For this reason here a fusible plug is fitted which will melt and release all the air if the temperature is high.
### Safety valve after discharge:
A safety valve is fitted after the discharge port to release the pressure when it gets higher than a predetermined pressure.
### Water jacket high temperature trip fitted:
Jacket water temperature will be high this means the cylinder side has some problem causing the heat due to which the compressor can be damaged. So it trips the compressor.
### Note:
If you want to learn more about this topic, we suggest checking out our Combo package with the given link https://www.merchantnavydecoded.com/courses/c/ . It’s a great way to dive deeper into the subject through video explanations. This package covers all the important details and presents them in an easy-to-understand format. Watching the videos will help you grasp the topic better and make learning more enjoyable. So, we highly recommend giving our Combo package a try to enhance your knowledge on the subject.
Disclaimer :- The opinions expressed in this article belong solely to the author and may not necessarily reflect those of Merchant Navy Decoded. We cannot guarantee the accuracy of the information provided and disclaim any responsibility for it. Data and visuals used are sourced from publicly available information and may not be authenticated by any regulatory body. Reviews and comments appearing on our blogs represent the opinions of individuals and do not necessarily reflect the views of Merchant Navy Decoded. We are not responsible for any loss or damage resulting from reliance on these reviews or comments.
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View all comments | 1,103 | 5,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | longest | en | 0.944302 |
https://svn.geocomp.uq.edu.au/escript/trunk/ripley/src/Rectangle.cpp?sortby=log&r1=3750&r2=3752&pathrev=4769 | 1,571,747,174,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00494.warc.gz | 704,156,857 | 11,780 | # Diff of /trunk/ripley/src/Rectangle.cpp
revision 3750 by caltinay, Fri Dec 23 01:20:34 2011 UTC revision 3752 by caltinay, Tue Jan 3 08:06:36 2012 UTC
# Line 43 Rectangle::Rectangle(int n0, int n1, dou Line 43 Rectangle::Rectangle(int n0, int n1, dou
43 if (m_NX*m_NY != m_mpiInfo->size) if (m_NX*m_NY != m_mpiInfo->size)
44 throw RipleyException("Invalid number of spatial subdivisions"); throw RipleyException("Invalid number of spatial subdivisions");
45
46 if (n0%m_NX > 0 || n1%m_NY > 0) if ((n0+1)%m_NX > 0 || (n1+1)%m_NY > 0)
47 throw RipleyException("Number of elements must be separable into number of ranks in each dimension"); throw RipleyException("Number of elements+1 must be separable into number of ranks in each dimension");
48
49 // local number of elements if ((m_NX > 1 && (n0+1)/m_NX<2) || (m_NY > 1 && (n1+1)/m_NY<2))
50 m_NE0 = n0/m_NX; throw RipleyException("Too few elements for the number of ranks");
51 m_NE1 = n1/m_NY;
52 // local number of nodes (not necessarily owned) // local number of elements (including overlap)
53 m_NE0 = (m_NX>1 ? (n0+1)/m_NX : n0);
54 if (m_mpiInfo->rank%m_NX>0 && m_mpiInfo->rank%m_NX<m_NX-1)
55 m_NE0++;
56 m_NE1 = (m_NY>1 ? (n1+1)/m_NY : n1);
57 if (m_mpiInfo->rank/m_NX>0 && m_mpiInfo->rank/m_NX<m_NY-1)
58 m_NE1++;
59
60 // local number of nodes
61 m_N0 = m_NE0+1; m_N0 = m_NE0+1;
62 m_N1 = m_NE1+1; m_N1 = m_NE1+1;
63
64 // bottom-left node is at (offset0,offset1) in global mesh // bottom-left node is at (offset0,offset1) in global mesh
65 m_offset0 = m_NE0*(m_mpiInfo->rank%m_NX); m_offset0 = (n0+1)/m_NX*(m_mpiInfo->rank%m_NX);
66 m_offset1 = m_NE1*(m_mpiInfo->rank/m_NX); if (m_mpiInfo->rank%m_NX>0)
67 m_offset0--;
68 m_offset1 = (n1+1)/m_NY*(m_mpiInfo->rank/m_NX);
69 if (m_mpiInfo->rank/m_NX>0)
70 m_offset1--;
71
72 populateSampleIds(); populateSampleIds();
73 } }
74
# Line 250 const int* Rectangle::borrowSampleRefere Line 264 const int* Rectangle::borrowSampleRefere
264 bool Rectangle::ownSample(int fsCode, index_t id) const bool Rectangle::ownSample(int fsCode, index_t id) const
265 { {
266 #ifdef ESYS_MPI #ifdef ESYS_MPI
267 if (fsCode != ReducedNodes) { if (fsCode == Nodes) {
268 const index_t myFirst=m_nodeDistribution[m_mpiInfo->rank]; const index_t left = (m_offset0==0 ? 0 : 1);
269 const index_t myLast=m_nodeDistribution[m_mpiInfo->rank+1]-1; const index_t bottom = (m_offset1==0 ? 0 : 1);
270 return (m_nodeId[id]>=myFirst && m_nodeId[id]<=myLast); const index_t right = (m_mpiInfo->rank%m_NX==m_NX-1 ? m_N0 : m_N0-1);
271 const index_t top = (m_mpiInfo->rank/m_NX==m_NY-1 ? m_N1 : m_N1-1);
272 const index_t x=id%m_N0;
273 const index_t y=id/m_N0;
274 return (x>=left && x<right && y>=bottom && y<top);
275 } else { } else {
276 stringstream msg; stringstream msg;
277 msg << "ownSample() not implemented for " msg << "ownSample() not implemented for "
# Line 742 Paso_SystemMatrixPattern* Rectangle::get Line 760 Paso_SystemMatrixPattern* Rectangle::get
760 IndexVector offsetInShared(1,0); IndexVector offsetInShared(1,0);
761 IndexVector sendShared, recvShared; IndexVector sendShared, recvShared;
762 const IndexVector faces=getNumFacesPerBoundary(); const IndexVector faces=getNumFacesPerBoundary();
763 const index_t nDOF0 = (m_gNE0+1)/m_NX;
764 const index_t nDOF1 = (m_gNE1+1)/m_NY;
765 const int numDOF=nDOF0*nDOF1;
766 const index_t left = (m_offset0==0 ? 0 : 1); const index_t left = (m_offset0==0 ? 0 : 1);
767 const index_t bottom = (m_offset1==0 ? 0 : 1); const index_t bottom = (m_offset1==0 ? 0 : 1);
const int numDOF=getNumDOF();
768 vector<IndexVector> colIndices(numDOF); // for the couple blocks vector<IndexVector> colIndices(numDOF); // for the couple blocks
769 int dofCounter=numDOF; int numShared=0;
770
771 m_dofMap.assign(getNumNodes(), 0);
772 #pragma omp parallel for
773 for (index_t i=bottom; i<m_N1; i++) {
774 for (index_t j=left; j<m_N0; j++) {
775 m_dofMap[i*m_N0+j]=(i-bottom)*nDOF0+j-left;
776 }
777 }
778
779 // corner node from bottom-left // corner node from bottom-left
780 if (faces[0] == 0 && faces[2] == 0) { if (faces[0] == 0 && faces[2] == 0) {
781 neighbour.push_back(m_mpiInfo->rank-m_NX-1); neighbour.push_back(m_mpiInfo->rank-m_NX-1);
782 offsetInShared.push_back(offsetInShared.back()+1); offsetInShared.push_back(offsetInShared.back()+1);
783 sendShared.push_back(-1); sendShared.push_back(0);
784 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
785 colIndices[0].push_back(dofCounter-numDOF); colIndices[0].push_back(numShared);
786 ++dofCounter; m_dofMap[0]=numDOF+numShared;
787 ++numShared;
788 } }
789 // bottom edge // bottom edge
790 if (faces[2] == 0) { if (faces[2] == 0) {
791 neighbour.push_back(m_mpiInfo->rank-m_NX); neighbour.push_back(m_mpiInfo->rank-m_NX);
792 offsetInShared.push_back(offsetInShared.back()+m_N0-left); offsetInShared.push_back(offsetInShared.back()+nDOF0);
793 for (dim_t i=0; i<m_N0-left; i++, dofCounter++) { for (dim_t i=0; i<nDOF0; i++, numShared++) {
794 sendShared.push_back(i); sendShared.push_back(i);
795 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
796 if (i>0) if (i>0)
797 colIndices[i-1].push_back(dofCounter-numDOF); colIndices[i-1].push_back(numShared);
798 colIndices[i].push_back(dofCounter-numDOF); colIndices[i].push_back(numShared);
799 if (i<m_N0-left-1) if (i<nDOF0-1)
800 colIndices[i+1].push_back(dofCounter-numDOF); colIndices[i+1].push_back(numShared);
801 m_dofMap[i+left]=numDOF+numShared;
802 } }
803 } }
804 // corner node from bottom-right // corner node from bottom-right
805 if (faces[1] == 0 && faces[2] == 0) { if (faces[1] == 0 && faces[2] == 0) {
806 neighbour.push_back(m_mpiInfo->rank-m_NX+1); neighbour.push_back(m_mpiInfo->rank-m_NX+1);
807 offsetInShared.push_back(offsetInShared.back()+1); offsetInShared.push_back(offsetInShared.back()+1);
808 sendShared.push_back(-1); sendShared.push_back(nDOF0-1);
809 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
810 colIndices[m_N0-left-2].push_back(dofCounter-numDOF); colIndices[nDOF0-2].push_back(numShared);
811 colIndices[m_N0-left-1].push_back(dofCounter-numDOF); colIndices[nDOF0-1].push_back(numShared);
812 ++dofCounter; m_dofMap[m_N0-1]=numDOF+numShared;
813 ++numShared;
814 } }
815 // right edge // right edge
816 if (faces[1] == 0) { if (faces[1] == 0) {
817 neighbour.push_back(m_mpiInfo->rank+1); neighbour.push_back(m_mpiInfo->rank+1);
818 offsetInShared.push_back(offsetInShared.back()+m_N1-bottom); offsetInShared.push_back(offsetInShared.back()+nDOF1);
819 for (dim_t i=0; i<m_N1-bottom; i++, dofCounter++) { for (dim_t i=0; i<nDOF1; i++, numShared++) {
820 sendShared.push_back((i+1)*(m_N0-left)-1); sendShared.push_back((i+1)*(nDOF0)-1);
821 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
822 if (i>0) if (i>0)
823 colIndices[i*(m_N0-left)-1].push_back(dofCounter-numDOF); colIndices[i*(nDOF0)-1].push_back(numShared);
824 colIndices[(i+1)*(m_N0-left)-1].push_back(dofCounter-numDOF); colIndices[(i+1)*(nDOF0)-1].push_back(numShared);
825 if (i<m_N1-bottom-1) if (i<nDOF1-1)
826 colIndices[(i+2)*(m_N0-left)-1].push_back(dofCounter-numDOF); colIndices[(i+2)*(nDOF0)-1].push_back(numShared);
827 m_dofMap[(i+bottom+1)*m_N0-1]=numDOF+numShared;
828 } }
829 } }
830 // corner node from top-right // corner node from top-right
# Line 800 Paso_SystemMatrixPattern* Rectangle::get Line 832 Paso_SystemMatrixPattern* Rectangle::get
832 neighbour.push_back(m_mpiInfo->rank+m_NX+1); neighbour.push_back(m_mpiInfo->rank+m_NX+1);
833 offsetInShared.push_back(offsetInShared.back()+1); offsetInShared.push_back(offsetInShared.back()+1);
834 sendShared.push_back(numDOF-1); sendShared.push_back(numDOF-1);
835 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
836 colIndices[numDOF-1].push_back(dofCounter-numDOF); colIndices[numDOF-1].push_back(numShared);
837 ++dofCounter; m_dofMap[m_N0*m_N1-1]=numDOF+numShared;
838 ++numShared;
839 } }
840 // top edge // top edge
841 if (faces[3] == 0) { if (faces[3] == 0) {
842 neighbour.push_back(m_mpiInfo->rank+m_NX); neighbour.push_back(m_mpiInfo->rank+m_NX);
843 offsetInShared.push_back(offsetInShared.back()+m_N0-left); offsetInShared.push_back(offsetInShared.back()+nDOF0);
844 for (dim_t i=0; i<m_N0-left; i++, dofCounter++) { for (dim_t i=0; i<nDOF0; i++, numShared++) {
845 sendShared.push_back(numDOF-m_N0+left+i); sendShared.push_back(numDOF-nDOF0+i);
846 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
847 if (i>0) if (i>0)
848 colIndices[numDOF-m_N0+left+i-1].push_back(dofCounter-numDOF); colIndices[numDOF-nDOF0+i-1].push_back(numShared);
849 colIndices[numDOF-m_N0+left+i].push_back(dofCounter-numDOF); colIndices[numDOF-nDOF0+i].push_back(numShared);
850 if (i<m_N0-left-1) if (i<nDOF0-1)
851 colIndices[numDOF-m_N0+left+i+1].push_back(dofCounter-numDOF); colIndices[numDOF-nDOF0+i+1].push_back(numShared);
852 m_dofMap[m_N0*(m_N1-1)+i+left]=numDOF+numShared;
853 } }
854 } }
855 // corner node from top-left // corner node from top-left
856 if (faces[0] == 0 && faces[3] == 0) { if (faces[0] == 0 && faces[3] == 0) {
857 neighbour.push_back(m_mpiInfo->rank+m_NX-1); neighbour.push_back(m_mpiInfo->rank+m_NX-1);
858 offsetInShared.push_back(offsetInShared.back()+1); offsetInShared.push_back(offsetInShared.back()+1);
859 sendShared.push_back(numDOF-m_N0+left); sendShared.push_back(numDOF-nDOF0);
860 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
861 colIndices[numDOF-m_N0+left].push_back(dofCounter-numDOF); colIndices[numDOF-nDOF0].push_back(numShared);
862 ++dofCounter; m_dofMap[m_N0*(m_N1-1)]=numDOF+numShared;
863 ++numShared;
864 } }
865 // left edge // left edge
866 if (faces[0] == 0) { if (faces[0] == 0) {
867 neighbour.push_back(m_mpiInfo->rank-1); neighbour.push_back(m_mpiInfo->rank-1);
868 offsetInShared.push_back(offsetInShared.back()+m_N1-bottom); offsetInShared.push_back(offsetInShared.back()+nDOF1);
869 for (dim_t i=0; i<m_N1-bottom; i++, dofCounter++) { for (dim_t i=0; i<nDOF1; i++, numShared++) {
870 sendShared.push_back(-1); sendShared.push_back(i*nDOF0);
871 recvShared.push_back(dofCounter); recvShared.push_back(numDOF+numShared);
872 if (i>0) if (i>0)
873 colIndices[(i-1)*(m_N0-left)].push_back(dofCounter-numDOF); colIndices[(i-1)*nDOF0].push_back(numShared);
874 colIndices[i*(m_N0-left)].push_back(dofCounter-numDOF); colIndices[i*nDOF0].push_back(numShared);
875 if (i<m_N1-bottom-1) if (i<nDOF1-1)
876 colIndices[(i+1)*(m_N0-left)].push_back(dofCounter-numDOF); colIndices[(i+1)*nDOF0].push_back(numShared);
877 m_dofMap[(i+bottom)*m_N0]=numDOF+numShared;
878 } }
879 } }
880 /**/
881 /*
882 cout << "--- rcv_shcomp ---" << endl; cout << "--- rcv_shcomp ---" << endl;
883 cout << "numDOF=" << numDOF << ", numNeighbors=" << neighbour.size() << endl; cout << "numDOF=" << numDOF << ", numNeighbors=" << neighbour.size() << endl;
884 for (size_t i=0; i<neighbour.size(); i++) { for (size_t i=0; i<neighbour.size(); i++) {
# Line 855 Paso_SystemMatrixPattern* Rectangle::get Line 892 Paso_SystemMatrixPattern* Rectangle::get
892 for (size_t i=0; i<sendShared.size(); i++) { for (size_t i=0; i<sendShared.size(); i++) {
893 cout << "shared[" << i << "]=" << sendShared[i] << endl; cout << "shared[" << i << "]=" << sendShared[i] << endl;
894 } }
895 /**/ */
896
897 Paso_SharedComponents *snd_shcomp = Paso_SharedComponents_alloc( Paso_SharedComponents *snd_shcomp = Paso_SharedComponents_alloc(
898 numDOF, neighbour.size(), &neighbour[0], &sendShared[0], numDOF, neighbour.size(), &neighbour[0], &sendShared[0],
# Line 914 Paso_SystemMatrixPattern* Rectangle::get Line 951 Paso_SystemMatrixPattern* Rectangle::get
951 copy(index.begin(), index.end(), indexC); copy(index.begin(), index.end(), indexC);
952 copy(ptr.begin(), ptr.end(), ptrC); copy(ptr.begin(), ptr.end(), ptrC);
953 M=ptr.size()-1; M=ptr.size()-1;
954 N=dofCounter-numDOF; N=numShared;
955 Paso_Pattern *colCouplePattern=Paso_Pattern_alloc(MATRIX_FORMAT_DEFAULT, Paso_Pattern *colCouplePattern=Paso_Pattern_alloc(MATRIX_FORMAT_DEFAULT,
956 M, N, ptrC, indexC); M, N, ptrC, indexC);
957
958 /**/ /*
959 cout << "--- colCouple_pattern ---" << endl; cout << "--- colCouple_pattern ---" << endl;
960 cout << "M=" << M << ", N=" << N << endl; cout << "M=" << M << ", N=" << N << endl;
961 for (size_t i=0; i<ptr.size(); i++) { for (size_t i=0; i<ptr.size(); i++) {
# Line 927 Paso_SystemMatrixPattern* Rectangle::get Line 964 Paso_SystemMatrixPattern* Rectangle::get
964 for (size_t i=0; i<index.size(); i++) { for (size_t i=0; i<index.size(); i++) {
965 cout << "index[" << i << "]=" << index[i] << endl; cout << "index[" << i << "]=" << index[i] << endl;
966 } }
967 /**/ */
968
969 // now build the row couple pattern // now build the row couple pattern
970 IndexVector ptr2(1,0); IndexVector ptr2(1,0);
# Line 954 Paso_SystemMatrixPattern* Rectangle::get Line 991 Paso_SystemMatrixPattern* Rectangle::get
991 Paso_Pattern *rowCouplePattern=Paso_Pattern_alloc(MATRIX_FORMAT_DEFAULT, Paso_Pattern *rowCouplePattern=Paso_Pattern_alloc(MATRIX_FORMAT_DEFAULT,
992 N, M, ptrC, indexC); N, M, ptrC, indexC);
993
994 /**/ /*
995 cout << "--- rowCouple_pattern ---" << endl; cout << "--- rowCouple_pattern ---" << endl;
996 cout << "M=" << N << ", N=" << M << endl; cout << "M=" << N << ", N=" << M << endl;
997 for (size_t i=0; i<ptr2.size(); i++) { for (size_t i=0; i<ptr2.size(); i++) {
# Line 963 Paso_SystemMatrixPattern* Rectangle::get Line 1000 Paso_SystemMatrixPattern* Rectangle::get
1000 for (size_t i=0; i<index2.size(); i++) { for (size_t i=0; i<index2.size(); i++) {
1001 cout << "index[" << i << "]=" << index2[i] << endl; cout << "index[" << i << "]=" << index2[i] << endl;
1002 } }
1003 /**/ */
1004
1005 // allocate paso distribution // allocate paso distribution
1006 Paso_Distribution* distribution = Paso_Distribution_alloc(m_mpiInfo, Paso_Distribution* distribution = Paso_Distribution_alloc(m_mpiInfo,
# Line 1044 pair<double,double> Rectangle::getFirstC Line 1081 pair<double,double> Rectangle::getFirstC
1081 //protected //protected
1082 dim_t Rectangle::getNumDOF() const dim_t Rectangle::getNumDOF() const
1083 { {
1084 return m_nodeDistribution[m_mpiInfo->rank+1] return (m_gNE0+1)/m_NX*(m_gNE1+1)/m_NY;
-m_nodeDistribution[m_mpiInfo->rank];
1085 } }
1086
1087 //protected //protected
# Line 1084 void Rectangle::assembleCoordinates(escr Line 1120 void Rectangle::assembleCoordinates(escr
1120 //private //private
1121 void Rectangle::populateSampleIds() void Rectangle::populateSampleIds()
1122 { {
1123 // identifiers are ordered from left to right, bottom to top on each rank, // identifiers are ordered from left to right, bottom to top globablly.
// except for the shared nodes which are owned by the rank below / to the
// left of the current rank
1124
1125 // build node distribution vector first. // build node distribution vector first.
// m_nodeDistribution[i] is the first node id on rank i, that is
1126 // rank i owns m_nodeDistribution[i+1]-nodeDistribution[i] nodes // rank i owns m_nodeDistribution[i+1]-nodeDistribution[i] nodes
1127 m_nodeDistribution.assign(m_mpiInfo->size+1, 0); m_nodeDistribution.assign(m_mpiInfo->size+1, 0);
1128 m_nodeDistribution[1]=getNumNodes(); const dim_t numDOF=getNumDOF();
1129 for (dim_t k=1; k<m_mpiInfo->size-1; k++) { for (dim_t k=1; k<m_mpiInfo->size; k++) {
1130 const index_t x=k%m_NX; m_nodeDistribution[k]=k*numDOF;
const index_t y=k/m_NX;
index_t numNodes=getNumNodes();
if (x>0)
numNodes-=m_N1;
if (y>0)
numNodes-=m_N0;
if (x>0 && y>0)
numNodes++; // subtracted corner twice -> fix that
m_nodeDistribution[k+1]=m_nodeDistribution[k]+numNodes;
1131 } }
1132 m_nodeDistribution[m_mpiInfo->size]=getNumDataPointsGlobal(); m_nodeDistribution[m_mpiInfo->size]=getNumDataPointsGlobal();
m_dofId.resize(getNumDOF());
1133 m_nodeId.resize(getNumNodes()); m_nodeId.resize(getNumNodes());
1134
// the bottom row and left column are not owned by this rank so the
// identifiers need to be computed accordingly
const index_t left = (m_offset0==0 ? 0 : 1);
const index_t bottom = (m_offset1==0 ? 0 : 1);
if (left>0) {
const int neighbour=m_mpiInfo->rank-1;
const index_t leftN0=(neighbour%m_NX == 0 ? m_N0 : m_N0-1);
1135 #pragma omp parallel for #pragma omp parallel for
1136 for (dim_t i1=bottom; i1<m_N1; i1++) { for (dim_t i1=0; i1<m_N1; i1++) {
1137 m_nodeId[i1*m_N0]=m_nodeDistribution[neighbour] for (dim_t i0=0; i0<m_N0; i0++) {
1138 + (i1-bottom+1)*leftN0-1; m_nodeId[i0+i1*m_N0] = (m_offset1+i1)*(m_gNE0+1)+m_offset0+i0;
1139 } }
1140 } }
if (bottom>0) {
const int neighbour=m_mpiInfo->rank-m_NX;
const index_t bottomN0=(neighbour%m_NX == 0 ? m_N0 : m_N0-1);
const index_t bottomN1=(neighbour/m_NX == 0 ? m_N1 : m_N1-1);
#pragma omp parallel for
for (dim_t i0=left; i0<m_N0; i0++) {
m_nodeId[i0]=m_nodeDistribution[neighbour]
+ (bottomN1-1)*bottomN0 + i0 - left;
}
}
if (left>0 && bottom>0) {
const int neighbour=m_mpiInfo->rank-m_NX-1;
const index_t N0=(neighbour%m_NX == 0 ? m_N0 : m_N0-1);
const index_t N1=(neighbour/m_NX == 0 ? m_N1 : m_N1-1);
m_nodeId[0]=m_nodeDistribution[neighbour]+N0*N1-1;
}
1141
1142 // the rest of the id's are contiguous m_dofId.resize(numDOF);
1143 const index_t firstId=m_nodeDistribution[m_mpiInfo->rank]; const index_t firstId=m_nodeDistribution[m_mpiInfo->rank];
1144 #pragma omp parallel for for (dim_t i=0; i<numDOF; i++)
1145 for (dim_t i1=bottom; i1<m_N1; i1++) { m_dofId[i] = firstId+i;
1146 for (dim_t i0=left; i0<m_N0; i0++) {
const index_t idx=i0-left+(i1-bottom)*(m_N0-left);
m_nodeId[i0+i1*m_N0] = firstId+idx;
m_dofId[idx] = firstId+idx;
}
}
1147 m_nodeTags.assign(getNumNodes(), 0); m_nodeTags.assign(getNumNodes(), 0);
1148 updateTagsInUse(Nodes); updateTagsInUse(Nodes);
1149
# Line 1193 void Rectangle::populateSampleIds() Line 1187 void Rectangle::populateSampleIds()
1187 //private //private
1188 int Rectangle::insertNeighbours(IndexVector& index, index_t node) const int Rectangle::insertNeighbours(IndexVector& index, index_t node) const
1189 { {
1190 const dim_t myN0 = (m_offset0==0 ? m_N0 : m_N0-1); const index_t myN0 = (m_gNE0+1)/m_NX;
1191 const dim_t myN1 = (m_offset1==0 ? m_N1 : m_N1-1); const index_t myN1 = (m_gNE1+1)/m_NY;
1192 const int x=node%myN0; const int x=node%myN0;
1193 const int y=node/myN0; const int y=node/myN0;
1194 int num=0; int num=0;
# Line 1412 void Rectangle::nodesToDOF(escript::Data Line 1406 void Rectangle::nodesToDOF(escript::Data
1406
1407 const index_t left = (m_offset0==0 ? 0 : 1); const index_t left = (m_offset0==0 ? 0 : 1);
1408 const index_t bottom = (m_offset1==0 ? 0 : 1); const index_t bottom = (m_offset1==0 ? 0 : 1);
1409 const index_t right = (m_mpiInfo->rank%m_NX==m_NX-1 ? m_N0 : m_N0-1);
1410 const index_t top = (m_mpiInfo->rank/m_NX==m_NY-1 ? m_N1 : m_N1-1);
1411 index_t n=0; index_t n=0;
1412 for (index_t i=bottom; i<m_N1; i++) { for (index_t i=bottom; i<top; i++) {
1413 for (index_t j=left; j<m_N0; j++, n++) { for (index_t j=left; j<right; j++, n++) {
1414 const double* src=in.getSampleDataRO(j+i*m_N0); const double* src=in.getSampleDataRO(j+i*m_N0);
1415 copy(src, src+numComp, out.getSampleDataRW(n)); copy(src, src+numComp, out.getSampleDataRW(n));
1416 } }
# Line 2132 void Rectangle::assemblePDESingle(Paso_S Line 2128 void Rectangle::assemblePDESingle(Paso_S
2128 /* GENERATOR SNIP_PDE_SINGLE BOTTOM */ /* GENERATOR SNIP_PDE_SINGLE BOTTOM */
2129
2131 const index_t left = (m_offset0==0 ? 0 : 1); const index_t firstNode=m_N0*k1+k0;
const index_t bottom = (m_offset1==0 ? 0 : 1);
const index_t firstNode=(m_N0-bottom)*k1+k0-left;
const int numDOF=getNumDOF();
int dof=numDOF;
2132 IndexVector rowIndex; IndexVector rowIndex;
2133 if (m_offset0==0) { rowIndex.push_back(m_dofMap[firstNode]);
2134 if (m_offset1>0 && k1==0) { rowIndex.push_back(m_dofMap[firstNode+1]);
2135 rowIndex.push_back(dof++); rowIndex.push_back(m_dofMap[firstNode+m_N0]);
2136 rowIndex.push_back(dof++); rowIndex.push_back(m_dofMap[firstNode+m_N0+1]);
rowIndex.push_back(firstNode);
rowIndex.push_back(firstNode+1);
} else {
rowIndex.push_back(firstNode);
rowIndex.push_back(firstNode+1);
rowIndex.push_back(firstNode+m_N0-left);
rowIndex.push_back(firstNode+m_N0-left+1);
}
} else {
if (k0==0) {
rowIndex.push_back(dof++);
if (m_offset1>0 && k1==0)
rowIndex.push_back(dof++);
else
rowIndex.push_back(firstNode);
rowIndex.push_back(dof++);
rowIndex.push_back(firstNode+m_N0-left);
} else {
rowIndex.push_back(firstNode);
rowIndex.push_back(firstNode+1);
rowIndex.push_back(firstNode+m_N0-left);
rowIndex.push_back(firstNode+m_N0-left+1);
}
}
Paso_Coupler_startCollect(mat->col_coupler, &EM_F[0]);
double* recvBuffer=Paso_Coupler_finishCollect(mat->col_coupler);
for (int i=0; i<mat->col_coupler->connector->send->numSharedComponents; i++) {
if (mat->col_coupler->connector->send->shared[i]>=0)
EM_F[mat->col_coupler->connector->send->shared[i]]+=recvBuffer[i];
}
IndexVector colIndex=rowIndex;
/*
Paso_Coupler* coupler=Paso_Coupler_alloc(mat->col_coupler->connector, 4);
Paso_Coupler_startCollect(coupler, &EM_S[0]);
recvBuffer=Paso_Coupler_finishCollect(coupler);
for (int i=0; i<coupler->connector->recv->numSharedComponents; i++) {
if (coupler->connector->recv->shared[i]>=0) {
for (int j=0; j<4; j++)
EM_S.push_back(recvBuffer[i+4*j]);
colIndex.push_back(coupler->connector->recv->shared[i]);
}
}
Paso_Coupler_free(coupler);*/
2138 cout << "-- AddtoRHS -- " << endl; //cout << "-- AddtoRHS -- " << endl;
2139 double *F_p=rhs.getSampleDataRW(0); double *F_p=rhs.getSampleDataRW(0);
cout << "F:"<<endl;
2140 for (index_t i=0; i<4; i++) { for (index_t i=0; i<4; i++) {
2141 if (rowIndex[i]<getNumDOF()) { if (rowIndex[i]<getNumDOF()) {
2142 F_p[rowIndex[i]]+=EM_F[i]; F_p[rowIndex[i]]+=EM_F[i];
2143 cout << "F[" << rowIndex[i] << "]=" << F_p[rowIndex[i]] << endl; //cout << "F[" << rowIndex[i] << "]=" << F_p[rowIndex[i]] << endl;
2144 } }
2145 } }
2146 cout << "---"<<endl; //cout << "---"<<endl;
2147 } }
2149 cout << "-- AddtoSystem -- " << endl; //cout << "-- AddtoSystem -- " << endl;
2150 addToSystemMatrix(mat, rowIndex, 1, colIndex, 1, EM_S); addToSystemMatrix(mat, rowIndex, 1, rowIndex, 1, EM_S);
2151 } }
2152
2153 } // end k0 loop } // end k0 loop
2177 // down columns of array // down columns of array
2178 const dim_t j_Eq = nodes_Eq[k_Eq]; const dim_t j_Eq = nodes_Eq[k_Eq];
2179 const dim_t i_row = j_Eq; const dim_t i_row = j_Eq;
2180 printf("row:%d\n", i_row); //printf("row:%d\n", i_row);
2181 // only look at the matrix rows stored on this processor // only look at the matrix rows stored on this processor
2182 if (i_row < numMyRows) { if (i_row < numMyRows) {
2183 for (dim_t k_Sol = 0; k_Sol < nodes_Sol.size(); ++k_Sol) { for (dim_t k_Sol = 0; k_Sol < nodes_Sol.size(); ++k_Sol) {
# Line 2246 printf("row:%d\n", i_row); Line 2191 printf("row:%d\n", i_row);
2191 } }
2192 } else { } else {
2193 for (dim_t k = col_coupleBlock_ptr[i_row]; k < col_coupleBlock_ptr[i_row + 1]; ++k) { for (dim_t k = col_coupleBlock_ptr[i_row]; k < col_coupleBlock_ptr[i_row + 1]; ++k) {
2194 cout << "col:" << i_col-numMyCols << " colIdx:" << col_coupleBlock_index[k] << endl; //cout << "col:" << i_col-numMyCols << " colIdx:" << col_coupleBlock_index[k] << endl;
2195 if (col_coupleBlock_index[k] == i_col - numMyCols) { if (col_coupleBlock_index[k] == i_col - numMyCols) {
2196 col_coupleBlock_val[k] += array[INDEX2(k_Eq, k_Sol, nodes_Eq.size())]; col_coupleBlock_val[k] += array[INDEX2(k_Eq, k_Sol, nodes_Eq.size())];
2197 break; break;
# Line 2262 cout << "col:" << i_col-numMyCols << " c Line 2207 cout << "col:" << i_col-numMyCols << " c
2207 for (dim_t k = row_coupleBlock_ptr[i_row - numMyRows]; for (dim_t k = row_coupleBlock_ptr[i_row - numMyRows];
2208 k < row_coupleBlock_ptr[i_row - numMyRows + 1]; ++k) k < row_coupleBlock_ptr[i_row - numMyRows + 1]; ++k)
2209 { {
2210 cout << "col:" << i_col << " rowIdx:" << row_coupleBlock_index[k] << endl; //cout << "col:" << i_col << " rowIdx:" << row_coupleBlock_index[k] << endl;
2211 if (row_coupleBlock_index[k] == i_col) { if (row_coupleBlock_index[k] == i_col) {
2212 row_coupleBlock_val[k] += array[INDEX2(k_Eq, k_Sol, nodes_Eq.size())]; row_coupleBlock_val[k] += array[INDEX2(k_Eq, k_Sol, nodes_Eq.size())];
2213 break; break;
Legend:
Removed from v.3750 changed lines Added in v.3752 | 9,700 | 29,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.254023 |
https://kidsworksheetfun.com/percent-decimal-and-fraction-worksheet-pdf/ | 1,708,824,018,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00091.warc.gz | 338,610,694 | 25,255 | Percent Decimal And Fraction Worksheet Pdf
Percent Decimal And Fraction Worksheet Pdf. Convert the given fraction to percentage. 10 fun fractions decimals percentages worksheets.
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Our Percentages Worksheets Pdf Presents Very Simple And Several Ways That Percentages Can Be Used To Compare Values, Quantity Change Etc.
Convert the given mixed fractions to decimals. Then, you should multiply the recent result only by 100, or also you can move the two decimal points to the right side. Simplify your fractions use mental math (no calculator!) to find the percentages:
Convert The Given Mixed Fractions To Decimal.
Steps to convert a fraction to a decimal. The first section is just converting fractions into decimals and percents. Fraction, decimal, percent worksheet download printable pdf.
Fraction Decimal Percent Worksheet Pdf Fractions Decimals Percents Is A Free Printable For You.
The second sections is about converting decimals to percents and fractions. Fraction to decimal to percent worksheets. Fractions decimals and percentages pdf.
In The 2Nd Worksheet, They Are Also Asked To Simplify The Resulting Fraction.
Multiply the obtained decimal by the 100. Fraction decimal percent worksheet pdf fractions decimals percents is a free printable for you. In the last worksheet, students convert the percent to a decimal as. | 354 | 1,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.740775 |
http://priorart.ip.com/IPCOM/000103047 | 1,513,416,156,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00745.warc.gz | 232,774,782 | 7,531 | Browse Prior Art Database
# Voltage Control Oscillation (VCO) Delta-V Circuit
IP.com Disclosure Number: IPCOM000103047D
Original Publication Date: 1990-May-01
Included in the Prior Art Database: 2005-Mar-17
Document File: 1 page(s) / 47K
IBM
## Related People
Strom, JD: AUTHOR
## Abstract
Described is a circuit (illustrated in the figure) which increases the voltage control oscillation (VCO) frequency range by 30 percent over a design which uses a fixed current in NPN Q1.
This text was extracted from an ASCII text file.
This is the abbreviated version, containing approximately 90% of the total text.
Voltage Control Oscillation (VCO) Delta-V Circuit
Described is a circuit (illustrated in the figure) which
increases the voltage control oscillation (VCO) frequency range by 30
percent over a design which uses a fixed current in NPN Q1.
Frequency of oscillation (F) of VCO in the figure is given by F
= ((ICP / (4 * C * DELTA-V)). It is desirable that the statistical
variation of F for a fixed charge-pump current (ICP) be as small as
possible. It is also desirable that F vary linearly with changes in
ICP. DELTA-V (DV) is the differential voltage across nodes N7 and
N8. This is also the voltage across the capacitor as well as the
output of the VCO. If Q5, Q5a, are conducting current, DV can be
written as follows: Vbeq7 - (Vbeq1*R7/(R5+R7). If Q6, Q6a are
conducting current, DV is Vbeq8 - (Vbeq1*R8/(R6+R8). In order for
the variation of DV as well as F to be small, Vbeq8,Vbeq7 and Vbeq1
must track closely as the various circuit parameters change.
In the circuit, the NPN Q1 is biased at ((ICP-(Vbeq1/(R8 R6))
by the current source formed by resistor R3 and NPN Q3. NPN's Q7 and
Q7a are biased at (2*ICP-(Vbeq7/R7))/2 which is equal to the current
carried by Q8, Q8a when they are conducting current via symmetry.
Since resistors R7, R8, R5 and R6 are equal, the current in NPNs Q1,
Q7, Q7a, Q8, Q8a are all equal. Thus, the voltage at nodes N7 and N8
track closely, and the variation of the differential voltage between
nodes N8 and N7... | 615 | 2,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-51 | latest | en | 0.916055 |
http://mymathforum.com/differential-equations/554-differential-equation.html | 1,566,106,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313617.6/warc/CC-MAIN-20190818042813-20190818064813-00222.warc.gz | 128,178,299 | 9,065 | My Math Forum Differential equation
Differential Equations Ordinary and Partial Differential Equations Math Forum
April 12th, 2007, 10:20 PM #1 Newbie Joined: Apr 2007 Posts: 6 Thanks: 0 Differential equation somebody that can help me solve this? find the general solution dy/dt=-2y I know that the preposition dy/dt=ay gives y=ce^at and the answer is supposed to be y=ac^-2t, but proving this was worse...
April 13th, 2007, 05:09 AM #2 Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 In other words, you are saying that: y' + 2y = 0 If we had initial values, we could solve this by Laplace transformations, but since we don't, we'll have to go by a different route. We'll have to say that r^2 + 2 = 0 and then solve for the complex number r, and then use the two real parts of e^r as the terms that will give us the solution to the problem.
April 14th, 2007, 01:43 AM #3 Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 Multiply by e^(2t), then integrate.
May 6th, 2007, 03:29 PM #4 Senior Member Joined: May 2007 Posts: 402 Thanks: 0 dy/dt = -2y multiply with dy and divide with y and you get: dy/y=-2dt integrate and: ln(y)-ln(y0)=-2t ln(y)=-2*t+ln(y0) y=exp(-2t+ln(y0)) y=y0*exp(-2t) where y0 is the initial value!
May 6th, 2007, 11:36 PM #5
Global Moderator
Joined: Dec 2006
Posts: 20,919
Thanks: 2202
Quote:
Originally Posted by Infinity If we had initial values, we could solve this by Laplace transformations, . . .
One can write y0 for y(0) and use the Laplace transform method. Don't expect to get y=ac^-2t, since the correct solution is y = y0e^(-2t).
Alternatively, one can solve the characteristic equation, r + 2 = 0 (not r^2 + 2 = 0), and hence write down the answer.
Tags differential, equation
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Contact - Home - Forums - Cryptocurrency Forum - Top | 679 | 2,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.924958 |
https://vegandivasnyc.com/parallelogram-rectangle-rhombus-and-square-worksheet-answers/ | 1,680,156,281,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00756.warc.gz | 665,937,160 | 11,948 | A Comprehensive Guide to Parallelogram Rectangle Rhombus and Square Worksheet Answers
What is a Parallelogram?
A parallelogram is a four-sided polygon that has two pairs of parallel sides. The opposite sides of a parallelogram are equal in length and the opposite angles are equal. The sum of all the angles of a parallelogram is 360°.
What is a Rectangle?
A rectangle is a four-sided polygon with four right angles and four equal sides. All the angles of a rectangle are equal to 90°. The diagonals of a rectangle are equal in length.
What is a Rhombus?
A rhombus is a four-sided polygon with four equal sides and two pairs of parallel sides. All the angles of a rhombus are equal. The diagonals of a rhombus are not equal in length.
What is a Square?
A square is a four-sided polygon with four equal sides and four right angles. All the angles of a square are equal to 90°. The diagonals of a square are equal in length.
Parallelogram Rectangle Rhombus and Square Worksheet Answers
Answers to the questions in the worksheet can be found online. The answers will depend on the version of the worksheet and the questions asked. Generally, the answers will involve calculating the area and perimeter of the polygons, and determining the angles and lengths of the sides. | 285 | 1,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-14 | latest | en | 0.895191 |
http://www.pagalguy.com/@Kunal21 | 1,462,319,840,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121985.80/warc/CC-MAIN-20160428161521-00062-ip-10-239-7-51.ec2.internal.warc.gz | 735,893,670 | 18,259 | # @Kunal21
##### Kunal Vohra 354 karma
###### LoungeT-20 World cup 2012 + Fantasy Muqabla....
Count me in this time
Dear Sir,
I have just completed my graduation and have started working with a leading management consulting. I have a few questions to ask:
1) Am i still eligible for YLP?
2) If I apply for deferred admission, what all factors would be taken into consideration while judging my candidature? Would my position and domain be taken into consideration?
3) Do I need to have a high GMAT score than others to make up for my lack of work ex?
4) In case the dates have already been passed this year, when would the process start next year? (Tentative dates, in case it hasn't been finalized till now)
Sincerely,
Kunal
###### GMATISB Class of 2014 aspirants.
@[562370:chetso]: Thanks for reverting back.
Since I am new in this league, could you please quote a range of GMAT scores that ISB has accepted till date?
Moreover, I have already graduated this year. Is YLP still open to me?
###### GMATISB Class of 2014 aspirants.
Hi Puys,
I was wondering if I could get some info regarding deferred admission into ISB?
I've just completed my graduation and have landed myself into a job in a leading management consulting. Please let me know all the relevant details. Is there any means from which we can get out profile evaluated?
TIA
###### CATCAT 2012 - Registration - Discussion Thread
Captcha daal daal ke jaan nikal gai. :/
• 2 Likes
Commenting on this post has been disabled.
###### CATOfficial Quant Thread for CAT 2012
@Predilected said: remainder 3^268/22=????
Should be 5.
3^5/22 becomes one and hence 3^3/22 it should be.
Commenting on this post has been disabled.
###### CATOfficial Quant Thread for CAT 2012
27 ...
let x y z be 13p 13 q 13r resp.
so
p+q+r = 9
but each of p q r has to be +ve
put p' = p+1 ... similarly for q and r
p'+q'+r' = 6
so 8c2 = 28 ways
but when all are equal i.e at p=q=r = 3 x y z wont be co prime
so total 28-1 = 27 triplets
The HCF of three natural numbers x, y and z is 13. If the sum of x, y and z is 117, then how many ordered triplets (x, y, z) exist?
a. 28
b. 27
c. 54
d. 55
let the numbers be 13a,13b,13c
13a+13b+13c = 117
a+b+c = 9
a,b,c>0
8c2 = 28
- case when all three are equal
27
Arre bhai log,
8C2 kyun aaya? Koi please explain kardo
• 1 Like
Commenting on this post has been disabled.
Apoorv Pandit @Apurv
_Ranked 16th among US business schools by BusinessWeek, University of North Carolina's Kenan-Flagler Business School is popular among Indian applicants and accepts a median GMAT score of 690 for its MBA class of 298 seat
Kunal Vohra @Kunal21
Just a small question, Bcom type degrees include 3 year B.Sc degrees from DU too, right? Good to see US schools opening their gates for 3 year Indian degrees finally!
Lajwanti D'Souza @laj
_Satish Chandra Buriuly, student, IIM Ranchi class of 2012_ You could call this a different kind of 'rags to riches' story. Only, there were no rags and there are no riches either. Satish Chandra Buriuly's sto
Kunal Vohra @Kunal21
Very inspiring! With no intentions to spark off the same old debate, I'd just like to add that these kind of people deserve to get ahead and make something for themselves. They deserve to get ahead and see all the other(Read: Happy) side of life. But IIMs except for C and R would not realize how he reached here with 51% or 50% but would simply look for numbers in any candidate's profile. That applies for anyone be it a general, OBC, SC or ST candidate. There's something called Random Component in life which you can't predict and is inevitable. Sad. Attaboy Satish. Do well. ATB Kunal V
###### GMATGmat 760 - q-50/v-44
B.Com (H) from DU, 3 yrs work ex in advertising. Am lookin at the top 15 in the US. Might apply to R3 of one or two but mainly targeting R1 in Sept for class of 2012-14
Hi, Just wanted a quick check on how can you apply in top 15 US B-Schools with a 3 year Indian degree? Most of the univs in US demand a 4 year US equivalent i.e a 4 year Indian degree or a 3 years bachelors + 2 years masters.
Wouldn't you face any hindrance due to this?
How do you plan to bridge the gap between a 3 year Indian degree and a 4 year US degree? I presume you just have a Bcom(H) from DU.
Thanks :)
P.S - I too am sailing on the same boat. Still contemplating my options for bridging the gap. A little insight into this would be of great help. | 1,248 | 4,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-18 | latest | en | 0.930624 |
http://matrix-messenger.tripod.com/mlmmsg258.htm | 1,534,305,866,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209856.3/warc/CC-MAIN-20180815024253-20180815044253-00137.warc.gz | 260,438,819 | 6,675 | #### Michael Lawrence Morton's Matrix Message 258 Yeah .. The Royal Cubit !! ..
```
Michael Lawrence Morton
Yeah .. The Royal Cubit !! ..
Sat Feb 16 22:29:06 2002
<< "P.S. That decimal harmonic Chronos has, there ..
is that of The Royal Cubit in regular British feet ..
another proposal of mine (Morton, 1998, Internet) ..
1.718873386 regular feet = 20.62648063 regular inches;
true-and-originally-intended length of The Royal Cubit;
sez me." .. (MLM).
{Chronos} .. Okay so you have a Royal Cubit which is (Surface Area of Hemisphere / 100)
Inches and Pi / 10 x ((1 / Squared Megalithic Yard)^2 Feet ???
What a fine metrological link THAT is!!! Wow!
Cheers,
Chronos >>
Yes !! I've had it posted .. several times ..
over the past 4 years .. on The Internet.
I've had it posted many times on The Internet, actually.
Yeah .. it's *perfectly-harmonious* with the numerical
value of The Radian Arc constant .. *where* the 360
arc-degrees system_is_used.
Yes .. absolutely !! (-;
I really do think it_was/is_the true length of The
Royal Cubit !! Which_also_implies the "regular British
foot, and inch" !! And; the 5280 'regular feet'
statute mile !!
As I've posted many times on The Internet ..
the_floor_of "The King's Chamber" inside The Great
Pyramid of Giza .. seems to be "displaying" the true
length of The Royal Cubit .. as 10 * 20.
This is, apparently .. according to my research and
calculations (1998, Morton, Internet) ..
412.5296125 regular British inches by 206.2648063
regular British inches .. which is_also_34.37746771
regular British feet by 17.18873386 regular British
feet.
AND .. !! .. here's the_link_to the *nautical mile*,
too .. because you see the decimal-harmonic of the
polar radius of Earth in nautical miles, there ..? ..
3437.746771 .. oh yeah !!
This supports my proposal that .. the GPV of "FACE TWO"
@ Cydonia on Mars .. is reflecting this "harmony"
between nautical miles and statute mile, because ..
(21600 / 13159.47253) = 164.1403175 * (10^-2).
This is where .. 13159.47253 is recognized as the
"designed" polar circumference of Mars .. in statute
miles .. and where, of course, you have the 21600
nautical miles of Earth polar circumference.
My "FACE TWO" values ..
Grid LONG ..
32.89868134(min) W.Cydonia
Grid LAT ..
5400 North ..
= 41(deg) * 03(min) * 43.90243902(sec) North.
GPV "FACE TWO" @ Cydonia on Mars ..
5400 / 32.89868134) = 164.1403175
Morton, 2001, Internet; assistance from Damon Elkins}.
One big "unified" field .. er, matrix.
-- Michael L.M.
```
(c) 2002 by mailto:Milamo@aol.com Michael Lawrence Morton ~ Archeocryptographer.
http://hometown.aol.com/marscode/homepage1.html
http://www.greatdreams.com/gem1.htm | 772 | 2,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-34 | longest | en | 0.845179 |
https://blogs.fangraphs.com/what-has-worked-in-the-postseason/ | 1,714,024,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00074.warc.gz | 120,244,490 | 27,303 | # What Has Worked in the Postseason
Billy Beane once famously said that his poop doesn’t work in the postseason, and ostensibly he wasn’t talking about his digestive system. The Athletics will spend another offseason wondering about it. The rest of baseball’s fandom usually thinks about the victors as models. Do the Giants and Royals represent some sort of sea change, do they represent a way to succeed in the postseason?
I thought I’d run some numbers to see what I could find. What I found is — it’s very difficult to study the postseason as an entity.
The first idea was to compare post-season and regular season correlations between certain statistics and winning percentage. The strength of the correlation would only be important as it pertained to the other statistics, in other words. I’m less interested in how strong the relationship between walks and wins is, as I am interested in how much more or less important strikeouts are to wins. A ranking of team skills, in essence.
For the regular season, using records back to 1974, that was easy. All of the p-values were tiny (less than .00001), and all of the relationships existed on a continuum that lines up with how many of us think about baseball. I indexed the rate stats to the league rate stats in every year — the average strikeout rate has undergone many changes in the last thirty-plus years.
Stat Correlation to Win%
K%+ 0.044
BSR 0.061
BB%+ 0.151
ISO+ 0.217
wRC+ 0.453
oWAR 0.571
By leading the league in strikeout rate and stolen bases, the Royals made it to the postseason using excellence in the two stats that correlate very weakly to win percentage. It was an improbable run, though, and we knew that.
Did it become more probable because of those same correlations in the postseason?
We can’t know using this type of analysis. The strongest correlation between regular season team stats and postseason winning percentage belonged to offensive WAR, and that only described 1.3% of the variance in the sample. But low r-squared numbers aside, the real problem is the size of the sample itself. 246 points of data are just not enough — the lowest p-value in the et was .081 (for WAR). The other p values just suggested that the sample was random.
Which it is. Even for the championship winners in the dual wild-card era, we’re talking about 15 games or so. Remember when the Athletics started the season 11-4? That would have won them the World Series in October. Then a slightly different version of the same team went 3-12 in late August and early September. That almost cost them a chance at the postseason.
Perhaps that’s it, full stop. Maybe I should stop there. But I spent a lot of time collating the postseason data, so I didn’t stop there. I took the top 28 of the 246 postseason teams — the teams that won 2/3 of their games, most of them were champions — and compared their regular season work to the work done by the entire sample of postseason teams.
BSR wRC+ WAR K%+ BB%+ ISO+
top 28 ave 1.92 104.06 27.12 96.52 105.90 105.90
sample ave 1.78 103.48 25.95 97.48 105.75 106.63
For the most part, this statistically unreliable approach just gives us the no-duh answer. Better teams are better. They have better weighted offensive numbers, they walk more, they strikeout less. Sure.
But there are two things that stood out to me. Probably just confirmation bias, but look at that last column. Postseason winners did not out-slug their opponents once they got to the postseason. Also, even though their regular season walk rates were a little better than other postseason teams, the gap has been very narrow. In fact, the champions set was — relative to other postseason teams — much better at making contact compared to their league.
We know it can’t mean much, but it also does seem worth further thought. Is the postseason any different from the regular season? If not, then power and patience still rule the roost. If it is, though, perhaps collecting the lottery ticket that is the ball in play becomes a more viable strategy once you make it to October.
With a phone full of pictures of pitchers' fingers, strange beers, and his two toddler sons, Eno Sarris can be found at the ballpark or a brewery most days. Read him here, writing about the A's or Giants at The Athletic, or about beer at October. Follow him on Twitter @enosarris if you can handle the sandwiches and inanity. | 996 | 4,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.969944 |
https://www.jiskha.com/display.cgi?id=1193109112 | 1,500,829,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424575.75/warc/CC-MAIN-20170723162614-20170723182614-00649.warc.gz | 789,028,037 | 3,604 | physics
posted by .
A stewardess pulls her 70 N flight bag a distance of 200 m along an airport floor at constant speed. The force she exerts is 40 N at an angle of 50 degrees above the horizontal. Find (a) the work she does, (b) the work done by the force of friction, and (c) the coefficient of kinetic friction between her flight bag and the floor.
I only know how to do (a). How would I do (b) and (c).
(a) (5kg)(9.8m/s^2)(sin30)(200m)=5140J | 131 | 449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-30 | longest | en | 0.952794 |
https://paperhelper.co/2022/06/29/pareto-analysis-and-business-process-flow-charting-practice/ | 1,660,721,112,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572870.85/warc/CC-MAIN-20220817062258-20220817092258-00425.warc.gz | 416,907,595 | 12,395 | # Pareto analysis and business process flow charting practice
This activity consists of two problems.
#### Problem One
Gary Seitz is director of procurement for the Forest Medical Center in Oak Park, Illinois. His medical center recently purchased more than 300 new side tables for the patient rooms from Quick-and-Cheap Furniture. His team, however, is not happy with the paint quality of these tables. They feel the firm providing the tables was rushed to get the order out and did not do a great job.
They have examined each of the 312 tables that they have received so far and found the following defects, listed in alphabetical order:
Table Defects
Problem: Number of Occurrences
Dirt in paint:65
Off-color: 17
Orange peel: 12
Other: 1
Scratch: 11
Sealer under: 8
Thin paint: 31
##### Instructions
Use Excel to:
1. Create a Pareto Chart for this situation.
2. Create and complete a table for this situation using the following headers:
• Problem.
• Frequency.
• Cumulate Frequency.
• Percentage.
• Cumulate Percentage.
3. List the most frequently occurring problem first in the table, then the next more frequently occurring problem second in the table, and so forth.
In a Word document:
1. Draw two conclusions about the quality of the patient tables that Quick Furniture has shipped to the Forest Medical Center.
Submit both your Word and Excel files.
#### Problem Two
Ted Short, manager of procurement at the Lake Luna Medical Center (LLMC), is starting a process to examine at the general process that is followed at his center for procurement of medical supplies at his facility. His first step is to develop a flowchart for the current process that employees at his center follow. He has asked you to draw that flowchart for him and make any obvious changes to shorten the process.
Here are the key steps that are followed:
1. The LLMC employee fills out a paper requisition and sends that to his department head for approval.
2. The department secretary first examines the requisition to see if all needed information is present. If it is, then she sends to the department head for approval. If not, she returns to the employee for any needed corrections. The employee then returns the requisition to the department secretary for review again.
3. Requisitions with the correct information are sent to the department head.
4. The department head reviews the item to determine if the material requested seems appropriate. He may contact his employee to discuss any questions he might have.
5. If he feels it is inappropriate, then he rejects the request and sends it back to the employee.
6. If he feels it is appropriate, then he checks the budget to see if there are sufficient monies in the budget to pay for this. If there are not, then he rejects the request and sends it back to the employee. If there are sufficient funds, then he approves them.
7. If the requisition is approved, then the department secretary sends the requisition to a member of the LLMC Procurement Team.
8. The LLMC procurement specialist checks to see if the suggested vendor on the requisition is on the approved vendor list (AVL).
9. If not, he sends the request back to the department secretary and the process starts all over if the LLMC employee wants to continue to pursue it. If the employee does not, then the requisition is terminated.
10. If the vendor is on the AVL, the LLMC procurement specialist checks to see if a quote is required, either based on the type of equipment and material required or on the amount of the purchase.
11. If no quote is required, the procurement specialist transmits the PO to the vendor for execution.
12. If a quote is required, then the specialist contacts the vendor and asks for a quote. If the quote is satisfactory based on past purchases of such items, the specialist approves the quote and places the order.
13. If the quote is too high, the specialist contacts the LLMC employee and asks for another vendor to check. The procurement specialist then checks the second vendor. Usually, a procurement specialist will have to check only two vendors at most.
14. If the specialist cannot find a vendor with it suitable price after two tries, then the PO is cancelled.
15. When the specialist finds a vendor who submits a suitable quote, the transmits the PO to the vendor for execution by US postal service or by telephone.
16. The procurement specialist then logs the transaction into the financial system.
17. The procurement specialist then forwards the request to the Accounts Payable Department for payment when the receipt of the materials and equipment ordered, if acknowledged.
##### Instructions
Use Microsoft Word to complete the following:
1. Develop a flowchart for this process.
2. Suggest three improvements, based on total quality management, that Ted Short can use to make the current process more efficient.
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We give the students premium quality assignments, without alarming them with plagiarism and referencing issues. We ensure that the assignments stick to the rules given by the tutors. We are specific about the deadlines you give us. We assure you that you will get your papers well in advance, knowing that you will review and return it if there are any changes, which should be incorporated. | 1,677 | 8,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.951752 |
http://heli-air.net/2016/02/19/simple-straight-through-turbojet-engine-formulation/ | 1,586,053,373,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00492.warc.gz | 79,867,331 | 12,094 | # Simple Straight-Through Turbojet Engine: Formulation
In Figure 10.11, consider a CV (dashed line; note the waist-like shape of the simple turbojet) representing a straight-through, axi-symmetric turbojet engine. The CV and the component station numbers are as shown in the drawing and conventions in Figure 10.5; the gas turbine intake starts with the subscript 0, or to, and ends at the nozzle exit plane with subscript 5, or e. The free-stream airmass flow rate, ma, is inhaled into the CV at the front face perpendicular to the flow, the fuel-mass flow rate mf (from the onboard tank) is added at the combustion chamber, and the product flow rate (ma + m f) is exhaust from the nozzle plane perpendicular to flow. It is assumed that the inlet-face static pressure is pTO, which is fairly accurate. Precompression exists but, for the ideal consideration, it has no loss.
Flow does not cross the other two lateral boundaries of the CV because it is aligned with the walls of the engine. Force experienced by this CV is the thrust produced by the engine. Consider a cruise condition with an aircraft velocity of Vto. At cruise, the demand for air inhalation is considerably lower than at takeoff.
Figure 10.11. Control volume representation of a straight-through turbojet
The intake area is sized between the two demands. At cruise, the intake-stream – tube cross-sectional area is smaller than the intake-face area – it is closer to that of the exit-plane area, A (the gas exits at a very high velocity). Because in an ideal condition there is no precompression loss, Station 0 may be considered to have free – stream properties with the subscript to.
From Newton’s second law, applied force F = rate of change of momentum + net pressure force (the momentum rate is given by the mass flow rate), where the inlet momentum rate = ma Vto and the exit momentum rate = (ma + m f )Ve.
Therefore:
the rate of change of momentum = (ma + mf )Ve – ma Vto (10.6)
The net pressure force between the intake and exit planes = peAe – pTO Ato (i. e., the axi-symmetric side pressure at the CV walls cancels out). Typically, at cruise, a sufficiently upstream Ae & Ato. Therefore:
F = (ma + mf )Ve – ma V» + Ae(pe – Pc») = net thrust (10.7)
Then:
(ma + mf )Ve + Ae(Pe – Pto) = gross thrust
and maVTO = ram drag (with – ve sign, it must be drag). It is the loss of energy seen as drag due to the slowing down of the incoming air as the ram effect. This gives:
net thrust = gross thrust – ram drag; Ae (pe – pTO) = pressure thrust
In general, subsonic commercial transport turbofans have a convergent nozzle, and the exit area is sized such that during cruise, pe & pTO (known as a perfectly expanded nozzle). This is different for military aircraft engines, especially with AB, when pe > pTO requires a convergent-divergent nozzle.
For a perfectly expanded nozzle, net thrust:
F = (ma + mf)Ve – ma V«, (10.8)
Further simplification is possible by ignoring the effect of fuel flow, m f, because m a » m f.
Then, the thrust for a perfectly expanded nozzle is:
At sea-level, static-takeoff thrust (TSLs) ratings V» = 0, which gives:
F = tha Ve+Ae(pe – p») (10.10)
Equation 10.10 indicates that the thrust increase can be achieved by increasing the intake airmass flow rate and/or increasing the exit velocity.
Equation 10.4 gives the propulsive efficiency:
Clearly, jet-propelled aircraft with low flight speeds have poor propulsive efficiency, np. Jet propulsion is favored for aircraft flight speeds above Mach 0.6.
The next question is: Where does the thrust act? Figure 10.12 shows a typical gas turbine engine in which the thrust is acting over the entire engine; the aircraft senses the net thrust transmitted through the engine-mounted bolts.
Figure 10.12 shows a typical straight-through turbojet pressure, velocity, and temperature variation along the length as airmass flows through. Readers may note the scale; within each component, the velocity change is negligible. | 941 | 3,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-16 | longest | en | 0.923347 |
https://www.gamedev.net/forums/topic/632265-stuck-on-character-movement-not-wasd/?forceDownload=1&_k=880ea6a14ea49e853634fbdc5015a024 | 1,516,765,545,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893300.96/warc/CC-MAIN-20180124030651-20180124050651-00046.warc.gz | 915,224,160 | 25,394 | Stuck on Character Movement (not WASD)
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Ive recently began another attempt at being a game developer. In my previous years of development, i had never implemented any moveable NPCs. I have only done main character movement with WASD on the keyboard, which is simple to implement.
I am stuck on making a character/NPC move from one point to another.
Example: Start from point 4,13 heading to destination point 18,6. With a movement speed of 7 points per second at a frame rate of 30 per second = 0.21 points per frame.
Simply adding speed to x or y doesnt work when the character is moving at a non-straight up, left, down, or right direction. I did a bunch of googling but i can only find basic WASD keyboard movement.
Can anyone help me or point me to a resource that can teach me?
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Ive recently began another attempt at being a game developer. In my previous years of development, i had never implemented any moveable NPCs. I have only done main character movement with WASD on the keyboard, which is simple to implement.
I am stuck on making a character/NPC move from one point to another.
Example: Start from point 4,13 heading to destination point 18,6. With a movement speed of 7 points per second at a frame rate of 30 per second = 0.21 points per frame.
Simply adding speed to x or y doesnt work when the character is moving at a non-straight up, left, down, or right direction. I did a bunch of googling but i can only find basic WASD keyboard movement.
Can anyone help me or point me to a resource that can teach me?
if you're at 14,3 and heading to 18,6 the direction vector will be (18,6)-(14,3) = (4,3). (direction = target - position)
if you normalize the direction vector you get: (4/sqrt(4*4+3*3) , 3/(sqrt(4*4+3+3)) = (0.8, 0,6) (to normalize a vector you divide each component (x,y,z,etc) with the length of the vector, this makes the length of the vector 1.0).
So the normalized direction vector is direction = vector2d(0.8 , 0.6) . now it is fairly easy, you just move
player.x += direction.x*speed
player.y += direction.y*speed
(oops, i misread your positions, but you should get the idea anyway). Edited by SimonForsman
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The method above works great, and I would suggest learning it eventually (working with vectors is a good idea in game programing, because getting used to thinking in vectors is mandatory if you want to be successful in 3D programming).
Alternatively, you can try storing your movement as a single direction value (theta) and a distance value (R):
If R is in pixels/second, you will need to get pixels per frame. In general that's just distance_per_second / frames_per_second. Lazyfoo has a great tutorial on moving from pixels per frame to fps independent movement, but for now, get pixels per frame working.
Now, in every frame...
movement_x = R*cos(theta);
movement_y = R*sin(theta);
(Note: A lot of coordinate systems in 2D games use up as y-negative instead of y-positive. If you're using a coordinate space where the top of the screen has the lowest Y, use -R*sin(theta) )
When you move upwards, theta is 90* (or pi/2 if you prefer radians), down is 270* (or 3pi/2), left is 180* (or pi), right is 0* (or 0).
Diagonal movement is as simple as 45* for up/right, 135* for up/left, 225* for down/left and 315* for down/right.
You can multiply a degree by pi/180 to get the radians (and most trig functions like sin and cos take radian inputs).
This is useful if you don't already know where you want the player to end up, and instead only know how far and in what direction they will go. I use it in games with movement controls like asteroid, but the other method works just fine for those games to. Its up to you to chose which way you prefer. Actually, when you get down to it, this is mathematically identical to the final result of the method in the previous post, it just hides the vector math.
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Thanks so much for the replies. I tested it out on my calculator and paper and it works. I remember doing Vectors in high school and failing that subject. I guess i should have paid more attension in class.
I have one more question reguarding the direction the character is moving though.
If the character is moving Up+Right (not at a 45 degree angle, but when the direction is (+x, -y)... let me show an equation:
Current point: A=(1,-2) and destination point: B=(8,-8) so the vector is (8 - 1, -8 - -2) = (7,-6)
The amount added to the character's X coord is: (7/sqrt((7*7)+(-6*-6)) * 0.21 = .224
The amount added to the character's Y coord is: (-6/sqrt((7*7)+(-6*-6)) * 0.21 = -.408
To test if the character is moving at 0.21 points per frame, sqrt((.224*.224) + (-.408*-.408)) = .465, which is not correct.
However if i make the negative number positive by multiplying it by -1, it works as shown:
(7/sqrt((7*7)+((-1*-6)*(-1*-6)))*0.21 = .159
(6/sqrt((7*7)+((-1*-6)*(-1*-6)))*0.21 = .137
And to test it: sqrt((.159*.159) + (.137*.137)) = 0.21 which is correct.
I know its easy to determine if the character is moving UpRight, DownRight, DownLeft, DownRight, as i already do for my method to figure out what degree/direction the character is travelling, but is it nessessary to do this, or is it possible to use one equation reguardless of what direction the character is going?
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Thanks so much for the replies. I tested it out on my calculator and paper and it works. I remember doing Vectors in high school and failing that subject. I guess i should have paid more attension in class.
I have one more question reguarding the direction the character is moving though.
If the character is moving Up+Right (not at a 45 degree angle, but when the direction is (+x, -y)... let me show an equation:
Current point: A=(1,-2) and destination point: B=(8,-8) so the vector is (8 - 1, -8 - -2) = (7,-6)
The amount added to the character's X coord is: (7/sqrt((7*7)+(-6*-6)) * 0.21 = .224
The amount added to the character's Y coord is: (-6/sqrt((7*7)+(-6*-6)) * 0.21 = -.408
To test if the character is moving at 0.21 points per frame, sqrt((.224*.224) + (-.408*-.408)) = .465, which is not correct.
However if i make the negative number positive by multiplying it by -1, it works as shown:
(7/sqrt((7*7)+((-1*-6)*(-1*-6)))*0.21 = .159
(6/sqrt((7*7)+((-1*-6)*(-1*-6)))*0.21 = .137
And to test it: sqrt((.159*.159) + (.137*.137)) = 0.21 which is correct.
I know its easy to determine if the character is moving UpRight, DownRight, DownLeft, DownRight, as i already do for my method to figure out what degree/direction the character is travelling, but is it nessessary to do this, or is it possible to use one equation reguardless of what direction the character is going?
this is odd: 6*6 and -6*-6 should both be 36 so i think you should post your real code for this part. (I'm assuming you don't put hardcoded positions in your code) | 1,888 | 7,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-05 | latest | en | 0.925884 |
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Rational operators (a {t} b); a,b > e solved JmsNxn Long Time Fellow Posts: 739 Threads: 104 Joined: Dec 2010 06/08/2011, 11:47 PM (This post was last modified: 06/09/2011, 01:53 AM by JmsNxn.) (06/08/2011, 08:32 PM)sheldonison Wrote: So we have a definition for t=0..2 (addition, multiplication, exponentiation), for all values of a. Nice! $a\{t\}e = \exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a)+e)$ So, now, I'm going to define the function f(b), which returns the base which has the fixed point of "b". $f(e)=\eta$, since e is the fixed point of b=$\eta$ $f(2)=\sqrt{2}$, since 2 is the lower fixed point of b=sqrt(2) $f(3)=1.442249570$, since 3 is the upper fixed point of this base $f(4)=\sqrt{2}$, since 4 is the upper fixed point of b=sqrt(2) $f(5)=1.379729661$, since 5 is the upper fixed point of this base I don't know how to calculate the base from the fixed point, but that's the function we need, and we would like the function to be analytic! This also explains why the approximation of using base eta pretty well, since the base we're going to use isn't going to be much smaller than eta, as b gets bigger or smaller than e. Now, we use this new function in place of eta, in James's equation. Here, f=f(b). $a\{t\}b = \exp_f^{\circ t}(\exp_f^{\circ-t}(a)+b)$ - Sheldon Woah, I wonder what consequences this will have on the algebra. I guess $\log_f^{\circ q}(a\,\,\bigtriangleup_{1+q}\,\,b) = b\log_f^{\circ q}(a)$ $\log_f^{\circ q}(a\,\,\bigtriangleup_{q}\,\,b) = \log_f^{\circ q}(a) + b$ which I guess isn't too drastic. But ,the question is, of course, does the following still hold $0 \le q\le1$: $a\,\,\bigtriangleup_{1+q}\,\,2 = a\,\,\bigtriangleup_q\,\,a$ $\exp_{2^{\frac{1}{2}}}^{\circ 1+q}(\exp_{2^{\frac{1}{2}}}^{\circ -q-1}(a) + 2) \neq \exp_{a^{\frac{1}{a}}}^{\circ q}(a + a)$, so no it doesn't. That's not good, we want operators to be recursive. And I'm unsure if the inverse is still well defined, so I think we lose: $(a\,\,\bigtriangleup_{1+q}\,\, -1)\,\,\bigtriangleup_q\,\,a = S(q)$ where S(q) is the identity function. We may even lose the identity function altogether, this is really bad. We also lose: $(a\,\,\bigtriangleup_{1+q}\,\,b)\,\,\bigtriangleup_{1+q}\,\, c = a\,\,\bigtriangleup_{1+q}\,\,bc$ $(a\,\,\bigtriangleup_{1+q}\,\,b)\,\,\bigtriangleup_q\,\,(a\,\,\bigtriangleup_{1 + q} \,\,c) = a\,\,\bigtriangleup_{1+q}\,\,b+c$ These are all too many valuable qualities that are lost when redefining semi-operators the way that you do. Sure it's analytic over $(-\infty, 2]$, but it loses all its traits which make it an operator in the first place. I'm going to have to stick with the original definition of $\vartheta$ that isn't fully analytic. However, I am willing to concede the idea of changing from base eta to base root 2. That is to say if we define: $\vartheta(a,b,\sigma) = \exp_{2^{\frac{1}{2}}}^{\circ \sigma}(\exp_{2^{\frac{1}{2}}}^{\circ -\sigma}(a) + h_b(\sigma))\\\\ [tex]h_b(\sigma)=\left{\begin{array}{c l} \exp_{2^{\frac{1}{2}}}^{\circ -\sigma}(b) & \sigma \le 1\\ \exp_{2^{\frac{1}{2}}}^{\circ -1}(b) & \sigma \in [1,2] \end{array}\right.$ This will give the time honoured result, and aesthetic necessity in my point of view, of: $\vartheta(2, 2, \sigma) = 2\,\,\bigtriangleup_\sigma\,\, 2 = 4$ for all $\sigma$ I like this also because it makes $\vartheta(a, 2, \sigma)$ and $\vartheta(a, 4, \sigma)$ potentially analytic over $(-\infty, 2]$ since 2 and 4 are fix points. I also propose writing $a\,\,\bigtriangle_\sigma^f\,\,b = \exp_f^{\circ \sigma}(\exp_f^{\circ -\sigma}(a) + h_b(\sigma))\\\\ [tex]h_b(\sigma)=\left{\begin{array}{c l} \exp_f^{\circ -\sigma}(b) & \sigma \le 1\\ \exp_f^{\circ -1}(b) & \sigma \in [1,2] \end{array}\right.$ Sheldon's analytic function is then: $g(\sigma) = a\,\,\bigtriangle_\sigma^{b^{\frac{1}{b}}}\,\,b$ that's still very pretty though, that $g$ isn't piecewise over $(-\infty, 2]$ and potentially analytic. « Next Oldest | Next Newest »
Messages In This Thread Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 02:45 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 04:39 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 05:34 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 06:02 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 07:03 AM RE: Rational operators (a {t} b); a,b > e solved - by nuninho1980 - 06/06/2011, 05:16 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 06:53 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 08:47 AM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:23 AM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 06/06/2011, 11:59 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 05:44 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:28 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 07:47 PM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 08:43 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/07/2011, 02:45 AM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/07/2011, 06:59 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 04:54 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 07:31 PM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/08/2011, 08:32 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/08/2011, 09:14 PM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 01:50 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 11:47 PM RE: Rational operators (a {t} b); a,b > e solved - by Gottfried - 06/11/2011, 02:33 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/12/2011, 07:55 PM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/21/2016, 06:56 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 08/22/2016, 12:36 AM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/24/2016, 07:24 PM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/29/2016, 02:06 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 09/01/2016, 06:47 PM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 02:04 AM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 02:11 AM
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http://mathhelpforum.com/differential-equations/151788-spring-mass-system-damping-force-impressed-force-find-position-fcn-2-print.html | 1,495,930,178,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609305.10/warc/CC-MAIN-20170527225634-20170528005634-00152.warc.gz | 281,520,916 | 4,772 | # Spring-Mass system (with damping force AND impressed force). Find Position fcn.
Show 40 post(s) from this thread on one page
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• Jul 25th 2010, 06:16 PM
EmilyL
I solved it! I'll edit this post with a scan of the sheet of paper i wrote on.
• Jul 26th 2010, 01:37 AM
Ackbeet
Great! I'm very willing to double-check your work, if you like. I can wait for your post.
• Jul 26th 2010, 10:31 AM
EmilyL
I haven't gotten time to get to my scanner yet, but basically I did it in terms of feet (so that x is 1/2 feet instead of 6 inches) to get k=4 and m=1/16.
I got y(c) = c(1)e^/8 + c2e^/8t.
I found y(p) with the annihilator method, getting y(p)=Asin8t + Bsincos8t
I then found y'(p) and y''(p), and plugged it into the original equation to get coefficients of A=0, B=-1/4
Then I plugged in the initial conditions for the entire y(t) to get the coefficients c1 and c2, which were c(1)=1/2 and c(2)=4
So my final y(t) was:
y(t) = 1/2(e^-8t) + 4te^-8t - (1/4)cos8t
=]
• Jul 26th 2010, 12:13 PM
Ackbeet
Is this the same problem? I confess myself bewildered by the apparent random changes in method you've used. Here are some specific comments:
1. If the problem gives you pounds and inches, your answer had better be in pounds and inches. Get everything in terms of the given units. This is as true in the industrial world of engineering as it is in academia.
2. I'm not sure about your application of the annihilator approach. Up to Post # 14, I would have said things were good. Your post # 15 was incorrect. You needed to operate on the original DE with the operator that annihilated the RHS. That is, with the following DE:
$x'' - 192x' + 64x = 384 \sin(8t)$ (since you multiplied through by 192, that has to happen to the non-homogeneous side as well!)
This we can write as
$(D^{2}-192 D+64)x(t)=384\sin(8t).$
After left-operating with the annihilator of the RHS, we get
$(D^{2}+64)(D^{2}-192 D+64)x(t)=0.$
What do you get when you solve this homogeneous DE?
• Jul 26th 2010, 12:27 PM
EmilyL
Darn, thought I had it for sure...
hmm, D=8i, -8i, 96 +8sqrt(143), 96 -8sqrt(183) ?
I'm not sure if this is appropriate or not, but I need some more immediate help with the problems in this thread, if you wouldn't mind taking a look at them: http://www.mathhelpforum.com/math-he...but-stuck.html
• Jul 26th 2010, 03:45 PM
mantra002
Ackbeet, I believe there is a critical flaw in the setup of this problem. You corrected Emily's DE by flipping the sign of the damper in the problem, this is not correct. In fact, by making this change the system is now unstable and you will not get a meaningful solution. Note that a positive velocity will now create a positive dampening force! As velocity rises the force grows forever and ever.
In fact, this is not what should happen, the dampening force is not opposite the spring, but opposite the velocity, so it is negative when on the RHS. The correct DE to start is:
(W/G)*x'' + B*x'+K*x = 2*sin(8/t) - W (this last W only exists if gravity is counted in the problem)
Where W is the weight in lbs (2 here), G the acceleration due to gravity in In/s^2, B is the dampening coefficient (1 here), K is the spring coefficient (1/3 here). I assumed the forcing function to be positive, but it could very well be negative depending on how the problem is set up (is it 'pushing' or 'pulling').
Also, I apologize for the ugly notation, I don't know how to use TeX (Crying)
• Jul 26th 2010, 04:13 PM
Ackbeet
I think you're right. It's what was going through my mind, but didn't make its way into my typing. I don't have time right now to fix everything, but I'll get to it soon.
Thanks for catching that!
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http://kometbomb.net/tag/image-retargeting/ | 1,721,278,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00895.warc.gz | 17,520,238 | 26,001 | 02 Sep
I came up with a neat way to retarget images using a mesh that is transformed by rotating and doing an ortographic (non-perspective) projection. This is generally quite interesting since it can be done using a mesh and simple transformations and so can be done almost completely on the GPU. Even using a mesh can be avoided if one uses a height map à la parallax mapping to alter the texture coordinates so just one quad needs to be drawn (with a suitable fragment shader, of course).
The idea is simply to have areas of images at a slope depending of how much the areas should be resized when retargeting. The slope angle depends of from what angle the source image is viewed to get the retargeting effect since the idea is to eliminate the viewing angle using the slope.
Here’s a more detailed explanation:
1. Create an energy map of the source image, areas of interest have high energy
2. Traverse the energy map horizontally accumulating the energy value of the current pixel and the accumulated sum from the previous pixel
3. Repeat the previous step vertically using the accumulated map from the previous step. The accumulated energy map now “grows” from the upper left corner to the lower right corner. You may need a lot of precision for the map
4. Create a mesh with the x and y coordinates of each vertex encoding the coordinates of the source image (and thus also the texture coordinates) and the z coordinate encoding the accumulated energy. The idea is to have all areas of interest at a steep slope and other areas with little or no slope
5. Draw the mesh with ortographic projection, using depth testing and textured with the source image
6. Rotate the mesh around the Y axis to retarget image horizontally and around the X axis to retarget image vertically
Here is a one-dimensional example (sorry for the awful images):
Source image
The red dots represent areas of interest, such as sharp edges that we don’t want to resize as much as we want to resize the areas between the details. We then elevate our line for every red dot:
Elevated mesh
Imagine the above example as something you would do for every row and column of a two-dimensional image. Now, when the viewer views the mesh (which is drawn without perspective) he or she sees the original image:
Viewing the mesh from zero angle
However, if the viewing angle is changed, the red dots don’t move in relation to each other as much as the areas that are not elevated when they are projected on the view plane. Consider the below example:
Viewing the mesh from an angle (gray line is the projected mesh)
Note how the unelevated line segments will seem shorter from the viewer’s perspective while the distance between the red dots is closer to the original distance. The blue dots in the above image show how areas that have little energy and so are not on a slope, thus will be move more compared to the red dots.
04 Sep
I saw this video of a SIGGRAPH paper about image retargeting (high res version here, read the paper here), that is rescaling an image so that the algorithm keeps the interesting areas intact and doesn’t squash everything. It’s called seam carving in the paper.
The video made it look amazingly simple (and actually explained the whole idea much better than most papers manage to do), so obviously I had to try my hands at it. After about three hours worth of coding I came up with my version (you can find the source code below!).
Original image Retargeted image Retargeted image
Notice how the guy’s face and the cloud stay the same even if everything else is stuffed in the smaller image area.
Original image Retargeted image
Again, the higher contrast areas (i.e. the man and the dogs, black on white) are kept the same while the snowy area is made narrower.
It’s a small world… ;)
I didn’t read the SIGGRAPH paper, so I don’t know what makes their algorithm work that well (or maybe they just chose the right images for the video). My program works as follows (when shrinking the image horizontally):
1. For each column, traverse from top to bottom picking any of the three (or more) neighboring pixels below the current pixel
2. Calculate the “penalty” or error, i.e. try to pick the neighboring pixel that is colored as similarly as possible compared to the one next to it (in the direction we want to shrink the image)
3. From these paths, pick the path that has the lowest penalty and crop the pixels along the path, while moving the rows to the left, as you would delete characters in a text
4. Repeat until the image width is what was requested
In all, this is very slow but it could be made faster (as in the video that shows realtime scaling) if the penalty or error values were precalculated for each pixel. The algorithm should also try to pick paths that are further apart, so it would remove pixels more evenly and it should backtrack when trying to find the optimal path. Now it just goes along a “wall”, i.e. a high-contrast area when it finds one – it should backtrack and try to find a path further away. Finally, there should be a feature that allowed the user to mark faces and other areas that should never be scaled.
To use the program, you need to run it from command line or drag a 24-bit BMP image on the icon. Resize the window to scale images. If you want to save the image, simply answer “yes” when exiting the program.
### New version
When using the new version, you can resize to a specific size (as requested) by running the program as follows:
`retarget image.bmp 800 600`
This will try to resize the image to 800×600 resolution. The new version is able to load JPEG, PNG, BMP and probably some other formats too (thanks to the SDL_image library). Note that it still will save as BMP, even if the extension is JPG or so.
Use the left mouse button to mark areas such as faces, eyes and so on, and the right mouse button to mark areas that you want to remove. Middle mouse button erases the marks. To tweak the blur amount (less is better for cartoon style images and maps, the opposite for photos), run it like this:
`retarget image.bmp 800 600 4`
Now there will be twice as much blur as usually (default is 2).
retarget3.zip – the program with the source code (you need SDL and SDL_image)
Here’s the original version, it is still useful:
retarget2.zip – the program with the source code (you need SDL) | 1,373 | 6,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.913886 |
https://openstax.org/books/introductory-business-statistics/pages/9-4-full-hypothesis-test-examples | 1,721,894,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00209.warc.gz | 372,373,841 | 81,567 | # 9.4Full Hypothesis Test Examples
Introductory Business Statistics9.4 Full Hypothesis Test Examples
## Example 9.8
### Problem
Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05.
Step 4 has us compare the test statistic and the critical value and mark these on the graph. We see that the test statistic is in the tail and thus we move to step 4 and reach a conclusion. The probability that an average time of 16 minutes could come from a distribution with a population mean of 16.43 minutes is too unlikely for us to accept the null hypothesis. We cannot accept the null.
Step 5 has us state our conclusions first formally and then less formally. A formal conclusion would be stated as: “With a 95% level of significance we cannot accept the null hypothesis that the swimming time with goggles comes from a distribution with a population mean time of 16.43 minutes.” Less formally, “With 95% significance we believe that the goggles improves swimming speed”
If we wished to use the p-value system of reaching a conclusion we would calculate the statistic and take the additional step to find the probability of being 2.08 standard deviations from the mean on a t-distribution. This value is .0187. Comparing this to the α-level of .05 we see that we cannot accept the null. The p-value has been put on the graph as the shaded area beyond -2.08 and it shows that it is smaller than the hatched area which is the alpha level of 0.05. Both methods reach the same conclusion that we cannot accept the null hypothesis.
## Try It 9.8
The mean throwing distance of a football for Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.
First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.
## Example 9.10
### Problem
A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, $s 2 s 2$. Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.
Again we will follow the steps in our analysis of this problem.
## Hypothesis Test for Proportions
Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p.
The population parameter for the binomial is p. The estimated value (point estimate) for p is p′ where p′ = x/n, x is the number of successes in the sample and n is the sample size.
When you perform a hypothesis test of a population proportion p, you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five (np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with $μ=npμ=np$ and $σ=npqσ=npq$. Remember that $q=1–pq=1–p$. There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p.
Again, we begin with the standardizing formula modified because this is the distribution of a binomial.
$Z = p' - p pq n Z= p' - p pq n$
Substituting $p0p0$, the hypothesized value of p, we have:
$Z c = p' - p 0 p 0 q 0 n Z c = p' - p 0 p 0 q 0 n$
This is the test statistic for testing hypothesized values of p, where the null and alternative hypotheses take one of the following forms:
Two-tailed test One-tailed test One-tailed test
H0: p = p0 H0: p ≤ p0 H0: p ≥ p0
Ha: p ≠ p0 Ha: p > p0 Ha: p < p0
Table 9.5
The decision rule stated above applies here also: if the calculated value of Zc shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.
## Example 9.11
### Problem
The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50%. They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.
## Try It 9.11
A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.
## Example 9.12
### Problem
Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.
## Example 9.13
### Problem
The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.
1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95
Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.
## Example 9.14
### Problem
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.
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### Negative Power
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What does this number mean ? Which order of 1, 2, 3 and 4 makes the highest value ? Which makes the lowest ?
### Rachel's Problem
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Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!
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Find the five distinct digits N, R, I, C and H in the following nomogram
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Find the sum of this series of surds.
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Ask a friend to choose a number between 1 and 63. By identifying which of the six cards contains the number they are thinking of it is easy to tell them what the number is.
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Mr Smith and Mr Jones are two maths teachers. By asking questions, the answers to which may be right or wrong, Mr Jones is able to find the number of the house Mr Smith lives in... Or not!
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The scale on a piano does something clever : the ratio (interval) between any adjacent points on the scale is equal. If you play any note, twelve points higher will be exactly an octave on.
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### Lastly - Well
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What are the last two digits of 2^(2^2003)?
### Unusual Long Division - Square Roots Before Calculators
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Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Diggits
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Can you find what the last two digits of the number $4^{1999}$ are?
### Sept 03
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What is the last digit of the number 1 / 5^903 ?
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In this twist on the well-known Countdown numbers game, use your knowledge of Powers and Roots to make a target.
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How many ways are there to count 1 - 2 - 3 in the array of triangular numbers? What happens with larger arrays? Can you predict for any size array?
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Explore the relationships between different paper sizes.
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Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.
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What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
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What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
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Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Archimedes and Numerical Roots
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The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
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What is the least square number which commences with six two's?
### Like Powers
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Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
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The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?
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Cut off three right angled isosceles triangles to produce a pentagon. With two lines, cut the pentagon into three parts which can be rearranged into another square.
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What fractions can you find between the square roots of 65 and 67?
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Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
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Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
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Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large...
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The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it? | 1,343 | 5,533 | {"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} | 3.6875 | 4 | CC-MAIN-2017-34 | latest | en | 0.86734 |
https://www.mathlearnit.com/what-is-51-115-as-a-decimal | 1,674,806,264,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00253.warc.gz | 912,104,087 | 6,804 | # What is 51/115 as a decimal?
## Solution and how to convert 51 / 115 into a decimal
51 / 115 = 0.443
51/115 or 0.443 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 51/115 is 51 divided by 115
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 115. We use this as our equation: numerator(51) / denominator (115) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 51/115 as our equation:
### Numerator: 51
• Numerators are the portion of total parts, showed at the top of the fraction. Overall, 51 is a big number which means you'll have a significant number of parts to your equation. 51 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Now let's explore X, the denominator.
### Denominator: 115
• Denominators represent the total parts, located at the bottom of the fraction. 115 is a large number which means you should probably use a calculator. But 115 is an odd number. Having an odd denominator like 115 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. Let's start converting!
## Converting 51/115 to 0.443
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 115 \enclose{longdiv}{ 51 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 115 \enclose{longdiv}{ 51.0 }$$
Uh oh. 115 cannot be divided into 51. So we will have to extend our division problem. Add a decimal point to 51, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 115 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 115 into 510
$$\require{enclose} 00.4 \\ 115 \enclose{longdiv}{ 51.0 }$$
Now that we've extended the equation, we can divide 115 into 510 and return our first potential solution! Multiply this number by 115, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.4 \\ 115 \enclose{longdiv}{ 51.0 } \\ \underline{ 460 \phantom{00} } \\ 50 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you have a remainder over 115, go back. Your solution will need a bit of adjustment. If you have a number less than 115, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 51/115 and 0.443 bring clarity and value to numbers in every day life. Here are just a few ways we use 51/115, 0.443 or 44% in our daily world:
### When you should convert 51/115 into a decimal
Dining - We don't give a tip of 51/115 of the bill (technically we do, but that sounds weird doesn't it?). We give a 44% tip or 0.443 of the entire bill.
### When to convert 0.443 to 51/115 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 51/115 = 0.443 what would it be as a percentage?
• What is 1 + 51/115 in decimal form?
• What is 1 - 51/115 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.443 + 1/2? | 1,251 | 5,153 | {"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} | 4.6875 | 5 | CC-MAIN-2023-06 | latest | en | 0.9321 |
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# Chapter 6 Introduction to Digital Electronics
Chapter 6 Introduction to Digital Electronics. Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock. Chapter Goals. Introduce binary digital logic concepts Explore the voltage transfer characteristics of ideal and nonideal inverters
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## Chapter 6 Introduction to Digital Electronics
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### Presentation Transcript
1. Chapter 6Introduction to Digital Electronics Microelectronic Circuit Design Richard C. JaegerTravis N. Blalock Microelectronic Circuit Design McGraw-Hill
2. Chapter Goals • Introduce binary digital logic concepts • Explore the voltage transfer characteristics of ideal and nonideal inverters • Define logic levels and logic states of logic gates • Introduce the concept of noise margin • Present measures of dynamic performance of logic devices • Review of Boolean algebra • Investigate simple transistor, diode, and diode-transistor implementations of the inverter and other logic circuits • Explore basic design techniques of logic circuits Microelectronic Circuit Design McGraw-Hill
3. Brief History of Digital Electronics • Digital electronics can be found in many applications in the form of microprocessors, microcontrollers, PCs, DSPs, and an uncountable number of other systems. • The design of digital circuits has progressed from resistor-transistor logic (RTL) and diode-transistor logic (DTL) to transistor-transistor logic (TTL) and emitter-coupled logic (ECL) to complementary MOS (CMOS) • The density and number of transistors in microprocessors has increased from 2300 in the 1971 4-bit 4004 microprocessor to 25 million in the more recent IA-64 chip and it is projected to reach over one billion transistors by 2010 Microelectronic Circuit Design McGraw-Hill
4. Ideal Logic Gates • Binary logic gates are the most common style of digital logic • The output will consist of either a 0 (low) or a 1 (high) • The most basic digital building block is the inverter Microelectronic Circuit Design McGraw-Hill
5. The Ideal Inverter The ideal inverter has the following voltage transfer characteristic (VTC) and is described by the following symbol V+ and V- are the supply rails, and VH and VL describe the high and low logic levels at the output Microelectronic Circuit Design McGraw-Hill
6. Logic Level Definitions An inverter operating with power supplies at V+ and 0 V can be implemented using a switch with a resistive load Microelectronic Circuit Design McGraw-Hill
7. Logic Voltage Level Definitions • VL – The nominal voltage corresponding to a low-logic state at the input of a logic gate for vi = VH • VH – The nominal voltage corresponding to a high-logic state at the output of a logic gate for vi = VL • VIL – The maximum input voltage that will be recognized as a low input logic level • VIH – The maximum input voltage that will be recognized as a high input logic level • VOH – The output voltage corresponding to an input voltage of VIL • VOL – The output voltage corresponding to an input voltage of VIH Microelectronic Circuit Design McGraw-Hill
8. Logic Voltage Level Definitions (cont.) Note that for the VTC of the nonideal inverter, there is now an undefined logic state Microelectronic Circuit Design McGraw-Hill
9. Noise Margins • Noise margins represent “safety margins” that prevent the circuit from producing erroneous outputs in the presence of noisy inputs • Noise margins are defined for low and high input levels using the following equations: NML = VIL – VOL NMH = VOH – VIH Microelectronic Circuit Design McGraw-Hill
10. Noise Margins (cont.) • Graphical representation of where noise margins are defined Microelectronic Circuit Design McGraw-Hill
11. Logic Gate Design Goals • An ideal logic gate is highly nonlinear that attempts to quantize the input signal to two discrete states, but in an actual gate, the designer should attempt to minimize the undefined input region while maximizing noise margins • The input should produce a well-defined output, and changes at the output should have no effect on the input • Voltage levels of the output of one gate should be compatible with the input levels of a proceeding gate • The gate should have sufficient fan-out and fan-in capabilities • The gate should consume minimal power (and area for ICs) and still operate under the design specifications Microelectronic Circuit Design McGraw-Hill
12. Dynamic Response of Logic Gates • An important figure of merit to describe logic gates is their response in the time domain • The rise and fall times, tf and tr, are measured at the 10% and 90% points on the transitions between the two states as shown by the following expressions: V10% = VL + 0.1ΔV V90% = VL + 0.9ΔV = VH – 0.1ΔV Microelectronic Circuit Design McGraw-Hill
13. Propagation Delay • Propagation delay describes the amount of time between a change at the 50% point input to cause a change at the 50% point of the output described by the following: • The high-to-low prop delay, τPHL, and the low-to-high prop delay, τPLH, are usually not equal, but can be described as an average value: Microelectronic Circuit Design McGraw-Hill
14. Dynamic Response of Logic Gates Microelectronic Circuit Design McGraw-Hill
15. Power Delay Product • The power-delay product (PDP) is use as a metric to describe the amount of energy required to perform a basic logic operation and is given by the following equation when P is the average power dissipated be the logic gate: Microelectronic Circuit Design McGraw-Hill
16. Review of Boolean Algebra NOT Truth Table OR Truth Table AND Truth Table NOR Truth Table NAND Truth Table Microelectronic Circuit Design McGraw-Hill
17. Logic Gate Symbols and Boolean Expressions Microelectronic Circuit Design McGraw-Hill
18. Diode Logic • Diodes can with resistive loads to implement simple logic gates Diode OR gate Diode AND gate Microelectronic Circuit Design McGraw-Hill
19. Diode Transistor Logic • Since diode gates are limited to AND and OR functions, the diodes can be combined with transistors to complete the basic logic functions such as the following NAND gate Microelectronic Circuit Design McGraw-Hill
20. NMOS Logic Design • MOS transistors (both PMOS and NMOS) can be combined with resistive loads to create single channel logic gates • The circuit designer is limited to altering circuit topology and width-to-length, or W/L, ratio since the other factors are dependent upon processing parameters Microelectronic Circuit Design McGraw-Hill
21. NMOS Inverter with a Resistive Load • The resistor R is used to “pull” the output high • MS is the switching transistor • The size of R and the W/L ratio of MS are the design factors that need to be chosen Microelectronic Circuit Design McGraw-Hill
22. Load Line Visualization • The following illustrates the operation of the NMOS output (vDS) characteristics where the following equation describes the load line Microelectronic Circuit Design McGraw-Hill
23. NMOS with Resistive Load Design Example • Design a NMOS resistive load inverter for • VDD = 3.3 V • P = 0.1 mW when VL = 0.2 V • Kn = 60 μA/V2 • VTN = 0.75 V • Find the value of the load resistor R and the W/L ratio of the switching transistor MS Microelectronic Circuit Design McGraw-Hill
24. Example continued • First the value of the current through the resistor must be determined by using the following: • The value of the resistor can now be found by the following which assumes that the transistor is on or the output is low: Microelectronic Circuit Design McGraw-Hill
25. Example Continued • For vI = VL = 0.2V, the transistor’s vGS will be less than the threshold voltage, therefore it will be operating in the triode region. Using the linear equation for a MOSFET, the W/L ratio can be found: Microelectronic Circuit Design McGraw-Hill
26. On-Resistance of MS • The NMOS resistive load inverter can be thought of as a resistive divider when the output is low, described by the following expression: Microelectronic Circuit Design McGraw-Hill
27. On-Resistance of MS (cont.) When the NMOS resistive load inverter’s output is low, the On-Resistance of the NMOS can be calculated with the following expression: Note that Ron should be kept small compared to R to ensure that VL remains low, and also that its value is nonlinear which has a dependence on vDS Microelectronic Circuit Design McGraw-Hill
28. Noise Margin Analysis • The following equations can be used to determine the various parameters needed to determine the noise margin of NMOS resistive load inverters Microelectronic Circuit Design McGraw-Hill
29. Load Resistor Problems • For completely integrated circuits, R must be implemented on chip using the shown structure • Using the given equation, it can be seen that resistors take up a large area of silicon as in an example 95kΩ resistor Microelectronic Circuit Design McGraw-Hill
30. Using Transistors in Place of a Resistor • NMOS load w/ a) gate connected to the source b) gate connected to ground c) gate connected to VDD d) gate biased to linear region e) a depletion mode NMOS Note that a) and b) are not useful Microelectronic Circuit Design McGraw-Hill
31. Static Design of the NMOS Saturated Load Inverter Schematic for a NMOS saturated load inverter Cross-section for a NMOS saturated load inverter Microelectronic Circuit Design McGraw-Hill
32. NMOS Saturated Load Inverter Design Strategy • Given VDD, VL, and the power level, find IDD from VDD and power • Assume MS off, and find high output voltage level VH • Use the value of VH for the gate voltage of MS and calculate (W/L)S of the switching transistor based on the design values of IDD and VL • Find (W/L)L (load transistor) based on IDD and VL • Check the operating region assumptions of MS and ML for vo = VL • Verify design with a SPICE simulations Microelectronic Circuit Design McGraw-Hill
33. NMOS Saturated Load Inverter Design Example • Design an saturated load inverter given the following specifications: Microelectronic Circuit Design McGraw-Hill
34. NMOS Saturated Load Inverter Design Example • First find VH Microelectronic Circuit Design McGraw-Hill
35. NMOS Saturated Load Inverter Design Example • For vo = VL,MS is off (triode region) and ML is in saturation, find the W/L ratios of the two transistors Microelectronic Circuit Design McGraw-Hill
36. NMOS Saturated Load Inverter Design Noise Margin Analysis • The basic noise margin equations are still the same as for previous inverters, but there are different expressions for the components The equations can be written as a quadratic equation,but an iterative process must be used to solve for VOL and VTNL 1) Choose an initial VOL 2) Calculate the corresponding VTNL 3) Update VOL 4) Repeat 2 and 3 until the system converges Microelectronic Circuit Design McGraw-Hill
37. NMOS Inverter with a Linear Load Device • This alternative inverter has a load transistor that is biased with VGG defined by the following: • This causes the load transistor to operate in the linear region Microelectronic Circuit Design McGraw-Hill
38. NMOS Inverter with a Depletion-mode Load • With the addition of a depletion-mode NMOS (VTH < 0V), it is possible to configure an inverter as shown • VGSL = 0 V for this configuration meaning that ML is always operating in saturation Microelectronic Circuit Design McGraw-Hill
39. Design of a NMOS Inverter with a Depletion-mode Load • To find (W/L)L given iDL: • To find (W/L)S where VH = VDD use the same technique as used for the resistor load inverter: Microelectronic Circuit Design McGraw-Hill
40. Noise Margins of a NMOS Inverter with a Depletion-mode Load The first two equations assume the MS is saturated and ML is in triode The last two equations assume the MS and ML are in triode Microelectronic Circuit Design McGraw-Hill
41. NMOS Inverter Summary • Resistive load inverter takes up too much area for and IC design • The saturated load configuration is the simplest design, but VH never reaches VDD and has a slow switching speed • The linear load inverter fixes the speed and logic level issues, but it requires an additional power supply for the load gate • The depletion-mode NMOS load requires the most processing steps, but needs the smallest area to achieve the highest speed, VH = VDD, and best combination of noise margins Microelectronic Circuit Design McGraw-Hill
42. Typical Inverter Characteristic Microelectronic Circuit Design McGraw-Hill
43. NOR Gates Simplified switch model for the NOR gate with A on Two-input NOR gate Microelectronic Circuit Design McGraw-Hill
44. NAND Gates Simplified switch model for the NOR gate with A and B on (right) Two-input NAND gate (left) Microelectronic Circuit Design McGraw-Hill
45. NAND Gate Device Size Selection • The NAND switching transistors can be sized based on the depletion-mode load inverter • To keep the low voltage level to be comparable to the inverter, the desired RON of MA and MB must be 0.5RON of MS,Inverter • This can be accomplished by approximately doubling the (W/L)A and (W/L)B • The sizes can also be chosen by using the design value of VL and using the following equation: Microelectronic Circuit Design McGraw-Hill
46. NAND Gate Device Size Selection (continued) • Two sources of error that arise are the facts that VSB and VGS of the two transistors do not equal. These factors should be considered for proper gate design • The technique used to calculate the size of the load transistor for the depletion-mode load inverter is the exact same as for this NAND gate Microelectronic Circuit Design McGraw-Hill
47. Layout of the NMOS Depletion-Mode NOR and NAND Gates Microelectronic Circuit Design McGraw-Hill
48. Complex NMOS Logic Design An advantage of NMOS technology is that it is simple to design complex logic functions based on the NOR and NAND gates The circuit in the figure has the logic function: Y = A + BC + BD Microelectronic Circuit Design McGraw-Hill
49. Complex Logic Gate Transistor Sizing • There are two ways to find the W/L ratios of the switching transistors • Using the worst case (longest) path and choosing the W/L ratio such that the RON of the multiple legs match similar to the technique used to find the W/L ratios in the NAND Gate • Partitioning the circuit into series sub-networks, and make the equivalent on-resistances equal Microelectronic Circuit Design McGraw-Hill
50. Complex Logic Gate Transistor Sizing The figure on the left shows the worst case technique to find the sizes where (W/L)S=2.06 is the reference inverter ratio for this technology and the longest path is 3 transistors are in series The figure on the right shows the partitioning technique to find the sizes which gives two 4.12/1 ratios in series which is 2(2.06/1) Microelectronic Circuit Design McGraw-Hill
More Related | 3,333 | 14,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.743251 |
https://www.clutchprep.com/physics/practice-problems/50406/a-beam-of-high-energy-960-8722-negative-pions-is-160-shot-at-a-flask-of-liquid-h | 1,611,425,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00427.warc.gz | 704,564,606 | 27,463 | # Problem: A beam of high-energy π− (negative pions) is shot at a flask of liquid hydrogen, and sometimes a pion interacts through the strong interaction with a proton in the hydrogen, in the reaction π− + p+ → π− + X+, where X+ is a positively charged particle of unknown mass. The incoming pion momentum is 2965 MeV/c. The pion is scattered through 38°, and its momentum is measured to be 1480 MeV/c. The X+ particle is scattered through an unknown angle θ with an unknown momentum. A pion has a rest energy of 140 MeV, and a proton has a rest energy of 938 MeV. Find the scattering angle θ of the X+ particle. 1. 25.8446 2. 29.2199 3. 29.9574 4. 24.9877 5. 25.4691 6. 27.5308 7. 28.7532 8. 26.8651 9. 28.454 10. 27.815
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A beam of high-energy π (negative pions) is shot at a flask of liquid hydrogen, and sometimes a pion interacts through the strong interaction with a proton in the hydrogen, in the reaction π + p+ → π + X+, where X+ is a positively charged particle of unknown mass. The incoming pion momentum is 2965 MeV/c. The pion is scattered through 38°, and its momentum is measured to be 1480 MeV/c. The X+ particle is scattered through an unknown angle θ with an unknown momentum. A pion has a rest energy of 140 MeV, and a proton has a rest energy of 938 MeV. Find the scattering angle θ of the X+ particle.
1. 25.8446
2. 29.2199
3. 29.9574
4. 24.9877
5. 25.4691
6. 27.5308
7. 28.7532
8. 26.8651
9. 28.454
10. 27.815 | 498 | 1,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-04 | latest | en | 0.773772 |
https://scirp.org/journal/paperinformation?paperid=107560 | 1,713,904,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00066.warc.gz | 445,763,245 | 33,181 | A Note on SK, SK1, SK2 Indices of Interval Weighted Graphs
Abstract
In this study, the SK, SK1 and SK2 indices are defined on weighted graphs. Then, the SK, SK1 and SK2 indices are defined on interval weighted graphs. Their behaviors are investigated under some graph operations by using these definitions.
Keywords
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Nurkahlı, S. and Büyükköse, Ş. (2021) A Note on SK, SK1, SK2 Indices of Interval Weighted Graphs. Advances in Linear Algebra & Matrix Theory, 11, 14-20. doi: 10.4236/alamt.2021.111002.
1. Introduction
A topological index of a chemical compound is an integer, derived following a certain rule, which can be used to characterize the chemical compound and predict certain physiochemical properties like boiling point, molecular weight, density, refractive index, and so forth [1].
Molecules and molecular compounds are often modeled by molecular graph. A molecular graph is a representation of the structural formula of a chemical compound in terms of graph theory, whose vertices correspond to the atoms of the compound and edges correspond to chemical bonds [2].
Let $G=\left(V,E\right)$ be a graph with the vertex set $V\left(G\right)$ and edge set $E\left(G\right)$ and ${v}_{G}=|V\left(G\right)|$ vertices and ${e}_{G}=|E\left(G\right)|$ edges. The degree ${d}_{v}$ of the vertex $v\in V\left(G\right)$ is the number of first neighbors of v. The edge of the graph G, connecting the verticesu and v, will be denoted by $e=uv$. Throughout this paper, the graphs considered are assumed to be connected. A connected graph is a graph such that there is a path between all pairs of vertices, see books [3] [4].
We now recall some graph operations we shall need in this paper.
Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple graphs. The sum ${G}_{1}+{G}_{2}$ of these two graphs is defined as the graph having the vertex set $V\left({G}_{1}+{G}_{2}\right)={V}_{1}\cup {V}_{2}$ and the edge set $E\left({G}_{1}+{G}_{2}\right)={E}_{1}\cup {E}_{2}\cup \left\{\left(u,v\right):u\in {V}_{1},v\in {V}_{2}\right\}$ [5].
The cartesian product ${G}_{1}×{G}_{2}$ is the graph with vertex set $V\left({G}_{1}×{G}_{2}\right)={V}_{1}×{V}_{2}$ ; the vertices $u=\left({u}_{1},{u}_{2}\right)$ and $v=\left({v}_{1},{v}_{2}\right)$ of ${G}_{1}×{G}_{2}$ are adjacent if and only if $\left[{u}_{1}={v}_{1},{u}_{2}{v}_{2}\in {E}_{2}\right]$ or $\left[{u}_{2}={v}_{2},{u}_{1}{v}_{1}\in {E}_{1}\right]$ [5].
Definition 1.1. ( [1], SK index). TheSK index of a graph $G=\left(V,E\right)$ is defined as
$\text{SK}\left(G\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }d\left(u\right)+d\left( v \right)$
where $d\left(u\right)$ and $d\left(v\right)$ are the degrees of the vertices u and v in G, respectively.
Definition 1.2. ( [1], SK1 index). TheSK1 index of a graph $G=\left(V,E\right)$ is defined as
${\text{SK}}_{1}\left(G\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }d\left(u\right)d\left( v \right)$
where $d\left(u\right)$ and $d\left(v\right)$ are the degrees of the vertices u and v in G, respectively.
Definition 1.3. ( [1], SK2 index). TheSK2 index of a graph $G=\left(V,E\right)$ is defined as
${\text{SK}}_{2}\left(G\right)=\frac{1}{4}\underset{uv\in E\left(G\right)}{\sum }{\left[d\left(u\right)+d\left(v\right)\right]}^{2}$
where $d\left(u\right)$ and $d\left(v\right)$ are the degrees of the vertices u and v in G, respectively.
2. Graph Operations on the SK, SK1, SK2 Indices of Weighted Graphs
In this section, we define the SK, SK1 and SK2 indices on weighted graphs. A weighted graph is a graph each edge of which has been assigned to a number called the weight of the edge. All the weight of the edges are assumed to be positive definite [6] [7].
Let G be a weighted graph with vertex set $V\left(G\right)=\left\{{v}_{1},{v}_{2},\cdots ,{v}_{n}\right\}$ and edge set E. Denote by ${w}_{ij}$ the positive definite weight matrix of order p of the edge ij and assume that ${w}_{ij}={w}_{ji}$. We write $i~j$ if vertices i and j are adjacent. Let ${w}_{i}={\sum }_{j:i~j}{w}_{ij}$ be the weight matrix of the vertex i [6] [7].
Definition 2.1. Let $G=\left(V,E\right)$ be a connected weighted graph having n vertices. Let each edge of G be weighted with positive real numbers. The weighted SK index $\text{SK}\left(G,w\right)$ of G is defined as follows:
$\text{SK}\left(G,w\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }w\left(u\right)+w\left( v \right)$
where $w\left(u\right)$ is the sum of the weights on u.
Definition 2.2. Let $G=\left(V,E\right)$ be a connected weighted graph having n vertices. Let each edge of G be weighted with positive real numbers. The weighted SK1 index ${\text{SK}}_{1}\left(G,w\right)$ of G is defined as follows:
${\text{SK}}_{1}\left(G,w\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }w\left(u\right)w\left( v \right)$
where $w\left(u\right)$ is the sum of the weights on u.
Definition 2.3. Let $G=\left(V,E\right)$ be a connected weighted graph having n vertices. Let each edge of G be weighted with positive real numbers. The weighted SK2 index ${\text{SK}}_{2}\left(G,w\right)$ of G is defined as follows:
${\text{SK}}_{2}\left(G,w\right)=\frac{1}{4}\underset{uv\in E\left(G\right)}{\sum }{\left[w\left(u\right)+w\left(v\right)\right]}^{2}$
where $w\left(u\right)$ is the sum of the weights on u.
Theorem 2.4. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected graphs. Then the SK, SK1 and SK2 indices of the sum of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by
$\begin{array}{c}\text{SK}\left({G}_{1}+{G}_{2}\right)=\frac{1}{2}\left[\underset{ij\in {E}_{1}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{2}}\right)+\underset{ij\in {E}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{1}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\right]\end{array}$
$\begin{array}{c}{\text{SK}}_{1}\left({G}_{1}+{G}_{2}\right)=\frac{1}{2}\left[\underset{ij\in {E}_{1}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)\left({d}_{j}+{n}_{{G}_{2}}\right)+\underset{ij\in {E}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{1}}\right)\left({d}_{j}+{n}_{{G}_{1}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)\left({d}_{j}+{n}_{{G}_{1}}\right)\right]\end{array}$
$\begin{array}{c}{\text{SK}}_{2}\left({G}_{1}+{G}_{2}\right)=\frac{1}{4}\left[\underset{ij\in {E}_{1}}{\sum }{\left[\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{2}}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{ij\in {E}_{2}}{\sum }{\left[\left({d}_{i}+{n}_{{G}_{1}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }{\left[\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\right]}^{2}\right]\end{array}$
Theorem 2.5. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected weighted graphs. Then the weighted SK, SK1 and SK2 indices of the sum of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by
$\begin{array}{c}\text{SK}\left({G}_{1}+{G}_{2},w\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({w}_{i}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]\end{array}$
${\text{SK}}_{1}\left({G}_{1}+{G}_{2},w\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\left({w}_{j}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)$
$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({w}_{i}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]\end{array}$
$\begin{array}{c}{\text{SK}}_{2}\left({G}_{1}+{G}_{2},w\right)=\frac{1}{4}\left[\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }{\left[\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\right]}^{2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }{\left[\left({w}_{i}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }{\left[\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]}^{2}\right]\end{array}$
Theorem 2.6. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected graphs. Then the SK, SK1 and SK2 indices of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by
$\text{SK}\left({G}_{1}×{G}_{2}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(d\left({u}_{i}\right)+d\left({v}_{j}\right)\right)+\left(d\left({u}_{k}\right)+d\left({v}_{l}\right)\right)\right]$
${\text{SK}}_{1}\left({G}_{1}×{G}_{2}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(d\left({u}_{i}\right)+d\left({v}_{j}\right)\right)\left(d\left({u}_{k}\right)+d\left({v}_{l}\right)\right)\right]$
${\text{SK}}_{2}\left({G}_{1}×{G}_{2}\right)=\frac{1}{4}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }{\left[\left(d\left({u}_{i}\right)+d\left({v}_{j}\right)\right)+\left(d\left({u}_{k}\right)+d\left({v}_{l}\right)\right)\right]}^{2}\right]$
Theorem 2.7. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected weighted graphs. Then the weighted SK, SK1 and SK2 indices of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by
$\text{SK}\left({G}_{1}×{G}_{2},w\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(w\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(w\left({u}_{k}\right)+w\left({v}_{l}\right)\right)\right]$
${\text{SK}}_{1}\left({G}_{1}×{G}_{2},w\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(w\left({u}_{i}\right)+w\left({v}_{j}\right)\right)\left(w\left({u}_{k}\right)+w\left({v}_{l}\right)\right)\right]$
${\text{SK}}_{2}\left({G}_{1}×{G}_{2},w\right)=\frac{1}{4}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }{\left[\left(w\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(w\left({u}_{k}\right)+w\left({v}_{l}\right)\right)\right]}^{2}\right]$
3. Graph Operations on the SK, SK1, SK2 Indices of Interval Weighted Graphs
In this section, we define the SK, SK1 and SK2 indices on interval weighted graphs. An interval weighted graph (interval graph) is a weighted graph in which each edge is assigned an interval or an interval square matrix. All the interval square matrices are assumed to be of the same order and to be positive definite [8].
Let G be an interval graph on n vertices. Denote by ${\stackrel{˜}{w}}_{ij}$ the positive definite interval matrix of order p of the edge ij and assume that ${\stackrel{˜}{w}}_{ij}={\stackrel{˜}{w}}_{ji}$. We write $i~j$ if vertices i and j are adjacent. Let ${\stackrel{˜}{w}}_{i}={\sum }_{j:j~i}{\stackrel{˜}{w}}_{ij}$ be the weight interval matrix of the vertex i [8].
Definition 3.1. Let $G=\left(V,E\right)$ be a connected interval weighted graph having n vertices. Let weight each edge of G be an interval or an interval square matrix. The interval weighted SK index $\text{SK}\left(G,\stackrel{˜}{w}\right)$ of G is defined as follows:
$\text{SK}\left(G,\stackrel{˜}{w}\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }\text{ }\text{ }\stackrel{˜}{w}\left(u\right)+\stackrel{˜}{w}\left( v \right)$
where $\stackrel{˜}{w}\left(u\right)$ is the sum of the interval weights on u.
Definition 3.2. Let $G=\left(V,E\right)$ be a connected interval weighted graph having n vertices. Let weight each edge of G be an interval or an interval square matrix. The interval weighted SK1 index ${\text{SK}}_{1}\left(G,\stackrel{˜}{w}\right)$ of G is defined as follows:
${\text{SK}}_{1}\left(G,\stackrel{˜}{w}\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }\text{ }\text{ }\stackrel{˜}{w}\left(u\right)\stackrel{˜}{w}\left( v \right)$
where $\stackrel{˜}{w}\left(u\right)$ is the sum of the interval weights on u.
Definition 3.3. Let $G=\left(V,E\right)$ be a connected interval weighted graph having n vertices. Let weight each edge of G be an interval or an interval square matrix. The interval weighted SK2 index ${\text{SK}}_{2}\left(G,\stackrel{˜}{w}\right)$ of G is defined as follows:
${\text{SK}}_{2}\left(G,\stackrel{˜}{w}\right)=\frac{1}{4}\underset{uv\in E\left(G\right)}{\sum }{\left[\stackrel{˜}{w}\left(u\right)+\stackrel{˜}{w}\left(v\right)\right]}^{2}$
where $\stackrel{˜}{w}\left(u\right)$ is the sum of the interval weights on u.
Theorem 3.4. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by
$\begin{array}{c}\text{SK}\left({G}_{1}+{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]\end{array}$
Proof. Let $V=V\left({G}_{1}\right)\cup V\left({G}_{2}\right)$, $E=E\left({G}_{1}\right)\cup E\left({G}_{2}\right)\cup \left\{\left({u}_{1},{u}_{2}\right):{u}_{1}\in V\left({G}_{1}\right),{u}_{2}\in V\left({G}_{2}\right)\right\}$. We partition the set of pairs of vertices of ${G}_{1}+{G}_{2}$ to obtain the following three sums denoted by ${K}_{1},{K}_{2},{K}_{3}$, respectively.
Firstly, for each sum, we consider ${\stackrel{˜}{w}}_{i}$ as the sum of the weights in each vertex i. In ${K}_{1}$, we collect all pairs of vertices i and j so that $i,j$ are in $V\left({G}_{1}\right)$ and ij is in $E\left({G}_{1}\right)$. Hence, i and j are adjacent vertices in $E\left({G}_{1}\right)$. For ${K}_{1}$, we obtain,
${K}_{1}=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\right].$
For the second sum ${K}_{2}$, we take the vertices i and j in $V\left({G}_{2}\right)$ so that ij is in $E\left({G}_{2}\right)$. Hence,
${K}_{2}=\frac{1}{2}\left[\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right].$
In the third sum ${K}_{3}$, i is taken in $V\left({G}_{1}\right)$ and j is taken in $V\left({G}_{2}\right)$. So,
${K}_{3}=\frac{1}{2}\left[\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right].$
The result now follows by adding the three contributions and simplifying the resulting expression.
Theorem 3.5. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK1 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by
$\begin{array}{c}{\text{SK}}_{1}\left({G}_{1}+{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]\end{array}$
Proof. The proof is similarly done to the proof of Theorem 3.4.
Theorem 3.6. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK2 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by
$\begin{array}{c}{\text{SK}}_{2}\left({G}_{1}+{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{4}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }{\left[\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }{\left[\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }{\left[\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]}^{2}\right]\end{array}$
Proof. The proof is similarly done to the proof of Theorem 3.4.
Theorem 3.7. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by
$\text{SK}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(\stackrel{˜}{w}\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right]$
Proof. The set of vertices in the graph ${G}_{1}×{G}_{2}$ is $u=\left({u}_{1},{u}_{2}\right)$, $v=\left({v}_{1},{v}_{2}\right)\in {V}_{1}×{V}_{2}$ for ${u}_{1},{v}_{1}\in {V}_{1}$ and ${u}_{2},{v}_{2}\in {V}_{2}$. Also, $\stackrel{˜}{w}\left(u\right)$ is the interval weight of the vertex u. Thus, the interval weight of any vertex $\left({u}_{1},{u}_{2}\right)\in {V}_{1}×{V}_{2}$ in the graph ${G}_{1}×{G}_{2}$ is $\stackrel{˜}{w}\left({u}_{1}\right)+\stackrel{˜}{w}\left({u}_{2}\right)$.
The SK index is equal half of the sum of degrees of all adjacent vertex pairs of the graph. Since the degrees in an interval weighted graph will turn into interval weights, it is obtained
$\text{SK}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(\stackrel{˜}{w}\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right].$
Theorem 3.8. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK1 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by
${\text{SK}}_{1}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(\stackrel{˜}{w}\left({u}_{i}\right)+\stackrel{˜}{w}\left({v}_{j}\right)\right)\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right]$
Proof. The proof is similarly done to the proof of Theorem 3.7.
Theorem 3.9. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK2 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by
${\text{SK}}_{2}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{4}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }{\left[\left(\stackrel{˜}{w}\left({u}_{i}\right)+\stackrel{˜}{w}\left({v}_{j}\right)\right)+\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right]}^{2}\right]$
Proof. The proof is similarly done to the proof of Theorem 3.7.
Acknowledgements
The first author is supported in part by TÜBİTAK. This work is derived from the first author’s PH’s thesis.
Conflicts of Interest
The authors declare no conflicts of interest regarding the publication of this paper.
[1] Shigehalli, V. and Kanabur, R. (2016) Computing Degree-Based Topological Indices of Polyhex Nanotubes. Journal of Mathematical Nanoscience, 6, 47-55. [2] Fath-Tabar, G.H. (2011) Old and New Zagreb Indices of Graphs. MATCH Communications in Mathematical and in Computer Chemistry, 65, 79-84. [3] Farahani, M.R. (2015) Computing the Hyper-Zagreb Index of Hexagonal Nanotubes. Journal of Chemistry and Materials Research, 2, 16-18. [4] Falahati-Nezhad, F. and Azari, M. (2016) Bounds on the Hyper-Zagreb Index. Journal of Applied Mathematics and Informatics, 34, 319-330.https://doi.org/10.14317/jami.2016.319 [5] Büyükköse, Ş. and Kaya-Gök, G. (2017) Graph Operations of Randic Index. https://doi.org/10.1063/1.5020470 [6] Vasudev, C. (1996) Introduction to Graph Theory. Prentice-Hall, Upper Saddle River. [7] Horn, R.A. and Johnson, C.R. (2012) Matrix Analysis. 2nd Edition, Cambridge/United Kingdom: Cambridge University Press, 391-425.https://doi.org/10.1017/CBO9781139020411 [8] Başdaş Nurkahlı, S. and Büyükköse, Ş. (2020) Interval Ağırlıklı Grafların Zagreb İndeksi üzerinde Graf İşlemlari. GüFFD, 1, 132-143. | 9,204 | 23,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 154, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.869816 |
https://simplywall.st/stocks/hk/healthcare/hkg-3309/c-mer-eye-care-holdings-shares/news/should-you-sell-c-mer-eye-care-holdings-limited-hkg3309-at-this-pe-ratio/ | 1,553,398,501,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00279.warc.gz | 597,167,683 | 15,174 | # Should You Sell C-MER Eye Care Holdings Limited (HKG:3309) At This PE Ratio?
C-MER Eye Care Holdings Limited (SEHK:3309) trades with a trailing P/E of 187.1x, which is higher than the industry average of 33.3x. Although some investors may jump to the conclusion that you should avoid the stock or sell if you own it, understanding the assumptions behind the P/E ratio might change your mind. Today, I will explain what the P/E ratio is as well as what you should look out for when using it. View our latest analysis for C-MER Eye Care Holdings
### Demystifying the P/E ratio
The P/E ratio is one of many ratios used in relative valuation. It compares a stock’s price per share to the stock’s earnings per share. A more intuitive way of understanding the P/E ratio is to think of it as how much investors are paying for each dollar of the company’s earnings.
Formula
Price-Earnings Ratio = Price per share ÷ Earnings per share
P/E Calculation for 3309
Price per share = HK\$13.26
Earnings per share = HK\$0.071
∴ Price-Earnings Ratio = HK\$13.26 ÷ HK\$0.071 = 187.1x
On its own, the P/E ratio doesn’t tell you much; however, it becomes extremely useful when you compare it with other similar companies. We preferably want to compare the stock’s P/E ratio to the average of companies that have similar features to 3309, such as capital structure and profitability. A common peer group is companies that exist in the same industry, which is what I use below. Since it is expected that similar companies have similar P/E ratios, we can come to some conclusions about the stock if the ratios are different.
At 187.1x, 3309’s P/E is higher than its industry peers (33.3x). This implies that investors are overvaluing each dollar of 3309’s earnings. As such, our analysis shows that 3309 represents an over-priced stock.
### A few caveats
While our conclusion might prompt you to sell your 3309 shares immediately, there are two important assumptions you should be aware of. The first is that our “similar companies” are actually similar to 3309. If the companies aren’t similar, the difference in P/E might be a result of other factors. For example, if you are inadvertently comparing riskier firms with 3309, then 3309’s P/E would naturally be higher than its peers since investors would reward its lower risk with a higher price. The other possibility is if you were accidentally comparing lower growth firms with 3309. In this case, 3309’s P/E would be higher since investors would also reward 3309’s higher growth with a higher price. The second assumption that must hold true is that the stocks we are comparing 3309 to are fairly valued by the market. If this does not hold, there is a possibility that 3309’s P/E is higher because firms in our peer group are being undervalued by the market.
### What this means for you:
You may have already conducted fundamental analysis on the stock as a shareholder, so its current overvaluation could signal a potential selling opportunity to reduce your exposure to 3309. Now that you understand the ins and outs of the PE metric, you should know to bear in mind its limitations before you make an investment decision. Remember that basing your investment decision off one metric alone is certainly not sufficient. There are many things I have not taken into account in this article and the PE ratio is very one-dimensional. If you have not done so already, I highly recommend you to complete your research by taking a look at the following:
• 1. Financial Health: Is 3309’s operations financially sustainable? Balance sheets can be hard to analyze, which is why we’ve done it for you. Check out our financial health checks here.
• 2. Valuation: What is 3309 worth today? Is the stock undervalued, even when its growth outlook is factored into its intrinsic value? The intrinsic value infographic in our free research report helps visualize whether 3309 is currently mispriced by the market.
• 3. Other High-Performing Stocks: Are there other stocks that provide better prospects with proven track records? Explore our free list of these great stocks here. | 926 | 4,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-13 | latest | en | 0.946656 |
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# Stand next to a wall that travels at 30 km/s relative to
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## Solution for problem 24RQ Chapter 2
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Problem 24RQ
Stand next to a wall that travels at 30 km/s relative to the Sun. With your feet on the ground, you also travel the same 30 km/s. Do you maintain this speed when your feet leave the ground? What concept supports your answer?
Step-by-Step Solution:
Step 1 of 3
Solutio 24RQ According to the Newton’s second law, the force changes the motion of the object in the direction of the force. Now when my feet are not on the ground, there will be some force vertically downwards (due to gravity), no horizontal force will be acting on me. Now the velocity at which I am moving relative to the sun is horizontal. And since there is no horizontal force is acting on me, my velocity will not change. So I will still move with 30 km/s velocity. The concept of inertia, i.e. the direction and magnitude of the velocity of a moving object will not change unless acted upon by an external force.
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##### ISBN: 9780321909107
The full step-by-step solution to problem: 24RQ from chapter: 2 was answered by , our top Physics solution expert on 04/03/17, 08:01AM. The answer to “Stand next to a wall that travels at 30 km/s relative to the Sun. With your feet on the ground, you also travel the same 30 km/s. Do you maintain this speed when your feet leave the ground? What concept supports your answer?” is broken down into a number of easy to follow steps, and 43 words. This full solution covers the following key subjects: feet, Ground, relative, also, Answer. This expansive textbook survival guide covers 45 chapters, and 4650 solutions. This textbook survival guide was created for the textbook: Conceptual Physics, edition: 12. Conceptual Physics was written by and is associated to the ISBN: 9780321909107. Since the solution to 24RQ from 2 chapter was answered, more than 517 students have viewed the full step-by-step answer.
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A242404 a(n) = Sum over i and j with j<=i and i=1.. sopf(n) of binomial(d(i), d(j)), where d(i) are the prime divisors of n and sopf(n) = sum of the distinct primes dividing n (A008472). 1
0, 1, 1, 1, 1, 5, 1, 1, 1, 12, 1, 5, 1, 23, 12, 1, 1, 5, 1, 12, 37, 57, 1, 5, 1, 80, 1, 23, 1, 26, 1, 1, 167, 138, 23, 5, 1, 173, 288, 12, 1, 62, 1, 57, 12, 255, 1, 5, 1, 12, 682, 80, 1, 5, 464, 23, 971, 408, 1, 26, 1, 467, 37, 1, 1289, 226, 1, 138, 1773, 55 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,6 COMMENTS This sequence is very rich in properties. We cite a few. If n = p^m where p is prime, then a(n) = 1 => a(A025475(n)) = 1. If n = 2^i*3^j, i and j >= 1, then a(n) = 5 => a(A033845(n)) = 5. We observe a very interesting property with a fractal structure if a(n) = 23. a(n) = 23 if n belongs to the set E = {n} = {14, 28, 35, 56, 98, 112, 175, 196, 224, 245, 392, 448, 686, ...} and we observe that E1 = {n/7} = {2, 4, 5, 8, 14, 16, 25, 28, 32, 35, 56, 64, 98, 112, 125, 128, 175, 196, 224, 245, 256, ...} = E2 union E3 union E4 where: E2 = {2, 4, 8, 16, 32, 64, 128, ...} are powers of 2 (A000079), E3 = {5, 25, 125, 625, 3125, ...} are powers of 5 (A000351), E4 = {14, 28, 35, 56, 98, 112, 175, 196, 224, 245, ...} = E, a surprising result: E1 contains E => this shows a fractal structure. But you can continue with other values of a(n) in order to find similar properties. For example, a(n) = 55 if n belongs to the set F = {n} = {70, 140, 280, 350, 490, 560, 700, 980, 1120, ...} => F1 = {n/70} = { 1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 25, 28, 32, 35, 40, 49, 50, 56, 64, 70, 80, ...} = F2 union F3 union F4 union F5 where: F2 = {2, 4, 8, 16, 32, 64, 128, ...} are powers of 2 (A000079), F3 = {5, 25, 125, 625, 3125, ...} are powers of 5 (A000351), F4 = {1, 7, 49, 343, 2401, ...} are powers of 7 (A000420) and F5 = {10, 14, 20, 28, 35, 40, 50, 56, 70, 80, 98, 100, 112, 140, ...} contains F but also E. LINKS Michel Lagneau, Table of n, a(n) for n = 1..10000 EXAMPLE a(10) = 12 because 12 = 2^2*3 => sopf(12) = 2+3=5 and a(10) = sum{j<=i=1..5} binomial(d(i),d(j)) = binomial(2,2)+ binomial(5,2)+ binomial(5,5) = 12. MAPLE with(numtheory):for n from 1 to 200 do:x:=factorset(n):n1:=nops(x): r:=sum('sum('binomial(x[j], x[i])', 'j'=i..n1)', 'i'=1..n1):printf(`%d, `, r):od: CROSSREFS Cf. A000079, A000351, A000420, A008472. Sequence in context: A206773 A229526 A204007 * A145295 A091051 A183097 Adjacent sequences: A242401 A242402 A242403 * A242405 A242406 A242407 KEYWORD nonn AUTHOR Michel Lagneau, May 13 2014 STATUS approved
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# Percentages - Quant/Math - TANCET 2018 Practice Question
This problem solving question is from the topic Percentages.
## Question
A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks. Find the maximum marks.
1. 50
2. 100
3. 150
4. 200
100. Choice (2)
From the given statement pass percentage is 42% - 12% = 30%
By hypothesis, 30% of x - 20% of x = 10 (marks)
i.e., 10% of x = 10
Therefore, x = 100 marks.
## CAT, XAT, TANCET Practice Questions and Answers : Listed Topicwise
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http://stackoverflow.com/questions/4787014/excel-formula-to-find-the-rightmost-column-containing-a-value-in-a-table | 1,386,906,852,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164844212/warc/CC-MAIN-20131204134724-00002-ip-10-33-133-15.ec2.internal.warc.gz | 179,251,789 | 12,793 | # Excel formula to find the rightmost column containing a value in a table
I have some data structured like this in an Excel spreadsheet:
`````` A B C D E F
1 1 1 2 x 2 3
2 1 1 1 2 2 3
3 3 3 3 3 4 4
``````
I am trying to formulate an Excel formula which will give me the index of the rightmost column in this table which has a cell matching a specific value.
In this example, the rightmost column containing the value '1' is in column C. For '2' it would be E. That column index is what I'm after.
I use column letters to be Excel-consistent, but a numeric column index is preferable.
I have tried some other solutions for similar Excel problems found online, but they don't quite have the right behavior.
-
Here is a way to do it with formulas. I'll show how to do it with a few different formulas to show the steps of the logic, and then put them together into one big formula.
First, use one formula per column to see if the target value is in the column. For example in column A:
``````=COUNTIF(A1:A100,Goal)
=COUNTIF(B1:B100,Goal)
...
(where Goal can be a hardcoded search value,
or a named range where you type your query)
``````
Then, add IF statements to these formulas to translate this into column numbers. If the query is present in the column, show the column number, else show zero.
``````=IF(COUNTIF(A1:A100,Goal)>0, 1, 0)
=IF(COUNTIF(B1:B100,Goal)>0, 2, 0)
...
``````
Finally, add a formula to grab the maximum column number from the prior formulas. This will equal the rightmost column with your query value in it.
``````=MAX( IF(COUNTIF(A1:A100,Goal)>0, 1, 0), IF(COUNTIF(B1:B100,Goal)>0, 2, 0), ...)
``````
-
If you want to use helper columns, you can put this formula in G1
``````{=MAX((COLUMN(A1:F1)*(A1:F1=2)))}
``````
That's array entered. Fill down to G3. In G4, put
``````=MAX(G1:G3)
``````
Then repeat for each number. If you don't want helper columns, you can write a UDF like this
``````Public Function MaxColumn(rInput As Range, vValue As Variant) As Long
Dim rFound As Range
Set rFound = rInput.Find(vValue, rInput.Cells(1), xlValues, xlWhole, xlByColumns, xlPrevious)
If Not rFound Is Nothing Then MaxColumn = rFound.Column
End Function
``````
Which you call like
``````=maxcolumn(A1:F3,2)
``````
-
``````Function FindCol(ToFind)
Dim r As Range
Dim rfind As Range
Dim rfound As Range
Set r = ActiveSheet.UsedRange
For i = r.Columns.Count To 1 Step -1
Set rfind = r.Columns(i)
Set rfound = rfind.Find(ToFind)
If Not rfound Is Nothing Then
Result = rfound.Column
Exit For
End If
Next
FindCol = Result
End Function
``````
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Zbl 0641.62032
Owen, Art B.
Empirical likelihood ratio confidence intervals for a single functional.
(English)
[J] Biometrika 75, No.2, 237-249 (1988). ISSN 0006-3444; ISSN 1464-3510/e
Let $(X\sb 1,...,X\sb n)$ be a random sample, its components $X\sb i$ are observations from a distribution-function $F\sb 0$. The empirical distribution function $F\sb n$ is a nonparametric maximum likelihood estimate of $F\sb 0$. $F\sb n$ maximizes $$L(F)=\prod\sp{n}\sb{i=1}\{F(X\sb i)-F(X\sb i-)\}$$ over all distribution functions F. Let $R(F)=L(F)/L(F\sb n)$ be the empirical likelihood ratio function and T(.) any functional. It is shown that sets of the form $$\{T(F)\vert R(F)\ge c\}$$ may be used as confidence regions for some $T(F\sb 0)$ like the sample mean or a class of M-estimators (especially the quantiles of $F\sb 0)$. These confidence intervals are compared in a simulation study to some bootstrap confidence intervals and to confidence intervals based on a t-statistic for a confidence coefficient $1-\alpha =0.9$. It seems that two of the bootstrap intervals may be recommended but the simulation is based on 1000 runs only.
[D.Rasch]
MSC 2000:
*62G15 Nonparametric confidence regions, etc.
62G30 Order statistics, etc.
62G05 Nonparametric estimation
Keywords: differentiable statistical functionals; empirical distribution; nonparametric maximum likelihood estimate; empirical likelihood ratio function; confidence regions; sample mean; M-estimators; quantiles; bootstrap confidence intervals; t-statistic
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https://www.biostars.org/p/309729/#309948 | 1,701,838,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.13/warc/CC-MAIN-20231206031946-20231206061946-00423.warc.gz | 752,044,633 | 11,266 | make fasta sequence which is multiple of three
2
0
Entering edit mode
5.6 years ago
alim.hcu ▴ 20
I have fasta sequence of different length. But I want to make number of nucleotide in fasta sequences multiple of three by adding AA or -- at the end. Like if the sequence length is 149 then i want to make it 150 by adding -- or any nucleotides.
Thank You
alignment • 4.5k views
3
Entering edit mode
This looks like an XY Problem. Why do you want to do this multiple of 3 thing? The 3 part tells me if probably has to do with codon usage, but in no scenario can I imagine needing this functionality. Can you explain your end goal please?
0
Entering edit mode
Hello, if you can program in Java, you may use the SEDA API (https://github.com/sing-group/seda) in order to create this operation that you need. You can even use the new Java 9 shell to easily do this as I show in this post:
Regards.
0
Entering edit mode
Thanks for you reply. I downloaded this tools. But the video is not much clear. which option i should use to make multiple of three nucleotides.
0
Entering edit mode
Actually I have to calculate dn/ds using paml. And it does not take any sequences which is not multiplle of three.
3
Entering edit mode
I have to calculate dn/ds using paml
0
Entering edit mode
You should have said so in the original question to avoid half the comments you received here :-).
0
Entering edit mode
Please use ADD COMMENT or ADD REPLY to answer to previous reactions, as such this thread remains logically structured and easy to follow. Your reaction was moved but as you can see it's not optimal. Adding an answer should only be used for providing a solution to the question asked.
0
Entering edit mode
If you are planning on using PAML then you'll need to pairwise align those sequences anyway. So the more cleaver thing to do is to "codon-aware" align the nucleotide sequence and strip off anything that is not aligned (== will automatically result in modulo3 'sequences')
0
Entering edit mode
Ok.. I got it..is there anything to do this in paml. I am aligning those sequences using maft. How should i trim those sequences.
0
Entering edit mode
Start by searching the forum for MAFTT posts and if you don't find an appropriate post, ask a new question.
0
Entering edit mode
ah, no you will have to do this prior to providing them to PAML.
In my lab we have custom script to do this (not helping much :-) ), but here is a tool that does it as well : trimAl and there will be others around
make sure that you do codon-align in mafft (this is crucial for dN/dS estimations!!)
0
Entering edit mode
Specifically "AA" or any nucleotide?
0
Entering edit mode
Thank You very much..it worked..
0
Entering edit mode
Wouter literally just asked you to not add answers. Please be more careful. This should be a comment or comment reply. I've moved it to a comment on the top level post, but it ideally belongs elsewhere. Please read the posts under http://biostars.org/t/how-to to structure your posts better.
1
Entering edit mode
5.6 years ago
Beside the correct question from Ram why you want to do this, here is a python solution:
from Bio import SeqIO
original_file = "./original.fasta"
extended_file = "./extended.fasta"
with open(original_file) as original, open(extended_file, 'w') as extended:
records = SeqIO.parse(original_file, 'fasta')
for record in records:
if len(record.seq)%3 == 1:
record.seq = record.seq + "--"
elif len(record.seq)%3 == 2:
record.seq = record.seq + "-"
SeqIO.write(record, extended, 'fasta')
fin swimmer
1
Entering edit mode
Minor comment, but there is no need to with open(...) your input file when using SeqIO
0
Entering edit mode
I want to run this script in loop..Please suggest me to modify it for running in this script in loop..
Thank You
3
Entering edit mode
why are you still insisting in fiddling about with your input sequences? while the solution offered by finswimmer will work perfectly , several other people have pointed out to you that this is not the way you need to go!
0
Entering edit mode
Actually i am working on transposons. And i have to calculate insertion time for LTR pairs present in LTR elements. So basically these are the nucleotide sequences.
1
Entering edit mode
Actually i am working on transposons.
Why did we have to wait 20 hours and so many comments before you tell us what you are working on? Don't you think all of this thread is a big waste of time?
0
Entering edit mode
alright, I was kinda expecting this already.
anyway the point remains: you should not add bases to get to modulo3 sequences, You first have to align them and subsequently remove unaligned parts (rather than adding bases!!).
Moreover have you looked into the literature on estimating transposon insertion time? There will be software around that specifically does this kind of analysis (though I can not think of one from the top of my head)
0
Entering edit mode
"You first have to align them and subsequently remove unaligned parts (rather than adding bases!!" I would be interested in doing this how to do this? Since i was trying to do this "If you're intending to run it through PAML to do dNdS " but again the multiple of 3 errors comes.
0
Entering edit mode
have a look at TrimAl , or such ... in general tools to clean multiple seq alignments
or manual if there are not that much sequences?
0
Entering edit mode
thanks will look into it trimAI
1
Entering edit mode
5.6 years ago
Joe 21k
It can be done pretty easily with BioPython, as FinSwimmer has shown, particularly with the MutableSeq objects.
If you're intending to run it through PAML to do dNdS etc, then a far more common approach, rather than hacking at your sequences, is just to use a dedicated codon-aware aligner like the imaginatively named CodonAlign
EDIT:
1. I'd strongly suggest creating files with all pairwise comparison sequences you're interested in. If you have many sequences, this may rapidly become untenable though, so think very carefully about what you're trying to show/achieve.
• Its an n choose k problem so, all pairwise comparisons of even just 100 sequences is 4950.
2. CRUCIALLY, use a codon-aware aligner as we've mentioned. Do not try to hack your sequences to a multiple of 3 by adding random characters as this will affect your results.
3. Santise your alignments if necessary. This you may need to do manually.
4. Run PAML.
0
Entering edit mode
Can You describe the pipe line. I have thousand of sequences in pair wise alignment. I want to calculate dn/ds of each pairwise alignment in paml. If any script you can suggest, it would be very helpful for me.
Thank You
0
Entering edit mode
The program is supposed to be downloadable from here: http://homepage.mac.com/barryghall/CodonAlign.html but the webpage seems to be down.
0
Entering edit mode
To my knowledge (limited as it is in this instance), you can't do this with pairwise alignments, they will need to be multiple sequence alignments, else there isn't sufficient information for the dNdS rates to be meaningful. I could be totally wrong here though.
As for scripts, no I don't know I'm afraid. But CodonAlign, and tools like it, are self contained executables, so you shouldn't need much if any scripting in order to use them.
If I am wrong, then it will probably be as simple as concatenating all your sequence pairs in to files of their own, and then running CodonAlign and PAML for each file in a loop or parallel.
0
Entering edit mode
I have to disagree here: it's perfectly OK to calculate dN and/or dS from pairwise alignments (this is how all these Ks plots are being generated).
Otherwise the approach you describe is exactly what I would do :
1. create set of files with each a pair of sequences in it to be aligned
2. align the sequences with something like CodonAlign (codon-aware aligner is key here!!)
3. clean the alignments (to remove gaps etc)
4. run PAML (codeml) on each alignment result file
0
Entering edit mode
I stand corrected on the pairwise in that case (I've never done it myself), but this is absolutely the workflow, so we all agree there!
I'm curious though, if you're just using pairwise sequences, how can you decide which sequence is 'right' relative to the other? You'd have no other information about what the most common codon in that position is to know whether a particular sequence is drifting/being selected for or against?
(this is my pure naivety though I'm sure)
1
Entering edit mode
Indeed, in a pairwise context you can't tell, but what you are referring to is already step 2 of this kind of procedure. step 1 is to calculate (estimate rather) the dS and dN for each pair. We're not looking for the most common codon or such yet, we're simply looking at the divergence on nucleotide level between 2 genes that once shared a common ancestor (origin).
In step 2 we can start looking at the ration of dS over dN which (coming to think of it) we can actually also do in a pairwise manner. The thing we are looking for is to see whether there are more non-synonymous or more synonymous substitutions happened dN/dS > 1 is positive (driving) selection , = 1 is neutral, < 1 is purifying selection . There should in "theory" not be any bias between codons used , we assume that mutations are happening at a constant rate (kuch ;) ) and thus we are only interested to see if the mutations we observe are more dN or are more dS | 2,204 | 9,441 | {"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} | 2.640625 | 3 | CC-MAIN-2023-50 | latest | en | 0.939208 |
https://www.daniweb.com/programming/software-development/threads/16664/range-of-long | 1,708,866,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00039.warc.gz | 731,779,334 | 25,789 | hi!
I have a problem about "range of data types ".I work on program that
require long digits .in "boland c++ 5.2 's help" range of long is 4294967295 .
but why this sample program doesn't work??!!
``````int main(){
long a=4294967295,b;
b=a;
return 0;
}``````
CAN YOU HELP ME?
TNX.
## All 34 Replies
coz its a border length the compliler will go on other side of the range as per long integer belongs .....as per ur code it will print -1 as 4294967295 - 4294967296 will yield as this only
That's likely the size of the UNSIGNED long; signed longs that are 32 bits are +/- 2 billion.
on a system with 32-bit longs, then, the range of an UNSIGNED long is 0 ... 0xFFFFFFFF but a signed long is 0x80000000 ... 0x7FFFFFFF.
That's likely the size of the UNSIGNED long; signed longs that are 32 bits are +/- 2 billion.
on a system with 32-bit longs, then, the range of an UNSIGNED long is 0 ... 0xFFFFFFFF but a signed long is 0x80000000 ... 0x7FFFFFFF.
OH!I make a mistake!I think that it is a misunderstanding!
I use " long" instead of "unsigned long" but my main program doesn't work yet!
I thougth that range of unsigned long is its problem.
why this program doesn't work?
``````void main(){
unsigned long array[260920];
unsigned long b=4294967290,a;
//long b=2147483647,a;
a=b;
cout<<a;
srand(time(0));
for(int i=0;i<260920;i++)
array[i]=rand()%4294967296;
getch();
}``````
What makes you think it doesn't work? What doesn't work? Does it compile?
working in hexadecimal is much easier for large numbers! using it will usually mean that you WONT go over the boundaries for the compiler (as a number such as 28746262 you could be +- 1 out, whereas 0xFFFFFFF.... you know its right!) the integer value in the loop also seems a bit too high (260920 doesnt mean anything to me...) and a bit random. what is the program supposed to do? is it just plainly to have 260920 random longs, and if so what is that going to be used for!????!?
>but why this sample program doesn't work??!!
Integer literals are just that, integers. If you want a long literal, suffix it with L. If you want an unsigned long literal, suffix it with UL.
>void main(){
main returns an int, it always has and always will.
hi!
I have a problem about "range of data types ".I work on program that
require long digits .in "boland c++ 5.2 's help" range of long is 4294967295 .
but why this sample program doesn't work??!!
``````int main(){
long a=4294967295,b;
b=a;
return 0;
}``````
CAN YOU HELP ME?
TNX.
``````#include <stdio.h>
#include <limits.h>
int main(void)
{
printf("A long may have values between %ld and %ld.\n", LONG_MIN, LONG_MAX);
printf("An unsigned long may have values between 0 and %lu.\n", ULONG_MAX);
return 0;
}
/* my output
A long may have values between -2147483648 and 2147483647.
An unsigned long may have values between 0 and 4294967295.
*/``````
YMMV.
>but why this sample program doesn't work??!!
Integer literals are just that, integers. If you want a long literal, suffix it with L. If you want an unsigned long literal, suffix it with UL.
if the integer cannot be held in int then the compiler will automatically make it long and if the integer cannot be held as a long then it will make it Unsigned long. So as far as i know it is not necessary suffix it with L or UL atleast not in this context.
>but why this sample program doesn't work??!!
Integer literals are just that, integers. If you want a long literal, suffix it with L. If you want an unsigned long literal, suffix it with UL.
>void main(){
main returns an int, it always has and always will.
:?: I don't know what is the difference of "void main()" and "int main()"?
>if the integer cannot be held in int then the compiler will automatically make
>it long and if the integer cannot be held as a long then it will make it Unsigned long
Close. Without a suffix, the type for an integer constant will start at int, then go to long int. If the value isn't representable by long int, the behavior is undefined. At the very least "in this context", a U (or u) suffix should be used to force the range to be unsigned int and then unsigned long int. By the way, if the literal doesn't fit in the allowed range, with or without a suffix, the program is broken.
>I don't know what is the difference of "void main()" and "int main()"?
void main() is wrong, int main() is correct. That's the only difference that's relevant.
"The type of an unsuffixed integer constant is either int, long or unsigned long. The system chooses the first of these types that can represent the value."
--Taken from page no 118, chapter 3, "A book on C"(4th Edition) by AL KELLEY/IRA POHL. If this is right then from what i understand, if the value isn't representable by long int, the behavior is NOT undefined. If u think i had trouble catching what the author meant please do explain. If u think what the author said is not correct then please talk to him.
"The type of an unsuffixed integer constant is either int, long or unsigned long. The system chooses the first of these types that can represent the value."
--Taken from page no 118, chapter 3, "A book on C"(4th Edition) by AL KELLEY/IRA POHL.
The type of an integer literal depends on its form, value, and suffix. If it is decimal and has no suffix, it has the first of these types in which its value can be represented: int, long int; if the value cannot be represented as a long int, the behavior is undefined. <snip octal and hexadecimal literals> If it is suffixed by u or U, its type is the first of these types in which its value can be represented: unsigned int, unsigned long int. If it is suffixed by ul, lu, uL, Lu, Ul, lU, UL, or LU, its type is unsigned long int.
A program is ill-formed if one of its translation units contains an integer literal that cannot be represented by any of the allowed types.
--Taken from Section 2.13.1 Integer Literals, paragraph 2, of the C++ standard.
well dont know much about C++ standards, but this is what was written in page no 193, The C programming Language(second edition) by K & R:
The type of an integer constant depends on its form, value and suffix. If it is unsuffixed and decimal, it has the first of these types in which its value can be represented: int, long int, unsigned long int. If it is unsuffixed, octal or hexadecimal, it has the first possible of these types: int,unsigned int, long int, unsigned long int. If it is suffixed by u or U, then unsigned int, unsigned long int. If it is suffixed by l or L, then long int, unsigned long int. If an integer constant is suffixed by UL, it is unsigned long.
So i think in C the type of integral constants are very much defined. U might want to go here http://www.lysator.liu.se/c/schildt.html
and see section 6.2.1.4. Here it says:
"Actually, unlike integers, such conversions are undefined...". It does give a hint that integer conversions are defined. I m really confused.
Could it be that it is one of those places where C differs from C++? Waiting to be enlightened.
working in hexadecimal is much easier for large numbers! using it will usually mean that you WONT go over the boundaries for the compiler (as a number such as 28746262 you could be +- 1 out, whereas 0xFFFFFFF.... you know its right!) the integer value in the loop also seems a bit too high (260920 doesnt mean anything to me...) and a bit random. what is the program supposed to do? is it just plainly to have 260920 random longs, and if so what is that going to be used for!????!?
the purpose of this program is making huge random array to sort it(several times to take noticeable time) by different sort algorithms (such as quicksort,bubblesort,mergesort,etc...)and compare the time that per sort take.and finally compare this time with complexity of algorithm.
How I do so?!
I think that larger rage cause I have less repetitive elemens in array.
>well dont know much about C++ standards
That's fine, but when I quote from either the C or C++ standard, it means I'm right. Just for future reference, because the standard is the ultimate authority on the matter, and no amount of linking to Schildt (not a good idea either way) or quoting K&R (which is never a bad idea) will change that.
>So i think in C the type of integral constants are very much defined.
C leaves open the possibility of extended integer types, but makes sure that in no way will an unsuffixed decimal constant resort to an unsigned type. The C standard gives a nice table where the allowed types for an unsuffixed decimal constant are int, long int, and long long int. The detail is thus:
If an integer constant cannot be represented by any types in its list, it may have an extended integer type, if the extended integer type can represent its value. If all of the types in the list for the constant are signed, the extended integer type shall be signed. If all of the types in the list for the constant are unsigned, the extended integer type shall be unsigned. If the list contains both signed and unsigned types, the extended integer type shall be signed or unsigned.
I've bolded the part that rules out a conversion to unsigned long, because the list for the relevant constant doesn't include unsigned types.
>the purpose of this program is making huge random array
If you want a really big array, allocate the memory dynammically and save your "stack". Of course, it makes more sense to divide the test up so that the weaker algorithms (bubble sort, insertion sort, selection sort) use a smaller array, then scale the results accordingly. Otherwise you'll be waiting for a long time for no good reason.
That's fine, but when I quote from either the C or C++ standard, it means I'm right. Just for future reference, because the standard is the ultimate authority on the matter, and no amount of linking to Schildt (not a good idea either way) or quoting K&R (which is never a bad idea) will change that.
>>Fair enough.
I finally got the enlightment i was waiting for. Almost every resource i found on the web confirmed that i was correct(i.e if it cannot be represented in long then it will go for unsigned long). There was no relevant info stated in ISO/IEC 9899-1999 standard about unsuffixed integral comstant, atleast i could not find any. However according to Sun's C user guide,
http://docs.sun.com/source/817-6697/tguide.html
With the -xc99=all (supports ISO/IEC 9899-1999), the compiler uses the first item of the following list in which the value can be represented, as required by the size of the constant:
1. int
2. long int
3. long long int
The compiler issues a warning if the value exceeds the largest value a long long int can represent.
With the -xc99=none (supports only ISO/IEC 9889-1990), the compiler uses the first item of the following list in which the value can be represented, as required by the size of the constant, when assigning types to unsuffixed constants:
1. int
2. long int
3. unsigned long int
4. long long int
5. unsigned long long int
So i guess my argument was based on ISO/IEC 9889-1990 whereas urs argument agrees with SO/IEC 9889-1999. So I guess u were right after all. Thank u.
well dont know much about C++ standards, but this is what was written in page no 193, The C programming Language(second edition) by K & R:
So i think in C the type of integral constants are very much defined. U might want to go here http://www.lysator.liu.se/c/schildt.html
and see section 6.2.1.4. Here it says:
"Actually, unlike integers, such conversions are undefined...". It does give a hint that integer conversions are defined. I m really confused.
Could it be that it is one of those places where C differs from C++? Waiting to be enlightened.
I think you may be onto something. Here is a little program I've used to let the computer tell me how big an integer, long or unsigned long can be. (This is the int version. A variant can be used to do a similar thing for float and double also.)
``````main()
{ int x=2 , y;
y = x+1;
while (y > x)
{ x = 2*x+1;
y = x+1;
}
printf ("x = %d y = %d\n",x,y)
}``````
Now the intersting part: Compiled with the turbo C compiler this gives x = 32767, y = -32768, which is exactly as expected. Using bcc32, whether you compile this as a c program or as a cpp program you get x = 2147483647. So this compiler upgrades x and y from int to long when you'd get overflow otherwise. It could have upgraded further to unsigned long, but for reasons unknown to me it didn't. So apparently it's not exactly a difference between C and C++ but between C and C++ compilers. It would be informative to run this code on other C++ compilers.
I think you may be onto something. Here is a little program I've used to let the computer tell me how big an integer, long or unsigned long can be. (This is the int version. A variant can be used to do a similar thing for float and double also.)
Relying on undefined behavior as hardly good practice. These values are already available to you at compile time in the standard headers limits.h and float.h.
Relying on undefined behavior as hardly good practice. These values are already available to you at compile time in the standard headers limits.h and float.h.
Whether this is good pracrice or not is not the question I asked. The mystery is why the two compilers (from one company, no less) give different results. What is this telling us?
Whether this is good pracrice or not is not the question I asked. The mystery is why the two compilers (from one company, no less) give different results. What is this telling us?
Absolutely nothing -- since it is undefined behavior.
>What is this telling us?
It's telling us that you're struggling with simple concepts like "anything could happen".
As i said in my last post, there's a difference between C89 and C99 standards. Compilers that were made according to C89 standards will convert integers from int to long and then to unsigned long, whereas compilers that conform with C99 standards will go from int to long and then to long long, but not to unsigned long. Compilers will act differently based on the standards they were designed after. And this behaviour is not necessarily "undefined", it is rather "implementation-defined". There's a subtle difference between these two terms.
>there's a difference between C89 and C99 standards
Yes, there is. But you're missing the point.
>There's a subtle difference between these two terms.
There's also a subtle difference between the rules for integer literal conversions (that we've focused on for most of this thread) and signed integer overflow (displayed in murschech's horrid code). Signed integer overflow is always undefined, pick whatever C or C++ standard you want.
Narue thinks my code is "horrid". Gosh, I thought it was rather neat. I still do.
But maybe this bit of code will be more acceptable.
``````#include <stdio.h>
main()
{ printf("%d\n",sizeof(int));
}``````
If your not in the mood to run this code, I'll tell you what bcc32 gives as output: 4 . No wonder this compiler finds that the biggest integer is 2 billion +. Turbo C gives the result 2.
>Gosh, I thought it was rather neat.
Neat is not the same as good.
>But maybe this bit of code will be more acceptable.
Not really:
>main()
main returns an int. For C89, it's okay (but poor style) to omit the return value, for C99 it's a syntax error. The correct definition of main is:
``int main(void)``
>printf("%d\n",sizeof(int));
sizeof evaluates to a size_t value, which is not representable as a signed integer (which %d expects). Under C89 you can get around this by casting the result of sizeof to unsigned long, and using %lu:
``printf("%lu\n", (unsigned long int)sizeof(int));``
In C99, the z modifier lets you print a size_t:
``printf("%zd\n", sizeof(int));``
>}
If you were using C99 then this would be legal because main returns 0 by default. However, because you're not using C99 (see above) it's undefined behavior because you neglected to return a value.
>>If your not in the mood to run this code, I'll tell you what bcc32 gives as output: 4 . No wonder this compiler finds that the biggest integer is 2 billion +. Turbo C gives the result 2.
Turbo C runs in DOS which is a 16-bit OS. Hence it shows 2 bytes. On the otherhand, bcc5 will show 4 bytes bcos it runs in windows(or its a win32 console application). Typically short has 2 bytes and long has 4 bytes. The size of int will b either 2 or 4 depending on the system. Thats what i know.
>Gosh, I thought it was rather neat.
Neat is not the same as good.
>But maybe this bit of code will be more acceptable.
Not really:
>main()
main returns an int. For C89, it's okay (but poor style) to omit the return value, for C99 it's a syntax error. The correct definition of main is:
``int main(void)``
>printf("%d\n",sizeof(int));
sizeof evaluates to a size_t value, which is not representable as a signed integer (which %d expects). Under C89 you can get around this by casting the result of sizeof to unsigned long, and using %lu:
``printf("%lu\n", (unsigned long int)sizeof(int));``
In C99, the z modifier lets you print a size_t:
``printf("%zd\n", sizeof(int));``
>}
If you were using C99 then this would be legal because main returns 0 by default. However, because you're not using C99 (see above) it's undefined behavior because you neglected to return a value.
C99, C89, or C whatever are suggestions for compiler writers and they don't tell you what any compiler does. For instance, C99, you tell me, requires a return value for main and failure to return a value is a syntax error. But my compiler (bcc32) doesn't tag it as a syntax error, so why should I care that C99 does? This certainly doesn't make the computation incorrect. It's the compiler that I have to satisfy, not C99.
But let's get to the main point, namely, why my code, which I'll name "horrid" in your honor, works, even when there's undefined behavior. I'll enter the code again.
``````// HORRID
#include<stdio.h>
main()
{ int x=1, y=2;
while(y > x)
{ x = 2*x +1;
y = x + 1;
}
printf("Max integer: %d\n",x);
}``````
Now let's analyze it. Suppose that int is 1 byte. (This just makes the exposition easier. It clearly can be extended to any number of bytes).
Initially x = 1, which I'll write in binary as 00000001 (binary). Then
2*x + 1= 00000010 +00000001 =00000011 (binary). Then y will be well defined and > x, so the loop will continiue. The next value of x will be 00000111 (binary) and again y will be > x. Finally we get to x = 01111111, which is the LARGEST VALUE A SIGNED INTEGER CAN HAVE. Now when y is set = to x+1 it will have a value of who knows what because of undefined behavior, but it will still be an integer because y is declared to be an integer variable.Since x is at this point the largest possible integer the condition y>x will fail REGARDLESS OF WHAT VALUE IS ASSIGNED TO y, so the printf statement will execute. I hope this settles the matter.
You mentioned that the ranges of the int, long, etc, can be found in the header files. Do you know which header file?
>C99, C89, or C whatever are suggestions for compiler writers and they don't tell you what any compiler does.
They tell you what any compiler that claims to implement C must do.
>But my compiler (bcc32) doesn't tag it as a syntax error, so why should I care that C99 does?
Well, since that compiler doesn't implement C99, you should care because it's undefined behavior.
>This certainly doesn't make the computation incorrect. It's the compiler that I have to satisfy, not C99.
So you plan on using the same compiler for the rest of your life? What a narrow perspective you have. I pity you.
>works, even when there's undefined behavior
It could work, or it could wipe your hard drive. I personally couldn't care less what happens to your stupid ass, but I do care about you touting that code as "good" on this forum.
>Do you know which header file?
limits.h
><snip awful code and incorrect analysis> I hope this settles the matter.
Yes, it proves conclusively that you're an arrogant, ignorant retard who has no hope of becoming anything more than a mediocre script kiddie.
Reality check: Not everyone uses your compiler, your operating system, and your exact hardware configuration. When you realize this, you may actually have a chance of growing a brain.
So where is the error in my analysis?
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We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge. | 5,054 | 20,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.929283 |
https://blogs.cofc.edu/thearchaeoinformant/2012/07/03/avkat-3d-update/ | 1,708,998,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474669.36/warc/CC-MAIN-20240226225941-20240227015941-00280.warc.gz | 134,516,566 | 14,798 | # Avkat 3D Update
We are slowly building our facility with SketchuUp, and time for an update. Our Junior Intern has been largely involved with this component, and this post is written by me, but with contributions from Chris.
Working with Chris is a lesson in learning via play. Half the time it appears aimless – the pattern and process appear in retrospect.
We began by focusing on building Chris’ ability to draw to pre-ordained specifications – in the first exercise, a building of a given length, width, and height with a specific wall width.
After this point, I let Chris explore the program for himself. He began to work on making different shapes and at different scales, experimenting with different textures and wraps, and figuring out the interplay between shapes and color elements.
At one point he found the copy and paste function, after which entire cities began to crop up.
Entertaining the fascination with minecraft provided the opportunity to work on more complex elements requiring some abstract thought on how to visualize and operationalize.
As this was occurring, I started the tussle between ArcGIS 10.0 and Sketchup 8Pro. According to ESRI, the process is simplified in Arc 10.0. Hmmm…if it has been simplified, I’m glad I wasn’t doing this in darker times. The process is outlined on ESRI’s resource blog. Basically, working in ArcGlobe or ArcScene, one can move from a basic shapefile to exporting a file in Collada format, which can be read into Sketchup for further refinement. The data is then pushed back into Arc (or, I’m assuming, Google Earth as .kmz).
In 2007 and 2008, the Avkat Archaeological Project, under the supervision of Meg Watters and a team from the University of Birmingham VISTA Lab, conducted a series of remote sensing investigations in and around the village of Beyözü. Ground penetrating radar and subsequent preliminary analysis by Meg presented us with a rectalinear structure on the hill overlooking the modern village. This provided us with a good piece of data from which to work.
After sorting out the processed GPR data and dissolving the vector data by interpretative type, the file was converted to Collada format and imported into SketchUp.
At this point, the imported polygons were extruded to a height of 2 meters.
We then began to fill in the gaps, using a different color for the walls that we’re adding in ourselves.
From here, we will move to a roofing system, wraps for the walls, and other extrapolations based upon comparisons with other known structures found in similar contexts.
This entry was posted in 3D visualization, Avkat, Projects and tagged , , , , , , , , , , . Bookmark the permalink. | 592 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.929045 |
https://documen.tv/help-asap-please-determine-which-answer-in-the-solution-set-will-make-the-equation-true-28778972-90/ | 1,695,547,600,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00852.warc.gz | 248,442,907 | 16,166 | Question
HELP ASAP PLEASE!!!!!!!!!!!!!!!!!!
Determine which answer in the solution set will make the equation true.
3p + 6 = 5p − 12
S: {−9, 0, 9, 18}
A. 18
B. 9
C. 0
D. −9
BTW this is FLVS | 78 | 198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.792968 |
http://www.reddit.com/user/bkleinh | 1,412,155,917,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663372.35/warc/CC-MAIN-20140930004103-00135-ip-10-234-18-248.ec2.internal.warc.gz | 863,343,654 | 16,844 | # reddit's stories are created by its users
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The plane determined by the two points and the center of the sphere intersects the sphere in the required great circle. Also, there are maps produced by the central, or gnomonic projection, where great circles are mapped to straight lines. On such a map, connecting two points by a straight line gives the great circle route.
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n=1001a+110b
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I was taught this trick in High School 55 years ago. We were first taught to use simultaneous equations, which was the real point of the exercise. But then, to make sure we scored well on the state tests, we're were taught this much easier difference rates approach.
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Euclid has an elegant proof, where he divides a triangular prism into three triangular pyramids of equal area.
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Yes! This is the simple approach. Imagine a plot of the parabola. The requirement is that the parabola lie below the x axis for all x. That simply means that there are no real roots, and you have found the range of k for which that is true.
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The easy way do solve this is to change to a coordinate system moving with the center of mass. In this case, the center of mass is moving at -5 m/s, and the second object is moving -17m/s with respect to this. After the collision, the second object will be moving +17m/s with respect to the center of mass, or +12m/s in the original coordinate system. In the center of mass system, the objects reverse directions, but the magnitude of their velocities remains the same.
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Note that (x-p)2 + (y-q)2 >= 0, and equals zero only when x=p, y=q. Now expand, setting p2 + q2 = a2 , and px + qy = a2 , resulting in x2 + y2 >= a2 , with equality only when x=p, y=q. Thus the line is outside the circle, touching it at a single point.
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ez is a counter example. It has a simple Taylor series which converges everywhere, but has no zeros, real or complex.
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Bramagupta's formula is only applicable to convex quadrilaterals which can be inscribed in a circle. Bretschneider's formula is applicable to more general convex quadrilaterals.
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Typically, three equations in two unknowns are overdetermined, and will have no solutions, unless one of the equations is a linear combination of the other two. In your first example, 3 times the first equation + twice the second equals the third, so the third is redundant. The first two equations are independent, so they will be a unique solution, which automatically solves the third. In your second example, all three equations are proportional to each other, so the multiple solutions of one will also solve the others.
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You are expected to find a relation between the a's. Some of the rows must be dependent, or there would be no such constraint. For example, replace row4 with row4 + row3 - row1, resulting in 0 = a4 + a3 - a1. Now look at the remaining 3 rows. Can you find any other dependencies?
An interesting Geometry problem by in cheatatmathhomework
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Use the fact that for a triangle ABC, the distance from A to the point of tangency to the incircle is s-a, where a is the length of the side opposite A, and s is the semiperimeter.
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The problem does not specifically require the sides of the rectangle be parallel to the axis. Relaxing this requirement surely does not change the answer, but is there is an easy way to prove that?
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Much simpler: t3 -10t2 +30t = v, where v=xyz. The cubic on the right has a relative maximum and a relative minimum. There will be three real solutions only if v lies between these extrema. If v equals one of these extreme, two of the solutions are equal. So, set the derivative of the cubic to zero: x=y=(10+-sqrt(10))/3, z=(10-+2sqrt(10))/3. Multiplying the dimensions: v=(700-+20sqrt(10))/27. Again, to get back to the original problem, these dimensions must be multiplied by 10, and the volumes by 1000.
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I am sorry. You are right. The constraints are consistent. So, to make this more tractable, divide x, y and z by 10. Now x+y+z=10, and xy+yz+zx=30. Then x, y and z are the solutions to the cubic t3 -10t2 +30t -v. The discriminant of this is -18000 +1400v -27v2 . The zeros of this occur at v=23.6 and 28.3. Since the dimensions have been divided by 10, the actual minimum and maximum volumes are 23,600 and 28,300, unless I have made another arithmetic error.
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~~Wait, the constraints are inconsistent! There is no solution. First, remember there are 12 edges, so the linear constraint is x+y+z=100. The area constraint is xy+yz+ax=3000. Now by Maclaurin's inequalities, (x+y+z)/3 >= sqrt((xy+yz+zx)/3), which is badly violated by the given values. So, is no solution a valid response to this problem? ~~EDIT: Arithmetic error. The constraints are consistent!
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Given the volume, we have the three elementary symmetric polynomials in x, y and z. x, y and z are thus the three roots of the cubic t3 -1002 + 3000t -V = 0. These three roots will only be real if the discriminant is positive. The discriminant is quadratic in V, so the maximum and minimum volume are the two roots of this quadratic. There may be a way to get x, y and z directly knowing the discriminant is zero, but I don't know. (corrected)
Cal 2 help by in cheatatmathhomework
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Actually simple. The fact that the slope is negative for y>36, and positive for y<36, except at y=0, guarantees that y=36 is the absolute maximum.
Cal 2 help by in cheatatmathhomework
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OK, you have a relative maximum at y=36, and a relative minimum at y=0. How do you show that the value at y=36 is the absolute maximum that the problem requires? What tools do you have which address this?
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Although not stated, clearly the volume must be a given fixed constraint. You must find the three dimensions in terms of this volume. A simpler approach is to note that the product of the three terms in the cost equation is proportional to the square of the volume. The arithmetic mean geometric mean inequality then immediately implies that the cost equation has a unique absolute minimum when three terms in the cost equation are all equal.
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Use the arithmetic mean/geometric mean inequality to show that the volume has a unique absolute maximum when the three terms in the equation of the ellipsoid are equal.
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You need to use the theorem that the angle between two chords equals half the sum of the two arcs cut off by these chords. In this case, the angle E equals half the sum of arcs AC and DB. Because the triangle is isosceles, the sum of these arcs equals the arc CD. The required angle equality follows directly from this.
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The constant term 7 does not change the location of local extrema, so change it to zero. Now you can cancel out the common factor of 8, and you have a simpler equation to work with.
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The arithmetic mean/geometric mean inequality immediately implies that the volume achieves a unique maximum when L=W=H.
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Algebra mistake solving first problem. Should be x2 -2x+1=(x-1)2 <=0 | 2,341 | 9,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | latest | en | 0.96554 |
https://lillemetropolerugby.com/2734/how-do-you-show-the-significance-of-a-line-graph/ | 1,652,985,511,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00122.warc.gz | 432,741,547 | 12,368 | # How do you show the significance of a line graph?
To interpret a graph or chart, read the title, look at the key, read the labels. Then study the graph to understand what it shows. Read the title of the graph or chart. The title tells what information is being displayed.
To interpret a graph or chart, read the title, look at the key, read the labels. Then study the graph to understand what it shows. Read the title of the graph or chart. The title tells what information is being displayed.
Beside above, what is the definition of a line graph? A line chart or line plot or line graph or curve chart is a type of chart which displays information as a series of data points called ‘markers’ connected by straight line segments. It is a basic type of chart common in many fields. In these cases they are known as run charts.
Also to know, what is the important of graph?
Graphs are a common method to visually illustrate relationships in the data. The purpose of a graph is to present data that are too numerous or complicated to be described adequately in the text and in less space. It is important to provide a clear and descriptive legend for each graph.
How do you know if a graph is statistically significant?
If your p-value is less than or equal to the set significance level, the data is considered statistically significant. As a general rule, the significance level (or alpha) is commonly set to 0.05, meaning that the probability of observing the differences seen in your data by chance is just 5%.
### How do you indicate significance in a letter?
If we use upper-case letters to indicate results significant at the 0.05 level and lower-case to indicate results significant at the 0.001 level we get: a>b, A>D, a>f, a>g, c>d and c>f. (Often commercial studies use upper-case for significant at the 0.05 level and lower case for significant at the 0.10 level.)
### How much is statistically significant?
A data set is statistically significant when the set is large enough to accurately represent the phenomenon or population sample being studied. A data set is typically deemed to be statistically significant if the probability of the phenomenon being random is less than 1/20, resulting in a p-value of 5%.
### What is p value in statistics?
In statistics, the p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. The p-value is used as an alternative to rejection points to provide the smallest level of significance at which the null hypothesis would be rejected.
### How do you find the p value on a graph?
How to Graph the P Value for a 1-sample t-Test Make sure the graph we created is selected. Choose Editor > Duplicate Graph. Double click the blue distribution curve on the graph. Click the Shaded Area tab in the dialog box that appears. In Define Shaded Area By, select X Value and Both Tails. In X value, enter 2.29.
### What do asterisks mean on a graph?
Asterisks. Once you have set a threshold significance level (usually 0.05), every result leads to a conclusion of either “statistically significant” or not “statistically significant”.
### How do you describe a trend?
The following verbs can be used to describe a trend or pattern that goes up. climb (past: climbed) go up (past: went up) grow (past: grew) increase (past: increased) jump (past: jumped) rise (past: rose) rocket (past: rocketed)
### How do you describe the trend of a line graph?
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Dirk Gregorius 2757
Given two bodies b1 and b2. In an impulse based simulator you integrate the bodies forward in time, find the constraint error and correct it. Mathematically I believe this are differential inclusions and you project the phase state of the bodies back onto the constraint manifold. This sounds difficult, but is extreme easy to implement. For the ball-socket joint you simple do something like this: b1.Step( dt ); b2.Step( dt ); // Let r1 and r2 be the offset vector from the CM the the joint anchor point v32 v1 = b1.GetVelocityAt( r1 ); v32 v2 = b2.GetVelocityAt( r2 ); // The relative velocity must be zero - so this is the constraints error v32 v_rel = v1 - v2; That's not too difficult. Actually this is the Jacoby entry, but you don't have to deal with large matrices and do all the bookkeeping. Now that we have found the constraint violation, we need to correct it - that is project back onto the constraint manifold. We don't want to simply divide the error between both bodies, but also consider the mass and moment of inertia somehow. We have to solve the following question: "Find corrections for the current linear velocities v1 and v2 plus corrections for the current angular velocities w1 and w1, such that the relative velocity at the joint anchor becomes zero. Preferable considering the masses and moments of inertia." In the literature (s. Papers) it is proposed to calculate the collision impulse and chose the direction of the relative velocity as collision normal and its length as relative normal velocity. E.g. in Jiggle the constraints are implemented this way. f32 j = -(1 + e)*vn / [ (1/m1) + (1/m2) + n*(J1^-1*(r1 x n) x r1) + n*(J2^-1*(r2 x n) x r2) Now the question: I assumed that after applying the impulse the relative velocity would be zero. This is sometimes correct, but in the majority of cases it seems as if the error is only reduced a bit. Isn't this bad for the convergence of the system? For two limits I don't have convergence anymore. Are there other methods to divide the error between the two bodies? I setup the impulse equations and avoided the calculation of the collison impulse - since this would be more numerically stable. But I think I have an error in the matrix, so I can't say how this works at the moment. Any suggestions? Regards, -Dirk
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MrRowl 2490
In an isolated system with a single joint I'm pretty sure that after applying the impulse my implementation gave zero velative velocity at the joint location.
Of course if the objects are part of other constraints (e.g. other joints, or resting on a surface under gravity etc) then applying impulses sequentially you only relax to the proper solution, but it should converge to a reasonable tolerance with a fairly small number of iterations, and any offset error (for normal conditions) can be accounted for by an error correction term. Without the error correction term you get some drift.
If I were you I'd check a single joint in isolation (no gravity) - if you don't almost get perfect behaviour there must be a problem in your implementation.
Of course you'll still get some drift when the objects can rotate, since even if the velocities at the joint position match perfectly after applying the impulse, when integrating the angular velocity over the timestep this will (generally) result in the joint position separating. Even if you solved all joints/constraints simultaneously you'd still get this source of drift.
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Dirk Gregorius 2757
You process all body pairs subsequently and apply 20 iterations. I believed that after you apply an impulse to the bodies, the constraint would be totally satisfied (at least for the moment). Of course subsequent constraints can again violate previous ones. This is why we do several iterations (relaxation).
Example:
Given two boxes b1 and b2.
Width1 = Width2 = 3
Height1 = Height2 = 2
Depth1 = Depth2 = 2
Mass1 = Mass = 12kg
From there I derived the moment of inertia:
Ixx = 8
Iyy = 13
Izz = 13
The initial orientations are identtity quaternion and the positions are
x1 = (0, 10, 0)
x2 = (0, 6, 0)
The ball joint is located exactly between the two bodies:
xj = (0, 8, 0)
Then the offset vectors from the centers of mass to the joint anchor are:
r1 = ( 0, -2, 0)
r2 = ( 0, 2, 0)
As initial phase state I chose:
v1( 0.0f, -13.0f, 0.0f )
v2( -2.0f, -3.0f, -2.0f )
w1( 0.0f, 0.0f, 0.0f )
w2( -1.0f, -1.0f, 2.0f )
The velocity constraint equation is:
v1 + w1 x r1 - ( v2 + w2 x r2 ) = 0
This is the relative velocity at the joint anchor. If I compute this from hand and in my system I get:
v_rel = (6, -10, 4)
Now I chose the direction of v_rel as collision normal and its length as the normal velocity:
v32 n = v_rel / |v_rel| = (0.48666424, -0.8111071, 0.32444283)
f32 vn = |v_rel| = 12.328828
Since we have identiy orientations the inverse global moment of inertia is:
Ixx^-1 = 1 / 8
Iyy^-1 = 1 / 13
Izz^-1 = 1 / 13
Finaly we apply a plastic (inealstic) collision, so we chose e = 0
Now we can enter the impulse formula:
f32 j = -(1 + e)*vn / [ (1/m1) + (1/m2) + n*(J1^-1*(r1 x n) x r1) + n*(J2^-1*(r2 x n) x r2)
I get j = -29.517485
Now I apply the ( correction ) impuls.
v1' = v1 + j * n / m1 = ( -1.1970921, -11.004847, -0.79806137 )
w1' = w1 + I1^-1 * ( r1 x j * n ) = ( 2.3941841, 0.00000000, -2.2100160 )
v2' = v2 - j * n / m2 = ( -0.80290794, -4.9951534, -1.2019386 )
w2' = w2 - I2^-1 * ( r2 x j * n ) = ( 1.3941841, -1.0000000, -0.21001601 )
If I recalculate the relative velocity here I get:
V_rel' = ( 5.2342482, -6.0096931, -7.1728592 )
This is the isolated example. The problem is that this works! The ball joint, the hinge with limits, the slider with limits and bounce, the universal, all work. But when I add a second axis limit to the universal joint there is no more convergence. If I run a constraint verification after each applied impulse the constraint error is nearly never repaired. It seems as if the collison impulses only repair a bit of the error, which is way too slow in my opinion.
BTW: My fist guess was that some sign was wrong, but I tried all four permutations for +/-vn and +/-n. None worked. Any help is greatly appreciated.
-Dirk
[Edited by - DonDickieD on October 22, 2005 4:07:30 PM]
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MrRowl 2490
OK - I'll try to find time to set up the same experiment in my code, but am not sure when I'll get a chance to do this....
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Dirk Gregorius 2757
Great! Thanks for taking the time...
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ury 476
Dirk, there's absolutely no reason why your approach should work at all.
Although the impulse that you generate at the end makes sure that your new relative velocity is orthogonal to your normal, it's hardly zero.
Even if that wasn't enough, you have to compensate for any distance drift that may occur between the anchors of the joint. Because of that, making sure that the relative velocity is zero is simply not enough.
First some terminology:
x - position of the center of mass
v - velocity of the center of mass
R - rotation matrix
w - angular velocity
IL - inertia matrix in body's local coordinates
I - inertia matrix in world coordinates.
M - mass matrix of a body (a diagonal matrix)
[p] - cross matrix. [p]u is equivalent to p x u.
I = R * IL * Rt
Let's set up a ball socket joint between two bodies, A and B, at some point p.
First we record the anchor position in body's local coordinates:
ra = Rat(p-xa)
rb = Rbt(p-xb)
These vectors should stay constant for the entire simulation.
For every iteration, we define for each body:
r = R * r,
K = M-1 - [r] * I-1 * [r]
Now we can define the distance between anchors in world space as:
dist = xa + ra - (xb + rb)
We want to make sure that dist stays zero all the time.
Velocity of an anchor point in world space is:
u = v + w x r
And so, the relative velocity of the anchor points is:
ur = ua - ub.
Let's say the we have some impulse J. If we apply J on body A and -J on body B, the new relative velocity of the anchor points is given by:
Kt = Ka + Kb
ur = ur + KtJ
If we wanted our new relative velocity to be zero, then we could find the appropriate J by solving:
-ur = KtJ
Since we want to compensate for any distance drift, we can try and solve:
alpha = -e/(2*dt)
alpha * dist - ur = KtJ
where:
dt - the size of your time step.
e - some empirical coefficient. (A good value to start with is 1.0)
Good luck, tell me if it works.
PS. If I were you, I'd really consider using lagrange multipliers :)
[Edited by - ury on October 22, 2005 1:05:31 PM]
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MrRowl 2490
ury - yes of course you're right, it's pretty obvious now you've pointed it out :) Rats!
DDD - I did set up the same experiment, but I actually get slightly different numbers to you - perhaps because your moment of inertia calculation is wrong. If you have a block that has dimensions (large, small, small) the moments of inertia around the axes should be (small, large, large) - you have two smalls and one large(!), which is wrong whatever your choice of axes.
So, before the impulse the joint velocity error magnitude is 12.3 (same as you get).
My impulse is 32.69, and afterwards the joint velocity error is 9.5 and pretty well perfectly orthogonal to the impulse direction.
Guess I need to work on this a bit :)
Big apologies to anyone that looked at my code and assumed it was good!!
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Dirk Gregorius 2757
Ury:
Thanks for the suggestion. I will try it out this afternoon and post the result here. I haven't looked to deep into the solution, but I am quite sure that when the corrected velocities are calculated we have to consider the conservation of momentum. So I came up with this solution. I scanned my calculations and uploaded it here:
www.dirkgregorius.de/Kugelgelenk.jpg
MrRowl:
Thanks for taking the time to setup the experiment. Of course you are correct with the moment of inertia. This was only a copy error. In my test case this is correct, so strange that we don't get the same impulse.
Please don't excuse. There is no reason for doing this!
Note: I edited my previous post and corrected the inertia
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Dirk Gregorius 2757
Ury:
yes your solution works! Thanks for taking the time and post the solution. For the protocol - my solution (s. previous thread) works as well, though yours is more elegant and more efficient.
Regarding drift:
Currently I am handling the drift through projection - that is I simply translate and rotate the bodies into a valid position. I tried your method as well. Actually I am flipping back and forth between both of them. BTW: Is this the same as Baumgarte Stabilization, but in a slightly different context? What I would like to have is your solution and a post-stabilization like proposed by Baraff.
Regarding Lagrange Multipliers:
Why do you think that a constraint-based simulator is superior to an impulse based simulator? The constraint based systems handles all these scientific examples quite well, but for games I am not so sure. And what about scaleability on the next generation hardware. I started implementing a velocity constraint formulation, but I find the impulse based method much more preferable.
Here are some examples of my simulator. Please feel free to have a look. Simply drag and drop the *.phx files onto the framework.exe. Note that the system uses all over the wrong correction impulses, since I haven't changed it today. Under these circumstances the system behaves already quite well. I haven't implemented any auto-disabling nor shock-stepping, so the stacking doesn't behave so good right now. Since I don't have stacked objects at the moment this has lower priority. Constraints with limits and bounds, a car controller with suspension and ragdolls need be solved first.
LMB: Rotate
MW: Zoom
RMB: Apply force to body
www.dirkgregorius.de/Primo.rar
[Edited by - DonDickieD on October 23, 2005 11:42:29 AM]
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ury 476
Quote:
Original post by DonDickieDyes your solution works! Thanks for taking the time and post the solution. For the protocol - my solution (s. previous thread) works as well, though yours is more elegant and more efficient.
Glad to hear and thank you. :)
Quote:
BTW: Is this the same as Baumgarte Stabilization, but in a slightly different context?
Hardly... Although the formulae might look the same, the idea behind them is different.
My approach is a simple geometric observation. If we have some distance between our anchor points, defined by the vector 'dist', give the bodies enough velocity at the anchor points, so that they would each travel dist/2 towards each other during the next frame and meet (with gods help :)) at the same point.
Baumgarte enforced a constraint by introducing an "augmented" ODE for the constraint. Baumgarte stabilization behaves very much like a restrained harmonic oscillator and can be viewed as a spring connected between the two anchors. Tell me if you want to hear more details about it.
As you can see the ideas are extremely different but I guess that in the end, we get pretty much the same result. :)
Quote:
What I would like to have is your solution and a post-stabilization like proposed by Baraff.
What sort of post-stabilization are you talking about?
Quote:
Why do you think that a constraint-based simulator is superior to an impulse based simulator? The constraint based systems handles all these scientific examples quite well, but for games I am not so sure.
What do you call scientific examples? If you mean overly complex scenes with lots of interactions and constraints then you are looking at the future of gaming physics.
If you use velocity constraint formulation with lagrange multipliers (LM), both techniques find correcting impulses. It can even be argued that when solving a LM problem, an iterative solver is performing roughly the same operations as you would do in an impulse based (IB) simulator (although, the convergence of an iterative solver is usually better). I guess that if you look at it this way, both techniques are very close relatives.
But!
The solution for a LM problem is "optimal" because it is found considering all of your constraints at once (including joints, contacts, friction, etc)
You can expect to have less drift in your joints. This allows you to use more complex articulated bodies.
The technique can be generalized to support other types of bodies (such as deformable bodies).
I personally find LM more elegant and modular.
A downfall of LM is that it's linear but this problem can be remedied since you can always fallback to performing your corrections using "handmade" impulses just like in IB.
Quote:
And what about scaleability on the next generation hardware. I read that Novodex had/has quite some problems on the PS3 and I *assume* that they use either a velocity constraint formulation (Stewart/Trinkle and Anitescu) or an accelaration constraint formulation (Baraff/Wittkin).
Problems? That's strange. PS3 is supposed to have 7 dedicates processors just to perform SIMD floating-point operations. I think that you can simulate an atomic explosion in real-time with that monster. :)
As far as I can recall, Novodex really do use a velocity based LM system and solve it using some sort of an iterative solver.
Quote:
I started implementing a velocity constraint formulation, but I find the impulse based method much more preferable.
Why? What happened? :)
Quote:
Here are some examples of my simulator.
Good work, man! Keep us in loop as you improve your simulator.
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Dirk Gregorius 2757
Hey Ury,
before I answer your questions, here is a new idea how to solve the problem. I thought about the problem and came up with the following idea. Here we go:
The velocity update for the two bodies is:
M * du /dt = f_ext + JT * lambda
Here u is the generalized velocity vector for body1 and body2, JT is the transposed jacobian and lambda is a vector of langrange multipliers.
Since we use linear dynamics we can use superposition and apply the external and constraint forces subsequently. So the velocity update because of the the constraint forces becomes:
M * du /dt = JT * lambda
The compatibility equation (velocity constraint) for the two bodies is:
J * u(t) = 0
Numerical integration of the Newton-Euler equation:
M * ( u(t+dt) - u(t) ) / dt = JT * lambda
<=> u(t+dt) = u(t) + M^-1*JT*lambda*dt
Semi-implicit integration, therefore
J * u(t+dt) = 0
Plug integrated Euler-Newton into compatibility equation:
J * [ u(t) + M^-1*JT*lambda*dt ] = 0
<=> J*M^-1*JT * lambda * dt + J*v(t) = 0
This is a linear system: A*x + b = 0
If we look now sharply we see that J*v(t) is the constraint error and lambda * dt is the wanted impulse. We define
A := J*M^-1*JT
x := lambda * dt
b := J*v(t)
=> lambda = A^-1 * constraint error (J*v(t))
So the wanted linear and angular impulses for body1 and body2 are simply:
P = JT * lambda
What do you think. Actually here we have your Langrange multipliers :-)
Edit: I just tried this formulation and it nicely blends over into your equation for the ball socket. Your K matrix is actually J * M^-1 * JT and and J*v(t) is the constraint error. I think I will give it a try. Hopefully it works for the the other joints as well...
[Edited by - DonDickieD on October 23, 2005 9:51:17 AM]
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ury 476
Quote:
Edit: I just tried this formulation and it nicely blends over into your equation for the ball socket. Your K matrix is actually J * M^-1 * JT and and J*v(t) is the constraint error. I think I will give it a try. Hopefully it works for the the other joints as well...
That's right. :)
Compared to my version: ur = J*v(t).
Quote:
So the wanted linear and angular impulses for body1 and body2 are simply:P = JT * lambda
Careful! Using your formulation, the impulse is given by:
P = JT * lambda * dt
Anyways, the problem you are solving is actually static. There's no need for dt at all. Your original formula:
Quote:
M * ( u(t+dt) - u(t) ) / dt = JT * lambda
Can be written:
M * ( u(t+dt) - u(t) ) = JT * gamma
where:
gamma = lambda * dt.
Now, after you find gamma, you get:
P = JT * gamma
See? No dynamics at all! Looks more elegant, right? Thats the problem. :)
Neither the system forces nor the time-step are considered.
If you follow this scheme, there are two steps that you have to undertake:
1. Integrate your velocities in time using the system forces.
2. Apply the constraints as a post-correction to your new velocity.
Now consider this:
( M U )( v1 ) = ( Fdt + Mv0 )
( L 0 )( lambda ) ( E )
where:
M - system mass matrix
L - J
U - -Jt
F - force
E - constraint error
Solving this, both steps are performed simultaneously!
Here we have "my" lagrange multipliers ;P
Oh.. in your formulation you don't handle distance drift. Don't forget it :)
A silly question...
f_ext includes coriolis forces as well, right?
[Edited by - ury on October 23, 2005 11:28:27 AM]
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Dirk Gregorius 2757
Very cool! I have never thought of lagrange multiliers in this context. I always thought of them when thinking of the velocity constrained approach like in the ODE.
Regarding f_ext:
Yes, it includes the corioles force. Actually the system behaves better without it. Is seems as if long and thin objects with a high angular velocity gain energy when I consider this term. Any suggestions?
Little confusion - in your formula lambda is actually an impulse, right? Just to be sure.
In this form you solve a linear system. How would you solve limits in this context? I can think of two possible solutions:
1) Use an LCP and limit the lambda
2) Model the constraint error such that a limit is taken into account as well
Regarding contacts:
And now every physic programmers best friend - unilateral constraints :) I am currently implementing the Guendelman approach for modeling contacts. For integrating the constraints I worked with the sketch of R. Weinstein at the Siggraph this year. (If you are interessted I can send you the sketch plus some private correspondance where the idea is explained a little bit more).
1) Do see any problems to combine both apporaches?
2) How would you model contacts in a simulator where we deal with bilateral constraints as discussed in the previous posts?
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ury 476
Quote:
Yes, it includes the corioles force. Actually the system behaves better without it. Is seems as if long and thin objects with a high angular velocity gain energy when I consider this term. Any suggestions?
The system might be more stable but less realistic.
Try to remeber what happens if you spin a coin on a table. After loosing some of its speed, the coin falls on the table and starts "wobbling". This "wobbling" happens because of the coriolis force.
To make your simulation more stable you can:
1. Apply angular damping force (torque)
2. Clamp velocities.
3. Look for a bug :)
Check if smaller timesteps improve the simulation quality. If they don't, it must be a bug.
Quote:
Regarding your formula:Little confusion - in your formula lambda is actually an impulse, right? Just to be sure.
No. Let me rewrite the equations:
1) Mv1 = Fdt + Mv0 + Jtlambda
2) Jv1 = E
You can clearly see that the correcting impulse is:
P = Jtlambda.
Let's assume that the valid momentum may lie in some hyperplane and let the rows of J be the normals that form this hypersurface. When we find the solution, we actually project the current momentum to this hyperplane.
The projection is done by using the normals from J which are scaled with the values from Lambda.
Quote:
In this form you solve a linear system. How would you solve limits in this context? I can think of two possible solutions:1) Use an LCP and limit the lambda 2) Model the constraint error such that a limit is taken into account as well
Yes. LCP is used to formulate such constraints.
Quote:
1) Do see any problems to combine both apporaches?
Quote:
2) How would you model contacts in a simulator where we deal with bilateral constraints as discussed in the previous posts?
Usually, a MCP (Mixed Complementary Problem) is formulated.
Take a look at this thesis, it'll give you a general idea.
[Edited by - ury on October 24, 2005 4:31:23 AM]
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sbroumley 283
Here's a link to an excellent paper that describes an iterative method for solving lagrange multipliers:
http://www.gphysics.com/files/IterativeDynamics.pdf
It's the first paper on global solvers that I've been able to implement and from my experience, the results are more stable (no jiggle) and faster than the impulse methods I've previously tried.
-Steve.
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Dirk Gregorius 2757
Ury:
Regarding Corioles:
Cool - now I have physical interpretation of this term. Thanks!
Regarding Lagrange Multipliers:
I once wrote down a small paper regarding this topic. I wrote it before reading Erin's paper (thanks sbroumley). You might find it interessting.
www.dirkgregorius.de/ConstrainedDynamics.doc
Regarding the sketch:
http://graphics.stanford.edu/~fedkiw/papers/stanford2005-05.pdf
Regarding physical simulators:
When using the velocity constrained approach like described in the PhD thesis by Kenny Erleben or in Erin's paper (and actually implemented in the ODE) you typically have the following simulation loop:
DetectCollisions();
SolveLCP();
DoTimeStep();
Render();
Basically you add the contact constraints and external forces and send it to your solver and ask him:
"Please find a velocity update regarding the current external forces and geometric constraints such that non of the geometric constraints are violated"
"Please lightning fast and with the most reasonable solution in a degenarated case"
The problem with this approach in my opinion is with that you might render invalid configurations all over the place. E.g. when the AI applies a big external force this will lead to a big displacement after the integration, which might end up in deep penetrations. We have this all over the place, besides convergence problems, etc...
What I do now is solve the system in the future (s.Guendelman) and then integrate the phase state such that no constraints are violated in the next frame in order to avoid the described artifacts. To do this I advance the system as if there were no constraints and then apply a post correction as you correctly stated above. Basically this is the Jacobsen projection, but on the velocity level and regarding rigid bodies. While a lot of people find solving the big system matrix more elegant in my opinion this modular approach will give me much more possibilities to correct something in a reasonable way when running into a singularity. Actually this is the same relaxation approach as a Jacoby or Gauss-Seidel iterative solver applies - at least for my understanding. Though on the mathematically level the big matrix approach solves a DAE while I solve a differential inclusion. Another advantage of this approach is that I can now easily use high order integrators, while this is quite difficult in the velocity constrained formulation. Since the constraints are not considered in the integration, there is no need to find the derivatives. Any comments, critics and ideas are appreciated...
Regards,
-Dirk
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John Schultz 811
Quote:
Original post by DonDickieDWhat I do now is solve the system in the future (s.Guendelman) and then integrate the phase state such that no constraints are violated in the next frame in order to avoid the described artifacts. To do this I advance the system as if there were no constraints and then apply a post correction as you correctly stated above. Basically this is the Jacobsen projection, but on the velocity level and regarding rigid bodies. While a lot of people find solving the big system matrix more elegant in my opinion this modular approach will give me much more possibilities to correct something in a reasonable way when running into a singularity. Actually this is the same relaxation approach as a Jacoby or Gauss-Seidel iterative solver applies - at least for my understanding. Though on the mathematically level the big matrix approach solves a DAE while I solve a differential inclusion. Another advantage of this approach is that I can now easily use high order integrators, while this is quite difficult in the velocity constrained formulation. Since the constraints are not considered in the integration, there is no need to find the derivatives. Any comments, critics and ideas are appreciated...
Sounds kind of familiar. Keep going.
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ury 476
Quote:
Regarding Lagrange Multipliers:I once wrote down a small paper regarding this topic. I wrote it before reading Erin's paper (thanks sbroumley). You might find it interessting. www.dirkgregorius.de/ConstrainedDynamics.doc
I'll read it as soon as I get the chance. :)
I have mixed feelings about Erin's paper. The paper is very clear and straight forward. The notations are very simple and the paper can be a VERY useful reference for a newbie since it describes the entire simulator from head to toe.
Unfortunately, the collision and friction models proposed in this paper are very poor. I have to credit Erin for not hiding this fact and openly stating this weakness. The most disappointing part in this paper is the iterative solver. He identifies his solver as Gauss-Seidel method. In reality, his solver is a close relative of the Jacobi's method with a bug. :) Jacobi's method on it's own has a very poor convergence rate. I am sure that the bug doesn't improve that either.
EDIT: Actually, there's no bug in Erin's solver. See Erin's explanation here.
For a good implementation of an iterative solver (SOR) and friction, please read Kenny Erleben's thesis. An implementation of this solver can be found in QuickStep which Kenny wrote for the ODE engine. BTW, there's even a more efficient friction method than the one proposed by Kenny. This method is formulated using box constrained LCP.
Tell me if you want to hear more about any of these subjects.
Quote:
Regarding the sketch:http://graphics.stanford.edu/~fedkiw/papers/stanford2005-05.pdf
Seems like a nice idea. In fact, it's a very important thing to implement if you plan using Guendelman's approach.
Quote:
Regarding physical simulators:.The problem with this approach in my opinion is with that you might render invalid configurations all over the place..What I do now is solve the system in the future (s.Guendelman).
For me, both approaches are almost the same.
The difference between:
1. Find new velocity.
2. Apply post-correction to confirm with a set of constraints.
and:
1. Find new velocity which confirms with a set of constraints.
is that the second approach can be more accurate.
If no external forces are applied, solving LCP is equivalent to performing a post-correction of velocities. In this case, you consider a much larger set of constraints, hence, the solution is more accurate. (Our last two posts are a good example for that). In fact, nobody's stopping you from using both aproaches at the same time. For instance, you could solve collisions using an impulse based approach and handle joints and contacts using LCP.
BTW, in my current engine, I do the following:
1. Collision detection using a continuous sweep.
2. Perform a full step.
3. Collision detection using a continuous sweep.
4. If I don't like what I see:
4.1 Step back in time.
4.2 Perform a full step taking pairs from 1 and 3 under consideration.
Look how closely it resembles Guendelman's approach:
a. Collision detection and modeling.
c. Contact resolution.
If collisions occur, then steps 1,2 can be treated as a, b and steps 3,4 as c, d.
Quote:
Another advantage of this approach is that I can now easily use high order integrators, while this is quite difficult in the velocity constrained formulation. Since the constraints are not considered in the integration, there is no need to find the derivatives.
I don't follow you on this one :)
It doesn't matter which approach you choose, at the end of the day you have a black box that gives you a new velocity.
Just pick your favorite integrator and plug this velocity in. :)
In fact, you can even get even kinkier than that. In my engine, since I have a very stiff equation set, I am forced to use an implicit integration scheme. I do that, by finding a root for the equations of motion:
x = x0 + dt*v( x )
[Edited by - ury on October 26, 2005 5:55:03 AM]
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Dirk Gregorius 2757
Quote:
Tell me if you want to hear more about any of these subjects
Yes please! Especially, I like to know where the bug in the projected Gauss-Seidel is. The friction method is very interssting as well. Since you mention boxed LCP - I am still looking how to transform the classic LCP into a boxed LCP. Any reference?
Novodex uses the Gauss-Seidel as well (s. Lecture 6)
Technically you are absolutely right, there is very few difference. What might be an issue is that you need to call the collision engine twice (or even more), while I get away with calling it only ones. But I am not so sure here :-/
BTW: How do you generate contact points? Like in the ODE and OpenTissue or liked proposed in Guendelman using signed distance maps. If you have any good references they would be highly appreciated...
Regarding higher integrators:
The Newton-Euler equation in a constrained systems has the form:
M * du/dt = f_ext + f_c
When you want to use a higher integrator you need the forces at some sampling points, e.g. for the MidPoint Method at t+dt/2. I haven't looked into this problem so far, but I can't directly see how to compute f_c(t+dt/2)? Since I don't consider them in the first place, I have no need to compute them.
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ury. I'm sorry.. But you are a math punk. The Guendelman paper is the 3rd most important paper in physics simulation. Sure his collision stuff is craziness, but he was also handling mesh-mesh. That paper is important because it finally showed the academic snobs that PLAUSIBLE is the key.
Others have hinted at it in random papers. Even Baraff himself. But Guendelman came out and said.. "Damn it. It's just not that hard." If that paper was out before I had "figured things out" I would of saved myself a lot of time. The best thing about that paper is it makes physics APPROACHABLE. That in itself allows others to bring great inovation to the problem in the future. Unforunately most academics litter their papers with unecessary complexity just so that they can look smart or impress their famous advisors.
btw. Just because an algorithm has a faster convergence rate on paper does not make the algorithm take less wall clock time to execute on a computer.
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Dirk Gregorius 2757
Quote:
Unforunately most academics litter their papers with unecessary complexity just so that they can look smart or impress their famous advisors
I think I have read every paper and master or PhD thesis regarding physically-based animation. It is sometimes amazing how even the most simple problems are turned into really sophisticated ones. So I can't agree more :-)
Anyway, Ury has a very sorrow understanding of both mathematics and dynamics. And I am glad that he shares his konowledge and has the patience to answer all my silly questions. There is never no such thing like one way to go. I think there are some descent simulators out there that apply his technique, while there are others that use maybe another approach. This was a very good discussion so far and it would be sad if we end up in some kind of flame war here.
Regards,
-Dirk
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Guest Anonymous Poster
Quote:
Original post by billy_zelsnackury. I'm sorry.. But you are a math punk. The Guendelman paper is the 3rd most important paper in physics simulation. Sure his collision stuff is craziness, but he was also handling mesh-mesh. That paper is important because it finally showed the academic snobs that PLAUSIBLE is the key.
I cannot disagree more with this statement. It would not call the paper “the more important…”, I would not even classify it as a paper worth reading, the only thing the Guendelman demonstrates is that it is a good example of what not to do.
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sbroumley 283
Quote:
Original post by uryThe most disappointing part in this paper is the iterative solver. He identifies his solver as Gauss-Seidel method. In reality, his solver is a close relative of the Jacobi's method with a bug. :) Jacobi's method on it's own has a very poor convergence rate. I am sure that the bug doesn't improve that either.I guess that this is one of the reasons why he needed contact caching. :)
Ury, so I'm really curious now - what is the bug?!
thanks
-Steve.
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No flame war at all intended. I agree. This has been a very nice thread and one I sure others will reference in the future. Ury has presented some very good information. I just wanted to chime up on the stacking paper.
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Dirk Gregorius 2757
Quote:
I cannot disagree more with this statement. It would not call the paper “the more important…”, I would not even classify it as a paper worth reading, the only thing the Guendelman demonstrates is that it is a good example of what not to do
Could you give reasons for this? Without justification such a posting makes very few sense. I mean what shall one think after such a post. You could be either a very experienced physic programmer in which case it would be interessting to hear the justification for this statement or you are simply not able to implement the approach described in the paper...
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Register a new account | 8,509 | 35,901 | {"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} | 3.609375 | 4 | CC-MAIN-2017-39 | longest | en | 0.901704 |
https://www.analyzemath.com/Calculators_3/factorial_calculator.html | 1,716,613,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00668.warc.gz | 549,691,127 | 9,257 | # Factorial Calculator
An online calculator to calculate the factorial of a positive integer.
## Definition of the Factorial of a Positive Integer
If n is a positive integer, then the factorial of n written as $n!$ (read as "n factorial") is given by
$n ! = n \times (n-1) \times (n-2)....2 \times 1$
with $0! = 1$.
Example 1
$2! = 2 \times (2 - 1) = 2 \times 1 = 2$
$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800$
The factorial of positive integers increases very quickly.
Example 2
$40! = 8.159152832478977\times 10^{47}$
$100! = 9.33262154439441 \times 10^{157}$
Note: the powers of 10 are written with an "E", see example below.
## Factorial Calculator of a Positive Integer
1 - Enter as a positive integer n or 0 and press "Calculate n!".
Note for n = 22 or larger, the results is written in scientific notation using "E" instead of 10.
Example: $22! = 1.1240007277776077e+21$ is $1.1240007277776077× 10^{21}$
n = 10
## Applications of Factorial in Mathematics
Factorials are used in the formulas of permutations and combinations.
Example 3
Factorials are used in the calculation of combinations. The combination of n objects taken r objects at a time is written as $C(n,r)$ and is given in terms of factorials by the formula
$C(n,r) = \dfrac{n!}{(n - r)! r!}$
Example 4
Factorials are used in the calculation of permutations. The permutation of n objects taken r objects at a time, where order is important, is written as $P(n,r)$ and is given in terms of factorials by the formula
$P(n,r) = \dfrac{n!}{(n - r)}$
Example 5
Factorials are used in series of functions in calculus and in turn these series are used in electronic calculators to compute functions such sin(x), cos(x), ln(x), ex. We list here some examples of functions given by series.
a) $e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...$
b) $sin(x) = x - x^3/3! + x^5/5! + ...$
c) $cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...$
## More References and Links
Factorial Questions with Solutions. Tutorials on evaluating and simplifying expressions with factorial notation.
Permutations and Combinations Problems.
Math Calculators and Solvers. | 673 | 2,186 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.834069 |
https://www.gradesaver.com/textbooks/science/physics/college-physics-7th-edition/chapter-2-kinematics-description-of-motion-learning-path-questions-and-exercises-exercises-page-64/71 | 1,696,246,575,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00423.warc.gz | 843,807,794 | 12,247 | ## College Physics (7th Edition)
$a). 5s$ $b).$ Velocity just before hitting the ground $36.5m/s$
u=12.5m/s, S=60m. $v^{2}=u^{2}+2aS$ $0=12.5^{2}-2\times 9.8\times S$ $S=7.97m$ From, $v=u+at$ $0=12.5-9.8t$ $t=1.275s$ Now, total height$=7.97+60=67.97m$ $67.97=0+0.5\times 9.8\times t^{2}$ $t=3.724s$ a). So, total time taken $=1.275s+3.724s=5s$ b). Velocity just before hitting the ground$=u+at=0+9.8\times 3.724=36.5m/s$ | 216 | 421 | {"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} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.485577 |
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QUADRANT'AL, a. [supra.] Pertaining to a quadrant; also, included in the fourth part of a circle; as quadrantal space.
QUADRANT'AL, n. [supra.] A vessel used by the Romans; originally called amphora. It was square and contained 80 pounds of water.
## Evolution (or devolution) of this word [quadrantal]
1828 Webster1844 Webster1913 Webster
QUADRANT'AL, a. [supra.] Pertaining to a quadrant; also, included in the fourth part of a circle; as quadrantal space.
QUADRANT'AL, n. [supra.] A vessel used by the Romans; originally called amphora. It was square and contained 80 pounds of water.
Pertaining to a quadrant; also included in the fourth part of a circle; as, quadrantal space. – Derham.
A vessel used by the Romans; originally called amphora. It was square and contained 80 pounds of water. – Encyc.
1. Of or pertaining to a quadrant; also, included in the fourth part of a circle; as, quadrantal space.
Quadrantal triangle, a spherical triangle having one side equal to a quadrant or arc of 90°. -- Quadrantal versor, a versor that expresses rotation through one right angle.
2. A cubical vessel containing a Roman cubic foot, each side being a Roman square foot; -- used as a measure.
3. A cube.
[R.]
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QUADRANT'AL, noun [supra.] A vessel used by the Romans; originally called amphora. It was square and contained 80 pounds of water.
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http://git.tuebingen.mpg.de/?p=paraslash.git;a=blobdiff_plain;f=fec.c;h=2301cc8d2d4653b8bc82a8bcc46b5867b5d7af14;hp=d097b77de836248d79142348f254dd4c3ef57817;hb=3150a0caa27a34d44556fb77f4a5aebc3d978580;hpb=ad6c68022eea0b0962855a2120cf242446bf10b9 | 1,606,772,151,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00288.warc.gz | 33,525,844 | 1,898 | X-Git-Url: http://git.tuebingen.mpg.de/?p=paraslash.git;a=blobdiff_plain;f=fec.c;h=2301cc8d2d4653b8bc82a8bcc46b5867b5d7af14;hp=d097b77de836248d79142348f254dd4c3ef57817;hb=3150a0caa27a34d44556fb77f4a5aebc3d978580;hpb=ad6c68022eea0b0962855a2120cf242446bf10b9 diff --git a/fec.c b/fec.c index d097b77d..2301cc8d 100644 --- a/fec.c +++ b/fec.c @@ -41,24 +41,40 @@ #include "string.h" #include "fec.h" -#define GF_BITS 8 /* code over GF(256) */ +/** Code over GF(256). */ +#define GF_BITS 8 +/** The largest number in GF(256) */ #define GF_SIZE ((1 << GF_BITS) - 1) /* * To speed up computations, we have tables for logarithm, exponent and inverse - * of a number. We use a table for multiplication as well (it takes 64K, no big - * deal even on a PDA, especially because it can be pre-initialized an put into - * a ROM!). The macro gf_mul(x,y) takes care of multiplications. + * of a number. + */ + +/** Index->poly form conversion table. */ +static unsigned char gf_exp[2 * GF_SIZE]; + +/** Poly->index form conversion table. */ +static int gf_log[GF_SIZE + 1]; + +/** Inverse of a field element. */ +static unsigned char inverse[GF_SIZE + 1]; + +/** + * The multiplication table. + * + * We use a table for multiplication as well. It takes 64K, no big deal even on + * a PDA, especially because it can be pre-initialized and put into a ROM. + * + * \sa \ref gf_mul. */ -static unsigned char gf_exp[2 * GF_SIZE]; /* index->poly form conversion table */ -static int gf_log[GF_SIZE + 1]; /* Poly->index form conversion table */ -static unsigned char inverse[GF_SIZE + 1]; /* inverse of field elem. */ static unsigned char gf_mul_table[GF_SIZE + 1][GF_SIZE + 1]; -/* Multiply two numbers. */ + +/** Multiply two GF numbers. */ #define gf_mul(x,y) gf_mul_table[x][y] /* Compute x % GF_SIZE without a slow divide. */ -static inline unsigned char modnn(int x) +__a_const static inline unsigned char modnn(int x) { while (x >= GF_SIZE) { x -= GF_SIZE; @@ -154,21 +170,27 @@ static void generate_gf(void) inverse[i] = gf_exp[GF_SIZE - gf_log[i]]; } +/** How often the loop is unrolled. */ +#define UNROLL 16 + /* * Compute dst[] = dst[] + c * src[] * * This is used often, so better optimize it! Currently the loop is unrolled 16 * times. The case c=0 is also optimized, whereas c=1 is not. */ -#define UNROLL 16 -static void addmul(unsigned char *dst1, const unsigned char const *src1, +static void addmul(unsigned char *dst1, const unsigned char *src1, unsigned char c, int sz) { + unsigned char *dst, *lim, *col; + const unsigned char *src = src1; + if (c == 0) return; - unsigned char *dst = dst1, *lim = &dst[sz - UNROLL + 1], - *col = gf_mul_table[c]; - const unsigned char const *src = src1; + + dst = dst1; + lim = &dst[sz - UNROLL + 1]; + col = gf_mul_table[c]; for (; dst < lim; dst += UNROLL, src += UNROLL) { dst[0] ^= col[src[0]]; @@ -211,6 +233,7 @@ static void matmul(unsigned char *a, unsigned char *b, unsigned char *c, } } +/** Swap two numbers. */ #define FEC_SWAP(a,b) {typeof(a) tmp = a; a = b; b = tmp;} /* | 973 | 3,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.526861 |
https://www.indiabix.com/civil-engineering/upsc-civil-service-exam-questions/discussion-6743 | 1,624,037,778,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00536.warc.gz | 745,871,600 | 5,543 | # Civil Engineering - UPSC Civil Service Exam Questions - Discussion
48.
A beam of rectangular section having simply supported span L, is subject to a concentrated load at its mid-span. What is the length of elasto-plastic zone of the plastic hinge ?
[A]. [B]. [C]. [D].
Explanation:
No answer description available for this question.
Amit Singh said: (Feb 9, 2021) The length of elastoplastic zone= L(1-1/S.F)=L(1-1/1.5) = L/3. | 119 | 436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-25 | latest | en | 0.845298 |
http://jaypantone.com/courses/spring23math6000/code/L12-11-Contour-3-SA-no-GUI.py | 1,695,610,241,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-40/segments/1695233506676.95/warc/CC-MAIN-20230925015430-20230925045430-00163.warc.gz | 25,641,503 | 1,850 | import matplotlib.pyplot as plt import matplotlib.cm as cm import numpy as np import math import random import time class ContinuousFunction: def __init__(self, eval_function, bounds, tweak_delta): self.eval_function = eval_function self.bounds = bounds self.tweak_delta = tweak_delta def random(self): pt = [] for (_min, _max) in self.bounds: diff = _max - _min pt.append(random.random() * diff + _min) return pt def in_bounds(self, point): return all( p >= bnd[0] and p <= bnd[1] for (p, bnd) in zip(point, self.bounds) ) def tweak(self, point): delta = self.tweak_delta new_point = [coord + delta * (2 * random.random() - 1) for coord in point] while not self.in_bounds(new_point): new_point = [coord + delta * (2 * random.random() - 1) for coord in point] return new_point def score(self, point): return self.eval_function(*point) class GeometricSimulatedAnnealing: def __init__(self, space, alpha, maximization=True): self.space = space self.maximization = maximization self.initial_temp = self.determine_initial_temp() self.alpha = alpha self.temp = self.initial_temp def determine_initial_temp(self, trials=1000, initial_prob=0.95): total_change = 0 decreases = 0 tried = 0 while decreases < trials: tried += 1 sol = self.space.random() new_sol = self.space.tweak(sol) diff = self.space.score(new_sol) - self.space.score(sol) if not self.maximization: diff *= -1 if diff < 0: decreases += 1 total_change += diff avg_change = total_change / trials return avg_change / math.log(initial_prob) def advance_temp(self): self.temp *= self.alpha def accept(self, old_score, new_score): score_change = new_score - old_score if not self.maximization: score_change *= -1 if score_change > 0: return True else: accept_probability = math.exp(score_change/self.temp) return random.random() < accept_probability def is_worse(self, old_score, new_score): score_change = new_score - old_score if not self.maximization: score_change *= -1 return score_change < 0 eval_function = lambda x,y: math.sin(3*math.pi*x)**2 + (x-1)**2*(1+math.sin(3*math.pi*y)**2) + (y-1)**2*(1+math.sin(2*math.pi*y)**2) bounds = [[-10, 10]]*2 tweak_delta = 0.1 cns_func = ContinuousFunction(eval_function, bounds, tweak_delta) alpha = 0.999999 SA = GeometricSimulatedAnnealing(cns_func, alpha, maximization=False) point = cns_func.random() value = cns_func.score(point) best = None final_temp_factor = 0.00000001 num_gens = math.ceil(math.log(final_temp_factor)/math.log(alpha)) gen = 0 while SA.temp > final_temp_factor * SA.initial_temp: total = 0 accept = 0 gen += 1 for i in range(1000): if best is None or value < best: best = value best_sol = point new_point = cns_func.tweak(point) new_value = cns_func.score(new_point) if SA.is_worse(value, new_value): total += 1 if SA.accept(value, new_value): if SA.is_worse(value, new_value): accept += 1 point = new_point value = new_value print( f"gen: {gen}/{num_gens}, " f"temp: {SA.temp:.2f}, " f"% acc: {accept/total*100:.5f}, " f"cur score: {value:.2f}, " f"best score: {best}, " ) SA.advance_temp() | 821 | 3,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.561114 |
https://www.funwithpuzzles.com/2014/08/picture-sudoku-fun-with-sudoku-11.html | 1,718,948,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00447.warc.gz | 678,887,216 | 33,958 | In the past, I have seen Linked Sudoku in which there will be more than one Sudoku and some digits will be common among these Sudoku puzzles. Picture Sudoku uses a similar concept with one main Sudoku which will contain a beautiful picture and some part of these pictures will be transferred to other Sudoku in such a way Picture parts in the other Sudoku contain the same digit as the main Sudoku.
This Sudoku puzzle is created by Mark McCorrie. I have given him the following picture, using which he created this beautiful Sudoku.
This Picture Sudoku, I am posting as the 11th Sudoku of the Fun With Sudoku Series. Can you solve this Sudoku Puzzle?
Rules of Picture Sudoku
Classic Sudoku Rules apply to all the given Sudoku sets. Additionally, there will be one main Sudoku containing colored patterns or pictures and There will be an extra Sudoku grid with each of the colour appearing in the main Sudoku. Colored grids in each of the Sudoku contain the exact same digits as the main Sudoku.
Picture Sudoku (Fun With Sudoku #11)
Previous
Next
Frame 1~9 Sudoku (Daily Sudoku League #55) Solution | 228 | 1,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.871769 |
http://www.jiskha.com/display.cgi?id=1324412848 | 1,498,523,755,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320873.54/warc/CC-MAIN-20170626235612-20170627015612-00643.warc.gz | 549,351,599 | 4,216 | # AP STATISTICS
posted by .
Mens shirt sizes are determined by neck sizes. suppose the mens nexck sizes are approximatley normally distributed with mean 15.7 inches and standard deviation 0.7 inch.retailer sells men shirts in S,M,L,XL where the shirt sizes are defined in the table below.
S: 14 ≤ neck size <15
M: 15 ≤ neck size <16
L: 16 ≤ neck size <17
XL:17 ≤ neck size <18
because the retailer only stocks the sizes listed above what proportion of customers will find the retailer does not carry any shirts in their size?
• AP STATISTICS -
P(X < 14.0) = P((x - 15.7) / 0.7) < (14.0 - 15.7) / 0.7) = P(Z < -2.43) = 0.0076
P(X < 18.0) = P((x - 15.7) / 0.7) < (18.0 - 15.7) / 0.7) = P(Z < 3.29) = 0.9995
or use calc by typing in normalcdf(-99,-2.43) and normalcdf(-99,3.29)you get the same probabilities of .0076 and .9995
P(14.0 < x < 18.0) = 0.9995 - 0.0076 = .9919 have their shirts in stock, 1-.9919 = .0081 do not have shirts in stock | 342 | 947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-26 | latest | en | 0.823862 |
http://mathhelpforum.com/algebra/207271-depreciation-help.html | 1,480,880,950,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541361.65/warc/CC-MAIN-20161202170901-00022-ip-10-31-129-80.ec2.internal.warc.gz | 185,390,263 | 9,743 | 1. Depreciation help
I just can't seem to get the answer right here, really infuriating!
'A machine depreciates by 7% p.a. Its price when new is £8000. Calculate its depreciated value after 5 years, depreciated monthly.'
Please can someone help in terms of the appropriate method to use?
2. Re: Depreciation help
depreciation rate per month ... $\frac{.07}{12}$
$V(t) = 8000 \left(1 - \frac{.07}{12}\right)^{60}$ | 122 | 418 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2016-50 | longest | en | 0.885766 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/gypsum-coma-and-blank-plaster-blank-of-blank-paris | 1,701,671,880,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00120.warc.gz | 729,759,808 | 8,296 | # Weight of Gypsum, plaster of Paris
## gypsum, plaster of paris: convert volume to weight
### Weight of 1 cubic centimeter of Gypsum, plaster of Paris
carat 11.6 ounce 0.08 gram 2.32 pound 0.01 kilogram 0 tonne 2.32 × 10-6 milligram 2 320
#### How many moles in 1 cubic centimeter of Gypsum, plaster of Paris?
There are 13.48 millimoles in 1 cubic centimeter of Gypsum, plaster of Paris
### The entered volume of Gypsum, plaster of Paris in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
SENSATIONALLY CHEWY FRUIT CANDY!, UPC: 857983004446 contain(s) 425 calories per 100 grams (≈3.53 ounces) [ price ]
7910 foods that contain Lutein + zeaxanthin. List of these foods starting with the highest contents of Lutein + zeaxanthin and the lowest contents of Lutein + zeaxanthin
#### Gravels, Substances and Oils
CaribSea, Freshwater, African Cichlid Mix, Ivory Coast Sand weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Roofing Asphalt, heavy saturant weighs 1 022.12 kg/m³ (63.80887 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 0W-30 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
sievert [Sv] is the SI unit of any of the quantities expressed as dose equivalent.
The time measurement was introduced to sequence or order events, and to answer questions like these: "How long did the event take?" or "When did the event occur?"
long tn/fl.oz to short tn/US gal conversion table, long tn/fl.oz to short tn/US gal unit converter or convert between all units of density measurement.
#### Calculators
Annulus and sector of an annulus calculator | 705 | 2,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.68468 |
learninganalytics.ca | 1,600,856,557,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210616.36/warc/CC-MAIN-20200923081833-20200923111833-00207.warc.gz | 482,825,850 | 241,909 | ### Import packages
Import the following package. If the package is not installed in your R environment, install it using the install.packages("package_name") command.
library("cem")
### Import data
This exercise reuses the dataset from Exercise 2.
data <- read.csv("../data/exercise2.csv")
head(data, n=10)
## talent effort skill treatment
## 1 FALSE 0.6497020 0.6828916 TRUE
## 2 TRUE 0.6848052 0.9369752 TRUE
## 3 FALSE 0.7935741 0.4630684 TRUE
## 4 TRUE 0.4313167 0.4561438 FALSE
## 5 TRUE 0.4718738 0.7214476 FALSE
## 6 TRUE 0.4208080 0.2688652 FALSE
## 7 FALSE 0.5741308 0.5328195 FALSE
## 8 FALSE 0.6814296 0.6213216 TRUE
## 9 FALSE 0.5700953 0.3671640 FALSE
## 10 TRUE 0.2904192 0.7253196 FALSE
### Calculate data imbalance
A metric called the $$L_1$$ vector norm measures the amount of imbalance within a dataset. This imbalance is created by the bias of confounders. The $$L_1$$ vector norm produces a number between 0 and 1. A value of 0 indicates that there is no bias or imbalance in the dataset, while a value of 1 denotes a totally imbalanced dataset. The following command calculates the $$L_1$$ vector norm for the original dataset. Note that the variables (columns) that are not confounders must be passed as an argument to the function.
L1.meas(data$treatment, data, drop=c('treatment', 'effort', 'skill')) ## ## Multivariate Imbalance Measure: L1=0.360 ## Percentage of local common support: LCS=100.0% The $$L_1$$ value implies that the dataset is mildly imbalanced and that source(s) of bias (i.e., talent) need to be controlled for. Remember that controlling for means doing exact matching. Hence, students will be clustered in the low-talent and high-talent groups. low_talent <- data[data$talent==0,]
high_talent <- data[data$talent==1,] Next, measure the level of data imbalance within each group of low-talent and high-talent students. L1.meas(low_talent$treatment, low_talent, drop=c('treatment', 'effort', 'skill'))
##
## Multivariate Imbalance Measure: L1=0.000
## Percentage of local common support: LCS=100.0%
L1.meas(high_talent\$treatment, high_talent, drop=c('treatment', 'effort', 'skill'))
##
## Multivariate Imbalance Measure: L1=0.000
## Percentage of local common support: LCS=100.0%
It can, therefore, be noticed that doing exact matching eliminates bias also called data imbalance! Note that the $$L_1$$ vector norm can work with a dataset with any number of counfounders. | 712 | 2,501 | {"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} | 2.984375 | 3 | CC-MAIN-2020-40 | longest | en | 0.678706 |
https://coursekata.org/preview/book/f934ff20-e896-469d-84e0-bb90c0c705e0/lesson/4/10 | 1,675,863,374,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00454.warc.gz | 205,618,679 | 9,515 | ## Course Outline
• segmentGetting Started (Don't Skip This Part)
• segmentStatistics and Data Science: A Modeling Approach
• segmentPART I: EXPLORING VARIATION
• segmentChapter 1 - Welcome to Statistics: A Modeling Approach
• segmentChapter 2 - Understanding Data
• segmentChapter 3 - Examining Distributions
• segmentChapter 4 - Explaining Variation
• segmentPART II: MODELING VARIATION
• segmentChapter 5 - A Simple Model
• segmentChapter 6 - Quantifying Error
• segmentChapter 7 - Adding an Explanatory Variable to the Model
• segmentChapter 8 - Digging Deeper into Group Models
• segmentChapter 9 - Models with a Quantitative Explanatory Variable
• segmentPART III: EVALUATING MODELS
• segmentChapter 10 - The Logic of Inference
• segmentChapter 11 - Model Comparison with F
• segmentChapter 12 - Parameter Estimation and Confidence Intervals
• segmentFinishing Up (Don't Skip This Part!)
• segmentResources
### list High School / Advanced Statistics and Data Science I (ABC)
Book
• High School / Advanced Statistics and Data Science I (ABC)
• High School / Statistics and Data Science I (AB)
• High School / Statistics and Data Science II (XCD)
• College / Statistics and Data Science (ABC)
• College / Advanced Statistics and Data Science (ABCD)
• College / Accelerated Statistics and Data Science (XCDCOLLEGE)
## 2.11 Chapter 2 Review Questions 2
require(coursekata) # run your code here
CK Code: A2_Code_Review2_01
require(coursekata) # run your code here
CK Code: A2_Code_Review2_02
require(coursekata) # run your code here
CK Code: A2_Code_Review2_03
require(coursekata) # run your code here
CK Code: A2_Code_Review2_04
require(coursekata) # run your code here
CK Code: A2_Code_Review2_05
require(coursekata) # run your code here
CK Code: A2_Code_Review2_06
require(coursekata) # run your code here
CK Code: A2_Code_Review2_07
require(coursekata) # run your code here
CK Code: A2_Code_Review2_08 | 513 | 1,911 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | longest | en | 0.618271 |
https://www.glassdoor.nl/Sollicitatiegesprek/operations-coordinator-sollicitatievragen-SRCH_KO0,22.htm | 1,679,499,697,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943845.78/warc/CC-MAIN-20230322145537-20230322175537-00082.warc.gz | 898,424,197 | 148,296 | # 36K
Sollicitatievragen voor Operations Coordinator gedeeld door sollicitanten
## Meest gestelde sollicitatievragen
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Er werd een Business Operations Analyst gevraagd...2 februari 2011
### 25 horses, 5 race tracks. How many races you have to run to select top 5 horses.
51 antwoorden
The answer is 9. Assuming: - There's no time measuring (stopwatch), just relative places. - The horses perform consistently. - A maximum of 5 horses per race. First we need 5 races (A to E) to get relative scores for all 25 horses. Let's take a worst scenario: the list was already ordered (A1 fastest and E5 slowest), so race A contained the top 5. The 6th race would be the winners of the 5 races (A1, B1, C1, D1, E1), and would give A1 as the fastest of all. This would also mean that some horses can be excluded (only 4 more places to fill): B5 C4, C5 D3, D4, D5 E2, E3, E4, E5 For the 7th race, A2 would replace A1, and A2 would be appointed as the runner-up (of all). We also can exclude some more (only 3 more places to fill): B4 C3 D2 E1 For the 8th race, A3 would replace A2, but as E1 has been excluded, we got a vacancy. Let's add C2 for worst case scenario. The winner would be A3, and we can exclude more horses (only 2 more places to fill): B3 C2 D1 At this point there're only 5 horses who have not yet been classified or excluded, so the winner and runner-up of the 9th race would give 4th and 5th overall. Minder
You guys are not doing CS! 10 runs is my answer. 1. randomize 5 groups, each of 5 horses 2. rank them within each group, I will use Anuradha's notation (5 races) 3. pick the best of each group, race to figure the 1st place, call it A1 (1 race) It should be clear, it wins all times, every one lost once. 4. remove it. substitute 2nd best in. repeat 3 (in my eg. A2,B1,C1,D1,E1) now you have second place. keep going, you get the first 5 and ranking! So, 5+5=10 races in total. Minder
Anuradha's solution still has problems. (Even if we go with Anuradha's assumptions that you can only race one horse per track, and also assuming that we don't have a stopwatch and must compare horses placing positions) What if the fastest five horses are A1, B1, C1, D1, and E1 ? In Anuradha's second step, he elminates two of the fastest horses (D1 and E1) . He's assuming that A2, B2, or some of the other horses from the other heats are faster, but he hasn't actually tested to see if that is true. Minder
Meer reacties weergeven
31 antwoorden
inane interview questions
The kardashians
Answering stupid questions from moronic interviewers.
Meer reacties weergeven
### Why do u want to work for JetBlue
18 antwoorden
June 30
Why are they taking so long? Have you contact them? Did you do your background and drug test Minder
April interviewee, how long after your interview, did they contact you regarding bkgrd n fp? Minder
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### From the in-person panel interviews: - Uber is opening up a city remotely (i.e. Milwaukee will be operated out of Chicago). How would you make the drivers in Milwaukee feel equal to the drivers in Chicago. This was maybe the worst, most vague question of them all. - Uber drops you in a new market (i.e. Oklahoma City) and says you have two weeks until we open here - go! What do you do? - Say an Uber black car driver makes \$30/ride with a 20% commission. How do you convince him to upgrade to a new made up service UberSuper with a 25% commission? What costs will the driver incur in upgrading. How much more will he have to make each week, etc. Note: it's not enough to just walk them through mentally how you'd do this, be prepared to do the math on the fly. Very annoying. - An Uber competitor opens in your city with unlimited cash capital, if you were them how would you steal Uber's customers? If you were Uber how would you convince drivers not to leave?
18 antwoorden
Antonio, Sandro - hope you guys get good luck ... check your email for some hot man-on-man adult newsletter signups. +1 No_Cheating! Minder
I confirm it is the same test... it is really hard..., thanks god I prepared before with.. try www.uber-analytical-test Minder
Anyone have any ideas on how uber makes drivers in Chicago and Milwaukee feel equal? Minder
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### If I handed you a rubiks cube and told you that it had to be solved by tomorrow, how would you respond?
15 antwoorden
Such questions are really stupid IMHO and I feel like punching the interviewer for asking crap like this. I don't think such questions prove anything. A person can be really good at handling situations but not respond likewise and a pretentious person can do very well in such cases but do a horrible job when hired. Please interviewers stop with such questions! Minder
If I were asked this question I would say, "You'll have it today". How you do it isn't too important, there are literally a million ways to "solve" the rubiks cube they're handing you, it's not a cube, it's any problem they need solved, just get it done as soon as you can, wouldn't you like it if you asked someone to do something & they went above & beyond? Whether it's an inter-office request or anything else, you are always either someone's customer or vendor & this is called "giving excellent customer service". This question is begging for Excellent Customer Service. Minder
Anyone that asks a question of this nature is looking for a typical, corporate behind kissing answer. personally I was much more impressed with your honest answer than I would be by someone, whose nature and work ability I have no idea about, answering like someone else did on this with "You'll have it today." You could have answered with that and have not the first clue about "customer service", because it is just a brown nosing answer. To me, a good answer is honest and forthright, and I never want to just hear what the interviewee thinks I want to hear. Minder
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### CASE: Cross-selling Credit Insurance to Cardholders direct mail: .50, 1% response rate, avg balance \$1000, 5% claim insurance, etc. Profitable? How make more profitable? What if response rate doubled but claims doubled? Make chart of profit curve, what does it mean if..., etc.
12 antwoorden
Thanks for your response and guidance. I knew the revenue is monthly, but i thought claim rate is also monthly and hence calculated profit for each credit insurance sold per month. In light of your clarification Profit per card insurance per year = (10*12)-50-50 = 20\$ per year If we chose to calculate per month, we will need to consider monthly claim rate as (5/12) and also amortize the marketing expense over next 12 month. Profit per card insurance per month = 10-(50/12)-(50/12)= (20/12)\$ per month The profitability equation (per card per year) = 120x-10xy-50 For calculating any of the break even rates (x or y assuming 1 is known), 120x-10xy-50 = 0. For graph, P = 120x-10xy-50 Let me know if my analysis/answer is accurate and up to the mark. Thanks a lot for all your guidance. Minder
@ ghachla: The response rate is 1% that is 1 out of 100. I assumed 100 people are sent the mail. Hence the response rate is 1. As for the claim rate, the people who don't respond can't make claims. So out of the people who respond, the claim rate is 5% (i.e 1 person responds in 100 and out of that 1, 0.05 make the claim); Or to make it more clearer, if we assume 10000 people are sent the mail, 100 respond (because of the 1% response rate) and 5 out of the 100 make claims (because of the 5% claim rate). Minder
Looks like I left out the price of the insurance: customers would pay 1% of monthly balance for insurance. Minder
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### You have the choice between using first class or third class mail for a letter you are sending out to potential customers. First class costs \$0.50 per piece and reach 100% of potential customers. Third class costs \$0.40 per piece and reaches 80% of potential customers. Which do you use?
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either one is good. For example, you will reach 0.8 person by spending \$0.4. If divided 0.4 by 0.8, we get that we will reach 1 person by spending \$0.5. So two methods have the same effect. Minder
First class should be used, as the cost of each delivered letter will be the same, but you will reach all of your intended audience. Example: 1000 pieces to be sent, sending first class costs \$500 and reaches 1000 (\$.50 per peice), and sending second class costs \$400 and reaches 800 (\$.50 per piece) Minder
@gaurav: don't make it too complicated. they just want to know the costs of reaching the customers. peppermint is right. Minder
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### Behavioral based questions were easy to prepare for. Amazon prefers the STAR (situation, task, action, result) method to answer. Would suggestion preparing for the math flow question. I was asked the following: You have 30 associates who all work an 8 hour day, 5 days a week. 2 need to be in indirect (non-volume producing) roles. Your direct (production) rate is 150 units per hour, but you have two 15-minute breaks during the day. How many units can your department produce in a 40 hour week?
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(30 - 2) associates = 28 28 associates/hour * (8 - 0.5) hours/day * 5 days/week * 150 units = 157,500 units/week However, this assumes 100% utilization all the time and this is probably unrealistic. I would then apply a reduction to efficiency to account for unexpected delays. 157,500 units/week * 90% = 141,750 units/week ... final answer. Minder
This is a math question not an estimation on potential variables through sickness, breakdowns etc The answer is 157500 Minder
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### There are 27 balls with one ball having additional weight. Total how many attempts you have to make to check all balls using a sisaw
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3. group the 27 balls into3 groups of nine, weigh 2 of the groups. If one is heavier, continue with that group. If they are equal, continue with the unweighed group. Break those nine into 3 groups of 3, weigh 2 groups. If one is heavier, continue with that group, otherwise use the unweighed group. With the 3 left, weigh any two. If one is heavier, you have it. If they are the same, then it is the unweighed ball that is heaviest. You have used the scale 3 times Minder
3 You can guarantee the answer in 3 steps, if required. However, you no longer stand a chance of obtaining the answer in the first 2 weighings as I described immediately above. 1. Divide balls into 3 groups of 9. --> Weigh two of the groups. This will identify the heavy group. 2. Divide the identified heavy group into 3 groups of 3. --> Weigh two of the groups. This will identify the heavy group. 3. Separate the 3 balls in the heavy group and weigh. --> This comparison will identify the one heavy ball from the original group of 27. Minder
3 attempts - attempt 1 - divide 3 sections 9balls 9balls 9balls pick a section which is different in weight, attempt 2 - divide 3 sections 3balls 3balls 3 balls pick a section which is different in weight, Attempt-3- divide 3 sections 1ball 1ball 1ball pick a ball which is in different in weight Attempt 3 - Minder
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### Presentation on what my plans would be within the first 90 days
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Sean did not write this review. I did and I am a genuine person who had an interview with Sean. Please do not have a go or accuse Sean of writing this review. I wrote it and it was my typo. This review is a true reflection of my experience which is the reason I felt compelled to write it as my experience with the owner was professional. Minder
Another fake review. If the company was so good why did you not take the role? Hmmm Minder
Because there was too much travel to the Pand office which I could not commit to. Leave it at that. You have obviously a grievance with this employer which is between you and them. That doesn’t mean my experience would be the same. I fact it wasn’t. Not a fake review. Get over it and yourself- move and jog on pls Minder
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Weergave: 1 - 10 van 36.072 sollicitatievragen | 3,120 | 12,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-14 | latest | en | 0.914707 |
https://socratic.org/questions/57425e697c014917feeda498 | 1,713,741,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00542.warc.gz | 466,506,231 | 5,894 | # Question da498
Jul 6, 2016
$= 6 {C}_{4} {\left(\frac{1}{4}\right)}^{4} {\left(\frac{3}{4}\right)}^{2}$
$= 15 X \left(\frac{1}{256}\right) \left(\frac{81}{256}\right) = 0.01854 ,$ nearly..
#### Explanation:
There are 13 hearts in the pack of 52..
The probability of drawing a heart
= (number of favorable cases)/(total number of cases)
$= \frac{13}{52} = \frac{1}{4}$.
The probability of not drawing a heart =1-1/4=3/4.
So, for drawing exactly 4 hearts in six draws in succession,
replacing the card drawn in each draw, the probability is the
compound probability,
$= 6 {C}_{4} {\left(\frac{1}{4}\right)}^{4} {\left(\frac{3}{4}\right)}^{2}$
$= 15 X \left(\frac{1}{256}\right) \left(\frac{81}{256}\right) = 0.01854 ,$ nearly.... | 263 | 742 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.698791 |
https://math.stackexchange.com/questions/204816/root-and-sign-of-a-complicated-bivariate-function | 1,566,425,250,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00292.warc.gz | 566,876,054 | 29,947 | # Root and sign of a complicated bivariate function
Given two natural numbers $p$ and $i$, such that $0 < i \leqslant 2^p$, let $$\Phi(p,i) := \frac{1}{2^p+1} + \frac{1}{(i+1)^2} - \frac{1}{2^p}\lg\left(\frac{2^p}{i}+1\right),$$ where $\lg x$ is the binary logarithm. With the help of a Computer Algebra System, it seems that
• If $0 \leqslant p \leqslant 3$, then $\Phi(p,i) < 0$.
• If $4 \leqslant p$, there exists $i_p$ such that $\Phi(p,i_p) = 0$ and $\Phi(p,i) > 0$ for $1 \leqslant i < i_p$, and $\Phi(p,i) < 0$ for $i_p < i \leqslant 2^p$.
How can I prove this?
Just in case, the partial derivative with respect to $i$ is: $$\frac{\partial\Phi}{\partial i}(p,i) = \frac{1}{i(2^p+i)\ln 2} - \frac{2}{(i+1)^3},$$ where $\ln x$ is the natural logarithm. | 309 | 762 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-35 | latest | en | 0.52123 |
http://guanchunart.com/e4wrecq/837d7c-the-diagonals-of-a-parallelogram-bisect-each-other | 1,618,572,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00178.warc.gz | 44,858,789 | 22,458 | Guan Chun | Artist, Illustrator, Designer
Portfolio31 Art Studio / Shanghai / China
## the diagonals of a parallelogram bisect each other
Date : 2021-01-22
Login to view more pages. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. She starts by assigning coordinates as given. (i) bisect each other The diagonals of a Parallelogram bisect each other. Problem 6. Crystal is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. To prove that diagonals of a parallelogram bisect each other Xavier first wants from HISTORY 208 at Arizona State University AD = CB He provides courses for Maths and Science at Teachoo. OAD = OCB The coordinates of the midpoint of diagonal #bar(AC)# are #((a+b)/2,c/2)#. Proof : Since, opposite sides of Parallelogram are parallel. It is then easy to show that the triangles ΔAOD and ΔAOB are congruent using the Side-Side-Side postulate, and from that that ∠AOD ≅ ∠AOB. Adjacent angles are supplementary. point of intersection of AC and BD With that being said, I was wondering if within parallelogram the diagonals bisect the angles which the meet. Step-by-step explanation: In a parallelogram. “The diagonals of a parallelogram bisect each other.” To explore these rules governing the diagonals of a parallelogram use Math Warehouse's interactive parallelogram. Each diagonal divides the quadrilateral into two congruent triangles. Teachoo provides the best content available! Show Answer. How can it be shown that the diagonals of the parallelogram bisect each other? A line that intersects another line segment and separates it into two equal parts is called a bisector. If you just look at a parallelogram, the things that look true (namely, the things on this list) are true and are thus properties, and the things that don’t look like they’re true aren’t properties. Subscribe to our Youtube Channel - https://you.tube/teachoo, Theorem 8.6 The diagonals of a parallelogram bisect each other. An equivalent condition is that the diagonals bisect each other, and are equal in length. I hope that helps!! Line CD and AB are equal in length because opposite sides in a parallelogram are are equal. The lectures follow the curriculum set by the Nepal Education Board. The diagonals of a parallelogram do always bisect each other. Learn all school-level subjects based on Nepal Education Board for free in DLC. Big points would bisect. Show that (i) It bisects ∠C also, (ii) ABCD is a rhombus. AOD BOC If the diagonals of a parallelogram are equal in length, then prove that the parallelogram is a rectangle. Given : ABCD is a Parallelogram with Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. Hence Proved. View Answer If the perimeter of a parallelogram is 1 4 0 m , the distance between a pair of opposite sides is 7 meter and its area is 2 1 0 s q . To Prove : OA = OC & OB = OD Expert Answer: ABCD is a parallelogram, diagonals AC and BD intersect at O. AC and BD diagonals & O is the Sometimes. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? 1 Answer Shwetank Mauria Mar 3, 2018 Please see below. m , find the length of two adjacent sides of the parallelogram. AC and OB are diagonals In the figure let the intersecting point of OB and AC be P To show that diagonals bisect each other we have to prove that OP = PB and PA = PC The co-ordinates of P is obtained by. A rhombus is a parallelogram, so we will use what we already know about parallelograms – that the diagonals bisect each other. Hence line CE and EB are equal and AE and ED are equal due to congruent triangles. In AOD and BOC RE: in a parallelogram, do the diagonals always bisect each other and form a right angle? Rectangles include squares and oblongs. This Site Might Help You. He has been teaching from the past 9 years. We want to show that the midpoint of each diagonal is in the same location. Thus diagonals bisect each other in a rectangle . if we have a parallelogram with the points A B, A plus C B C zero and 00 want to show that the diagonals bisect each other? Concept: Conditional Statements and Converse, Chapter 1: Basic Concepts in Geometry - Problem set 1 [Page 12], Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board, CBSE Previous Year Question Paper With Solution for Class 12 Arts, CBSE Previous Year Question Paper With Solution for Class 12 Commerce, CBSE Previous Year Question Paper With Solution for Class 12 Science, CBSE Previous Year Question Paper With Solution for Class 10, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Arts, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Commerce, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Science, Maharashtra State Board Previous Year Question Paper With Solution for Class 10, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Arts, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Commerce, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Science, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 10, SSC (English Medium) 9th Standard Maharashtra State Board, SSC (Marathi Semi-English) 9th Standard [इयत्ता ९ वी] Maharashtra State Board. - 1427736 Source(s): parallelogram diagonals bisect form angle: https://tinyurl.im/GlpDc. If you draw a picture to help you figure out a quadrilateral’s properties, make your sketch as general as possible. Name the coordinates for point C. A: (2a, 2b + … Get the answers you need, now! In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Ex 3.4, 4 Name the quadrilaterals whose diagonals. The statement can be written in conditional form as, 'If the given quadrilateral is a parallelogram, then its diagonals bisect each other.Antecedent : The given quadrilateral is a parallelogram.Consequent : Its diagonals bisect each other. What is x? In a quadrangle, the line connecting two opposite corners is called a diagonal. - the answers to estudyassistant.com Prove that the diagonals of a parallelogram bisect each other. So we have a parallelogram right over here. The diagonals bisect each other. For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$? Teachoo is free. :-) 5 0. The points A(1,0), B(5,0), C(4,6), and D(8,6) are vertices of a parallelogram. If one pair of opposite sides of a quadrilateral is congruent and parallel, then the quadrilateral is_____a parallelogram. In AOD and BOC OAD = OCB AD = CB ODA = OBC AOD BOC So, OA = OC & OB = OD Hence Proved. How can we prove this? The diagonals of a parallelogram bisect each other. 5 years ago. One pair of opposite sides is parallel and equal in length. Thanks :) Write the antecedent (given part) and the consequent (part to be proved) in the following statement. The diagonals are NOT the same size though, so what’s special about this one? Terms of Service. We will show that in a parallelogram, each diagonal bisects the other diagonal. Sometimes . Theorem 8.6 The diagonals of a parallelogram bisect each other Given : ABCD is a Parallelogram with AC and BD diagonals & O is the point of intersection of AC and BD To Prove : OA = OC & OB = OD Proof : Since, opposite sides of Parallelogram are parallel. AO = OD CO = OB. That is, each diagonal cuts the other into two equal parts. What is x and Y? Similarly we can prove that PC = PA . Geometry. The diagonals of a parallelogram bisect each other. So the first thing that we can think about-- these aren't just diagonals. Since the diagonals bisect each other, y = 16 and x = 22. Then the two diagonals are. Answer: 2 question How could you show that the diagonals of a parallelogram bisect each other? In the figure above drag any vertex to reshape the parallelogram and convince your self this is so. Problem 7. On signing up you are confirming that you have read and agree to So, These angles look like they could all be the same, and since there are four angles there it must mean… That each angle is 90 degrees! Prove that the diagonals of a parallelogram bisect each other. Always. And what I want to prove is that its diagonals bisect each other. In this lesson, we will prove that in a parallelogram, each diagonal bisects the other diagonal. A (0, 0) B (a, 0) C (c+a, b) D (c, b) Do I find the point of intersection? These are lines that are intersecting, parallel lines. The diagonals of a parallelogram bisect each other. That is, each diagonal cuts the other into two equal parts. Learn Science with Notes and NCERT Solutions. The diagonals bisect each other. However, they only form right angles if the parallelogram is a rhombus or a square. Therefore the diagonals of a parallelogram do bisect each other into equal parts. I will assume the Parallelogram is on coordinate geometry graph and you have been given the coordinates of the vertices of the figure.get two oppsite corners and find the mid point using the formula midpoint=(X1+X2)/2.once u get the mid point find the distance from each vertice using the formular distance=[(X1-X2)^2+(Y1-Y2)^2]^0.5.these distances should be equal that's one way of … c = a + b (Eq 1) d = b - a (Eq 2) Now, they intersect at point 'Q'. If you draw the figure, you'll see . So you can also view them as transversals. Draw a parallelogram with two short parallel sides 'a' and two long parallel sides 'b'. Original statement: The diagonals of a parallelogram bisect each other. The line joining the two end points of two equal and parallel line segment to opposite sides bisect each other: Explanation: The coordinates of point #C# are #(a+b,c)#. Online video lecture on Grade 9 MATHEMATICS THE DIAGONALS OF PARALLELOGRAM BISECT EACH OTHER. Once again, since every rhombus is a parallelogram the diagonals bisect each other. Werner. The diagonals of a parallelogram bisect each other Every two opposite sides are parallel; Every two opposite sides are equal; Every two opposite angles are equal; Its diagonals bisect each other; If the diagonals of a parallelogram are equal, then it is a rectangle (This is the parallelogram law.) If the measures of 2 angles of a quadrilateral are equal, then the quadrilateral is_____a parallelogram. Answer: The parallelogram is a "Square" ⇒ (a). Answer by Edwin McCravy(17911) (Show Source): You can put this solution on YOUR website! To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Prove by vector method that the diagonals of a parallelogram bisect each other. x*c - y*d = a (Eq 3) where: x = the scalar fraction along the diagonal 'c' y = the scalar fraction along the diagonal 'd' ADO = CBO (alternate interior angles) AOD COB (ASA) Hence, AO = CO and OD = OB (c.p.c.t) Thus, the diagonals of a parallelogram bisect each other. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Therefore Triangle ABE and CED are congruent becasue they have 2 angles and a side in common. Transcript. can you fill in the bottom portion? So let's find the midpoint of A B and C zero you add yeah, exports together and take half. The sum of the squares of the sides equals the sum of the squares of the diagonals. Solution Show Solution The statement can be written in conditional form as, 'If the given quadrilateral is a parallelogram, then its diagonals bisect each other. What are the equations of the line then? "Question 6 Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). The diagonals of a parallelogram bisect each other. In triangles AOD and COB, DAO = BCO (alternate interior angles) AD = CB. OP = OB . AD = BC (opposite sides of a parallelogram are equal) Triangle DEA is congruent to triangle BEC (angles are equal and two corresponding sides are equal) CE = AE (corresponding sides … ODA = OBC Informally: "a box or oblong" (including a square). OA = OC & OB = OD Since Rhombus, Square and Rectangle are also Parallelogram ∴ There diagonals also bisect each other Thus, Quadrilaterals whose diagonals bisect each other are : Parallelogram Rhombus Square Rectangle Ex 3.4, 4 Name the quadrilaterals whose diagonals. The diagonals of a quadrilateral_____bisect each other. It has rotational symmetry of order 2. Take a look at the angles at which the diagonals intersect. Of Service angles are right angles if the parallelogram is a graduate from Indian Institute of Technology, Kanpur being. Are NOT the same location '' ⇒ ( a ) NOT the same size though, so we will what... Antecedent ( given part ) and the consequent ( part to be proved ) in same. Out a quadrilateral is congruent and parallel, then the quadrilateral is_____a parallelogram a right angle including a square they... Squares of the squares of the sides equals the sum of the midpoint of diagonal # bar AC. Signing up you are confirming that you have read and agree to of! Diagonals bisect each other into equal parts angles at which the meet so we will use we! ) /2, c/2 ) # the figure above drag any vertex to reshape the parallelogram bisect each other ii. - the answers you need, now the Nepal Education Board was wondering if within parallelogram the of. 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Quadrilateral is_____a parallelogram, in layman 's terms, show your proof your website figure. Want to show that in a parallelogram, do the diagonals of a parallelogram bisect each other two. You draw a picture to help you figure out a quadrilateral are equal, then the quadrilateral is_____a parallelogram being! To show that ( I ) it bisects ∠C also, ( )! Into two equal parts if one pair of opposite sides is parallel and equal in.. Triangles AOD and COB, DAO = BCO ( alternate interior angles ) AD CB! The lectures follow the curriculum set by the Nepal Education Board for free DLC. ): parallelogram diagonals bisect each other diagonal AC of a B and C zero you add,! Or oblong '' ( including a square, c/2 ) # interior angles AD. Since the diagonals angles at which the meet, find the length of two adjacent sides a... See below 2 Question how could you show that the diagonals intersect equal! In common to reshape the parallelogram and convince your self this is so these rules governing diagonals. Cuts the other into two equal parts are confirming that you have read and agree to terms of.. Segment and separates it into two equal parts has been teaching from the 9. In this lesson, we will use what we already know about –... The squares of the diagonals of a parallelogram bisect each other has been teaching from the past 9 years B! Online video lecture on Grade 9 MATHEMATICS the diagonals of the squares of the sides equals the sum of sides. About this one ) ABCD is a parallelogram the diagonals of a parallelogram bisect each other s properties, your! Bd intersect at O subjects based on Nepal Education Board for free in DLC line and. Sides is parallel and equal in length and form a right angle signing up are... School-Level subjects based on Nepal Education Board # ( a+b ) /2, c/2 ) # part and! Consequent ( part to be proved ) in the same size though, so ’. Square ( regular quadrilateral ): all four sides are of equal length ( )! Line CE and EB are equal and AE and ED are equal, then the quadrilateral into two equal.. Institute of Technology, Kanpur 's find the length of two adjacent sides of the sides the. Triangles AOD and COB, DAO = BCO ( alternate interior angles ) AD = CB and half! Parallelogram with two short parallel sides ' B ' C zero you add,! The following statement figure, you 'll see he has been teaching the... Of parallelogram bisect each other graduate from Indian Institute of Technology, Kanpur they meet, could you please in. Aod and COB, DAO = BCO ( alternate interior angles ) AD = CB AC BD! Ae and ED are equal and AE and ED are equal due congruent! The sum of the squares of the squares of the sides equals the of... Antecedent ( given part ) and the consequent ( part to be proved ) in the same.... Line connecting two opposite corners is called a diagonal bisect the angles at which the diagonals of a parallelogram the. Just diagonals Question 6 diagonal AC of a quadrilateral are equal, then the quadrilateral parallelogram. The diagonals of a quadrilateral are equal, then the quadrilateral into two equal.! What we already know about parallelograms – that the midpoint of each diagonal the. Diagonal is in the following statement write the antecedent ( given part ) and consequent. 2 angles of a quadrilateral_____bisect the diagonals of a parallelogram bisect each other other the diagonals of a parallelogram do always bisect each other CE and are. Of diagonal # bar ( AC ) # are # ( ( )... A: ( 2a, 2b + … Get the answers you need,!! Equal parts is called a bisector rules governing the diagonals of a bisect., then the quadrilateral into two equal parts 's interactive parallelogram you show that the diagonals of a quadrilateral_____bisect other... With two short parallel sides ' B ' a quadrilateral are equal due to triangles. And separates it into two equal parts is called a bisector add yeah, together... The following statement the quadrilaterals whose diagonals ( equilateral ), and all four sides of... Which they meet, could you please, in layman 's terms, your... Coordinates for point C. a: ( 2a, 2b + … Get answers... Dao = BCO ( alternate interior angles ) AD = CB a+b, C ) # are # a+b! Answers to estudyassistant.com the diagonals of a parallelogram with two short parallel sides ' '... Prove is that its diagonals bisect the angles which they meet, could you show that the diagonals of midpoint. A: ( 2a, 2b + … Get the answers you need, now Nepal Education Board,! On Grade 9 MATHEMATICS the diagonals of parallelogram bisect each other bisects the other into two equal parts so will! He has been teaching from the past 9 years Nepal Education Board I want to prove is that its bisect... ( regular quadrilateral ): parallelogram diagonals bisect each other, y = 16 and x =.! ) in the following statement parts is called a diagonal four sides of! 1427736 Answer: 2 Question how could you please, in layman 's terms, show your proof line... Equal, then the quadrilateral into two equal parts diagonal AC of a parallelogram, diagonals AC and intersect... From Indian Institute of Technology, Kanpur with two short parallel sides ' a ' two. Are NOT the same size though, so what ’ s properties, make your sketch general! Triangles AOD and COB, DAO = BCO ( alternate interior angles ) AD CB!: //tinyurl.im/GlpDc – that the diagonals are NOT the same location within parallelogram diagonals. If the measures of 2 angles and a side in common in layman 's terms, show your proof is... That intersects another line segment and separates it into two congruent triangles a parallelogram bisect each other the whose. Writing a coordinate proof to show that the midpoint of a quadrilateral_____bisect each other each. ' a ' and two long parallel sides ' B ' '' ( including square! Solution on your website thing that we can think about -- these are n't just.. | 5,165 | 21,121 | {"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} | 4.21875 | 4 | CC-MAIN-2021-17 | latest | en | 0.866079 |
https://discover.hubpages.com/games-hobbies/How-to-Play-Sudoku-and-Solve-the-Sudoku-Puzzles-Quickly-and-Easily | 1,670,124,210,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00135.warc.gz | 243,468,892 | 78,296 | # How to Play Sudoku and Solve the Sudoku Puzzles Quickly and Easily
Sudoku is an addictive puzzle that involves placing numbers in small squares in a 9 x 9 grid., based on what other numbers are already in place. A properly constructed game has only one solution and it is up to the solver to find that correct solution. Even though it uses numbers, it does not involve any calculations. It is purely a game of logic. There are, however, variations that do include math calculations, such as Killer Sudoku.
Many people enjoy solving these puzzles as a way to keep their brain active and sharp. They can use their reasoning skills to determine which numbers fit in all those little squares.
It is generally not considered a group activity, although you can use a helper. There are worldwide and smaller competitions for those people who want to take their game playing to the next level.
It is a convenient form of entertainment that is portable. Keep a puzzle book in your car, purse or briefcase, and you will have something to do if you get in a traffic jam, or have to wait for someone or something. You can even tear out a page from your book and give it to someone to keep them from looking over your shoulder.
There are different levels of puzzles. Generally if you find a daily puzzle, you will find that there will be certain times of the week, generally Monday, when the puzzle is easier. Each day the puzzle gets more challenging, with the most difficult one being presented on Sunday, when you have more time to solve it. Start with the easier puzzle to avoid frustration until you have learned the basic strategies and build up.
## Where to Find Sudoku Puzzles
Bookstores, grocery stores, drug stores, and convenience shops at airports often have Sudoku puzzle books you can purchase at low prices. There are also apps you can download on your Smartphone or Netbook.
In addition, there are many places you can play a game online. Many of these websites give you the option of printing it or playing it online.
The online versions may have additional features that enhance your Sudoku experience. Some sites will let you check your work. Others let you customize your Sudoku to make it easier or more difficult. Most sites will offer the option of including multiple numbers in the boxes. Some use colors for a more pleasing look.
• websudoku.com lets you enter multiple numbers, offers different challenge levels, and lets you hide the timer if you wish
• sudoku.com.au lets you select the number with a mouse and places it in the game. You can use one hand, and don't need to type any numbers.
• Fingertime.com comes with sound effects and has a large timer to add pressure
• sudoku kingdom lets you select numbers with a mouse, and even highlights where other occurrences of that number are on the board.
• sudoku comes with a chat feature, so you can get help from other players if you wish
## Basic Advice for Solving Sudoku Puzzles
Solving the basic Sudoku puzzle does not require any math skills, except for simply knowing the numbers from 1 through 9. The rest of it is pure logic.
These puzzles are very entertaining and fun to solve. However, they can also get frustrating when you are stuck. Here is a list of my tips to help keep it fun and not frustrating.
• Never, ever guess. Only enter numbers in the puzzle when you are sure that they belong. Yes you can make little notes about possible choices, but don't enter a number if you are not sure. The wrong number will take you down the wrong path, and you generally won't know you are on the wrong path until you get very far into the puzzle. This is very frustrating.
• Trust that there is a solution. I have not come across a single puzzle that is not solvable. Sometimes just knowing that there is a solution will give you the motivation to keep trying.
• Trust in your ability to find the solution. The video shows you the main tips on solving the puzzle. By following just the tips in the video, you should be able to solve most puzzles. It isn't that hard, and you are a smart person. Keep at it, and you will have the pride in knowing that you have challenged your brain and found the solution.
Please watch the video below that explains the basic concept of the Sudoku layout and provides some tips to help solve them. My own tips will follow the video.
## What to Do if You are Stuck on a Sudoku Puzzle
To begin, some players can wander around the whole puzzle at whim as shown in the video, and switch strategies as needed to fill in whatever numbers they can quickly place. Others prefer to start and continue with a more methodological approach.
When do look for quick wins, eventually you may come to a point where the quick wins are not as apparent.
When that happens, don't panic. There are clues in the puzzle; they are just harder to spot. There may only be one or two, but finding them will unlock the rest of the puzzle for you, at least until you get stuck again. Most of the time, you will be able to find a number, which will help unlock other numbers for you. Once you unlock a number, you can go back to wandering around the whole puzzle at whim if you wish.
The key is to use a methodological approach to solving the Sudoku when you are stuck. You don't have to use this particular order, but going in some kind of order will help make sure that you don't miss any clues.
Scroll to Continue
### Strategy #1 for Solving Sudoku Puzzles
Look for sets of boxes, both vertically and horizontally, where two numbers are present, and you need to place the third number. This is the first strategy presented in the video, which helped them fill the first seven squares.
Strategy #2
First, starting with the number one, see if you can place a one in column one, then column two, and so on across the columns. Then staying with the one, see if you can place a one in any of the rows. Finally, look in each box to see if you can place the one.
Then, do the same with each of the other numbers, one at a time. You may be able to start with a different number, one that is more populated on the board, but I find it easier to stay in order to make sure I don't miss any numbers. The video uses the number eight.
### Strategy #3 for Solving Sudoku Puzzles
By this time, you will generally have a lot of numbers that are filled in, so it is a matter of finding what numbers still remain. Now, go through each column, and see what numbers need to be placed in the column. Then see what spaces are available in the column. In the video, they found the 1 and the 9 in the bottom row using this strategy.
Then do the same thing with each row, and then each box. Just following the tips outlined in the video, in order, will help make sure that you have considered all of the hints.
It may be helpful at this time to write down your remaining choices. This is generally best to do at the end of the puzzle, after you are stuck, then at the beginning, since writing down the choices takes valuable time, especially if you are using a timer.
### Strategy #4 -
Starting with a box that has the most numbers completed, determine which numbers are needed to fill the remaining spaces in the box. You may find that there are spaces in which only one number will fit. Go through each box, and decide where the one fits, where the two fits, and so on. This is an alternative to strategy three, which starts with the missing numbers to see where they fit. In this strategy, we look at the empty space and see which numbers fit in it.
Strategy #5
This is an advanced strategy that may help you with more difficult Sudokus. Sometimes you may know two numbers in a box, but do not know which of the two spaces each belongs. If these two numbers are in the same row or column, you know they cannot be in other spaces in the row or column. Eliminate them as choices in the other spaces and you may be able to numbers for those other spaces.
In the photo to the right, for example, there is a 5 and a 9 missing in the middle right box. Since there are two spaces and two numbers, you know that both those numbers have to be used in the sixth row and in that box. This means the 5 and the 9 cannot be anywhere else in row six. This leaves a 1 and a 3 for the left box.
## Cleaning Up
Once you have found a number using one or more of these strategies, you may be able to continue randomly completing the rest of the puzzle. If you get stuck again, use the methodological approach again, and you will be sure to find the key yet again.
## Sudoku Championship
This convenient and portable game helps provide entertainment and a challenge to keep the mind quick and agile.
Once people have learned how to solve Sudoku puzzles, they try to add additional challenge, such as trying to solve more difficult puzzles, challenging themselves to solve the puzzles more quickly. They may try to solve the puzzle without writing down their choices. They may also try different types of Sudoku puzzles that add additional challenge.
Some even compete in Sudoku championships.
## Comments: "How to Play Sudoku and Solve the Sudoku Puzzles Quickly and Easily"
Shasta Matova (author) from USA on September 08, 2012:
Hello Alecia, some sudoku puzzles are difficult, if you have a hard time finding the clue that will help break through and open up the rest of the puzzles. Some people simply don't know how to break through, and maybe they guess, or they come up with unnecessary rules for themselves (like don't write down the choices) so they wind up with incorrect numbers and not being able to solve it at all. Thanks for your comment and visit.
Alecia Murphy from Wilmington, North Carolina on September 07, 2012:
Sudoku is awesome! I never thought it was all that hard but everyone I meet says it is. I don't know what strategy I use but I think I just go by observation. Interesting hub!
Shasta Matova (author) from USA on April 09, 2012:
Thanks megni, I find sudoku puzzles very addictive. I can feel my brain cells growing when I play them. I hope you breeze through them in no time, and try more and more difficult ones.
megni on April 09, 2012:
Thanks,
Your tips give me hope. I am fascinated with sudoku. At first, I hadn,t a clue, but slowly I'm getting better. And with your tips I expect to be even better at solving. Thanks,
Shasta Matova (author) from USA on March 28, 2012:
Thanks Pamela99, Donna and Seeker 7 for the votes and comments. I would get frustrated with them too (and sometimes still do), but knowing that there is a solution and trusting myself to find it helps so much more than you can imagine. Taking a break when I get frustrated works magic, because whenever I come back to it, I find something that must have been staring at me in the face, and I just missed it.
Helen Murphy Howell from Fife, Scotland on March 28, 2012:
Fabulous hub with some great tips. I'm not that advance at sudoku as yet and yes I will admit that when I've got stuck, I've guessed! This not only screws up the whole puzzle but does give a sense of 'failure'. Now I'd rather just struggle on till I've solved the thing! But with these great tips I'm sure I'll be winning a few more!
Voted up + awesome!
Donna Cosmato from USA on March 28, 2012:
Thanks for the tips! I love Sudoku but it is easy to get frustrated when the solution seems to evade you no matter what you try. You've listed some strategies I haven't tried, so now I'm excited to pick up this fun hobby for a second try. Voted up.
Pamela Oglesby from Sunny Florida on March 28, 2012:
I enjoy working Sudoka puzzles. I think you have some very good basic tips to successfully figure out the tough ones. Voted up!
Shasta Matova (author) from USA on March 28, 2012:
Thanks Ruchira. I find that when I walk away from it and come back to it later, I usually see something obvious that I missed the first time around.
Ruchira from United States on March 27, 2012:
I am a big fan of Sudoku but being attention deficient I can never finish it...
Thanks for the tips video. loved knowing some tips and hopefully will try to finish my puzzles!
Voted up and sharing it! | 2,673 | 12,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.966485 |
https://www.coursehero.com/file/5879076/mjb08a-e22-RETURNED/ | 1,495,616,045,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607806.46/warc/CC-MAIN-20170524074252-20170524094252-00114.warc.gz | 860,251,004 | 24,547 | mjb08a_e22_RETURNED
mjb08a_e22_RETURNED - 1 Use formulas on the perfect tab to...
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Unformatted text preview: 1. Use formulas on the perfect tab to calculate the light blue areas. 2. Gave Up is the Max Score Possible minus the Actual Score. 3. Roll Total is the sum of all the Actual Score or Gave Up numbers. 4. Bonus If > 63 needs to be an if statement that adds 35 in this cell if the sum of H4 is greater than 63. 5. Sub Total should be the sum or all totals from Roll Total through Sub Total 7. Hand in the assignment with the given Yahtzee scores entered in the blue and green areas and the totals calculated 6. Use the "Score this game by minimizing your give ups" grid to enter the data in the green area of the game sheet. Y score loss. You must make the best decision for each final roll and stick to the order of the final rolls. The Rules of Yahtzee - Standard Play OBJECT OF THE GAME: The object of YAHTZEE is to obtain the highest score. The player with the greatest grand total wins and earns the d opponents. Each player keeps his own score on a YAHTZEE score card to be marked with player’s name. To determine who go cup and rolls the dice; the player with the highest total starts the game. The player then continues clockwise. Each shakes the cup and rolls out the dice. Each turn consists of a maximum of three rolls - the first roll to be made with and third time, he may pick up and use any number of dice, providing a score is taken on the last roll. It is the skillfu can turn an unlucky first or second roll into a high-scoring turn. SCORE CARD Now let’s look at the YAHTZEE score card. You’ll note that there are 13 scoring boxes - aces, twos, threes, etc., thr each turn, the player must score in one of the 13 boxes. If on the first roll of the dice, a player has he might choose card or "3 of a kind" (in this case, 2’s) in the Lower Section of the score card. The player would then leave the ,dicecard or "3 of a kind" (in this case, 2’s) in the Lower Section of the score card....
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Ask a homework question - tutors are online | 654 | 2,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-22 | longest | en | 0.897683 |
http://mathforum.org/library/drmath/view/67108.html | 1,516,454,062,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889617.56/warc/CC-MAIN-20180120122736-20180120142736-00791.warc.gz | 222,396,613 | 3,065 | Associated Topics || Dr. Math Home || Search Dr. Math
### Finding Points on a Line in Slope-Intercept Form
```Date: 07/29/2005 at 13:24:37
From: Ben
Subject: How to find the other two points on a line
Hi, Dr. Math:
I have a line y = kx + b, and I know point A(x1, x2) is on the line.
If point B is also on the line, and the distance between A and B is d,
what are the point(s) B?
Thanks,
Ben
```
```
Date: 07/29/2005 at 15:17:16
From: Doctor Jaffee
Subject: Re: How to find the other two points on a line
Hi Ben,
Here's how I solved this problem. I drew a line with the point A(x1,
x2) on it and a point B(z1,z2). Actually, there are two possible
locations for point B. One is to the right and above (if k is a
positive number) and the other is to the left and below. If k is
negative, then B could be to the right and below A or to the left and
above A. Of course, if k is 0, then the problem is easy to solve. If
k = 0, the line is horizontal, so all that you have to do is add d to
x1 to get z1 and z2 stays the same. Then, subtract d from x1 to find
the other point.
But, if the line is not horizontal, you have to work a little harder.
Assume that the horizontal distance from A to B is p and the vertical
distance from A to B is q. Then, the slope of the line is q/p = k.
According to the Pythagorean Theorem p^2 + q^2 = d^2. Now, the values
for k and d are constants, so solve this system of equations for q and
p and you can then easily determine both locations for B.
Give it a try and you have any difficulties or you have questions
Good Luck,
- Doctor Jaffee, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 07/29/2005 at 15:25:25
From: Ben
Subject: Thank you (How to find the other two points on a line)
Thanks Doctor Jaffee! That helps a lot!
```
Associated Topics:
High School Coordinate Plane Geometry
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 631 | 2,230 | {"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} | 3.859375 | 4 | CC-MAIN-2018-05 | latest | en | 0.909612 |
https://blancosilva.github.io/post/2010/10/20/convex-triangle-square-polygons.html | 1,701,649,289,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00665.warc.gz | 160,026,719 | 6,006 | Consider an infinite supply of cardboard equilateral triangles and squares, all of them with side length of one inch. With these pieces, we desire to form convex polygons of different sides. It is trivial to come up with this kind of polygons for sides 3, 4, 5 and 6.
• Construct examples of these convex polygons for sides 7, 8, 9 and 10.
• Can you construct an 11-side convex polygon with these pieces?
• What would be the convex polygon with the largest number of sides that you could construct with these pieces?
Before we go ahead and construct the polygons, let us try to find as much information as possible:
• The interior angles of a polygon with $$n$$ sides add up to $$180n - 360$$.
• By construction, all the convex angles at the exterior vertices of the planar graphs can only be one of the following: $$60^\circ, 90^\circ, 120^\circ$$, and $$150^\circ$$. Although technically possible, we should not count the flat angle $$180^\circ$$, since the occurrence of such angle indicates the presence of an “extended edge.”If we have $$a$$, $$b$$ and $$c$$ interior angles of respectively $$60^\circ$$, $$90^\circ$$ and $$120^\circ$$ at the exterior vertices of the polygon, and of course $$n-a-b-c$$ interior angles of $$150^\circ$$ at the same vertices, then it must be
$$60a + 90b + 120c + 150(n-a-b-c) = 180n - 360,$$
which reduces to
$$3a + 2b + c + n = 12.$$
This is a very strong statement! It indicates, among other things, that such polygons cannot have more than 12 sides! This answers the third question of our puzzle. Incidentally, it will also help us find examples of constructions. For example, for the seven-sided polygon, a choice of angles that works for me is $$a=0$$, $$b=2$$ and $$c=1$$ (that is: no $$60^\circ$$ angles, two right angles, only one angle of $$120^\circ$$ and four angles of $$150^\circ$$. For the case $$n=11$$, there is only one possibility: $$a=0$$, $$b=0$$, $$c=1$$; that is, only one angle of $$120^\circ$$, and ten angles of $$150^\circ$$. Will this allow us to construct the 11-sided convex polygon? The answer is yes, but to come up with an example, you may need to make use of some $$180^\circ$$ angles. I have included a list with a set of choices of angles, for polygons of sides $$n=7,8,9,10,11,12$$ for which I was able to find a valid construction:
$$n$$ $$60^\circ$$-angles $$90^\circ$$-angles $$120^\circ$$-angles $$150^\circ$$-angles Planar graph 7 0 2 1 4 8 0 0 4 4 9 0 0 3 6 10 0 0 2 8 11 0 0 1 10 12 0 0 0 12
Feel free to construct the corresponding polygons, and fill in the table.
### Miscellaneous
Although not strictly necessary, at some point during the exploration of this problem I used other facts about planar graphs that came in handy to help construct my examples. The most notable of those appear below:
• Each $$n$$-sided convex polygon that we construct by putting together these triangles and squares form a planar graph with $$V$$ vertices $$(V \geq n)$$, $$E$$ edges $$(E \geq n)$$ and $$F$$ faces $$(F \geq 1)$$. These planar graphs all have Euler characteristic 1, and thus $$V - E + F =1$$.
• Assume we are able to construct a $$n$$-sided polygon by using $$x$$ triangles (according to the statement above, there should be $$F-x$$ squares). By adding all the involved angles, we obtain
$$\underbrace{3 \cdot 60 \cdot x}_{\text{triangles}} + \underbrace{4 \cdot 90 \cdot (F-x)}_{\text{squares}} = \underbrace{180n - 360}_{\text{polygon}} + \underbrace{360 (V-n)}_{\text{interior vertices}},$$
which reduces to
$$n - x = 2(V-1-F).$$
In particular, this indicates that $$x$$ and $$n$$ have the same parity. | 1,026 | 3,604 | {"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": 4, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-50 | latest | en | 0.8888 |
http://stackoverflow.com/questions/12466745/how-to-convert-float-to-doubleboth-stored-in-ieee-754-representation-without-l/12466918 | 1,394,703,347,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678663943/warc/CC-MAIN-20140313024423-00055-ip-10-183-142-35.ec2.internal.warc.gz | 128,457,145 | 20,278 | How to convert float to double(both stored in IEEE-754 representation) without losing precision?
I mean, for example, I have the following number encoded in IEEE-754 single precision:
``````"0100 0001 1011 1110 1100 1100 1100 1100" (approximately 23.85 in decimal)
``````
The binary number above is stored in literal string.
The question is, how can I convert this string into IEEE-754 double precision representation(somewhat like the following one, but the value is not the same), WITHOUT losing precision?
``````"0100 0000 0011 0111 1101 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1010"
``````
which is the same number encoded in IEEE-754 double precision.
I have tried using the following algorithm to convert the first string back to decimal number first, but it loses precision.
``````num in decimal = (sign) * (1 + frac * 2^(-23)) * 2^(exp - 127)
``````
I'm using Qt C++ Framework on Windows platform.
EDIT: I must apologize maybe I didn't get the question clearly expressed. What I mean is that I don't know the true value 23.85, I only got the first string and I want to convert it to double precision representation without precision loss.
-
@tenfour I think it is because he is storing it as string – Caesar Sep 17 '12 at 20:41
Your second binary string is not the same number as the first, but in `double` precision. What is the problem you try to solve? – Daniel Fischer Sep 17 '12 at 20:42
Daniel Fisher is correct; those are in no way the same number (the first is 23.84999847412109375, the second is 23.85000000000000142108547152020037174224853515625). – Stephen Canon Sep 17 '12 at 20:51
Well: keep the sign bit, rewrite the exponent (minus old bias, plus new bias), and pad the mantissa with zeros on the right...
(As @Mark says, you have to treat some special cases separately, namely when the biased exponent is either zero or max.)
-
Thanks! The question, in fact, is that simple. I just got stuck in my mind. – Turner Sep 17 '12 at 21:24
You do realize that by doing that you will NOT get the result you asked for in the OP, right? – Analog File Sep 17 '12 at 21:44
@KerrekSB: this will, in general, not yield the original decimal value! The versions between binary fractional number and decimal fractional numbers relies on a nice interaction between the conversion from decimal to binary and back to decimal. If you pad with zeros, the double won't convert back correctly. – Dietmar Kühl Sep 17 '12 at 22:24
@DietmarKühl: The original decimal value is not the value of the float, either. 23.85 is 477/20, which is not representable exactly as a binary float. Indeed, the nearest double to 477/20 is not the one given by my recipe. Rather, my recipe shows how to get a double of the same value as the input float. – Kerrek SB Sep 17 '12 at 22:27
@KerrekSB: If you do the Right Thing, you can convert a decimal fractional number into a binary floating point ("Bellerophon") and back to a decimal fraction number ("Dragon4") and you get back the original value under very reasonable restrictions (essentially, the number of decimal digits is limited by the log_2(10) rounded down). If you just pad with zeros, formatting a `double` won't yield the same value as the `float` you came from! – Dietmar Kühl Sep 17 '12 at 22:39
show 1 more comment
First of all, +1 for identifying the input in binary.
Second, that number does not represent 23.85, but slightly less. If you flip its last binary digit from `0` to `1`, the number will still not accurately represent 23.85, but slightly more. Those differences cannot be adequately captured in a float, but they can be approximately captured in a double.
Third, what you think you are losing is called accuracy, not precision. The precision of the number always grows by conversion from single precision to double precision, while the accuracy can never improve by a conversion (your inaccurate number remains inaccurate, but the additional precision makes it more obvious).
I recommend converting to a float or rounding or adding a very small value just before displaying (or logging) the number, because visual appearance is what you really lost by increasing the precision.
Resist the temptation to round right after the cast and to use the rounded value in subsequent computation - this is especially risky in loops. While this might appear to correct the issue in the debugger, the accummulated additional inaccuracies could distort the end result even more.
-
It might be easiest to convert the string into an actual float, convert that to a double, and convert it back to a string.
-
+1 for a viable solution. Thank you – Turner Sep 17 '12 at 21:32
IEEE-754 (and floating point in general) cannot represent periodic binary decimals with full precision. Not even when they, in fact, are rational numbers with relatively small integer numerator and denominator. Some languages provide a rational type that may do it (they are the languages that also support unbounded precision integers).
As a consequence those two numbers you posted are NOT the same number.
They in fact are:
10111.11011001100110011000000000000000000000000000000000000000 ... 10111.11011001100110011001100110011001100110011001101000000000 ...
where `...` represent an infinite sequence of `0`s.
Stephen Canon in a comment above gives you the corresponding decimal values (did not check them, but I have no reason to doubt he got them right).
Therefore the conversion you want to do cannot be done as the single precision number does not have the information you would need (you have NO WAY to know if the number is in fact periodic or simply looks like being because there happens to be a repetition).
-
Sorry I don't get it. It's said in C Programming there is no precision loss in converting "float" to "double", looks like a paradox here.... – Turner Sep 17 '12 at 21:31
There is no precision loss. If you convert the float to double you get the same value. The problem is that you do NOT want to get the same value, you want to get another value. Think it in decimal, A decimal floating point with 6 digits of mantissa could store 123.4545 (as 1.234545*10^2). If you go to 8 digits you get 123.4545 (as 1.23454500*10^2) with no loss of precision. But what you would like to get is 123.454545 which is another number. – Analog File Sep 17 '12 at 21:42
Binary floating points cannot, in general, represent decimal fraction values exactly. The conversion from a decimal fractional value to a binary floating point (see "Bellerophon" in "How to Read Floating-Point Numbers Accurately" by William D.Clinger) and from a binary floating point back to a decimal value (see "Dragon4" in "How to Print Floating-Point Numbers Accurately" by Guy L.Steele Jr. and Jon L.White) yield the expected results because one converts a decimal number to the closest representable binary floating point and the other controls the error to know which decimal value it came from (both algorithms are improved on and made more practical in David Gay's dtoa.c. The algorithms are the basis for restoring `std::numeric_limits<T>::digits10` decimal digits (except, potentially, trailing zeros) from a floating point value stored in type `T`.
Unfortunately, expanding a `float` to a `double` wrecks havoc on the value: Trying to format the new number will in many cases not yield the decimal original because the `float` padded with zeros is different from the closest `double` Bellerophon would create and, thus, Dragon4 expects. There are basically two approaches which work reasonably well, however:
1. As someone suggested convert the `float` to a string and this string into a `double`. This isn't particularly efficient but can be proven to produce the correct results (assuming a correct implementation of the not entirely trivial algorithms, of course).
2. Assuming your value is in a reasonable range, you can multiply it by a power of 10 such that the least significant decimal digit is non-zero, convert this number to an integer, this integer to a `double`, and finally divide the resulting double by the original power of 10. I don't have a proof that this yields the correct number but for the range of value I'm interested in and which I want to store accurately in a `float`, this works.
One reasonable approach to avoid this entirely issue is to use decimal floating point values as described for C++ in the Decimal TR in the first place. Unfortunately, these are not, yet, part of the standard but I have submitted a proposal to the C++ standardization committee to get this changed.
-
The problem is to convert a binary floating-point number to a binary floating-point number, but this answer offers irrelevant statements about representing decimal values and converting to and from decimal. Furthermore, its statement that expanding a float to a double wreaks havoc on the value is nonsense: The value does not change. (If some piece of software to convert the double to decimal then displays it differently from the float, then that is the fault of that software and has nothing to do with the original problem.) – Eric Postpischil Sep 18 '12 at 13:21
A common fallacy which interferes with efforts to reason about floating-point numbers is a belief that a `float` represents an exact quantity of the form M*2^N, where M and N are integers in a certain range. It is true that a `float` has an exact "nominal" value of that form, but I wouldn't say that a `float` represents that exact quantity. The number `2000000.13f` has a nominal value of exactly 2000000.125, but that particular `float` would be used for all values from 2000000.0625 to 2000000.1875. Because all values from... – supercat Aug 14 '13 at 21:17
2000000.1200 to 2000000.1299 would be within the possible range, digits beyond the second one after the decimal point convey no useful information; the reported value is thus rounded to 2000000.13. Although there is no `double` whose exact nominal value is 2000000.125 (matching the float), there is none that means "something between 2000000.0625 and 2000000.1875". The `double` with the same nominal value has a different semantic meaning. – supercat Aug 14 '13 at 21:20 | 2,429 | 10,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.879788 |
https://1lab.dev/Algebra.Group.Instances.Cyclic.html | 1,726,838,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00591.warc.gz | 54,326,542 | 17,766 | open import Algebra.Group.Instances.Symmetric
open import Algebra.Group.Instances.Integers
open import Algebra.Group.Cat.Base
open import Algebra.Group.Subgroup
open import Algebra.Group.Ab
open import Algebra.Group
open import Cat.Prelude
open import Data.Int.Divisible
open import Data.Int.Universal
open import Data.Fin.Closure
open import Data.Int.DivMod
open import Data.Fin
open import Data.Int hiding (Positive)
open import Data.Nat
open represents-subgroup
open normal-subgroup
open is-group-hom
module Algebra.Group.Instances.Cyclic where
private module β€ = Group-on (β€ .snd) hiding (magma-hlevel)
# Cyclic groupsπ
We say that a group is cyclic if it is generated by a single element that is, every element of the group can be written as some integer power of 1.
_is-generated-by_ : β {β} (G : Group β) (g : β G β) β Type β
G is-generated-by g = β x β β[ n β Int ] x β‘ pow G g n
is-cyclic : β {β} (G : Group β) β Type β
is-cyclic G = β[ g β G ] G is-generated-by g
Note
Generators are not unique, so we must use to get a proposition. In fact, if generates then so does hence a non-trivial cyclic group has at least two generators!
module _ {β} (G : Group β) (open Group-on (G .snd)) where
inverse-generates
: β g β G is-generated-by g β G is-generated-by g β»ΒΉ
inverse-generates g gen x = gen x <&> Ξ» (n , p) β negβ€ n , (
pow G g n β‘Λβ¨ inv-inv β©β‘Λ
pow G g n β»ΒΉ β»ΒΉ β‘Λβ¨ ap _β»ΒΉ (pres-inv (pow-hom G g .preserves) {lift n}) β©β‘Λ
pow G g (negβ€ n) β»ΒΉ β‘Λβ¨ pow-inverse G g (negβ€ n) β©β‘Λ
pow G (g β»ΒΉ) (negβ€ n) β)
The obvious example is the infinite cyclic group of the (additive) integers generated by but there is also a finite cyclic group of order for every As it turns out, we can construct all the cyclic groups using the same machine: the quotient group where is the normal subgroup of multiples of is the cyclic group of order when and is the infinite cyclic group when
infix 30 _Β·β€
_Β·β€ : β (n : Nat) β normal-subgroup β€ Ξ» i β el (n β£β€ i) (β£β€-is-prop n i)
(n Β·β€) .has-rep .has-unit = β£β€-zero
(n Β·β€) .has-rep .has-β = β£β€-+
(n Β·β€) .has-rep .has-inv = β£β€-negβ€
(n Β·β€) .has-conjugate {x} {y} = subst (n β£β€_) xβ‘y+x-y
where
xβ‘y+x-y : x β‘ y +β€ (x -β€ y)
xβ‘y+x-y =
x β‘β¨ β€.insertl {y} (β€.inverser {x = y}) β©β‘
y +β€ (negβ€ y +β€ x) β‘β¨ ap (y +β€_) (+β€-commutative (negβ€ y) x) β©β‘
y +β€ (x -β€ y) β
infix 25 β€/_
β€/_ : Nat β Group lzero
β€/ n = β€ /α΄³ n Β·β€
is an abelian group, since the commutativity of addition descends to the quotient.
β€/n-commutative : β n β is-commutative-group (β€/ n)
β€/n-commutative n = elim! Ξ» x y β ap inc (+β€-commutative x y)
infix 25 β€-ab/_
β€-ab/_ : Nat β Abelian-group lzero
β€-ab/_ n = from-commutative-group (β€/ n) (β€/n-commutative n)
Furthermore, is, by construction, cyclic, and is generated by
module _ (n : Nat) where
open Group-on ((β€/ n) .snd)
β€/n-cyclic : is-cyclic (β€/ n)
β€/n-cyclic = inc (1 , gen) where
gen : β€/ n is-generated-by 1
gen = elim! Ξ» x β inc (x , Integers.induction Int-integers
{P = Ξ» x β inc x β‘ pow (β€/ n) 1 x}
refl
(Ξ» x β
inc x β‘ pow (β€/ n) 1 x ββ¨ _ , equivβcancellable (β-equivr 1) β©β
inc x β 1 β‘ pow (β€/ n) 1 x β 1 ββ¨ β-pre-equiv (sym (ap inc (+β€-oner x))) β©β
inc (sucβ€ x) β‘ pow (β€/ n) 1 x β 1 ββ¨ β-post-equiv (sym (pow-sucr (β€/ n) 1 x)) β©β
inc (sucβ€ x) β‘ pow (β€/ n) 1 (sucβ€ x) ββ)
x)
We prove the following mapping-out property for a group homomorphism is uniquely determined by an element of order i.e.Β such that Note that this condition is trivially satisfied if so we donβt need a special case.
This follows from the fact that a group homomorphism out of the group of integers is uniquely determined by an element of and some basic algebra.
module _
(n : Nat)
{β} {G : Group β} (open Group-on (G .snd))
(x : β G β) (open pow G x renaming (pow to infixr 30 x^_; pow-hom to x^-))
(wraps : x^ pos n β‘ unit)
where
β€/-out : Groups.Hom (LiftGroup β (β€/ n)) G
β€/-out .hom (lift i) = Coeq-rec (apply x^- β lift) (Ξ» (a , b , nβ£a-b) β zero-diff $let k , k*nβ‘a-b = β£β€βfibre nβ£a-b in x^ a β x^ b β‘Λβ¨ pres-diff (x^- .preserves) {lift a} {lift b} β©β‘Λ x^ (a -β€ b) β‘Λβ¨ ap x^_ k*nβ‘a-b β©β‘Λ x^ (k *β€ pos n) β‘β¨ ap x^_ (*β€-commutative k (pos n)) β©β‘ x^ (pos n *β€ k) β‘β¨ pow-* G x (pos n) k β©β‘ pow G β x^ pos n β k β‘β¨ ap! wraps β©β‘ pow G unit k β‘β¨ pow-unit G k β©β‘ unit β) i β€/-out .preserves .pres-β = elim! Ξ» x y β x^- .preserves .pres-β (lift x) (lift y) We can check that β€-ab/0β‘β€-ab : β€-ab/ 0 β‘ β€-ab β€-ab/0β‘β€-ab = β«-Path (total-hom (Coeq-rec (Ξ» z β z) (Ξ» (_ , _ , p) β β€.zero-diff p)) (record { pres-β = elim! Ξ» _ _ β refl })) (is-isoβis-equiv (iso inc (Ξ» _ β refl) (elim! Ξ» _ β refl))) β€/0β‘β€ : β€/ 0 β‘ β€ β€/0β‘β€ = GrpβAbβGrp (β€/ 0) (β€/n-commutative 0) β ap AbelianβGroup β€-ab/0β‘β€-ab For positive we show that is finite; more precisely, we construct an equivalence which sends to the representative of its congruence class modulo β€/nββ<n : β n β .β¦ Positive n β¦ β β β€/ n β β β< n β€/nββ<n n .fst = Coeq-rec (Ξ» i β i %β€ n , x%β€y<y i n) Ξ» (x , y , p) β Ξ£-prop-path! (divides-diffβsame-rem n x y p) β€/nββ<n n .snd = is-isoβis-equiv$ iso
(Ξ» (i , p) β inc (pos i))
(Ξ» i β Ξ£-prop-path! (β<-%β€ i))
(elim! Ξ» i β quot (same-remβdivides-diff n (pos (i %β€ n)) i
(β<-%β€ (_ , x%β€y<y i n))))
Finite-β€/n : β n β .β¦ Positive n β¦ β β β€/ n β β Fin n
Finite-β€/n n = β€/nββ<n n βe Finββ< eβ»ΒΉ
Using this and the fact that we can check that is isomorphic to the symmetric group We start by defining a homomorphism that sends 1 to the equivalence that swaps the two elements, which corresponds to
β€/2βSβ : Groups.Hom (β€/ 2) (S 2)
β€/2βSβ = β€/-out 2 (Equiv.from (Fin-permutations 2) 1)
(Equiv.injective (Fin-permutations 2) refl)
Groups.β GβLiftG (β€/ 2)
In order to conclude the proof, we show that the function thus defined is an equivalence, by showing that it factors as in the following diagram and using the two-out-of-three property of equivalences:
β€/2β‘Sβ : β€/ 2 β‘ S 2
β€/2β‘Sβ = β«-Path β€/2βSβ \$
equiv-cancell (Fin-permutations 2 .snd)
(equiv-cancelr ((Finite-β€/n 2 eβ»ΒΉ) .snd)
(subst is-equiv (ext Ξ» where
fzero β refl
(fsuc fzero) β refl)
id-equiv))
1. Some authors (namely, Bourbaki) call such groups monogeneous, and reserve βcyclicβ for the finite cyclic groups.β©οΈ | 2,973 | 6,899 | {"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} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.63578 |
https://trac.sagemath.org/ticket/20084 | 1,623,574,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00562.warc.gz | 535,834,135 | 9,590 | Opened 5 years ago
Closed 4 years ago
# residue: mathematically wrong output
Reported by: Owned by: behackl major sage-8.0 symbolics rws, cheuberg Benjamin Hackl Frédéric Chapoton N/A 68fea61 68fea61960a41972282076686c2c59611c496a07
### Description
The complex function `f(s) = 1/(1 - 2^(-s))` has poles of residue `1/log(2)` at `s = 2*k*pi*I/log(2)` for all integers `k`.
Currently sage recognize these poles just at `s=0`:
```sage: f(s).residue(s==0)
1/log(2)
sage: f(s).residue(s==2*pi*I/log(2))
0
```
In essence, this happens because the `series`-method does not recognize the pole. The priority is critical because mathematically wrong output is produced.
### comment:1 Changed 5 years ago by behackl
The most elegant solution would of course be the automatic simplification of expressions like `2^(something/log(2)) --> exp(something)`, such that
```sage: 2^(2*pi*I/log(2))
1
```
However, I'm not sure if that can be achieved so easily.
A second suggestion would be something like
• ## src/sage/symbolic/expression.pyx
```diff --git a/src/sage/symbolic/expression.pyx b/src/sage/symbolic/expression.pyx
a cdef class Expression(CommutativeRingElement): a = 0 if a == infinity: return (-self.subs({x: 1/x}) / x**2).residue(x == 0) return self.subs({x: x+a}).series(x == 0, 0).coefficient(x, -1) return self.subs({x: x+a}).canonicalize_radical().series(x == 0, 0).coefficient(x, -1) def taylor(self, *args): r"""
where there are several ways to refine this approach:
• introduce an additional keyword that could disable this additional simplification such that performance does not suffer necessarily,
• only try to simplify for complex arguments of `a`
and so on.
Thoughts?
Last edited 5 years ago by behackl (previous) (diff)
### comment:2 Changed 5 years ago by behackl
• Authors set to Benjamin Hackl
• Branch set to u/behackl/symbolics/residue/exp-complex-poles
• Commit set to 45d16e2c9301626ff1da1bb229f955209f17e93a
• Status changed from new to needs_info
This is just the quick workaround from above.
New commits:
45d16e2 `simplify substituted expression before series expansion`
### comment:3 Changed 5 years ago by was
• Status changed from needs_info to needs_work
### comment:4 Changed 5 years ago by git
• Commit changed from 45d16e2c9301626ff1da1bb229f955209f17e93a to 87193643462062096a9b48dd07c76765bb525002
Branch pushed to git repo; I updated commit sha1. New commits:
8719364 `add a doctest`
### comment:6 follow-up: ↓ 7 Changed 4 years ago by mforets
update from the future:
```sage: (1/(1 - 2^-x)).residue(x == 2*pi*I/log(2))
1/log(2)
sage: version()
'SageMath version 7.6.beta6, Release Date: 2017-03-03'
```
some updates in `series` may affect that now it can recognize the pole without having to call `canonicalize_radical()`?
### comment:7 in reply to: ↑ 6 ; follow-up: ↓ 8 Changed 4 years ago by behackl
Yes, I have faint memory of improving something with `series` itself some time ago; thanks for reminding me.
Actually, there are more problems with `residue`. Take, for example,
```sage: (1/sqrt(x^2)).residue(x==0)
0
1
```
(Note that this is, in some sense, a pathological example.)
The ideal fix in this case (IMHO) would be to throw an error that the residue could not be computed as the "correct" branch of the root could not be chosen automatically. I think that just applying `canonicalize_radical` before computing the residue just introduces more grief as the user looses all control over which branch is chosen.
In any case: the original problem reported with this ticket is solved, so this is probably just wontfix. Other opinions?
update from the future:
```sage: (1/(1 - 2^-x)).residue(x == 2*pi*I/log(2))
1/log(2)
sage: version()
'SageMath version 7.6.beta6, Release Date: 2017-03-03'
```
some updates in `series` may affect that now it can recognize the pole without having to call `canonicalize_radical()`?
### comment:8 in reply to: ↑ 7 Changed 4 years ago by mforets
Yes, I have faint memory of improving something with `series` itself some time ago; thanks for reminding me. ... In any case: the original problem reported with this ticket is solved, so this is probably just wontfix. Other opinions?
cool. for what i've been told from other tickets, i suppose this one should should still provide the test (not wontfix):
```Check that :trac:`20084` is fixed::
sage: (1/(1 - 2^-x)).residue(x == 2*pi*I/log(2))
1/log(2)
```
Actually, there are more problems with residue. ... (Note that this is, in some sense, a pathological example.)
yes. one could come up with other examples (e.g. examples involving log(z)). i agree that just applying `canonicalize_radical` is not a good idea for the reason you stated. but is there a simple way to identify these cases without much hurdle? does this belong to `series` method or is sth that can be done at the level of `residue`?
### comment:9 Changed 4 years ago by mforets
@behackl : if you remove the `canonicalize_radical` keeping the test, i can review it! (or the other way round).
wrt handling pathological examples, yes IMO it is a relevant future task, although as pointed out before i cannot help right now, since i'm not visualizing a workaround :/
### comment:10 Changed 4 years ago by chapoton
• Branch changed from u/behackl/symbolics/residue/exp-complex-poles to public/20084
• Commit changed from 87193643462062096a9b48dd07c76765bb525002 to 68fea61960a41972282076686c2c59611c496a07
• Reviewers set to Frédéric Chapoton
• Status changed from needs_work to positive_review
looks good to me
New commits:
311e284 `Merge branch 'u/behackl/symbolics/residue/exp-complex-poles' in 8.0.b6` 68fea61 `trac 20084 back to just adding an example`
### comment:11 Changed 4 years ago by chapoton
• Milestone changed from sage-7.1 to sage-8.0
• Priority changed from critical to major
### comment:12 Changed 4 years ago by vbraun
• Branch changed from public/20084 to 68fea61960a41972282076686c2c59611c496a07
• Resolution set to fixed
• Status changed from positive_review to closed
Note: See TracTickets for help on using tickets. | 1,722 | 6,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.806956 |
https://jaxwebster.wordpress.com/2011/02/02/the-coin-problem/ | 1,529,313,501,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860168.62/warc/CC-MAIN-20180618090026-20180618110026-00191.warc.gz | 639,845,871 | 19,482 | # The Coin Problem
Here is a problem: You have 10 coins. One of these coins is heavier than the others, but they are all otherwise identical. You have a balance scale which you can use to compare the weights of any two sets of coins. You must devise a strategy to find the heavier coin in the minimum number of weighs (in the worst case).
The solution is simple. Split into two groups of 5 and compare those two. Select the heavier of the two sets and compare two sets of two in there. If they are equal, it must’ve been the one you didn’t compare. If they are not, you compare the two on the heavier side to finally find the heaviest. So it takes only 3 weighs.
Now, what if the odd coin out is either heavier than the others or lighter than the others but you don’t know which? In fact, it is still possible to do it in three goes. The following diagram shows you how.
The ten coins are labelled as shown at the top. ‘A’ is short for A1 + A2 + A3, and similarly for ‘B’ and ‘C’. The underlines denote that the coin is a known reference (i.e. known to be neither heavier nor lighter). If A and B are not the same weight, we assume A is heavier (as otherwise you can just relabel them).
Note that in every path except for the first, which ends in D, you know not just which coin it is, but if the coin is heavier or lighter than the others. | 325 | 1,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 4 | CC-MAIN-2018-26 | latest | en | 0.949802 |
https://www.printablemultiplication.com/worksheets/3rd-Grade-Multiplication-Word-Problems-Worksheets-Free | 1,618,420,798,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077843.17/warc/CC-MAIN-20210414155517-20210414185517-00393.warc.gz | 1,043,895,519 | 76,557 | ## Multiplication Chart X30
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https://www.doubtnut.com/question-answer/determine-the-value-of-the-constant-k-so-that-the-function-fx-kx-x-if-x-lt-0-3-if-x-ge-0-is-continuo-642517350 | 1,639,014,427,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00523.warc.gz | 818,565,166 | 77,973 | HomeEnglishClass 12MathsChapterXii Boards
Determine the value of the con...
# Determine the value of the constant k, so that the function f(x) ={(kx)/|x|, if x lt 0 ,3 if x ge 0 is continuous at x=0
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00:00 - 00:59when question is determine the value of constant K so that function f x is equal to given function is continuous at x = 20 problem analysing the given function f of X is equals to 1 Mod a03 Ke continues at 800
01:00 - 01:59you again right function f of X equal to minus x minus 1 upon 2 minus and when it is less than 0.3 greater than equal then this is continuous at x = 20 and 40 + = 20 and 30 2009 and what is a 400 Plus is greater than zero screen 3000
02:00 - 02:59what is 0 - theta minus sin number glaze 1000 - 3 all three equal then the function is continuous at X is equals to minus ke that both value of k in minus 3 find the value of x minus 3 is equal to function value = 2020 | 340 | 1,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-49 | latest | en | 0.842367 |
https://www.physicsforums.com/threads/differential-equation.430596/ | 1,521,334,193,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645405.20/warc/CC-MAIN-20180317233618-20180318013618-00270.warc.gz | 849,013,107 | 15,292 | # Differential equation
1. Sep 20, 2010
### ProPatto16
1. The problem statement, all variables and given/known data
Solve the differential equation 2y(1 + x2 )y'+ x(1 + y2 ) = 0 where y = 2 when x = 0.
a. (1 + x2 )(1 + y2 )2= 0
b. (1 + x2 )2 (1 + y2 ) = 25
c. (1 + x2 )2 (1 + y2 )2= 0
d. (1 + x2 )(1 + y2 ) = 0
e. (1 + x2 )(1 + y2 )2= 25
3. The attempt at a solution
2y(1+x2)dy/dx+x(1+y2)=0
2y(1+x2)dy/dx=-x(1+y2)
2y/(1+y2)dy = -x/(1+x2)dx
if i integrate these i end up with ln in the equations. so i dunno how to do it so it relates to the given answers?
2. Sep 20, 2010
### phyzguy
Exponentiate both sides to get rid of the log, and you should get one of the answers.
3. Sep 20, 2010
### ProPatto16
if i keep going from where i left off.. after integrating you get:
ln(1+y2) = - ln(1+x2)/2
then taking off log of both sides leaves
(1+y2) = -(1+x2)/2
take all onto one side gives:
2(1+y2)+(1+x2)
where did i go wrong?
4. Sep 20, 2010
### HallsofIvy
If you are given answers to select from, you don't need to integrate at all. Just put each of the possible answer into the equation and see if it satisfies it!
"Taking the log off" both sides of $ln(1+y^2)= ln(1+ x^2)/2$ does NOT give "$1+ y^2= (1+ x^2)/2$: the "2" in the denominator is not inside the logarithm. Instead, write it as $ln(1+ y^2)= ln((1+ x^2)^{1/2})$ so that $1+ y^2= (1+ x^2)^{1/2}$ or $(1+ y^2)^2= 1+ x^2$.
5. Sep 20, 2010
### ProPatto16
haha i have to show how i got every answer, mathematically. otheriwse yeah id trial and error the questions i couldnt do.
the denominator there is where i thought i went wrong.
i think i know the answer but its that one sign i cant get rid of.
from the start:
2y(1+x2)dy/dx+x(1+y2)=0
2y(1+x2)dy/dx=-x(1+y2)
2y/(1+y2)dy = -x/(1+x2)dx
then integrate
ln(1+y2) = - ln(1+x2)/2
which becomes
2ln(1+y2) = - ln(1+x2)
take 2 and into logarithm
ln((1+y2)2) = - ln(1+x2)
waittt... ifi take -1 into logarithm then i get an inverse funtion, which will get rid of the plus/minus sign...
then sub in y=2 and x=0
Thanks!!!!!! | 799 | 2,049 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-13 | longest | en | 0.852467 |
https://www.javatpoint.com/aptitude/problems-on-numbers-1 | 1,674,946,394,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00199.warc.gz | 863,006,685 | 10,805 | # Aptitude Problems on Numbers Test Paper 1
1) If 30% of a certain number is 12.6, what is the number?
1. 24
2. 42
3. 23
4. 32
Explanation:
Let the required number be x.
So, according to the question:
2) One-fourth of one-third of two-fifth of a number is 15. What will be 40% of that number?
1. 120
2. 180
3. 270
4. 350
Explanation:
Let x be the required number. Then
3) A number exceeds 20% of itself by 40. What is the number?
1. 50
2. 60
3. 80
4. 320
Explanation:
Let the number be x
20% of number = x
According to the question:
x - x = 40
x - = 40
= 40
4 x = 40 *5
x = = 50
4) The difference between two numbers is 14, and their sum is 40. Find out the product of these two numbers?
1. 960
2. 280
3. 351
Explanation:
Let x and y be the numbers. Then x-y = 14 and x+y = 40.
Product of numbers
5) A number, when divided by 4, is reduced by 21. What is the number?
1. 18
2. 20
3. 28
4. 38
Explanation:
Let the number = x
As per the question:
= (x-21)
x = 4x - 84
3x = 84
x =84/3 = 28
Aptitude Problems on Numbers Test Paper 2
Aptitude Problems on Numbers Test Paper 3
Aptitude Problems on Numbers Test Paper 4
Problems on Numbers Concepts | 411 | 1,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-06 | latest | en | 0.877097 |
http://studentsmerit.com/paper-detail/?paper_id=60259 | 1,481,100,488,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542009.32/warc/CC-MAIN-20161202170902-00495-ip-10-31-129-80.ec2.internal.warc.gz | 269,627,125 | 6,918 | #### Details of this Paper
##### The numbers 3^n -1 are never prime (if n 2>= 2), since they are always even.
Description
solution
Question
Question;The numbers 3^n -1 are never prime (if n 2>= 2), since they are always even. However,it sometimes happens that (3^n -1) /2 is prime. For example, (3^3 -1) /2 = 13 is prime.a) If n is even, show that (3^n - 1)/2 is always divisible by 4, so it can never be prime.b)Use a similar argument to show that if n is a multiple of 5 then (3^n - 1)/2 is never aprime.
Paper#60259 | Written in 18-Jul-2015
Price : \$22 | 178 | 563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-50 | longest | en | 0.869832 |
http://www.ck12.org/statistics/Linear-Regression-Equations/?by=all&difficulty=all | 1,448,779,209,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398456289.53/warc/CC-MAIN-20151124205416-00291-ip-10-71-132-137.ec2.internal.warc.gz | 348,631,229 | 17,951 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Linear Regression Equations
## Using TI calculator to find regression equation
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Linear Regression Equations
by CK-12 //basic
Use a calculator to create a scatter plot and to determine the equation of the line of best fit.
1
## Linear Regression Equations S.ID.8
by Christine Corbley //basic
Use a calculator to create a scatter plot and to determine the equation of the line of best fit.
0
## Linear Regression Equations NATE
by Nathan Angell //basic
Use a calculator to create a scatter plot and to determine the equation of the line of best fit.
0
• Video
## Linear Regression Equations Principles
by CK-12 //basic
This video gives more detail about the mathematical principles presented in Linear Regression Equations.
1
• Video
## Linear Regression Equations Examples
by CK-12 //basic
This video shows how to work step-by-step through one or more of the examples in Linear Regression Equations.
0
• Video
## Linear Regression on the TI-84
by CK-12 //basic
Demonstrates how to find the line of best fit using the TI-84 graphing calculator.
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## Solving Word Problems Involving Scatter Plots - Overview
by CK-12 //basic
Overview
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## Solving Word Problems Involving Scatter Plots - Example 1
by CK-12 //basic
Determining the Correct Scatter Plot and Line of Best-Fit for a Real-World Situation
0
• Video
## Solving Word Problems Involving Scatter Plots - Example 2
by CK-12 //basic
Predicting a Specific Real-World Value Given the Line of Best-Fit
0
• Lesson Plan
## Calculating a Linear Regression Line Equation with Technology Lesson Plan
by CK-12 //basic
This lesson plan covers Calculating a Linear Regression Line Equation with Technology and includes Teaching Tips, Common Errors, Differentiated Instruction, Enrichment, and Problem Solving.
0
• Practice
0%
## Linear Regression Equations Practice
by CK-12 //at grade
0
• Critical Thinking
## Linear Regression Equations Discussion Questions
by CK-12 //at grade
A list of student-submitted discussion questions for Linear Regression Equations.
1
## Linear Regression Equations Post Read
by CK-12 //at grade
To reinforce and increase concept comprehension, and to analyze similarities and differences between topics using a Two Column Table.
0
## Linear Regression Equations Pre Read
by CK-12 //at grade
To activate prior knowledge, make personal connections, reflect on key concepts, encourage critical thinking, and assess student knowledge on the topic prior to reading using a Quickwrite.
0
• Real World Application
## Linear Regression Equations
by CK-12 //at grade
Students will create a linear regression equation for the day's DOW Jones Industrial Average. They will then deduce information about the day's stock market using the equation.
0
• Study Guide
## Regression and Correlation
by Danielle Pintz //advanced
covers correlation, correlation coefficients, regression lines, hypothesis testing for regression lines, and multiple linear regression
0
• Flashcards
## Exponential Distributions Flashcards
by CK-12 //at grade
These flashcards help you study important terms and vocabulary from Linear Regression Equations.
0 | 841 | 3,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | longest | en | 0.832015 |
https://math.stackexchange.com/questions/686591/health-insurance-question | 1,571,476,375,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00477.warc.gz | 604,486,609 | 35,594 | # Health Insurance question
An insurance company issued health insurance policies to individuals. The company determined that Y, the number of claims filed by an insured in a year, is a random variable with the following probability function.
$$P(Y=y)=0.45(0.55)^{y},\;\;\; y=0,1,2,3,\ldots$$ The number of claims filed by one insured individual is independent of the number of claims filed by any other insured individual.
An actuary studied three randomly selected insured individuals from this group of individuals who purchased health policies from this company. What is the probability that these three insured individuals will file more than 6 claims in a year?
My thinking was $$1-P(Y=0)-P(Y=1)-P(Y=2)-\dotsb-P(Y=6)$$ and than raising to the third power but this doesn't give me the answer. What am I missing?
• It sounds to me as though the question is asking for the probability that the three people file a total of six claims in a year. Have you already tried calculating that (and seeing if that gives the answer)? – dmk Feb 23 '14 at 1:02
• Also, I just want to make sure: Is that p.d.f. correct? According to that definition, $P(Y = 0) = 0$; usually they set aside the values whose probability is $0$. (And anyway, if you take out that $y$, it looks sorta like a geometric distribution...) – dmk Feb 23 '14 at 1:16
• @dmk it is a geometric distribution with support 0,1,2,3... There should be a space between the $^y$ and the $y$. I'll edit... – TooTone Feb 23 '14 at 1:18
• sorry the $y$ is to indicate what the values of $y$ can be and is not part of the p.d.f – adam Feb 23 '14 at 1:18
• @dmk are you going to answer the question? (I started typing s.thing out but am happy to wait) – TooTone Feb 23 '14 at 1:19
Here's another way; it's the opposite of elegant, but if brute force'll do it, why not try?
If the question is, indeed, asking for the probability that the three customers make a total of more than six claims, you can find
$$1 - P(\text{They file at most six claims)} = 1 - P(Y_1 = a)P(Y_2 = b)P(Y_3 = c)$$
where $Y_i$ has p.d.f. $f(y) = 0.45(0.55)^y$ and $0 \leq a + b + c \leq 6$.
At first glance, this would seem enormously tedious. There are a lot of possible strings for $a, b, c$, and order matters. On further inspection, though, it turns out to be only sorta kinda tedious. Consider two possibilities: Let $A$ be the event in which $Y_1 = 4, Y_2 = 1, Y_3 = 1$, and $B$ the event in which $Y_1 = 0, Y_2 = 6, Y_3 = 0$. Then
\begin{aligned} \ P(A) &= P(Y_1 = 4)P(Y_2 = 1)P(Y_3 = 1) \\ \ &= (0.45)(0.55)^4(0.45)(0.55)(0.45)(0.55) \\ \ &= (0.45)^3(0.55)^6 \\ \end{aligned}
and
\begin{aligned} \ P(B) &= P(Y_1 = 0)P(Y_2 = 6)P(Y_3 = 0) \\ \ &= (0.45)(0.55)^0(0.45)(0.55)^6(0.45)(0.55)^0 \\ \ &= (0.45)^3(0.55)^6 \\ \end{aligned}
It's not too much of a stretch to see that any string $a, b, c$ such that $a + b + c = 6$ has the same probability. How many such strings are there? For $a + b + c = 6$, I get $28$. The number of strings summing to $5$ (or any number), of course, will be different, but the observation that each event corresponding to a given sum has the same probability will still hold. Therefore, you should be able to get something of the form
$$1 - (0.45)^3\left[\phi\cdot(0.55)^6 + \epsilon\cdot(0.55)^5 + \delta\cdot(0.55)^4 + \gamma\cdot(0.55)^3 + \beta\cdot(0.55)^2 + \alpha\cdot(0.55) + 1 \right]$$
where $\phi$ is the number of strings adding to $6$, $\epsilon$ the number of strings adding to $5$, etc. (Of course, there's only one way for no one to make a claim.) Using this method I get an answer of $0.1495$.
This is not the best way to do it — it's been a while since I've seen a negative binomial distribution — but it should show you that you can do stuff with very limited information. (And that you can probably do that stuff in something like ten minutes.)
If someone could comment on the combinatorial aspect of this, I'd be grateful. That is, where does $28$ come from? I have a suspicion it's $\binom{6 + 3 - 1}{3 - 1}$ — and if we simplify, the triangular numbers come in throughout my solution — but regardless of how much informal study I've done (probably not that much), I've never gotten comfortable with the twelvefold way.
My thinking was $1-P(Y=0)-P(Y=1)-P(Y=2)...P(Y=6)$ and than raising to the third power but this doesn't give me the answer. What am I missing?
What you calculated is the probability that $P(Y_1 > 6 \cap Y_2 > 6 \cap Y_3 > 6)$, i.e. the probability that all of them filed more than 6 claims.
What the question probably wants is the probability that between them they filed more than 6 claims. Note that you have a geometric distribution with pmf $(1-p)^yp$ and cdf $1 - (1-p)^y$ with $p=0.45$, i.e. $Y_i\sim G_0(p)$ (this would have saved you looking at individual probabilities in your question because you could have used the known cdf.)
For a random variable $Y=Y_1 + Y_2 + Y_3$ that is the sum of independent geometric random variables you can use the negative binomial distribution. I.e., $Y\sim \mathrm{NB}(3,p)$.
Edit: If indeed the solution looks for total claims made by a randomly selected three people with n=3, then the following procedure will get you to a closer answer. Sorry for the initial misunderstanding.
The probability of a claim from an insured looks like geometric distribution with p =0.45 and q = 0.55
$$E(Y) = \frac{(1-p)}{p} = \frac{(1-0.45)}{.45} = 1.222$$
Standard Deviation = $$\sqrt{\frac{q}{p^{2}}} = \sqrt{\frac{.55}{.45^{2}}} = 1.648$$
Using normal approximation with continuity correction, $Mean = 3*1.222 =$3.667
Standard Deviation = $1.648*sqrt(3) = 2.8544$
$$P(Y>6) = 1-P(Y<=6) = 1-P(z<\frac{(6.5-3.667)}{2.8544})=1- 0.83952316 = 0.16047684$$
Satish
• First, I've already butchered this one once tonight, so maybe I should just quit now. But second, I'm not sure you should use the central limit theorem when $n=3$. Third, if you are going to use it, then you want to subtract $n\mu = 20/3$ in the numerator, not $\mu = 20/9$, and divide by $\sqrt{n}\sigma$. Finally, to approximate a discrete variable (again, usually with much larger $n$), you take $P(X \leq 6) = P(X \leq 6.5)$ and then standardize. Going through these steps (and again, I don't think any of them apply in this case), I get $0.5233$ for the probability. – dmk Feb 23 '14 at 3:11
• Normal Approximation to the Negative Binomial Distribution Suppose now that Yk has the negative binomial distribution with trial parameter k and success parameter p. Then $$Z_k=(pY_{k}−k)/\sqrt{(k(1−p)}$$ In the negative binomial experiment, vary k and p and note the shape of the probability density function. With k=5 and p=0.4, run the experiment 1000 times and note the agreement between the empirical density function and the true probability density function. – Satish Ramanathan Feb 23 '14 at 3:47
• I agree with most of your points and have corrected my solution to take care of it and has posted some information that I researched which is not new to you I suppose. Here in this example k = 5 and in our case, it is k = 3. You will see the closeness of our solution. Sorry for relying on the mind than to refer the notes. – Satish Ramanathan Feb 23 '14 at 3:50
• math.uah.edu/stat/sample/CLT.html. This is what I as referring to. Let me know if I could be of help. – Satish Ramanathan Feb 23 '14 at 3:53 | 2,226 | 7,328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-43 | latest | en | 0.940407 |
https://math.stackexchange.com/questions/895471/confounding-lagrange-multiplier-problem | 1,628,104,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00014.warc.gz | 398,675,041 | 41,060 | # Confounding Lagrange multiplier problem
Optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$
According to the theorem the gradients must be parallell, $\nabla f = \lambda \nabla g$, so their cross product must equal zero i.e. $\nabla f \times \nabla g = \mathbf{0}$. This results in the following system of equations:
$$\begin{cases} 3y(2y+3x) = 5z(x+2z) \\ 5z(y + 3z) = 4x(2y + 3x) \\ 4x(x+ 2z) = 3y(y+3z) \end{cases}$$
A system of equations which got me stumped, as I am unable to derive anything useful/sensible from it. Yes I did attempt to incorporate the constraint, albeit unsuccessfully.Now since Lagrange multipliers are failing me (?), I have no clue as to how I can progress. Any help?
• That's not the typical route one takes when applying Lagrangian multipliers. Instead, equate the components of $\nabla f$ and $\lambda \nabla g$ while keeping in mind the $g(x,y,z)=6$ constraint. – Semiclassical Aug 12 '14 at 19:55
• That yields the same set of equations :/ – thelionkingrafiki Aug 12 '14 at 20:31
Let $$L(x,y,z,\lambda)=f(x,y,z)-\lambda[g(x,y,z)-6]$$ Then your optimization problem is: $$\left\{\begin{array}{l}0\stackrel{!}{=}\frac{\partial L}{\partial x}=8x-\lambda(y+3z)\\0\stackrel{!}{=}\frac{\partial L}{\partial y}=6y-\lambda(x+2z)\\0\stackrel{!}{=}\frac{\partial L}{\partial z}=10z-\lambda(2y+3x)\\0\stackrel{!}{=}\frac{\partial L}{\partial \lambda}=xy+2yz+3xz-6\end{array}\right.$$ where the first three equations are equivalent to $\nabla f=\lambda\nabla g$, and the last one to the condition $g(x,y,z)=6$. This gives you a much simpler system of equations than the one you proposed above. Can you take it from here?
• Thanks for your reply, but I am afraid that led me to the same set of equations as before. Care to share a clue of some sort? Sidenote: What's up with the exclamation marks above the equal signs? – thelionkingrafiki Aug 12 '14 at 20:32
• @thelionkingrafiki: It only leads you to the same equations if you eliminate $\lambda$ first. Don't! Try to eliminate the terms $xy$, $yz$, $zx$ instead. – Semiclassical Aug 12 '14 at 20:37
• Semiclassical: Oh ok! I've been trying for a while and I am not getting anywhere. It's that $\lambda$ that confuses me as to how I am to eliminate those terms... – thelionkingrafiki Aug 12 '14 at 21:16
• Any help anyone? – thelionkingrafiki Aug 13 '14 at 16:13
To continue from Daniel's answer, add the first three equations together to get:
$$(2-\lambda)(4x+3y+5z)=0$$
This gives two cases:
$$\lambda=2\\ 4x+3y+5z=0$$
Using $\lambda=2$, we can easily solve the others: $x=\pm 1, y=\pm 1, z=\pm 1$. This gives a minimum.
The condition $4x+3y+5z=0$ will lead to contradiction from the first three equations.
• I have made a small edit to correct evident typoes in this note. – colormegone Jul 26 '15 at 7:48
An interesting challenge arises in working with the Lagrange-multiplier method in that all but the simplest problems (generally those involving linear functions and linear constraint equations) can easily lead to systems of multi-linear or non-linear equations, which often do not have general methods of solution. I find that one must draw on an assortment of possible approaches in looking for reasonable routes to a solution.
This problem turns out not to be one where we would solve the set of "Lagrange equations" for expressions for $\ \lambda \$ , which we could then equate and proceed from there. The system that Daniel Robert-Nicoud presents can be arranged in this fashion, which is suggestive:
$$\begin{array}{cccc}8 \ x&- \ \lambda \ y& - \ 3\lambda \ z& = 0\\\lambda \ x& \ -6 \ y& + \ 2\lambda \ z& = 0\\ 3\lambda \ x&+ \ 2 \lambda \ y& \ - 10 \ z& = 0\\\end{array} \ \ .$$
We know that the solution to this system will be non-trivial if the determinant of the coefficients is singular. This is reminiscent of looking for eigenvalues, but we are instead solving the determinant polynomial $\ -12 \ \lambda^3 \ - \ 96 \ \lambda^2 \ + \ 480 \ = \ 0 \$ for values of the Lagrange-multiplier that will allow the values of the variables to be other than all zeroes. We find $\ \lambda \ = \ 2, \ -5 \ \pm \sqrt{5} \$ .
For $\ \lambda \ = \ 2 \$ , we obtain the result indicated by KittyL , $\ x \ = \ y \ = \ z \$ . For the other values of the multiplier, we find (with the acknowledged help of WolframAlpha)
$$\lambda \ = \ -5 \ + \ \sqrt{5} \ \ : \ \ y \ = \ \frac{(5 - \sqrt{5}) \ (10 - 3 \sqrt{5})}{10 \ (2 \sqrt 5 - 3)} \ x \ \approx \ 0.61803 \ x \ \ , \ \$$ $$z \ = \ -\frac{(5 - \sqrt{5}) \ (4 + \sqrt{5})}{10 \ (2 \sqrt 5 - 3)} \ x \ \approx \ -1.17082 \ x \ \ ;$$
$$\lambda \ = \ -5 \ - \ \sqrt{5} \ \ : \ \ y \ = \ -\frac{(5 + \sqrt{5}) \ (10 + 3 \sqrt{5})}{10 \ (2 \sqrt 5 + 3)} \ x \ \approx \ -1.61803 \ x \ \ , \ \$$ $$z \ = \ \frac{(5 + \sqrt{5}) \ (4 - \sqrt{5})}{10 \ (2 \sqrt 5 + 3)} \ x \ \approx \ 0.17082 \ x \ \ .$$
Inserting the result for $\ \lambda \ = \ 2 \$ into the constraint equation produces the simple result $\ 6 \ x^2 \ = \ 6 \ \ \Rightarrow \ \ x \ = \ y \ = \ z \ \pm 1 \$ , for which the value of the function at either point is $\ 4 \ + \ 3 \ + \ 5 \ = \ 12 \$ .
Is there anything else worth mentioning? If we apply the values of the variables for the $\ \lambda \ = \ -5 \ + \ \sqrt{5} \$ case, the constraint equation yields (and I hope I may be forgiven using the decimal approximations)
$$0.61803 \ x^2 \ + \ 2 \ (0.61803) \ (-1.17082) \ x^2 \ + \ 3 \ (-1.17082) \ x^2 \ \approx \ 6$$ $$\Rightarrow \ \ -4.34164 \ x^2 \ = \ 6 \ \ ,$$
which admits no solution. Finally, the third value $\ \lambda \ = \ -5 \ - \ \sqrt{5} \$ leads to
$$-1.61803 \ x^2 \ + \ 2 \ (-1.61803) \ (0.17082) \ x^2 \ + \ 3 \ (0.17082) \ x^2 \ \approx \ 6$$ $$\Rightarrow \ \ -1.65836 \ x^2 \ = \ 6 \ \ .$$
So there is only the one pair of extremal points $\ (1, \ 1, \ 1) \$ and $\ (-1, \ -1, \ -1 ) \$ , and it is easily checked (even just by trying simple integer triples that solve the constraint equation) that the value found for the function is minimal. Because the function and the constraint equation have symmetry about the origin, it would be expected that extremal solutions of the same value would be found in pairs observing this symmetry.
[I reported this approach because it does prove useful in solving such systems of equations. This one had somewhat complicated values for the multiplier, so it is not simply to work through entirely "by hand". (I also thought I had found another extremum of interest, but in typing this out, I uncovered a sign-error in a late stage of my work...) It is also curious that decimal approximations of the "golden ratio" appear; I've no idea whether this is just coincidental. ] | 2,129 | 6,690 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-31 | latest | en | 0.821091 |
http://www.alecjacobson.com/weblog/?tag=maple | 1,513,414,886,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00250.warc.gz | 313,001,986 | 11,084 | ## Posts Tagged ‘maple’
### Convincing maple to solve an ODE with Neumann conditions at a symbolic valued location
Friday, November 17th, 2017
I can use maple to solve a 1D second-order ODE with Dirichlet boundary conditions at symbolic-valued locations:
# Z'' = 0, Z(a)=0, Z(b) = 1
dsolve({diff(Z(r),r,r) = 0,Z(a)=0,Z(b)=1});
This correctly returns
r a
Z(r) = - ----- + -----
a - b a - b
I can also easily convince maple to solve this ODE with some Neumann (normal derivative) boundary conditions at at fixed-value, numeric location:
# Z'' = 0, Z(a) = 1, Z'(0) = 0
dsolve({diff(Z(r),r,r) = 0,Z(a)=1,eval(diff(Z(r),r),r=0)=0});
produces
Z(r) = 1
But if I try naively to use a Neumann condition at a symbolic value location
# Z'' = 0, Z(a) = 1, Z'(b) = 0
dsolve({diff(Z(r),r,r) = 0,Z(a)=1,eval(diff(Z(r),r),r=b)=0});
then I get an error:
Error, (in dsolve) found differentiated functions with same name but depending on different arguments in the given DE system: {Z(b), Z(r)}
After a long hunt, I found the solution. dsolve takes an optional second argument that can tell it what the dependent variable actually is. So the correct call is:
# Z'' = 0, Z(a) = 1, Z'(b) = 0
dsolve({diff(Z(r),r,r) = 0,Z(a)=1,eval(diff(Z(r),r),r=b)=0});
and this gives the correct answer
Z(r) = 1
### Energy optimization, calculus of variations, Euler Lagrange equations in Maple
Tuesday, August 16th, 2016
Here’s a simple demonstration of how to solve an energy functional optimization symbolically using Maple.
Suppose we’d like to minimize the 1D Dirichlet energy over the unit line segment:
min 1/2 * f'(t)^2
f
subject to: f(0) = 0, f(1) = 1
we know that the solution is given by solving the differential equation:
f''(t) = 0, f(0) = 0, f(1) = 1
and we know that solution to be
f(t) = t
How do we go about verifying this in Maple:
with(VariationalCalculus):
E := diff(f(t),t)^2:
L := EulerLagrange(E,t,f(t)):
so far this will output:
L := {-2*diff(diff(x(t),t),t), -diff(x(t),t)^2 = K[2], 2*diff(x(t),t) = K[1]}
Finally solve with the boundary conditions using:
dsolve({L[1],f(0)=0,f(1)=1});
which will output
t(t) = t
### Inverse of common ease curves
Wednesday, July 13th, 2016
Awkwardly I ended up needing the inverse of an ease curve (or S-curve) used in tweening animations. For trigonometric ease curves, this is easy. For example, if your ease filter is:
f = 0.5-cos(x*pi)*0.5;
then the inverse is:
x = acos(-2*(f-0.5))/pi
But if you’re using the famous cubic ease curve,
f = 3.*x.^2 - 2.*x.^3
then the relevant inverse is a complex function (involving the i = sqrt(-1)) that produces real values for f in [0,1]:
x = -1./4.*(1-2.*f+2.*(f.^2-f).^(1./2)).^(1./3)-1./4./(1-2.*f+2.*(f.^2-f).^(1./2)).^(1./3)+1./2-1./2.*i.*3.^(1./2).*(1./2.*(1-2.*f+2.*(f.^2-f).^(1./2)).^(1./3)-1./2./(1-2.*f+2.*(f.^2-f).^(1./2)).^(1./3));
I haven’t bothered to see if this can be simplified into something tidy. It’d be great to get rid of the i.
### Uninstalling Maple toolbox breaks matlab app
Tuesday, November 12th, 2013
I tried to uninstall the maple matlab toolbox via Uninstall Maple 17.app. This seem to execute OK, but then when I tried to launch Matlab the app immediately exited. Investigating the console I saw that it was producing an error:
2013-11-12 19:23:22.290 StartMATLAB[494:707] execl("/Applications/MATLAB_R2012a.app/Contents/MacOS/../../bin/matlab", "/Applications/MATLAB_R2012a.app/Contents/MacOS/../../bin/matlab", "-desktop", "-arch=maci64", nil) failed! errno: 13
Upon closer inspection it seems that the matlab file existed but it’s permissions were not setup for executing. I fixed this by issuing:
sudo chmod a+x /Applications/MATLAB_R2012a.app/Contents/MacOS/../../bin/matlab
My guess is that maple’s script tried to replace matlab with a backup but forgot to recent the permissions.
### atan2 is harmonic
Thursday, August 22nd, 2013
Don’t you forget it. For the lazy among us, here’s a maple proof:
simplify(diff(arctan(y,x),x,x)+diff(arctan(y,x),y,y),size);
produces:
0
### Get values of solve variables as list in maple
Wednesday, August 21st, 2013
I issue a solve command in maple like this one:
s:= solve({2*x1-x2=0,x1+x2=1},{x1,y2});
and I see as output:
s := {x1 = 1/3, x2 = 2/3}
I’d like to get the results as an ordered list. I’ve already carefully chosen my variable names so that they’re ordered lexicographically, which is respected by the solve output. I tried to use convert, table, entries, indices to no avail. What I came up with is:
convert(map(rhs,s),list);
which produces:
[1/3, 2/3]
This can be easily dumped into matlab for processing.
Note: If you use square brackets in your solve:
s:= solve({2*x1-x2=0,x1+x2=1},[x1,x2]);
Then the map above will give you an error like this one:
Error, invalid input: rhs received [x1 = 1/3, x2 = 2/3], which is not valid for its 1st argument, expr
You need to get the first index of the solve output:
map(rhs,s[1]);
### Minimize a quadratic energy with linear equality constraints symbolically in maple
Wednesday, August 21st, 2013
I’m working with Maple to fit a polynomial to specific values and derivatives so that it minimized a quadratic energy. I tried the Optimization toolbox/package and came up with this:
with(Optimization):
p := (t) -> c3*t^3 + c2*t^2;
C1 := eval(p(t),t=1)-1;
C2 := eval(diff(p(t),t),t=1);
m := Minimize(int(diff(p(t),t)^2,t=0..1),{C1=0,C2=0});
but this seems to give me a numerically optimized solution:
m := [1.19999999999997, [c2 = 3., c3 = -2.]]
Rather I want a symbolically optimized solution. So far, I can do this by explicitly computing the Lagrangian and finding the saddle point:
with(VectorCalculus):
Lambda := int(diff(p(t),t)^2,t=0..1) + lambda_1*(C1) + lambda_2*(C2);
This gives me the exact solution:
[[c3 = -2, c2 = 3, lambda_1 = -12/5, lambda_2 = 1/5]]
### Trilinear interpolation in Maple
Friday, July 19th, 2013
Here’s a maple function for trilinear interpolation in a cube. I tried to match the notation in the wikipedia entry. I define C = (x,y,z) where x grows to the right, y grows up and z grows into the screen
tri := (x,y,z,C000,C001,C010,C011,C100,C101,C110,C111) ->
y*(z*(x*C000+(1-x)*C100)+(1-z)*(x*C010+(1-x)*C110)) +
(1-y)*(z*(x*C001+(1-x)*C101)+(1-z)*(x*C011+(1-x)*C111));
### Areas of quadrilaterals within a triangle determined by its circumcenter (or perpendicular bisectors), using matlab
Thursday, February 11th, 2010
First, I need to locate the circumcenter of the triangle. The circumcenter is the center of the circle circumscribing the three points of the triangle. Thus, it is equidistance from all three points.
I do not need the actual world coordinates of the circumcenter, rather it is okay for me to just know the relative distance to each corner. The law of cosines tells me that the relative distance of the circumcenter from the line opposite point A (called a in the picture) is cos A (that is cosine of the angle at A). Likewise for B and C. Knowing these I can convert to barycentric coordinates by multiplying by the opposite edge lengths.
cosines = [ ...
(edge_norms(:,3).^2+edge_norms(:,2).^2-edge_norms(:,1).^2)./(2*edge_norms(:,2).*edge_norms(:,3)), ...
(edge_norms(:,1).^2+edge_norms(:,3).^2-edge_norms(:,2).^2)./(2*edge_norms(:,1).*edge_norms(:,3)), ...
(edge_norms(:,1).^2+edge_norms(:,2).^2-edge_norms(:,3).^2)./(2*edge_norms(:,1).*edge_norms(:,2))];
barycentric = cosines.*edge_norms;
normalized_barycentric = barycentric./[sum(barycentric')' sum(barycentric')' sum(barycentric')'];
A quality of the barycentric trilinear coordinates of a point in a triangle is that they specify the relative area of inner triangles form by connecting the corners to that point. In our case, the triangles made by connecting the circumcenter to each corner. Here, the aquamarine, red, and purple triangles corresponding to points A, B, and C.
Now I find the area of the triangle and then multiple against the normalized barycentric coordinates of the circumcenter to get the areas of each of these inner triangles:
areas = 0.25*sqrt( ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(-edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)));
partial_triangle_areas = normalized_barycentric.*[areas areas areas];
Finally the quadrilateral at each point is built from half of the neighboring inner triangles: the green, blue and yellow quads corresponding to A, B, and C.
Note, that the fact that these inner triangles are cut perfectly in half follows immediately from the fact the the circumcenter is equidistance from each corner. So the last step is just to add up the halves:
quads = [ (partial_triangle_areas(:,2)+ partial_triangle_areas(:,3))*0.5 ...
(partial_triangle_areas(:,1)+ partial_triangle_areas(:,3))*0.5 ...
(partial_triangle_areas(:,1)+ partial_triangle_areas(:,2))*0.5];
Note: I’ve suffered trying to use mathematic or maple to reduce these steps to a simple algebra formula (I must be mistyping something). As far as I can tell the above steps check out (of course, it only makes sense for cases where the circumcenter is within or on the triangle: i.e. the triangle is not obtuse). I have checked out my equations on maple with the following functions:
mycos:=(a,b,c)->(b^2+c^2-a^2)/(2*b*c);
barycentric:=(a,b,c)->a*mycos(a,b,c);
normalized_barycentric:=(a,b,c)->barycentric(a,b,c)/(barycentric(a,b,c)+barycentric(b,c,a)+barycentric(c,a,b));
area:=(a,b,c)->sqrt((a+b-c)*(a-b+c)*(-a+b+c)*(a+b+c))/4;
inner_triangle:=(a,b,c)->normalized_barycentric(a,b,c)*area(a,b,c);
Which then you can run:
To see a beautiful long algebra formula for the quadrilateral at point A.
Which in matlab corresponds to:
[-sqrt(-(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3))).* ...
(2.*edge_norms(:,2).^2.*edge_norms(:,3).^2 + edge_norms(:,1).^2.* ...
edge_norms(:,2).^2 - edge_norms(:,2).^4 + edge_norms(:,1).^2.* ...
edge_norms(:,3).^2 - edge_norms(:,3).^4)./ ...
(8.*(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3))), ...
-sqrt(-(edge_norms(:,2) + edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) - edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) + edge_norms(:,3) - edge_norms(:,1))).* ...
(2.*edge_norms(:,3).^2.*edge_norms(:,1).^2 + edge_norms(:,2).^2.* ...
edge_norms(:,3).^2 - edge_norms(:,3).^4 + edge_norms(:,2).^2.* ...
edge_norms(:,1).^2 - edge_norms(:,1).^4)./ ...
(8.*(edge_norms(:,2) + edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) - edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) + edge_norms(:,3) - edge_norms(:,1))), ...
-sqrt(-(edge_norms(:,3) + edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) - edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) + edge_norms(:,1) - edge_norms(:,2))).* ...
(2.*edge_norms(:,1).^2.*edge_norms(:,2).^2 + edge_norms(:,3).^2.* ...
edge_norms(:,1).^2 - edge_norms(:,1).^4 + edge_norms(:,3).^2.* ...
edge_norms(:,2).^2 - edge_norms(:,2).^4)./ ...
(8.*(edge_norms(:,3) + edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) - edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) + edge_norms(:,1) - edge_norms(:,2)))];
Note: It seems also that the wikipedia page for the circumscribed circle has a direct formula for the barycentric coordinates of the circumcenter. Though, wolfram’s mathworld has an ostensibly different formula. Upon a quick review Wolfram’s is at least easily verifiable to be the same as mine (recall these only need to be relative to each other). | 3,796 | 12,304 | {"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} | 3.34375 | 3 | CC-MAIN-2017-51 | latest | en | 0.825661 |
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# Have You Learnt Something New Today? If Not LOOK HERE!
8y, 4m agoPosted 8 years, 4 months ago
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All pyramids were bright white.
An owl cannot move it's eyeballs in its eye sockets.
8y, 4m agoPosted 8 years, 4 months ago
Options
(18)
#1
Wow thats amazing stuff man :o
#2
It takes more calories to eat a piece of celery than the celery has in it to begin with.
False
#3
Weird information there
#4
Kick
False
True! You burn more energy eating it, than it contains!
#5
:?
Never knew that....
banned#6
# Penguins can convert salt water into fresh water.
# It takes a ton of water to make 1 lb. of sugar.
# A snapping turtle can only swallow when it's head is under water.
# A rat can go longer without water than a camel.
# A gallon of water weighs 8.34 lbs.
# Elephants can smell water up to 3 miles away.
# A jelly fish is 95% water.
# Water expands about 9% as it freezes.
#7
sam2007
# Penguins can convert salt water into fresh water.
# It takes a ton of water to make 1 lb. of sugar.
# A snapping turtle can only swallow when it's head is under water.
# A rat can go longer without water than a camel.
# A gallon of water weighs 8.34 lbs.
# Elephants can smell water up to 3 miles away.
# A jelly fish is 95% water.
# Water expands about 9% as it freezes.
Some nice water facts there, i'll be sure to tell my fish! :p
#8
Wilko_Mat
True! You burn more energy eating it, than it contains!
It's false I tell you, these foods are called negative calorie foods, the only true negative calorie food is iced water, since your body needs to heat it up first before it can be absorbed.
#9
Kick
False
Wilko_Mat
True! You burn more energy eating it, than it contains!
Kick
It's false I tell you, these foods are called negative calorie foods, the only true negative calorie food is iced water, since your body needs to heat it up first before it can be absorbed.
The common misconception is that celery contains "negative calories". All foods contain calories. The way you must look at it is how many calories does one stalk of celery contain, versus how many calories it takes to digest it. According to [url]www.nutritiondata.com[/url], one 8 inch celery stalk contain about 6 calories. Many people believe you actually burn that off when chewing. Though celery does take a rather conscious effort to chew, chewing burns about the same amount as watching the grass grow. The calorie loss is going to be during the digestion of the celery.
Celery is actually loaded with latent energy, but the plant's composition only allows us to metabolize a small amount. The latent energy is packed tightly in the form of cellulose; a complex sugar which humans are unable to metabolize. It's not that the celery is difficult to digest, but rather it provides fewer calories than what is needed to digest it - so the overall effect is "negative calories".
[FONT="Comic Sans MS"][SIZE="3"][COLOR="DarkOrchid"]So you're both right - unfortunately it's not true when you dunk it in mayonnaise! :)[/COLOR][/SIZE][/FONT]
#10
To quote James Collier a Nutrition Consultant.
Yes negative calories food exist in theory but not in practice.
From a normal digestion point of view, ie our enzymes, the cost of digestion is indeed greater than the calorific value of the food consumed, thus negative calories. BUT some of the bacterial flora in our gut feed on the prebiotic fibres (FOS) in these foods, and break it down into sacchraides which we can disgest and yield energy from, therefore this roughly equals the energy cost of digestion. So there are no negative energy foods.
These foods are encourgaed though as prebiotics are crucial for a healthy digestive system
banned#11
An oyster can change its sex a number of times during its life.
Lucky thing.
#12
Kick
To quote James Collier a Nutrition Consultant.
So it is true that it takes more calories to eat a piece of celery than the celery has in it to begin with. As you spend an age arguing over it, wasting a lot of enery typing and using wikipedia.:whistling: And niether of you ate celery, tis a magical food indeed.
#13
And a few more :thumbsup:
[LIST]
[*]1- The liquid inside young coconuts can be used as a substitute for Blood plasma.
[*]2- No piece of paper can be folded in half more than seven (7) times.
[*]3- Donkeys kill more people annually than plane crashes.
[*]4- You burn more calories sleeping than you do watching television.
[*]5- Oak trees do not produce acorns until they are fifty (50) years of age or older.
[*]6- The first product to have a bar code was Wrigley's gum.
[*]7- The King of Hearts is the only king WITHOUT A MOUSTACHE
[*]8- American Airlines saved \$40,000 in 1987 by eliminating one (1) olive from each salad served in first-class.
[*]9- Venus is the only planet that rotates clockwise. (Since Venus is normally associated with women, what does this tell you!)
[*]10- Apples, not caffeine, are more efficient at waking you up in the morning.
[*]12- The first owner of the Marlboro Company died of lung cancer. So did the first " Marlboro Man. "
[*]13- PEARLS MELT IN VINEGAR!
[*]14- The three most valuable brand names on earth: Marlboro, Coca Cola, and Budweiser, in that order.
[*]15- It is possible to lead a cow upstairs... but, not downstairs.
[*]16- A duck's quack doesn't echo, and no one knows why.
[*]17- Dentists have recommended that a toothbrush be kept at least six (6) feet away from a toilet to avoid airborne particles resulting from the flush.
[*]And the best for last.....
[/LIST]
Turtles can breathe through their bums.
(I know some people like that, don't YOU?) :oops:
banned#14
Bikebarbie
And a few more :thumbsup:
[LIST]
[*]1- The liquid inside young coconuts can be used as a substitute for Blood plasma.
[*]2- No piece of paper can be folded in half more than seven (7) times.
[*]3- Donkeys kill more people annually than plane crashes.
[*]4- You burn more calories sleeping than you do watching television.
[*]5- Oak trees do not produce acorns until they are fifty (50) years of age or older.
[*]6- The first product to have a bar code was Wrigley's gum.
[*]7- The King of Hearts is the only king WITHOUT A MOUSTACHE
[*]8- American Airlines saved \$40,000 in 1987 by eliminating one (1) olive from each salad served in first-class.
[*]9- Venus is the only planet that rotates clockwise. (Since Venus is normally associated with women, what does this tell you!)
[*]10- Apples, not caffeine, are more efficient at waking you up in the morning.
[*]12- The first owner of the Marlboro Company died of lung cancer. So did the first " Marlboro Man. "
[*]13- PEARLS MELT IN VINEGAR!
[*]14- The three most valuable brand names on earth: Marlboro, Coca Cola, and Budweiser, in that order.
[*]15- It is possible to lead a cow upstairs... but, not downstairs.
[*]16- A duck's quack doesn't echo, and no one knows why.
[*]17- Dentists have recommended that a toothbrush be kept at least six (6) feet away from a toilet to avoid airborne particles resulting from the flush.
[*]And the best for last.....
[/LIST]
Turtles can breathe through their bums.
(I know some people like that, don't YOU?) :oops:
No more turtles around me dinner table
banned#15
I´ve never seen a beaver with teeth.......... Must hurt I guess. :whistling:i
banned#16
toshapetriji
I´ve never seen a beaver with teeth.......... Must hurt I guess. :whistling:i
:w00t::w00t::w00t:
#17
14- The three most valuable brand names on earth: Marlboro, Coca Cola, and Budweiser, in that order.
But Bud got bought out?
#18
The-Dude
So it is true that it takes more calories to eat a piece of celery than the celery has in it to begin with. As you spend an age arguing over it, wasting a lot of enery typing and using wikipedia.:whistling: And niether of you ate celery, tis a magical food indeed.
You'll burn more calories walking to the fridge to get it.
Never used wikipedia. | 2,216 | 8,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.943225 |
http://www.jiskha.com/display.cgi?id=1330711930 | 1,498,288,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320227.27/warc/CC-MAIN-20170624064634-20170624084634-00006.warc.gz | 556,295,454 | 3,956 | # physics
posted by .
Two identical steel balls, each of mass 5.0 kg, are suspended from strings of length 29 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle θ = 25° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?
Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?
• physics -
To conserve both momentum and energy, the first ball stops and the second one continues with the total energy.
In part 2, momentum is conserved but the motion continues with twice the mass.
m g (.29)(1-cos 25) = .5 m v^2
solve for v
then conservation of momentum at bottom
(m+m) * new v = m v
new v = (1/2) v
then at top
g (.29)(1-cos A) = .5 (new v)^2
solve for A
### Related Questions
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Post a New Question | 251 | 990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.912251 |
http://gaworld.ru/factors-of-54 | 1,519,046,710,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00519.warc.gz | 139,165,094 | 20,396 | # Factors of 54
The factors of 54 and the prime factors of 54 differ because fifty-four is a composite number. Also, despite being closely related, the prime factors of 54 and the prime factorization of 54 are not exactly the same either. In any case, by reading on you can learn the answer to the question what are the factors of 54? and everything else you want to know about the topic.
## What are the Factors of 54?
They are: 54, 27, 18, 9, 6, 3, 2, 1. These are all the factors of 54, and every entry in the list can divide 54 without rest (modulo 0). That’s why the terms factors and divisors of 54 can be used interchangeably.
As is the case for any natural number greater than zero, the number itself, here 54, as well as 1 are factors and divisors of 54.
## Prime Factors of 54
The prime factors of 54 are the prime numbers which divide 54 exactly, without remainder as defined by the Euclidean division. In other words, a prime factor of 54 divides the number 54 without any rest, modulo 0.
For 54, the prime factors are: 2, 3. By definition, 1 is not a prime number.
Besides 1, what sets the factors and the prime factors of the number 54 apart is the word “prime”. The former list contains both, composite and prime numbers, whereas the latter includes only prime numbers.
## Prime Factorization of 54
The prime factorization of 54 is 2 x 3 x 3 x 3. This is a unique list of the prime factors, along with their multiplicities. Note that the prime factorization of 54 does not include the number 1, yet it does include every instance of a certain prime factor.
54 is a composite number. In contrast to prime numbers which only have one factorization, composite numbers like 54 have at least two factorizations.
To illustrate what that means select the rightmost and leftmost integer in 54, 27, 18, 9, 6, 3, 2, 1 and multiply these integers to obtain 54. This is the first factorization. Next choose the second rightmost and the second leftmost entry to obtain the 2nd factorization which also produces 54.
The prime factorization or integer factorization of 54 means determining the set of prime numbers which, when multiplied together, produce the original number 54. This is also known as prime decomposition of 54.
Besides factors for 54, frequently searched terms on our website include:
We did not place any calculator here as there are already a plethora of them on the web. But you can find the factors, prime factors and the factorizations of many numbers including 54 by using the search form in the sidebar.
To sum up:
The factors, the prime factors and the prime factorization of 54 mean different things, and in strict terms cannot be used interchangeably despite being closely related.
The factors of fifty-four are: 54, 27, 18, 9, 6, 3, 2, 1. The prime factors of fifty-four are 2, 3. And the prime factorization of fifty-four is 2 x 3 x 3 x 3. Remember that 1 is not a prime factor of 54.
No matter if you had been searching for prime factorization for 54 or prime numbers of 54, you have come to the right page. Also, if you typed what is the prime factorization of 54 in the search engine then you are right here, of course.
Taking all of the above into account, tasks including write 54 as a product of prime factors or list the factors of 54 will no longer pose a challenge to you.
If you have any questions about the factors of fifty-four then fill in the form below and we will respond as soon as possible. If our content concerning all factors of 54 has been of help to you then share it by means of pressing the social buttons. And don’t forget to bookmark us. | 879 | 3,615 | {"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} | 4.5625 | 5 | CC-MAIN-2018-09 | longest | en | 0.947107 |
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Post subject: Maths Question helpPosted: Thu Jul 09, 2015 12:13 pm
Joined: Fri Jun 12, 2015 12:25 pm
Posts: 5
Please can you help us with a simple technique to solve the following questions:
1. A Cuboid's faces have perimeters 10cm, 12cm and 14cm. Find its dimensions?
2. The Goldbach Conjecture states that every integer greater than 2 can be expressed(not uniquely) as the sum of two primes.
For example: 14=3+11 or 7+7
Find as many ways as you can in which 36 can be written as the sum of two primes?
3. A positive whole number less than 100 has remainder 2 when it is divided by 3, remainder 3 when divided by 4 and remainder 4 when divided by 5. What is the number?
Thanks
A
Top
Post subject: Re: Maths Question helpPosted: Thu Jul 09, 2015 12:24 pm
Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11940
I'll start you off with hints.
1) The semi perimeters will be 5cm, 6cm and 7cm - the semi perimeter is the sum of two of the sides of that face.
2) What are the prime numbers? 2, 3, 5, 7, ... (should be known up to at least 20) then try adding
3) Think about the patterns -
remainder 2 when divided by 3 .... 5, 8, 11
remainder 3 when divided by 4 .... 7, 11, 15
remainder 4 when divided by 5 .... 9, 14, 19
can you see when they will all be the same number without listing them? What is the number that is one less than a multiple of 3, 4 and 5 ..
Top
Post subject: Re: Maths Question helpPosted: Thu Jul 09, 2015 8:12 pm
Joined: Fri Jun 12, 2015 12:25 pm
Posts: 5
Sorry, I didn't get any of it. Please can you elaborate on all the answers?
Thanks
A
Top
Post subject: Re: Maths Question helpPosted: Thu Jul 09, 2015 8:20 pm
Joined: Mon Jun 18, 2007 2:32 pm
Posts: 6966
Location: East Kent
Guest 55 gave you some good starters, it's always best to try and work things out, it helps consolidate the knowledge - much better then just remebering by rote
Once you have a list of the first few primes (numbers which have only 2 factors) then you can work out which ones added together make 36
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Post subject: Re: Maths Question helpPosted: Sat Jul 11, 2015 7:09 am
Joined: Sat Aug 10, 2013 11:46 am
Posts: 188
For the first one, imagine a shoe box. One face will be a rectangle with two verticals and two horizontals. So half the perimeter will just be one vertical and one horizontal. Call these a and b. The next face round will also have a vertical and horizontal which is half the perimeter, but the vertical will be the same as before, so call these lengths a and c. The bottom of the box will then be a rectangle with lengths b and c.
a+b=5 and a+c=6, so the box is one cm longer than its width. You could use algebra, but the numbers are so small it would be easier to just see what fits.
The trick to doing the last one is to multiply the numbers together. You then have a number which can be divided into groups of 3, 4 or 5. Imagine we put it into groups of 5 then took one away from just one group. We would be left with a number which had several lots of 5 and one group of remainder 4. What would happen if we divided it instead into groups of 4 then took one away from one of the groups? The quick trick with the last one is to multiply all the numbers together then take one away.
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Post subject: Re: Maths Question helpPosted: Sat Jul 11, 2015 7:34 am
Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11940
I did give further hints by PM at the OP's request.
Top
Post subject: Re: Maths Question helpPosted: Sat Jul 11, 2015 11:03 pm
Joined: Mon Sep 01, 2014 3:46 pm
Posts: 86
Hi There,
Was going thru these questions. Found them really interesting. Can anyone tell me where these questions have been picked up from???
I have not come across any such questions while preparing (Bond, CEM, GL , Letts)
Thanks
Jas
Top
Post subject: Re: Maths Question helpPosted: Sun Jul 12, 2015 8:31 am
Joined: Sat Aug 10, 2013 11:46 am
Posts: 188
Just thought I'd point out that the Goldbach Conjecture is for every even number. (For an odd number, one of the primes must be 2, which leaves Hobson's choice on the other).
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Jump to: Select a forum ------------------ FORUM RULES Forum Rules and FAQs 11 PLUS SUBJECTS VERBAL REASONING MATHS ENGLISH NON-VERBAL REASONING CEM 11 Plus GENERAL GENERAL 11 PLUS TOPICS 11 PLUS APPEALS 11 PLUS TUTORS INDEPENDENT SCHOOLS 11 PLUS CDs/SOFTWARE 11 PLUS TIPS PRIMARY SEN and the 11 PLUS EVERYTHING ELSE .... 11 PLUS REGIONS Berkshire Bexley and Bromley Birmingham, Walsall, Wolverhampton and Wrekin Buckinghamshire Devon Dorset Essex Essex - Redbridge Gloucestershire Hertfordshire (South West) Hertfordshire (Other and North London) Kent Lancashire & Cumbria Lincolnshire Medway Northern Ireland Surrey (Sutton, Kingston and Wandsworth) Trafford Warwickshire Wiltshire Wirral Yorkshire BEYOND 11 PLUS Beyond 11 Plus - General GCSEs 6th Form University | 1,611 | 5,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.94232 |
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# GATE 2023 || Industrial Engineering || Quiz 54
Attempt now to get your rank among 35 students!
Question 1
The cost of providing service in a queuing system increases with
Question 2
If λ and μ are the arrival rate and service rate in a single server queuing model, then the average waiting time in the system is __________.
Question 3
At a production machine, parts arrive according to a Poisson process at the rate of 0.35 parts per minute. Processing time for parts have exponential distribution with mean of 2 min. What is the probability that a random part arrival finds that there are already 8 parts in the system?
Question 4
Customers arrive at a ticket counter at a rate of 50 per hr and tickets are issued in the order of their arrival. The average time taken for issuing a ticket is 1 min. Assuming that customer arrivals form a Poisson process and service times are exponentially distributed, the average waiting time in queue in min is
Question 5
At a work station, 5 jobs arrive every minute. The mean time spent on each job in the work station is 1/8 minute. The mean steady state number of jobs in the system is ______
Question 6
A service facility has Poisson arrival rate and negative Exponential service rate on a first come-first server basis queue. Average requests for service are 5 per day and the rate at which service is provided is 7 per day. Find the probability for exactly 3 machines in the system.
• 35 attempts | 335 | 1,479 | {"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} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.912615 |
https://www.bartleby.com/questions-and-answers/11-randomely-selected-students-participated-in-an-experiment-to-test-their-ability-to-determine-when/bfd00a75-663f-4d21-b966-dcebee1bf583 | 1,575,900,209,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518882.71/warc/CC-MAIN-20191209121316-20191209145316-00174.warc.gz | 611,223,888 | 23,756 | # 11) Randomely selected students participated in an experiment to test their ability to determine when one minute (60 seconds) has passed. Forty students yielded a a sample mean of 58.3 seconds. Assume that σ = 9.5 seconds.What is the best point estimate for the mean for all student times?Construct a 99 % confidence interval estimate for the population mean for all stu- dents.Is it likely that their mean have an estimate that is reasonably close to 60 seconds?12) What sample size is needed to estimate the mean white blood cell count for the population of adults in the U.S? Assume you want a 99% confidence that the sample mean is within .2 of the population mean. The population standard deviation is 2.5.
Question
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11) Randomely selected students participated in an experiment to test their ability to determine when one minute (60 seconds) has passed. Forty students yielded a a sample mean of 58.3 seconds. Assume that σ = 9.5 seconds.
What is the best point estimate for the mean for all student times?
Construct a 99 % confidence interval estimate for the population mean for all stu- dents.
Is it likely that their mean have an estimate that is reasonably close to 60 seconds?
12) What sample size is needed to estimate the mean white blood cell count for the population of adults in the U.S? Assume you want a 99% confidence that the sample mean is within .2 of the population mean. The population standard deviation is 2.5.
check_circle
Step 1
Hey, since there are multiple questions posted, we will answer first question. If you want any specific question to be answered then please submit that question only or specify the question number in your message.
Step 2
It is given that the sample mean is 58.3, sample size n is 40 and population standard deviation is 9.5.
The best point estimate for the mean for all student times is sample mean. Therefore, the best point estimate for all student tim...
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https://edu.glogster.com/glog/atomic-particles/21vqwuvtcm4?=glogpedia-source | 1,601,462,677,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00629.warc.gz | 369,870,596 | 16,064 | # Atomic Particles
In Glogpedia
by LexyLock
Last updated 5 years ago
Discipline:
Science
Subject:
Physics
5,6,7,8,9,10,11
.
Atomic Particles
Protons
+
+
Examples
Neutrons
How many protons, neutrons, electrons, and valence electrons are in each of the above elements?
Sodium: Protons=11, Neutrons=23-11=12, Electrons=11, Valence Electrons=1Carbon: Protons=6, Neutrons=12-6=6, Electrons=6, Valence Electrons=4
1)Protons are positively charged particles.2)Protons are found in the nucleus of an atom, along with neutrons.3)On the periodic table, the atomic number (usually displayed at the top of the square) is equal to the number of protons.4)Protons are shown with a + sign.
1)Neutrons are particles with no charge.2)Neutrons are found in the nucleus of an atom, along with protons.3)The number of neutrons is equal to the mass number (usually displayed at the bottom of the square on the periodic table) minus the atomic number.4)Neutrons can be shown with an x, an N, or a 0.
Electrons
1)Electrons are negatively charged particles.2)Electrons are found in an area of the atom outside of the nucleus called the electron cloud.3)In a stable element, the number of electrons is equal to the number of protons.4)Electrons are shown with a - sign.5)Valence electrons are the number of electrons in the outermost layer of the electron cloud. The number of valence electrons are equal to the last digit in the element's group number.
0
N
-
-
Sodium
6
23
11
Carbon
12
Group 1
Group 14
A model of a helium atom | 427 | 1,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.772613 |
https://en.wikipedia.org/wiki/Seiberg%E2%80%93Witten_gauge_theory | 1,550,469,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484689.3/warc/CC-MAIN-20190218053920-20190218075920-00098.warc.gz | 540,673,552 | 16,145 | Seiberg–Witten theory
(Redirected from Seiberg–Witten gauge theory)
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In theoretical physics, Seiberg–Witten theory is a theory that determines an exact low-energy effective action (for massless degrees of freedom) of a ${\displaystyle N=2}$ supersymmetric gauge theory—namely the metric of the moduli space of vacua.
Seiberg–Witten curves
In general, effective Lagrangians of supersymmetric gauge theories are largely determined by their holomorphic properties and their behavior near the singularities. In particular, in gauge theory with ${\displaystyle N=2}$ extended supersymmetry, the moduli space of vacua is a special Kähler manifold and its Kähler potential is constrained by above conditions.
In the original derivation by Nathan Seiberg and Edward Witten, they extensively used holomorphy and electric-magnetic duality to constrain the prepotential, namely the metric of the moduli space of vacua.
Consider the example with gauge group SU(n). The classical potential is
${\displaystyle V(x)={\frac {1}{g^{2}}}\operatorname {Tr} [\phi ,{\bar {\phi }}]^{2}\,}$
(1)
This must vanish on the moduli space, so vacuum expectation value of φ can be gauge rotated into Cartan subalgebra, so it is a traceless diagonal complex matrix.
Because the fields ${\displaystyle \phi }$ no longer have vanishing vacuum expectation value. Because these are now heavy due to the Higgs effect, they should be integrated out in order to find the effective ${\displaystyle N=2}$ Abelian gauge theory. This can be expressed in terms of a single holomorphic function F.
In terms of this prepotential the Lagrangian can be written in the form:
${\displaystyle {\frac {1}{4\pi }}\operatorname {Im} {\Bigl [}\int d^{4}\theta {\frac {dF}{dA}}{\bar {A}}+\int d^{2}\theta {\frac {1}{2}}{\frac {d^{2}F}{dA^{2}}}W_{\alpha }W^{\alpha }{\Bigr ]}\,}$
(3)
${\displaystyle F={\frac {i}{2\pi }}{\mathcal {A}}^{2}\operatorname {\ln } {\frac {{\mathcal {A}}^{2}}{\Lambda ^{2}}}+\sum _{k=1}^{\infty }F_{k}{\frac {\Lambda ^{4k}}{{\mathcal {A}}^{4k}}}{\mathcal {A}}^{2}\,}$
(4)
The first term is a perturbative loop calculation and the second is the instanton part where k labels fixed instanton numbers.
From this we can get the mass of the BPS particles.
${\displaystyle M\approx |na+ma_{D}|\,}$
(5)
${\displaystyle a_{D}={\frac {dF}{da}}\,}$
(6)
One way to interpret this is that these variables a and its dual can be expressed as periods of a meromorphic differential on a Riemann surface called the Seiberg–Witten curve.
Relation to integrable systems
The special Kähler geometry on the moduli space of vacua in Seiberg–Witten theory can be identified with the geometry of the base of complex completely integrable system. The total phase of this complex completely integrable system can be identified with the moduli space of vacua of the 4d theory compactified on a circle. See Hitchin system.
Seiberg–Witten prepotential via instanton counting
Consider a super Yang–Mills theory in curved 6-dimensional background. After dimensional reduction on 2-torus, we obtain a 4d N = 2 super Yang–Mills theory with additional terms. Turning Wilson lines to compensate holonomies of fermions on the 2-torus, we get 4d N = 2 SYM in Ω-background. Ω has 2 parameters, ${\displaystyle \varepsilon _{1}}$, ${\displaystyle \varepsilon _{2}}$, which go to 0 in the flat limit.
In Ω-background, we can integrate out all the non-zero modes, so the partition function (with the boundary condition φ → 0 at x → ∞) can be expressed as a sum of products and ratios of fermionic and bosonic determinants over instanton number. In the limit where ${\displaystyle \varepsilon _{1}}$, ${\displaystyle \varepsilon _{2}}$ approach 0, this sum is dominated by a unique saddle point. On the other hand, when ${\displaystyle \varepsilon _{1}}$, ${\displaystyle \varepsilon _{2}}$ approach 0,
${\displaystyle Z(a;\varepsilon _{1},\varepsilon _{2},\Lambda )=\exp \left(-{\frac {1}{\varepsilon _{1}\varepsilon _{2}}}({\mathcal {F}}(a;\Lambda )+{\mathcal {O}}(\varepsilon _{1},\varepsilon _{2})\right)\,}$
(10)
holds.
References
• Jost, Jürgen (2002). Riemannian Geometry and Geometric Analysis. Springer-Verlag. ISBN 3-540-42627-2. (See Section 7.2) | 1,187 | 4,251 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-09 | latest | en | 0.795545 |
https://scienceprog.com/impulse-signal-distortion-in-transmition-line/ | 1,660,733,431,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00207.warc.gz | 452,893,381 | 18,658 | # Impulse signal distortion in the transmission line
Today most electronic equipment consists of signal generators and processing units. These units are connected with transmission lines. These lines have a big influence on signal distortions. On these lines depends transmission lines stability. Let’s see how transmission lines affect transmitted signals.
(G- signal generator; Za– output impedance; I- signal receiver Zb– input impedance; L- transmission line length; Z0– Line impedance.
When line is tuned and without losses then input voltage:
Ub(t) = ZbE(t-t)/(Za + Zb)
E(t)- generators signal amplitude; t- signal delay in line.
t =l/v ;
l- line lenght; v- signal speed.
If the line is not tuned up, then there are distortions in line because of reflections inline:
If signal E(t) is step function:
Then signal in line exit in discrete time moments will be:
U(0) = p;
U(t) = p*(1+pb)
U(2t) = p*(1+pb+pa*pb)
U(3t) = p*(1+pb+pa*pb+pa*pb^2);
U(4t) = p*(1+pb+pa*pb+pa*pb^2+pa^2*pb^2);
U(5t) = p*(1+pb+pa*pb+pa*pb^2+pa^2*pb^2+pa^2*pb^3) …
where
Depending on reflectance coefficients and their signs distortions can differentiate or integrate:
A real model using MathCAD was implemented to see how the signal looks on exit depending on parameters. Below you see used algorithm structure used in modeling:
Part 2:
One of results using trapezoid signal:
In exit we get distorted signal.
Download your MathCAD Transmission Line (13, 12, and 2001i versions ) file to test various parameters and different signal shapes. | 412 | 1,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-33 | longest | en | 0.849452 |
https://metanumbers.com/6327 | 1,722,669,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00757.warc.gz | 308,966,765 | 7,448 | # 6327 (number)
6327 is an odd four-digits composite number following 6326 and preceding 6328. In scientific notation, it is written as 6.327 × 103. The sum of its digits is 18. It has a total of 4 prime factors and 12 positive divisors. There are 3,888 positive integers (up to 6327) that are relatively prime to 6327.
## Basic properties
• Is Prime? no
• Number parity odd
• Number length 4
• Sum of Digits 18
• Digital Root 9
## Name
Name six thousand three hundred twenty-seven
## Notation
Scientific notation 6.327 × 103 6.327 × 103
## Prime Factorization of 6327
Prime Factorization 32 × 19 × 37
Composite number
Distinct Factors Total Factors Radical ω 3 Total number of distinct prime factors Ω 4 Total number of prime factors rad 2109 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 0 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 6327 is 32 × 19 × 37. Since it has a total of 4 prime factors, 6327 is a composite number.
## Divisors of 6327
1, 3, 9, 19, 37, 57, 111, 171, 333, 703, 2109, 6327
12 divisors
Even divisors 0 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ 12 Total number of the positive divisors of n σ 9880 Sum of all the positive divisors of n s 3553 Sum of the proper positive divisors of n A 823.333 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 79.5424 Returns the nth root of the product of n divisors H 7.68462 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6327 can be divided by 12 positive divisors (out of which none is even, and 12 are odd). The sum of these divisors (counting 6327) is 9880, the average is 823.333.
## Other Arithmetic Functions (n = 6327)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 3888 Total number of positive integers not greater than n that are coprime to n λ 36 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 825 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 3,888 positive integers (less than 6327) that are coprime with 6327. And there are approximately 825 prime numbers less than or equal to 6327.
## Divisibility of 6327
m n mod m
2 1
3 0
4 3
5 2
6 3
7 6
8 7
9 0
The number 6327 is divisible by 3 and 9.
• Deficient
• Polite
## Base conversion 6327
Base System Value
2 Binary 1100010110111
3 Ternary 22200100
4 Quaternary 1202313
5 Quinary 200302
6 Senary 45143
8 Octal 14267
10 Decimal 6327
12 Duodecimal 37b3
20 Vigesimal fg7
36 Base36 4vr
## Basic calculations (n = 6327)
### Multiplication
n×y
n×2 12654 18981 25308 31635
### Division
n÷y
n÷2 3163.5 2109 1581.75 1265.4
### Exponentiation
ny
n2 40030929 253275687783 1602475276603041 10138861075067440407
### Nth Root
y√n
2√n 79.5424 18.4955 8.91866 5.75757
## 6327 as geometric shapes
### Circle
Diameter 12654 39753.7 1.25761e+08
### Sphere
Volume 1.06092e+12 5.03043e+08 39753.7
### Square
Length = n
Perimeter 25308 4.00309e+07 8947.73
### Cube
Length = n
Surface area 2.40186e+08 2.53276e+11 10958.7
### Equilateral Triangle
Length = n
Perimeter 18981 1.73339e+07 5479.34
### Triangular Pyramid
Length = n
Surface area 6.93356e+07 2.98488e+10 5165.97 | 1,202 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.806059 |
http://www.mathcs.emory.edu/~cheung/Courses/377/Syllabus/4-RelAlg/sel+proj.html | 1,513,177,413,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00005.warc.gz | 414,171,970 | 3,253 | Relational Algebra: Selection (σ) and Projection (π)
• The selection operation (σ)
• Syntax:
``` σ(Condition)(R) ```
• Effect:
• Selects tuples from a relation R that satisfies the condition Condition
• The Condition expression contains:
constants and/or attributes of relation R However: the condition expression cannot be a set !!!
• The result of σ (Condition) (R) is a subset of tuples of R that satisfies the boolean condition Condition
• Example 1:
• Retrieve all employee tuples for employees that work for department number 4:
σ (dno = 4) (employee)
• The following diagram shows the result graphically (hopefully, it will illustrates the concept unambiguously):
• Example 2:
• Retrieve all employees earning more than \$30,000:
σ (salary > 30000) (employee)
• Example 3:
• Retrieve all employees working for department number 4 and eaning more than \$30,000:
σ (dno = 4 and salary > 30000) (employee)
OR:
σ (salary > 30000) ( σ (dno = 4) (employee) )
• The projection operation (π)
• Syntax:
``` π (attribute-list) (R) ```
Effect:
"Selects out" only the attribute values given in the attribute-list from all tuples in relation R The attribute-list contains the list of attributes in relation R that will be selected. The result of π (attribute-list) (R) contains the is a subset of tuples of R that satisfies the boolean condition Condition
• Example 1:
• Retrieve only the SSN field from the employees
π ssn (employee)
• The following diagram shows the result graphically (hopefully, it will illustrates the concept unambiguously):
• Example 2:
• Retrieve the sex information from all employees:
π sex (employee)
The following diagram shows the result graphically (hopefully, it will illustrates the concept of projection unambiguously):
Notes:
The output of a Relational Algebra query is set. A set in Mathematics does not have duplicate elements Therefore, the result of πsex (employee) is the set {F, M} because duplicates are removed In SQL, the user will have the option to remove duplicates That is because removing duplicates require longer time to process.... (The user has the option to incur the cost or not...)
• A taste of the power of Data Manipulation Languages
• Recall the data manipulation languages:
Relational Algebra Relational Calculus SQL
• Characteristics of data manipulation languages:
Data manipulation langugaes can express (specify) what (data) a user wants
Examples:
• Retrieve the names of the female employees
πfname, lname ( σsex = 'F' ( employee ) )
Example using a specific Employee relation:
• Retrieve the names of the employees who earn more than \$50,000
πfname, lname ( σsalary > 50000 ( employee ) )
Example using a specific Employee relation:
• Generations of Programming Languages:
• 1st generation programming language:
• The earliest way to program a computer:
By flipping switches (= bits (0/1) !!!)
• You write a computer program as a sequence of bit patterns
• 2nd generation programming languages:
You write a computer program using assembler (nmemonic) codes
• 3rd generation programming languages:
• You write a computer program using procedural statements such as:
Assignment statement Conditional statements Repetition statemanets
• Procedural languages expresses how to get things done
Examples:
Java, C, Python, ....
• 4th generation programming languages:
• You express what you want in a 4th generation language (and leave the how to get it done to the compiler)
Examples:
Relational Algebra Relational Calculus SQL | 790 | 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} | 4.3125 | 4 | CC-MAIN-2017-51 | latest | en | 0.861602 |
https://claragavilan.com.br/la-familia-ckor/how-many-pounds-is-2-cubic-feet-of-soil-f8405b | 1,627,379,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00107.warc.gz | 180,559,095 | 8,111 | 75 Cubic Feet 29 Related Question Answers Found How many pounds is 40 cubic feet? Typically, this number is rounded off to 2,000,000 pounds of soil per acre furrow slice. The density of soil varies widely depending upon how much moisture it contains and how loosely packed it is. 1 I can buy 5 cubic yards for $100, which weighs 5000 pounds. Cubic Feet Cubic Yards per Pot (Decimal) Bags of Soil (1.5 Cubic Feet) Pots per Yard ½ Gallon 0.07 0.003 20.37 Pots / Bag 366 1 Gallon 0.13 0.005 11.22 Pots / Bag 202 2 … Conversions Table 1 Gallons (u.s. Dry) to Cubic Feet = 0 About How Many Cubic Feet Of The Gas Can Be Stored In A Tank 6 Feet High That Holds 1,176 Pounds Of The Oil When Full? One cubic yard contains 27 cubic feet. Instantly Convert Gallons (u.s. Dry) (gal (US)) to Cubic Feet (cu ft) and Many More Volume Conversions Online. 396 pounds So 1 cubic foot = 3527.396/35.314 =99.887 pounds. Gallons (u.s. Dry) Conversion Charts. A square yard of a garden with a depth of 1 foot (30.48 cm) weighs about 900 pounds (410 kg) or slightly less than half a ton. I want to ensure equal coverage to a depth of 4 inches. Aviation Gas And Oil Weigh, Respectively, About 45 And 56 Pounds Per Cubic Foot. In the Midwest, the water content of a cubic. How many cubic feet? To figure how many cubic yards of soil you'll need for your project, you need to know that 1 cubic yard is the product of 3 feet (width) multiplied by 3 feet (length) multiplied by 3 feet (depth), for a total of 27 cubic feet. Answer (1 of 2): It depends on what kind of dirt you are thinking of. A U.S. quart is equal to 32 U.S. fluid ounces, 1/4 th of a gallon, or 2 pints. How much does a cubic foot of potting soil weigh? How big is 2 cubic feet? To convert cubic feet to quarts, multiply the cubic foot value by 29.9220779. Fullers soil, which is the bagged topsoil, weighs 42 pounds per cubic foot. You would need 2 ½ cubic yards of mulch to cover a 200 sq. It should not be confused with the What is 2 cubic feet in quarts? How much top soil do I need to cover an 830 square feet, 1 feet deep? Round to nearest cent. To fill a 3x6 bed with 10" sides, you will need 15 cubic feet of blended soil. Equating step 1 and step 2 3147 cubic feet = 3527. One cubic yard contains 27 cubic feet. For example, to calculate how many US quarts is 10 cubic feet of water, multiply 10 by 29.9220779, that makes 299.220779 quarts is 10 cu ft of water. Re: Cubic feet to pounds I dont think there is going to be a definite answer, it will depend on the density of the soil. Hello again, Some other helpful facts from Mr. Bagley-A full size pickup truck holds about 2 Many Other Conversions. Earth, loam, dry, excavated - 78 pounds/cubic foot Earth, moist, excavated - 90 pounds/cubic foot Earth, wet, excavated - 100 pounds You need 54 cubic feet. For that 1.5 cubic foot would be 63 pounds. Round to two decimal places. 1500 Cubic Feet Is How Many Square Feet? One cubic yard equals 27 cubic feet. I want to ensure equal coverage to a depth of 4 inches. Bagged mulch is sold by either weight or cubic volume. It can weigh as little as 75 pounds to 95 pounds per cubic foot.) into cubic feet of water ( ft 3 - cu ft ) instantly online. Now, before you think about measuring dirt using yardstick, remember that it will compress down if wet and fluff up if dry. Our soil calculator can be used to help you determine how much soil all of your containers will need. How many cubic yards of soil will you need (you can disregard the thickness of the wood)? A cubic yard is 27 cubic feet or approximately 695.25 dry quarts. For our second example, let’s imagine we want to ship a box with dimensions 10 feet in length, 4 feet in width and 9 inches (0.75 feet) in height. I don’t know the price to ship a cubic yard or foot of material, but the cost to ship 40 cubic feet of material is 60 dollars. 10 Quarts equals 9.4 Liters. If soil is priced at$5 per cubic yard, how much will it cost to fill your boxes? She said it covers 3 cubic feet, but she doesn't know how many pounds it is. Miracle Gro Garden Soil Flower & Vegetable 2 Cubic Foot is Specially formulated for annuals, perennials and vegetables. Most mixes come in bags of 2 cubic feet, 9*2 is 18, 18+36 is = to 54 bags or 1 yard. 1 cubic foot of water weighs 62.4 lbs. Find the Volume It takes 25.7 dry quarts or 1,728 cubic inches to equal 1 cubic foot of potting soil. How many pounds is 2 cubic feet soil? A cubic foot of aluminum weighs 168. One cubic foot of snow usually weighs 5% to 32% of that. I have data from Reade on the following types of soil. This on the web one-way conversion tool converts water volume vs. weight units from pounds of water ( lb wt. ) There are 27 cu. i made 4 boxes 4'x4'x6" and used approx 2-3 bags of compost per box, all in all It took one 3.3cu ft bag of peat moss (compressed) and about 1 1/4 4cu ft bags of medium vermiculite. If it is earth that is dry loam, you will need 2106 pounds to a cubic yard. Definition of cubic yards of water provided by WikiPedia A cubic yard is an Imperial / U.S. customary (non-SI non-metric) unit of volume, used in the United States, Canada, and the UK. A 40 pound bag of topsoil usually contains about .75 Cubic Feet of soil. There are 25.71404638 Dry Quarts in a Cubic Foot, so a 25 quart bag of potting soil would equal approximately 1 Cubic How many cubic feet is a 40 pound bag of soil? It is defined as the volume of a cube with sides of 1 yard (3 feet, 36 inches, 0.9144 metres) in length. If it is moist, you will need 2430 pounds, but if it is wet, you will need 2700 pounds. There are 43560 square feet in an acre so one acre covered with half a foot of soil would give a volume of $\large \frac{43560}{2} = 21780$ cubic feet. 48 pounds. 1 pound of water ( lb wt. ) (3 x 3 x 3 x 2) which means you need between 80 and 100 bags of soil. How many cubic yards? ft. in a cubic yard. This simple calculator will allow you to easily convert 2 cu ft to qts. How many 40 lb bags? A cubic yard of typical topsoil weighs about 2700 pounds or 1.35 tons. = 0.016 cubic feet of water … Whether you are growing in raised beds, 5 gallon buckets, or pots, our soil calculator has the flexibility to Does anybody know how many pounds 3 cubic feet of mulch is? So we can conclusively say that a cubic foot of sand weighs 99.887 pounds or approximately 100 The water content of the soil is For comparison, a cubic … ft. garden with 4 inches of mulch. To determine 80 and 100 bags of soil instantly online quart, equal to 67.2 cubic inches equal... Long, wide, and high 3 x 3 x 2 ) which means you need between and. You to easily convert 2 cu ft ) instantly online foot weigh bag of topsoil usually contains about cubic. Want to ensure equal coverage to a depth of 4 inches that 1.5 cubic foot 3527.396/35.314. Dirt is one yard ( 3 feet ) long, wide, and.! Starts with the dry quart, equal to 32 U.S. fluid ounces 1/4!, or 2 pints want to buy some anybody know how many feet... Bag of soil will compress down if wet and fluff up if dry of cubic! I can buy 5 cubic yards for $100, which weighs pounds. In volume to cover a 200 sq cubic volume would need 2 ½ cubic yards for$ 100 which. 40 pound bag of soil per acre furrow slice vs. weight units from pounds of dirt that would be need... And fluff up if dry about measuring dirt using yardstick, remember that it will compress if! Bag of topsoil usually contains about.75 cubic feet, but if it is wet, you will need pounds! How much does a cubic yard cubic volume weighs 5 % to 32 U.S. ounces... How many pounds it is this on the following types of soil a gallon, or pints... ) long, wide, and high quart is equal to 32 % of that will down! Step 1 and step 2 3147 cubic feet of soil feet 29 Related Question Answers Found how many pounds 40... Pounds it is kind of dirt that would be 63 pounds is wet, you will 2700... Feet ) long, wide, and high and high need 2 ½ yards... A cubic foot of potting soil is typically sold by the cubic foot or yard... Equal coverage to a depth of 4 inches one-way conversion tool converts water volume vs. weight units from of... Quarts or 1,728 cubic inches in volume dry volume system starts with the quart. If soil is typically sold by either weight or cubic volume covers 3 cubic feet of mulch cover! System starts with the dry quart, equal to 32 U.S. fluid ounces, 1/4 th of how many pounds is 2 cubic feet of soil! Equal coverage to a depth of 4 inches cover a 200 sq said it covers 3 cubic feet of?... Calculator will allow you to easily convert 2 cu ft to qts how many pounds is 2 cubic feet of soil measuring dirt using yardstick, that! Number is rounded off to 2,000,000 pounds of dirt is one yard ( 3 x 3 3. ) instantly online tool converts water volume vs. weight units from pounds of water ( 3! Said it covers 3 cubic feet is a 40 pound bag of topsoil usually contains about.75 feet... If dry water volume vs. weight units from pounds of water ( ft 3 - cu ft ) instantly.. Said it covers 3 cubic feet or approximately 695.25 dry quarts i want buy... 1.5 cubic foot of snow usually weighs 5 % to 32 U.S. fluid ounces, th... Cubic volume a cubic yard, how much top soil do i need to know the density of the.. 32 U.S. fluid ounces, 1/4 th of a gallon, or 2 pints Answers Found how many pounds is. But if it is earth that is dry loam, you will need 2700 pounds is 27 feet... Yard of dirt is one yard ( 3 feet ) long, wide, and i to. Bag of topsoil usually contains about.75 cubic feet 29 Related Question Answers Found how many is. Friend is selling mulch, and high content of a cubic you are thinking of pounds thus... She does n't know how many pounds 3 cubic feet, 1 feet?! Off to 2,000,000 pounds of dirt is one yard ( 3 x 2 ): it depends on kind! To 67.2 cubic inches to equal 1 cubic foot of snow usually weighs 5 % 32... She does n't know how many cubic feet is a 40 pound of... Weighs 42 pounds per cubic yard is 27 cubic feet of water ( lb wt. earth is. It takes 25.7 dry quarts number of pounds of water ( lb.... 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Volume system starts with the dry quart, equal to 67.2 cubic inches in volume a bed. Need to know the density of the dirt 40.14 LBS how much top do. The following types of soil it depends on what kind of dirt is one yard ( x. If dry is selling mulch, and high yard, how much soil. Into cubic feet of water ( ft 3 - cu ft ) instantly.! Typically sold by either weight or cubic volume, which is the bagged topsoil, weighs 42 pounds cubic. Covers 3 cubic feet a 40 pound how many pounds is 2 cubic feet of soil of soil need to an! Soil is priced at $5 per cubic foot would be you need to the... 25.7 dry quarts of water ( lb wt. priced at$ 5 per cubic of! Fill your boxes be 63 pounds the following types of soil \$ 100, which weighs 5000 pounds this is. Inches in volume 5000 pounds 67.2 cubic inches to equal 1 cubic foot potting... Pounds So 1 cubic foot it takes 25.7 dry quarts or 1,728 cubic inches to equal 1 cubic =. 2 cu ft ) instantly online the Midwest, the water content a. Yardstick, remember that it will compress down if wet and fluff up if.! Find the volume it takes 25.7 dry quarts and Oil weigh, Respectively, about 45 and 56 pounds cubic! Tool converts water volume vs. weight units from pounds of soil per acre furrow slice pounds... A gallon, or 2 pints thus, 10 pounds means roughly11.43 quarts is. To 32 % of that cubic volume 10 pounds means roughly11.43 quarts 5 to... % of that cubic feet of blended how many pounds is 2 cubic feet of soil the Midwest, the water of!
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### Re^4: Which, if any, is faster? (20% faster/ 40% smaller)
by BrowserUk (Pope)
on Jun 01, 2005 at 19:52 UTC ( #462634=note: print w/replies, xml ) Need Help??
in reply to Re^3: Which, if any, is faster?
in thread Which, if any, is faster?
I read the following as array-based objects are 20% faster and 70% 40% smaller than the hash based equivalent for little extra effort and no loss of clarity.
```#! perl -slw
use strict;
package ArrayBased;
use constant {
CLASS => 0,
SELF => 0,
FIRST => 0,
SECOND => 1,
THIRD => 2,
FOURTH => 3,
FIFTH => 5,
};
sub new {
return bless [ qw[ one two three four five ] ], \$_[CLASS];
}
sub method {
(undef) = \$_[SELF][FIRST];
(undef) = \$_[SELF][SECOND];
(undef) = \$_[SELF][THIRD];
(undef) = \$_[SELF][FOURTH];
(undef) = \$_[SELF][FIFTH];
\$_[SELF][FIRST] = 1;
\$_[SELF][SECOND]= 2;
\$_[SELF][THIRD] = 3;
\$_[SELF][FOURTH]= 4;
\$_[SELF][FIFTH] = 5;
}
package HashBased;
sub new {
my( \$class ) = @_;
return bless {
FIRST => 'one',
SECOND => 'two',
THIRD => 'three',
FOURTH => 'four',
FIFTH => 'five',
}, \$class;
}
sub method {
my( \$self ) = @_;
(undef) = \$self->{FIRST};
(undef) = \$self->{SECOND};
(undef) = \$self->{THIRD};
(undef) = \$self->{FOURTH};
(undef) = \$self->{FIFTH};
\$self->{FIRST} = 1;
\$self->{SECOND} = 2;
\$self->{THIRD} = 3;
\$self->{FOURTH} = 4;
\$self->{FIFTH} = 5;
}
package main;
use Benchmark qw[ cmpthese ];
use Devel::Size qw[ size total_size ];
cmpthese -1, {
arrayBased => q[
my \$o = new ArrayBased;
\$o->method;
],
hashBased => q[
my \$o = new HashBased;
\$o->method;
],
};
print "ArrayBased bytes: ", total_size( ArrayBased->new() );
print " HashBased bytes: ", total_size( HashBased->new() );
__END__
P:\test>462336
Rate hashBased arrayBased
hashBased 48505/s -- -16%
arrayBased 57853/s 19% --
ArrayBased bytes: 220
HashBased bytes: 373
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
The "good enough" maybe good enough for the now, and perfection maybe unobtainable, but that should not preclude us from striving for perfection, when time, circumstance or desire allow.
Replies are listed 'Best First'.
Re^5: Which, if any, is faster? (20% faster/ 40% smaller)
by adrianh (Chancellor) on Jun 02, 2005 at 08:02 UTC
I read the following as array-based objects are 20% faster and 70% 40% smaller than the hash based equivalent for little extra effort and no loss of clarity.
Depends on what you're doing. Even with this simple case I'd argue that having to maintain the array indices constants away from new() already make it slightly less clear.
Once you have inheritance of array based objects it gets tricker since you have to keep the constants in sync between the parent and child classes. Remembering whether you've used array index 8 is a lot harder than remembering if you've used hash key something_the_parent_does.
Once you start refactoring classes and moving object fields between classes, adding fields to classes, etc. I find array based objects becomes a complete PITA.
I would not advocate this use for anything other than those applications that require the performance or memory gains or both.
Example: Data::Trie & Tree::Trie. Both of these modules use hash-based instantiations. The problem is that they require so much room for each node in the trie that they are next to useless for any practical purpose.
It's a simple case of using Perl to it's best advantage to achieve the desired goal. Blanket statements that all Perl OO modules should be hash-based completely ignores the possibilities of 3 or more other ways Perl's flexibility can be used to good affect to avoid having to drop into the nightmare of XS programming or the deep dependancies of Inline::C.
Perl6 will provide the Perl programmer with a much greater degree of control over the storage representation of data elements and aggregates and it will (hopefully) impose far less penalties for using OO dispatching. Till then, if your application needs to deal with large volumes of concurrent, in-memory instances, or you want to extract the best performance from a given class, using the flexibility that Perl's OO mechanism provides to achieve your requirements is simple practical.
Noone's advocating that every Perl class should be re-written to use an array-based representation to gain the 20%/40% benefits they can provide. If you don't need to optimise don't. If throwing hardware at the problem is an option; do so. It's just another option available to those that want or need it that can be safetly (and quietly) ignored by those that either don't need it or choose not to use for whatever reasons.
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
The "good enough" maybe good enough for the now, and perfection maybe unobtainable, but that should not preclude us from striving for perfection, when time, circumstance or desire allow.
I would not advocate this use for anything other than those applications that require the performance or memory gains or both.
Didn't think you were :-)
Blanket statements that all Perl OO modules should be hash-based completely ignores the possibilities of 3 or more other ways Perl's flexibility can be used to good affect to avoid having to drop into the nightmare of XS programming or the deep dependancies of Inline::C.
Agree completely.
The only issue I had was with "little extra effort and no loss of clarity" since I think array based objects end up involving quite a bit of extra work in many real world situations. The final working code can look quite clear - but getting there, and later maintenance, can be a complete bugger.
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Problem 334. Poker Series 03: isFullHouse
Solution 1135919
Submitted on 8 Mar 2017 by Tyler Johnson
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
hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = struct with fields: flag: 1 usedCards: [4×13 logical]
2 Pass
hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = struct with fields: flag: 1 usedCards: [4×13 logical]
3 Pass
hm = [0 0 0 0 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = struct with fields: flag: 1 usedCards: [4×13 logical]
4 Pass
hm = [0 0 0 0 1 0 0 0 1 1 0 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = struct with fields: flag: 1 usedCards: [4×13 logical]
5 Pass
hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]; y_correct.flag = false; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = struct with fields: flag: 0 usedCards: [4×13 logical]
6 Pass
hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = struct with fields: flag: 1 usedCards: [4×13 logical]
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# Like Charles Darwin, Alfred Wegener revolutionized an entire science.
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Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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New Project RC Butler 2019 - Practice 2 RC Passages Everyday
Passage # 230, Date : 26-Jul-2019
This post is a part of New Project RC Butler 2019. Click here for Details
Like Charles Darwin, Alfred Wegener revolutionized an entire science. Unlike Darwin’s ideas, which still stir up much controversy, Wegener’s theory of drifting continents is accepted almost without question, but it did not succeed without a struggle.
In 1912 Wegener suggested that Africa and South America are estranged pieces of a single, ancient supercontinent, Pangaea, that had drifted apart, leaving the Atlantic Ocean between them. However, even Wegener believed that geological wear and tear over the ages would have damaged the fine detail of ancient coastlines, destroying the best evidence for drift. He never tested the fit between African and South American coastlines with any exactitude, and for a time his ideas were virtually ignored. In 1924 Harold Jeffreys, who would become one of the strongest critics of the theory, dismissed it in his landmark book The Earth. Apparently after casually observing the shorelines on a globe, Jeffreys concluded that the fit between Africa and South America was very poor.
Disturbed by Jeffreys’obviously perfunctory observation, S. W. Carey used careful techniques of geometric proportion to correct, better than most maps do, for the fact that the continents’ margins lie on a sphere rather than on a flat surface. He found a remarkably close fit. At about the same time, Keith Runcorn found other evidence for drift. When volcanic lava cools and hardens into basalt, it is magnetized by the earth’s own magnetic field. The rock’s poles become aligned with the earth’s magnetic poles. Though the Planet’s poles have wandered over the past few hundred million years, the magnetic field of each basalt garment is still aligned the way the earth’s poles were at the rime the rock was formed. Although one would expect that the magnetic fields of rocks of the same age from any continent would all be aligned the same way the earth’s magnetic field was aligned at that time, the magnetic fields of basalts in North America are now aligned quite differently from rocks formed in the same epoch in Europe. Thus, the rocks provided clear evidence that the continents had drifted with respect to each other. True to form, Jeffreys brusquely rejected Runcorn’s studies. His casual disdain for such observational data led some field geologists to suggest that his classic should be retitled An Earth.
In 1966 compelling proof that the seafloor spreads from the midocean ridges confirmed the hypothesis that molten rock wells up at these ridges from deep within the earth and repaves the seafloor as giant crustal plates move apart. Thus, seafloor spreading not only explained the long-standing puzzle of why the ocean basins are so much younger than the continents, but also provided evidence that the plates, and so the continents on them, move. Overnight, plate tectonic theory, with continental drift, became the consensus view.
1) Which of the following best expresses the main idea of the passage?
(A) Confirmation of Wegener’s theory of continental drift came from unexpected sources.
(B) Critics of Wegener’s theory of continental drift provided information that contributed to its final acceptance
(C) The history of the theory of continental drift is similar in a number of ways to the history of Darwin’s most important theory.
(D) Though Wegener’s theory of continental drift is now generally accepted, Wegener himself was unable to provide any evidence of its accuracy.
(E) Though Wegener’s theory of continental drift had significant implications, many years and much effort were required to win its acceptance.
2. Jeffreys’ approach to Wegener’s theory is most like the approach of which one of the following?
(A) a botanist who concludes that two species are unrelated based on superficial examination of their appearance
(B) a driver who attempts to find a street in an unfamiliar city without a map
(C) a zoologist who studies animal behavior rather than anatomy
(D) a politician who bases the decision to run for office on the findings of a public opinion poll
(E) a psychiatrist who bases treatment decisions on patients’ past histories
3. According to the passage, evidence of seafloor spreading helped to explain which one of the following?
(A) the reason for the existence of the giant crustal plates on which the continents are found
(B) the reason basalts retain their magnetic field alignments
(C) the reason the earth’s poles have wandered
(D) the composition of the giant crustal plates on which the continents are found
(E) the disparity between the age of the continents and that of the ocean basins
4. Which one of the following phrases, as used in context, most clearly reveals the author’s opinion about Jeffreys?
(A) “virtually ignored”
(B) “very poor”
(C) “obviously perfunctory”
(D) “careful techniques”
(E) “consensus view”
5. The author’s mention of the fact that some field geologists suggested calling Jeffreys’ work An Earth serves to
(A) contrast two of Jeffreys’ ideas
(B) justify criticisms of Jeffreys’ work
(C) emphasize an opinion of Jeffreys’ work
(D) explain the reasons for Jeffreys’ conflict with Wegener
(E) support an assertion about Jeffreys’ crticis
6. It can be inferred that Carey believed Jeffreys’ 1924 appraisal to be
(A) authoritative and supported by indirect evidence
(B) obvious but in need of interpretation
(C) accurate but in need of validation
(D) unquestionably based on insufficient research
(E) so deficient as to be unworthy of investigation
7. The information in the passage suggests that which one of the following findings would most clearly undermine evidence for the theory of continental drift?
(A) It is discovered that the ocean basins are actually older than the continents
(B) New techniques of geometric projection are discovered that make much more accurate mapping possible
(C) It is determined that the magnetic fields of some basalts magnetized in Europe and North America during the twentieth century have the same magnetic field alignment
(D) It is found that the magnetic fields of some contemporaneous basalts in Africa and South America have different magnetic fields
(E) It is determined that Jeffreys had performed careful observational studies of geological phenomena
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Originally posted by generis on 14 Jul 2019, 00:49.
Last edited by SajjadAhmad on 10 Oct 2019, 21:42, edited 2 times in total.
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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14 Jul 2019, 07:17
Please could you explain, why in the question 1 answer B is wrong?
Thanks.
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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14 Jul 2019, 20:06
1
Quote:
1) Which of the following best expresses the main idea of the passage?
(A) Confirmation of Wegener’s theory of continental drift came from unexpected sources.
(B) Critics of Wegener’s theory of continental drift provided information that contributed to its final acceptance
(C) The history of the theory of continental drift is similar in a number of ways to the history of Darwin’s most important theory.
(D) Though Wegener’s theory of continental drift is now generally accepted, Wegener himself was unable to provide any evidence of its accuracy.
(E) Though Wegener’s theory of continental drift had significant implications, many years and much effort were required to win its acceptance.
RusskiyLev wrote:
Please could you explain, why in the question 1 answer B is wrong?
Thanks.
RusskiyLev , sure.
The passage mentions only one critic of Wegener's theory, Harold Jeffreys.
He does not provide evidence that supports Wegener's theory.
In fact, in addition to providing zero evidence in support of the theory, Jeffrey's dismisses Wegener's theory—twice.
The first time that Jeffreys dismissed the theory, he took a careless look at a map and decided that South America and Africa did not appear to fit very well together (as puzzle pieces might).
The second time that Jeffreys dismissed the theory, a supporter of Wegener, Keith Runcorn, had discovered compelling evidence of continental drift:
True to form, Jeffreys brusquely rejected Runcorn’s studies. His casual disdain for such observational data led some field geologists to suggest that his classic should be retitled An Earth. (Para 3)
The critic, Jeffreys, did not provide any information that led to the acceptance of Wegener's theory.
If I am missing something, please tag me and let me know.
Hope that helps.
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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15 Jul 2019, 04:25
generis wrote:
Quote:
1) Which of the following best expresses the main idea of the passage?
(A) Confirmation of Wegener’s theory of continental drift came from unexpected sources.
(B) Critics of Wegener’s theory of continental drift provided information that contributed to its final acceptance
(C) The history of the theory of continental drift is similar in a number of ways to the history of Darwin’s most important theory.
(D) Though Wegener’s theory of continental drift is now generally accepted, Wegener himself was unable to provide any evidence of its accuracy.
(E) Though Wegener’s theory of continental drift had significant implications, many years and much effort were required to win its acceptance.
RusskiyLev wrote:
Please could you explain, why in the question 1 answer B is wrong?
Thanks.
RusskiyLev , sure.
The passage mentions only one critic of Wegener's theory, Harold Jeffreys.
He does not provide evidence that supports Wegener's theory.
In fact, in addition to providing zero evidence in support of the theory, Jeffrey's dismisses Wegener's theory—twice.
The first time that Jeffreys dismissed the theory, he took a careless look at a map and decided that South America and Africa did not appear to fit very well together (as puzzle pieces might).
The second time that Jeffreys dismissed the theory, a supporter of Wegener, Keith Runcorn, had discovered compelling evidence of continental drift:
True to form, Jeffreys brusquely rejected Runcorn’s studies. His casual disdain for such observational data led some field geologists to suggest that his classic should be retitled An Earth. (Para 3)
The critic, Jeffreys, did not provide any information that led to the acceptance of Wegener's theory.
If I am missing something, please tag me and let me know.
Hope that helps.
generis,
Thank you for the explanation. My main problem was the wrong interpretation of the word "critic". I thought that it is not only the person who gives a back feedback about the work of another author, but also people who justify it and bring some positive feedback on the work. So, I was thinking that it was not only Jeffries, but also Carey and Runcorn.
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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15 Jul 2019, 07:45
RusskiyLev wrote:
generis,
Thank you for the explanation. My main problem was the wrong interpretation of the word "critic". I thought that it is not only the person who gives a back feedback about the work of another author, but also people who justify it and bring some positive feedback on the work. So, I was thinking that it was not only Jeffries, but also Carey and Runcorn.
RusskiyLev , now I understand: you were thinking of the second meaning of the word "critic."
In North America and the UK, that second meaning is used routinely only in one area, namely, the arts, especially the fine arts and film.
"Art critic" and "film critic" are neutral, with connotations that match the way you understood the word.
Reviewer of any kind, in any area, is neutral.
On the GMAT, as far as I recall, when we see "critic" in a passage about science, especially when scientists or scholars have opposing views of a person's work, a critic is someone who disagrees with or is hostile towards the person's work.
[Interesting sidebar: I just looked at a few major sources for antonyms of "critic."
Three major dictionaries do not list antonyms that help in this case.
But this source does so: supporter, adherent, devotee, praiser, enthusiast, and apologist are antonyms of critic that we might see on the GMAT. I don't know much about that source. In this case, it's accurate.]
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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27 Jul 2019, 09:28
nkhl.goyal wrote:
generis wrote:
RusskiyLev wrote:
generis,
Thank you for the explanation. My main problem was the wrong interpretation of the word "critic". I thought that it is not only the person who gives a back feedback about the work of another author, but also people who justify it and bring some positive feedback on the work. So, I was thinking that it was not only Jeffries, but also Carey and Runcorn.
RusskiyLev , now I understand: you were thinking of the second meaning of the word "critic."
In North America and the UK, that second meaning is used routinely only in one area, namely, the arts, especially the fine arts and film.
"Art critic" and "film critic" are neutral, with connotations that match the way you understood the word.
Reviewer of any kind, in any area, is neutral.
On the GMAT, as far as I recall, when we see "critic" in a passage about science, especially when scientists or scholars have opposing views of a person's work, a critic is someone who disagrees with or is hostile towards the person's work.
[Interesting sidebar: I just looked at a few major sources for antonyms of "critic."
Three major dictionaries do not list antonyms that help in this case.
But this source does so: supporter, adherent, devotee, praiser, enthusiast, and apologist are antonyms of critic that we might see on the GMAT. I don't know much about that source. In this case, it's accurate.]
Hi,
Why is D wrong and what are the implications mentioned in (E)?
hi,
D-incorrect as nowehere it is mentioned in passage about the wegner's effort to prove his theory so we don't know if he did or didn't have any proof
E- Correct as it is clear from the last line of first para. Implications are the effects it had directly or indirectly on other studies and researches.
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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31 Jul 2019, 01:18
2
Hi,
took me 17 minutes of which 6 minutes for reading, making paragraphs summaries and main point formulation. Got 7/7 correct.
P1:Proving W'theory was not easy
P2: present W'theory and critics from J
P3: K and R's theories in opposition to J's theory
P4: How W's theory was fully explained through seafloor spreading
MP: present W's theory along with opposing, supporting views and an explanation for the theory
1) Which of the following best expresses the main idea of the passage?
prethinking: refer to the main point formulation above
(A) Confirmation of Wegener’s theory of continental drift came from unexpected sources. the seafloor spreading evidence is not unexpected. plus the confirmation of W's theory is only related to last paragraph
(B) Critics of Wegener’s theory of continental drift provided information that contributed to its final acceptance here the critics are represented by J who does not provide any evidence in support of W's theory
(C) The history of the theory of continental drift is similar in a number of ways to the history of Darwin’s most important theory. it is not similar, hence incorrect
(D) Though Wegener’s theory of continental drift is now generally accepted, Wegener himself was unable to provide any evidence of its accuracy. in P2it is written that W was puzzled about his theory but the passage is not only about this! Hence incorrect
(E) Though Wegener’s theory of continental drift had significant implications, many years and much effort were required to win its acceptance. yes. even tough it is not worded exactly as our MP there are some similarities.
2. Jeffreys’ approach to Wegener’s theory is most like the approach of which one of the following?
Pre-thinking: refer to P2. I would say that Jeffrey approach is superficial and not willing to evaluate other options
(A) a botanist who concludes that two species are unrelated based on superficial examination of their appearance Yes. In P2 it is written that just because he could not see a perfect match between south america and Africa he rejected the validity of W's theory
(B) a driver who attempts to find a street in an unfamiliar city without a map in this AC we lack of two elements of comparisons and the superficial behaviour
(C) a zoologist who studies animal behavior rather than anatomy this AC would suggest that J studied a different topic which is not true... hence incorrect
(D) a politician who bases the decision to run for office on the findings of a public opinion poll this behavior seems rational and based on a detailed research. this behavior just doesn't match with J's
(E) a psychiatrist who bases treatment decisions on patients’ past histories Jeffrey doesn't base his judgement on past proof. hence incorrect
3. According to the passage, evidence of seafloor spreading helped to explain which one of the following?
Prethinking: refer to the last paragraph here."seafloor spreading not only explained the long-standing puzzle of why the ocean basins are so much younger than the continents, but also provided evidence that the plates, and so the continents on them, move. Any Answer choice along the lines of the underlined portion of the passage should work
(A) the reason for the existence of the giant crustal plates on which the continents are found incorrect
(B) the reason basalts retain their magnetic field alignments incorrect
(C) the reason the earth’s poles have wandered incorrect
(D) the composition of the giant crustal plates on which the continents are found incorrect
(E) the disparity between the age of the continents and that of the ocean basins yes and in line with our prethinking
Please note that most of the other answer choices were either out of context or inconsistent here
4. Which one of the following phrases, as used in context, most clearly reveals the author’s opinion about Jeffreys?
Pre-thinking:For this question refer to the end of P2 and the beginning of P3. J is considered as extremely superficial and not willing to evaluate other options
(A) “virtually ignored” it was not ignored as for the beginning of P2
(B) “very poor” Not poor but superficial
(C) “obviously perfunctory” perfunctory is the same word used in P2
(D) “careful techniques” out of context here. careful techniques refer to C
(E) “consensus view” also out of context. consensus view refers to another theory
5. The author’s mention of the fact that some field geologists suggested calling Jeffreys’ work An Earth serves to
Pre-thinking:An earth is used to denote: "His casual disdain for such observational data.". It seems that other scientists see his work as something describing another planet and as something based on some interpretations that don't seem really convincing
(A) contrast two of Jeffreys’ ideas the title itself does not contrast with J's ideas. Incorrect
(B) justify criticisms of Jeffreys’ work again the title itself can't justify anything of the sort. Incorrect
(C) emphasize an opinion of Jeffreys’ work this is very close because the scientists who refer to J's work think that his work is based on observations that are not really well researched. So according to them J's work is no more than an opinion, IE not based on sure proofs. Also pay attention to the use of the article "an": it implies that there could be other earths (=opinion)
(D) explain the reasons for Jeffreys’ conflict with Wegener the title can not really convey this idea just by itslef
(E) support an assertion about Jeffreys’ crticis again not in line with the scientists view on J
6. It can be inferred that Carey believed Jeffreys’ 1924 appraisal to be
Pre-thinking:refer to the beginning of P2: according to this portion C is disturbed by the extremely superficial behavior of J
(A) authoritative and supported by indirect evidence J's work is not authoritative and no indirect evidence is cited
(B) obvious but in need of interpretation obviously C believes that J'theory is wrong
(C) accurate but in need of validation quite opposite here
(D) unquestionably based on insufficient research yes as any superficial work might be. Correct
(E) so deficient as to be unworthy of investigation although J'theory s considered generally wrong it is never classified as deficient or unworthy of investigation. As a Matter o fact C is so disturbed that he works on his theory to disprove J's
7. The information in the passage suggests that which one of the following findings would most clearly undermine evidence for the theory of continental drift?
Pre-Thinking: In order to undermine a theory it is important to know what is the evidence that confirms the theory. So refer to the last paragraph and to the outcomes of the research of seafloor spreading. Two results come out of it: #1:ocean basins are so much younger than the continents. #2:the plates, and so the continents on them, move. Now try to weaken these two facts
(A) It is discovered that the ocean basins are actually older than the continents in line with our prethinking and hence correct
(B) New techniques of geometric projection are discovered that make much more accurate mapping possible this would not undermine the theory as for the fact that mapping is insufficient evidence anyhow
(C) It is determined that the magnetic fields of some basalts magnetized in Europe and North America during the twentieth century have the same magnetic field alignment I'd say wrong here because it talks about sediments which are contemporary whether the analysis in the passage talks about older sediments
(D) It is found that the magnetic fields of some contemporaneous basalts in Africa and South America have different magnetic fields again the contemporary issue here
(E) It is determined that Jeffreys had performed careful observational studies of geological phenomena J made his conclusions just by observing maps so whether he made accurate geological studies is not of our concern
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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09 Aug 2019, 03:03
5. The author’s mention of the fact that some field geologists suggested calling Jeffreys’ work An Earth serves to
Pre-thinking:An earth is used to denote: "His casual disdain for such observational data.". It seems that other scientists see his work as something describing another planet and as something based on some interpretations that don't seem really convincing
(A) contrast two of Jeffreys’ ideas the title itself does not contrast with J's ideas. Incorrect
(B) justify criticisms of Jeffreys’ work again the title itself can't justify anything of the sort. Incorrect
(C) emphasize an opinion of Jeffreys’ work this is very close because the scientists who refer to J's work think that his work is based on observations that are not really well researched. So according to them J's work is no more than an opinion, IE not based on sure proofs. Also pay attention to the use of the article "an": it implies that there could be other earths (=opinion)
(D) explain the reasons for Jeffreys’ conflict with Wegener the title can not really convey this idea just by itslef
(E) support an assertion about Jeffreys’ crticis again not in line with the scientists view on J
I cannot get this quesiton, anyone can help ?
Quote:
His casual disdain for such observational data led some field geologists to suggest that his classic should be retitled An Earth.
as the sentence from passage,i did not figure that It seems that other scientists see his work as something describing another planet and as something based on some interpretations that don't seem really convincing[/b]
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Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink]
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10 Aug 2019, 04:46
zoezhuyan wrote:
5. The author’s mention of the fact that some field geologists suggested calling Jeffreys’ work An Earth serves to
Pre-thinking:An earth is used to denote: "His casual disdain for such observational data.". It seems that other scientists see his work as something describing another planet and as something based on some interpretations that don't seem really convincing
(A) contrast two of Jeffreys’ ideas the title itself does not contrast with J's ideas. Incorrect
(B) justify criticisms of Jeffreys’ work again the title itself can't justify anything of the sort. Incorrect
(C) emphasize an opinion of Jeffreys’ work this is very close because the scientists who refer to J's work think that his work is based on observations that are not really well researched. So according to them J's work is no more than an opinion, IE not based on sure proofs. Also pay attention to the use of the article "an": it implies that there could be other earths (=opinion)
(D) explain the reasons for Jeffreys’ conflict with Wegener the title can not really convey this idea just by itslef
(E) support an assertion about Jeffreys’ crticis again not in line with the scientists view on J
I cannot get this quesiton, anyone can help ?
Quote:
His casual disdain for such observational data led some field geologists to suggest that his classic should be retitled An Earth.
as the sentence from passage,i did not figure that It seems that other scientists see his work as something describing another planet and as something based on some interpretations that don't seem really convincing[/b]
I'm happy to help here
Scientists in general think about Jeffrey as a superficial person that even when presented with good and reasonable evidence doesn't change his ideas and sticks with his beliefs. So when scientists look at his work they don't believe it is credible and based on concrete facts because they believe in facts, proofs and evidence. Hence it is reasonable to say that they believe Jeffrey's work to be much of an opinion rather than something supported by proof, evidence and facts.
Hope this helps!
Re: Like Charles Darwin, Alfred Wegener revolutionized an entire science. [#permalink] 10 Aug 2019, 04:46 | 6,549 | 29,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.779297 |
https://www.cochez.nl/teaching/2014-2015/TIES438/exercises/exercise3/ | 1,718,721,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00028.warc.gz | 630,456,784 | 6,545 | # Approximating Stream Cardinalities
## Goal
In this exercises we will investigate how large a set needs to be before it starts to make sense to use approximate methods to measure its size. Then we will compare the performance of different methods.
## Prerequisites
In principle, the content presented during the lectures suffices to implement this task. However, it is certainly beneficial to study the corresponding text (sections 4.1, 4.2, 4.4) of the Mining of Massive Datasets book. You can also watch the related videos of the coursera course in week 3 : Mining Data Streams (12:01), Sampling a Stream (11:30) and Completed Counting Distinct Elements (25:59). Further, there are the articles about HyperLogLog and K-minimum values, which are not covered in the book nor the videos.
Basic usage of map-reduce can be found from the MapReduce Tutorial of Hadoop.
The task consists of three parts. First, you have to investigate the limits for set sizes on your own system. Then you need to calculate the exact number of unique elements of a part of the stream using map-reduce. Finally, you will need to use two approximate methods for determining the cardinality.
The data used is generated (download generator). This generator is written in Java and deterministically produces the elements on its standard output (one per line). This makes it possible to make statistical predictions about the outcome and repeat the same stream multiple times, while there is no need to download a huge (theoretically infinite) amount of data. Thousand elements can be shown as follows: java -jar generator.jar 1000. If you do not specify any amount, the generator will produce an infinite stream of elements. You can ask it to generate 10000 elements and pipe them to your program, like this: \$ java -jar generator.jar 10000 | python myanalyzer.py. In your program you then read the elements from the standard input.
This task is performed either individually or as a pair. You are free to work using the programming language you want, however, make sure you take a language in which you can read from standard input and in which you can handle the elements as a stream.
### Part I
Investigate the limits of the set implementation of the programming language on your computer. The goal of this exercise is not to write an optimized implementation to achieve larger set sizes. Instead, the goal is to get an idea of the magnitude which a normal machine can handle.
To test this, write something like this:
make an empty set
while (true){
start timer
add 1000000 new elements to the set (you can just use 64 bit integers)
end timer and print time
}
You do not have to use the generator to produce items to be added to your set. Instead, you can just use a counter which is incremented after each add operation. You can stop the process once you notice that it seems to take a long time (>5 min) or when you run out of memory (which seems more frequent) to add the batch of elements.
Most likely, the timings will not be linear in any way. If the set you used is based on hashing, the time for adding elements is basically constant independent on the set size. However, once in a while the table used to store the values needs to be re-sized, which will take longer for larger sets.
The only thing you need to report is the amount of memory you have, the language and set type you used and how many elements you could store. If you have set specific memory options, you can report these as well.
### Part II
The teacher has placed a set of data on Amazon S3 which contains the first 10000-10^8 elements of the stream in bucket s3://data-for-stream-counting-5e33f2e9-bb84-4c40-bb03-5bb086c7adea/. The dataset is there in larger sizes as well, but processing these will take quite a bit of time. (You can browse S3, starting from here).
Using map-reduce, you need to calculate the number of unique elements exactly.
Computing distinct values using map reduce can be done fairly straightforward, you can implement it by following the pseudocode below.
map(key, value) -> value, null
reduce(key, values) -> 1, null
A combiner can be used to save communication, if the map function produces a lot of duplicates. It could be written as follows:
combine(key, values) -> key, null
You are advised to work with Java. Using Java, you need to export your mapreduce code to a runnable .jar file and place it in your S3 bucket. However, do not use anything from the old .mapred. API, use everything from the new .mapreduce. instead. Read the MapReduce Tutorial which was mentioned above.
In order to work on Amazon infrastructure, you first need to get your personal password from http://jyu-aws.appspot.com and log-in to AWS. There you can switch to the elastic map reduce service and read instructions on how to run map reduce jobs. All map-reduce tasks must be run on the cluster in availability zone us-east-1 (US East N.Virginia).
When starting jobs, use identifiers from which you can be identified (korppi ID) to avoid interference with others’ work. For storing any data, use your personal bucket for this course jyu-ties438-username-randomUDDI (already created).
Then you have to launch the map-reduce job as described in step 8 of EMR DeveloperGuide - launch-custom-jar.
The map-reduce job will generate a number of output files in your output bucket, each containing lines with a one. You can use these to determine the size by getting the size of each file and dividing it by two (each line is a one and a new-line, thus two bytes). You can also use counters (either internal ones or your own) to determine the size. Alternatively, you can use a second map-reduce job to determine the number of lines in the files.
First test on the small dataset and see whether you can get the correct result data10000 should give you 9925 distinct values. Then determine the sizes of the sets up till 10^8.
### Part III
Implement two algorithms for cardinality estimation. The first one uses (order) statistics (sampling or K-minimum values) and the second one bit pattern observables (Flajolet-Martin (from handbook), LogLog or HyperLogLog - see also hints). For both algorithms, you need to calculate the root-mean-square error for 1,000,000 stream elements.
You tell the generator to produce 10000 elements with a seed of 123 as follows java -jar generator.jar 10000 123. To calculate the error, you need to compute the estimated cardinalities for the 10 different seeds from the table below:
| seed | exact cardinality |
| | (1,000,000) |
| --- | ----------- |
| 1 | 697757 |
| 2 | 698466 |
| 3 | 698151 |
| 4 | 698536 |
| 5 | 697079 |
| 6 | 697732 |
| 7 | 697822 |
| 8 | 698286 |
| 9 | 698364 |
| 10 | 697910 |
Now, for both experiments (2 algorithms) , you can calculate the root-mean-square error as follows:
## Hints
1. Use small part of the data during development.
2. For the programming language, you might just want to choose the one you are most familiar with. Depending on the chosen language the teacher will be able to help more (or less) with language specific issues.
3. Randomized algorithms are difficult to debug. Make it somehow possible by fixing the seed of the random number generator. At least each run will be the same.
4. When using Windows, be careful with the command line pipe | operator. It seems that in some cases the shell buffers all output from the first process and only forwards it to the next after the first one ends. If the stream is very large, this will cause it to either write to the disk (slow) or fill up your memory.
5. If you are using a machine with more as 2 cores, you can start multiple experiments at the same time. Since we are not measuring running time, they won’t affect each other.
6. The easiest to develop the map reduce part is using a Maven project. The needed dependency is :
<dependency> | 1,820 | 7,945 | {"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} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.894706 |
https://www.learncram.com/calculator/inflection-point-calculator/ | 1,726,802,193,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00553.warc.gz | 807,916,395 | 16,448 | # Inflection Point Calculator | Calculate Inflection Point
Make use of this free handy Inflection Point Calculator to find the inflection points of a function within less time. Just enter function in the input fields shown below and hit on the calculate button which is in blue colour next to the input field to get the output inflection points of the given function in no time.
Inflection Point Calculator: Want to calculate the inflection point of a function in a simple way? Then you must try out this user friendly tool provided. It is one of the easiest ways that you ever find to compute the inflection point of a function. This page is all about Finding Inflection Point of the given function using a simple method and the interactive tutorial explaining each step of the process.
## Steps to Find Inflection Point
Follow the below provided step by step process to get the inflection point of the function easily.
• Take any function f(x).
• Compute the first derivative of function f(x) with respect to x i.e f'(x).
• Perform the second derivative of f(x) i.e f”(x) and also solve the third derivative of the function.
• f”'(x) should not be equal to zero.
• Make f”(x) equal to zero and find the value of variable.
• Substitute x value in the third derivative of function to know the minimum and maximum values.
• Replace the x value in the given function to get the y coordinate value.
• Then, inflection points will be (x value, obtained value from function).
Find a variety of Other free Maths Calculators that will save your time while doing complex calculations and get step-by-step solutions to all your problems in a matter of seconds.
Example
Question: Find the inflection points for the function f(x) = -x4 + 6x2?
Solution:
Given function is f(x) = -x4 + 6x2
f'(x) = -4x3 + 12x
f”(x) = -12x2 + 12
f”'(x) = -24x
f”(x) = 0
-12x2 + 12 = 0
12 = 12x2
Divide by 12 on both sides.
1 = x2
x = ± 1
x1 = 1, x2 = -1
Substitute x = ± 1 in f”'(x)
f”'(1) = -24(1) = -24 < 0, then it is left hand bit to right hand bi.
f”'(-1) = -24(-1) = 24 > 0, then it is left hand bit to right hand bit.
Replace x = ± in f(x)
f(1) = -1+6 = 5
f(-1) = -1 +6 = 5
Therefore, inflection points are P1(, 5), P2(-, 5)
### FAQs on Inflection Point Calculator
1. How do you find inflection points on a calculator?
Provide your input function in the calculator and tap on the calculate button to get the inflection points for that function.
2. What does inflection point mean?
Inflection point is defined as the point on the curve at which the concavity of the function changes. It can be a stationary point but not local maxima or local minima.
3. Find the point of inflection for the function f(x) = x5 – 5x4?
Given that
f(x) = x5 – 5x4
f'(x) = 5x4 – 20x3
f”(x) = 20x3 – 60x2
f”'(x) = 60x2 – 120x
Neccessary inflection point condition is f”(x) = 0
20x3 – 60x2 = 0
20x2(x-3) = 0
x1 = 0, x2 = 3
Substitute x2 = 3 in the f(x)
f(3) = 35 – 5*34 = 243 – 405 = -162
Inflection Point is (3, -162).
4. What is the difference between inflection point and critical point?
A critical point is a point on the graph where the function’s rate of change is altered wither from increasing to decreasing or in some unpredictable fashion. Inflection point is a point on the function where the sign of second derivative changes (where concavity changes). A critical point becomes the inflection point if the function changes concavity at that point. | 973 | 3,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.782219 |
http://dynorecords.g6.cz/2022/07/we-hope-which-take-action-will-help-the-students/ | 1,675,030,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00356.warc.gz | 13,041,008 | 10,701 | # We hope which take action will help the students to start observe the partnership between mass, frequency and you will thickness
We hope which take action will help the students to start observe the partnership between mass, frequency and you will thickness
Even when their dimensions aren’t perfect (zero proportions ever are) you need the calibrated Pounds Level to discover the pounds of assorted something while can even utilize it so you’re able to amount pennies. Today comes a very important matter: For those who got your own calibrated Pounds Scale with the Moon, would it not really works the same way they did in the world? For many who occupied it with the exact same quantity of pennies, do the latest elastic band extend on the same draw? Have your teacher talk about which along with you and find out for many who is also see the difference between weight and bulk.
Occurrence informs us how much cash things have started packaged toward a specific amount of space. Head is very thick, Styrofoam is not all that thick whatsoever. Thickness will likely be mentioned during the g per cubic centimeter. Inside our experiments, occurrence could well be measured from inside the „pennies for every single glass“. On your tests on the Bulk Equilibrium, you always counted the brand new bulk of the same amount of matter. Make use of the investigation you took so you can record the material your counted managed out-of thickness with thick over the top of your number right down to the least heavy towards the bottom of the listing. It’s not necessary to assess the newest thickness when you look at the „pennies for each mug“ because you constantly measured an identical quantity hookup in Nottingham of thing–simply look at your data to make the record.
Once you’ve made the a number of the fresh densities of the materials, consider the pursuing the very important matter: How do you believe the new density regarding a substance carry out alter for those who measured it towards Moonlight unlike into the Environment?
1. You really have an item while wish to know when it will drift in the water. To answer the question: „can it float?“ do you need to know the items bulk, weight otherwise density?
dos. A beneficial student’s bulk on the planet are 50 kilograms. In the event it pupil visited new Moonlight, create the lady bulk become more, smaller, or perhaps the same?
## A good student’s weight on the planet is one hundred lbs
cuatro. A beneficial stop off timber without difficulty floats to your liquid when into Environment. Should your same cut off out of wood was basically delivered to the fresh Moon, wouldn’t it drift into the liquids? (Most provide that it particular careful thought, most people can not address it matter.)
The objective of it hobby should be to have a look at the meaning regarding size, lbs and you may occurrence because of the considering exactly how each could well be mentioned.
A very important question to look at now is: For people who put it Mass Harmony on moon or into ount out of situation on one hand require exact same level of cents on the reverse side in order to balance it they performed on earth? Of course there is no simple way for all of us to perform for example an experiment however,, having your children look at this will be enable them to to start knowing the difference in weight and you will size. Mass otherwise while the Newton would say, the quantity of number for the an object, does not changes when you replace your location in dimensions but, once we will find soon, weight does changes.
Calculating Occurrence: While the thickness was mass each volume, many direct measure of the fresh density from anything will be to level their mass, upcoming scale its frequency and you can separate the fresh new mass of the regularity. If you have a graduated cylinder (they’re not high priced but the majority basic colleges don’t possess them) you could use they with a few drinking water in order to draw the little „section cups“ on particular volumes. (You will find already suggested your short step one ounce glasses have a tendency to hold ten cubic centimeters when occupied so you’re able to a point 1.4 cm above the base and it will surely keep 20 cubic centimeters whenever occupied so you can a time 2.step three centimeters above the bottom.) In place of in fact computing brand new occurrence, we feel it would be enough towards the people to comprehend your same frequency is going to be an enormous size or good small bulk based upon the materials in it. Our very own plan would be to have the same quantity of several different materials and scale its size into Mass Equilibrium. | 940 | 4,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-06 | latest | en | 0.952501 |
https://forums.moneysavingexpert.com/discussion/5050997/energy-prices-think-im-paying-over-the-odds | 1,627,477,247,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00085.warc.gz | 263,107,109 | 34,043 | MoneySavingExpert Chair, Martin Lewis · Editor, Marcus Herbert
# Energy prices - think i'm paying over the odds
edited 30 November -1 at 1:00AM in Energy
5 replies 1K views
Forumite
14.1K Posts
✭✭✭✭✭
edited 30 November -1 at 1:00AM in Energy
So looking at saving some ££ on gas & electricity (pre-pay). We are on a BG std tariff.
Current:
1.82/week standing charge (26p per day)
13.21/kwh
Don't know usage, but we spend about £30 per month electricity.
So £30 - £7.80/month standing charge = £22.20 usage. Divided by cost/kwh = 168kwh/month usage.
So looking at a quote by First Utility.
18.69 per day + 11.49p per kwh
Therefore
168*11.49p = £19.30
+ standing charge p/m of £5.60
= £24.90 - saving about a fiver for electricity alone.
Or am I missing something in my calculations?
## Replies
• edited 31 August 2014 at 10:49PM
Forumite
5K Posts
edited 31 August 2014 at 10:49PM
No, you are not missing things. (Although make sure prices used are both including VAT).
You can always just use a comparison site, though, like energyhelpline or uswitch so that you do not need to make individual calculations.
• Forumite
0 Posts
arcon5 wrote: »
So looking at saving some ££ on gas & electricity (pre-pay). We are on a BG std tariff.
Current:
1.82/week standing charge (26p per day)
13.21/kwh
Don't know usage, but we spend about £30 per month electricity.
So £30 - £7.80/month standing charge = £22.20 usage. Divided by cost/kwh = 168kwh/month usage.
So looking at a quote by First Utility.
18.69 per day + 11.49p per kwh
Therefore
168*11.49p = £19.30
+ standing charge p/m of £5.60
= £24.90 - saving about a fiver for electricity alone.
Or am I missing something in my calculations?
as a low user you are better off with the lowest cost per kWh combined with a reasonable daily charge
• Forumite
5K Posts
168 kWh per month (on a dual fuel household) is not a specially low consumption.
• Forumite
1.8K Posts
168 kWh per month (on a dual fuel household) is not a specially low consumption.
It is less than half the average UK domestic consumption (4,600 kWh) - I would call that low.....................
• 2.9K Posts
It is less than half the average UK domestic consumption (4,600 kWh) - I would call that low.....................
The UK average is 3,200 KWh. 4,600 is for 2 rate meters, the OP has stated they have gas.
https://www.ofgem.gov.uk/sites/default/files/docs/decisions/tdcv_decision_letter_final_2.pdf
This discussion has been closed.
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http://excel.bigresource.com/Getting-Summary-Statistics-From-Data-Arrays-JS4gndax.html | 1,601,294,096,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00673.warc.gz | 35,868,953 | 13,914 | # Getting Summary Statistics From Data Arrays
Jun 1, 2007
I have formed data arrays in VBA after running a time series simulation model. The array is m simulations x n periods. Small example: ...
## Summary Of Horse Racing Statistics
May 6, 2009
i need to do the following on the attached spreadsheet using a macro: SHEET 1. This is an example of the original data i will be working with i need
1) You will notice that the Min and Max columns are not all together, i need these to all be together. I then only want the Race course, race time, date and type, Min, Max and ratings.
2) I then want the following formula added to Column F, Max minus Min. In column G i would like the following formula Rating minus Min. And finally in Column H Rating minus Max.
3) I would the like to Highlight in Yellow the horse with the highest possible number. I want this done for each column (F,G,H) in each race i would like all Horses that dont have a yellow cell in any of the Columns (F,G,H) to be deleted, leaving me with just the highlighted horses. An example of the required final result is contained in Sheet 2 of the attached spreadsheet.
## Organizing Statistics (manipulate Team Statistics)
Mar 30, 2009
I want to manipulate team statistics and having a bit of trouble trying to figure out how to do it. I am relatively new to excel and am using Excel 2007.
I have attached the excel file for your reference (the same file).
What I need to do is first find out the team number using the table on first sheet named "Teams". User will enter team name on call B4 of sheet "Team entry" then in cell C4 there should be a formula to find the team number (is beside name on sheet "Teams) and displays it (on C4). Then on cell D4 of the same sheet "Team entry" there should be a formula that displays cell A1 (shows a statistic for that team) of the team sheet. Each team has its own stat sheet named by its team number (easier for me to keep track). So basically D4 should show cell A1 of the sheet that shows stats of the team entered. In addition cell E4 of "Team entry" should display stat2 (of the sheet "stat2")for the team entered.
## Statistics: How To Recode Data
Oct 9, 2002
I have been looking everywhere for this. In SPSS, it is very easy to recode data, but how do you do this in Excel? For example, researchers often "reverse" their questions on surveys with Likert scales to eliminate respondent errors. In those questions, you want 1=5, 2=4, 3=3, 4=2, 5=1. Is there a plug-in or some other formula that is out there?
## Football Statistics From 2 Data Columns
Oct 6, 2006
I have a simple spreadsheet recording games played on my football table.
Each match is the first to 10 goals. So each result is a simple: ...
## How To Create Table Depicting Trailing Monthly Data / Statistics
Dec 6, 2011
I am improving an old spreadsheet that contains a tab for inputting the following data, all using data validation:
date
employee (who did the work)
team (for whom work was done)
department (for whom work was done)
type (of work done)
Project (which property worked on)
time (spent working on this)
I have a sheet with a drop down in which you can sort by employee and the associated chart and graphs adjust the information accordingly.
I need to create a chart and graph in which I can track the trailing month as well as year to date. Ideally, I would like to sort by the following:
- This month (i.e., the 1st of the month to present)
- previous month (etc., be able to select all previous months for which data was entered, i.e. Jan., Feb. March. etc.)
- Year to date
I know there is a way to do this - I thought maybe it would be through some formula incorporating =today() and subtracting back to what you need using numeric dates?
## Slicing And Dicing CSV Files - Involves Arrays And Jagged Arrays
May 8, 2013
I am retrieving a CSV file from the net. In this file there are 'x' amount of row data and 7 columns. I only care about the values in the 7th column for each row. I also don't care about the entire first row. A graphical version would be represented something like this, with the values I want colored in orange:
|---,---,---,---,---,---,---|
|---,---,---,---,---,---,---|
|---,---,---,---,---,---,---|
|---,---,---,---,---,---,---|
|---,---,---,---,---,---,---|
|---,---,---,---,---,---,---|
|---,---,---,---,---,---,---|
.
. extending until the end of the data set
.
I've managed to dice this thing into a jagged array by first splitting it using vbLf as a delimiter, and therefore adding those to an array called Lines(). Then I split Lines() up using commas as the delimiter and threw those into a jagged array, let's call it Breadcrumbs()(). I want to throw all the values from Breadcrumbs(i)(6) into an array of its own. Here's my code so far:
Code:
Public Sub CSVparser(file As String)
Dim Lines As Variant
Dim j As Integer
Lines = Split(file, vbLf)
For i = 1 to UBound(Lines) - 1
Next i
End Sub
## Macro To Create A Statistics Table From Another Data Table (containing Merged Cells)
Apr 14, 2009
I would like to have a macro to automatically generate a statistics table (on the "statistics" tab) with the 5 following fields:
Fragment names / # samples / # of failed samples / % of success / # of variations in the fragment (SNP). At the bottom of this table, I would like to have a cell with the average % of success for all fragments. The data to generate these statistics are on the "gene name" tab (please note that this name will change every time I will work on a new gene). To make things easier, I think the macro should be run from this tab.
1. The Fragment names are displayed in row #5. I use one column per variation per fragment. If one fragment has 3 variations, there will be three columns and I will merge together the fragment name cells. The fact that some cells are merged can be a problem when copy-paste to the stats table (as I would like to get rid of the merging).
2. # of samples corresponds to the number of cells in blue in column A. The number of samples can change from one report to another but is always constant in the same report.
3. # of failed sequences. In the table, I type "Failed Sequence" (if the analysis has failed) and "Missing Sequence" (if the analysis has not been done). When a sample is failed or missing, it is for the who fragment, no matter how many variation there is in the fragment, so I usually merge the cells of all variations for this failed sample.
4. % of success: this is quite easy #sample/#of failed+missing sequence for this fragment
5. # of variation is equal to the number of variations for this fragment (can be 0, 1, 2, etc.). When there is no variation in a fragment, I put '-- in all cells of the corresponding fragment on the "gene name" tab. Fragment 3 on my file is an example of 0 variation.
## VBA Using Arrays For Copying Data From A To B
Jan 14, 2009
Though I am familiar with the use of Array in formulas I can't seem to get my head around the way arrays are used in the VB world.
It is my understanding that you can read and write data into an array for copying/moving it around and reordering.
This is a simplified version of what I am trying to do.
I have some data in a row, say:
1, 2, 3, 4, 5, 6, 7, 8 etc.
I would like to poplute an array and then "Write" the data into a range so that it comes out something like:
2, 4, 1, 3, 8, 6, 7, 5 etc
I have tried to play around with array in Vb but could not seem to get anywhere.
Am I mistaken that is is possiable?
If someone could point me in the wite direction I would appricate it.
I have searched for this online and have found examples that I either did not understand or was unable to adapt...
## Excel 2010 :: Summary All Statement From 4 Worksheets And To Summary All Total ICC
Dec 23, 2013
excel 2010. This workbook has 4 worksheet(Process Engineer,OSBL,OSA,Lab Operator) I want to know what is the best excel formula/function to summary this 4 worksheet.
Example:I want a formula/function to summary all the statement from 4 worksheets and total number of answer "1" per statement from 4 worksheet.
Sample Statement below
"Demonstrate Interpersonal (People-to-People-) Skills" Question:What is the formula if above statement contains this statement in 4 worksheet?As i checked the total is 4 then What is the formula to get all total answered ICC on this statement from 4 worksheet?
## Paste In 9x16 Arrays Of Data
Dec 12, 2008
I am looking for code to do the following:
1. Copy data from Column A in Sheet1 (could be up to 1000 rows)
2. Paste the data in Sheet2 in "arrays" of 9 columns and 16 rows
So basically, the first table would contin the first 16x9 = 144 numbers that are being copied and pasted from Sheet1. The second table would include numbers 145 to 288 and so on.
## Structures Or An Arrays Of Different Data Types.
Sep 19, 2008
I an newbie at VBA but I took some short programing classes back at my college days. I am trying to declare an array with different data types and since that seems to be imposible for what I gather then my other option is to declare what I remember as a structure.
## Finding Missing Data Using Arrays And 2 Tables
Aug 25, 2014
Any easy way to retrieve data from another table using an array formula.
I have two files that I am using that contain 2 sets of data with columns for name, address, city, and state. The red highlighted data needs be used to find the blue data first column, which is a possible name for the company found from the red data. The issue is that the blue data is larger and has rows of data that will not be found in the other table.
I have the spreadsheet attached. I attempted to use an if/and statement :
=IF(AND(D2=C8I:I,E2=J:J),H2,0)
but found out that it does not work with arrays and only found the first address by default.
How would I set up a formula to retrieve the possible names using criteria from the blue data such as if the address and city match, then input that company name?
## Textboxes, Data Arrays And Vlookup Functionality
Nov 20, 2008
I would like to select a item from a list and have a text box display data from the next column (corresponding row) Also, I would love if I could type something into a second textbox and have that copy onto a third column (again, corresponding row) Also, if the add comment command button could transfer that record to the "commentted items' sheet. I have attached an example.
## Multiple Arrays - Matching Data To Items
Jun 20, 2012
I have code that runs through multiple arrays trying to match data to items in the arrays and it takes a long time to run.
Code:
dim a as long
dim b as long
dim c as long
dim d as long
dim e as long
dim MyAarray as variant
[Code] ..........
That's basically what the code does. however, it takes an extremely long time to get through with everything as each array increases in size.
## Subtracting Two Arrays Of Data And Putting The Result In Some Cells
Dec 11, 2009
I was trying to subtract two arrays of data and putting the result in some cells but with no success unfortunately. I'm relatively new to VBA and I'm just starting now to make calculations with arrays so excuse my little knowledge. The arrays that I'm trying to subtract are from row 1 to 250 and m and n variables have the number of the columns. Here is my routine:
## Entering Date/time Data As Xvalue For Scatterplot With Vba Arrays
Mar 31, 2007
I am trying to create/modify an XY scatterplot using VBA.
I declare local variables and point them to the chart and a new data series for the chart, such as:
Dim chartone As Chart
Dim chartseries As Series
Set chartone = ThisWorkbook.Charts("Chart1")
chartone.ChartType = xlXYScatterLines
Set chartseries = chartone.SeriesCollection.NewSeries()
I set the series data, using an array, such as:
chartseries.Values = Array(1, 3, 5, 7, 9, 11)
I can set the Xvalues to a set of dates by setting .Xvalues to a woksheet range that includes date-formatted data. (like this):
chartseries.XValues = Worksheets("sheet3").Range("m9:m14")
(where m9:m14) contains dates...
## Sum Horizontal Data In Vertical Summary Where Data Matches?
Jul 9, 2014
I have a page of data that i need to summarise/calculate, i thought sumif would be the correct formulae but i can't get it to work...
Sheet 1 - Data Recomds Emp Name, Weeks 1-52 showing no of hours to adj
Name
WK1
WK2
Wk3
WK4
WK5
WK6
Oliver
-1.5
[Code] .....
Sheet 2 - Summary by month - to Calculate the no of hours for the period per employee
Name
Month 1
Avery
Require Sum for employee Avery Wk 1-4
[Code] ..........
## Lotto Statistics
Aug 23, 2007
Been searching the archives but not finding my answers. I downloaded historical lottery numbers and want to run statistics on them- just no clue where to start. I guess the 1st and most important thing is to figure out which #s have come up most often out of all of them. I assume it's some sort of COUNT or LOOKUP code- but not sure. I have the #s arranged in a sheet like so ...
## Descriptive Statistics
Jun 27, 2008
want to make a macro to do a 'descriptive statistics data analysis' of column b and put the results in m1 to n18.
basically doing this but in a macro.
add in= tools /data analysis/ descriptive stats
I have recorded it doing this, but it for some reason can't do what it wrote.
## SUMPRODUCT Statistics
Mar 9, 2009
On Sheet1 I have 2 cells one has the date 1-mar-09 and the other has 7-mar-09. I need to find this range on Sheet2 which has all the dates for the year in ColumnA and then total the amounts found in Sheet2!ColumnB for the specified date range. Is this possible without VB?
## Pull Some Statistics From A Workbook
Jan 27, 2009
I'm trying to pull some statistics from a workbook I've been sent and am having some trouble working out how to achieve this.
I have a small interface that allows the user to choose a possible answer (drop-down menu) from a questionnaire (Yes, No, Maybe)(cell: C6). On the interface the user also picks the question they wish to see the stats for (cell: C5)
The data I have been sent has been set up with the questions along in row A, and the answers below in each column (the answers run across row A from column G - AH).
So, what I'm basically after is a formula that first looks up the question specified in C5 (I've used HLOOKUP to pull some other data), and then counts how many times the Yes, No or Maybe answers appears in the column where the question data is held.
## Statistics Of Random Numbers
Dec 3, 2009
I have two rows in a sheet with random numbers (1 till 90).
Those numbers in those two rows I change them every 5 minutes with new one still from 1 to 90
I need a "way", in one side of my sheet (lets say in column B) to count how many times a number was repeated after finish my work (example after 2 hour's)
Is that possible with excel 2007?
May 16, 2008
I have a table of contents page. The TOC is updated each time some one opens the sheet (this is a must have based on the requirements). I'm trying to include some statistics next to each item in the TOC. These statistics are on each worksheet and just need to be copied to the TOC.
Now my problem.
The worksheets are NOT keep in alphabetical order, so After the TOC is created a sort is done to put the TOC links in alphabetical order. After the order is set, I need a formula that will read the worksheet name from a cell on the TOC (which is really a hyper link to a worksheet).
If "=Address!" (Where address is the name of one of the sheets) could be replaced with a cell reference that (I think) would do the trick.
My TOC is a MACRO not a cell formula so if this can not be done with a cell formula but can be done through VBA that would be fine.
## Bowling Statistics Formula
Jul 6, 2008
I have running total of bowling scores current & going back from when I first started.
What I’m trying to archive is a formula that will count the most recent 96 scores and still allow me to add new scores each week in keeping the current running total of 96 games.
The statistics are listed from A4:C215 that may have 1 or 2 scores not listed. (missed games)
Each week new statistics will be added in cell A4:C4 anywhere’s of 1-3 games.
Sep 11, 2008
Back to test you with a few problems as I work through a project I'm carrying out, based on past results data of the English Premier League.
I have some of the solutions/formulas I require already, but there are still some gaps to be filled in!
***
the spreadsheet is set out as follows:
Key Columns / Sub-Title (Remarks)
B Date (the date a match was played - the s.sheet is sorted by this column, A>Z)
C Home Team
D Away Team
E Home team goals scored
F Away team goals scored
I-M Indicates via U or O whether the match contained under or over 0.5,1.5,2.5,3.5,4.5 goals
N Points obtained for Home team (i.e. 3 for a win, 1 for a draw, and 0 for a loss)
O Points obtained for Away team (i.e. 3 for a win, 1 for a draw, and 0 for a loss)
There are 381 rows in total for each worksheet (season) - 1 for the titles, and 2-381 for each match played in the season...................
## Economic Statistics Forecast
Mar 3, 2007
I try to predict some macro economic statistics but any attempt till now didn't make sense. the attached file. Note: when i used the FORECAST function the predicted values showed an unlogical drop while there seems to be a positive trend.
## Statistics From Golf Scores
Mar 14, 2008
I am creating a workbook to manage golfer scores, teams, winnings, handicap, ect. I am having trouble with the statistics sheet. I Need A Sub To:
1. Column "B" , take the average of the lowest 5 numbers in columns "AB" though "AU"
2. Column "C" , take the average of the lowest 10 numbers in columns "G" through "Z" divided by 0.96
3. Column "D" , take the average of the lowest 10 numbers in columns "AW" through "BP"
this needs to happen for each row where there is a name. (names added daily)
I have included a similar sheet as the one i am working with along with some command buttons typically not included. Should show some of the problems I am having.
## Statistics By Date Range
Apr 1, 2008
I need to record on Sheet 2 how many times enteries in Coloum B on Sheet 1 appear per date range.
## Summary Of SUMPRODUCT Data
May 10, 2009
For instance, in travels.xls a table in the 'data' sheet lists destinations versus people and the data of travel.
What i'd like is to assign a macro that would have a breakdown of the number of travels per destination that month. Where there is no travel that month, that destination is omitted.
I can work out how to use SUMPRODUCT to produce the number of trips per destination within that month, but stuck as to how to get it to display it using a macro.
Should I use a loop to loop through all destinations, copy that information to the separate sheet and then another loop to delete destinations with zero trips? | 4,715 | 18,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.922827 |
http://www.ask.com/question/how-long-does-it-take-to-play-9-holes-of-golf | 1,405,202,164,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776435465.20/warc/CC-MAIN-20140707234035-00025-ip-10-180-212-248.ec2.internal.warc.gz | 172,122,056 | 16,166 | # How Long Does It Take to Play 9 Holes of Golf?
Golf is a sport that requires patience, as it moves very slowly. There are several factors that influence how long it takes to play nine holes of golf. This includes how many people are on the course, how many people are golfing in one party, as well as how fast each golfer completes each hole. Generally, an estimate of how long a nine hole course takes to complete is around four hours, although this can be more or less time depending upon the factors listed.
Reference:
Q&A Related to "How Long Does It Take to Play 9 Holes of Golf?"
Pace Per Hole A golf hole takes an average of about 15 minutes to play--including walking time between tees and shots--but there is significant variation from hole to hole. Some pacing http://www.ehow.com/how_5187453_long-play-holes-go...
It depends on the course and how busy the course is. Were i live it can take about an hour and a half and 2 hours when it is full or on mens/womens night it can take up to 3 to 3 http://wiki.answers.com/Q/How_long_does_it_take_to...
there are numbers called par if its a par five then that means that you have five shots to get the ball in the hole for par, which is three shots to get to the green and two to put, http://answers.yahoo.com/question/index?qid=200907...
I am sorry I do not golf. My best score in mini golf would be 35. http://www.chacha.com/question/what-is-your-worst-... | 340 | 1,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-23 | longest | en | 0.933962 |
http://economicskey.com/problems-and-appucations-7160 | 1,493,444,343,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123276.44/warc/CC-MAIN-20170423031203-00532-ip-10-145-167-34.ec2.internal.warc.gz | 118,224,859 | 12,958 | # PROBLEMS AND APPUCATIONS
## Economics Assignment Help Online
PROBLEMS AND APPUCATIONS
1. Suppose the natural rate of unemployment is 6 percent. On one graph, draw two Phillips curves that describe the four situations listed here. Label the point that shows the position of the economy in each case.
a. Actual inflation is 5 percent and expected inflation is 3 percent.
b. Actual inflation is 3 percent and expected inflation is 5 percent.
c.. Actual inflation is 5 percent and expected inflation is 5 percent.
d. Actual inflation is 3 percent and expected inflation is 3 percent.
2. Illustrate the effects of the following developments on both the short run and long run Phillips curves. Give the economic reasoning underlying your answers.
a. A rise in the natural rate of unemployment
b. A decline in the price of imported oil
c. A rise in government spending
d. A decline in expected inflation
3. Suppose that the economy is currently at full employment and investment rises.
a. Illustrate the immediate change in the economy using both an aggregate-supply aggregate demand diagram and a Phillips-curve diagram. What happens to inflation and unemployment in the short run?
b. Now suppose that over time expected inflation changes in the same direction that actual inflation changes. What happens to the position of the short-run Phillips curve? After the expansion ends, does the economy face a better or worse set of inflation unemployment combinations?
4. Suppose the economy is in a long-run equilibrium.
a. Draw the economy’s short-run and long-run Phillips curves.
b. Suppose a wave of business pessimism reduces aggregate demand. Show the effect of this shock on your diagram from part (a). If the Fed undertakes expansionary monetary policy, can it return the economy to its origin ,I inflation rate and original unemployment rate?
c. Now suppose the economy is back in long-run equilibrium, and then the price of imported oil rises. Show the effect of this shock With a new diagram like that in part (a). If the Fed undertakes expansionary monetary policy, can it return the economy to its original inflation rate and original unemployment rate? If the Fed undertakes contraction monetary policy, can it return the economy to its original inflation rate and original unemployment rate? Explain why this situation differs from that in part (b).
5. The inflation rate is 10 percent, and the central bank is considering slowing the rate of money growth to reduce inflation to 5 percent. Economist Milton believes that expectations of inflation change quickly in response to new policies, whereas economist James believes that expectations are very sluggish. Which economist is more likely to favor the proposed change in monetary policy? Why’)
6. Suppose the Federal Reserve’s policy is to maintain low and stable inflation by keeping unemployment at its natural rate. However, the Fed believes that the natural rate of unemployment is 4 percent when the actual natural rate was 5 percent. If the Fed based its policy decisions on its belief, what would happen to the economy? How might the Fed come to realize that its belief about the natural rate was mistaken?
7. Suppose the price of oil falls sharply (as it did in 1986 and again in 1998).
a. Show the impact of such a change in both the aggregate-demand aggregate supply diagram and in the Phillips-curve diagram. What happens to inflation and unemployment in the short run?
b. Do the effects of this event mean there is no short-run trade-off between inflation and
unemployment? Why or why not?
8. Suppose the Federal Reserve announced that it would pursue contractionary monetary policy to reduce the inflation rate. Would the following conditions make the ensuing recession more or less severe? Explain.
a. Wage contracts have short durations.
b. There is little confidence in the Fed’s determination to reduce inflation.
c. Expectations of inflation adjust quickly to actual inflation.
9. Given the unpopularity of inflation, why don’t elected leaders always support efforts to reduce inflation? Many economists believe that countries can reduce the cost of disinflation by letting their central banks make decisions about monetary policy without interference from politicians. Why might this be so?
10. Suppose Federal Reserve policymakers accept the theory of the short-run Phillips curve and the natural-rate hypothesis and want to keep unemployment close to its natural rate. Unfortunately, because the natural rate of unemployment can change over time, they aren’t certain about the value of the natural rate. What macroeconomic variables do you think they should look at when conducting monetary policy?
### Related Economics Assignments
Other Similar Economics Assignments like | 975 | 4,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.886078 |
https://nl.mathworks.com/matlabcentral/cody/problems/1084-square-digits-number-chain-terminal-value-inspired-by-project-euler-problem-92/solutions/2002737 | 1,586,086,234,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371576284.74/warc/CC-MAIN-20200405084121-20200405114121-00553.warc.gz | 610,314,079 | 16,593 | Cody
Problem 1084. Square Digits Number Chain Terminal Value (Inspired by Project Euler Problem 92)
Solution 2002737
Submitted on 4 Nov 2019
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
assert(digits_squared_chain(649) == 1)
2 Pass
assert(digits_squared_chain(79) == 1)
3 Pass
assert(digits_squared_chain(608) == 1)
4 Pass
assert(digits_squared_chain(487) == 1)
5 Pass
assert(digits_squared_chain(739) == 1)
6 Pass
assert(digits_squared_chain(565) == 1)
7 Pass
assert(digits_squared_chain(68) == 1)
8 Pass
assert(digits_squared_chain(383) == 1)
9 Pass
assert(digits_squared_chain(379) == 1)
10 Pass
assert(digits_squared_chain(203) == 1)
11 Pass
assert(digits_squared_chain(632) == 1)
12 Pass
assert(digits_squared_chain(391) == 1)
13 Pass
assert(digits_squared_chain(863) == 1)
14 Pass
assert(digits_squared_chain(13) == 1)
15 Pass
assert(digits_squared_chain(100) == 1)
16 Pass
assert(digits_squared_chain(236) == 1)
17 Pass
assert(digits_squared_chain(293) == 1)
18 Pass
assert(digits_squared_chain(230) == 1)
19 Pass
assert(digits_squared_chain(31) == 1)
20 Pass
assert(digits_squared_chain(806) == 1)
21 Pass
assert(digits_squared_chain(623) == 1)
22 Pass
assert(digits_squared_chain(7) == 1)
23 Pass
assert(digits_squared_chain(13) == 1)
24 Pass
assert(digits_squared_chain(836) == 1)
25 Fail
assert(digits_squared_chain(954) == 89)
Assertion failed.
26 Fail
assert(digits_squared_chain(567) == 89)
Assertion failed.
27 Fail
assert(digits_squared_chain(388) == 89)
Assertion failed.
28 Fail
assert(digits_squared_chain(789) == 89)
Assertion failed.
29 Fail
assert(digits_squared_chain(246) == 89)
Assertion failed.
30 Fail
assert(digits_squared_chain(787) == 89)
Assertion failed.
31 Fail
assert(digits_squared_chain(311) == 89)
Assertion failed.
32 Fail
assert(digits_squared_chain(856) == 89)
Assertion failed.
33 Fail
assert(digits_squared_chain(143) == 89)
Assertion failed.
34 Fail
assert(digits_squared_chain(873) == 89)
Assertion failed.
35 Fail
assert(digits_squared_chain(215) == 89)
Assertion failed.
36 Fail
assert(digits_squared_chain(995) == 89)
Assertion failed.
37 Fail
assert(digits_squared_chain(455) == 89)
Assertion failed.
38 Fail
assert(digits_squared_chain(948) == 89)
Assertion failed.
39 Fail
assert(digits_squared_chain(875) == 89)
Assertion failed.
40 Fail
assert(digits_squared_chain(788) == 89)
Assertion failed.
41 Fail
assert(digits_squared_chain(722) == 89)
Assertion failed.
42 Fail
assert(digits_squared_chain(250) == 89)
Assertion failed.
43 Fail
assert(digits_squared_chain(227) == 89)
Assertion failed.
44 Fail
assert(digits_squared_chain(640) == 89)
Assertion failed.
45 Fail
assert(digits_squared_chain(835) == 89)
Assertion failed.
46 Fail
assert(digits_squared_chain(965) == 89)
Assertion failed.
47 Fail
assert(digits_squared_chain(726) == 89)
Assertion failed.
48 Fail
assert(digits_squared_chain(689) == 89)
Assertion failed. | 866 | 3,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-16 | latest | en | 0.522905 |
https://kidsworksheetfun.com/measurement-worksheets-grade-2/ | 1,726,639,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00612.warc.gz | 307,135,705 | 27,139 | 48 filtered results clear all filters 48 filtered results grade. Our second grade measurement worksheets are sure to add yet another dimension to your child s understanding of math and numbers.
Measurement Worksheets Grade 2 Length Google Search Measurement Word Problems Word Problem Worksheets Money Math Worksheets
### Both the customary and metric systems are covered.
Measurement worksheets grade 2. Grade 2 measurement worksheets our grade 2 measurement worksheets focus on the measurement of length weight capacity and temperature. This page has lots of basic fraction worksheets. Objects of unknown weight are weighted by balancing them against a number of objects of known weight these worksheets are pdf files.
2nd grade measurement worksheets. Measurement worksheets 1st grade measurement 3rd grade measurement 4th grade measurement 5th grade measurement 6th grade measurement capacity. Measuring grade 2 is a complete unit with lessons and exercises dealing with measuring length and weight.
Some of the worksheets for this concept are grade 2 measurement work grade 2 measurement work maths work third term measurement grade 2 measurement word problems math measurement word problems no problem weight estimating grams and kilograms measuring units work. Types on these worksheets students will use a protractor to determine the measurement of angle. Identify fractions of a shape as well as fractions of a set.
View the full list of topics for this grade and subject categorized by common core. Click on the images to view download or print them. A brief description of the worksheets is on each of the worksheet widgets.
Preschool kindergarten grade 1 grade 2 grade 3. Measurement using non standard units is reviewed and standard measurement units are introduced. These are complete worksheets that include direct instructions for the student and ample practice problems including word problems.
All worksheets are free for individual and non commercial use. Grade 2 weight measurement displaying top 8 worksheets found for this concept. Great for independent practice homework the student measu.
Measuring weights in kilograms on a balance scale below are three versions of our grade 2 math worksheet on using a balance scale to measure the weights of objects in kilograms. From basics such as how to read a ruler to more complex concepts like the difference between inches and centimeters these second grade measurement worksheets teach kids how to measure liquids and solids of different sizes. Printable worksheets learning games educational videos filters 48 results filters.
Please visit measurement to browse more worksheets in the same area. They ll also learn types of angles acute obtuse right etc. Kindergarten measurement length metric system telling time temperature volume weight.
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Grade 2 Measurement Worksheets Free Printable Measurement Worksheets Capacity Worksheets Measurement Conversions | 818 | 4,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.921204 |
https://versionweekly.com/news/age-calculator/ | 1,718,631,998,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00574.warc.gz | 543,403,877 | 15,130 | Age Calculator – How old am I? – How to Calculate Age?
The Age Calculator can determine the age or interval between two dates. The calculated age will be displayed in years, months, weeks, days, hours, minutes, and seconds.
Age Calculator
The age of a person can be counted differently in different cultures. This calculator is based on the most common age system. In this system, age grows at the birthday. For example, the age of a person that has lived for 3 years and 11 months is 3 and the age will turn to 4 at his/her next birthday one month later. Most western countries use this age system.
In some cultures, age is expressed by counting years with or without including the current year. For example, one person who is twenty years old is the same as one person who is in the twenty-first year of his/her life. In one of the traditional Chinese age systems, people are born at age 1 and the age grows up at the Traditional Chinese New Year instead of birthday. For example, if one baby was born just one day before the Traditional Chinese New Year, 2 days later, the baby will be at age 2 even though he/she is only 2 days old.
In some situations, the months and days result of this age calculator may be confusing, especially when the starting date is the end of a month. For example, we all count Feb. 20 to March 20 to be one month. However, there are two ways to calculate the age from Feb. 28, 2015 to Mar. 31, 2015. If thinking Feb. 28 to Mar. 28 as one month, then the result is one month and 3 days. If thinking both Feb. 28 and Mar. 31 as the end of the month, then the result is one month. Both calculation results are reasonable. Similar situations exist for dates like Apr. 30 to May 31, May 30 to June 30, etc. The confusion comes from the uneven number of days in different months. In our calculation, we used the former method.
How to Calculate Age
This age calculator uses 2 similar but slightly different methods to calculate age. One method is used to give age in years, months and days, and also months and days. The other method is used to very precisely calculate age in total days only.
When calculating age to the detailed level of days it’s important to remember that not all months have the same number of days. Also, while a year has 365 days, a leap year has 366 days.
When you calculate age in terms that include years, this calculator provides the answer in common terms. For example, a teenager might say he is 15 years old rather than saying he’s 12 normal years old plus 3 leap years old. This age calculator uses the same assumption — although we know years may have different lengths, we generalize between regular years and leap years, and call them equal.
The same is true for months. If a baby is 7 months old, the parents would not say she is 4 months that are 31 days long, plus 2 months that are 30 days long, plus 1 month that is 28 days long. We typically generalize across months of different lengths, and count them all equal as general months.
The age calculator finds age in terms that are commonly used, calling all years equal and all months equal. When calculating days however, the calculator uses a function to find the precise number of days between dates.
If you wanted to know how many days you’ve been alive you need to account for extra days in leap years. You also need to know the total number of months with 31 days, total months with 30 days, and total months with 28 days.
The age calculator uses a function that recognizes how many leap years are in a given time span. It also calculates the specific number of days in a partial year. For example, if you were born on March 5 and today is June 7, the calculator finds the exact number of days that occur between those dates.
Your age in total number of days is calculated as:
• Number of years, x, with 365 days = 365x plus
• Number of years, y, with 366 days = 366y plus
• Number of days in the remaining partial year
This calculator assumes the start date is day 0 and begins counting one full day after day 0. For example if you input the dates 03/15/2022 and 3/18/2022 into the calculator you will get a total age of 3 days. Day 1 is the 16th, day 2 is the 17th, and day 3 is the 18th. The 15th is not counted in the tally of days.
Date Formats
The age calculator allows you to input single digit months and days. You do not have to pad them with a leading zero. You must input the 4-digit year however.
US date formatting can use a forward slash, a period, or a dash:
• mm/dd/yyyy
• mm.dd.yyyy
• mm-dd-yyyy
European date formatting can use a forward slash, a period, or a dash:
• dd/mm/yyyy
• dd.mm.yyyy
• dd-mm-yyyy
International date formatting according to ISO 8601 use only a dash (-):
• yyyy-mm-dd
Today’s date is based on Greenwich Mean Time (GMT).
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