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http://geeksplanet.net/tag/data-structures/ | 1,547,699,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658702.9/warc/CC-MAIN-20190117041621-20190117063621-00235.warc.gz | 94,917,095 | 8,493 | Archives for
# Data Structures
## simple Implementation of stack using a linked list
Stacks are linear data structures. This means that their contexts are stored in what looks like a line (although vertically). This linear property, however, is not sufficient to discriminate a stack from other linear data structures. For example, an array is a sort of linear data structure in which you can access any element directly. In contrast, in a stack, you can only access the element at its top. In a stack Last thing In is the First thing Out. Thus, we say that a stack enforces LIFO order.
One disadvantage of implementing stack using an array is wastage of space. Most of the times most of the array is left unused. A better way to implement stack is by using a linked list. By using a single linked list to implement a stack there is no wastage of space.
Implementation of stack using a linked list
```/*
A simple implementation of linked list in C plus plus (CPP/C++)
*/
/*
Compiler used: g++
*/
#include<iostream>
using namespace std;
struct node{
int data;
node *next;
};
//This function pushes data(node) to the stack
node * push(node*,node*);
//This function pops data (node) and returns pointer to the poped node
node * pop(node*);
//This function shows the stack
void show_stack(node*);
///These two global variables will help us track the start and end of the list
node *f,*l;
main()
{
int i,ch;
f=l=NULL;
cout<<"\t 1 to push"<<endl;
cout<<"\t 2 to pop"<<endl;
cout<<"\t 3 to show stack"<<endl;
cout<<"\t 0 to exit"<<endl;
while(1)
{
cin>>ch;
switch(ch)
{
case 1:
{
if(f==NULL)
l=f=push(f,l);
else
l=push(f,l);
break;
}
case 2:
{
cout<<"popped:"<<pop(f)->data<<endl;
break;
}
case 3:
{
show_stack(f);
break;
}
case 0:
break;
default:
}
if(ch==0)
break;
}
}
node * push(node *f,node *l)
{
node *n;
n=new node;
cout<<"Enter data:";
cin>>n->data;
n->next=NULL;
if(f==NULL)
{f=l=n;return f;}
else
{
l->next=n;
l=n;return l;
}
}
void show_stack(node *f)
{
cout<<"showing data"<<endl;
node *guest=f;
while(guest!=NULL)
{
cout<<"\t"<<guest->data<<endl;
guest=guest->next;
}
}
node * pop(node *f)
{
node *guest,*lb;
guest=lb=f;
while(guest->next!=NULL)
{
lb=guest;
guest=guest->next;
}
lb->next=NULL;
l=lb;
return guest;
}
```
## Implementation of Singly linked list in C plus plus
The simplest kind of linked list is a singly-linked list , which has one link per node. This link points to the next node in the list, or to a null value or empty list if it is the final node.A singly linked list’s node is divided into two parts. The first part holds or points to information about the node, and second part holds the address of next node. A singly linked list travels one way.
A singly-linked list containing two values: the value of the current node and a link to the next node
```/*
A simple implementation of linked list in C plus plus (CPP/C++)
*/
/*
For any help, post comment here
Compiler used: g++
*/
#include<iostream>
using namespace std;
struct node{
int data;
node *next;
};
/*
This function adds a node at the end of the
list and returns pointer to the added node,
This takes pointer to the first node and
pointer to the last node as argument. By
giving the last node as argument we are saving
some computer labour as it need not travel
from the start to the end to find the last node
*/
/*
This function just takes the first node as
the argument and traverses the entire list
displaying the data in each node.
*/
void shownodes(node*);
/*
This function takes the first node as argument
and traverses the list until it finds the last
node and deletes it and returns pointer to the
new last node.
*/
node * delete_node(node*);
/*
*f and *l are the global variables keeping track
of first and last nodes.
*/
node *f,*l;
main()
{
int i,ch;
f=l=NULL;
cout<<"\t 1 to add a node"<<endl;
cout<<"\t 2 to see the nodes"<<endl;
cout<<"\t 3 to delete a node"<<endl;
cout<<"\t 0 to exit"<<endl;
while(1)
{
cin>>ch;
switch(ch)
{
case 1:
{
if(f==NULL)
l=f=addnode(f,l); //Since first node is the last node
else
break;
}
case 2:
shownodes(f);break;
case 3:
l=delete_node(f);break; // Last node has been changed to the last but one.
case 0:
break;
default:
}
if(ch==0)
break;
}
}
{
node *n;
n=new node;
//Allocating memory for the new node
cout<<"Enter data:";
cin>>n->data;
//This is going to be the last node, so its next point to NULL
n->next=NULL;
//If there is no first node, then the new node is the first and last node.
if(f==NULL)
{f=l=n;return f;}
else
{
// Pointing the last node to the new node
l->next=n;
//Setting the new node as the last node
l=n;return l;
}
}
void shownodes(node *f)
{
cout<<"showing data"<<endl;
node *guest=f;
while(guest!=NULL)
{
cout<<"\t"<<guest->data<<endl;
guest=guest->next;
}
}
node * delete_node(node *f)
{
node *guest,*lb;
guest=lb=f;
while(guest->next!=NULL)
{
lb=guest;
guest=guest->next;
}
lb->next=NULL;
cout<<"node deleted"<<endl;
return lb;
}
```
## Program to convert infix expression to postfix in C | Shunting yard algorithm
This program is a implementation of shunting yard algorithm to convert an infix expression to post fix expression, This is a extension of the stack program published earlier
Algorithm:(Taken from wikipedia)
• While there are tokens to be read:
• If the token is a number, then add it to the output queue.
• If the token is a function token, then push it onto the stack.
• If the token is a function argument separator (e.g., a comma):
• Until the topmost element of the stack is a left parenthesis, pop the element from the stack and push it onto the output queue. If no left parentheses are encountered, either the separator was misplaced or parentheses were mismatched.
• If the token is an operator, o1, then:
• while there is an operator, o2, at the top of the stack, and either
o1 is associative or left-associative and its precedence is less than (lower precedence) or equal to that of o2, or
o1 is right-associative and its precedence is less than (lower precedence) that of o2,
pop o2 off the stack, onto the output queue;
• push o1 onto the stack.
• If the token is a left parenthesis, then push it onto the stack.
• If the token is a right parenthesis:
• Until the token at the top of the stack is a left parenthesis, pop operators off the stack onto the output queue.
• Pop the left parenthesis from the stack, but not onto the output queue.
• If the token at the top of the stack is a function token, pop it and onto the output queue.
• If the stack runs out without finding a left parenthesis, then there are mismatched parentheses.
• When there are no more tokens to read:
• While there are still operator tokens in the stack:
• If the operator token on the top of the stack is a parenthesis, then there are mismatched parenthesis.
• Pop the operator onto the output queue.
• Exit.
```#include <stdio.h>
#define size 10
char stack[size];
int tos=0,ele;
void push();
char pop();
void show();
int isempty();
int isfull();
char infix[30],output[30];
int prec(char);
int main()
{
int i=0,j=0,k=0,length;
char temp;
printf("\nEnter an infix expression:");
scanf("%s",infix);
printf("\nThe infix expresson is %s",infix);
length=strlen(infix);
for(i=0;i<= prec(stack[tos-1]) )
{
temp=pop();
printf("\n the poped element is :%c",temp);
output[j++]=temp;
push(infix[i]);
printf("\n The pushed element is :%c",infix[i]);
show();
}
else
{
push(infix[i]);
printf("\nThe pushed element is:%c",infix[i]);
show();
}
}
else
{
if(infix[i]=='(')
{
push(infix[i]);
printf("\nThe pushed-- element is:%c",infix[i]);
}
if(infix[i]==')')
{
temp=pop();
while(temp!='(')
{output[j++]=temp;
printf("\nThe element added to Q is:%c",temp);
//temp=pop();
printf("\n the poped element is :%c",temp);
temp=pop();}
}
}
}
}
printf("\nthe infix expression is: %s",output);
}
while(tos!=0)
{
output[j++]=pop();
}
printf("the infix expression is: %s\n",output);
}
//Functions for operations on stack
void push(int ele)
{
stack[tos]=ele;
tos++;
}
char pop()
{
tos--;
return(stack[tos]);
}
void show()
{
int x=tos;
printf("--The Stack elements are.....");
while(x!=0)
printf("%c, ",stack[--x]);
}
//Function to get the precedence of an operator
int prec(char symbol)
{
if(symbol== '(')
return 0;
if(symbol== ')')
return 0;
if(symbol=='+' || symbol=='-')
return 1;
if(symbol=='*' || symbol=='/')
return 2;
if(symbol=='^')
return 3;
return 0;
} | 2,224 | 8,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | longest | en | 0.629281 |
https://stats.stackexchange.com/questions/511224/can-you-sum-principal-components | 1,726,314,325,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00216.warc.gz | 505,620,615 | 40,248 | # Can you sum principal components?
I am currently reviewing principal components from my recent output. For each principal component I know you get an eigenvalue which represents how much of the variance is explained. If I was to take the cumulative sum of the eigenvalues such that I wanted say 75% of variance explained would it make sense to sum the respective principal components?
E.g. if cumulative sum of pc1, pc2, pc3 accounted for 75% of variance (from summing eigenvalues) could I sum pc1 + pc2 + pc3?
• For what purpose?
– chl
Commented Feb 25, 2021 at 19:43
• The first principal component is the direction along which the data has maximum variance. In your example, if your hypothesis is correct, the data would have greater variance in the direction pc1 + pc2 + pc3. Commented Feb 25, 2021 at 20:10 | 197 | 816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.936074 |
https://www.distancesto.com/fuel-cost/sk/trencianska-to-cieszyn/history/107206.html | 1,695,344,906,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00772.warc.gz | 815,012,750 | 14,667 | # INR4.66 Total Cost of Fuel from Trenčianska to Cieszyn
Your trip to Cieszyn will consume a total of 1.86 gallons of fuel.
Trip start from Trenčianska, SK and ends at Cieszyn, PL.
Trip (74.5 mi) Trenčianska » Cieszyn
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Trenčianska, 020 01 Púchov, Slovakia and end at Cieszyn, Poland.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Trenčianska to Cieszyn plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Trenčianska to Cieszyn.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Trenčianska to Cieszyn.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Trenčianska to Cieszyn.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Cieszyn are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Trenčianska to Cieszyn.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Trenčianska to Cieszyn.
Speaking of travel time, a flight to Cieszyn takes up a lot less. How much less? Flight time from Trenčianska to Cieszyn.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Trenčianska to Cieszyn.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
Recent Fuel Calculations for Trenčianska SK:
Fuel Cost from Trenčianska to Hamburg
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Fuel Cost from Trenčianska to Košice
Fuel Cost from Trenčianska to Prešov
Fuel Cost from Trenčianska to Bratislava | 688 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.933703 |
http://mathhelpforum.com/trigonometry/59133-trigonometry-minutes.html | 1,524,611,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947421.74/warc/CC-MAIN-20180424221730-20180425001730-00493.warc.gz | 196,332,524 | 10,117 | # Thread: Trigonometry with minutes
1. ## Trigonometry with minutes
Hi guys, i have a maths worksheet that is due in tomorrow and i cant remember how to do some of the questions. The question reads: Find the size of the angle θ, correct to the nearest minute, in cach of the following.
I have a right-angled triangle with the value of the opposite angle as 27m, and the adjacent angle as 32m, with the hypotenuse having no value. this indicates that i have to use the Tangent function. How would i go about solving this?
(I dont actually need the answer, just the method.)
2. Originally Posted by mcaelli
Hi guys, i have a maths worksheet that is due in tomorrow and i cant remember how to do some of the questions. The question reads: Find the size of the angle θ, correct to the nearest minute, in cach of the following.
I have a right-angled triangle with the value of the opposite angle as 27m, and the adjacent angle as 32m, with the hypotenuse having no value.
No. You don't. An angle of "27m" makes no sense, that's a length measurement. You have a triangle with [b]legs[b] of length 27 m and 32 m. The hypotenuse certainly DOES have a value- you just aren't given it. You could calculate it by using the Pythagorean theorem but don't, it is not necessary for this problem.
this indicates that i have to use the Tangent function. How would i go about solving this?
(I dont actually need the answer, just the method.)
Do you at least remember that "tan= opposite side/adjacent side"? Since you are given those lengths, divide to find the tangent of the angle and then take the inverse tangent or 'arctangent", presumably on your calculator. Be sure your calculator is in "degree" mode. Your calculator will probably give you a decimal answer. Since there are 60 minutes in one degree, multiply the decimal part by 60 to find the minutes.
3. Right, sorry im a bit tired at the moment. I knew pretty much all of that except for the fact about there being 60 minutes in one degree, i think i must have missed that lesson. thanks for the help, i know what to do now. | 492 | 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} | 3.796875 | 4 | CC-MAIN-2018-17 | latest | en | 0.95587 |
imagine.kicbak.com | 1,503,433,617,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00030.warc.gz | 185,944,244 | 12,351 | Feeds:
Posts
# Introduction:
Check here for Part 1 or Part 3
Last time, in part-one, we established the assumptions that we are going to make about our raindrops. These assumptions included the shape, the different rays we will consider, refraction, reflection, total internal reflection and also the Fresnel effect. This time we will discuss our first approach to solving the problem.
So our first attempt to optimize the rendering of raindrops is an interpolation based approach. Interpolation may sound like a complex word at first, but its really just weighted averages. If you have two points along an axis, points A and B, and a third point in the middle point M. Point M is affected by A and B each, equally. In this case the value of M is equal to the value of A * 0.5 + B * 0.5. When M is only 10% of the way to point B, so much closer to point A. The value of M is equal to A * 0.9 + B * 0.1.
So when we first wanted to address rendering thousands of raindrops, we wanted to use this idea, of interpolation, to interpolate the points of where rays intersected the environment map. Below is a series of images and explanations to hopefully explain it in better detail.
# Solution Explained:
When we first began, the idea was that we could interpolate XYZ values, (3-dimensional space), vectors of the refracted rays of light as they intersect the environment map. The illustration below will hopefully capture the idea, in a 2-dimensional space.
Figure 1: The camera is at the bottom in dark-purple, with a view vector. The raindrop is in blue. The orange is the environment map.
The first step is depicted above. Basically we are going to ray-trace a raindrop. The key here is that for this specific View vector we end up with a certain I(x,y,z). I is the point in which this view vector for this raindrop intersects the environment map. N1 and N2 are the surface normals where the light vector intersects the surface of the raindrop. For our research we treated all raindrops as spheres.
Continuing on from Figure 1, we can calculate other view vectors and how they refract with the raindrop.
Figure 2: The camera is at the bottom in dark-purple, with a view vector. The raindrop is in blue. The orange is the environment map.
Looking at Figure 2, we can calculate all the vectors that intersect the raindrop, and this is done through ray-tracing. The next task we want to perform is to store all of the I(x,y,z) data into a map, for the raindrop.
Figure 3: Same as Figure 2, but this time we have the mask where we are going to store our data from I(x,y,z).
What we have done in Figure 3, above, is defined the “screen” for the raindrop, depicted by the green section. In the primary view we are looking at the scene from above, looking down. The green-grid is the view from the camera. These green screens are always aligned to the camera. Within each “pixel” of the green-screen we are going to store I(x,y,z) depending on where it intersects the environment map. Normally an environment map is interested in storing color data that is RGB information, however for us, we want to store X,Y,Z data that is positional data. Later we will grab the color data that correlates to the position of the intersection with the environment map. Let’s take a look at what we are interpolating.
Figure 4: Here we have raindrops 1 and 2 with their view-aligned screens. Raindrop 3 is going to have it's values interpolated.
So for Raindrops 1 and 2 we go ahead and calculate their values for their screens using ray-tracing. We store I(x,y,z) data in each “pixel”. Raindrop 3 is going to have it’s values of its screen interpolated from the values found in raindrop’s 1 and 2.
Figure 5: The view from the camera of each raindrop.
In figure 5, we have the same “pixel” in each screen. That pixel for raindrop-3 is going to have its value interpolated between raindrops 1 and 2. The result should be that we can ray-trace raindrops 1 and 2 and for raindrop 3 we do not need to perform the expensive activity of ray-tracing, primarily calculating intersection, refraction and reflection.
Unfortunately we get mixed results. When the raindrops are along the X axis we receive decent results. However as the raindrops move along the Y-axis, especially near the poles the solution collapses. Here are some examples of the results.
# Results:
When we interpolate along the X-axis as shown in this image:
Figure 6: Interpolating along the X-axis.
We achieve the following results.
Figure 7: The image on the right is directly to the right of the camera, the image on the left is to the left of the camera. The image in the middle is the drop with interpolated intersection (X,Y,Z) positional data used to access the environment map. The drop on top is the desired result which was ray-traced and acts as our ground-truth.
Looking at this image, you might think, hey not bad, looks pretty accurate. However there is an issue that arises when you then decide to incorporate the Y-axis. So we put two drops in the scene one directly above the camera and one directly below the camera, and then interpolated the same middle drop. Here is what results:
Figure 8: The image above is located above the camera, and the bottom image, below the camera. The drop on the right was ray-traced. The drop on the left is the interpolated result. Looks like we might have some artifacts...
Yeah, so umm, not quite right. Look like we have some issues with the result. Obviously there is a problem with how the values are being interpolated but although we know what the problem is, we never found a way to solve it. In addition I am not aware of the mathematical cause of the problem. My guess is that we are using a spherical environment map, but our interpolation process ignores this fact.
# Conclusions:
Of course there is a mathematical explanation for the cause of the interpolation results. Sadly, I was never able to discover what that explanation is and continue to be very interested in one day figuring it out. I am hoping that these posts will help to bring those intelligent individuals forward so that they can explain it to me. This is one example how doing research raises more questions than it answers. Nevertheless the fight continues and you will see how we solve this problem next time.
There are some other issues that this approach does not solve. What happens when the raindrops are not along the same axis, or if one drop is placed behind the camera. Another challenge is that refraction is based on the vector between the viewpoint (the camera) and the raindrop, so raindrops at different distances will have different refraction vectors calculated, in turn affecting the results of the interpolated drops. All in all it can become quite complicated. So for the time being we abandoned interpolating intersection points with the environment map.
Next time, you’ll see what we did to solve this problem and get wonderful raindrops, and it begins with none other than the power of the GPU.
Be Sociable, Share! | 1,562 | 7,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.932836 |
https://hackage.haskell.org/package/statistics-0.13.1.1/src/tests/Tests/ApproxEq.hs | 1,591,294,309,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00487.warc.gz | 367,736,471 | 1,869 | {-# LANGUAGE FlexibleContexts, FlexibleInstances, TypeFamilies #-} module Tests.ApproxEq ( ApproxEq(..) ) where import Data.Complex (Complex(..), realPart) import Data.List (intersperse) import Data.Maybe (catMaybes) import Numeric.MathFunctions.Constants (m_epsilon) import Statistics.Matrix hiding (map, toList) import Test.QuickCheck import qualified Data.Vector as V import qualified Data.Vector.Generic as G import qualified Data.Vector.Unboxed as U import qualified Statistics.Matrix as M class (Eq a, Show a) => ApproxEq a where type Bounds a eq :: Bounds a -> a -> a -> Bool eql :: Bounds a -> a -> a -> Property eql eps a b = counterexample (show a ++ " /=~ " ++ show b) (eq eps a b) (=~) :: a -> a -> Bool (==~) :: ApproxEq a => a -> a -> Property a ==~ b = counterexample (show a ++ " /=~ " ++ show b) (a =~ b) instance ApproxEq Double where type Bounds Double = Double eq eps a b | a == 0 && b == 0 = True | otherwise = abs (a - b) <= eps * max (abs a) (abs b) (=~) = eq m_epsilon instance ApproxEq (Complex Double) where type Bounds (Complex Double) = Double eq eps a@(ar :+ ai) b@(br :+ bi) | a == 0 && b == 0 = True | otherwise = abs (ar - br) <= eps * d && abs (ai - bi) <= eps * d where d = max (realPart \$ abs a) (realPart \$ abs b) (=~) = eq m_epsilon instance ApproxEq [Double] where type Bounds [Double] = Double eq eps (x:xs) (y:ys) = eq eps x y && eq eps xs ys eq _ [] [] = True eq _ _ _ = False eql = eqll length id id (=~) = eq m_epsilon (==~) = eql m_epsilon instance ApproxEq (U.Vector Double) where type Bounds (U.Vector Double) = Double eq = eqv (=~) = eq m_epsilon eql = eqlv (==~) = eqlv m_epsilon instance ApproxEq (V.Vector Double) where type Bounds (V.Vector Double) = Double eq = eqv (=~) = eq m_epsilon eql = eqlv (==~) = eqlv m_epsilon instance ApproxEq Matrix where type Bounds Matrix = Double eq eps (Matrix r1 c1 e1 v1) (Matrix r2 c2 e2 v2) = (r1,c1,e1) == (r2,c2,e2) && eq eps v1 v2 (=~) = eq m_epsilon eql eps a b = eqll dimension M.toList (`quotRem` cols a) eps a b (==~) = eql m_epsilon eqv :: (ApproxEq a, G.Vector v Bool, G.Vector v a) => Bounds a -> v a -> v a -> Bool eqv eps a b = G.length a == G.length b && G.and (G.zipWith (eq eps) a b) eqlv :: (ApproxEq [a], G.Vector v a) => Bounds [a] -> v a -> v a -> Property eqlv eps a b = eql eps (G.toList a) (G.toList b) eqll :: (ApproxEq l, ApproxEq a, Show c, Show d, Eq d, Bounds l ~ Bounds a) => (l -> d) -> (l -> [a]) -> (Int -> c) -> Bounds l -> l -> l -> Property eqll dim toList coord eps a b = counterexample fancy \$ eq eps a b where fancy | la /= lb = "size mismatch: " ++ show la ++ " /= " ++ show lb | length summary < length full = summary | otherwise = full summary = concat . intersperse ", " . catMaybes \$ zipWith3 whee (map coord [(0::Int)..]) xs ys full | '\n' `elem` sa = sa ++ " /=~\n" ++ sb | otherwise = sa ++ " /=~" ++ sb (sa, sb) = (show a, show b) (xs, ys) = (toList a, toList b) (la, lb) = (dim a, dim b) whee i x y | eq eps x y = Nothing | otherwise = Just \$ show i ++ ": " ++ show x ++ " /=~ " ++ show y | 962 | 3,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.477792 |
https://stats.stackexchange.com/questions/217972/r-survival-analysis-cox-regression-and-cumulative-time-dependent-covariate | 1,712,963,855,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00405.warc.gz | 512,643,386 | 40,567 | # R Survival Analysis - Cox Regression and Cumulative Time Dependent Covariate
Hoping for some help related to a survival analysis using R and the survival package. I've been relying heavily on a series of blog posts done by Dayne Batten, particularly this portion: [http://daynebatten.com/2015/12/survival-analysis-customer-churn-time-varying-covariates/]
I've collected and merged the data as instructed using the tmerge function. My model relies on a cumulative time-dependent covariate, which strays away from the example provided. So my first question is does a cumulative covariate affect the validity of the Cox Regression? Here is my code at the moment:
fit <- coxph( Surv(tstart, tstop, had_event) ~ review_event, data = newdatatestcum)
My second question pertains to the lack of an ID being assigned within this model. For each customer ID within this data I have up to a few hundred lines of events with my covariate. I don't see how this regression could possibly be accounting for that.
• Why do you think it would affect the validity of the Cox regression? Jun 9, 2016 at 12:54
• Being as new to this as I am I feared any deviation from the post and there is little documentation on cumulative covariates as it relates to the Cox Regression. Really just hoping to hear of others' experiences and possible pitfalls in trusting the results. Jun 9, 2016 at 13:17
• @Chapin23 Think of doing regular linear regression $y \sim X \beta$ where each row in the predictor $X$ varies from a sample to the next one. Time doesn't play a role here because the model is Markov.
– wsw
Jun 29, 2016 at 21:52
In the logic of the (extended) Cox model, the hazard $h(t)$ is defined as $$h(t) dt = P(T = t| T \geq t, \mathcal{H}(t_-))$$
In other words, the hazard at time $t$ depends on the probability of the event to happen at time $t$, given that it has not happened so far ($T\geq t$) and given the past ($\mathcal{H}(t_-)$). This past includes information up to time $t$. In particular, if you have a covariate $x(t)$ that you want to use, you may as well use any functional form of $x(t)$, like $\int_0^t x(s)ds$ (the cumulative version), or $\log x(t)$, or whatever. | 564 | 2,173 | {"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} | 2.859375 | 3 | CC-MAIN-2024-18 | longest | en | 0.918261 |
http://experiment-ufa.ru/319/100_as_a_decimal | 1,529,481,406,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00599.warc.gz | 108,412,617 | 6,175 | # 319/100 as a decimal
## 319/100 as a decimal - solution and the full explanation with calculations.
If it's not what You are looking for, type in into the box below your number and see the solution.
## What is 319/100 as a decimal?
To write 319/100 as a decimal you have to divide numerator by the denominator of the fraction.
We divide now 319 by 100 what we write down as 319/100 and we get 3.19
And finally we have:
319/100 as a decimal equals 3.19
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http://hismath.blogspot.com/2009/05/jsh-pells-equation-and-question-of-math.html | 1,521,692,619,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647768.45/warc/CC-MAIN-20180322034041-20180322054041-00797.warc.gz | 144,719,592 | 7,155 | ## JSH: Pell's equation and question of math fraud
I've been pondering the situation where I found that what is commonly called Pell's Equation, has this parametric rational solution that math people don't talk about, and also has this connection to these alternate equations that make solving them easy, or vice versa, you can solve Pell's Equation with them, and now I can get some resolution to troubling questions.
Like, I have various math discoveries, which other people say are not math discoveries, and I've considered that hey, maybe I'm evolutionarily more advanced, so I can understand things that less advanced people can't, which is more simply captured by noting that dogs can't learn calculus. Their brains make that impossible. If I were evolutionarily advanced then I could have math discoveries simple to me that others could no more understand than a dog could understand calculus.
That was a scary scenario, and I'm happy to put it to rest, which is what Pell's Equation has allowed.
Pell's Equation is just x^2 - Dy^2 = 1, where math people are looking for integers that fit into all those boxes, so D is a positive integer, and x and y are integers. 9^2 - 5*4^2 = 1, is an example of such an outcome.
The equation has been known for thousands of years or something.
Well in our greatly advanced times, it seems mainstream math people have not bothered to note that if D is a prime number such that D = 1 mod 4, then another equation often called the negative Pell's Equation is solved by Pell's Equation:
j^2 - Dk^2 = -1, j = sqrt((x-1)/2)
So with my example above with D=5, notice that j = sqrt((9-1)/2) = 2, works, as 2^2 - 5*1^2 = -1.
And you could go the OTHER way, and find the solution to Pell's Equation, as j = sqrt((x-1)/2) means that
x = 2j^2 + 1, and x = 2*2^2 + 1 = 9, as required.
That always works. It's actually a rather trivial result mathematically which you can figure out from the main Pell's Equation as
x^2 - Dy^2 = 1, means that x^2 - 1 = Dy^2, so (x-1)(x+1) = Dy^2,
and if D is prime then it can only be a factor of x-1 or x+1, meaning that one of them must be either a square or 2 times a square as if x is odd then they both are even. You can work out all the alternates by noting how D can divide across. Trivial algebra.
That's not hard to understand, but if you go on the web and look at math texts on Pell's Equation you will not find that result:
http://en.wikipedia.org/wiki/Pell's_equation
or
http://mathworld.wolfram.com/PellEquation.html
You may think there is a trick in D being prime, where D = 1 mod 4, but turns out there are two more equations like the negative Pell's Equation if D does not equal 1 mod 4, and one more on top of those if D is a composite in a special way.
So there are 4 other alternates to Pell's Equation, where any of those will solve Pell's Equation if they exist, and the solution is smaller.
Here's a fun example to show the size difference:
1766319049^2 - 61*226153980^2 = 1
but, consider
j^2 - Dk^2 = -1
as you get a solution to x, when x^2 - Dy^2 = 1, with x = 2j^2 + 1, and with D=61, notice an astounding difference in the size of the solution—
29718^2 - 61*3805^2 = -1
and a quick check with your computer's calculator will show that, yes:
1766319049 = 2*297182 + 1.
You can use continued fractions to solve in either direction which is neat because the way math people teach continued fractions you may naively think there is only one answer they give, but math people traditionally go ONE WAY, using all positives.
The reality of the simpler solution for the negative Pell's Equation indicates they're going the wrong way, so using continued fractions properly—using negatives—could greatly simplify solving Pell's Equation by working to give one of the alternates sooner, and then allowing one to pull its answer from them.
Ok. That's not rocket science. It's relatively simple algebra that I assume you can all understand. I'm not trying to teach calculus to dogs here. I'm not some super advanced mutant freak who can understand things you cannot, which means, fraud is now on the table, as I've talked about the above for a while now.
The Wikipedia has not been updated. Wolfram hasn't updated. Far as I know math people are still teaching Pell's Equation like they always have, and emails on this subject to people like Arjen Lenstra, have not been answered. He's an expert in this area.
Near as I can tell math people have no intentions of ever acknowledging the above. Ever.
Now then, are they dogs who can't learn calculus, or are they academics who are playing a dangerous game, certain they can win, and damn humanity in the process?
You try and figure it out. It's your world as well as mine. If there is fraud on this level, you can be very certain it's very bad, and the reasons are terrifying and not at all good for the future of the entire human race.
After all, they're being rather daring at this point. | 1,235 | 4,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-13 | latest | en | 0.979798 |
https://noncommutativeanalysis.wordpress.com/category/d-shift-space/ | 1,500,706,051,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423903.35/warc/CC-MAIN-20170722062617-20170722082617-00150.warc.gz | 650,798,305 | 35,941 | ## Category: d shift space
### Spaces of Dirichlet series with the complete Pick property (or: the Drury-Arveson space in a new disguise)
John McCarthy and I have recently uploaded a new version of our paper “Spaces of Dirichlet series with the complete Pick property” to the arxiv. I would like to advertise the central discovery of this paper here.
Recall that the Drury-Arveson space $H^2_d$ is the reproducing kernel Hilbert space on the open unit ball of a $d$ dimensional Hilbert space, with reproducing kernel
$k(z,w) = \frac{1}{1 - \langle z, w \rangle}$.
It has the remarkable universal property that every Hilbert function space with the complete Pick property is naturally isomorphic to the restriction of $H^2_\infty$ to a subset of the unit ball (see Theorem 6 and its corollary in this post), and consequently, every complete Pick algebra is a quotient of the multiplier algebra $\mathcal{M}_\infty = Mult(H^2_\infty)$. To the best of my knowledge, no other Hilbert function spaces with such a universal property have been studied.
John and I discovered another reproducing kernel Hilbert space that turns out to be “the same” as the Drury-Arveson space $H^2_\infty$. Since the space $H^2_\infty$ as been so well studied, it interesting to discover a new incarnation. The really interesting part is that the space we discovered is a space of analytic functions on a half plane (that is, a space of functions in one complex variable), rather than a space of analytic functions in infinitely many variables on the unit ball of a Hilbert space.
To be precise, the spaces we consider are spaces of Dirichlet series $\mathcal{H}$, of the form
$\mathcal{H} = \{f(s) = \sum_{n=1}^\infty \gamma_n n^{-s} : \sum |\gamma_n|^2 a_n^{-1} < \infty \}$.
(Here $a_n$ is a sequence of positive numbers). These are Hilbert function spaces on some half plane that have a kernel of the form $k(s,u) = \sum a_n n^{-s-\bar u}$.
We first answer the question which of these spaces $\mathcal{H}$ have the complete Pick property. This problem has a simple solution (which has been anticipated by similar results on spaces on the disc): if we denote by $g(s) = \sum a_n n^{-s}$ the “generating function” of the space, and if we write
$\frac{1}{g(s)} = \sum c_n n^{-s}$,
then $\mathcal{H}$ is a complete Pick space if and only if $c_n \leq 0$ for all $n \geq 2$.
After we know to tell when these spaces are complete Pick, it is natural to ask which complete Pick spaces arise like this? We do not give a complete answer, but our surprising discovery is that things can easily be cooked up so to obtain the Drury-Arveson space $H^2_d$, where $d$ can be any cardinal number in $\{1,2,\ldots, \infty\}$. For example, $\mathcal{H}$ turns out to be “the same” as $H^2_\infty$ if the kernel $k$ is given by
$k(s,u) = \frac{P(2)}{P(2) - P(2+s+\bar u)}$,
where $P(s) = \sum_{p} p^{-s}$ is the prime zeta function (the sum is taken over all primes $p$).
Now, I have been a little vague about what it means that $\mathcal{H}$ is “the same” as $H^2_\infty$. In fact, this is a subtle question, and we devote a part of our paper what it means for two Hilbert function spaces to be the same — something that has puzzled us for a while.
What does this appearance of Drury-Arveson space as a space of Dirichlet series mean? Can we use this connection to learn something new on multivariable operator theory, or on Dirichlet series? How did the prime zeta function smuggle itself into this discussion? This requires further thought.
### A corrigendum
Matt Kennedy and I have recently written a corrigendum to our paper “Essential normality, essential norms and hyperrigidity“. Here is a link to the corrigendum. Below I briefly explain the gap that this corrigendum fills.
A corrigendum is correction to an already published paper. It is clear why such a mechanism exists: we want the papers we read to represent true facts, so false claims, as well as invalid proofs or subtle gaps should be pointed out to the community. Now, many many papers (I don’t want to say “most”) have some kind of mistake in them, but not every mistake deserves a corrigendum – for example there are mistakes that the reader will easily spot and fix, or some where the reader may not spot the mistake, but the fix is simple enough.
There are no rules as to what kind of errors require a corrigendum. This depends, among other things, on the authors. Some mistakes are corrected by other papers. I believe that very quickly some sort of mechanism – say google scholar, or mathscinet – will be able to tell if the paper you are looking up is referenced by another paper pointing out a gap, so such a correction-in-another-paper may sometimes serve as legitimate replacement for a corrigendum, when the issue is a gap or minor mistake.
There is also a question of why publish a corrigendum at all, instead of updating the version of the paper on the arxiv (and this is exactly what the moderators of the arxiv told us at first when we tried to upload our corrigendum there. In the end we convinced them that the corrigendum can stand by itself). I think that once a paper is published, it could be confusing to have a version more advanced than the published version; it becomes very clumsy to cite papers like that.
The paper I am writing about (see this post to see what its about) had a very annoying gap: we justified a certain step by citing a particular proposition from a monograph. The annoying part is that the proposition we cite does not exactly deal with the situation we deal with in the paper, but our idea was that the same proof works in our situation. We did not want to spell out the details because we considered that to be very easy, and in any case it was not a new argument. Unfortunately, the same proof does work when working with homogeneous ideals (which was what first versions of the paper treated) but in fact it is not clear if they work for non-homogeneous ideals. The reason why this gap is so annoying, is that it leads the reader to waste time in a wild goose chase: first the reader goes and finds the monograph we cite, looks up the result (has to read also a few extra pages to see he understands the setting and notation in the monograph), realises this is is not the same situation, then tries to adapt the method but fails. A waste of time!
Another problem that we had in our paper is that one requires our ideals to be “sufficiently non-trivial”. If this were the only problem we would perhaps not bother writing a corrigendum just to introduce a non-triviality assumption, since any serious reader will see that we require this.
If I try to take a lesson from this, besides a general “be careful”, it is that it is dangerous to change the scope of the paper (for us – moving form homogeneous to non-homogeous ideals) in late stages of the preparation of the paper. Indeed we checked that all the arguments work for the non-homogneous case, but we missed the fact that an omitted argument did not work.
Our new corrigendum is detailed and explains the mathematical problem and its solutions well, anyone seriously interested in our paper should look at it. The bottom line is this as follows.
Our paper has two main results regarding quotients of the Drury-Arveson module by a polynomial ideal. The first is that the essential norm in the non selfadjoint algebra associated to a the quotient module, as well as the C*-envelope, are as the Arveson conjecture predicts (Section 3 in the paper) . The second is that essential normality is equivalent to hyperrigidity (Section 4 in the paper).
Under the assumption that all our ideals are sufficiently non-trivial (and some other standing assumptions stated in the paper), the situation is as follows.
The first result holds true as stated.
For the second result, we have that hyperrigidity implies essential normality (as we stated), but the implication “essential normality implies hyperrigidity” is obtained for homogeneous ideals only.
### Essential normality, essential norms and hyper rigidity
Matt Kennedy and I recently posted on the arxiv a our paper “Essential normality, essential norms and hyper rigidity“. This paper treats Arveson’s conjecture on essential normality (see the first open problem in this previous post). From the abstract:
Let $S = (S_1, \ldots, S_d)$ denote the compression of the $d$-shift to the complement of a homogeneous ideal $I$ of $\mathbb{C}[z_1, \ldots, z_d]$. Arveson conjectured that $S$ is essentially normal. In this paper, we establish new results supporting this conjecture, and connect the notion of essential normality to the theory of the C*-envelope and the noncommutative Choquet boundary.
Previous works on the conjecture verified it for certain classes of ideals, for example ideals generated by monomials, principal ideals, or ideals of “low dimension”. In this paper we find results that hold for all ideals, but – alas! – these are only partial results.
Denote by $Z = (Z_1, \ldots, Z_d)$ the image of $S$ in the Calkin algebra (here as in the above paragraph, $S$ is the compression of the $d$-shift to the complement of an ideal $I$ in $H^2_d$). Another way of stating Arveson’s conjecture is that the C*-algebra generated by $Z$ is commutative. This would have implied that the norm closed (non-selfadjoint) algebra generated by $Z$ is equal to the sup-norm closure of polynomials on the zero variety of the ideal $I$. One of our main results is that we are able to show that the non-selfadjoint algebra is indeed as the conjecture predicts, and this gives some evidence for the conjecture. This is also enough to obtain a von Neumann inequality on subvarieties of the ball, what would have been a consequence of the conjecture being true.
Another main objective is to connect between essential normality and the noncommutative Choquet boundary (see this and this previous posts). A main result here is we have is that the tuple $S$ is essentially normal if and only if it is hyperrigid (meaning in particular that all irreducible representations of $C^*(S)$ are boundary representations).
### The remarkable Hilbert space H^2 (part III – three open problems)
This is the last in the series of three posts on the d–shift space, which accompany/replace the colloquium talk I was supposed to give. The first two parts are available here and here. In this post I will discuss three open problems that I have been thinking about, which are formulated within the setting of $H^2_d$.
### The remarkable Hilbert space H^2 (Part II – multivariable operator theory and model theory)
This post is the second post in the series of posts on the d–shift space, a.k.a. the Drury–Arveson space, a.k.a. $H^2_d$ (see this previous post about the space $H^2$).
### The remarkable Hilbert space H^2 (Part I – definition and interpolation theory)
This series of posts is based on the colloquium talk that I was supposed to give on November 20, at our department. As fate had it, that week studies were cancelled.
Several people in our department thought that it would be a nice idea if alongside the usual colloquium talks given by invited speakers which highlight their recent achievements, we would also have some talks by department members that will be more of an exposition to the fields they work in. So my talk was supposed to be an exposition to the setting in which much of the research I do goes on.
The topic of the “talk” is the Hilbert space $H^2_d$. There will be three parts to this series:
1. Definition and interpolation theory.
2. Multivariate operator theory and model theory
3. Current research problems | 2,796 | 11,642 | {"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": 48, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.86071 |
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• Turing bifurcations , Chaos, Population dynamics | 1,187 | 5,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-09 | latest | en | 0.860096 |
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### MEAP Preparation - Grade 4 Mathematics2.19 Convert Units of Capacity - US Customary System (WIZ Math)
Example: Convert 3 gallons to quarts. 1 gallon = 4 quarts Hence 3 gallons = 4 x 3 quarts Answer: 12 quarts Directions: Answer the following. Also write at least 10 examples of your own.
Q 1: Convert 6 gallon to cup:144 cup960 cup192 cup96 cup Q 2: Convert 10 gallon to fl oz:1,920 fl oz2,560 fl oz12,799.99999 fl oz1,280 fl oz Q 3: Convert 2 quart to quart:20 quart3 quart4 quart2 quart Q 4: Convert 6 cup to quart:1.5 quart3 quart15 quart2.25 quart Q 5: Convert 7 pint to cup:14 cup28 cup140 cup21 cup Question 6: This question is available to subscribers only! Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 302 | 1,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.806818 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=18480 | 1,369,198,439,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701314683/warc/CC-MAIN-20130516104834-00023-ip-10-60-113-184.ec2.internal.warc.gz | 318,944,858 | 784 | ```Question 29985
For csc^2 a- 1/cotA cscA--what is the simplest equivalent trigonometric expression
HOPE YOU MEAN
(csc^2 a- 1)/cotA cscA
=COT^2(A)/COT(A)COSEC(A)=COT(A)/COSEC(A)=(COS(A)/SIN(A))/(1/SIN(A))=COS(A)``` | 95 | 215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.431522 |
https://www.numbersaplenty.com/454580 | 1,701,342,610,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00462.warc.gz | 1,039,335,937 | 3,377 | Search a number
454580 = 225717191
BaseRepresentation
bin1101110111110110100
3212002120022
41232332310
5104021310
613424312
73602210
oct1567664
9762508
10454580
11290595
1219b098
1312bba9
14bb940
158ea55
hex6efb4
454580 has 48 divisors (see below), whose sum is σ = 1161216. Its totient is φ = 145920.
The previous prime is 454579. The next prime is 454603. The reversal of 454580 is 85454.
It is a nialpdrome in base 14.
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 2285 + ... + 2475.
It is an arithmetic number, because the mean of its divisors is an integer number (24192).
2454580 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 454580, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (580608).
454580 is an abundant number, since it is smaller than the sum of its proper divisors (706636).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
454580 is a wasteful number, since it uses less digits than its factorization.
454580 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 224 (or 222 counting only the distinct ones).
The product of its (nonzero) digits is 3200, while the sum is 26.
The square root of 454580 is about 674.2254815713. The cubic root of 454580 is about 76.8900437653.
The spelling of 454580 in words is "four hundred fifty-four thousand, five hundred eighty". | 495 | 1,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.885009 |
fouroaksprimary.net | 1,610,852,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00522.warc.gz | 40,870,745 | 10,497 | # Wednesday’s Maths – Mrs Elvins
Good morning everyone! Here is your Maths work for Wednesday!
Remember to start with a column on your Daily Practice sheet! Now practise counting in multiples of 2, 3, 4 and 5 up to the twelfth multiple (ie. 2 x 12). Can you count the multiples forwards and backwards? If there is someone else you can practise with, can you play ‘Ping Pong’ like we do in class (you say one multiple and someone else says the next one etc).
How did you get on with the true or false questions yesterday? The first one was true – the number sentences did match the place value grids! The second one was also true! The missing digit in each calculation was 9!
For the next week or two, we are now going to look at money! Today and tomorrow we are going to be reinforcing our current understanding of money before we apply that understanding to our objectives next week.
First watch today’s video.
Video
Now complete the activity on counting money in pence. Everyone should be able to have a go at all the questions today!
Activity sheet
If you want to try something else, ask an adult to give you different handfuls of coins to count and total or try this challenge! | 273 | 1,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-04 | latest | en | 0.928182 |
http://at.metamath.org/ileuni/releldmb.html | 1,610,974,687,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514796.13/warc/CC-MAIN-20210118123320-20210118153320-00577.warc.gz | 10,127,716 | 3,846 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > releldmb GIF version
Theorem releldmb 4514
Description: Membership in a domain. (Contributed by Mario Carneiro, 5-Nov-2015.)
Assertion
Ref Expression
releldmb (Rel 𝑅 → (A dom 𝑅x A𝑅x))
Distinct variable groups: x,A x,𝑅
Proof of Theorem releldmb
StepHypRef Expression
1 eldmg 4473 . . 3 (A dom 𝑅 → (A dom 𝑅x A𝑅x))
21ibi 165 . 2 (A dom 𝑅x A𝑅x)
3 releldm 4512 . . . 4 ((Rel 𝑅 A𝑅x) → A dom 𝑅)
43ex 108 . . 3 (Rel 𝑅 → (A𝑅xA dom 𝑅))
54exlimdv 1697 . 2 (Rel 𝑅 → (x A𝑅xA dom 𝑅))
62, 5impbid2 131 1 (Rel 𝑅 → (A dom 𝑅x A𝑅x))
Colors of variables: wff set class Syntax hints: → wi 4 ↔ wb 98 ∃wex 1378 ∈ wcel 1390 class class class wbr 3755 dom cdm 4288 Rel wrel 4293 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 629 ax-5 1333 ax-7 1334 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-8 1392 ax-10 1393 ax-11 1394 ax-i12 1395 ax-bndl 1396 ax-4 1397 ax-14 1402 ax-17 1416 ax-i9 1420 ax-ial 1424 ax-i5r 1425 ax-ext 2019 ax-sep 3866 ax-pow 3918 ax-pr 3935 This theorem depends on definitions: df-bi 110 df-3an 886 df-tru 1245 df-nf 1347 df-sb 1643 df-clab 2024 df-cleq 2030 df-clel 2033 df-nfc 2164 df-ral 2305 df-rex 2306 df-v 2553 df-un 2916 df-in 2918 df-ss 2925 df-pw 3353 df-sn 3373 df-pr 3374 df-op 3376 df-br 3756 df-opab 3810 df-xp 4294 df-rel 4295 df-dm 4298 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 788 | 1,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.143189 |
https://techcommunity.microsoft.com/t5/excel/creating-a-column-that-excludes-values-if-they-appear-in-both-of/td-p/2352913 | 1,638,041,161,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00107.warc.gz | 663,450,304 | 72,198 | Visitor
Creating a column that excludes values if they appear in both of two other columns
I'm in need of some assistance, and am turning to the community here for any help that can be offered. I am trying to set up a column (D) that will omit/exclude the value for a column (A) if that same value appears in another column (B). The formula we've been utilizing
=IF(ROWS(\$1:1)*COUNT(C:C),"",INDEX(A:A,SMALL(C:C,ROWS(\$1:1)))) i
s giving us no results a. The spreadsheet that we have been working on (which is failing miserably) is attached.
2 Replies
Re: Creating a column that excludes values if they appear in both of two other columns
Try this:
``=FILTER(A2:A18,1-COUNTIFS(B2:B6,A2:A18))``
Re: Creating a column that excludes values if they appear in both of two other columns
=IF(IF(ISNUMBER(MATCH(A2,\$B\$2:\$B\$6,0)),"",A2)=0,"",IF(ISNUMBER(MATCH(A2,\$B\$2:\$B\$6,0)),"",A2))
you have to adjust the range \$B\$2:\$B\$6 to match that range in your application.
Cheers | 286 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.863595 |
https://proofwiki.org/wiki/Definition:Primitive_Recursion/One_Variable | 1,618,189,626,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00439.warc.gz | 567,806,532 | 10,242 | # Definition:Primitive Recursion/One Variable
## Definition
Let $a \in \N$ be a natural number.
Let $g: \N^2 \to \N$ be a function.
Then the function $h: \N \to \N$ is obtained from the constant $a$ and $g$ by primitive recursion if and only if:
$\forall n \in \N: \map h n = \begin {cases} a & : n = 0 \\ \map g {n - 1, \map h {n - 1} } & : n > 0 \end{cases}$
## Also see
It can be seen that this is a special case of primitive recursion on several variables, with $k = 0$ and $f$ replaced by the constant function $f_a$. | 175 | 529 | {"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} | 3.203125 | 3 | CC-MAIN-2021-17 | longest | en | 0.734297 |
http://www.fixya.com/support/t1012441-weather_casio_fx_115es_solve_5_5_matrix | 1,519,312,783,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814124.25/warc/CC-MAIN-20180222140814-20180222160814-00326.warc.gz | 469,879,037 | 33,329 | Question about Casio FX-115ES Scientific Calculator
# Weather casio FX-115ES can solve 5*5 matrix
Hello,
i would like to know whether this model casio FX-115ES can solve 5 by 5 matrices and if not please suggest me which can solve it.
Posted by on
• mndymtchll Apr 24, 2009
I can't get an answer for my problem
×
• Level 1:
An expert who has achieved level 1.
Mayor:
An expert whose answer got voted for 2 times.
• Contributor
According to the manual for the Casio FX115ES, the largest matrix it solves is a 3x3.
Posted on Sep 11, 2008
Hi,
a 6ya Technician can help you resolve that issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repair professionals here in the US.
Goodluck!
Posted on Jan 02, 2017
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my-video-file.mp4
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## Related Questions:
### Calculate the inverse of a 4X4 matrix on fx-570ms
The Casio FX-570MS does not handle matrices. So you cannot create matrices, let alone invert them. Sorry.
The Casio FX115 ES handles matrices for up to 3X3 dimension. Create the matrix.Have its name displayed on the command line and press the [x^-1] (reciprocal ) key.
Apr 16, 2014 | Casio FX-115ES Scientific Calculator
### Is fx 115ES supports matrix with complex numbers as elements..if not please suggest one...
The calculator barely handles matrices of up to 3X3. Matrices with complex coefficients are well beyond its capabilities.
Oct 13, 2013 | Casio FX-115ES Scientific Calculator
### How do i multiply to 4x4 matrices using fx-991 es ?
The enclosed screen capture shows all the possible dimensions for matrices on the Casio FX-991ES. As you can see the maximum dimension of any matrix is 3. You can only create matrices with dimensions less than or equal to 3.
Dec 01, 2010 | Casio FX-115ES Scientific Calculator
### How to solve 7*3 matrices on casio scientific calculator
The enclosed screen capture shows all the possible dimensions for matrices on the Casio FX-119ES/991ES. As you can see the maximum dimension of any matrix is 3. You can only create matrices with dimensions less than or equal to 3.
Nov 30, 2010 | Casio FX-115ES Scientific Calculator
### On which calculator can i solve 6*6 matrix
If you want to manipulate 6x6 matrices, you can use one of the CASIO Graphing calculators, such as theCFX-9850GPlus, or the FX-9750G plus.
Graphing calculators from other manufacturers can also handle such matrices. The FX-115ES can handle matrices with dimensions up to 3x3.
Nov 25, 2010 | Casio FX-115ES Scientific Calculator
### How to input numbers to solve a matrix using a casio fx-115es
Press MODE 6 to get into matrix mode. Note: do this only once. If you do it again, it will erase all your matrices. Pick the matrix, set its dimension, then the calculator will put up a screen allowing you to enter/edit the matrix elements. At any time you're in matrix mode, press SHIFT 4 to bring up the matrix menu.
All this is explained in the manual beginning on page E-57. If you've misplaced the manual, you can download a copy from
http://support.casio.com/manualfile.php?rgn=1&cid=004001004
Nov 08, 2010 | Casio FX-115ES Scientific Calculator
### I have a Casio fx-115ES; I have followed the manual for creating and multiplying matrices. The matrices have been simple, 2x2 for each, but it returns with "Dimension Error". Any suggestions? Thank you for...
Calculator should be in [MATRIX] MODE
Try the matrices 2X2 matrices matA [1,2,3,4] matB [5,6,7,8], Add them, subtract them, multiply them, take the square of each. If that works for these matrices, then it must be your data. Be careful with negative numbers. To enter those, you must use the change sign (-).
There is also the possibility that the instructions were misread.
Dec 14, 2009 | Casio FX-115ES Scientific Calculator
### How do i solve a matric problem with a TI- 83+
Hello,
You should try to use the correct heading.
Press [2nd][Matrix] right arrow to [EDIT][ENTER] to create the [A]
matrix. Exemple of a 3x3 Matrix [A].
Once the matrix [A] is created you press [2nd][MATRIX] -->[MATH]
You can add similar matrices, you can multiply matrices, subbtract them, raise a matrix to a power, extract the square root, etc...
Here are the other operations you can pefrorm on matrices. Sorry I cannot go into the detail of things;A minimun of theory is needed to be able to use the commands.
Oct 14, 2009 | Casio FX-115ES Scientific Calculator
### How to do matrix multiplication on Casio 115ES?
Hello
First you must set Matrix calculation
[MODE][6:Matrix]. Then By entering one of the numbers [1:MatA] or [2:Matb] or [3:MatC] you get to choose the dimensions of the matrix
(mxn]. Once finished entering the matrix you clear the screen.
The operations on matrices are available by pressing [Shift][Matrix]
[1:Dim] to change the dimension of a matrix (in fact redefining the matrix)
[2:Data] enter values in a matrix
[3:MatA] access Matrix A
[4:Matb] access Matrix B
[5:MatC] access matrix C
[6:MatAns] access the Answer Matrix (the last matrix calculated)
[7:det] Calculate the determinant of a matrix already defined
[8:Trn] The transpose of a matrix already defined
To subtract MatA-MatB
To multiply MatAxMatB
To raise a matrixe to a power 2 [x2], cube [x3]
To obtain inverse of MatA already defined MatA[x-1] [x-1] is the x to the power -1 key
Dimensions of matrices involved in operations must match (see you textbook for the rules)
Hope helps
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Level 3 Expert | 1,482 | 5,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.869492 |
https://physics.stackexchange.com/questions/338469/in-a-four-mass-six-spring-vibration-how-is-the-kinetic-energy-represented | 1,713,360,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00376.warc.gz | 402,045,924 | 39,635 | # In a four mass six spring vibration, how is the kinetic energy represented
This is from Hobson, Riley, Bence Mathematical Methods, p 322. A spring system is described as follows (they are floating in air like molecules):
The equilibrium positions of four equal masses M of a square with sides 2L are $R_n=\pm L_i\pm L_j$ and displacements from equilibrium are $q_n=x_ni+y_nj$. According to the text,
"The coordinates for the system are thus x1, y1, x2, . . . , y4 and the kinetic energy of matrix A is given trivially by $MI_8$ where $I_8$ is the 8x8 identity".
What does that mean? The velocity doesn't even appear. How does that relate to energy?
The origin of the mass matrix lies in a change of coordinates from Cartesian to generalized coordinates.
The kinetic energy of a system of $N$ particles in terms of Cartesian coordinates is $$K=\frac 12\sum_{a=1}^Nm_a\dot{\vec r}_a\cdot\dot{\vec r}_a.$$ If the system is scleronomic, i.e., the relation between Cartesian and generalized coordinates do not involve time explicitly, $\vec r_i=\vec r_i(q_1,\ldots,q_n)$, then by the chain rule $$K=\frac 12\sum_{a=1}^Nm_a\sum_{i=1}^n\frac{\partial \vec r_a}{\partial q_i}\dot q_i\cdot\sum_{j=1}^n\frac{\partial \vec r_a}{\partial q_j}\dot q_j.$$ By rearranging terms this can be written as $$K=\frac12\sum_{i,j}A_{ij}\dot q_i\dot q_j=\frac 12\dot q^TA\dot q,$$ where $A_{ij}\equiv \sum_am_a\frac{\partial \vec r_a}{\partial q_i}\cdot\frac{\partial\vec r_a}{\partial q_j}$ are the components of the mass matrix.
Sometimes it is safer calculate $A$ by using its definition. In simpler systems however it is better to write the kinetic energy explicitly in terms of the generalized coordinates and then compare with the quadratic form $\frac 12\dot q^TA\dot q$ to obtain $A$.
An engineer would call it the "mass matrix" not the "kinetic energy matrix". The KE is given by $\frac 1 2 \mathbf{v}^T \mathbf{M} \mathbf{v}$ where $\mathbf{v}$ is the vector of the velocity components $\dot x_1, \dots, \dot x_4, \dot y_1, \dots, \dot y_4$ and $\mathbf{M} = M \mathbf{I}_8$ - i.e. an $8\times8$ diagonal matrix with all the diagonal terms equal to $M$.
"Kinetic energy matrix" seems a silly name IMO, because as you said it doesn't fully represent the kinetic energy of the system.
The "stiffness matrix" (or whatever the mathematicians who wrote your book call it!) can similarly be written as an $8\times8$ matrix $\mathbf{K}$, though it's not so simple as the mass matrix. The potential energy stored in the springs is then given by $\frac 1 2 \mathbf{x}^T \mathbf{K} \mathbf{x}$
Reading a textbook or web page on matrix methods for modelling multi-degree-of-freedom (MDOF) systems, written for engineers or physicists rather than for mathematicians, might help to understand the basic ideas. | 801 | 2,795 | {"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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.825913 |
https://www.physicsforums.com/threads/applied-forces.109644/ | 1,524,449,533,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00077.warc.gz | 833,582,306 | 15,029 | # Homework Help: Applied Forces
1. Feb 6, 2006
### Huskies213
Can anyone help with what formulas to apply/use here ?
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 3.5 N and points in the negative y direction. Find the direction ° (counterclockwise from theC +x axis) and magnitude (N) of the third force acting on the object.
2. Feb 6, 2006
### Astronuc
Staff Emeritus
The fact that the object moves with 'constant' velocity means that it is not accelerating, which would imply a change in velocity. The fact that it is not accelerating indicates that the net force or sum of forces is nil (0).
Therefore the third force must have a component equal to but opposite of the force in the +ve x-direction, and a y component which is equal to but oppositve the force in the -ve y-direction.
Then the angle is simply determined by an inverse trig function based on ratio of either the x or y component for the third force and the total or resultant force.
3. Feb 6, 2006
### Huskies213
Re
i got the answer of 7.38 for the first part (which is correct) but i'm still unsure about the second part. can anyone help ?
4. Feb 6, 2006
### Astronuc
Staff Emeritus
The vector is in the upper left quadrant (of the Cartesian coordinate system) since its x-component must be - to offset the + x-force, and its y-component must be + to offset the - y-force.
7.38 is correct! This should simply be $\sqrt{3.5^2\,+\,6.2^2}$.
Think of a vector F which has two components, Fx and Fy, and both components are orthogonal.
The magnitude of F = $\sqrt{(F_x^2\,+\,F_y^2)}$, and Fx = F cos$\theta$ and Fy = F sin$\theta$, where $\theta$ is the angle between F and the x-axis.
Think of the definition of the cos of an angle and how it relates to the legs and hypotenuse of a right triangle.
Last edited: Feb 6, 2006 | 514 | 1,953 | {"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} | 4.03125 | 4 | CC-MAIN-2018-17 | longest | en | 0.912766 |
https://www.answers.com/Q/Can_force_change_mass | 1,701,623,214,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00172.warc.gz | 731,458,579 | 44,985 | 0
# Can force change mass
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Related questions
### What is needed to change the direction of a moving mass?
The answer is force because Force is needed to change the direction of a moving mass.
no
no
### How can you find force when mass and velocity are given?
You cannot. Force = Mass*Acceleration or Mass*Rate of change of Velocity.
### Does The Law of Applied Force state that a body's change in mass is proportional to the amount of force applied to it?
The Law of Applied Force states that a body's change in mass is proportional to the amount of force applied to it.
### How do you calculate force when mass and velocity are given?
Force equals the mass times the rate of change of the velocity.
### How does mass affects the force?
Mass and Force have no relationship except the gravitational force that is dependent on the mass of the body. Mass affects only the inertial force. Inertial force is the force required to change a state of rest or motion of a body. Greater the mass greater the inertial force required.
### How does the force of graviti change with mass and distance?
as distance increases gravity's force decreases as mass increases gravity's force increases
### What two things can change the acceleration of an object?
The two factors are 1. the mass of the object, and 2. the force exerted on it.
### Is acceleration constant or does it change?
Going back to the equation F=m·a you can see that if the force changes but the mass does not, accelleration will change as well. If mass and force do not change, accelleration will be constant.
### How does the force change as the mass changes?
The correct question if Force due to Gravity varies directly with mass. As mass increase the Force due to gravity increases linearly.
### How do we Show how change in speed is related to force and mass?
F=ma F- Force m- mass a- acceleration | 447 | 2,028 | {"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} | 3.859375 | 4 | CC-MAIN-2023-50 | latest | en | 0.916482 |
https://www.engageny.org/ccls-math/1oa7 | 1,571,615,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00299.warc.gz | 896,842,107 | 13,309 | 1.OA.7 | EngageNY
## CCLS - Math: 1.OA.7
Category
Operations And Algebraic Thinking
Sub-Category
Work With Addition And Subtraction Equations.
State Standard:
Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2. | 126 | 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} | 3.203125 | 3 | CC-MAIN-2019-43 | latest | en | 0.858464 |
https://www.codebymath.com/index.php/welcome/lesson/divergent-series | 1,643,056,365,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00576.warc.gz | 741,474,949 | 6,380 | ## Lesson goal: A series that diverges
Previous: What kinds of numbers? | Home | Next: Numbers ending in 1
Here's a lesson that uses the for-loop to study a series (a series is a bunch of numbers added together). In pre-calculus or calculus, you might have learned that $\Sigma\frac{1}{n}$ diverges (or gives an infinite value if you run the sum over an infinite number of terms). If you take the first few terms in the sum, you'll get $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$. Each term gets smaller and smaller, so does it really diverge? Let's write some code here to test it.
As you work on this, notice the similarities between the $\Sigma$ from math, and the for-loop in programming. In this case $\Sigma_{n=a}^{b}$ runs $n$ from $a$ to $b$, while for n=a,b do runs n from a to b.
#### Now you try. Fix the for-loop and sum=sum+ line to see if $\Sigma\frac{1}{n}$ diverges.
Type your code here:
See your results here: | 258 | 932 | {"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} | 4.03125 | 4 | CC-MAIN-2022-05 | longest | en | 0.895252 |
https://betterlesson.com/community/lesson/12474/volume-of-rectangular-prism | 1,498,703,038,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00147.warc.gz | 716,751,116 | 13,737 | Lesson: Volume of Rectangular Prism
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Lesson Objective
SWBAT find the volume of a rectangular prism.
Lesson Plan
Element/Time Lesson Steps Do Now (8:30 – 8:40)/(10:00 –10:10) · T displays 4 review problems · S complete problems independently on do now sheet · T walks around to check hw/help students with do now questions · T/S review problems together · S correct their work to reflect the correct answers HW Check · T displays correct answers and work · S correct work with colored pencil T walks around to review specific students work · T/S review 3 questions Mental Math (8:40 – 8:50) /(10:10 – 10:20) · Mini Lesson (8:50 – 9:25)/(10:20 – 10:55) · T will tell students that Volume is a measure for 3D shapes. · T will tell student volume is the number of cubic units that can fit inside of a solid. (The number of cubes that fit in a shape) · T/S will show an example using cubes as manipulatives o Give students a rectangular solid and several one inch cubes o Ask students to fill the solid with cubes and to count how many cubes fit inside of the solid o Explain that this is the volume of the solid · T will tell students that while we can always find volume by filling a solid with cubes, we might not always have that available so there is a formula to help us. · T will share formula V = lwh · T will demonstrate how to find the volume of a shape using the formula o Demonstrate with the solid used with manipulatives o Show that the length of the shape is 6 because 6 cubes fit across the length o Show that the width of the shape is 4 because 4 cubes fit across the width o Show that the shape is 2 units high because two cubes are the same height as the solid o Multiply 6 x 4 x 2 to get 48 same as what we did when we filled the shape o Demonstrate another example without manipulatives § Identify the height, width and length but putting a H, L and W next to those measures § Multiply the three measures together to get the volume § Stop and discuss the units for volume – ask students to show the fingers for the exponent that represents “cubed” § Explain to students that since we are multiplying 3 dimensions together our units are cubic units (Relate to how area is square units) · S will find the volume of the shape · Connect volume with filling a shape Skills Time (9:35- 9:50) /(11:05-11:25) ·
Lesson Resources
Geometry Notes Volume of Rectangular Prisms 2,081 21 4volume hw 1,181 21 4volume practice 1,243 U9.L16.SHAPES.Volume.docx 1,350 | 646 | 2,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-26 | longest | en | 0.831926 |
http://mathmatik.com/math-answers/online-college-math/have-a-value-but-approaches.html | 1,521,891,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650262.65/warc/CC-MAIN-20180324112821-20180324132821-00333.warc.gz | 180,234,081 | 12,911 | We Promise to Make your Math Frustrations Go Away!
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Prev Next | 920 | 4,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.958341 |
https://math.stackexchange.com/questions/2859191/monoid-in-general-dynamic-system-definition | 1,660,635,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572221.38/warc/CC-MAIN-20220816060335-20220816090335-00556.warc.gz | 372,732,844 | 64,405 | # Monoid in general dynamic system definition
I am a newbie in this field but what difference does taking monoid or group in the following definition of dynamic system make?
A tuple $$(T,M,\phi)$$
is called dynamic system, where $T$ is additively written monoid (time), $M$ is a phase space and $\phi$ is an evolution operator
$$\phi = U\subseteq T\times M \rightarrow M$$
of the system.
I have found another stronger definiton in which $T$ is said to be additive group.
Does it matter? Is the addition necessarily commutative?
• I am at odds with $T$ being a group. How do you turn back the clock? Jul 22, 2018 at 7:54
## 1 Answer
The difference is whether the time $t$ can only forward or backward as well. A group has inverses, a monoid does not need them. In particular, $\mathbb R$ is a group, $[0, \infty)$ (with addition) is not. If your dynamical system is reversible, you might want to use $\mathbb R$, otherwise only $[0,\infty)$. | 257 | 950 | {"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": 2, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-33 | latest | en | 0.928966 |
https://www.lesswrong.com/posts/PhKZgz5Gxw9soHtng/optimization-speculations-on-the-x-and-only-x-problem | 1,670,200,908,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711001.28/warc/CC-MAIN-20221205000525-20221205030525-00070.warc.gz | 923,077,356 | 21,107 | # Optimization, speculations on the X and only X problem.
3 min read30th Mar 20215 comments
# Ω 4
Review
Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.
Epistemic status: Speculative. Rambling stream of ideas.
How do we tell if an arbitrary system is an optimizer or not?
Well "optimiser" is a set of hypothesis. We can take an AIXItl like structure, and use the universal prior over utility functions. The other hypothesis set is the default solomonov prior. All computer programs weighted by complexity.
Note that each hypothesis set contains a shrunken copy of the other. Within the space of all computer programs, some of those programs will be optimisers. Within the space of all possible optimizers, some will have the goal of making their output match the output of an arbitrary computer program. Thus the odds ratio in either direction is bounded. We call this the odds ratio approach.
The intuition is that this odds ratio provides a good measure if a program is optimizing.
(There is another measure of optimization, namely the weighted average utility over all Turing machine environments. We call this the weighted reward approach)
I will discuss a few programs, and how much optimization each formalism thinks they display.
### Trivial constant programms
``````While True:
print(0)``````
and
``````While True:
print(1)``````
Under the odds ratio approach, both these programs are on the not optimizer end of the scale and right next to each other.
Under the weighted reward approach, the answer is much more sensitive on choice of universal Turing machine. An environment that just gives reward for each 1 outputted is very simple. It might even account for say 5% (random number) of weight across all computable environments. If the environment that just rewards 0's isn't given a correspondingly high weight, then according to the weighted reward framework, one of these programs can be much more optimizing than the other.
### Conditional AIXI
``````if input.read(100)==long_random_string:
behave_like_AIXI()
else:
while True:
print(0)``````
and
``````if input.read(100)!=long_random_string:
behave_like_AIXI()
else:
while True:
print(0)``````
In the odds ratio formulation, both of these are pretty optimizerry, and almost identical. In either formalization they need to take the complexity of describing the long random string. In the optimizer formulation, they need to take the complexity of constantly outputing 0's. In the all programs formulation, they need the complexity of describing AIXI like behaviour.
In the weighted reward formulation, the first is basically the constant 0 program, while the other is nearly as much of an optimizer as AIXI.
## Optimising X and only X
The basic idea here is that once you compensate for the X optimization then the system isn't an optimizer.
One straightforward way of doing this is a rejection of low X proposed action. You have an arbitrary Turing machine T. Whenever T proposes an action A, the utility/ expected utility of that action is computed. If A is one of the top m actions when it comes to maximizing X, then A is taken. Otherwise T is told to try again. A system is said to be an X and only X maximizer if it is better modelled as being any old Turing machine behind a low X rejection gate. As opposed to a Y optimizer behind a low X rejection gate.
For example, a system that outputs whichever action in the million X-best actions contains the most 0's is an X and only X best optimizer. Here m=1000000, and the simple Turing machine behind the low X rejection gate is just a TM that cycles through all strings, starting with the ones full of 0's.
A system that outputs whichever action in the million X-best actions maximizes the number of paperclips in existence is not an X and only X maximiser. It is best modelled as a paperclip maximiser behind a low X rejector. And this comes back to an implicit choice of distribution over optimizers at the start of this article. But presumably the paperclip maximizer is prominent in that set.
Note that at the start of this, we used a set of optimisers. One way to formally specify that set would be to take an AIXI like design, and put a solomonov prior over utility functions (as functions of observations)
However, this fails to capture AI's that optimise for something that is not a simple function of observations. (Like some real world object in a world where the AI's senses can be fooled.) However, if we had a fomalization of an agent that did this kind of optimization, we could just plug in all possible utility functions, weighted by simplicity, into our set of optimizers. Likewise for other decision theories, compute limited AI's ect.
# Ω 4
New Comment
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Thanks for trying to clarify "X and only X", which IMO is a promising concept.
One thing we might want from an only-Xer is that, in some not-yet-formal sense, it's "only trying to X" and not trying to do anything else. A further thing we might want is that the only-Xer only tries to X, across some relevant set of counterfactuals. You've discussed the counterfactuals across possible environments. Another kind of counterfactual is across modifications of the only-Xer. Modification-counterfactuals seem to point to a key problem of alignment: how does this generalize? If we've selected something to do X, within some set of environments, what does that imply about how it'll behave outside of that set of environments? It looks like by your definition we could have a program that's a very competent general intelligence with a slot for a goal, plus a pointer to X in that slot; and that program would count as an only-Xer. This program would be very close, in some sense, to programs that optimize competently for not-X, or for a totally unrelated Y. That seems counterintuitive for my intuitive picture of an "X and only X"er, so either there's more to be said, or my picture is incoherent.
My picture of an X and only X er is that the actual program you run should optimize only for X. I wasn't considering similarity in code space at all.
Getting the lexicographically first formal ZFC proof of say the Collatz conjecture should be safe. Getting a random proof sampled from the set of all proofs < 1 terabyte long should be safe. But I think that there exist proofs that wouldn't be safe. There might be a valid proof of the conjecture that had the code for a paperclip maximizer encoded into the proof, and that exploited some flaw in computers or humans to bootstrap this code into existence. This is what I want to avoid.
Your picture might be coherent and formalizable into some different technical definition. But you would have to start talking about difference in codespace, which can differ depending on different programming languages.
The program if True: x() else: y() is very similar in codespace to if False: x() else: y()
If code space is defined in terms of minimum edit distance, then layers of interpereters, error correction and holomorphic encryption can change it. This might be what you are after, I don't know.
Well, a main reason we'd care about codespace distance, is that it tells us something about how the agent will change as it learns (i.e. moves around in codespace). (This is involving time, since the agent is changing, contra your picture.) So a key (quasi)metric on codespace would be, "how much" learning does it take to get from here to there. The if True: x() else: y() program is an unnatural point in codespace in this metric: you'd have to have traversed the both the distances from null to x() and from null to y(), and it's weird to have traversed a distance and make no use of your position. A framing of the only-X problem is that traversing from null to a program that's an only-Xer according to your definition, might also constitute traversing almost all of the way from null to a program that's an only-Yer, where Y is "very different" from X.
I don't think that learning is moving around in codespace. In the simplest case, the AI is like any other non self modifying program. The code stays fixed as the programmers wrote it. The variables update. The AI doesn't start from null. The programmer starts from a blank text file, and adds code. Then they run the code. The AI can start with sophisticated behaviour the moment its turned on.
So are we talking about a program that could change from an X er to a Y er with a small change in the code written, or with a small amount of extra observation of the world?
To clarify where my responses are coming from: I think what I'm saying is not that directly relevant to your specific point in the post. I'm more (1) interested in discussing the notion of only-X, broadly, and (2) reacting to the feature of your discussion (shared by much other discussion) that you (IIUC) consider only the extensional (input-output) behavior of programs, excluding from analysis the intensional properties. (Which is a reasonable approach, e.g. because the input-output behavior captures much of what we care about, and also because it's maybe easier to analyze and already contains some of our problems / confusions.)
From where I'm sitting, when a program "makes an observation of the world", that's moving around in codespace. There's of course useful stuff to say about the part that didn't change. When we really understand how a cognitive algorithm works, it starts to look like a clear algorithm / data separation; e.g. in Bayesian updating, we have a clear picture of the code that's fixed, and how it operates on the varying data. But before we understand the program in that way, we might be unable to usefully separate it out into a fixed part and a varying part. Then it's natural to say things like "the child invented a strategy for picking up blocks; next time, they just use that strategy", where the first clause is talking about a change in source code. We know for sure that such separations can be done, because for example we can say that the child is always operating in accordance with fixed physical law, and we might suspect there's "fundamental brain algorithms" that are also basically fixed. Likewise, even though Solomonoff induction is always just Solomonoff induction plus data, it can be also useful to understand SI(some data) in terms of understanding those programs that are highly ranked by SI(some data), and it seems reasonable to call that "the algorithm changed to emphasize those programs". | 2,304 | 10,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.916312 |
https://www.physicsforums.com/threads/compact-hausdorff-space-with-continuous-function.94772/ | 1,544,985,559,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827963.70/warc/CC-MAIN-20181216165437-20181216191437-00569.warc.gz | 998,623,499 | 14,729 | # Homework Help: Compact Hausdorff space with continuous function
1. Oct 15, 2005
### Oxymoron
Question
Let $X$ be a compact Hausdorff space and let $f:X\rightarrow X$ be continuous. Show that there exists a non-empty subset $A \subseteq X$ such that $f(A) = A$.
At the moment I am trying to show that $f$ is a homeomorphism and maybe after that I can show that $f(A) = A$. But Im not sure if this is the right tactic.
I know that $f$ is a continuous function, and since $f:X \rightarrow X$ it is bijective (?) I thought it might be since the domain and range coincide.
Anyway, if I take a closed subset $A \subseteq X$ then $A$ is automatically compact (since every closed subset of a Hausdorff space is compact). Then since $f$ is continuous it maps closed compact sets to closed compact sets.
Since $f$ is bijective, its inverse $f^{-1}$ exists, and $f^{-1}$ will also map closed compact sets to closed compact sets.
Therefore $f$ is a homeomorphism. Im not sure if I can conclude from this that $A$ is homeomorphic to $f(A)$ which implies $f(A) = A$.
PS. What about the open sets in $X$? I need help!
Last edited: Oct 15, 2005
2. Oct 15, 2005
### fourier jr
just because a continuous map goes from a set to itself doesn't mean it's a bijection. consider f:X->X with f(x)=0 for every x in X; that's continuous but not a bijection. off the top of my head i would try using the fact that a) every closed subspace of a Hausdorff space is compact and b) that every compact subspace of a Hausdorff space is closed. maybe also use the fact that a continuous image of a compact space is compact. i'm not sure if that will help or not. i'd have to think about it some more.
3. Oct 15, 2005
### Oxymoron
You're right. I don't know what I was thinking.
I have ventured down that avenue. I can show that $f$ maps closed compact sets to closed compact sets.
Im thinking that the easiest way to show that $f(A) = A$ is to show that $f$ is the identity function.
4. Oct 15, 2005
### Hurkyl
Staff Emeritus
Maybe playing with examples would help. You have things like rotations of disks and rings, contractions of intervals, and even the silly map of the unit circle that maps the point with an angle of t to the point with an angle of 2t.
It seems clear that you'll somehow have to actually use the fact X is compact. For example, you should be able to find a map of R that serves as a counterexample.
Last edited: Oct 15, 2005
5. Oct 15, 2005
### Oxymoron
Forget the identity function for now. I have another idea.
Let $A_0 \subseteq X$ be a non-empty closed set such that $f(A_0) \subset A_0$. Then let $A_1 = f(A_0)$ and after iterating we have $f(A_n) \subset A_n$ and
$$A_n = f(A_{n-1}) \quad \forall\, n \in \mathbb{N}$$
From this we see that $\{A_n\}_{n\in\mathbb{N}}^{\infty}$ is a decreasing sequence of non-empty closed sets. Now let
$$A = \bigcap_{n\in\mathbb{N}}A_n$$
So $A$ is also non-empty and closed. If we then observe $f(A)$ we can see that $f(A)$ is certainly contained within $f(A_n)$ and in fact
$$f(A) \subset f(A_n) \backslash A_n$$
for each $n\in\mathbb{N}$. But this simply says that
$$f(A) \subset A$$
Now lets take some point $a \in A$ and let $B = f^{-1}(a)$. Since $a \in A_{n+1} = f(A_n)$ we know that $B \cap A_n \neq \oslash$. Therefore $B \cap A_n$ is a decreasing sequence of non-empty closed sets in a compact space $X$. Therefore
$$\bigcap_{n\in\mathbb{N}} B \cap A_n \neq \oslash$$
Now choose some $y \in \bigcap_{n\in\mathbb{N}} B \cap A_n$. Then obviously $y \in A$ and $f(y) = a$. That is $f(A) = A$.
How does this look?
6. May 9, 2010
### complexnumber
Why is $\{A_n\}_{n\in\mathbb{N}}^{\infty}$ non-empty? Do you need to prove it or is it obvious? I am working on the same problem and cannot figure out this part. | 1,140 | 3,775 | {"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} | 3.328125 | 3 | CC-MAIN-2018-51 | latest | en | 0.967253 |
https://www.jiskha.com/display.cgi?id=1381979910 | 1,503,456,375,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.82/warc/CC-MAIN-20170823020201-20170823040201-00511.warc.gz | 895,764,523 | 3,814 | # Math
posted by .
is -1.59 written as a rational number in a/b form - is it -1 59/100??
• Math -
no, you have shown a mixed number. Ditch the space to get
-159/100
• Math -
thank you
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More Similar Questions | 581 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-34 | latest | en | 0.900452 |
https://www.geeksforgeeks.org/set-theory/?ref=outind | 1,709,621,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00362.warc.gz | 761,577,366 | 61,588 | Related Articles
# Set Theory – Definition, Types of Sets, Symbols & Examples
Set Theory is a branch of Mathematics that studies the collection of objects and operations based on it. Sets are studied under the logic division of mathematics.
A set is simply a collection of objects. The words collection, aggregate, and class are synonymous to set. On the other hand element, members, and objects are synonymous and stand for the members of the set of which the set is comprised.
In this article on Set Theory, we will learn about sets, types of sets, representation of sets, operations on sets, and application of sets in detail.
## History of Set Theory
The concept of Set Theory was propounded in the year 1874 by Georg Cantor in his paper name ‘On a Property of Collection of All Real Algebraic Numbers‘.
His concept of Set Theory was later used by other mathematicians in giving various other theories such as Klein’s Encyclopedia and Russell Paradox. Sets Theory is a foundation for a better understanding of topology, abstract algebra, and discrete mathematics.
## Set Definition
Sets are defined as ”a well-defined collection of objects”. Let’s say we have a set of Natural Numbers then it will have all the natural numbers as its member and the collection of the numbers is well defined that they are natural numbers.
A set is always denoted by a capital letter. For Example:
A set of Natural Numbers is given by N = {1, 2, 3, 4…..}. The term ‘well defined’ should be taken care of, as if we try to make a set of best players, then the term ‘best’ is not well defined. The concept of best, worst, beautiful, powerful, etc. varies according to the notions, assumptions, likes, dislikes, and biases of a person.
This explains what is a set, now let’s look at some set terms:
### Elements of a Set
The objects contained by a set are called the elements of the set. For Example:
In the set of Natural Numbers 1, 2, 3, etc. are the objects contained by the set, hence they are called the elements of the set of Natural Numbers. We can also say that 1 belongs to set N. It is represented as 1 ∈ N, where ∈ is the symbol of belongs to.
### Cardinal Number of a Set
The number of elements present in a set is called the Cardinal Number of a Set.
For Example
Let’s say P is a set of the first five prime numbers given by P = {2, 3, 5, 7, 11}, then the Cardinal Number of set P is 5. The Cardinal Number of Set P is represented by n(P) or |P| = 5.
## Examples of Set
A Set is a well-defined collection of objects. The objects inside a set are called members of the set. Some examples of sets are mentioned below:
Set of Natural Numbers: N = {1, 2, 3, 4….}
Set of Even Numbers: E = {2, 4, 6, 8…}
Set of Prime Numbers: P = {2, 3, 5, 7,….}
Set of Integers: Z = {…, -4, -3, -2, -1, 0, 1, 2,….}
Some Standard Sets used in Set Theory
• Set of Natural Numbers is denoted by N
• Set of Whole Numbers is denoted by W
• Set of Integers is denoted by Z
• Set of Rational Numbers is denoted by Q
• Set of Irrational Numbers is denoted by P
• Set of Real Numbers is denoted by R
## Representation of Set
Sets are primarily represented in two forms
• Roster Form
• Set Builder Form
### 1. Roster Form
In the Roster Form of the set, the elements are placed inside braces {} and are separated by commas.
Let’s say we have a set of the first five prime numbers then it will be represented by P = {2, 3, 5, 7, 11}. Here the set P is an example of a finite set as the number of elements is finite, however, we can come across a set that has infinite elements then in that case the roster form is represented in the manner that some elements are placed followed by dots to represent infinity inside the braces.
Let’s say we have to represent a set of Natural Numbers in Roster Form then its Roster Form is given as N = {1, 2, 3, 4…..}.
Set does not contain duplicate elements. For Example:
If A represents a set that contains all letter of word TREE, then correct roster form representation will be:
A = {T,R,E}
A≠{T,R,E,E}
### 2. Set Builder Form
In Set Builder Form, a rule or a statement describing the common characteristics of all the elements is written instead of writing the elements directly inside the braces. For Example:
A set of all the prime numbers less than or equal to 10 is given as P = {p : p is a prime number ≤ 10}. In another example, the set of Natural Numbers in set builder form is given as N = {n : n is a natural number}.
Read More on Representation of Sets
## Types of Set
There are different types of sets categorized on various parameters. Some type of Sets are mentioned below:
• Empty Set
• Singleton set
• Finite Set
• Infinite set
• Power set
• Universal set
• Equivalent set
• Disjoint set
• Overlapping set
Some of the sets are explained below:
### 1. Empty Set
A set that has no elements inside it is called an Empty Set. It is represented by Φ or {}. For Example A = {x : x ∈ N and 2 < x < 3}. Here, between 2 and 3, no natural number exists, hence A is an Empty Set. Empty Sets are also known as Null Sets.
### 2. Singleton Set
A set that has only one element inside it is called a Singleton Set. For Example, B = {x : x ∈ N and 2 < x < 4} = {3}. Here between 2 and 3 only one element exists, hence B is called a Singleton Set.
### 3. Finite Set
A set that has a fixed or finite number of elements inside it is called a Finite Set. For Example A = {x : x is an even number less than 10} then A = {2, 4, 6, 8}. Here A has 4 elements, hence A is a finite set. The number of elements present in a finite set is called the Cardinal Number of a finite set. It is given as n(A) = 4.
### 4. Infinite Set
A set that has an indefinite or infinite number of elements inside it is called a Finite Set. For Example A = {x : x is an even number l} then A = {2, 4, 6, 8……}. Here A has unlimited elements, hence A is an infinite set.
### 5. Equivalent Sets
If the number of elements present in two sets is equal i.e. the cardinal number of two finite sets is the same then they are called Equivalent Sets. For Example, A = {x : x is an even number up to 10} = {2, 4, 6, 8, 10} and B = {y : y is an odd number less than 10} = {1, 3, 5, 7, 9}. Here, the cardinal number of set A is n(A) = 5 and that of B is given as n(B) = 5 then we see that n(A) = n(B). Hence A and B are equivalent sets.
### 6. Equal Sets
If the number of elements and also the elements of two sets are the same irrespective of the order then the two sets are called equal sets. For Example, if set A = {2, 4, 6, 8} and B ={8, 4, 6, 2} then we see that number of elements in both sets A and B is 4 i.e. same and the elements are also the same although the order is different. Hence, A and B are Equal Sets. Equal Sets are represented as A = B.
### 7. Unequal Sets
If at least any one element of one set differs from the elements of another set then the two sets are said to be unequal sets. For Example, if set A = {2, 4, 6, 8} and B = {4, 6, 8, 10} then set A and B are unequal sets as 2 is present in set A but not in B and 10 is present in set B but not in A. Hence, one element differs between them thus making them unequal. However, the cardinal number is the same therefore they are equivalent sets.
### 8. Overlapping Sets
If at least any one element of the two sets are the same then the two sets are said to be overlapping sets. For Example, if set A = {1, 2, 3} and set B = {3, 4, 5} then we see that 3 is the common element between set A and set B hence, set A and set B are Overlapping Sets.
### 9. Disjoint Sets
If none of the elements between two sets are common then they are called the Disjoint Sets i.e., for two sets A and B if A. For Example, set A = {1, 2, 3} and set B = {4, 5, 6} then we observe that there is no common element between set A and set B hence, set A and B are Disjoint Sets.
Apart from the above-mentioned sets, there are other sets called, Subsets, Supersets, Universal Sets, and Power Sets. We will learn them below in detail.
Read, Types of Sets
## Subset
If A and B are two sets such that every element of set A is present in set B then A is called the subset of B. It is represented as A ⊆ B and read as ‘A is a subset of B’. Mathematically it is expressed as
A ⊆ B iff
a ∈ A ⇒ a ∈ B
If A is not a subset of B we write it as A ⊄ B.
For Example
if A = {1, 2} and B = {1, 2, 3} then we see that all the elements of A are present in B, hence A ⊆ B.
## Types of Subset
There are two main types of subset – Proper subset and Improper subset.
### 1. Proper Subset
If a subset doesn’t contain all the elements of the set or has fewer elements than the original set then it is called the proper subset. For example:
in set A = {1, 2} and B = {1, 2, 3}, the subset A doesn’t contain all the elements of the original set B, hence A is a proper subset of B. It is represented as A ⊂ B. Empty set is a proper subset of a given set as it has no elements.
### 2. Improper Subset
If a subset contains all the elements that are present in the original set then it is called an Improper Subset. For Example
if set A = {1, 2, 3, 4} and set B = {1, 2, 3, 4} then A is the improper subset of set B. It is mathematically expressed as A ⊆ B. Thus we deduce that two sets are equal iff A ⊆ B and B ⊆ A. It should be noted that an empty set is an improper subset of itself.
### Some Important Results on Subset
• Every set is a subset of itself
• An empty Set is a subset of every set.
• The number of possible subsets for a given finite set with ‘n’ number of elements is equal to 2n.
• N ⊂ W ⊂ Z ⊂ Q ⊂ R and T ⊂ R where N is a set of Natural Numbers, W is a set of Whole Numbers, Z is a set of integers, Q is a set of Rational Numbers, T is a set of irrational numbers and R is set of real numbers.
## Superset
If all the elements of set A are present in set B then set B is called the Superset of set A. It is represented as B ⊇ A.
Let’s say if A = {2, 3, 4} and B = {1, 2, 3, 4} then we see that all elements of set A are present in set B, hence B ⊇ A. If a superset has more elements than its subset then it is called a proper or strict superset.
A Proper Superset is represented as B ⊃ A. Some of the Properties of Supersets are mentioned below:
• Every set is a superset of itself.
• Every set is a superset of an empty set.
• Total number of possible supersets for a given subset is infinite
• If B is a superset of A then A is a subset of B
## Universal Set
The set that contains all the sets in it is called a Universal Set.
Let’s say set A = {1, 2, 3}, set B = {4, 5}, and set C = {6, 7} then Universal Set is given as U = {1, 2, 3, 4, 5, 6, 7}. Another Example of a Universal Set is U = {Set of All Living Beings} then which includes both floras and faunas. Flora and fauna are the subsets of Universal Sets U.
## Power Set
A set that contains all the subsets as its element is called the Power Set. For Example:
if set A = {1, 3, 5} then its subsets are {Φ}, {1}, {2}, {3}, {1, 3}, {3, 5}, {1, 5} and {1, 3, 5} then its Power Set is given as P(A) = {{Φ}, {1}, {2}, {3}, {1, 3}, {3, 5}, {1, 5}, {1, 3, 5}}. As we know the number of possible subsets for a finite set containing n elements is given by 2n then the number of elements in the power set is also 2n
## Set Theory Symbols
There are various symbols that are used in Sets Theory. The notations and their explanation are tabulated below:
SymbolExplanation
{}Set
x ∈ Ax is an element of set A
x ∉ Ax is not an element of set A
∃ or ∄There exist or there doesn’t exist
ΦEmpty Set
A = BEqual Sets
n(A)Cardinal Number of Set A
P(A)Power Set
A ⊆ BA is a subset of B
A ⊂ BA is the Proper subset of B
A ⊈ BA is not a subset of B
B ⊇ AB is the superset of A
B ⊃ AB is a proper superset of A
B ⊉ AB is not a superset of A
A ∪ BA union B
A ∩ BA intersection B
A’Complement of Set A
Some of the most common used set symbols:
Symbol
Related Set
N
set of natural numbers. examples : 2,43,56, etc.
Z
set of integers. examples : 4, -23, 0, etc.
R
set of real numbers. examples : 6.9 , 2√3, etc.
C
set of complex numbers. examples : 2 + 3i, i , etc.
## Set Operation
The sets undergo various operation which includes
We will learn them briefly below:
### 1. Union of Sets
Union of Sets basically refers to uniting two sets and writing their elements in a single set without repeating elements if common elements are present. The union of sets is given by A ∪ B.
For Example if Set A = {2, 4} and Set B = {4, 6} then A ∪ B = {2, 4} ∪ {4, 6} = {2, 4, 6}
### 2. Intersection of Sets
Intersection of sets refers to finding the common elements between two sets. It is given by A ∩ B. For Example if set A = {2, 4} and B = {4, 6} then A ∩ B = {2, 4} ∩ {4, 6} = {4}.
### 3. Difference of Sets
Difference of Sets refers to the deletion of common elements of two sets and writing down the remaining elements of two sets. It is represented as A – B. For Example if et A = {2, 4} and B = {4, 6} then A – B = {2}
### 4. Complement of Set
Compliment of Set refers to the set of elements from the universal set excluding the elements of the set of which we are finding the compliment. It is given by A’. For Example:
if we have to find out the complement of the set of Natural Numbers then it will include all the numbers in the set from the Real Numbers except the Natural Numbers. Here Real Number is the Universal set of Natural Numbers.
### 5. Cartesian Product of Sets
Cartesian Product of Sets refers to the product between the elements of two sets in ordered pair. It is given as A ✕ B. For Example if set A = {2, 4} and B = {4, 6} then A ✕ B = {(2,4), (2,6), (4,4), (4,6)}.
## Properties of Set Operations
The various properties followed by sets are tabulated below:
PropertyExpression
Commutative Property
A ∪ B = B ∪ A
A ∩ B = B ∩ A
Associative Property
(A ∪ B) ∪ C = A ∪ (B ∪ C)
Distributive Property
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Identity Property
A ∪ Φ = A
A ∩ U = A
Complement PropertyA ∪ A’ = U
Idempotent PropertyA ∪ A = A ∩ A = A
## Set Theory Formulas
The set theory formulas are given for two kinds of sets – overlapping and disjoint sets. Let’s learn them separately
### 1. Overlapping Set Formulas
Given that two sets A and B are overlapping, the formulas are as follows:
n(A ∪ B) n(A) + n(B) – n(A ∩ B) n(A ∩ B) n(A) + n(B) – n(A ∪ B) n(A) n(A ∪ B) + n(A ∩ B) – n(B) n(B) n(A ∪ B) + n(A ∩ B) – n(A) n(A – B) n(A ∪ B) – n(B) n(A – B) n(A) – n(A ∩ B) n(A ∪ B ∪ C) n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)
### 2. Disjoint Set Formula
If two sets A and B are disjoint sets
n(A ∪ B) n(A) + n(B) (A ∩ B) Φ n(A – B) n(A)
## De Morgan’s Laws
De Morgan’s Law is applicable in relating the union and intersection of two sets via their complements. There are two laws under De Morgan’s Law. Let’s learn them briefly
### 1. De Morgan’s Law of Union
De Morgan’s Law of Union states that the complement of the union of two sets is equal to the intersection of the complement of individual sets. Mathematically it can be expressed as
(A ∪ B)’ = A’ ∩ B’
### 2. De Morgan’s Law of Intersection
De Morgan’s Law of Intersection states that the complement of the intersection of two sets is equal to the union of the complement of individual sets. Mathematically it can be expressed as
(A ∩ B)’ = A’ ∪ B’
## Venn Diagram
Venn Diagram is a technique for representing the relation between two sets with the help of circles, generally intersecting.
For Example:
Two circles intersecting with each other with the common area merged into them represent the union of sets, and two intersecting circles with a common area highlighted represents the intersection of sets while two circles separated from each other represents the two disjoint sets.
A rectangular box surrounding the circle represents the universal set. The Venn diagrams for various operations of sets are listed below:
## Solved Examples on Set Theory
Example 1: If A and B are two sets such that n(A) = 17 and n(B) = 23 and n(A ∪ B) = 38 then find n(A ∩ B).
Solution:
We know that n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 38 = 17 + 23 – n(A ∩ B)
⇒ n(A ∩ B) = 40 – 38 = 2
Example 2: If X = {1, 2, 3, 4, 5}, Y = {4, 5, 6, 7, 8}, and Z = {7, 8, 9, 10, 11}, find (X ∪ Y), (X ∪ Z), (Y ∪ Z), (X ∪ Y ∪ Z), and X ∩ (Y ∪ Z)
Solution:
(X ∪ Y) = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
(X ∪ Z) = {1, 2, 3, 4, 5} ∪ {7, 8, 9, 10, 11} = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}
(Y ∪ Z) = {4, 5, 6, 7, 8} ∪ {7, 8, 9, 10, 11} = {4, 5, 6, 7, 8, 9, 10, 11}
(X ∪ Y ∪ Z) = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} ∪ {7, 8, 9, 10, 11} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
X ∩ (Y ∪ Z) = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8, 9, 10, 11} = {4, 5}
We have covered all the concepts required to learn the set theory. We have covered history, operations, formulas, symbols of set theory.
Set theory is an important topic and many questions come from set theory in many competitive Exams. Students should focus on set theory and practice with some questions provided in this article.
## FAQs on Set Theory
### 1. What is Set Theory?
Set Theory is the study of a collection of objects which are called sets and the relation between different sets.
### 2. How do you define Set?
Set is a well-defined collection of objects. These objects are called the element of sets.
### 3. Who Invented Set Theory?
Set Theory was invented by German Mathematician Georg Cantor.
### 4. What is Algebra of Sets?
Algebra of Sets deals with the following laws:
• Commutative Laws
• Associative Law
• Distributive Law
• Identity Law
• Idempotent Law
These laws are covered in the article under Properties of Sets.
### 5. What is a Subset?
A subset is a set that has fewer or equal elements of another set. The other set is called a Superset.
### 6. What is De Morgan’s Law?
De Morgan’s Law is a law that deals with the union and intersection of sets with the intersection and union of individual sets respectively. There are two laws under it namely De Morgan’s Law of Union and De Morgan’s Law of Intersection. These are covered under the section of De Morgan’s Law in this article. | 5,504 | 18,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-10 | latest | en | 0.956467 |
https://sites.engineering.ucsb.edu/~jbraw/mpc/fig-html/ch3/fig-3-5.html | 1,695,879,319,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00468.warc.gz | 574,234,995 | 3,722 | ## Code for Figure 3.5
### main.m
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83% Example Constraint tightening for a linear system.
A = [1, 1; 0, 1];
B = [0; 1];
K = [-0.4, -1.2];
% State constraints. u is unconstrained.
xmax = 1;
umax = 1;
Z = struct();
[GH, Z.psi] = hyperrectangle([xmax*[-1; -1]; -umax], [xmax*[1; 1]; umax]);
Z.G = GH(:,1:2);
Z.H = GH(:,3);
% Disturbance set. Because all sets are convex, we only have to keep track of
% the vertices.
infnorm = @(x) max(abs(x(:)));
wmax = 0.1;
W = [wmax, wmax, -wmax, -wmax;
wmax, -wmax, -wmax, wmax];
normW = infnorm(W);
KW = K*W;
normKW = infnorm(KW);
% For each value of N, see how far you need to squeeze.
Nmax = 25;
AkW = cell(Nmax + 1, 1);
alphaW = cell(Nmax + 1, 1);
normZbars = NaN(3, Nmax + 1);
Nkeep = [2, 3, 4, 5, 6, 7, 8, 10, 15];
Ak = eye(size(A));
alphas = NaN(Nmax + 1, 1);
thetas = cell(Nmax + 1, 1);
for n = 1:(Nmax + 1)
AkW{n} = Ak*W;
normAkW = infnorm(AkW{n});
normKAkW = infnorm(K*AkW{n});
alphas(n) = max(normAkW/normW, normKAkW/normKW);
alphaW{n} = alphas(n)*W;
if alphas(n) < 1
thetas{n} = tightenconstraints(Z, A, B, K, wmax*[1; 1], n);
else
thetas{n} = NaN(size(Z.psi));
end
psibar = Z.psi - thetas{n}/(1 - alphas(n) + eps());
normZbars(:,n) = psibar([1; 3; 5]);
Ak = (A + B*K)*Ak;
end
figure();
semilogy(0:Nmax, alphas, '-ok', [1, Nmax], [1, 1], '--r');
xlabel('N')
ylabel('Minimum \alpha');
figure();
hold('on');
Nplots = ceil(sqrt(length(Nkeep)));
for i = 1:length(Nkeep)
subplot(Nplots, Nplots, i);
n = Nkeep(i) + 1;
Vin = AkW{n};
Vout = alphaW{n};
plot(Vin(1,[1:end,1]), Vin(2,[1:end,1]), '-or', ...
Vout(1,[1:end,1]), Vout(2,[1:end,1]), '-sb');
title(sprintf('N = %d: \\alpha = %g', n, alphas(n)));
end
figure();
N = 0:Nmax;
plot(N, normZbars(1,:), '-or', N, normZbars(2,:), '-og', ...
N, normZbars(3,:), '-ob');
xlabel('N')
ylabel('Bound');
legend('x_1', 'x_2', 'u');
% Save data.
data = struct('N', 0:Nmax, 'alpha', alphas, 'normZbar', normZbars);
save('-v7', 'calcalpha.mat', '-struct', 'data');
### hyperrectangle.m
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19function [A, b] = hyperrectangle(lb, ub)
% [A, b] = hyperrectangle(lb, ub)
%
% Returns halfspace representation of hyperrectangle with bounds lb and ub.
% Any infinite or NaN bounds are ignored.
narginchk(2, 2);
if ~isvector(lb) || ~isvector(ub) || length(lb) ~= length(ub)
error('Inputs must be vectors of the same size!');
end
A = kron(eye(length(lb)), [1; -1]);
b = reshape([ub(:)'; -lb(:)'], 2*length(lb), 1);
goodrows = ~isinf(b) & ~isnan(b);
A = A(goodrows,:);
b = b(goodrows);
end%function
### tightenconstraints.m
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38function theta = tightenconstraints(Z, A, B, K, W, N)
% theta = tightenconstraints(Z, A, B, K, W, N)
%
% Tightens state and input constraints for a box W.
%
% Z should be a struct with fields G, H, and psi to define the feasible set Z as
%
% Gx + Hu <= psi
%
% A, B, and K should be the system model and controller gain. W should be a
% vector defining the maximum absolute values for w. The system evolves as
%
% x^+ = Ax + Bu + w, u = v + Kx, -W <= w <= W
%
% This means that W must be a symmetric box in R^n (we make this restriction
% because optimization becomes particularly easy).
%
% Returns theta such that the tighter constraints
%
% Gx + H(v + Kx) <= psi - theta
%
% are satisfied for any N realizations of W.
narginchk(6, 6);
if ~isstruct(Z) || ~all(isfield(Z, {'G', 'H'}))
error('Invalid input for Z!');
end
C = Z.G + Z.H*K;
A = A + B*K;
theta = zeros(size(size(C, 1)));
Ak = eye(size(A));
W = abs(W);
for i = 0:N
theta = theta + abs(C*Ak)*W;
Ak = A*Ak;
end
end%function | 1,572 | 3,905 | {"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} | 3.328125 | 3 | CC-MAIN-2023-40 | latest | en | 0.401935 |
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Visual StudioPast WinnersGeometry worksheets kuta software llc software graphing points.Consolidation | 1,927 | 10,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.908041 |
http://math.stackexchange.com/questions/149689/how-many-digits-are-there-in-2100 | 1,469,730,924,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828313.74/warc/CC-MAIN-20160723071028-00005-ip-10-185-27-174.ec2.internal.warc.gz | 158,123,702 | 19,703 | # How many digits are there in $2^{100}$?
This question was similarly answered at How many digits does $2^{1000}$ contain? . I saw in a book asking to do the same for $2^{100}$ without taking logaritms. There is a hit as following: " $2^{10}=1024$ and note that if $0<r<b<a$ then $\frac{a}{b}<\frac{a-r}{b-r}$ ". Honestly, I couldn't use this hint to solve the problem. Thanks.
-
Which book was that? – lhf May 26 '12 at 2:10
which book? why wouldn't you provide a reference to this :/ – baxx Jul 7 at 14:46
Write $2^{100}=(10^3+24)^{10}=10^{30}(1+0.024)^{10}$.
Consider $f(x)=x^{10}$. By the mean value theorem, $f(1+h)=f(1)+f'(\xi)h$, for some $\xi\in(1,1+h)$. Let $u=(1+h)^{10}$. Then $u=1+10\xi^9h<1+10(1+h)^9h=1+10\dfrac{u}{1+h}h$ and so $u<\dfrac{1+h}{1-9h}$.
When $h=0.024$, we get $u<\dfrac{1.024}{0.784}<10$, which implies that $10^{30}<2^{100}=10^{30}u<10^{31}$.
This means that $2^{100}$ has 31 digits. (By computing $\dfrac{1.024}{0.784}$, we get $2^{100}<1.31\cdot 10^{30}$.)
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Note that $2^{10} = 1024$ so $1000^{10} < 2^{100} = 1024^{10} < 1100^{10} = 1.1^{10}\cdot 1000^{10}$. We can estimate \begin{equation*} 1.1^{10} = 1.21^5 < 1.3^5 = 1.69^{2.5} < 2^3 = 8, \end{equation*} so $10^{30}<2^{100}<8\cdot 10^{30}$ and $2^{100}$ has 31 digits. On the other hand this doesn't use the given hint and the chain of equalities amounts to showing that $\log_{10}(1.1)<0.1$, so in some sense logarithms are involved.
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Consider $\frac{2^{100}}{10^{30}} = \left( \frac{1024}{1000} \right)^{10}$.
Use your hint with $a=1024$, $b=1000$ and $r=500$, then $$\frac{1024}{1000} < \frac{524}{500} = 1 + \frac{6}{125}$$
Now, use $1+a < \mathrm{e}^a$,
$$\frac{2^{100}}{10^{30}} < \left(1+\frac{6}{125}\right)^{10} < \exp\left(\frac{60}{125}\right) < \sqrt{\mathrm{e}} < 2$$
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It doesn't seem like the hint is buying you anything: if you skip it the $6$ changes to a $3$ and the rest of the argument still works. – Noah Stein May 25 '12 at 16:43
@NoahStein Quite true! – Sasha May 25 '12 at 16:46
right now it is been some times that I not on maths rails but I will suggest it with a simple log function: $$\log_{10}2^{100} = 100*log_{10}2 = 31$$
where 10 is our decimal digit base. Maybe I am wrong?
Sorry Just saw not without taking logaritms
-
OK. Your help was Welcome. :-) – Babak S. May 25 '12 at 20:54 | 917 | 2,321 | {"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} | 4.21875 | 4 | CC-MAIN-2016-30 | latest | en | 0.790169 |
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# Please rate my argument essay. Thanks.
Author Message
Manager
Joined: 10 Sep 2010
Posts: 133
Followers: 2
Kudos [?]: 30 [0], given: 7
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03 Nov 2010, 11:37
The following appeared in the editorial section of a local newspaper:
“This past winter, 200 students from Waymarsh State College traveled to the state capitol building to protest against proposed cuts in funding for various state college programs. The other 12,000 Waymarsh students evidently weren’t so concerned about their education: they either stayed on campus or left for winter break. Since the group who did not protest is far more numerous, it is more representative of the state’s college students than are the protesters. Therefore the state legislature need not heed the appeals of the protesting students.” Discuss how well reasoned . . . etc.
The author of the editorial section indicates that only 200 students from Waymarsh State College traveled to the state capitol building to protest against proposed cuts in funding, while majority of students stayed on campus or went on winter brake. The author concludes that state legislature need not heed the appeals of the protesting students.
The conclusion may well have merit. However, this poorly reasoned argument is based on questionable assumptions and premises. It lacks statistical data and is the result of generalization. I cannot accept the conclusion as valid.
First, it is not clear from the newspaper article how far is the state capitol building from Waymarsh State College. It could be that the trip requires substantial amount of money which majority of students do not have. In fact, difficult financial situation is the reason for the protest. Therefore, author's conclusion is wrong. Instead, it could be that all the students protest against proposed cuts in funding, while only 200 of them have sufficient funds to take a trip to capitol building. Newspaper article suggests that students protest happens during winter brake. Some students stayed on campus, while others left for winter brake. One can argue that students actually do have extra money, because they can leave for winter brake. However, the author of the article does not provide any information about relative numbers of students who stayed on campus and who left during winter brake.
Second, the article indicates that state legislature is planning to cut funding of several state college programs. It is not clear whether the funding of all college programs will be cut or only several selected programs. Thus it could be that 200 students who traveled to capitol building are enrolled in these selected college programs, which can potentially face the funding cuts. Thus, the author needs to provide specific information about the programs that can face funding cuts and about the number of students enrolled.
In sum, the conclusion may look appealing at first. The author looked over a number of points. If the author addresses the above points, he or she would have a better argument. As it stands, the logic has flaws and the argument is weak.
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Joined: 30 Nov 2010
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### Show Tags
13 Jan 2011, 13:47
Fijisurf wrote:
The following appeared in the editorial section of a local newspaper:
“This past winter, 200 students from Waymarsh State College traveled to the state capitol building to protest against proposed cuts in funding for various state college programs. The other 12,000 Waymarsh students evidently weren’t so concerned about their education: they either stayed on campus or left for winter break. Since the group who did not protest is far more numerous, it is more representative of the state’s college students than are the protesters. Therefore the state legislature need not heed the appeals of the protesting students.” Discuss how well reasoned . . . etc.
The author of the editorial section indicates that only 200 students from Waymarsh State College traveled to the state capitol building to protest against proposed cuts in funding, while majority of students stayed on campus or went on winter brake. The author concludes that state legislature need not heed the appeals of the protesting students.
The conclusion may well have merit. However, this poorly reasoned argument is based on questionable assumptions and premises. It lacks statistical data and is the result of generalization. I cannot accept the conclusion as valid.
First, it is not clear from the newspaper article how far is the state capitol building from Waymarsh State College. It could be that the trip requires substantial amount of money which majority of students do not have. In fact, difficult financial situation is the reason for the protest. Therefore, author's conclusion is wrong. Instead, it could be that all the students protest against proposed cuts in funding, while only 200 of them have sufficient funds to take a trip to capitol building. Newspaper article suggests that students protest happens during winter brake. Some students stayed on campus, while others left for winter brake. One can argue that students actually do have extra money, because they can leave for winter brake. However, the author of the article does not provide any information about relative numbers of students who stayed on campus and who left during winter brake.
Second, the article indicates that state legislature is planning to cut funding of several state college programs. It is not clear whether the funding of all college programs will be cut or only several selected programs. Thus it could be that 200 students who traveled to capitol building are enrolled in these selected college programs, which can potentially face the funding cuts. Thus, the author needs to provide specific information about the programs that can face funding cuts and about the number of students enrolled.
In sum, the conclusion may look appealing at first. The author looked over a number of points. If the author addresses the above points, he or she would have a better argument. As it stands, the logic has flaws and the argument is weak.
It's a good essay I'd give it a 4, but please keep in mind that I'm no GMAT awa guru or anything like that...
Nevertheless here are a few things that I would suggest you consider when you write your essays.
1) "Therefore, author's conclusion is wrong."I would refrain from saying that what would be better is if you said that the arguement is weak, or that the conclusion is invalid.
2) "Instead, it could be that all the students protest against proposed cuts in funding", what does it refer to exactly. Being specific can improve your scores incredibly.
3) "In sum, the conclusion may look appealing at first. The author looked over a number of points." You can merge these two sentences together using 'but' to create a well-structured essay.
Other than that you point out good points! IMO
HTH
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01 Sep 2012, 17:08
Good stuff, I would agree with the rating of around 4.
One really obvious issue is that you have misspelled "break" and spelt brake numerous times. This mistake is avoidable since the word break is spelt correctly in the original text. Also, "in sum" sounds awkward to me, maybe "In summary.." would be better.
Re: Please rate my argument essay. Thanks. [#permalink] 01 Sep 2012, 17:08
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Display posts from previous: Sort by | 1,853 | 8,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-36 | longest | en | 0.932669 |
http://tulyn.com/videotutorials/distributive_property.htm | 1,369,481,722,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705939136/warc/CC-MAIN-20130516120539-00079-ip-10-60-113-184.ec2.internal.warc.gz | 263,446,663 | 3,867 | # Distributive Property Videos
Many students find distributive property difficult. At TuLyn, we created video tutorials on distributive property to help you better understand distributive property. This page is for math students who need help, and for teachers and tutors who are looking for video tutorials on distributive property.
# Distributive Property Videos
Now finding distributive property help is easier. Below find the list of all video tutorials we have on distributive property. Each video clip helps you better understand the subject matter. Seeing different problems makes you familiar with the topic. Our distributive property video tutorials give brief but to-the-point review on distributive property. You no longer need to feel overwhelmed with distributive property homework and tests.
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# Simplifying Using Distributive Property With Negative Coefficients Video Clip
Simplifying Using Distributive Property With Negative Coefficients
Video Clip Length: 1 minute 59 seconds
Video Clip Views: 19563
This math video tutorial clip shows how to simplify algebraic expressions by following the rules of order of operations, distributive property and combining like terms.
Distributive property helps us get rid of the paranthesis first and then we combine the like terms to reach the solution.
addition, arithmetic operations, distributive property video, expressions, multiplication, operations, properties of addition, properties of multiplication, simplifying expressions
Learn English | 352 | 1,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.871121 |
thixotropic.blogspot.com | 1,490,849,244,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191986.44/warc/CC-MAIN-20170322212951-00357-ip-10-233-31-227.ec2.internal.warc.gz | 336,661,054 | 7,375 | ## 6.2.07
### doodle schooled
my mom always said, no matter how good you are at something, there will always be somebody else who will be better than you are.
and so, i present the "why" of math doodling, courtesy of the triumvirate of smarty pants in my personal orbit.
mvb:
Its a symbol of the multiplication you have to do to arrive at the final answer. There is really nothing to the graph itself. For instance write out the problem the traditional way:
42
X 12
--------
the steps you need to take are:
2X2 (symbolized by the upper right)
2X4 (upper left)
2X1 (lower right)
4X1 (lower left)
if you do the multiplication out longhand you'll see the way the are put together- when you do the 2X2 that is the right side of the answer, the 4X1 is the left side. But you have to add those middle 2 up (the 2X1 and 2X4) which is why the middle of the diagram is added. and the 1 is carried to the left side of the equation, just like in the long hand method.
If you reversed the order in which you draw the multiplication factors (i.e the 2 starting on the top, instead on the 2 on the bottom), your just reversing the order in which you are multiplying, getting the answer to 42 X 21.
i also think it works because when you multiply, you are just dealing with 1's which can be represented by simple tick marks. the process of multiplying you learn in grammer school is all about breaking a problem of multiplying 10s and 100s into the simple task of just multiplying with 1's, and then adding them up to get the 10s and 100s. This all goes back to binary theory!
Hey, I did this with 42 x 23 and got 966.
So did Matlab. I am matlab.
That's logic, not math.
*** but really ***
Do you know how an abacus works? this is a similar accounting system - it's a means of keeping track of the digits (in powers of ten) - then the intersections are the products. The problem is you have to be moving in the same direction on the 90 degree plane, so it's not very intuitive (i.e. the 10s columns in our numbers don't both go at the 'top', they go at the 'left' in their respective orientation).
OR, you can think of the numbers as sort of in a matrix, and you're doing inner and outer products. It's not as though the lines help with the accounting once the entry exceeds 9.
the flyshower:
this is basically a variation on lattice multiplication, which is basically the same general 2D format but with numbers. i'll show you sometime. yhat's pretty dope though thanks for sharing. sharing is caring, you know.
so if nothing else - i do care. really. | 636 | 2,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-13 | longest | en | 0.952394 |
https://www.dsprelated.com/freebooks/mdft/Periodic_Interpolation_Spectral_Zero.html | 1,558,406,383,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00199.warc.gz | 777,236,067 | 12,400 | ### Periodic Interpolation (Spectral Zero Padding)
The dual of the zero-padding theorem states formally that zero padding in the frequency domain corresponds to periodic interpolation in the time domain:
Definition: For all and any integer ,
(7.7)
where zero padding is defined in §7.2.7 and illustrated in Figure 7.7. In other words, zero-padding a DFT by the factor in the frequency domain (by inserting zeros at bin number corresponding to the folding frequency7.21) gives rise to periodic interpolation'' by the factor in the time domain. It is straightforward to show that the interpolation kernel used in periodic interpolation is an aliased sinc function, that is, a sinc function that has been time-aliased on a block of length . Such an aliased sinc function is of course periodic with period samples. See Appendix D for a discussion of ideal bandlimited interpolation, in which the interpolating sinc function is not aliased.
Periodic interpolation is ideal for signals that are periodic in samples, where is the DFT length. For non-periodic signals, which is almost always the case in practice, bandlimited interpolation should be used instead (Appendix D).
#### Relation to Stretch Theorem
It is instructive to interpret the periodic interpolation theorem in terms of the stretch theorem, . To do this, it is convenient to define a zero-centered rectangular window'' operator:
Definition: For any and any odd integer we define the length even rectangular windowing operation by
Thus, this zero-phase rectangular window,'' when applied to a spectrum , sets the spectrum to zero everywhere outside a zero-centered interval of samples. Note that is the ideal lowpass filtering operation in the frequency domain. The cut-off frequency'' is radians per sample. For even , we allow to be passed'' by the window, but in our usage (below), this sample should always be zero anyway. With this notation defined we can efficiently restate periodic interpolation in terms of the operator:
Theorem: When consists of one or more periods from a periodic signal ,
In other words, ideal periodic interpolation of one period of by the integer factor may be carried out by first stretching by the factor (inserting zeros between adjacent samples of ), taking the DFT, applying the ideal lowpass filter as an -point rectangular window in the frequency domain, and performing the inverse DFT.
Proof: First, recall that . That is, stretching a signal by the factor gives a new signal which has a spectrum consisting of copies of repeated around the unit circle. The baseband copy'' of in can be defined as the -sample sequence centered about frequency zero. Therefore, we can use an ideal filter'' to pass'' the baseband spectral copy and zero out all others, thereby converting to . I.e.,
The last step is provided by the zero-padding theorem7.4.12).
#### Bandlimited Interpolation of Time-Limited Signals
The previous result can be extended toward bandlimited interpolation of which includes all nonzero samples from an arbitrary time-limited signal (i.e., going beyond the interpolation of only periodic bandlimited signals given one or more periods ) by
1. replacing the rectangular window with a smoother spectral window , and
2. using extra zero-padding in the time domain to convert the cyclic convolution between and into an acyclic convolution between them (recall §7.2.4).
The smoother spectral window can be thought of as the frequency response of the FIR7.22 filter used as the bandlimited interpolation kernel in the time domain. The number of zeros needed in the zero-padding of in the time domain is simply length of minus 1, and the number of zeros to be appended to is the length of minus 1. With this much zero-padding, the cyclic convolution of and implemented using the DFT becomes equivalent to acyclic convolution, as desired for the time-limited signals and . Thus, if denotes the nonzero length of , then the nonzero length of is , and we require the DFT length to be , where is the filter length. In operator notation, we can express bandlimited sampling-rate up-conversion by the factor for time-limited signals by
(7.8)
The approximation symbol ' approaches equality as the spectral window approaches (the frequency response of the ideal lowpass filter passing only the original spectrum ), while at the same time allowing no time aliasing (convolution remains acyclic in the time domain).
Equation (7.8) can provide the basis for a high-quality sampling-rate conversion algorithm. Arbitrarily long signals can be accommodated by breaking them into segments of length , applying the above algorithm to each block, and summing the up-sampled blocks using overlap-add. That is, the lowpass filter rings'' into the next block and possibly beyond (or even into both adjacent time blocks when is not causal), and this ringing must be summed into all affected adjacent blocks. Finally, the filter can `window away'' more than the top copies of in , thereby preparing the time-domain signal for downsampling, say by :
where now the lowpass filter frequency response must be close to zero for all . While such a sampling-rate conversion algorithm can be made more efficient by using an FFT in place of the DFT (see Appendix A), it is not necessarily the most efficient algorithm possible. This is because (1) out of output samples from the IDFT need not be computed at all, and (2) has many zeros in it which do not need explicit handling. For an introduction to time-domain sampling-rate conversion (bandlimited interpolation) algorithms which take advantage of points (1) and (2) in this paragraph, see, e.g., Appendix D and [72].
Next Section:
Why a DFT is usually called an FFT in practice
Previous Section: | 1,171 | 5,743 | {"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} | 3.3125 | 3 | CC-MAIN-2019-22 | longest | en | 0.860095 |
https://www.toolmenow.com/7/Decimal-To-Hexadecimal/7299-decimal-in-hex | 1,723,308,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00824.warc.gz | 785,808,507 | 4,453 | # 7299 Decimal in Hex
Start convert the decimal number 7299 to hexadecimal.
Step 1: Divide by 16
Start by dividing 7299 by 16:
`7299 ÷ 16 = 456 (Quotient) with a remainder of 3`
Note: In hexadecimal 10 is represented as A, 11 as B, 12 as C, 13 as D, 14 as E, and 15 as F.
Step 2: Divide the Quotient
Now, divide the quotient (456) by 16:
`456 ÷ 16 = 28 (Quotient) with a remainder of 8`
Step 3: Divide the Quotient
Now, divide the quotient (28) by 16:
`28 ÷ 16 = 1 (Quotient) with a remainder of C`
Step 4: Final actions
The Quotient is less than 16 (1), so we will transfer it to the beginning of the number as a reminder.
Step 5: Write Remainders in Reverse Order
Now, write down the remainders obtained in reverse order:
`1C83`
So, the hexadecimal representation of the decimal number 7299 is 1C83.
Decimal To Hex Converter
Other examples of Decimal to Hexadecimal conversion | 276 | 896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-33 | latest | en | 0.89502 |
http://www.physicsforums.com/showthread.php?t=487465&page=3 | 1,386,845,960,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164580801/warc/CC-MAIN-20131204134300-00043-ip-10-33-133-15.ec2.internal.warc.gz | 486,688,243 | 10,439 | # Help me do this assignment. My grade depends on it!! *Cure a high schooler's anxiety*
by fphysicsclass
Tags: energy, kinetic, physics, potential, power
P: 37 I'd just like to let you know that the actual q's for six are the letters. Any of the numbers could match up with it.
PF Patron P: 7,122 For the work problem, remember, when you do work to something, you store some sort of potential energy into it. What description has an increase in potential energy?
PF Patron
P: 7,122
Quote by fphysicsclass I'd just like to let you know that the actual q's for six are the letters. Any of the numbers could match up with it.
Yes, I realize that. I should have said why should d) be 2.
P: 37 In c, the elastic PE is decreasing and the KE is increasing as the pole vaulter gains speed and passes over the pole. For d, I thought the KE increased because he begins to move downwards since the direction of the movement changes.
PF Patron P: 7,122 The key for c) is that the statement says he is slowing and the pole is unbending. In other words, his kinetic energy is decreasing and the pole's elastic potential energy is decreasing as well which looks more like 5. For d), yes, the kinetic energy will eventually increase, but when you just clear the bar, you're almost at a standstill and your gravitational potential energy is at it's highest which sounds more like 1.
P: 37 Oh! I was NOT thinking of 'instantaneous' descriptions of energy. Maybe I should remember that for the exam in the future. For 9, why is walking upstairs not positive work? For 8, where do I start?
PF Patron P: 7,122 For 9, it actually is! Bringing an object to a greater height gives it more potential energy. Giving an object energy is the result of doing a positive amount of work to it. The object now has the ability to perform work on something else because it has an energy associated with it.
P: 37 Thanks for that description! Yay! I got something right. Now help me with 8! jkjk. But seriously, I need help
P: 2,568 Jesus Christ, you can't even do number 4? Here, think about what the WORK is being done.
P: 37 No flyingpig. I can do number 4. If you read the comments I'm stuck on number 8. Just help me on this last one.
P: 37 Here it is if you don't want to go back to page 1. 8. A marble rolls down a ramp that sits on table. The marble leaves the base of the ramp with a horizontal velocity. The base of the ramp is 2 meters above the floor. The marble strikes the floor at a distance of 1.8 meters from where it left the ramp. Give the height of the ramp with respect to the floor. Choose one answer. 2.4 m 0.4 m 4.4 m 1.4 m 2.2 m
PF Patron P: 7,122 8 is actually a bit involved compared to the other problems. You really are looking for how high the ramp is here. So you again use your energy conservation laws and the ol $$v^2 = 2gh$$ guy. You want to find the height of the ramp. However you don't immediately know the velocity but you do have two things that can help you. Whatever this velocity is, it allowed the marble to travel 1.8 meters in the horizontal direction while simultaneously falling 2 meters. Use your kinematic equations to figure how what velocity would allow this to happen.
P: 37 Oh god. I got a 95 on the kinematics test, studied forever hoping that this stuff wouldn't come back to haunt me.... and now it has. We just started the work/energy/optics unit. I want to understand this. So from deriving several equations I got v = deltax/sqrt(deltay/.5g) Assuming that the initial horiz. speed was 5.9 m/s, I got deltay = .45m
P: 37 My answer is closest to .4 m. Do you think that's right?
P: 2,568 Huh, interesting. Try this x = vt (x is horizontal distance) y = -gt^2/2 + 2 (vertical distance, check that t = 0, y = 2) There is no velocity before because it is freefalling. Your answer is wrong, because the base of the ramp is 2m high, why would it be 0.4m...? But I am not really reading your posts so I am not much of help, but I have set up some eqtns for you
P: 37 I mean 2.4 m because you must add 2 to the .4m.
P: 37 See, my derived equation allows me to neglect time! :D I found it by myself... so proud. However, I think I neglected to add 2 m to .4 m because it asks for with respect to the floor. Look: 8. A marble rolls down a ramp that sits on table. The marble leaves the base of the ramp with a horizontal velocity. The base of the ramp is 2 meters above the floor. The marble strikes the floor at a distance of 1.8 meters from where it left the ramp. Give the height of the ramp with respect to the floor. Choose one answer. 2.4 m 0.4 m 4.4 m 1.4 m 2.2 m
PF Patron P: 7,122 Yup, so remember they want the height from the floor so the answer should be 2.4m
Related Discussions General Discussion 8 Academic Guidance 1 Nuclear Engineering 31 Academic Guidance 11 Nuclear Engineering 5 | 1,265 | 4,833 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2013-48 | longest | en | 0.966848 |
https://puzzling.stackexchange.com/questions/126307/whats-changing | 1,718,804,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00099.warc.gz | 421,984,037 | 39,213 | What's changing?
A friend of mine has the following puzzle(original post link: https://www.geocaching.com/geocache/GC46JBF_whats-changing) which we can't figure out how to solve:
ADEFBCDGBCDGABC
DGABCDADEFGDEFA
BCBCAEFGCDEADEF
ABDFGACFGACFGAD
EFACDEFCFGADEAC
DEGBFADEFGBCG A
DEGACDEFBCEFABC
DFGCADEFGACDEG.
Hint 1)
The first hint is the title "What's changing"
Hint 2)
The second hint is "7-segment".
Hint 3)
Using the the 7-segment system, the first character apparently is "C"(using ADEF). We can't however get the following characters due to not correctly applying the title hint.
Hint 4)
This is a geocaching puzzle of which the result should be something like this: N51° 11.542 E 004° 23.867
Can anyone help us solve this puzzle?
• It's actually pretty simple to solve. Commented Apr 16 at 13:35
• Just feed it through a 7 digit cipher decoder was what I was thinking
– PDT
Commented Apr 16 at 13:36
• @pdt Have you looked at the hints? I don't think this will be found in an online decoding tool. Commented Apr 16 at 13:53
• dcode.fr/7-segment-display? Is this not correct
– PDT
Commented Apr 16 at 14:06
• @PDT It doesn't quite do this code (see hint 1). It should decode to "CACHE LOC=" etc. Commented Apr 16 at 14:07
1 Answer
First split the encoded text into blocks of letters that are in alphabetical order.
ADEF BCDG BCDG ABCDG ABCD ADEFG DEF ABC BC AEFG CDE ADEF ABDFG ACFG ACFG ADEF ACDEF CFG ADE ACDEG BF ADEFG BCG ADEG ACDEF BCEF ABCDFG C ADEFG ACDEG.
The letters represent the
segments in a 7-segment display. They are labelled in this order:
-A-
F| |B
|-G-|
E| |C
-D-
Next apply the hint from the title to figure out how to decode this.
The blocks of letters represent those segments that are changed when the display goes from one character to the next. The display starts off showing nothing, so when segments ADEF are changed you see the letter C. If the next block of segments, BCDG is changed, that means BCG are switched on and D is switched off, then the result is ABCEFG, which represents the letter A.
This results in the following:
First column is the code, second column the current state of the display, and the last column the symbol shown on the display.
ADEF ADEF C
BCDG ABCEFG A
BCDG ADEF C
ABCDG BCEFG H
ABCD ADEFG E
ADEFG -
DEF DEF L
ABC ABCDEF O
BC ADEF C
AEFG DG =
CDE CEG n
ADEF ACDFG 5
ABDFG BC 1
ACFG ABFG degrees
ACFG BC 1
ADEF ABCDEF 0
ACDEF B '
CFG BCFG 4
ADE ABCDEFG 8
ACDEG BF "
BF -
ADEFG ADEFG E
BCG ABCDEF 0
ADEG BCFG 4
ACDEF ABDEG 2
BCEF ACDFG 5
ABCDFG B '
C BC 1
ADEFG ABCDEFG 8
ACDEG BF "
So the decrypted text is:
CACHE LOC=n51o10'48" E0425'18"
Note that the degree symbol is missing after E04, but that can be remedied by splitting the subsequent ACDEF block into AC DEF, to give the final answer:
CACHE LOC=n51o10'48" E04o25'18" | 941 | 2,963 | {"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} | 3.359375 | 3 | CC-MAIN-2024-26 | latest | en | 0.909736 |
https://pt.scribd.com/doc/228296201/Avaliacao-Geometria-Ano-9 | 1,524,207,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937161.15/warc/CC-MAIN-20180420061851-20180420081851-00169.warc.gz | 691,933,558 | 24,938 | # Avaliação – 2º Bimestre – Geometria – 9º ano – Professora F.
Meireles
1) Calcule as incógnitas:
A
CÁLCULOS
C
_________________________________________
_________________________________________
_________________________________________
_________________________________________
C
_________________________________________
_________________________________________
_________________________________________
_________________________________________
C
_________________________________________
_________________________________________
_________________________________________
_________________________________________
C
_________________________________________
_________________________________________
_________________________________________
_________________________________________
C
_________________________________________
_________________________________________
_________________________________________
_________________________________________
h
B
16
25
A
12
y
B
24
A
h
B
16
25
A
6
B
3
x
A
x
B
12
4
2) Determine o valor de x nos triângulos retângulos:
CÁLCULOS
20
_________________________________________
_________________________________________
_________________________________________
_________________________________________
4x
3x
x
x
_________________________________________
_________________________________________
_________________________________________
_________________________________________
6
20
x
_________________________________________
_________________________________________
_________________________________________
_________________________________________ | 241 | 1,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-17 | latest | en | 0.257345 |
http://www.qimacros.com/hypothesis-test/f-test-two-sample/ | 1,410,902,862,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657119965.46/warc/CC-MAIN-20140914011159-00237-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 751,035,292 | 7,688 | # F-Tests Determine if Variances are the Same or Different
## You Don't Have to be a Statistician to do an F-Test
### F-Test Example
If you're producing rubber made with two different recipes, you might want to know if the variances in tensile strengths are the same or different (Juran's QC Handbook 4th pg 23.74):
### The QI Macros for Excel Makes F-Tests as Easy as 1-2-3
1. Enter your data in Excel, then click and drag over the data to select it. Next click on the QI Macros Menu, Statistical Tools and then select the F-Test:
2. The QI Macros will prompt for a significance level (default = 0.05):
Show less
The QI Macros will perform the F-Test calculations and interpret the results for you.
### What's Cool about the QI Macros F-Test?
Unlike other statistical software, the QI Macros is the only SPC software that interprets the results for you. It compares the p-value (0.152) to the significance level (0.05) tells you: "Cannot Reject the Null Hypothesis because p>0.05" and that the "Variances are the same." Based on the F-test results we now know which t-Test to run.
### Interpreting F-Test results
• The null hypotheis (H0) - variances are the same
• The alternate hypothesis (Ha) - variances are different
Hypothesis Test Compare Result Classical Method test statistic > critical value (i.e. F > F crit) Reject the null hypothesis Classical Method test statistic < critical value (i.e. F < F crit) Cannot Reject the null hypothesis p value Method p value < a Reject the null hypothesis p value Method p value > a Cannot Reject the null hypothesis
Since F < Fcrit (.33 < 6.39) and p value > a ( .152 > 0.05), we cannot reject the null hypothesi - the Variances are the Same.
Excel Note:
If you run the test using, Excel's Data Analysis toolpak, Excel requires that the recipe with the largest variance be first to ensure correct calculations. The QI Macros does not have this requirement to perform the calculations correctly.
Show less | 497 | 1,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2014-41 | latest | en | 0.789033 |
https://rdrr.io/cran/metafor/man/influence.rma.uni.html | 1,603,348,493,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00400.warc.gz | 495,862,679 | 55,726 | influence.rma.uni: Outlier and Influential Case Diagnostics for 'rma.uni'... In metafor: Meta-Analysis Package for R
Description
The functions compute various outlier and influential case diagnostics (some of which indicate the influence of deleting one case/study at a time on the model fit and the fitted/residual values) for objects of class `"rma.uni"`.
Usage
``` 1 2 3 4 5 6 7 8 9 10 11 12``` ```## S3 method for class 'rma.uni' influence(model, digits, progbar=FALSE, ...) ## S3 method for class 'infl.rma.uni' print(x, digits=x\$digits, infonly=FALSE, ...) ## S3 method for class 'rma.uni' cooks.distance(model, progbar=FALSE, ...) ## S3 method for class 'rma.uni' dfbetas(model, progbar=FALSE, ...) ## S3 method for class 'rma.uni' hatvalues(model, type="diagonal", ...) ```
Arguments
`model` an object of class `"rma.uni"`. `x` an object of class `"infl.rma.uni"` (for `print`). `digits` integer specifying the number of decimal places to which the printed results should be rounded (if unspecified, the default is to take the value from the object). `progbar` logical indicating whether a progress bar should be shown (the default is `FALSE`). `infonly` logical indicating whether only the influential cases should be printed (the default is `FALSE`). `type` character string indicating whether to return only the diagonal of the hat matrix (`"diagonal"`) or the entire hat matrix (`"matrix"`). `...` other arguments.
Details
The `influence` function calculates the following leave-one-out diagnostics for each study:
• externally standardized residual,
• DFFITS value,
• Cook's distance,
• covariance ratio,
• the leave-one-out amount of (residual) heterogeneity,
• the leave-one-out test statistic for the test of (residual) heterogeneity,
• DFBETAS value(s).
The diagonal elements of the hat matrix and the weights (in %) given to the observed effects or outcomes during the model fitting are also provided (except for their scaling, the hat values and weights are the same for models without moderators, but will differ when moderators are included).
For details on externally standardized residuals, see `rstudent.rma.uni`.
The DFFITS value essentially indicates how many standard deviations the predicted (average) effect for the ith study changes after excluding the ith study from the model fitting.
Cook's distance can be interpreted as the Mahalanobis distance between the entire set of predicted values once with the ith study included and once with the ith study excluded from the model fitting.
The covariance ratio is defined as the determinant of the variance-covariance matrix of the parameter estimates based on the dataset with the ith study removed divided by the determinant of the variance-covariance matrix of the parameter estimates based on the complete dataset. A value below 1 therefore indicates that removal of the ith study yields more precise estimates of the model coefficients.
The leave-one-out amount of (residual) heterogeneity is the estimated value of τ² based on the dataset with the ith study removed. Note that this is always equal to `0` for fixed-effects models.
Similarly, the leave-one-out test statistic for the test of (residual) heterogeneity is the value of the test statistic of the test for (residual) heterogeneity calculated based on the dataset with the ith study removed.
Finally, the DFBETAS value(s) essentially indicate(s) how many standard deviations the estimated coefficient(s) change(s) after excluding the ith study from the model fitting.
A study may be considered to be ‘influential’ if at least one of the following is true:
• The absolute DFFITS value is larger than 3√(p/(k-p)), where p is the number of model coefficients and k the number of studies.
• The lower tail area of a chi-square distribution with p degrees of freedom cut off by the Cook's distance is larger than 50%.
• The hat value is larger than 3(p/k).
• Any DFBETAS value is larger than 1.
Studies which are considered influential with respect to any of these measures are marked with an asterisk. Note that the chosen cut-offs are (somewhat) arbitrary. Substantively informed judgment should always be used when examining the influence of each study on the results.
Value
An object of class `"infl.rma.uni"`, which is a list containing the following components:
`inf` an element of class `"list.rma"` with the externally standardized residuals, DFFITS values, Cook's distances, covariance ratios, leave-one-out τ² estimates, leave-one-out (residual) heterogeneity test statistics, hat values, weights, and an indicator whether a study is influential or not. `dfbs` an element of class `"list.rma"` with the the DFBETAS values. `...` some additional elements/values.
The results are printed with `print.infl.rma.uni` and plotted with `plot.infl.rma.uni`.
Note
Right now, leave-one-out diagnostics are calculated by refitting the model k times. Depending on how large k is, it may take a few moments to finish the calculations. There are shortcuts for calculating at least some of these values without refitting the model each time, but these are currently not implemented (and may not exist for all of the leave-one-out diagnostics calculated by the function).
It may not be possible to fit the model after deletion of the ith study from the dataset. This will result in `NA` values for that study.
Certain relationships between the leave-one-out diagnostics and the (internally or externally) standardized residuals (Belsley, Kuh, & Welsch, 1980; Cook & Weisberg, 1982) no longer hold for the meta-analytic models. Maybe there are other relationships. These remain to be determined.
Author(s)
Wolfgang Viechtbauer wvb@metafor-project.org http://www.metafor-project.org/
References
Belsley, D. A., Kuh, E., & Welsch, R. E. (1980). Regression diagnostics. New York: Wiley.
Cook, R. D., & Weisberg, S. (1982). Residuals and influence in regression. London: Chapman and Hall.
Hedges, L. V., & Olkin, I. (1985). Statistical methods for meta-analysis. San Diego, CA: Academic Press.
Viechtbauer, W. (2010). Conducting meta-analyses in R with the metafor package. Journal of Statistical Software, 36(3), 1–48. https://www.jstatsoft.org/v036/i03.
Viechtbauer, W., & Cheung, M. W.-L. (2010). Outlier and influence diagnostics for meta-analysis. Research Synthesis Methods, 1, 112–125.
`plot.infl.rma.uni`, `rstudent.rma.uni`, `weights.rma.uni`
Examples
``` 1 2 3 4 5 6 7 8 9 10``` ```### meta-analysis of the log risk ratios using a mixed-effects model ### with two moderators (absolute latitude and publication year) res <- rma(measure="RR", ai=tpos, bi=tneg, ci=cpos, di=cneg, mods = ~ ablat + year, data=dat.bcg) influence(res) plot(influence(res)) cooks.distance(res) dfbetas(res) hatvalues(res) ```
Example output
```Loading required package: Matrix
and introduction to the package please type: help(metafor).
\$inf
rstudent dffits cook.d cov.r tau2.del QE.del hat weight inf
1 0.2978 0.1785 0.0348 1.8003 0.1317 28.3142 0.1725 3.3664
2 -0.4303 -0.2368 0.0620 1.9207 0.1308 27.5744 0.2367 4.8106
3 -0.5100 -0.1094 0.0125 1.2348 0.1191 27.7572 0.0487 2.7920
4 -1.4032 -2.9415 7.3179 3.5225 0.0906 23.1836 0.8082 11.2312 *
5 -0.1490 -0.0263 0.0032 2.6341 0.1497 27.2543 0.2483 9.0681
6 1.0551 0.8926 0.7205 1.3621 0.0994 21.2875 0.4061 12.4817
7 -2.5961 -0.6815 0.4173 0.2379 0.0544 19.1240 0.0766 4.4008
8 0.4793 0.3703 0.1899 2.9984 0.1498 24.1266 0.3627 12.8020
9 0.2027 0.1305 0.0237 2.2071 0.1501 28.2874 0.1030 8.7848
10 -0.9872 -0.3870 0.1470 1.0702 0.1072 24.7567 0.1310 7.9919
11 -0.1197 -0.0030 0.0052 2.8336 0.1583 25.5103 0.2214 11.9238
12 1.4677 0.2171 0.0469 0.9274 0.1059 26.1197 0.0235 2.2836
13 2.1302 0.8150 0.4994 0.2178 0.0498 21.4920 0.1612 8.0630
\$dfbs
intrcpt ablat year
1 0.1492 -0.0622 -0.1491
2 -0.0949 -0.1001 0.0963
3 -0.0280 -0.0403 0.0283
4 2.2248 -2.5380 -2.2161
5 0.0350 0.0063 -0.0354
6 0.5935 -0.0008 -0.5954
7 -0.2659 0.5368 0.2591
8 -0.0394 -0.2110 0.0431
9 0.0728 -0.0865 -0.0718
10 -0.1188 -0.0973 0.1194
11 0.0631 -0.0343 -0.0631
12 -0.0413 0.0465 0.0419
13 -0.8276 0.6269 0.8279
1 2 3 4 5 6
0.034763870 0.061956371 0.012503192 7.317914072 0.003236974 0.720524896
7 8 9 10 11 12
0.417263935 0.189852980 0.023722929 0.146959602 0.005232795 0.046915399
13
0.499370066
intrcpt ablat year
1 0.14917967 -0.0622252054 -0.14912286
2 -0.09490912 -0.1001134132 0.09631548
3 -0.02800023 -0.0402975197 0.02831386
4 2.22478959 -2.5380258050 -2.21609317
5 0.03503418 0.0063184438 -0.03539793
6 0.59345585 -0.0008194425 -0.59535883
7 -0.26591994 0.5367554905 0.25909857
8 -0.03940376 -0.2109707986 0.04309572
9 0.07275483 -0.0864500483 -0.07183699
10 -0.11880193 -0.0972914132 0.11940279
11 0.06308093 -0.0342764351 -0.06312338
12 -0.04127415 0.0464523571 0.04192960
13 -0.82760563 0.6269221266 0.82787287
1 2 3 4 5 6 7
0.17250070 0.23668273 0.04870101 0.80821407 0.24826116 0.40606407 0.07657599
8 9 10 11 12 13
0.36274386 0.10304034 0.13102564 0.22140605 0.02353859 0.16124580
```
metafor documentation built on April 14, 2020, 7:40 p.m. | 3,067 | 9,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-45 | latest | en | 0.686546 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/payton1.html | 1,675,398,752,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00408.warc.gz | 35,270,322 | 2,882 | SEARCH HOME
Math Central Quandaries & Queries
Question from Payton, a student: how many ten thousands makes one million?
Hi Payton,
ten thousand is
10,000
and one million is
1,000,000
so to get from 10,000 to 1,000,000 you need to append two zeros. In other words you need to multiply 10,000 by 100.
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | 109 | 426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.917959 |
https://www.physicsforums.com/threads/i-dont-get-scalar-product.338212/ | 1,579,337,185,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592394.9/warc/CC-MAIN-20200118081234-20200118105234-00307.warc.gz | 1,038,884,714 | 15,365 | # I don't get scalar product?
## Homework Statement
Find the scalar product of the 2 vectors.
Vector A is north of east at 70 degrees with a magnitude of 3.60m
Vector B is south of west at 30 degrees with a magnitude of 2.40m
ABcosx
## The Attempt at a Solution
I did dot product using the formula, 3.60x2.40xcos(-100) and got the wrong answer.. The answer is 13.00. What the hell did I do wrong? I did as the formula said I should do and failed. I don't get it??
Last edited:
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
tiny-tim
Science Advisor
Homework Helper
Vector A is north of east at 70 degrees …
Vector B is south of west at 30 degrees …
I did dot product using the formula, 3.60x2.40xcos(-100) …
Hi elpermic!
It ain't 100.
Doc Al
Mentor
I did dot product using the formula, 3.60x2.40xcos(-100) and got the wrong answer.. The answer is 13.00.
(1) The formula calls for the angle between the two vectors. That angle is not 100°. (Draw a diagram.)
(2) The answer cannot be 13. | 299 | 1,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-05 | longest | en | 0.91768 |
https://www.asknumbers.com/FeetToPointsConversion.aspx | 1,719,126,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00334.warc.gz | 576,043,350 | 15,655 | # Feet to Points Converter
Feet to points converter. 1 Foot is equal to 864 points.
conversion table ←→
ft 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 500 1000 pt 864 1728 2592 3456 4320 5184 6048 6912 7776 8640 9504 10368 11232 12096 12960 13824 14688 15552 16416 17280 18144 19008 19872 20736 21600 22464 23328 24192 25056 25920 26784 27648 28512 29376 30240 31104 31968 32832 33696 34560 35424 36288 37152 38016 38880 39744 40608 41472 42336 43200 44064 44928 45792 46656 47520 48384 49248 50112 50976 51840 52704 53568 54432 55296 56160 57024 57888 58752 59616 60480 61344 62208 63072 63936 64800 65664 66528 67392 68256 69120 69984 70848 71712 72576 73440 74304 75168 76032 76896 77760 78624 79488 80352 81216 82080 82944 83808 84672 85536 86400 432000 864000
Round:
Feet:
Points:
The feet to points converter is a simple tool to convert from feet to points and vice versa. It works by taking the input value in feet and applying the conversion factor of 864 points per foot. The converter calculates the equivalent value in points and displays the result.
Below, you will find information on how many points are in a foot and how to accurately convert feet to points and vice versa.
## How many points in a foot and how to convert feet to points?
There are 864 points in a foot. To convert feet to points, multiply the feet value by 864.
For example, to convert 2 feet to points, you can use the following formula:
point = feet * 864
multiply 2 by 864:
point = 2 * 864 = 1728
Therefore, 2 feet equal to 1728 points.
Using the simple formula below, you can easily convert feet to points.
feet to points conversion formula:
point = feet * 864
## How to convert points to feet?
1 Point (pt) is equal to 0.00115741 foot (ft). To convert points to feet, multiply the point value by 0.00115741 or divide by 864.
For example, to convert 1000 points to feet, you can use the following formula:
feet = point / 864
divide 1000 by 864:
feet = 1000 / 864 = 1.1574
Therefore, 1000 points equal to 1.1574 feet.
Using the simple formula below, you can easily convert points to feet.
points to feet conversion formulas:
feet = point * 0.00115741
feet = point / 864
What is a Foot?
Foot is an imperial and United States Customary systems length unit. 1 foot = 864 points. The symbol is "ft".
What is a Point?
Point is a length unit and mostly used in typography, computer font sizes and printers. 1 point = 0.00115741 feet. The symbol is "pt".
Please visit all length units conversion to convert all length units.
Below, you have the option to create your own customized feet to points conversion table to meet your specific needs. This feature allows you to set the starting value, choose the increments between each entry, and select the desired level of accuracy. By tailoring the feet to points table according to your preferences, you can generate precise and personalized conversion results.
Create Conversion Table
Click "Create Table". Enter a "Start" value (5, 100 etc). Select an "Increment" value (0.01, 5 etc) and select "Accuracy" to round the result. | 1,104 | 3,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.49499 |
http://www.sudokudaddy.com/sudoku-tips/2011/tips-for-solving-sudoku-puzzles-page2.php?controller=pjLoad&action=pjActionLogin | 1,454,782,850,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701147492.21/warc/CC-MAIN-20160205193907-00269-ip-10-236-182-209.ec2.internal.warc.gz | 665,453,791 | 7,196 | # Sudoku Terminology
## To understand the different techniques for solving sudoku puzzles you must first understand the terminology
### Sudoku Terminology
##### Backdoor
A backdoor more or less refers to a scenario in which the puzzle can be solved through the use of a single candidate, and all puzzles have a few of these backdoors. Using a backdoor is considered a trial and error solving technique.
##### Big Number
A big number is a number that is the solved value for a cell. Cells without a big number have candidates, or small numbers.
##### Bilocation (unit)
A unit (two cells) with the same two candidates remaining. These units allow for the implementation of various solving techniques.
##### Bivalue (cell)
A single cell with two possible candidates remaining. Like the above units, this cell becomes useful for trying out different solving techniques.
##### Box
A standard Sudoku board is comprised of 9 boxes, each with 9 cells arranged in a 3x3 formation. Each box must contain one of each of the digits from 1-9.
##### Candidate
A candidate is any number that could still be a possible solution for a cell. Eliminating candidates is one of the prime objectives of many Sudoku solving techniques and strategies.
##### Cell
A single square on the Sudoku board. A standard board is comprised of 81 of these cells. Each cell is part of one row, one column, and one box.
##### Column
There are 9 vertical columns on each Sudoku board, with 9 cells in each. Along with rows, these columns are often referred to as lines. Each column must contain one of each of the digits from 1-9.
##### Coordinates
The standard way to refer to a specific cell is to use coordinates in the format R#C#, which means the row number followed by the column number.
So if I was referring to the cell in the fifth row from the top, and the 4th column from the left, it would be cell R5C4. This guide will use similar terminology when explaining strategies, and Sudoku software programs also use this numbering system.
##### Given
The cells which have values placed them to get the player started are called givens. These values cannot be altered, and set up the difficult of the entire puzzle. Most Sudoku puzzles have a minimum of 17 givens, as this is the minimum amount necessary to provide the puzzle with a unique solution.
##### Guess
Unlike trial and error, guessing is not considered logical in any way, and is frowned upon by most Sudoku players. A guess is often used by players when they’ve run out of solving techniques to utilize, including trial and error.
The player would then fill in one or more boxes with guesses and continue to solve the puzzle based on those guesses, in an attempt to see if a solution can be reached. As Sudoku is considered a game of logic, and each board can be solved through logic alone, guessing is often seen as cheating and/or giving up and admitting defeat at the hands of the board.
##### House
Each group of 9 cells is considered a house, whether it be a row, column, or box. The standard Sudoku board contains 27 houses, and each must contain one of the numbers from 1-9.
##### Improper Sudoku
A sudoku board which has multiple possible solutions.
##### Line
Rows and columns are often called lines. There are 18 of these lines, 9 rows and 9 columns, on a standard Sudoku board.
##### Peer
A peer is any cell located within the same house as the cell in question. Each cell has 20 different peers, 8 within the box, and 6 along that cell’s row and column, outside the box. None of a cell’s peers may contain the same number as it.
##### Proper Sudoku
This is a board which only has one possible correct solution.
##### Row
There are 9 horizontal rows on each Sudoku board, with 9 cells in each. Along with columns, these rows are often referred to as lines. Each row must contain one of each of the digits from 1-9.
##### Small Number
Small numbers are also known as candidates or pencilmarks. They are numbers which are possible big numbers for that cell, though it isn’t certain which of the small numbers is the actual big number yet.
##### Strong Inference
A strong inference is a deduction that can be drawn from two candidates which are linked. For example, both candidates in such a scenario could not both be false without causing a conflict, allowing the player to surmise that one is true and the other false.
Weak inference can be used in a similar way, though in its case, it implies that both scenarios cannot be true, so if one is true, the other is automatically false.
Link between 2 different candidates in a cell or unit with only 2 remaining candidates. These are great for solving techniques, as they assure that one of them must be true, and the other false. Weak links on the other hand, in which more than 2 candidates remain, are still useful for techniques, but take more work to uncover.
##### Trial and Error
A viable solving method, though more serious players may not use it, and consider just a step above guessing. Still, trial and error is a logical method to reach conclusions about candidates or values, so should not be discarded. Many other solving techniques use some form of trial and error in their own methodology.
Sudoku Beginner and Intermediate Strategies
###### Sudoku Playing Guide
The allure of Sudoku lies in the broad difficulty the game can present, from simple to annoying, which allows users of all skill levels to play at an suitable level. Sudoku in addition offers a wide variety of approaches that can be used to solve puzzles, from simple to complex.
And that's where the real charm of the game can be found, as every board can be solved through nothing but logical deductions. There's no guesswork or trial and error needed, although these are frequently employed tactics as well.
Sudoku Playing Guide
### New Printable Sudoku Puzzles
We have uploaded some new printable sudoku puzzles for you to enjoy. You can find the new puzzles by clicking the following link Printable sudoku puzzles
### Free Online Tetris Game
Play the classic puzzle rotatation game Tetris. See if you can beat the highest score! Play Free Tetris
### Free Maths Puzzle Game
Random numerical questions, try and accurately answer the sums as quickly as possible. Free Maths Game | 1,325 | 6,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2016-07 | latest | en | 0.912464 |
https://userforum.dhsprogram.com/index.php?t=msg&th=1579&start=0& | 1,660,909,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573667.83/warc/CC-MAIN-20220819100644-20220819130644-00792.warc.gz | 532,736,741 | 7,075 | The DHS Program User Forum
Discussions regarding The DHS Program data and results
Home » Topics » Fertility » Calculating WTFR
Calculating WTFR Thu, 02 October 2014 13:53
DHS user Messages: 104Registered: February 2013 Senior Member
I am wondering how you classify a wanted birth (specifically what variables you use). I am trying to calculate state level WTFR for Nigeria 2008, and I cannot get my national number to match the one on stat compiler.
Re: Calculating WTFR [message #3001 is a reply to message #3000] Thu, 02 October 2014 13:58
Bridgette-DHS Messages: 2554Registered: February 2013 Senior Member
Following is a response from Senior Data Processing Specialist, Noureddine Abderrahim:
The computation of the WTFR is a little bit more complicated than the TFR, since it excludes children whose birth order is above the ideal number declared by the respondent. The exposure used in the calculation of the WTFR, however, remains the same.
Attached is the logic used to compute the WTFR using the statistical package CSPro distributed by the Bureau of the Census.
• Attachment: wtfr.txt
Re: Calculating WTFR [message #4322 is a reply to message #3001] Fri, 08 May 2015 04:04
schoumaker Messages: 58Registered: May 2013 Location: Belgium Senior Member
Hello,
Thank you for sharing the CSPro program. I tried to adapt it in Stata and compute WTFR in some countries. I got the same results as the published results in Niger 2012 for instance, but not in Niger 2006.
if V613 in 0:90 & chi < V613 then
colt606 = 1;
xtab( t606b, rweight )
endif;
colt606 = 2;
It seems children are declared as wanted if the number of surviving children at the time of conception is lower than the desired number of children (chi < V613) AND if v613 is between 0 and 90. The first condition is fine, but I am mot sure what this second condition means. This looks like children are considered unwanted in case V613>90, including if V613 is a non-numeric responses (v613=96).
Is my interpretation correct ? I also read in the Guide to DHS Statistics that "For ideal number of children, non-numeric and "don't know" responses are considered to be high numbers, so that all births are considered wanted. Births to women with missing information on ideal
number of births are considered as unwanted. " (GUIDE TO DHS STATISTICS, 2006, p.88); this seems to be different from the CSPro code.
Thank you in advance for looking at this.
Bruno
Bruno Schoumaker
Centre for Demographic Research
Université catholique de Louvain
Re: Calculating WTFR [message #4373 is a reply to message #4322] Fri, 15 May 2015 15:37
Liz-DHS Messages: 1516Registered: February 2013 Senior Member
Dear Bruno,
Noureddine is temporarily unavailable, I am referring your query to another one of our experts. Thank you!
Re: Calculating WTFR [message #4384 is a reply to message #4322] Mon, 18 May 2015 11:30
Trevor-DHS Messages: 774Registered: January 2013 Senior Member
There is a mistake in the code below that was recently corrected. Instead of using 0:90, it should use 0:98.
This was a problem for most surveys conducted under DHSVI, and errata have been released for most surveys. Please see:
http://www.dhsprogram.com/publications/publication-FR277-DHS -Final-Reports.cfm
for the errata for the Niger 2012 survey.
If you make the correction to your program to include up to 98, then you should match Niger 2006 and the corrected Niger 2012 table in the errata.
Re: Calculating WTFR [message #4390 is a reply to message #4384] Mon, 18 May 2015 15:10
schoumaker Messages: 58Registered: May 2013 Location: Belgium Senior Member | 927 | 3,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.911178 |
http://www.mathisfunforum.com/viewtopic.php?pid=391395 | 1,540,046,012,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512836.0/warc/CC-MAIN-20181020142647-20181020164147-00177.warc.gz | 502,524,307 | 4,656 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2016-12-10 00:34:11
Hannibal lecter
Member
Registered: 2016-02-11
Posts: 225
Website
### about something called "discrete mathematics"
Hi,
I'm asking about discrete mathematics, we take this course in our studies but I didn't saw it in math is fun.com
it is about if p then q and if and only if.... and it is about True or False,...etc
can anyone tell me where can I find an introduction about discrete mathematics in math is fun?
Wisdom is a tree which grows in the heart and fruits on the tongue
Offline
## #2 2016-12-10 02:49:31
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,858
Website
### Re: about something called "discrete mathematics"
How about you play some Games instead? I guess a lot of these games are discrete rather than continuous
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
## #3 2016-12-10 02:57:15
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: about something called "discrete mathematics"
Hmmm, okay, you got me. Why should he do that? I know you have a reason.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #4 2016-12-10 09:43:39
Hannibal lecter
Member
Registered: 2016-02-11
Posts: 225
Website
### Re: about something called "discrete mathematics"
Hi, I didn't understand what you said, can you explain your words please, I'll try to play these games later but I need to study discrete mathematics like (if and only if)
I did it in university with our techer but I need to study it from math is fun.com
#bobbym
he told us there is a relationship between discrete mathematics and computer science,
so can you help me find any website or link in math is fun.com
Last edited by Hannibal lecter (2016-12-10 09:46:31)
Wisdom is a tree which grows in the heart and fruits on the tongue
Offline
## #5 2016-12-10 12:20:56
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: about something called "discrete mathematics"
Hi;
I was asking Agnishom why he answered the way he did. I did not find anything under the heading of Discrete Mathematics at Mathisfun. Have you googled for something else?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #6 2016-12-10 20:34:52
bob bundy
Registered: 2010-06-20
Posts: 8,399
### Re: about something called "discrete mathematics"
hi Hannibal lecter
I think this is one area of maths that is not covered by Maths Is Fun. It is also known as mathematical logic, propositional logic or even propositional calculus so you might find something under an alternative name. It is the gateway to computer science because logic deals in TRUE and FALSE and computers work with binary arithmetic, 0s and 1s. In logic we have operations like AND, OR and NOT. From these you can make up new operations like NAND (= NOT AND) and computer circuits are constructed out of components that behave like NAND. For example, you can build a circuit based on NANDs that adds two binary numbers together.
It is a long time since I last did this, but I might be able to remember and I'm sure it will be good for me, so post if you have difficulties with the subject, and I'll do my best.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline | 1,073 | 3,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-43 | latest | en | 0.934227 |
https://math.stackexchange.com/questions/2044973/is-there-an-algorithm-for-solving-magic-squares-puzzles | 1,619,038,926,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00214.warc.gz | 485,252,031 | 38,503 | # Is there an algorithm for solving “Magic Squares” puzzles?
In case you don't know what I'm referring to, it would be those "lights on, lights off" puzzles where you are free to toggle the status of any light in the grid, but doing so also toggles adjacent lights (but not diagonal).
The goal is usually to toggle all of the lights on, or to match a certain pattern of lights (i.e. middle light off).
The puzzle I'm currently stumped on begins as such:
O X O
O O O
X X O
The goal is to make a T shape, or:
O O O
X O X
X O X
I can never figure out a "tried and true" method to solve these, and am wondering if there is some lovely match trick that can be used to figure them out. More often than not, I end up just toggling at random and hoping for the best.
A solution to this particular problem would be nice but I'm also looking for an "algorithm" of sorts for future puzzles.
Thanks! | 224 | 898 | {"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} | 2.625 | 3 | CC-MAIN-2021-17 | longest | en | 0.968196 |
https://www.airmilescalculator.com/distance/bdl-to-mia/ | 1,653,315,408,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00254.warc.gz | 669,995,956 | 55,132 | # Distance between Windsor Locks, CT (BDL) and Miami, FL (MIA)
Flight distance from Windsor Locks to Miami (Bradley International Airport – Miami International Airport) is 1194 miles / 1922 kilometers / 1038 nautical miles. Estimated flight time is 2 hours 45 minutes.
Driving distance from Windsor Locks (BDL) to Miami (MIA) is 1417 miles / 2280 kilometers and travel time by car is about 26 hours 46 minutes.
1194
Miles
1922
Kilometers
1038
Nautical miles
2 h 45 min
161 kg
## How far is Miami from Windsor Locks?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 1194.412 miles
• 1922.220 kilometers
• 1037.916 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1196.620 miles
• 1925.773 kilometers
• 1039.834 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Windsor Locks to Miami?
Estimated flight time from Bradley International Airport to Miami International Airport is 2 hours 45 minutes.
## What is the time difference between Windsor Locks and Miami?
There is no time difference between Windsor Locks and Miami.
## Flight carbon footprint between Bradley International Airport (BDL) and Miami International Airport (MIA)
On average flying from Windsor Locks to Miami generates about 161 kg of CO2 per passenger, 161 kilograms is equal to 355 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Windsor Locks to Miami
Shortest flight path between Bradley International Airport (BDL) and Miami International Airport (MIA). | 453 | 1,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.85913 |
https://socratic.org/questions/how-do-you-solve-and-graph-11w-77 | 1,639,026,226,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363659.21/warc/CC-MAIN-20211209030858-20211209060858-00505.warc.gz | 563,147,998 | 5,859 | # How do you solve and graph 11w ≤ 77 ?
Jun 17, 2017
$w \le 7$
#### Explanation:
Solve for $w$ first by dividing both sides by $11$
$11 w \le 77$
$w \le 7$
A number line graph would have a closed circle on $11$ and a line extending to infinity to the left. | 86 | 263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 3.921875 | 4 | CC-MAIN-2021-49 | latest | en | 0.95509 |
https://www.cleanslateeducation.co.uk/t/problem-42/1423 | 1,657,013,972,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00504.warc.gz | 746,621,745 | 3,768 | # Problem 42 */**
Show that \displaystyle \sum_{r = 0}^{88} \frac 1 {\cos r \cdot \cos (r + 1)} = \frac {\cos 1} {\sin^2 1}
(all arguments in degrees)
[originally posted by Felix Felicis on TSR] | 72 | 197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.561804 |
https://betterlesson.com/browse/common_core/standard/80/ccss-math-content-1-g-a-1-distinguish-between-defining-attributes-e-g-triangles-are-closed-and-three-sided-versus-non-defining-a?from=cc_lesson_core | 1,540,072,894,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513441.66/warc/CC-MAIN-20181020205254-20181020230754-00271.warc.gz | 632,302,697 | 31,488 | ## Math
### HS Statistics & Probability
Back
1.G.A.1
1.G.A.1
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size); for a wide variety of shapes; build and draw shapes to possess defining attributes.
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View All 30 Lessons
1.G.A.1
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size); for a wide variety of shapes; build and draw shapes to possess defining attributes.
1.G.A.2
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.
1.G.A.3
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. | 278 | 1,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-43 | latest | en | 0.863882 |
http://docplayer.net/20520868-Two-state-option-pricing.html | 1,542,725,062,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746398.20/warc/CC-MAIN-20181120130743-20181120152743-00366.warc.gz | 87,428,022 | 24,282 | # Two-State Option Pricing
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## Transcription
1 Rendleman and Bartter [1] present a simple two-state model of option pricing. The states of the world evolve like the branches of a tree. Given the current state, there are two possible states next period. Using an arbitrage argument, one prices the option by working backwards from the future to the present. 1
2 Notation Let the subscript 0 denote the current state. Subscripts 1 and 2 denote the two possible states that may occur in the future. A stock worth s 0 in the current period is worth either s 1 or s 2 in the future period. A call worth c 0 in the current period is worth either c 1 or c 2 in the future period. There is also a risk-free asset. 2
3 Numerical Example We illustrate the model by a numerical example. A stock worth s 0 = 100 in the current period is worth either s 1 = 120 or s 2 = 90 in the future period. For simplicity, suppose that the risk-free rate of return is zero. 3
4 Call Suppose that the call expires in the future period, with exercise price 105. Thus the call is worth c 1 = s = 15 if state 1 occurs (the call is exercised), and c 2 = 0 if state 2 occurs (the call is not exercised). We use an arbitrage argument to find the current price c 0. 4
5 Black-Scholes Argument Black and Scholes argue that one can form a risk-free portfolio from the stock and the call. The difference in stock value between the two future states is s 1 s 2 = = 30, and the difference in call value between the two future states is c 1 c 2 = 15 0 = 15. 5
6 Hence the hedge ratio is Hedge Ratio = 1 2. In the current period, form a risk-free portfolio by buying one share of stock and selling two calls. The net cost of this portfolio is s 0 2c 0. 6
7 Risk-Free Portfolio If state 1 occurs, the portfolio is worth s 1 2c 1 = = 90. If state 2 occurs, the portfolio is worth s 2 2c 2 = = 90. The portfolio is indeed risk-free. 7
8 Call Price for No Arbitrage If there is no opportunity for arbitrage profit, the rate of return on this portfolio must equal the risk-free rate of return: (s 0 2c 0 ) (1 + risk-free return)=90. Hence so (100 2c 0 ) (1 + 0)=90, c 0 = 5. 8
9 Pricing Kernel One obtains the same call price via the pricing kernel (stochastic discount factor). Let p i denote the state price of one dollar in state i. Since there are two states and two assets (the stock and the risk-free asset), there exists a unique portfolio of these assets having payoff 1 in state i and payoff 0 in the other state. The state price p i is the cost of this portfolio. For an asset having payoff \$ i in state i, its price must be p 1 \$ 1 + p 2 \$ 2, to avoid an opportunity for profitable arbitrage. 9
10 For the stock, so s 0 = p 1 s 1 + p 2 s 2, 100 = 120p p 2. For the risk-free asset, 1 = p 1 + p 2, as the risk-free rate of return is zero. 10
11 Solving yields the state prices p 1 = 1 3 p 2 = 2 3. Pricing the call via the state prices gives c 0 = p 1 c 1 + p 2 c 2 = = 5, identical to the result found above. 11
12 Probability and Mean Return The probability of each state is irrelevant to the call price. This result is natural, since the call pricing is based on an arbitrage argument, and whether there is an arbitrage opportunity is independent of the probability of each state. The irrelevance of the probability corresponds to the property of the Black-Scholes model that the mean rate of return on the stock is irrelevant. In the two-state model, changing the probability of the two states changes the mean rate of return but has no effect on the call price. 12
13 The Black-Scholes Model as a Limit Wiener-Brownian motion can be derived as the limit of a binomial random walk. Using this relationship, by a complicated argument it is possible to derive the Black-Scholes formula for the call price by taking the limit of the call price derived in the two-state model. 13
14 References [1] Richard J. Rendleman, Jr. and Brit J. Bartter. Two-state option pricing. Journal of Finance, XXXIV(5): , December HG1J6. 14
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### Computational Finance Options
1 Options 1 1 Options Computational Finance Options An option gives the holder of the option the right, but not the obligation to do something. Conversely, if you sell an option, you may be obliged to | 7,627 | 31,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-47 | latest | en | 0.89094 |
http://www.reddit.com/r/nottheonion/comments/1540wj/christian_parents_group_to_sue_school_over_yoga/c7j6hbq?context=3 | 1,408,713,250,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500823598.56/warc/CC-MAIN-20140820021343-00268-ip-10-180-136-8.ec2.internal.warc.gz | 556,719,019 | 15,064 | you are viewing a single comment's thread.
[–] 378 points379 points (13 children)
sorry, this has been archived and can no longer be voted on
And next a group of concerned Christian parents is going to sue a school board because of their indoctrination of their kids with the Hindu & Arab discipline known as mathematics.
[–] 114 points115 points (12 children)
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Don't take kindly to Arabic numerals in these parts.
[–] 6 points7 points (11 children)
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It might be a good opportunity to introduce the base 12 counting system.
[–] 12 points13 points (8 children)
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In all honesty if we were going to change it base 16 is the way to go.
[–] 1 point2 points (2 children)
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Nahbro, base 2.
[–] 3 points4 points (0 children)
sorry, this has been archived and can no longer be voted on
[–] 2 points3 points (0 children)
sorry, this has been archived and can no longer be voted on
Base-8, because then everyone can hate themselves.
[–] 1 point2 points (4 children)
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But base 16 is absolutely terrible for decimals. Every fraction that isn't a power of two will have an infinitely repeating decimal.
[–] 4 points5 points (2 children)
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base 16 is terrible for decimals
base 16
decimals
[–] 2 points3 points (1 child)
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Oh, you know what I mean. Decimal points.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
So what you mean is called "decimal fractions" (thanks for that quick math theory refresher, wikipedia). If it's base 16, then it'd be "hexadecimal fractions."
It's been a while since I took discrete math, but it seems like you'd have a similar number of ugly repeating fractions in decimal as in hex -- maybe slightly more, because you lose out on fractions that are powers of 5 -- but it's nowhere near the order of magnitude of yuckiness you're making it out to be.
edit: thinking a little harder about this, though, you do make a good point. Any nice decimal fraction with a finite number of digits probably wouldn't translate well into hexadecimal fractions.
[–] 0 points1 point (0 children)
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What about base 30? Would be fun.
[–] 0 points1 point (1 child)
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Why base 12? Why not base 16?
[–] 1 point2 points (0 children)
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More factors | 720 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2014-35 | latest | en | 0.967129 |
https://www.jbigdeal.in/data-sufficiency-question-answers/ds4/ | 1,537,388,910,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156305.13/warc/CC-MAIN-20180919200547-20180919220547-00076.warc.gz | 782,530,943 | 30,209 | # Data Sufficiency Question Answers Test
## Data Sufficiency Online Mock Test
Congratulations - you have completed Data Sufficiency Online Mock Test.You scored %%SCORE%% out of %%TOTAL%%.Your performance has been rated as %%RATING%%
Question 1
Directions (Qns. 1 to 15): Each question below is followed by two labelled facts [labelled as (I) and (II)]. You are to determine whether the data given in the statement are sufficient for answering the questions. Use the data given, plus your knowledge of Mathematics and every day facts, to choose amongst possible answer from (A) to (E), Is the number 15 an odd integer? (You may assume that 15 is an integer)(I) N = 3K, where K is an integer(II) N = 6J + 3, where J is an integer
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 2
What was the value of sales of ABC Company in 1980?(I) The sales of ABC Company increased by Rs. 1, 00, 000 each year from 1970 to 1980.(II) The value of the sales of ABC Company doubled between 1970 and 1980.
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 3
If x6 – y6 = 0, what is the value of x3 – y3?(I) x is positive(II) y is greater than 1
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 4
If a and b are the both positive numbers, then which is larger, 2a or 3b?(I) a is greater than 2b(II) a is greater than or equal to b + 3
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 5
How far is it from town A to town B? Town C is 12 km east of town A (I) Town C is South of town B(II) It is 9 km from town B to town C
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 6
Is x greater than y?(I) x’ y = - 15 (II) x + y = 2
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 7
Which of the four numbers w, x, y and z is the largest?(I) The average of w, x, y and z is 25 (II) The numbers w, x and y are each less than 24
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 8
How much does Susan weigh?(I) Susan and John together weight 100 kg.(II) John weighs twice as much as Susan
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 9
Find x + y(I) x – y = 6 (II) – 2x + 2y = - 12
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 10
What percentage of families in a state have annual income over Rs. 2, 25, 000 and own a car?(I) 28% of the families in the state have an annual income over Rs. 2, 25, 000(II) 40% of the families in the state with an annual income over Rs. 2, 25, 000 own a car
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 11
Does every bird fly?(I) Tigers do not fly(II) Ostriches do not fly
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 12
A piece of wood 5 feet long is cut into three smaller pieces. How long is the longest of the three pieces?(I) One piece is 2 feet 7 inches long(II) One piece is 7 inches longer than another piece and the remaining piece is 5 inches long
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 13
How much is John’s weekly salary?(I) John’s weekly salary is twice as much as Fred’s weekly salary.(II) Fred’s weekly salary is 40% of the total of Chuck’s weekly salary and John’s weekly salary
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 14
On a certain race track, motorbikes average speed in 260 kmph. What is the length of the track?(I) On straight sectons, motorbikes can go at 200 kmph(II) Average lap time (once round the track) is 2 minutes 8 seconds
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Question 15
A sequence of numbers a1, a2, a3 ……. is given by the rule 12 = an + 1. Does 3 appear in the sequence?(I) a1 = 2(II) a3 = 16
A If you can get the answer from (I) alone but not from (II)alone. B If you can get the answer from (II) alone but not from (I) alone. C If you can get the answer from both (I) and (II) but not from (I) alone or (II) alone. D If either statement (I) or (II) is sufficient to answer the question asked. E If you cannot get the answer from statement (I) and (II) together, but need even more data.
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 15 questions to complete.
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Rating | 2,672 | 9,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | longest | en | 0.939225 |
https://www.got-it.ai/solutions/excel-chat/excel-tutorial/miscellaneous/get-last-entry-by-month-and-year | 1,723,068,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00811.warc.gz | 609,004,048 | 17,903 | Get instant live expert help with Excel or Google Sheets
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# Get last entry by month and year
We can use the LOOKUP function and the TEXT function to get the last entry by Month and Year in a data. The steps below will walk through the process.
Figure 1: How to Get Last entry by Month and Year
## General Formula
`=LOOKUP(2,1/(TEXT(dates,"mmyy")=TEXT(A1,"mmyy")),values)`
## Formula
`=LOOKUP(2,1/(TEXT(\$B\$4:\$B\$11,"mmyy")=TEXT(E6,"mmyy")),\$C\$4:\$C\$11)`
## Setting up the Data
• We will set up data by inputting our Dates (month and year) in Column B and Sales in Column C
• Our result will be returned in Column F by the formula. We intend to know the last SALE VALUE that was added to the table for each month
Figure 2: Setting up the Data
Note- We have formatted the dates by doing the following. First, we will type the dates in this format into COLUMN B: 1/1/2019. We will highlight the dates from Cell B4 to Cell B11. We will right-click and click on format cells.
Figure 3: Formatting the Dates
• We will click on Custom and select mmm-yy. We will press OK
• We will do the same thing for the LOOKUP dates. However, we will simply type the month and year instead. For instance, Jan 2019
## Get Last entry by Month and Year
• We will click on Cell F6
• We will insert the formula below into the cell
`=LOOKUP(2,1/(TEXT(\$B\$4:\$B\$11,"mmyy")=TEXT(E6,"mmyy")),\$C\$4:\$C\$11)`
• We will press the enter key
Figure 4: Last entry for January 2019
• We will click on Cell F6 again
• We will double click on the fill handle tool which is the small plus sign you see at the bottom right of Cell F6. Select and drag down to copy the formula to other cells.
Figure 5: Last entries for January, February, and March 2019
We can confirm our result by looking at the exact dates in figure 2.
## Instant Connection to an Expert through our Excelchat Service
Most of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes. | 660 | 2,541 | {"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} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.878499 |
https://pupman.com/listarchives/2002/May/msg00315.html | 1,585,841,119,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00285.warc.gz | 644,296,367 | 2,507 | # Re: Directions for tesla coil research
```Original poster: "rob by way of Terry Fritz <twftesla-at-qwest-dot-net>" <rob-at-pythonemproject-dot-com>
Tesla list wrote:
>
> Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<evp-at-pacbell-dot-net>
>
> > How does a bigger top load increase Q. if Q = Xl / R and Xl =
> > 2 * pi * f * l.
> >
> > from these equations, one would assume that if frequency goes down, due
to a
> > shift in resonant frequency, then Q decrease because Xl decrease. I know
> > I'm missing something important here, right?
> >
> > Shaun Epp
>
> I think so. The discussion was about reducing the frequency by
> increasing the top loading, but you have the wrong expression for Q in
> this case. The Q is the parallel resistance divided by the reactance
> so, for a given resistance (same power in the streamers) the Q will
> increase. Your expression is for an inductor with a resistance in
> series with it. If you look at the equivalent circuit you will see that
> increasing the parallel resistance does lower the series resistance,
> with the same effect of increasing the Q a bit. For coils with a Q much
> over 10, as are all the ones under discussion here, the product of
> series resistance and parallel resistance equals the square of the
> reactance.
>
> Rp Rs = X^2
>
> if I didn't make another typo.
>
> Ed
>
> Ed
Something I forgot to mention about all of this. There is unloaded Q,
and loaded Q. I assume that once the air starts to break down, you are
in the realm of loaded Q. There is a very good discussion of these
types of Q in the Amateur Radio Handbook, and probably lots elsewhere on
the web. For example, when I design matching networks for LNAs at work,
its best to have a circuit with a low loaded Q. The reason is that the
Q factor multiplies the current in the inductors (circulating current),
increasing the I*R loss. Thus high Q circuits can in some cases greatly
magnify the losses. Still, though, the circuit with the highest
unloaded Q elements has the lowest loss. I hope I am not being
confusing :) Rob.
--
-----------------------------
The Numeric Python EM Project
www.pythonemproject-dot-com
``` | 569 | 2,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.9284 |
https://rdrr.io/rforge/Rcgmin2/src/demo/broydt_test.R | 1,580,039,087,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251688806.91/warc/CC-MAIN-20200126104828-20200126134828-00064.warc.gz | 629,851,917 | 13,473 | # demo/broydt_test.R In Rcgmin2: EXPERIMENTAL Conjugate gradient minimization of nonlinear functions with box constraints
```rm(list = ls())
library(Rcgmin)
broydt.f <- function(x) {
n <- length(x)
res <- rep(NA, n)
res[1] <- ((3 - 0.5 * x[1]) * x[1]) - 2 * x[2] + 1
tnm1 <- 2:(n - 1)
res[tnm1] <- ((3 - 0.5 * x[tnm1]) * x[tnm1]) - x[tnm1 - 1] -
2 * x[tnm1 + 1] + 1
res[n] <- ((3 - 0.5 * x[n]) * x[n]) - x[n - 1] + 1
sum(res * res)
}
broydt.g <- function(x) {
n <- length(x)
gg[1] <- -2 + 2 * x[1] + 4 * x[3] + (6 - 2 * x[1]) * (1 -
2 * x[2] + x[1] * (3 - 0.5 * x[1])) - 2 * x[2] * (3 -
0.5 * x[2])
gg[2] <- -6 + 4 * x[4] + 10 * x[2] + (6 - 2 * x[2]) * (1 -
x[1] - 2 * x[3] + x[2] * (3 - 0.5 * x[2])) - 4 * x[1] *
(3 - 0.5 * x[1]) - 2 * x[3] * (3 - 0.5 * x[3])
tnm2 <- 3:(n - 2)
gg[tnm2] <- -6 + 4 * x[tnm2 - 2] + 4 * x[tnm2 + 2] + 10 *
x[tnm2] + (6 - 2 * x[tnm2]) * (1 - x[tnm2 - 1] - 2 *
x[tnm2 + 1] + x[tnm2] * (3 - 0.5 * x[tnm2])) - 4 * x[tnm2 -
1] * (3 - 0.5 * x[tnm2 - 1]) - 2 * x[tnm2 + 1] * (3 -
0.5 * x[tnm2 + 1])
gg[n - 1] <- -6 + 4 * x[n - 3] + 10 * x[n - 1] + (6 - 2 *
x[n - 1]) * (1 - x[n - 2] - 2 * x[n] + x[n - 1] * (3 -
0.5 * x[n - 1])) - 4 * x[n - 2] * (3 - 0.5 * x[n - 2]) -
2 * x[n] * (3 - 0.5 * x[n])
gg[n] <- -4 + 4 * x[n - 2] + 8 * x[n] + (6 - 2 * x[n]) *
(1 - x[n - 1] + x[n] * (3 - 0.5 * x[n])) - 4 * x[n -
1] * (3 - 0.5 * x[n - 1])
return(gg)
}
ni <- c(1, 2, 3)
times <- matrix(NA, nrow = 3, ncol = 4)
for (ii in ni) {
n <- 10^ii
cat("n=", n, "\n")
afname <- paste("ansbroydt", n, "UG", sep = "")
x0 <- rep(pi, n)
ut <- system.time(ans <- Rcgmin(x0, broydt.f, broydt.g, control = list(trace = 1)))[1]
times[ii, 1] <- ut
sink(afname, split=TRUE)
print(ans)
sink()
afname <- paste("ansbroydt", n, "UN", sep = "")
ut <- system.time(ans <- Rcgmin(x0, broydt.f, control = list(trace = 1)))[1]
times[ii, 2] <- ut
sink(afname, split=TRUE)
print(ans)
sink()
lower <- rep(1, n)
upper <- rep(Inf, n)
afname <- paste("ansbroydt", n, "BG", sep = "")
x0 <- rep(pi, n)
ut <- system.time(ans <- Rcgmin(x0, broydt.f, broydt.g, lower = lower,
upper = upper, control = list(trace = 1)))[1]
times[ii, 3] <- ut
sink(afname, split=TRUE)
print(ans)
sink()
afname <- paste("ansbroydt", n, "BN", sep = "")
ut <- system.time(ans <- Rcgmin(x0, broydt.f, lower = lower,
upper = upper, control = list(trace = 1)))[1]
times[ii, 4] <- ut
# ans
sink(afname, split=TRUE)
print(ans)
sink()
}
cat("Cols are UG, UN, BG, BN times\n")
print(times)
```
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Rcgmin2 documentation built on May 2, 2019, 4:41 p.m. | 1,253 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-05 | latest | en | 0.204986 |
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Note by Keshav Tiwari
5 years ago
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Sort by:
231... mine (MARKS)
- 5 years ago
I got AIR-89
- 5 years ago
- 5 years ago
Wow...that is great BTW,Congrats...amazing performance!!!
- 5 years ago
Bhaiya your rank is awesome :) ; can I know your marks to get an idea abt relation between rank and marks @SHASHANK GOEL
- 5 years ago
I Think around 250 . because at my centre a person got 82 rank on 252 marks
- 5 years ago
Bravo prakhar its 249
- 5 years ago
Congrats bhaiya for amazing score.Can you tell me from where did you practiced chemistry (all 3)
- 5 years ago
Well i was from VMC . The best environment i have ever went to.their modules were brilliant.For organic and physical chemistry the class notes were the best.for inorganic chemistry both modules and classnotes were good.for jee mains ncert is enough.however you must revise the inorganic chapters for the advance as well
- 5 years ago
81
- 5 years ago
Congrats Bhaiya! You were from VMC right? What was your general rank in HRTs? Also was module sufficient?
- 5 years ago
It was generally in top 10 but dont Judge yourself from that.the modules were sufficient as a matter of theory . However the understanding and the feel came from my classnotes by my brilliant bhaiyas(teachers)
- 5 years ago
Thank you bhaiyya. DPS Ghaziabad is on a roll lol. First Shobhit bhaiyya and now you.
- 5 years ago
MY RANK COMES AROUND 1000 IN VMC IN JEE MAINS AND AROUND 1500 IN JEE ADVANCE IN VMC DO YOU THINK I CAN GET IN IIT IF I WORK HARD NOW?
- 4 years, 11 months ago
Yes of course
- 4 years, 11 months ago
should i quit vmc ( i am a student of vmc - vsat classes) they do not reply to my doubts nor they care about marks student are getting , they just need their fee and nothing else, plz suggest.
- 4 years, 10 months ago
gr8
- 5 years ago
Congrats fellow VMCian !
- 5 years ago
184...........................................rank
- 5 years ago
@Keshav Tiwari From Kota Highest score is around 300 of the great Utkarsh Gupta.
- 5 years ago
That is a very good score, my score has dropped w.r.t official answer key. Hoping for the best. $:)$
- 5 years ago
I think the exam centre matters.....
E.g.- at my centre...we were not allowed transparent water bottles......they didnt supply drinking water too.....and tap water smelled aweful......now my seat was just beside window...(without anything to lose them up)........from where very aweful smell of toilet was coming..........
Dont u agree with me??
- 5 years ago
Sorry for the late reply. The environment has a role to play. Your score is really good( in my view) , not many people are scoring that much. Good luck !
- 5 years ago
Thank u very much......
- 5 years ago
From chaitanya-narayana someone is getting 311.........there is a rumor that a boy from Catalyser Indore is scoring 321.....
- 5 years ago
I am getting 212.....(bad score ).......how much r u all getting ??
- 5 years ago
Utkarsh Gupta of Resonance ?????
- 5 years ago
TOMORROW IS THE FINAL DAY
ALL THE BEST>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
- 5 years ago
What rank did you get? I got 232 marks and still haven't been able to access the results.
- 5 years ago
U will under 500 since some one with score of 218 was in top 1000.
- 5 years ago
I got my result right now!! I got 176 rank :-)
- 5 years ago
gr8
- 5 years ago
Woah that's too good!!!!
Congratulations!
- 5 years ago
Congrates,.......nitish and mayank.... Mine 210 marks.....406 rank
- 5 years ago
I got 130 marks and my AIR is 5695.
- 5 years ago
Mine was not very well. How was yours? How many marks are you expecting?
- 5 years ago
Same. Around 200, yours ?
- 5 years ago
I am getting around 230-235. Do you know any high scores from Kota/Hyderabad? Also which rank predictor do you think is most appropriate?
- 5 years ago
I have no idea about it. I don't thing any of them is appropriate. Let's wait and see. ;)
- 5 years ago
As per Fiit jee Rank Predictor yor AIR must be174.
- 5 years ago
Really ...what has happened to the community? Has the no. of members appearing in JEE dropped?:(
- 5 years ago
Someone from fiitjee south delhi scoring about 290
- 5 years ago
How are the overall results at FIITJEE? Do you know how many are above 250 or 200?
- 5 years ago
From my centre about 2 people are scoring more than 245
- 5 years ago
actually 1
- 5 years ago
Do u know what his name is?(who is scoring 290)...is he rupesh??
- 5 years ago
No i think soumya sharma
- 5 years ago
Ok...
- 5 years ago
fiitjee highest is 298
- 5 years ago
Who is , name , from where if U know ??
- 5 years ago
idk, my sir told me
- 5 years ago
I'm getting 130 (really a bad score) !
- 5 years ago
hey i got to know @ishan tarunesh is getting 33 rank ,, btw i saw this in newspaper ,, afterall congrats for all rankers .
- 5 years ago
Hi..
- 4 years, 8 months ago
I am getting 289 in advance
- 5 years ago
well I got 360/360 in JEE-Main.
- 5 years ago
Yeah right ;)
1st rank holder in India.
- 5 years ago
Hi I'm Shah Rukh Khan.
- 5 years ago
so,what???
- 5 years ago
Nothing, just saying. :')
- 5 years ago | 1,984 | 6,823 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "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} | 2.84375 | 3 | CC-MAIN-2021-25 | latest | en | 0.850684 |
https://yourwiseadvices.com/what-is-the-size-of-a-palm/ | 1,718,680,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00173.warc.gz | 959,119,747 | 14,453 | # What is the size of a palm?
## What is the size of a palm?
A palm, when used as a unit of length, is usually four digits (fingers) or three inches, i.e. 7.62 cm (for the international inch).
How many palms are in a pace?
How many Paces are in a Palm? The answer is one Palm is equal to 0.05 Paces.
### How do you measure inches in your palm?
The first joint of an index finger is about 1 inch long. When a hand is spread wide, the span from the tip of the thumb to the tip of the pinkie is about 9 inches; from the tip of the thumb to the tip of the index finger, around 6 inches.
What is palm of your hand?
The palm comprises the underside of the human hand. Also known as the broad palm or metacarpus, it consists of the area between the five phalanges (finger bones) and the carpus (wrist joint). Its symptoms include the formation of a firm nodule in the palm that later develops into a thick band.
#### How long is the palm of your hand?
The widest part of your palm is your palm size. The average palm size for a male is 3.3″ and 2.91′ for a female. In photo 3, for example, the palm size is about 3 inches.
What is considered a small hand?
A ‘small hand’ is defined as one with a thumb to fifth finger span of less than 8.5 inches (21.6 cm) and/or a second to fifth finger span of less than 6 inches (16.2 cm).
## How wide is the average palm?
How do you read a girl’s hand?
Choose a hand. In palmistry, it is thought that: For females, the right hand is what you’re born with, and left is what you’ve accumulated throughout your life. For males, it is the other way around. The left hand is what you’re born with, and the right is what you’ve accumulated throughout your life.
### What is upper side of palm called?
The front, or palm-side, of the hand is referred to as the palmar side. The back of the hand is called the dorsal side. There are 27 bones within the wrist and hand.
How long is the average pinky?
It turns out that many of us humans have pinky tips that are all around the same length, and handily, that length is about 1 inch. I mentioned this one day in the test kitchen at a tasting, and everyone immediately wanted to measure their pinkies.
#### What is the average hand size for a 13 year old?
Average hand span in centimetres, by age
Age1 Girls Boys
years cm
11 17.23 17.36
12 18.04 18.52
13 18.24 19.27
How wide is the average finger?
An MIT Touch Lab study of Human Fingertips to investigate the Mechanics of Tactile Sense found that the average width of the index finger is 1.6 to 2 cm (16 – 20 mm) for most adults.
## What is the length of a palm tree?
The palm was divided into four digits (digitus) of about 1.85 cm (0.7 in) or three inches (uncia) of about 2.47 cm (1.0 in). Three made a span (palmus maior or “greater palm”) of about 22.2 cm (9 in); four, a Roman foot; five, a hand-and-a-foot (palmipes) of about 37 cm (1 ft 3 in); six, a cubit (cubitus) of about 44.4 cm (1 ft 5.5 in).
How many digits are there in a palm?
The palm was divided into four digits (digitus) of about 1.85 cm (0.7 in) or three inches (uncia) of about 2.47 cm (1.0 in).
### What is the length of an ancient Greek palm?
The Ancient Greek palm ( Greek: παλαιστή, palaistḗ, δῶρον, dō̂ron, or δακτυλοδόχμη, daktylodókhmē) made up ¼ of the Greek foot ( poûs ), which varied by region between 27–35 cm (11 in–1 ft 2 in). This gives values for the palm between 6.7–8.8 cm (2.6–3.5 in), with the Attic palm around 7.4 cm (2.9 in).
What is the length of 4 fingers?
palm width of 4 extended fingers (palmus) (“palm” is also know as “hand”.) 7 cm, 2 3/4 in(at middle joint) 8 cm, 3 in(at knuckles) hand length hand length, heel to fingertip 19 cm, 7.5 in hand span hand width, from outspread thumb to little-finger | 1,078 | 3,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.96314 |
https://scienceblogs.com/builtonfacts/2009/08/11/maxwells-equations-2 | 1,603,929,212,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00250.warc.gz | 520,225,825 | 14,405 | Maxwell's Equations #2
In our examination of the first of Maxwell's four equations, we saw that magnetic charge doesn't exist as far as we can tell. On the other hand, electric charge permeates every aspect of our existence. The motion of charged electrons is one of the central pillars of modern civilization. The way that electric charge creates an electric field is the subject of the second of Maxwell's equations. In its full mathematical glory:
The triangle and the dot represent the divergence, exactly as in the first Maxwell equation. The letter E is the electric field. Unlike the magnetic field divergence equation, the divergence of the electric field is not zero everywhere. Instead, it's equal to the charge density denoted by the Greek letter rho, divided by the electric constant, Greek letter epsilon.
Time for a quick review of divergence. It's a property of a vector field (such as an electric field) that describes the sources of field lines. Field lines are not physically real but instead serve as a schematic representation, with their direction pointing along the field and their closeness representing the field strength. Draw a little box around a region of space, and if there's more field lines leaving, you have positive divergence. If there's more entering than leaving, you have negative divergence. If they're the same, you have zero divergence.
All three possibilities are pictured in this snazzy Wikipedia image:
There's two charges, one positive and one negative. In the space outside the charges, there's no charge density. Therefore according to our equation, the divergence has to be zero. Sure enough if you draw a little box there's just as many field lines entering as there are leaving. But draw a box around one of the charges and the situation is entirely different. There's an excess of field lines leaving the positive charge (positive divergence) and an excess of field lines entering the negative charge (negative divergence).
This result was known well before Maxwell, and is in fact separately named Gauss' Law in honor of the great Carl Gauss who did pioneering work in this kind of mathematics. If you know the appropriate calculus, it's a snappy two-line calculation to derive the more elementary Coulomb's Law version of the field of a point particle. One minor technicality of Gauss' Law is that it gives the electric field divergence in terms of a continuous charge density. While this is a near-perfect approximation for macroscopic charge distributions, point particles technically don't have a finite charge density by virtue of not having volume. This can be mathematically fixed by the method of the delta function.
You might notice that in the sample picture we've picked the field lines are closer together in the area between the charges. This is a typical feature of many charge configurations, related to the close mathematical correspondence between the "flow" of field lines down voltage gradients and the flow of water down a hill. If you build up charges on your body by rubbing the floor with your socks, you can feel this process in the form of a zap when the charges in your body come near to the charges in a door knob. The high field from the close proximity breaks down the resistance of the air and makes a spark.
All right, that's two equations down, two to go. In fact at this point we're done with electrostatics. However our world is one where charges move and currents flow, so we and Maxwell have two more equations to go before we've covered the whole sweeping panorama of electromagnetism.
More like this
Maxwell's Equations #1
All this week we're going to be briefly looking at James Clerk Maxwell's greatest contribution to physics - his theory of electromagnetism. The consequences and applications of the theory fill many volumes, but the conceptual and mathematical foundations of the theory can be expressed as four…
The Advent Calendar of Physics: Gauss and Maxwell
As the advent calendar moves into the E&M portion of the season, there are a number of possible ways to approach this. I could go with fairly specific formulae for various aspects, but that would take a while and might close out some other areas of physics. In the end, all of classical E&M…
Maxwell's Equations #4
Fundamentally Maxwell's equations describe the origins of electric and magnetic fields. Given a set of conditions on the right hand side of the equations, you'll have fields described by the left hand side. Between the four equations the fields are uniquely specified, and there is nothing more to…
Gauss' Law PROVED WRONG!
Just though I'd try writing a post title in the style of a crank. Kinda fun! Gauss' law, of course, is not wrong. But I got a question from a reader that deceptively simple and an interesting example of a theorem not quite working the way you'd expect. I've gone over Gauss' law before, so as a…
First, a comment directed toward Jason@6 of the previous post: The concept here is best visualized by considering fluids, which are also governed by a similar equation as Matt mentioned. Let water run onto a flat surface, and there will be a positive divergence at the point where the stream of water hits and spreads out. If there is a drain nearby, there will be a negative divergence at the point where it enters the drain. The divergence is zero for the points where there is no source or sink (drain) for the vector field.
This last statement leads to the "minor technicality" Matt alludes to: divE=0 almost everywhere in the picture he shows. It is zero where the background is yellow because there is no charge in those areas. It is only non-zero in the purple areas where the + and - charge is located. That makes it instructive to compare this picture to the one in the previous article. The solenoid produces a magnetic field that does not have any "source" or "sink" points. The magnetic field lines form closed loops ... while the electric field lines shown here do not.
CCPhysicist: Thanks. I know how to calculate divergence but it never really clicked to me what divergence is. Now maybe you can explain curl :P
Just a historical note: these four 'Maxwell's Equations' are actually due to Oliver Heaviside. He reformulated the cumbersome way Maxwell had presented his equations (20 equations with 20 unknowns) into the differential equation form that is being presented here.
http://en.wikipedia.org/wiki/Oliver_Heaviside
I don't think it is entirely productive to think of the two charge equations as electrostatics, or describing such situations. It is, as you will undoubtedly cover later, true by virtue of choosing suitable reference frames. However the magnetic field is, as you also probably will cover, a (low speed) relativity effect, so electrodynamics is already implicit in the equations.
It is AFAIU the actual choice of reference frames that can produce situations where electrostatics applies. It's a "simplest analysis" choice, as far as relativity goes.
"source" or "sink" points
One reason I like to think of EM in english (besides the text books) is these beautiful pair of words. Not so nice in my native language.
the cumbersome way Maxwell had presented his equations
I think there is a history there that, if known, ought to deepen one's understanding.
IIRC odds and ends of it, Maxwell, at least at first, managed to visualize his equations by a mechanical model of many a virtual gear and axis in place of the then non-existing field concept. Maybe he wasn't geared [sic!] to simplify the notation away from an explicit 3D representation.
By Torbjörn Lars… (not verified) on 17 Aug 2009 #permalink | 1,589 | 7,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-45 | longest | en | 0.958002 |
https://www.jiskha.com/similar?question=What+is+the+kinetic+energy+of+a+29+kg+dog+that+is+running+at+a+speed+of+7.9+m%2Fs+%28about+18+mi%2Fh%29%3F&page=441 | 1,563,451,696,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00154.warc.gz | 754,137,545 | 22,110 | # What is the kinetic energy of a 29 kg dog that is running at a speed of 7.9 m/s (about 18 mi/h)?
45,322 questions, page 441
1. ## Sat Math
The bottom of a ski slope is 6,500 feet above sea level, the top of the slope is 11,000 feet above sea level, and the slope drops 5 feet vertically for every 12 feet traveled in the horizontal direction. From the top of the slope, Kayla skis at an average
asked by Amy on August 1, 2018
2. ## physics help 3!! **
Based on the following data about planet X (which orbits around the Sun): Planet X's distance from Sun = 3.6*1012 m Planet X's radius = 2*106 m Planet X's mass = 8.2*1022 kg a.) Find gx, the size of the acceleration due to gravity on the surface of Planet
asked by kelsey on October 31, 2008
star*** by my answer 1. When you must handle several hazards at the same time, the best tactic is to (1 point) compromise the hazards. adjust speed to separate the hazards. minimize, then separate the hazards. **** 2. How many zones of space surrounding
asked by Anonymous on March 31, 2016
4. ## dynamics of rigid bodies
7. A 1800 kg car drives up an inclined of 8% from rest at uniform acceleration to a velocity of 60 kph after traveling 75 m. Its wheels are 3 m apart with its center of gravity midway of the wheels and 500 mm from the ground. If all frictional force is
asked by warren on July 7, 2015
5. ## Physics
Why does the force of gravity change the speed of a satellite in an elliptical orbit? 1)The force of gravity is always tangent to the satellite motion. 2)Gravity speeds up the satellite as it moves away and slows it on its return. 3)The force of gravity is
asked by dave on October 7, 2016
6. ## Physics
Why does the force of gravity change the speed of a satellite in an elliptical orbit? 1) The force of gravity is always tangent to the satellite motion. 2)Gravity speeds up the satellite as it moves away and slows it on its return. 3)The force of gravity
asked by dave on October 7, 2016
7. ## Physics
Why does the force of gravity change the speed of a satellite in an elliptical orbit? 1)The force of gravity is always tangent to the satellite motion. 2)Gravity speeds up the satellite as it moves away and slows it on its return. 3)The force of gravity is
asked by dave on October 7, 2016
8. ## Physics _ College
A 50 cm long hollow glass cylinder is open at both ends and is suspended in air. A source of sounds that produces a pure frequency is placed close to one end of the tube. The frequency is placed close to one end of the tube. The frequency of the sound
asked by Sonya on August 10, 2011
9. ## Chels
What can you say about the motion of a body if its veloctiy-time graph is a straight line inclined with the time axis? a. the body moves with uniform velocity b.the body moves with uniform acceleration c. the body moves with non uniform acceleration d. the
asked by Physics on April 29, 2008
10. ## physical science
assume that a parcel of air is forced to rise up and over a 6000 foot high mountain. The intial temperature of the parcel at sea level is 76.5 F, and the lifting condensation level (lvl) of the parcel is 3000 feet. The DAR is 5.5 F/1000 and the SAR is 3.3
asked by nissa on October 28, 2011
11. ## Physics
ok Boxes are moved on a converyor blet from where they are filled to the packing station 10m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accerlates for half the distance, then
asked by QUESTION on August 19, 2009
12. ## physics
ok Boxes are moved on a converyor blet from where they are filled to the packing station 10m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accerlates for half the distance, then
asked by QUESTION on August 20, 2009
13. ## Physics
Larry leaves home at 2:08 and runs at a constant speed to the lamppost. He reaches the lamppost at 2:12, immediately turns, and runs to the tree. Larry arrives at the tree at 2:28. What is Larry's average velocity during his trip from home to the lamppost,
asked by Sarah on September 29, 2012
14. ## Science
Please Check my Work:pls check my work asap! 1. A ___ has a definite shape and a definite volume. (1 point) 1. solid (My Choice) 2. liquid 3. gas 4. molecule 2. Why is a gas able to flow? (1 point) 1. Its particles have melted and can move around. 2. Its
asked by A.J on April 8, 2014
15. ## physics
A small airplane with a wingspan of 13.5 m is flying due north at a speed of 70.8 m/s over a region where the vertical component of the Earth's magnetic field is 1.20 µT downward. (a) What potential difference is developed between the airplane's wingtips?
asked by Charles on December 2, 2011
16. ## Physics, finding harmonics
A tuning fork with a frequency of 440 Hz is held above a resonance tube that is partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what three smallest heights of the air column will resonance occur? Where will the nodes
asked by Robbie on February 9, 2009
17. ## physical
You are driving along a highway at 27.7 m/s when you hear a siren. You look in the rear-view mirror and see a police car approaching you from behind with a constant speed. The frequency of the siren that you hear is 1830. Hz. Right after the police car
asked by itufizik on July 16, 2012
18. ## math
A airplane crosses the Atlantic Ocean (3000 miles) with an airspeed of 600 miles per hour. The cost C (in dollars) per passenger is given by... C(x)= 150 + x/20 + 36,000/x Where x is the ground speed (airspeed plus or minus wind). a. What is the cost per
asked by v on October 9, 2015
19. ## physics
What two facts did Newton need to be able to calculate the acceleration of the Moon? 1) the distance to the Moon and the diameter of the Moon. 2)the time it takes the Moon to make one revolution and the distance from the Earth to the sun. 3)the diameter of
asked by Betty on October 5, 2016
20. ## physics
What two facts did Newton need to be able to calculate the acceleration of the Moon? 1) the distance to the Moon and the diameter of the Moon. 2) the time it takes the Moon to make one revolution and the distance from the Earth to the sun. 3) the diameter
asked by Sue on October 8, 2016
21. ## physics
What two facts did Newton need to be able to calculate the acceleration of the Moon? A) the distance to the Moon and the diameter of the Moon B) the time it takes the Moon to make one revolution and the distance from the Earth to the sun C) the diameter of
asked by Sue on October 3, 2016
22. ## Physics help
A light rope is wrapped several times around a large wheel with a radius of 0.435 m. The wheel? A light rope is wrapped several times around a large wheel with a radius of 0.435 m. The wheel rotates in frictionless bearings about a stationary horizontal
asked by BB on February 27, 2012
23. ## Statistics
An important part of the customer service responsibilities of a natural gas utility company concerns the speed with which calls relating to no heat in a house can be serviced. Suppose that one service variable of importance refers to whether or not the
asked by Matt on February 10, 2007
24. ## physics
A car is moving at 25.0m/s on a straight highway lane. 10.0m in front of the car is a van that is travelling with the same speed. The van’s driver suddenly hits the brakes to avoid an obstruction, slowing down at a rate of 9.00m/s². The car’s driver
asked by allyson on September 7, 2016
25. ## physics
A helium balloon has a mass of 0.022 kg, which includes both the helium and the balloon itself. On this balloon, there is 0.290 N of buoyant force pushing the balloon upward. If I let go of the balloon, find the balloon's acceleration rate in the upward
asked by Nick on May 12, 2013
26. ## Physic
Two wave pulses on a string approach one another at the time t = 0, as shown in the figure below, except that pulse 2 is inverted so that it is a downward deflection of the string rather than an upward deflection. Each pulse moves with a speed of 1.0 m/s.
asked by Anonymous on November 22, 2014
27. ## Mathematics question
It was...the subject is FOILs,I thought we needed to put what we were doing in the subject... Subject area needs to be properly stated. This is not FOIL. That is multiplying of two binaries as in (a+b)(c+d) Here you are to factor. take the first:
asked by Margie... on December 4, 2006
28. ## Physics
In the figure below, a runaway truck with failed brakes is moving downgrade at 114 km/h just before the driver steers the truck travel up a frictionless emergency escape ramp with an inclination of 15°. The truck's mass is 5000 kg. (a) What minimum length
asked by Rachel on February 27, 2013
29. ## Physics
A playground is on the flat roof of a city school, 7.00 m above the street below. The vertical wall of the building is 8.00 m high, forming a 1.00 m high railing around the playground. A ball has fallen to the street below, and a passerby returns it by
asked by Chloe on October 15, 2009
One of the methods used to train astronauts for the effects of "zero gravity" in space is to put them in a specially equipped plane which has been stripped of seats and fitted with padded walls. The pilot then takes the plane up to an altitude of typically
asked by archi on September 25, 2011
31. ## Math
can you help i dnt know what to do so could you tell me how to solve it 1. every weekend, Phil drives from his home to his favorite amusement park, a distance of 60 miles. a. If he averages 40 mph for the first half of the trip, what must his average speed
asked by howard on December 30, 2012
A certain CD has a playing time of 80 minutes. When the music starts, the CD is rotating at an angular speed of 4.8 x 10^2 revolutions per minute (rpm). At the end of the music, the CD is rotating at 2.1 x 10^2 rpm. Find the magnitude of the average
asked by Mary on March 14, 2007
33. ## Physics
A rubber ball is dropped onto a ramp that is tilted at 20 degrees. A bounding ball obeys the "law of reflection," which says that the ball leaves the surface at the same angle it approached the surface. The ball's next bounce is 3.0 m to the right of its
asked by Anonymous on September 9, 2014
34. ## PHYSICS
Please help and let me know if I did something wrong. If a cart is released from rest at a 10 meter incline and takes 28 seconds to travel down this incline. 1. Calculate the average speed of the cart? I wrote s=ad/at=10m/28 sec = 357 m/s 2. calculate the
asked by GRISSELL on January 23, 2012
35. ## Spanish
Can somebody help with my Spanish review? Combina lógicamente las palabras de vocabulario de las dos columnas. a. la caza b. el boxeo c. penalizar d. el/la jinete e. la cancha f. transnochar g. el campamento h. los naipes 1. el arco y la flecha (A) 2. la
asked by Anon on January 17, 2014
36. ## Algebra
1. Evaluate. (8 – 3) (12 ÷ 2 • 2) – 20 Answer: I'm not sure if its -5 or 40 2. 6x – 7 = 5(x – 2) Answer: -3 3. Solve for y. 3x – 2y = 12 Answer: 3x-12/2 4. Solve. –3(x – 3) + 8(x – 3) = x – 9 Answer: 1 1/2 or 1.5 5. Find the slope of
asked by Loreal on December 23, 2007
37. ## science
A stone is dropped into a river from a bridge 43.9 m above the water. Another stone is dropped vertically down 1s after the first is dropped. Both stones strike the water at the same time. a) What is the initial speed of the second stone? Write an equation
asked by sara on September 17, 2006
38. ## English
HIDDEN LESSONS: BY DAVID SUZUKI In spite of the vast expanse of wilderness in this country, most Canadian children grow up in urban settings. In other words, they live in a world conceived, shaped and dominated by people. Even the farms located around
asked by No Name on July 2, 2015
39. ## Physics
Analyze the flight of a projectile moving under the influence of gravity (mg) and linear, i.e., laminar- flow drag (fD = bv, with b > 0 constant). Neglect \Phase I"; assume the projectile is launched from the origin at speed v1 and angle 1 above the
asked by Alyssa on March 2, 2015
40. ## physics
An astronaut is traveling in a space vehicle moving at 0.520c relative to the Earth. The astronaut measures her pulse rate at 74.0 beats per minute. Signals generated by the astronaut's pulse are radioed to the Earth when the vehicle is moving in a
asked by sara on October 10, 2014
41. ## Physics
Two eagles fly directly toward one another, the first at 13 m/s and the second at 21 m/s. Both screech, the first emitting a frequency of 3200 Hz and the second, a frequency of 3800 Hz. What frequencies do they hear if the speed of sound is 345 m/s? (Tip:
asked by Cindy on December 6, 2010
42. ## English
Can you tell me which of the following statements are correct? Thank you very much. 1) For my main course I'd like grilled chicken breast (and not a grilled chickne breast) with boiled potatoes and peas. 2)For my starter I'll have a soup of the day and for
asked by Franco on February 23, 2010
43. ## Physics
The flywheel of a steam engine runs with a constant angular speed of 112 rev/ min . When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.2 h. What is the magnitude of the constant angular acceleration of the wheel
asked by Ted on April 2, 2014
44. ## physics
true or false If a net force acts on an object, the object's speed will change. An object's velocity will change if a net force acts on the object. If two object have the same acceleration, they are under the influence of equal forces. The net force which
asked by molly on October 4, 2010
45. ## physics
A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel, of radius 0.363 m, and observes that drops of water fly off tangentially. She measures the height reached by drops moving vertically (Fig. P10.63). A drop
asked by Weiyi on June 27, 2014
46. ## Pysics
A 0.480 kg, 37.5 cm long metal rod is sliding down two metal rails that are inclined 42 degrees to the horizontal. The rails are connected at the bottom so that the metal rod and rails form a loop that has a resistance of 52 ohm. There is a 2.0 T vertical
asked by LS on December 6, 2014
47. ## Physics
A particular Ferris wheel (a rigid wheel rotating in a vertical plane about a horizontal axis) at a local carnival has a radius of 20.0 m and it completes 1 revolution in 9.84 seconds. (a) What is the speed (m/s) of a point on the edge of the wheel? Using
asked by waqas on August 6, 2011
48. ## RE: PHYSICS
A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to
asked by COFFEE on February 26, 2007
49. ## Physics
A trapeze artist weighs 8.00 x 10^2 N. The artist momentarily held to one side of a swing by a partner so that both of the swing ropes are at an angle of 30º with the vertical. In such a condition of static equilibrium, what is the horizontal force being
asked by Becky on April 13, 2011
50. ## Physics
A child pushes a merry go-round from rest to a final angular speed of 0.78 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 1.6 revolutions. What is the angular acceleration of the merry go-round? v=d/t .78rev/s =
asked by Joe on February 21, 2012
51. ## Physics
Malaan is travelling from Makhado to Polokwane by taxi at a speed of 120km/h.A Nissan bakkie passes the taxi while it is on its way to Polokwane and is moving at 150km/h. a) Determine Malaan's velocity with respect to the Nissan bakkie b) what is the taxi
52. ## to ms.sue-project
After reading Holes by Louis Sachar, you will complete the book reporting bingo assignment. You will select any five boxes in a row from the board, in any direction across, down, or diagonally, to create a bingo. Each assignment must: •Be neat •Be
asked by Celest on June 5, 2012
53. ## Precalc
Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the the main
asked by Zoe Cameron on April 2, 2015
54. ## math
Power companies typically bill customers based on the number of kilowatt-hours used during a single billing period. A kilowatt is a measure of how much power (energy) a customer is using, while a kilowatt-hour is one kilowatt of power being used for one
asked by HELLO on March 9, 2013
55. ## ME
Problem 3: Consider water as the substance. (a) Find the specific internal energy (Btu/lbm) when T = 500F and P = 375 psia (b) Find the pressure (psia) when T = 500F and v = 0.02024 ft3/lbm (c) Find the specific volume (ft3/lbm) when T = 500F and
asked by sara on September 19, 2015
56. ## PHYSICS (most is worked)
A glass ball, ball A, of mass 6.0 g moves at a velocity of 18.0 cm/s. It collides with a second ball, ball B, of mass 8.0 g, moving along the same line of velocity of 11.0 cm/s. After the collision, ball A is still moving, but with a velocity of 10.0 cm/s.
asked by Jessica on January 28, 2013
57. ## Math
paul can type 60 words per minute and jen can type 80 words per minute. how dose puls typing speed compare to jen? 1.Paul can type 1/4 as fast as Jen. 2.Paul can type 1/3 as fast as Jen. 3.Paul can type 3/4 as fast as Jen. 4.Paul can type 1 1/3 times as
asked by AK on July 16, 2008
58. ## Novel
Heres what I have so far for the intro. of the novel. Let me know what you think and what needs to be fixed! For as long as I or anyone else could remember, I was always an excellent girl. All the neighbors invited me for tea or lemonade when I passed by
asked by mysterychicken on February 22, 2009
59. ## Chemistry
In the cchemistry lab we conducted an experiment called atomic spectroscopy. We had to find the color, energy, wavelength, and 1/lamda for Hydrogen, neon, and mercury. For hydrogen: red 1.90ev 650nm 1.54e^-3 red 2.02ev 610nm 1.64e^-3 green 2.30ev 545nm
asked by Hannah on October 31, 2011
60. ## Physics
Both drawings show the same square, each of which has a side of length L=0.75 m. An observer O is stationed at one corner of each square. Two loudspeakers are locate at corners of the square, as in either drawing 1 or drawing 2. The speakers produce the
asked by Vivien on July 9, 2013
61. ## physics
Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting
asked by shahi on September 22, 2011
62. ## physics
Traumatic brain injury such as concussionn results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s^2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s^2
asked by mb on September 18, 2008
63. ## physics
Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting
asked by greg on September 19, 2010
64. ## Physics
Robin Hood is walking through Sherwood Forest when he comes upon Sheriff of Nottingham. The Sheriff tells Robin Hood that he is going to arrest hum for numerous misdeeds. Robin pulls out his bow, loads an arrow, puls the bow back 0.70 m, and shoots an
asked by Jennifer on December 13, 2007
65. ## Physics Classical Mechanics
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m=
asked by Teresa on December 8, 2013
66. ## PHYSICS(HELP!!)
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m=
asked by Anonymous on December 8, 2013
67. ## Classical Mechanics Physics
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m=
asked by Anonymous on December 6, 2013
68. ## physics(HELP)
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m=
asked by Anonymous on December 8, 2013
69. ## PHYSICS(ELENA
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m=
asked by Anonymous on December 8, 2013
70. ## PHYSICS!!! HELP
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m=
asked by Anonymous on December 8, 2013
71. ## Physics-Mechaincs
A rowboat crosses a river with a velocity of 1.01 m/s at an angle 30◦ North of West relative to the water. Water’s frame of reference: speed of boat: 1.01 m/s direction from current: 30◦ The river is 322mwide and carries a current of 1.63 m/s due
asked by Genevieve on September 19, 2010
72. ## Physics
A conical pendulum is constructed from a rope of length l and negligible mass, which is suspended from a fixed pivot attached to the ceiling. A small ball of mass m is attached to the lower end of the rope. The ball moves in a circle with constant speed in
asked by Anon on October 10, 2013
73. ## Physics Help!!!! Please check
A 3.00 kHz tone is being produced by a speaker with a diameter of 0.150 m. The air temperature changes from 0 to 29°C. Assuming air to be an ideal gas, find the change in the diffraction angle . Please tell me where I am going wrong. first I converted the
asked by Mary on May 2, 2007
74. ## physics
1)A hypothetical neutral atom has 77 electrons. One of its electrons is held in circular orbit of radius r = 0.244 nm by electrostatic attraction between the electron and the nucleus. Assume: the interaction between electrons is negligible. Calculate the
asked by physicshand on August 31, 2015
75. ## physics
A 2.00 kg ball moving to the right at 10.0 m/s makes an off-center collision with a stationary 3.00 kg ball. After the collision, the 2.00 kg ball is deflected upward at an angle of 30o from its original direction of motion and the 3.0 kg ball is moving at
asked by brett on March 25, 2016
76. ## Science
Daltons idea that atoms cannot be divided into smaller parts was was disproved by the discovery of______? The description of the structure of the atom is called the________? In a water molecule, two atoms of hydrogen and one atom of oxygen shared
asked by Summer Moriarty on April 18, 2016
77. ## PHYSICS- PROJECTILE MOTION
A physics book slides off a horizontal table top with a speed of 1.90 m/s. It strikes the floor after a time of 0.410 s. Ignore air resistance. a) Find the magnitude of the horizontal component of the book's velocity just before the book reaches the floor.
asked by SAM on October 20, 2016
78. ## Physics
A ball rolls on a circular track of radius 0.65 m with a constant angular speed of 1.2 rad/s in the counterclockwise direction. Part A If the angular position of the ball at t= 0 is theta= 0, find the x component of the ball's position at the time 2.6 s.
asked by Cooper on November 17, 2011
79. ## physics
A curve of radius 60m is banked so that a car traveling with uniform speed 70km/hr can round the curve without relying on friction to keep it from slipping to its left or right. The acceleration of gravity is 9.8m/s^2. What the the Angle of the curve? im
asked by puff puff on September 17, 2006
80. ## physics
The physics of satellite motion around Jupiter. A satellite of mass 2.00 x 104 kg is placed in orbit 6.00 x 105 m above the surface of Jupiter. Please refer to the data table for planetary motion included in this lesson. a) Determine the force of
asked by aly on April 24, 2011
81. ## Physics
Two carts, one of mass 2kg and another of mass 4kg are driving towards each other with a speed 3m/s. Assume the mass 2kg is moving to the right. Which cart has the most momentum? What is the total momentum of the system? After the collision, what is the
asked by Physics on February 13, 2018
82. ## Physics
A curve of radius 60m is banked so that a car traveling with uniform speed 70km/hr can round the curve without relying on friction to keep it from slipping to its left or right. The acceleration of gravity is 9.8m/s^2. What the the Angle of the curve? im
asked by puff puff on September 17, 2006
83. ## physics
a hot-air balloon having initial velocity vi as it ascends. Then, a water balloon is dropped from the basket when it is 135 meters high and would hit the ground after 9.00 seconds. a) find the velocity of the hot-air balloon upon release of the water
asked by lib on October 2, 2012
84. ## math
An airplane leaves Skyharbor airport at 8:30 PM. His average flight speed is 585 mph. He travels at 43 degrees east of north. The first half of his flight he encounters a cross wind of 72 mph at a direction of 70 degrees west of north. The second half of
asked by Joe on May 10, 2019
85. ## physics
1. It has been suggested that rotating cylinders about 11 mi long and 5.2 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall
asked by Elisa on July 11, 2008
86. ## physics
An object in the vacuum of space orbits the earth at a fixed speed in a circular orbit several hundred miles above the earth. What can we conclude about the reaction force? a) That there is no reaction force-the net force on the object is zero, so the
asked by Maggie on November 22, 2009
87. ## physics
An object in the vacuum of space orbits the earth at a fixed speed in a circular orbit several hundred miles above the earth. What can we conclude about the reaction force? a) That there is no reaction force-the net force on the object is zero, so the
asked by Maggie on November 22, 2009
88. ## physics
An object in the vacuum of space orbits the earth at a fixed speed in a circular orbit several hundred miles above the earth. What can we conclude about the reaction force? Is A the correct answer? a) That there is no reaction force-the net force on the
asked by ronnie on November 23, 2009
89. ## math
A fishing boat leaves port at 11 miles per hour at a bearing of 210 degrees for 2 hours, then turns to a bearing of 250 degrees at 8 miles per hour for 4 hours, and finally changes to a bearing of 280 degrees at 7 miles per hour for 2 hours. At this point,
asked by Shruti on November 7, 2012
90. ## physics
One of the methods used to train astronauts for the effects of "zero gravity" in space is to put them in a specially equipped plane which has been stripped of seats and fitted with padded walls. The pilot then takes the plane up to an altitude of typically
asked by afia on September 22, 2014
91. ## College Physics
A uniform rod of mass M = 437.0 g and length L = 37.1 cm stands vertically on a horizontal table. It is released from rest to fall. A)Which of the following forces are acting on the rod? B)Calculate the angular speed of the rod as it makes an angle θ =
asked by Kimmy on November 3, 2013
92. ## Help!! With 2 math questions!!
There are 36 inches in one yard. How many yards and inches are there in 3785 inches? I need the yards, and inches! The next question is....Find the number of quarts in 73,458 cubic inches of water. There are 231 cubic inches in 1 gallon. Help~*Nikki
asked by Nichoel on August 17, 2005
93. ## mayh
A ship travels at a constant speed of 25 kilometres per hour (Kph) in a straight line from port A located at position (x_A, y_A) =(-100, -100) to port B at (x_B ,y_B )=(300 ,100). A) Find parametric equation for the line of travel of the ship .your
asked by emy on February 16, 2011
The infamous Wile E. Coyote has been chasing the Roadrunner again. His latest idea for a trap involved launching an enormous boulder (about 16ft in diameter) up to a height of 750 ft, hoping that it would come down crashing on the unsuspecting Roadrunner.
asked by Kim on March 12, 2013
95. ## Psychology
Could someone see if I got any of the questions wrong. I feel that some questions have multiple answers but there is only one that is correct. Please help. thanks a lot Cognitive Development Respond to the following statements using Piaget’s Stages of
asked by Mike on November 28, 2009
96. ## Physics
A pilot of mass 65 kg in a jet aircraft makes a complete vertical circle in mid-air. The vertical circle has a radius of 1.70 km at a constant speed of 215 m/s. Determine the force of the seat on the pilot at (a) the bottom of the loop and (b) the top of
asked by Gina on October 28, 2009
97. ## Physics
A pilot of mass 65 kg in a jet aircraft makes a complete vertical circle in mid-air. The vertical circle has a radius of 1.70 km at a constant speed of 215 m/s. Determine the force of the seat on the pilot at (a) the bottom of the loop and (b) the top of
asked by Rachel on October 29, 2009
98. ## CULTURAL DIVERSITY
Can anyone please read over my assignment and let me know if there are any grammar or spelling errors? CheckPoint: Modern Challenges in Immigration o Should United States government policy favor certain kinds of immigrants? o Should [citizenship]
asked by Christina on June 15, 2007
99. ## Physics - check
A ball is thrown at a speed of 39.05m/s at an angle of 39.8degrees above the horizontal. Determine: V1perpendicular component: V1Sin(theta) = 25m/s V1paralell component: V1Cos(theta) = 30m/s a) the ball's location 2s after being thrown. V2paralell =
asked by Anonymous on March 13, 2008
100. ## Physics - check
A ball is thrown at a speed of 39.05m/s at an angle of 39.8degrees above the horizontal. Determine: V1perpendicular component: V1Sin(theta) = 25m/s V1paralell component: V1Cos(theta) = 30m/s a) the ball's location 2s after being thrown. V2paralell =
asked by Anonymous on March 15, 2008 | 8,349 | 30,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-30 | latest | en | 0.929605 |
https://discuss.codechef.com/t/doubt-in-monster-jan-long/17835 | 1,632,707,430,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00155.warc.gz | 257,128,425 | 7,213 | # doubt in MONSTER jan long.
we can do SOS DP and then apply sqrt decomposition. (https://discuss.codechef.com/convert_to_question/121319/?node_type=comment) but my doubt is isn’t it exceeding the time limit? (q^(1/2)* (17 * 2^(17))) ~ 2^30?
I have read sos dp through cf blog O(n*2^n).
UPD : got accepted https://www.codechef.com/viewsolution/17055899 : i just change the implementation from dp[i][j] to h[j] with same complexity! Now, more curious to know what’s actually happening?
The sos function is the slowest part of the solution, and it outweigths the other operations done in one bucket. Thus you should call sos function less times, and do the other calculations more, so change the size of the bucket from \sqrt{Q}=512 to something bigger, like 2-3000.
I didn’t go through your entire code, but it made my solution pass, and probably it will make yours too.
1 Like
That was the problem that I was facing. This is what I did :
Instead of choosing the block size ‘b’ as sqrt(q), I tried to find a value that will minimise the complexity (this is what I do usually for choosing the best block size for problems, if the TL is strict). First, we will write down the expression for the time complexity in terms of the block size ‘b’.
Lets see what are all the operations that we will be doing. We have to use SOS DP for each of the blocks. The complexity for this part is (17 * q * n)/b. Then, we need to go through all the blocks for each of the elements, and see in which block it gets killed. The complexity of this part is n * (n/b). And, if a monster gets killed in some block, you need to go through each query in the block to find exactly where it gets killed. The complexity for this part is n * b. Overall the expression for the time complexity is :
(17 * q * n)/b + n * (n/b) + n * b. If you differentiate this expression with respect to ‘b’, and equate it to 0, you will find that b = sqrt(17 * q + n). With this as the block size, the number of operations will be around 2^28.
One thing to keep in mind is that you can never be sure of a solution getting TLE based on the number of operations that you have calculated. It also depends on the type of the operations. In case of this problem, I think it was fast because there were bitwise operations involved, they are fast. I wasn’t sure that my solution was going to get accepted, but luckily it did.
Another thing is that sometimes, the worst case that you calculate might never be encountered (reasons might be you calculating the complexity incorrectly or the test cases being weak). This happened to me when I was trying to solve this problem:https://www.codechef.com/JAN17/problems/DIGITSEP/, The complexity I calculated should have given me a TLE for sure, but I was shocked to find that it was the second fastest submission for the problem. So, when in doubt, just submit and see.
2 Likes
Yeah the same thing happened with me (accessing 2d array takes much more time than accessing 1 d),I too am curious to know the reason.
cash miss occurs more in 2d array.
Everytime I am not able to crack the question
I saw that SOS DP takes O(n*2^n).
The complexity I calculated should have given me a TLE for sure so I was thinking thinking thinking… and implemented.
BTW can it be solved using BIT?
can anyone pls tell me how sos dp is helping in solving the problem ?after doing sos what r we getting?
i have no idea about sos dp…
@pk301 @avi224
@worldunique have a look : http://codeforces.com/blog/entry/45223
1 Like
gives me TLE : https://www.codechef.com/viewsolution/17054790
no, it’s not working i have changed it to 3000 but still TLE
thanks for your answer but i don’t know why my 2D array submission is still not passing. If possible please have a look : https://www.codechef.com/viewsolution/17056174
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I think its because of cache hits. I remember reading somewhere that in case of 2D arrays, access time is fast if consecutive accesses to memory are made to elements within the same row (it seems that this is because they are in the same block of memory). Try changing your array from dp[(1 << 18) + 2][18] to dp[18][(1<<18)+2] and let me know if it gets accepted.
1 Like
The SOS DP complexity O(n*2^n) is for array of size 2^n, not n. And no, it can’t be solved using BIT.
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thanks @hemanth_1 it works pretty well (even faster than my 1D array solution lol) :https://www.codechef.com/viewsolution/17060096
If possible please explain this a little more " access time is fast if consecutive accesses to memory are made to elements within the same row." in context to this solution.(as i am using dp[j][i-1] and dp[j][i] so, how access time differs? as in both cases we need to access both the rows!).
2D arrays are stored in memory in row major order. In your previous submission, your array was :dp[mask][i]. You loop over all the masks in the i-th and i-1th layer in a single iteration of SOS DP right ? when looping over masks, you go through all the rows of your 2D array, ie., for every consecutive access to memory, the locations are 17 positions apart. But if you change it to dp[i][mask], you will be accessing memory sequentially.
1 Like
Although, you do make a valid point, even I’m not sure about why its fast even though we are using the i-1th row. I’m guessing that since the entire previous row was used in the previous iteration, most of its locations are in the cache. | 1,334 | 5,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-39 | latest | en | 0.936499 |
https://stats.stackexchange.com/questions/218055/covariance-of-two-variables-that-are-products-of-shared-random-variables/218151 | 1,566,677,774,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00327.warc.gz | 635,934,966 | 31,648 | # Covariance of two variables that are products of shared random variables
How to analytically express cov(X,Y), when:
X=C*A/(A+B)
and
Y=C*B/(A+B)
Here C, A and B are independent variables with normal distributions.
More specifically I would like to express cov(X,Y) using the expected means of each of the input variables C, A and B, as well as their standard deviations or variances.
A is a normal distribution with an expected mean E(A) and a variance var(A)
B is a normal distribution with an expected mean E(B) and a variance var(B)
C is a normal distribution with an expected mean E(C) and a variance var(C)
Thank you so much in advance!
• if this is HW please add the self-study tag – Antoine Jun 9 '16 at 9:42
• Thank you for asking but it is for my research at work. I have really searched a lot the last few days for the answer, but I can only find examples where X and Y are linear combinations of random variables like X=aA+bB+cC (with small a, b and c as constants) but I cannot find solutions where X is a function that is the product of a combination of A, B and C. – Astrid Marie Jun 9 '16 at 10:18
• I think that would help if you could show some of those examples you refer to and then explain how it is different in your case. – Antoine Jun 9 '16 at 10:27
• Thanks for the tip. The question ressembles a bit this stats.stackexchange.com/questions/137571/covariance-of-product , but the example in the link is in matrix notation and it is not clear for me if Xi and Xj in the example are correlated, so I cannot used this directly. – Astrid Marie Jun 9 '16 at 10:36
• Given that this involves the ratio of Normals, convergence may be an issue – wolfies Jun 9 '16 at 11:24
One can find a general solution, without assuming Normality. In particular, if $A$, $B$ and $C$ are independent, and noting that the Covariance operator is the {1,1} central moment, then $\text{Cov}( \frac{C A}{A+B},\frac{C B}{A+B})$ is:
where I am using a developmental version of the CentralMomentToCentral function in mathStatica (alas, not in any public release yet).
The reason the solution does not converge (assuming Normal parents) is because $A+B$ is Normal, and thus $E[\frac{1}{A+B}]$ is the expectation of the inverse of a Normal variable, which is known not to converge (see, for instance, Mean and variance of inverse of a normal RV) | 606 | 2,350 | {"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} | 3.453125 | 3 | CC-MAIN-2019-35 | latest | en | 0.914687 |
https://forums.wolfram.com/mathgroup/archive/2008/Jul/msg00277.html | 1,624,441,783,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00434.warc.gz | 239,307,088 | 7,221 | Solving a DE using Mathematica
• To: mathgroup at smc.vnet.net
• Subject: [mg90424] Solving a DE using Mathematica
• From: Greg <starwar636 at aol.com>
• Date: Wed, 9 Jul 2008 04:52:25 -0400 (EDT)
```I'm having problems solving this problem although it should appear pretty straightfoward:
(-l^2 - m^2 + n^2/(r + 0.016 z)^2) Z[z] + Z''[z] == 0
I am solving for Z[z]. These are the lines I use:
DSolve[Above Equation, Z[z], z]
I get an odd solution so I do a solution check plugging back in Z[z] and Z''[z] yet I don't get 0. In the above, l,m,n,r are all constants.
```
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https://asicdigitaldesign.wordpress.com/2008/01/31/ultimate-technical-interview-question-take-2/ | 1,553,260,355,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202658.65/warc/CC-MAIN-20190322115048-20190322141048-00335.warc.gz | 418,716,245 | 15,597 | Ultimate Technical Interview Question – Take 2
January 31, 2008
Allow me to quote from Martin Gardner’s excellent, excellent book Mathematical Carnival (chapter 17):
`When a mathematical puzzle is found to contain a major flaw - when the answer is wrong, when there is no answer, or when, contrary to claims, there is more than one answer or a better answer - the puzzle is said to be "cooked". `
From the number of hits, it looks like the last post was quite popular. Therefore, I decided to give the problem some more thought and to try to find more minimal solutions – or as defined in the above quote “to cook this problem”.
My initial hunch was to try and utilize an SR latch somehow. After all it is a memory element for the price of only two gates. I just had a feeling there is someway to do it like that.
I decided to leave the count-to-3 circuitry, cause if we want to do a divide by 3, we somehow have to count…
Here is what I first came up with:
The basic idea is to use the LSB of the counter to set the SR flop and to reset the SR flop with a combination of some states and the low clock.
Here is the timing diagram that corresponds to the circuit above.
But! not everything is bright. The timing diagram is not marked red for nothing.
In an ideal world the propagation time through the bottom NOR gate would be zero. This would mean that exactly when the S pin of the SR latch goes high the R pin of the flop goes low – which means both pins are never high at the same time. Just as a reminder, if both inputs of an SR latch are high, we get a race condition and the outputs can toggle – not something you want on your clock signal. Back to the circuit… In our case, the propagation time through the bottom NOR gate is not zero, and the S pin of the latch will first go high, then – only after some time, the R pin will go low. In other words we will have on overlap time where both R and S pin of the latch will be high.
Looking back at the waveform, it would be nice if we could eliminate the second pulse in each set of two pulses on the R pin of the latch (marked as a * on the waveform). This means we just have to use the pulse which occurs during the “00” state of the counter.
This is easy enough, since we have to use the “00” from the counter and the “0” from the clock itself – this is just the logic for a 3 input NOR gate!
The complete and corrected circuit looks like this now:
And the corresponding waveform below. Notice how the S and R inputs of the SR latch are not overlapping. | 573 | 2,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-13 | latest | en | 0.962128 |
http://www.algebra.com/algebra/homework/Human-and-algebraic-language/Human-and-algebraic-language.faq.question.57092.html | 1,386,215,673,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163039753/warc/CC-MAIN-20131204131719-00007-ip-10-33-133-15.ec2.internal.warc.gz | 219,555,263 | 4,889 | # SOLUTION: Translate the following statement into an algebraic equation. Let x represent the number. 5 more than 13 times a number is 8 times that same number.
Algebra -> Human-and-algebraic-language -> SOLUTION: Translate the following statement into an algebraic equation. Let x represent the number. 5 more than 13 times a number is 8 times that same number. Log On
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Click here to see ALL problems on Human-and-algebraic-language Question 57092: Translate the following statement into an algebraic equation. Let x represent the number. 5 more than 13 times a number is 8 times that same number.Answer by funmath(2926) (Show Source): You can put this solution on YOUR website!Translate the following statement into an algebraic equation. Let x represent the number. 5 more than 13 times a number is 8 times that same number. "5 more than" is 5+ "13 times a number" is 13*x=13x "is" means = "8 times that same number" is 8*x=8x The equation is: 13x+5=8x Happy Calculating!!! | 304 | 1,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2013-48 | latest | en | 0.870039 |
https://www.doubtnut.com/qna/237056851 | 1,716,470,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00414.warc.gz | 650,918,886 | 49,670 | # Given the following equations and values, determine the enthalpy of reaction at 298 K for reaction: C2H4(g)+6F2(g)→2CF4(g)+4HF(g) H2(g)+F2(g)→2HF(g) ΔH∘1=−537kJ C(s)+2F2(g)→CF4(g) ΔH∘2=−680kJ 2C(s)+2H2(g)→C2H4(g) ΔH∘3=52kJ
A
1165kJ
B
+1165kJ
C
2486kJ
D
+2486kJ
Video Solution
Text Solution
Verified by Experts
## H2(g)+F2(g)→2HF(g)ΔH01=−537kJC(s)+2F2(g)roCF4(g)ΔH02=−680kJ2C(s)+2H2(g)→C2H4(g)ΔH03=52kJC2H4(g)+6F2(g)→2CF4(g)+4HF(g)ΔH={−ΔH03+2ΔH02+2ΔH01} ΔH∘=−52+2×−680+2×−537=−2486kJ.
|
Updated on:21/07/2023
### Similar Practice Problems
• Question 1 - Select One
## Given the following equations and ΔH∘ values, determine the enthalpy of reaction at 298 K for the reaction : C2H4(g)+6F2(g)→2HF4(g)+4HF(g) H2(g)+F2(g)→2HF(g)+2HF(g), ΔH∘1=−537 kJ C(s)+2F2(g)→CF4(g), ΔH∘2=−680 kJ 2C(s)+2H2(g)→C2H4(g), ΔH∘3=52 kJ
A1165
B2382
C+1165
D+2382
• Question 1 - Select One
## Calculate in kJ for the following reaction : C(g)+O2(g)→CO2(g) Given that, H2O(g)+C(g)+H2(g),ΔH=+131kJ CO(g)+12O2(g)→CO2(g), ΔH=−242kJ H2(g)+12O2(g)→H2O(g), ΔH=−242kJ
A393
B+393
C+655
D655
• Question 1 - Select One
## Which is the heat of reaction for the following reaction: CH4(g)+NH3(g)→3H2(g)+HCN(g) Use the following thermodynamic data in kJ/mol. N2(g)+3H2(g)→2NH3(g),ΔrH∘=−91.8 C(s)+2H2(g)→CH4(g),ΔrH∘=+74.9 H2(g)+2C(s)+N2(g)→2HCN(g),ΔrH∘=261.0
A299.3 kJ
B256.0 kJ
C149.5kJ
D101.5kJ
• Question 1 - Select One
## For the reaction C2H4(g)+3O2(g)→2CO2(g)+2H2O(l), Delta E=-1415 kJ.TheDeltaHat27^(@)C is
A1410kJ
B1420kJ
C+1420kJ
D+1410kJ
• Question 1 - Select One
## For the reaction C3H8(g)+5O2)(g)→2CO2(g)+4H2O(l) at constant temperature , ΔH−ΔE is
ART
B+RT
C3RT
D+3RT
• Question 1 - Select One
## Calculate the enthalpy change for the reaction C2H4(g)+H2(g)→C2H6(g) using the data given below : C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)ΔH=−1415kJ C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)ΔH=−1566kJ H2(g)+12O2(g)→H2O(l)ΔH=−286kJ
A437 kJ
B+35 kJ
C135 kJ
Dnone of these
• Question 1 - Select One
## For the equations C(diamond)+2H2(g)→CH4(g)ΔH1 C(g)+4H(g)→CH4(g)ΔH2 Predict whther
AΔH1=ΔH2
BΔH1>ΔH2
CΔH1<ΔH2
DΔH1=ΔH2+ΔvapH(C)+ΔdissH(H2)
• Question 1 - Select One or More
## Heat of formation of CH4 are: If given heat: C(s)+O2(g)→CO2(g) ΔH=−394KJ 2H2(g)+O2(g)→2H2O(l)→2H2O(l) ΔH=−394KJ CH4(g)+2O2(g)→CO2(g)+2H2O(l) ΔH=−394KJ
A70KJ
B16.7Kcal
C244KJ
D50Kcal
### Similar Questions
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 1,453 | 3,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-22 | latest | en | 0.498023 |
https://math.stackexchange.com/questions/1595132/how-to-prove-that-series-sum-1n-sin4n-sqrtn-converges | 1,571,881,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987838289.72/warc/CC-MAIN-20191024012613-20191024040113-00458.warc.gz | 602,585,140 | 31,861 | # How to prove that series $\sum (-1)^n\sin^4n /\sqrt{n}$ converges?
I have a series: $$\sum_{n=1}^\infty(-1)^n\frac{\sin^4n}{\sqrt n}.$$ How can we prove that it converges?
Usually, with $\sin^4n$ we would use Comparison Test, but it only applies when the terms are nonnegative.
• tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx – TomGrubb Dec 31 '15 at 14:35
• @bburGsamohT Is that function strictly decreasing? You need that for the AST. – Gregory Grant Dec 31 '15 at 14:36
• @GregoryGrant Compare it with the series $\sum(-1)^n\frac{1}{\sqrt{n}}$ – TomGrubb Dec 31 '15 at 14:37
• If $\alpha$ is not a multiple of $2\pi$, then the partial sums $\sum_{n = 1}^N e^{i\alpha n}$ are bounded. $(-1)^n\sin^4 n = \frac{e^{i\pi n}}{16}(e^{in} - e^{-in})^4$. Ask Dirichlet. – Daniel Fischer Dec 31 '15 at 14:43
• @bburGsamohT we can compare series that have only non-negative terms. – niar_q Dec 31 '15 at 14:47
Hint: Noting that $$\sin^4n=\frac{1}{8}(3-4\cos(2n)+\cos(4n))$$ you have \begin{eqnarray} \sum_{n=1}^\infty(-1)^n\frac{\sin^4n}{\sqrt n}&=&\frac{3}{8}\sum_{n=1}^\infty(-1)^n\frac{1}{\sqrt n}-\frac{1}{2}\sum_{n=1}^\infty(-1)^n\frac{\cos(2n)}{\sqrt n}+\frac{1}{8}\sum_{n=1}^\infty(-1)^n\frac{\cos(4n)}{\sqrt n}. \end{eqnarray} Now you can do the rest to show that $\sum_{n=1}^\infty(-1)^n\frac{\cos(2n)}{\sqrt n}$ and $\sum_{n=1}^\infty(-1)^n\frac{\cos(4n)}{\sqrt n}$ are convergent. | 570 | 1,405 | {"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} | 4.15625 | 4 | CC-MAIN-2019-43 | latest | en | 0.60638 |
https://www.meritnation.com/ask-answer/question/find-the-probability-that-a-number-selected-at-random-from-t/probability/3285990 | 1,618,923,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00263.warc.gz | 985,317,534 | 11,088 | # Find the probability that a number selected at random from the numbers 1, 2,3,……… 34, 35 is a) a prime number b ) is a multiple of 7 c ) a multiple of 2 and 5 d) a multiple of 2 or 3
Given set of numbers are: 1, 2, 3, 4, ........, 34, 35.
a)
Let A be the event of selecting a prime number out of given set of numbers.
Favourable outcomes to event A = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31
Therefore, number of favourable outcomes = 11
Total number of outcomes = 35
b)
Let B be the event of selecting a multiple of 7.
Favourable outcomes to event B = 7, 14, 21, 28 and 35
Therefore, number of favourable outcomes = 5
Total number of outcomes = 35
c)
Let C be the event of selecting a number which is multiple of 2 and 5.
Favourable outcomes to event C = 10, 20 and 30
Therefore, number of favourable outcomes = 3
Total number of outcomes = 35
d)
Let D be the event of selecting a number which is multiple of 2 or 3.
Favourable outcomes to event D = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 3, 9, 15, 21, 27 and 33.
Therefore, number of favourable outcomes = 23
Total number of outcomes = 35
• 17
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axelrose has asked for the wisdom of the Perl Monks concerning the following question:
I'd like to solve a special sorting task with Perl and wonder if this is a standard computer science problem.
Say I have three slots 1..3 and an empty slot 0
slot 1 => c
slot 2 => a
slot 3 => b
slot 0 => 0
The slots hold a labeled object (it's a tape library with magazine slots)
The goal is
slot 1 => a
slot 2 => b
slot 3 => c
and a list of move actions where ideally the number of moves is minimal.
A pragmatic solution is for instance
move( 1, 0 ); # ( 0, a, b, c )
move( 2, 1 ); # ( a, 0, b, c )
move( 3, 2 ); # ( a, b, 0, c )
move( 0, 3 ); # goal order achieved
where the function move( <from-slot>, <to-slot> ) moves an object from one slot to a different slot.
The question is not so much how to implement this in Perl but whether there exists an known algorithm. Of course I'm happy to post a Perl solution here.
Many thanks, Axel.
Replies are listed 'Best First'.
Re: sort with fewest moves
by chipmunk (Parson) on Feb 10, 2002 at 17:02 UTC
Here's an algorithm to find a solution using the shortest possible number of moves. It assumes that you know the proper order for the tapes, which you should, because figuring that out requires zero moves.
1. Starting at slot 1, find the first tape that is not in the right slot.
2. Move that tape to slot 0.
3. Since that tape was not in the right slot, there is another tape which belongs in that slot. Find that tape and move it into the right slot.
4. Repeat the previous step until the tape in slot 0 is moved into the right slot.
If there are N tapes that are not in the right slots, clearly each tape must be moved at least once, for a minimum of N moves. However, in order to move a tape, the destination slot must be empty, so one extra move is required to first move a tape into slot 0. Therefore, the above algorithm, which takes N+1 steps, finds a shortest solution.
This algorithm is not sufficent. Take for example, this arrangement of tapes: d c b a 0. Following your algorithm the tapes would be moved like so:
1. 0 c b a d
2. a c b 0 d
3. a c b d 0
... then it would stop since the tape that was in slot 0 was just moved to the right slot.
You could work around this, by after moving the tape in slot 0 to the right slot, starting from step 1 again (until all tapes are in order).
I apoligize in advance if any of the following contains an error, which it very well may.
If the removal of tape T causes the movement of a certain set of tapes "M" before tape T is replaced in the correct location, than the movement of any item in "M" must cause the movement of tape T and every other item in "M" before the replacement of tape T. (Attempt at proof if you don't find this obvious: Every item in set M either determines the movement of another tape in set M or tape T. Since each tape determines the movement of one and only one tape, let us order the set M starting with the tape that tape T determines the removal of, and continuing with the tape that determines the removal of, etc. The last element in set M must determine the removal of tape T, since that would stop the building of the set after starting from tape T (since tape T would then be moved back from the "free" slot to which it was removed). Thus, if we remove a tape in set M, it determines the removal of every tape after it in set M, and then determines the removal of tape T, which then determines the removal of every item before it in set M, which then determines the replacement of that tape in set M.)
(update: In case it's not obvious to you, note that this means that it doesn't matter which element we start at. The entire set of tapes can be divided into groups that we can remove one item to the "free" slot and organize by moving items around in the sequence dictated. To sort the whole sequence using our method of always moving things to where they belong, we must move any one item from each of these groups to the free slot and proceed. But, since these groups are independent it doesn't matter what we group we start with, and since we will not tapes already in the right place (if we didn't that would obviously add extra, unnecessary moves), we will pick each group exactly once...)
Also, moving any tape to a slot that is not where it should end up, besides the movement to the free slot to allow the initial movement of a bunch of tapes will not result in less moves being needed. If we move a tape T into a slot where it does not belong, either:
1. the set of tapes, M, whose movement is determined by T will all be allowed to move (without incuring an extra move to move something in/out of the free slot), but after any of this movement then require us to move tape T to the slot where it belongs which would be equivilent to moving tape T to the free slot, thus not saving any moves).
2. or, tape T has been moved to a place in the correct location for one of the tapes in set M, and thus, tape T less tapes then are in set M will allowed to be moved (without incuring an extra move, as before).
All other algorithms to solve this problem would either choose to not move tapes to the position where they will eventually end up or will choose a different tape to move initially. By the stuff above, those other ways eithe result in the same number of moves, or more moves, so this is the shortest way to solve the problem.
Re: sort with fewest moves
by jlongino (Parson) on Feb 10, 2002 at 17:21 UTC
If this were homework (but of course it's not), I imagine a teacher would be very impressed if one of their students developed a uniquely Perl solution. Instead of trying to move elements around one at a time, come up with an algorithm to swap at least two elements.
For example, with Perl you could solve this specific problem with the following two statements:
```(\$slots[1], \$slots[2]) = (\$slots[2], \$slots[1]);
(\$slots[2], \$slots[3]) = (\$slots[3], \$slots[2]);
and, of course, with one statement:
```(\$slots[1], \$slots[2], \$slots[3]) = (\$slots[2], \$slots[3], \$slots[1]);
Coming up with an algorithm for two in-place swaps shouldn't be too difficult (I've already shown you the Perl idiom), ++ if you can handle more than two elements at a time.
--Jim
Update: Well, that's the algorithm part I alluded to, your list of moves for the 2-element swap would look like this:
• swap(1, 2)
• swap(2, 3)
Granted trying to notate a 3 at-a-time swap would be difficult, if even possible (thus the ++).
If we'd known that you had to use a one-armed robot to begin with, the replies might have been more useful. ;)
Do you mean something like
```@slots = @slots[2,1,4,3]
It looks nice but I really need a list of move actions for the robot.
Re: sort with fewest moves (Bose-Nelson algorithm)
by grinder (Bishop) on Feb 11, 2002 at 10:42 UTC
Algorithms that do this that have been known about since the 1960s. The most well-known one is the Bose-Nelson sort (although trudging through web pages, I seem to be coming across Batcher's Odd-Even Merge Sort more often).
Interestingly enough, there seems to be scant details available on the net. Google doesn't turn up much. I read about this technique in a long-lost issue of Computer Language.
The only link halfway useful that I have found is www.cs.brandeis.edu/~hugues/sorting_networks.html ... I think you're going to have to dig out a copy of Knuth volume III - Searching and Sorting. <update> Hmm, I just did. The Knuth, as usual, is heavy on mathematics and short on algorithms. I still can't find any code to help you. It's in section 5.3.1 (Minimum-comparison sorting) for what it's worth.</update>
The algorithm essentially accepts a single number (the number of elements to sort), and then spits out a series of pairs, which are the indices of the elements to compare. And it turns out that as some of the comparison (a,b) and (c,d) don't affect each other, you can run them in parallel, thereby reducing the overall time taken.
It is apparently very hard to generate a minimal number of comparisons. These days people are attacking the problem through Genetic Algorithm (GA) techniques.
Some more links. Using sorting networks reveals more hits.
print@_{sort keys %_},\$/if%_=split//,'= & *a?b:e\f/h^h!j+n,o@o;r\$s-t%t#u'
I don't think sorting networks will work here.
If I remember correctly, (and I'm not certain I ever really understood them) they're built on the principle of swaps, and while you can definitely think of a move as a swap where one of the items is empty, a sorting network would suggest solutions that aren't possible.
For example a if the input was (0,2,1) the solution you would get is swap(1,2) -- but that's not possible. in this case only swaps in which one parameter is currently "0" are valid.
It really sounds like a Game Playing problem ... here's a recursive solution that tries all the "smart" moves and figures out which one leads to the correct order in the minimal number of moves.
If a "close to optimal" solution is good enough, then pick_move could be modified to use Alpha Beta Pruning to figure out which of the "smart" moves looks like it's the smartest. I would guess a good scoring method would award one point to for each tape in the correct slot, and half a point if there's a tape in slot 0. (in which case something else is empty, and can be filled directly)
```#!/usr/bin/perl -wl
use strict;
sub smart_moves {
my @slots = @{ shift(@_) };
my @from;
my \$to;
for (my \$i = 0; \$i < scalar(@slots); \$i++) {
if (0 != \$i and 0 == \$slots[\$slots[\$i]]) {
# only one smart move if the home for the tape
# in slot i is empty
return ( [ \$i, \$slots[\$i] ] );
}
if (0 == \$slots[\$i]) {
\$to = \$i;
next;
}
# don't move anything that's allready 'home'
push @from, \$i unless \$i eq \$slots[\$i];
}
return map { [\$_ , \$to ] } @from;
}
sub make_move {
# returns the new @slots after the move
my @slots = @{shift(@_)};
my @move = @{shift(@_)};
\$slots[\$move[1]] = \$slots[\$move[0]];
\$slots[\$move[0]] = 0;
return @slots;
}
sub pick_move {
my @slots = @{shift(@_)}; # current configuration
my @history = @{shift(@_)}; # moves made so far
my @moves = smart_moves(\@slots);
return @history if 0 == scalar @moves;
my @best;
foreach (@moves) {
my @s = make_move \@slots, \$_;
my @h = @history; # copy it
push @h, \$_;
my @result = pick_move(\@s, \@h);
if (0 == scalar(@best) || scalar(@result) <= scalar(@best)) {
@best = @result;
}
}
return @best;
}
my @slots = @ARGV;
my @done = pick_move(\@slots, []);
foreach (@done) {
print join(",", @slots) . "\t\$_->[0] => \$_->[1]";
@slots = make_move(\@slots,\$_);
}
print join(",", @slots)
__END__
laptop:~> monk.pl 0 2 1
0,2,1 2 => 0
1,2,0 1 => 2
1,0,2 0 => 1
0,1,2
laptop:~> monk.pl 0 1 2
0,1,2
laptop:~> monk.pl 0 2 1 4 5 3
0,2,1,4,5,3 5 => 0
3,2,1,4,5,0 4 => 5
3,2,1,4,0,5 3 => 4
3,2,1,0,4,5 2 => 3
3,2,0,1,4,5 1 => 2
3,0,2,1,4,5 3 => 1
3,1,2,0,4,5 0 => 3
0,1,2,3,4,5
laptop:~> monk.pl 0 2 1 7 8 9 5 4 3 6
0,2,1,7,8,9,5,4,3,6 9 => 0
6,2,1,7,8,9,5,4,3,0 5 => 9
6,2,1,7,8,0,5,4,3,9 6 => 5
6,2,1,7,8,5,0,4,3,9 8 => 6
6,2,1,7,8,5,3,4,0,9 4 => 8
6,2,1,7,0,5,3,4,8,9 7 => 4
6,2,1,7,4,5,3,0,8,9 3 => 7
6,2,1,0,4,5,3,7,8,9 6 => 3
6,2,1,3,4,5,0,7,8,9 2 => 6
6,2,0,3,4,5,1,7,8,9 1 => 2
6,0,2,3,4,5,1,7,8,9 6 => 1
6,1,2,3,4,5,0,7,8,9 0 => 6
0,1,2,3,4,5,6,7,8,9
Many thanks to you and all the others for the many and sound responses!
After all I understand that I should have named it "minimal move algorithm"
I need some time to digest all of this. For the moment I'll leave the monastery for the German Perl workshop which starts tomorrow evening:)
Cheers, Axel.
Re: sort with fewest moves
by belg4mit (Prior) on Feb 10, 2002 at 18:27 UTC
For some reason this reminded me of the [id://Hanoi|Towers of Hanoi]. There are of course some differences (slots<n in Hanoi, no swapping either {jlongino++}), but they do seem to be rather similar.
--
perl -pe "s/\b;([st])/'\1/mg"
Re: sort with fewest moves
by hossman (Prior) on Feb 10, 2002 at 22:15 UTC
The way i'm reading your post is: "I'm looking for an algorithm that will let me sort elements 1-N, given that the only memory available is an array from 1-N, plus a single temp variable, and the only allowed opperation is move(n, m)".
This sounds alot like Bubble Sort to me.
The distinction being that Bubble Sort is usually defined in terms of a swap(a, b) method which is considered atomic. Since swap) can be (and frequently is) defined in terms of two moves that use a temp variable, Bubble Sort can solve your problem given your constraints -- uut it will never come close to a solution with a minimal number of moves. It will only ever call move(a, 0) and move(0, a).
I can't think of any other constant space sorting algorithms (that operate on arrays). But one extremely important question that needs to be answered before you can even try to for finding an optimal algorithm is: how do you define optimal? is move(n,m) the only operation with a "cost" ? what about doing comparispons or slots? Would a solution that analyzed all of the slots in detail first,then built up a list of moves be considered optimal?
(I ask, because you're post only refers to the ideal situation being one in which "the number of moves is minimal.")
In this case, it appears that yes, move(n,m) is the only operation with cost. (I'm not sure if it's cost is constant over all n,m. I think we're supposted to assume that it is.)
In purticular, this is going to be applied to a tape-library servicing robot, which only has one operation: switch the tapes in positions N and M. (move(n,m))
You're right about bubble-sort being a possiblity... but I don't think it's a good one. Remember that it isn't finding the sequence of moves that has to be done in constant space, it's the acautual movement of physical tapes.
TACCTGTTTGAGTGTAACAATCATTCGCTCGGTGTATCCATCTTTG ACACAATGAATCTTTGACTCGAACAATCGTTCGGTCGCTCCGACGC
That's why I'm not clear on the goal, is it:
Find an algorithm, whose performance is inconsequential, that can determine the minimal number of moves to sort the tapes.
Or is it something like:
Find an optimal algorithm for sorting the tapes such that the only atomic operations move(m,n) and examine(n) -- which tells you what tape is in slot n.
If any amount of preprocessing and analysis is allowed, then any number of hueristics could be useful for find a path from the starting order to sorted order.
If nothing else, you can do a breadth first search of all the possible permutations of moves untill you achieve the desired ordering.
> is move(n,m) the only operation with a "cost" ?
Practically yes. Finding the right order is easily achieved with Perl in memory. The cost mainly results from time a robot arm needs to pick and move a tape. (about 30 seconds)
I neglect the time difference for tape moves betwenn different slots. (a maximum of 60 tapes)
> Would a solution that analyzed all of the slots in detail first,then built up a list of moves be considered optimal?
Yes - that's how I want to solve it.
Re: sort with fewest moves
by jlongino (Parson) on Feb 12, 2002 at 08:33 UTC
Rats! I figured someone would beat me to a solution. I haven't looked closely at hossman's. I came up with the solution without looking at any references. Mine has configurable parameters, produces some elementary stats, and is shorter :P. It can be improved I'm sure, but I was just happy to finish it. Specifically, instead of rebuilding the entire hash, the changed key/values could be updated. TIMTOWTDI!
```use strict;
my @slots;
my \$max = 10;
my \$empty = 0;
my \$sum_moves = 0;
my %rlookup;
my \$trials = 3;
my \$min_moves = \$max*\$max;
my \$max_moves = 0;
# initialize @slots
\$slots[\$_] = \$_ for (0..\$max);
srand time;
for my \$ct (1..\$trials) {
# randomize @slots
for (1..\$max) {
my \$rand = int(rand \$max) + 1;
(\$slots[\$_], \$slots[\$rand]) = (\$slots[\$rand], \$slots[\$_]);
}
%rlookup = ();
build_hash();
my \$moves = 0;
my \$unordered = inorder();
while (\$unordered || \$empty) {
print join " ", @slots, " \\$empty: \$empty ";
if (\$slots[0] != 0) {
print "move(\$rlookup{\$empty}, \$empty);\n";
\$slots[\$empty] = \$rlookup{\$slots[\$empty]};
\$slots[\$rlookup{\$empty}] = 0;
\$empty = \$rlookup{\$empty};
} else {
print "move(\$unordered, \$empty);\n";
(\$slots[\$empty], \$slots[\$unordered]) = (\$slots[\$unordered], \$
+slots[\$empty]);
\$empty = \$unordered;
}
\$moves++;
build_hash();
\$unordered = inorder();
}
print join " ", @slots, " \\$empty: \$empty\n";
print "Trial: ", pack("A5", \$ct), " Moves: \$moves\n\n";
\$min_moves = \$moves if \$moves < \$min_moves;
\$max_moves = \$moves if \$moves > \$max_moves;
\$sum_moves += \$moves;
}
print "Number of slots: \$max\n";
print "Trials: \$trials\n";
print "Average Moves: ", \$sum_moves / \$trials, "\n";
print "Minimum Moves: ", \$min_moves, "\n";
print "Maximum Moves: ", \$max_moves, "\n";
sub inorder {
my \$i;
for (1..\$max) {
\$i = \$_;
last if \$_ != \$slots[\$_];
}
return \$i % \$max;
}
sub build_hash {
\$rlookup{\$slots[\$_]} = \$_ for (0..\$max);
}
Sample Output
```0 2 3 7 6 4 10 5 8 9 1 \$empty: 0 move(1, 0);
2 0 3 7 6 4 10 5 8 9 1 \$empty: 1 move(10, 1);
2 1 3 7 6 4 10 5 8 9 0 \$empty: 10 move(6, 10);
2 1 3 7 6 4 0 5 8 9 10 \$empty: 6 move(4, 6);
2 1 3 7 0 4 6 5 8 9 10 \$empty: 4 move(5, 4);
2 1 3 7 4 0 6 5 8 9 10 \$empty: 5 move(7, 5);
2 1 3 7 4 5 6 0 8 9 10 \$empty: 7 move(3, 7);
2 1 3 0 4 5 6 7 8 9 10 \$empty: 3 move(2, 3);
2 1 0 3 4 5 6 7 8 9 10 \$empty: 2 move(0, 2);
0 1 2 3 4 5 6 7 8 9 10 \$empty: 0
Trial: 1 Moves: 9
0 5 8 10 6 2 9 3 7 1 4 \$empty: 0 move(1, 0);
5 0 8 10 6 2 9 3 7 1 4 \$empty: 1 move(9, 1);
5 1 8 10 6 2 9 3 7 0 4 \$empty: 9 move(6, 9);
5 1 8 10 6 2 0 3 7 9 4 \$empty: 6 move(4, 6);
5 1 8 10 0 2 6 3 7 9 4 \$empty: 4 move(10, 4);
5 1 8 10 4 2 6 3 7 9 0 \$empty: 10 move(3, 10);
5 1 8 0 4 2 6 3 7 9 10 \$empty: 3 move(7, 3);
5 1 8 3 4 2 6 0 7 9 10 \$empty: 7 move(8, 7);
5 1 8 3 4 2 6 7 0 9 10 \$empty: 8 move(2, 8);
5 1 0 3 4 2 6 7 8 9 10 \$empty: 2 move(5, 2);
5 1 2 3 4 0 6 7 8 9 10 \$empty: 5 move(0, 5);
0 1 2 3 4 5 6 7 8 9 10 \$empty: 0
Trial: 2 Moves: 11
0 6 7 4 5 2 3 1 8 9 10 \$empty: 0 move(1, 0);
6 0 7 4 5 2 3 1 8 9 10 \$empty: 1 move(7, 1);
6 1 7 4 5 2 3 0 8 9 10 \$empty: 7 move(2, 7);
6 1 0 4 5 2 3 7 8 9 10 \$empty: 2 move(5, 2);
6 1 2 4 5 0 3 7 8 9 10 \$empty: 5 move(4, 5);
6 1 2 4 0 5 3 7 8 9 10 \$empty: 4 move(3, 4);
6 1 2 0 4 5 3 7 8 9 10 \$empty: 3 move(6, 3);
6 1 2 3 4 5 0 7 8 9 10 \$empty: 6 move(0, 6);
0 1 2 3 4 5 6 7 8 9 10 \$empty: 0
Trial: 3 Moves: 8
Number of slots: 10
Trials: 3
Average Moves: 9.33333333333333
Minimum Moves: 8
Maximum Moves: 11 | 6,048 | 18,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-21 | latest | en | 0.913098 |
https://todaynetnews.com/representing-automata-graphically-22398 | 1,702,148,579,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00230.warc.gz | 631,849,158 | 11,524 | # Representing Automata Graphically
Begin
Q‟: = {q 0 };
:=
Figure 2.14 is the block diagram of the algorithm for determining Q' and '. Suppose we number the symbols of ‟ as a 1 , a 2 ,..., a n
T Q‟ unmarked
S
End
D
S
i n
D
B:= (T, a i )
Q‟:= Q‟ {B};
:= { ‟( T, a i ) = B }
i := i+1
Mark T
i := 1
Figure 2.14. Algorithm diagram
Example 2.21:
a) For a finite automaton NFA M = < Q, , , q 0 , F > whose transition graph is
Start
0
a
first
a
2
b
3
b
b
a
2.15. Let's build a deterministic finite Automaton D that recognizes the same language as M. a
Figure 2.15. Represent automata graphically
We can see that M predicts language N(M) = b * (a| b)a + (a| b) Applying the algorithm, we have:
D= ‟, ‟, q‟ 0 , F‟>. In there:
- ‟ = {a, b};
- q‟ 0 = {0} = A; Q‟ = {A};
- Consider A:
+ (A, a) = ({0}, a) = (0, a) = {1} = B
Q‟ = {A, B}; ‟ = {‟(A, a) = B}
+ (A, b) = ({0}, b)) = (0, b) = {0, 1} = C
Q‟ = {A, B, C}; ‟ = {‟(A, a) = B; ‟(A, b) = C}
- Consider B:
+ (B, a) = (1, a) = {1} = B
Q‟ = {A, B, C}; ‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B}
+ (B, b)) = (1, b) =
Q‟ = {A, B, C};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B}
- Consider C:
+ (C, a)) = ({0, 1}, a) = (0, a) (1, a) = {1, 2} = D
Q‟ = {A, B, C, D};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D}
+ (C, b)) = ({0, 1}, b) = (0, b) (1, b) = {0, 1} = C
Q‟ = {A, B, C, D};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C}
- Consider D:
+ (D, a)) = ({1, 2}, a) = (1, a) (2, a) = {1, 2} {3}= {1, 2, 3 }=E
Q‟ = {A, B, C, D, E};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C; ‟(D, a) = E}
+ (D, b)) = ({1, 2}, b) = (1, b) (2, b) = {3} = F
Q‟ = {A, B, C, D, E, F};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C; ‟(D, a) = E; ‟(D, b) = F}
- Consider E:
+ (E, a)) = ({1, 2, 3}, a) = (1, a) (2, a) (3, a) = {1, 2, 3} =E
Q‟ = {A, B, C, D, E, F};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C; ‟(D, a) = E; ‟(D, b) = F; ‟(E, a) = E}
+ (E, b)) = ({1, 2, 3}, b) = (1, b) (2, b) (3, b) = {3}= F
Q‟ = {A, B, C, D, E, F};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C; ‟(D, a) = E; ‟(D, b) = F; ‟(E, a) = E; ‟(E, b) = F}
- Consider F:
(F, a)) =({3}, a) =(3, a) =
Q‟ = {A, B, C, D, E, F};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C; ‟(D, a) = E; ‟(D, b) = F; ‟(E, a) = E; ‟(E, b) = F}
(F, b)) =({3}, b) =(3, b) =
Q‟ = {A, B, C, D, E, F};
‟ = {‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D;
‟(C, b) = C; ‟(D, a) = E; ‟(D, b) = F; ‟(E, a) = E; ‟(E, b) = F}
- F' = {E, F}.
So D = ‟, ‟, q‟ 0 , F‟>. In there:
Q' = {A, B, C, D, E, F} ;
‟ = {a, b}; q' 0 = A;
‟:{‟(A, a) = B; ‟(A, b)) = C; ‟(B, a) = B; ‟(C, a) = D; ‟(C, b) = C;
‟(D, a) = E; ‟(D, b) = F; ‟(E, a) = E; ‟(E, b) = F}
F' = {E, F}.
Transfer graph of automaton D
Start
A
a
B
a
C
a
D
a
E
b
F
b
b
ab a
Figure 2.16. Represent automata graphically
We can see that D predicts the language L(M) = b * a + (b | ) We can summarize the results of implementing the above algorithm as follows:
- q‟ 0 = {q 0 } = A;
- ‟ = {a, b};
- Determine Q' and '
Step
Mark T a i B = (T, a i ) Add to Q' Add to ' Kt {0} = A A first A a {1} = B B ‟(A, a) = B b {0, 1}= C C '(A, b) = C 2 B a {1} = B ‟(B, a) = B b 3 C a {1, 2}= D D ‟(C, a) = D b {0, 1}= C ‟(C, b) = C 4 D a {1, 2, 3}= E E ‟(D, a) = E b {3} = F F ‟(D, b) = F 5 E a {1, 2, 3}= E ‟(E, a) = E b {3}= F ‟(E, b) = F 6 F a b
Maybe you are interested!
Table 2.9. Algorithm simulation
- Determine F' = {B, C}.
2.6. Thomson's algorithm
For the purpose of using Finite Automat for lexical analysis. Thomson proposed an algorithm to solve the following problem: Given a regular expression. Let's build an Automat that recognizes the language represented by that regular expression. The ideology of the algorithm is based on the idea that every regular expression can be decomposed into simpler regular expressions. From there, build simple Automata that recognize simple regular expressions. Then synthesize it to get the Automat you need to build.
1) Method for building automata to predict simple regular expressions
a) Build an Automat to guess the empty word . Figure 2.17 is an Automat that accepts the empty word with i as the starting state, f as the ending state.
Start
i
f
Figure 2.17. Automat guesses the expression r=
b) Build an Automat to guess the character a .
Start
i
a
f
Figure 2.18. Automat guesses the expression r = a
Figure 2.18 is a diagram representing Automata guessing and receiving a , with i being the starting state, f being the ending state.
2) Method for building synthetic automata from automata
a) Suppose s and t are two regular expressions and N(s), N(t) are two Automata guessing s and t respectively, then the Automator guessing the regular expression s + t looks like Figure 2.19 .
N(s)
Start
i
f
N(t)
Figure 2.19. Automat guesses the expression r = s+t
Here i, f are new states that are not yet in N(s), N(t), from the starting state i Automat sees the empty word; it can go to the starting state of N(r) or the starting state of N(t). Conversely, from the final state of N(s) or N(t), it can move to the final state f when encountering an empty word.
b) Build an Automator to recognize regular expressions st. Suppose N(s), N(t) are two automatons that accept s and t respectively, then automata accepts the regular expression st as shown in figure 2.20.
Start
i
N(s)
N(t)
f
Figure 2.20. Automat guesses the expression r=st
Here i is a new state that does not exist in N(s) and N(t). When encountering an empty word, Automat moves to the starting state of N(s), the ending state of N(s) and the starting state. of N(t) are combined into one state, the final state of N(t) becomes the final state of the composite Automat.
c) Assuming s is a regular expression, and N(s) is an Automator that accepts it, we build an Automata that accepts s * and s + . Figure 2.21a depicts Automat N(s * ), Figure 2.21 b
describes Automat N(s + )
Start
i
N(s)
f
Figure 2.21a. Automat guesses the expression r = s *
Start
i
N(s)
f
Figure 2.21b. Automat guesses the expression r = s +
Here i, f are new states, respectively the starting state and final state of Automat.
3) Thomson's algorithm
Input: r – regular expression; Output: NFA M ; N( M) = N(r). Process:
Step 1: Parse r into simple regular expressions
- Find in the regular expression the operation with the lowest priority
- If so, parse the regular expression into operands according to that operation. Each row – a simpler regular expression.
- Go back to step 1 until you have analyzed all regular expressions and simple regular expressions of the form r = a.
Step 2: Construction
- Build simple automata that recognize simple regular expressions.
- Build synthesis automata in reverse order until building a synthesis automaton that recognizes the regular expression r.
Comment:
- With a regular expression r, Automat N(r) predicts that it will have a maximum number of states not exceeding twice the number of characters in r and the number of operations in it. This observation is due to the fact that in the Thomson algorithm, for each symbol or operation in r, we add no more than two states.
- Each Automat N(r) built according to the Thomson algorithm has only one initial state and one final state, and the final state never switches to any other state.
- Each state of Automat N(r) has only one transition state according to the symbol of the set and a maximum of two transition states according to the empty symbol , so in general, an Automat built according to the Thomson algorithm is an Automata. not deterministic.
- Because each time we build a new Automat, we add a new start and end state, so an Automat built according to the Thomson algorithm does not have two overlapping states.
Example 2.22:
Let r = (a b) * abb. Let's build NFA to predict regular expression r according to Thomson's algorithm.
Step 1: Analyze r = (a + b) * abb
- r = r 1 r 2 ;
r 1 = (a + b) * ab; r 2 = b;
- r 1 = r 3 r 4 ;
r 3 = (a + b) * a; r 4 = b;
- r 3 = r 5 r 6 ;
r 5 = (a + b) * ; r 6 = a;
- r 5 = (r 7 ) * ; r 7 = a + b;
- r 7 = r 8 r 9 ;
R 8 = a ; r 9 = b;
Step 2: Construction
- Build Automate to guess r 8 = a; r 9 = b
Start
2
a
3
Start
4
b
5
Figure 2.22. The automata guesses a and b
- Build a Automator to guess r 7 = r 8 + r 9
2
a
3
Start
first
6
4
b
5
Figure 2.23. Automata guesses r 7 = a+b | 3,417 | 8,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.582012 |
https://canthisevenbecalledmusic.com/advanced-mathematics-1-morse-code/ | 1,722,769,534,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640398413.11/warc/CC-MAIN-20240804102507-20240804132507-00648.warc.gz | 120,909,267 | 13,851 | # Advanced Mathematics 1: Morse Code
I recently wrote an article, on Heavy Blog Is Heavy, about twelve-tone technique, dodecaphony, tone row system, whatever you want to call it (read it here). And while it was fun, I only could go so deep into theory, not to alienate the diversity of readers, each with their own proficiency in music theory. So, this exercise got me into gears and I now want to do more of this, and go as deep as I personally can, so I’m beginning the Advanced Mathematics series, where I’ll pick a topic that takes from both math and music and try to explain it and where we can go with it. Enjoy.
Morse code in music, especially metal, for some reason, is so overdone. When I notice that something is “hidden” within the music as a Morse code message, I just roll my eyes. That’s far from my reaction when I found out this thing existed. Man, when I learned on some music forum that Dream Theater had hidden “eat my ass and balls” in the middle of the song In the Name of God, I was floored. However, when Haken did the same trick to play “Affinity Haken 1985”, on the first song of their latest album, I rolled my eyes and said “of course…” Evidently, it’s not a new thing, the first riff of Rush’s YYZ is Y-Y-Z in Morse code, and examples probably precede that, and new ones come up almost daily. One that I remember quite fondly is the S-O-S message in Watchtower’s song Mayday in Kiev, which takes some artistic liberties with it, such as beginning on the O instead of the S (see 1:47 in this video).
Despite all the overdoneness of it, crafting a riff, a rhythm section, a solo or simply adding Morse code to the background of your track can prove a fun and interesting challenge for any musician or composer. Indeed, it’s what we call creative limitation, and the concept states that restricting ourselves in any way can boost our creativity. Forcing yourself to base your song on Morse code messages, for example!
First of all, let’s learn the basics of the Morse code alphabet, numbers, and punctuation, which you can do here on your own. Got it? From there, the path is pretty straightforward: you make up some words and translate them into music. Want to write CTEBCM? If you choose the International Morse Code, it’ll be -.-./-/./-…/-.-./–! I used slash bars to separate the letters for better visibility. Basically, the hyphens (-) are long notes, and the dots (.) short ones. Usually, this is represented by the hyphens being eighth notes and the dotes being sixteenths, or it can be quarter notes and eighths, or quarter triplets and quarter quintuplets, but I’m not the one resolving this measure. As for the pitch of these notes, you can play it safe (and boring) by aggressively palm-muting open strings, or you can put some life into it and apply the Morse rhythm to a melody, any one, or apply a chord (or scale) to each letter and choose notes within that chord (or scale) to play! Oh… how the possibilities are endless if you have the faintest spark of imagination.
Using Morse code in music gives off a very recognizable sound: long and/or short notes separated by short silences. I mean, it can go almost unnoticed in the context of a djent song, but it’s very striking when it’s part of almost any other genre of music. If you want to be super strict, however, the long ticks (-) are three times as long as the short ones (.); the hyphens and dots are separated by a silence lasting one unit (as long as a dot); the letters are separated by three units of silence, and the words, by seven. However, I haven’t heard anybody in music do this, except when including telegram samples, like the Dream Theater and Haken examples mentioned at the start of this article. Even the Watchtower example doesn’t separate the letters’s elements by short silences, but at least their hyphens last three dots.
We’ve spoken about its rhythmic facet, but not about how we could apply morse code melodically or tonally. Now, you could come up with different systems and they would be as valid as mine, but I’m just making examples here. You’ve basically got two states in which your element can be: hyphen or dot. This means that you could assign each one to an instrument; let’s say guitar and keyboard. Thus, you’ve removed the rhythmic side of the code, but you need to be a bit creative. In order to fully understand what letter you play without the rhythmic pattern of the letters, you’d have to assign one note to each position of characters. There can be no more than five characters to encode a single letter (all five-tick characters are actually numbers), so you could easily apply a pentatonic scale to their positions. Let’s take aeolian A. The first unit of a character’s code would be A, the second, C; the third, D; the fourth, E; and the fifth, G. If we assign dot marks to the guitar and hyphen marks to the keyboard, and we tell them to play the letter X, the guitar will play the notes C and D while the keyboard will play the notes A and E. This chord can last any amount of time, and it won’t affect our understanding of the encoded letter, as we’ve removed entirely the need for rhythm, there. Of course, this is a bit contrived and complicated, but it’s just an example to show you that you can push the boundaries. I didn’t think about this beforehand, I just imagined that system here, on the spot, because we all can. In any case, I’m sure this is a far better way of encrypt your messages – no one might ever decrypt them!
In conclusion, Morse code is somewhat foreseeable and gimmicky nowadays, but it’s a fun way to develop your compositional skills by putting a restriction on them. It’s one way of putting hidden messages into music, either by sampling a telegram or by crafting your music around it. You can take that concept and move it, rotate it, transform it – basically use it in your own personal creative way -, and start applying it to your playing or composition right now. We didn’t really talk about mathematics, today. At least, it was very light material. Consider this a warmup because stuff is slowly going to become more and more complicated.
On June 16 2016, this entry was posted and tagged: | 1,409 | 6,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.944889 |
https://ishikawa.math.keio.ac.jp/QLEJ/index027.html | 1,675,684,663,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00239.warc.gz | 332,772,258 | 3,496 | The argument in $\S4.3$ is due to refs [1, 8] in $\S$0.0 (my home page).
4.3.1: Heisenberg's uncertainty principle is doubtful
Heisenberg's uncertainty principle is as follows.
Proposition 4.10 [Heisenberg's uncertainty principle]
$\mbox{(i):}$ The position $x$ of a particle $P$ can be measured exactly. Also similarly, the momentum $p$ of a particle $P$ can be measured exactly. However, the position $x$ and momentum $p$ of a particle $P$ can not be measured simultaneously and exactly, namely, the both errors $\Delta_x$ and $\Delta_p$ can not be equal to $0$. That is, the position $x$ and momentum $p$ of a particle $P$ can be measured simultaneously and approximately, (ii): And, $\Delta_x$ and $\Delta_p$ satisfy Heisenberg's uncertainty principle as follows. \begin{align} \Delta_x \cdot \Delta_p \; {\doteqdot} \; \hbar (= \mbox{Plank constant}/2\pi {\doteqdot} 1.5547 \times 10^{-34} Js ). \tag{4.20} \end{align}
This was discovered by Heisenberg's thought experiment due to $\gamma$-ray microscope. It is
$(A):$ one of the most famous statements in the 20-th century.
But, we think that it is doubtful in the following sense.
$\fbox{Note 4.1}$I think that Heisenberg's uncertainty principle(Proposition 4.10) is meaningless. That is because, for example,
$(\sharp):$ The approximate measurement and "error" in Proposition 4.10 are not defined.
This will be improved in Theorem 4.15 (in $\S$4.3.3) in the framework of quantum mechanics. That is, Heisenberg's thought experimentis an excellent idea before the discovery of quantum mechanics. Some may ask that
$(\sharp):$ If it be so, why is Heisenberg's uncertainty principle (Proposition 4.10) famous?
I think that
$(\sharp):$ Heisenberg's uncertainty principle (Proposition 4.10) was used as the slogan for advertisement of quantum mechanics in order to emphasize the difference between classical mechanics and quantum mechanics.
And, this slogan was completely successful. This kind of slogan is not rare in the history of science. For example, recall "cogito proposition (due to Descartes)", that is,
\begin{align} \mbox{ I think, therefore I am. } \end{align}
which is also meaningless (cf. $\S$8.4). However, it is certain that the cogito proposition built the foundation of modern science.
$\fbox{Note 4.2}$ Heisenberg's uncertainty principle (Proposition 4.10) may include contradiction, if we think as follows
$(\sharp):$ it is "natural" to consider that \begin{align} \Delta_x = | x - \widetilde{x}|, \quad \Delta_p = | p - \widetilde{p}|, \end{align} where \begin{align} \left\{\begin{array}{ll} \mbox{Position:}\quad & [x:\mbox{exact measured value (=true value)}, \widetilde{x}: \mbox{measured value}] \\ \mbox{Momentum:}\quad & [p:\mbox{exact measured value (=true value)}, \widetilde{p}: \mbox{measured value}] \end{array}\right. \end{align}
However, this is in contradiction with Heisenberg's uncertainty principle (4.20). That is because (4.20) says that the exact measured value $(x, p)$ can not be measured. | 846 | 3,002 | {"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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.882883 |
https://www.careerride.com/mchoice/formula-for-shear-stress-concentration-factor-10359.aspx | 1,679,993,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00204.warc.gz | 776,329,711 | 5,566 | # Formula for shear stress concentration factor
Q. The shear stress concentration factor (Ks) in mechanical springs is given as _____
- Published on 14 Sep 15
a. (1 + 0.5 / C)
b. 0.615 / C
c. (1 + 0.615 / C)
d. [(4C – 1) / (4C + 1)] + [0.615 / C]
ANSWER: (1 + 0.5 / C)
#### Discussion
• Sravanthi -Posted on 06 Nov 15
- Shear stress concentration factor (Ks): This factor accounts both direct shear stress and torsional shear stress. This factor is used only in case of static loading.
- Static load: It is a mechanical force which is applied slowly to an object.
- The formula to calculate shear stress concentration factor is : Ks = [1 + (0.5/C)], here C is the spring index.
- Shear stress (τ) in spring wire is given by: τ = Ks [8 FC / π d2]
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Question
# A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. If the value of g be 10 m/s2, the work done by the girl against the gravitational force will be : (a) 6000 J (b) 0.6 J (c) 0 J (d) 6 J
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Solution
## The girl is moving with the bag on a levelled road. Her direction of motion is perpendicular to the direction of gravitational force. So, work done by the girl against gravity is zero. Therefore, the answer is 0 J (c) 0 J
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# 11.8: Resumen
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Términos Clave
Evaluación 360
Una herramienta de evaluación que recopila comentarios de gerentes, pares, informes directos y clientes.
9-caja
Una herramienta matricial utilizada para evaluar el acervo de talentos de una organización en función del desempeño y factores potenciales.
Competencias
Un conjunto de comportamientos definidos que una organización podría utilizar para definir estándares de éxito.
Ciclo de vida de los empleados
Las diversas etapas de compromiso de un empleado: atracción, reclutamiento, incorporación, desarrollo, retención, separación.
La relación laboral; el vínculo jurídico entre empleadores y empleados que existe cuando una persona realiza un trabajo o servicios en condiciones específicas a cambio de pago.
Capital humano
Las habilidades, conocimientos y experiencia de un individuo o grupo, y ese valor para una organización.
Gestión de recursos humanos
La gestión de las personas dentro de las organizaciones, enfocándose en los puntos de contacto del ciclo de vida de los empleados.
Cumplimiento de recursos humanos
El rol de RRHH para asegurar el cumplimiento de las leyes y reglamentos que rigen la relación laboral.
Matriz de mérito
Una tabla de cálculo que proporciona un marco para los aumentos de mérito en función de los niveles de desempeño.
Modelo de pago por desempeño
El proceso y la estructura para relacionar los niveles de desempeño individuales con los niveles de recompensa
Gestión del desempeño
El proceso mediante el cual una organización asegura que sus objetivos generales se están cumpliendo evaluando el desempeño de los individuos dentro de esa organización.
Sociedad para la Gestión de Recursos Humanos
La sociedad profesional de RRHH más grande del mundo, con más de 285 mil miembros en más de 165 países. Es un proveedor líder de recursos atendiendo las necesidades de los profesionales de RRHH.
Planeación de sucesión
El proceso de identificación y desarrollo de nuevos líderes y empleados de alto potencial para reemplazar a los empleados actuales en un momento futuro.
El proceso de búsqueda y adquisición de candidatos calificados para el empleo dentro de una empresa; generalmente se refiere a una visión a largo plazo de construir ductos de talento, en lugar de reclutamiento a corto plazo.
Desarrollo de talento
Procesos integrados de RRHH que se crean para atraer, desarrollar, motivar y retener a los empleados.
Proceso de calibración de revisión de talento
La reunión en la que se revisa y discute la matriz de 9 cajas de una organización, con aportes y participación desde el liderazgo organizacional.
Estrategia de recompensas totales
Tal como lo acuñó World at Work, incluye compensación, beneficios, efectividad en la vida laboral, reconocimiento, gestión del desempeño y desarrollo del talento.
Capacitación, asignaciones de estiramiento, evaluaciones individuales, planes de desarrollo individual
Estas son herramientas que pueden ser utilizadas en el desarrollo del talento:
Formación: un foro para aprender en persona o en línea
Estirar asignaciones: desafiar roles para empleados de alto potencial
Evaluaciones individuales: inventarios de personalidad y estilo de trabajo de los empleados
Planes de desarrollo individual: documentos que destacan las oportunidades de crecimiento y el camino de acción de un empleado individual
Guerra por el talento
Acuñado por McKinsey & Company en 1997, se refiere a la creciente competencia para reclutar y retener empleados talentosos.
## Resumen de Learning Outcomes
11.2 Una introducción a la gestión de recursos humanos
1. ¿Cuál ha sido la evolución de la gestión de recursos humanos a lo largo de los años y cuál es el valor actual que aporta a una organización?
La gestión de recursos humanos comenzó en su primera “ola” como una función principalmente de cumplimiento, con el personal de recursos humanos encargado de hacer cumplir el cumplimiento de los empleados y ejecutar los procesos administrativos en curso. En la segunda ola, HR se centró en el diseño de áreas de práctica de recursos humanos, que podrían construirse sobre modelos de mejores prácticas. La ola 3 de RRHH trajo consigo el concepto de que HR debe ser un verdadero socio para el negocio y debe apoyar la estrategia empresarial a través de sus programas y servicios. Por último, en la cuarta ola, RRHH sigue siendo un socio para el negocio, pero mira fuera del negocio a clientes, inversionistas y comunidades para ver cómo puede ser competitivo en términos de participación de clientes, confianza de los inversores y reputación de la comunidad.
Algunas áreas clave que RRHH apoya dentro del proceso del ciclo de vida de los empleados incluyen: cumplimiento de recursos humanos, selección y contratación de empleados, gestión del desempeño, recompensas de compensación y desarrollo de talento y planificación de sucesión.
11.3 Gestión de Recursos Humanos y Cumplimiento
2. ¿Cómo aporta valor a una empresa el rol de cumplimiento de recursos humanos de RRHH?
Los recursos humanos ayudan a proteger a la empresa y a sus empleados para garantizar que se adhieran a las numerosas regulaciones y leyes que rigen la relación laboral. El impacto del incumplimiento puede ser muy costoso y puede ser en forma de costo financiero, legal o reputacional. Algunas de las leyes clave en las que HR maneja el cumplimiento incluyen la Ley de Normas Laborales Justas (FLSA), la Ley de Discriminación por Edad en el Empleo (ADEA), la Ley de Estadounidenses con Discapacidades (ADA) y la Ley de Licencia Familiar y Médica (FMLA), entre otras.
Algunas de las mejores prácticas para informar y responsabilizar a los empleados son brindar educación y capacitación para explicar las regulaciones, proporcionar documentación de referencia para orientar con las regulaciones y programar auditorías periódicas de cumplimiento para garantizar que se sigan los procesos. La programación de auditorías internas regulares de recursos humanos ayuda a la organización a planificar y sentirse cómoda con su nivel de preparación e ilustra el valor que un grupo sólido de recursos humanos puede aportar a la organización.
11.4 Gestión del desempeño
3. ¿Cómo afectan las prácticas de gestión del desempeño al desempeño de la empresa
La gestión del desempeño es un proceso crítico de negocio que el grupo de recursos humanos administra para el negocio. La gestión del desempeño alinea el trabajo de grupos individuales con los objetivos generales del negocio y permite que el negocio trabaje hacia sus metas. La gestión del desempeño también debería ayudar a la empresa a diferenciar entre los diferentes niveles de desempeño de los empleados a través de la gestión de retroalimentación y una estructura de recompensas.
La gestión del desempeño también permite a una empresa identificar a sus malos resultados y proporciona un proceso consistente para rastrear y administrar el bajo desempeño de una manera justa y consistente con la ley. Ha habido mucha discusión sobre las mejores prácticas para un proceso de gestión del desempeño más allá de un proceso formal y anual que a menudo resulta engorroso para el negocio. Por formal o informal, la administración de recursos humanos necesita garantizar que el proceso ayude a diferenciar los diferentes niveles de desempeño, administre el flujo de retroalimentación y sea consistente y justo para todos los empleados.
11.5 Influenciar en el desempeño y la motivación
4. ¿Cómo utilizan las empresas estrategias de recompensas para influir en el desempeño y motivación de los empleados?
Las empresas utilizan estrategias de recompensas para influir en el desempeño y la motivación de los empleados diferenciando entre los distintos niveles de desempeño. Esta estrategia se llama pago por desempeño, y vincula el nivel de desempeño del empleado con un marco consistente de recompensas en cada nivel. La investigación indica que la razón principal por la que las empresas implementan el pago por el desempeño es poder reconocer y recompensar a sus altos rendimientos.
Para implementar una estructura de pago por desempeño, RRHH y la organización primero necesitan definir una filosofía de compensación, luego realizar una revisión de las implicaciones financieras de dicho sistema. Deben identificarse las brechas en el sistema actual, y las prácticas de compensación deben actualizarse de acuerdo con el diseño de pago por desempeño determinado. Por último, la comunicación y la formación son claves para ayudar a los empleados a comprender el contexto y la filosofía, así como la metodología específica.
11.6 Construcción y organización para el futuro
5. ¿Qué es la adquisición de talento y cómo puede crear una ventaja competitiva para una empresa?
La gestión de recursos humanos juega el papel importante de gestionar los procesos de talento para una organización, y es fundamental en el proceso de adquisición de talento desde el exterior. La adquisición de talento es el proceso de determinar qué roles aún se necesitan en la organización, dónde encontrar personas y a quién contratar. Contratar a los mejores talentos es una fuente clave de ventaja competitiva para una empresa, y no todas las organizaciones son buenas para hacerlo.
El impacto de la contratación se magnifica especialmente cuando se habla de talento de liderazgo superior. El candidato de liderazgo adecuado puede marcar la diferencia en el crecimiento, el desempeño y la trayectoria de una organización a lo largo de los años. Los recursos humanos deben trabajar con el negocio para evaluar la necesidad y los detalles del trabajo, desarrollar un grupo de candidatos y luego evaluar a los candidatos para la persona adecuada para incorporar a la organización.
11.7 Desarrollo del Talento y Planificación de Sucesión
6. ¿Cuáles son los beneficios del desarrollo del talento y la planificación de la sucesión?
Los procesos de desarrollo de talento y planificación de sucesión proporcionan a las organizaciones los sistemas necesarios para evaluar y desarrollar a los empleados y tomar las decisiones adecuadas sobre su movimiento y desarrollo interno. Un importante proceso de desarrollo del talento implica una revisión del talento, en la que el liderazgo discute a los empleados de sus grupos en términos de su desempeño y potencial. El desempeño se basa en las evaluaciones actuales de la gestión del desempeño sobre el rol actual. El potencial se basa en indicaciones conductuales que predicen un futuro alto desempeño y promotabilidad en una organización. Luego se discute sobre las acciones de seguimiento y planes de desarrollo para los empleados, con base en dónde caen en la matriz de rendimiento/potencial. El beneficio de este proceso es que la organización obtiene una mejor comprensión de dónde se encuentra el mejor talento dentro de la organización y puede hacer planes para gestionar el desarrollo de ese talento.
Otro proceso clave para la gestión del talento es la planificación de sucesión. En este proceso, el liderazgo y los recursos humanos se reúnen para identificar roles de liderazgo y otros roles críticos en la organización, y luego discuten una cartera potencial de candidatos sucesores internos y externos en diferentes niveles de preparación para el rol. El resultado de la planificación de sucesión es que una organización llega a comprender la profundidad de su banco de talentos y conoce las áreas de brecha donde puede necesitar enfocarse en desarrollar o adquirir candidatos adicionales.
Preguntas de revisión de capítulos
1. ¿Cuáles son las cuatro “olas” de la evolución de la gestión de recursos humanos?
2. ¿Cuáles son algunas de las normativas clave con las que los recursos humanos deben gestionar el cumplimiento?
3. ¿Cuáles son algunas de las consecuencias no deseadas de un sistema de clasificación forzado?
4. ¿Cuáles son algunos de los retos de gestión del desempeño que se deben abordar, sin importar cuál sea el sistema?
5. ¿Por qué muchas empresas están interesadas en pasar a una estrategia de pago por desempeño?
6. ¿Cuáles son los principales pasos del proceso para implementar el pago por desempeño?
7. ¿Cuáles son algunas de las mejores prácticas para reclutar nuevos candidatos de liderazgo?
8. Describir los pasos de una sesión de revisión de talento.
9. ¿Cuál es la diferencia entre performance y potencial?
10. ¿Cómo se puede saber si un candidato tiene potencial?
## Ejercicios de aplicación de habilidades de gestión
1. ¿Cómo ha ayudado la evolución de la gestión de recursos humanos a lo largo de los años a convertirla en un mejor socio del negocio? ¿De qué manera esperarías que la gestión de recursos humanos siga evolucionando a lo largo de los años?
2. ¿Cree que es necesario un proceso formal y anual de gestión del desempeño para ayudar a una organización a alcanzar sus metas? ¿Por qué o por qué no? ¿Cuáles son los requisitos mínimos de proceso que deben cumplirse para evaluar con éxito el desempeño?
3. ¿Es posible que una organización recompense a las personas de manera justa sin implementar un proceso de pago por desempeño? ¿Por qué o por qué no? ¿Ves algún escollo en un proceso de pago por desempeño?
4. ¿Cómo impacta la “guerra por el talento” en los procesos de adquisición de talento? ¿Cómo puede ser más exitoso RRHH trabajando con el negocio para navegar por el panorama del talento competitivo?
5. ¿Cuáles son los beneficios de tener procesos de calibración de revisión de talento? ¿Cuál es la desventaja del proceso? ¿Debería una organización informar a los empleados cuál es su “calificación” de revisión de talento? ¿Por qué o por qué no?
## Ejercicios de decisión gerencial
1. Te han contratado como nuevo VP de Finanzas, y supervisas a un equipo de casi 30 personas. Tu gerente de RRHH te ha informado recientemente que ha habido varias relaciones con los empleados en tu grupo en el pasado reciente, y te preocupa el nivel de conocimiento que tu equipo directivo tiene sobre cómo lidiar con estos temas. ¿Qué podrías hacer para cerrar la brecha de conocimiento y mitigar el riesgo de problemas en tu grupo?
2. Su empresa ha decidido abandonar su proceso formal y anual de gestión del desempeño y pasar a un sistema basado en la retroalimentación continua y la comunicación con los empleados. Te preocupa porque siempre has tenido cuidado de diferenciar a tus empleados por nivel de desempeño, y te preocupa que esto lastime a tus empleados más fuertes. ¿Cómo puede usar asegurar que sus comentarios y comunicación con los empleados brinden una gestión del desempeño, a pesar de la falta de un sistema formal?
3. Su empresa ha implementado recientemente un modelo de pago por desempeño para compensación. Esto te preocupa porque sabes que tus empleados estarán aún más molestos con sus calificaciones de desempeño si saben que están vinculados a la compensación. ¿Qué acciones puedes tomar para comenzar a prepararte para este cambio?
4. Eres el director de una organización de ingeniería y llevas un tiempo luchando la “guerra por el talento”. Parece que cada vez que tienes una vacante de puesto, le haces saber a RRHH pero lleva una eternidad encontrar a alguien, y el candidato a menudo rechaza el trabajo. ¿Cuáles son algunas formas de asociarse mejor con Recursos Humanos para adelantarse a la curva para la próxima vez?
5. Usted es el vicepresidente de una línea de negocio en una empresa manufacturera internacional. Usted y varios de sus colegas de mucho tiempo se jubilarán en los próximos años, y es necesario comenzar a pensar en el talento y la planificación de la sucesión. Vas a entrar en una discusión de revisión de talento la próxima semana, y te estás dando cuenta de que tienes escasez de potencial dentro de tu organización. ¿Cuáles son algunas acciones que usted (y RRHH) pueden tomar ahora para asegurarse de que su unidad de negocio no se tambalea cuando se va a jubilar?
## Caso de Pensamiento Crítico
Zappos, Holacracia y Gestión de Recursos Humanos
En 2013, Zappos se desempeñaba bien bajo el liderazgo de Tony Hsieh y se preparaba para asumir un nuevo reto que, entre otras cosas, empujaría los límites de la gestión tradicional de recursos humanos. A pesar de que los negocios estaban en auge, Tony Hsieh no era un hombre que quisiera estar en modo status quo por demasiado tiempo, por lo que se propuso implementar un cambio organizacional y cultural llamado Holacracia. Zappos era la más grande y conocida de las 300 empresas a nivel mundial que habían adoptado Holacacy, una nueva forma de jerarquía que es una “estructura flexible y autónoma, donde no hay empleos fijos sino simplemente roles funcionales temporales”.
En una Holacracia, la unidad principal se llama el “círculo”, que es un equipo distinto pero fluido. El liderazgo se volvió igualmente fluido con los círculos cambiantes. Los círculos están diseñados para cumplir ciertos objetivos y se crean y se desintegran a medida que las necesidades del proyecto cambian. La intención es que las personas se autoseleccionen para trabajar en proyectos en los que quieran trabajar y para los que tengan las habilidades. Tony también eliminó todos los títulos anteriores. El rol de gerente se fue y fue reemplazado por tres roles: “enlaces principales” se enfocarían en guiar el trabajo en los círculos; “mentores” trabajarían en el crecimiento y desarrollo de los empleados; y “tasadores de compensación” trabajarían en determinar los salarios de los empleados. En 2015, decidió desglosar aún más las divisiones entre muchas de las funciones, cambiándolas todas a círculos centrados en los negocios. Hubo cambios en casi todas las estructuras de gestión de recursos humanos que se te ocurran, y hubo bastantes dolores de crecimiento dentro de la organización. Zappos comenzó a mirar el salario de los empleados, y la Holacracia parecía tener una curva de aprendizaje empinada para muchas personas, a pesar de que se creó una “constitución” para brindar orientación. Zappos también enfrentaba un 14% de desgaste, ya que algunos de los cambios rápidos y excesivos llevaban puestos en los empleados. Tony era un visionario, pero para mucha gente era difícil ponerse al día y ver la misma visión.
Desde una perspectiva de gestión de recursos humanos, podría haber algunos atributos positivos de una Holacracia si tuviera éxito, como construir compromiso y ayudar a construir talentos y habilidades. También hubo algunos riesgos que debían abordarse con detenimiento. Cuando se crea una organización en la que las personas no tienen equipos o proyectos establecidos sino que determinan en qué quieren trabajar, uno de los grandes desafíos va a ser determinar el nivel y la naturaleza de su rol, así como la compensación por ese rol. Si se compara la Holacracia con una organización consultora, en la que los consultores son llevados a diferentes proyectos con diferentes requerimientos, es fundamental determinar primero el nivel de su rol de consultor (con base en su educación, habilidades, experiencia, etc.) para que puedan pasar adecuadamente de proyecto a proyecto proyecto pero mantener un papel de cierto nivel. Ese nivel está entonces ligado a una escala salarial específica, por lo que el mismo consultor recibirá el mismo salario sin importar en qué proyecto se encuentre. Si ese consultor está “en el banquillo”, o no colocado en un proyecto (o autocolocado, en el caso de la Holacracia), entonces después de cierto periodo definido ese consultor puede estar en riesgo de terminación.
La holacracia es de alguna manera un concepto desafiante en el que pensar, y la autogestión puede no ser capaz de funcionar en todos los entornos. Una empresa que está implementando una Holacracia puede encontrar que son capaces de dominar el proceso de autoselección de trabajo en los “círculos”. La parte “tarea” de la ecuación puede no ser un gran problema una vez que la gente descubre cómo navegar por los círculos. Sin embargo, la parte “gente” de la ecuación puede necesitar algo de trabajo. El mayor desafío puede estar en las estructuras y procesos de gestión de recursos humanos que finalmente definen la relación empleador-empleado.
Preguntas de Pensamiento Crítico
1. ¿Cuáles son algunos de los procesos de gestión de recursos humanos que podrían ser potenciados por una Holacracia? ¿Qué procesos serán desafiados?
2. ¿Crees que una Holacracia puede compararse con una consultora? ¿Cómo son similares, s y en qué se diferencian? ¿Se te ocurre áreas de trabajo o industrias en las que la Holacracia sería muy difícil de implementar?
Fuentes: Askin y Petriglieri, “Tony Hsieh en Zappos: estructura, cultura y cambio”, INSEAD Business School Press, 2016.
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See you!
Op het einde van deze les kun je opsommen welke de voor- en de nadelen zijn wanneer de grootouders de kleinkinderen opvangen.
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Ryan's Toybox: "Ryan's bedroom has a toybox inside Ryan's toybox there are some cars and motorbikes. He sits down and counts how many wheels are in the toybox, Ryan counts 12!"
Children are shown on the next slides cars and motorbikes and can use the smartboard presentation to workout how many cars and motorbikes he has. This resource is fantastic for children to use numeracy for a real purpose.
Problems to Solve: "Animals in fields, ducks in a pond, pigs on a mountain and many more problems to solve."
Children are shown very visual pictures and images of senarios to solve. The problem to solve is written at the bottom of each page and the children are able to move the images around the screen and draw on any images which may help solve the problems! After the problem is solved a well done screen appears!
Sum Up: "Here a 4 numbers, use them in anyway you like to make the answer. You can add, takeaway, divide or multiply but you must find the answer."Children can move around the large colourful numbers on the screen and create sums to see if they can solve the problem and answer the sum. There are a series of different answers for children to find using the same senario.
Word Problems: "Add the number of bears to the number of Dwarves Snow White met?" "Double the number of Billy Goats Gruff to the number of trolls under the bridge..."A series of fantastic, interesting word problems to solve. All the problems involve well known story characters which children are familair with. Children can use the recording sheet to solve the problems, to show their working out. A great fun activity to get children thinking. Very visual images to inspire and create an interest.
* An accomplying worksheet for children to complete individually is included!
Ice Cream Challenge: "You only have 3 flavours of ice cream; rainbow, chocolate and strawberry. You must fill up the ice creams on the following slide using only these 3 flavours no ice cream should be the same!"A fun activity to be shown together on the Interactive Whiteboard and tried out, before children work on their own to solve this problem. This activity helps children to think methodically, carefully, thoughtfully demonstrating perservence and motivation in their work.
* An accomplying worksheet for children to complete individually is included!
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https://testbook.com/question-answer/errors-which-may-be-variable-both-in-magnitude-and--624beee0384390f4880322d0 | 1,669,684,883,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710684.84/warc/CC-MAIN-20221128235805-20221129025805-00581.warc.gz | 620,600,575 | 73,688 | # Errors which may be variable both in magnitude and nature are classified as __________.
This question was previously asked in
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1. interference error
2. calibration error
3. random error
4. instrument error
Option 3 : random error
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## Detailed Solution
Explanation:
Random Error:
• Random errors are statistical fluctuations (in either direction) in the measured data due to the precision limitations of the measurement device.
• It may be variable both in magnitude and nature.
• It can be evaluated through statistical analysis and can be reduced by averaging over a large number of observations
Following are the parameters that are used in statistical analysis to evaluate the random error.
• Mean
• Median
• Variance
• Deviation
• Standard Deviation
Explanation:
(i) Hysteresis error The hysteresis error of an instrument is the maximum difference in output at any measurement value within the specified range when approaching the point first with loading and then with unloading. (ii) Random errors Random errors are statistical fluctuations in the measured data due to the precision limitations of the measurement device. It may be variable both in magnitude and nature. (iii) Systematic errors Systematic error (also called systematic bias) is consistent, repeatable error associated with faulty equipment or a flawed experimental design. These errors are usually caused by measuring instruments that are incorrectly calibrated or are used incorrectly.
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MP Vyapam Sub Engineer Prelims Answer Key has been released on 22nd November 2022. The objections can be raised in the answer key till 24th November 2022. The exam was held from 6th November 2022. Check out and attempt the MP Vyapam Sub Engineer Previous Year Papers for the enhancement of your score. The candidates must also note that this MP Vyapam Sub Engineer recruitment is ongoing for a total of 2557 vacancies and the candidates who will clear the written exam will have to then appear for the document verification. | 477 | 2,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-49 | latest | en | 0.893197 |
https://bookriff.com/how-many-cards-do-you-get-in-speed/ | 1,653,170,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00600.warc.gz | 194,118,974 | 8,323 | # BookRiff
If you don’t like to read, you haven’t found the right book
## How many cards do you get in speed?
speed is a card game were each player tries to get rid of his or her cards. It is a very fast game wich you can play in a matter of minutes. It is played with a 52 card deck and no jokers.
## How do 3 people deal with speed?
Speed can be played with more than just two people. With three players, it is often unnecessary to have extra cards; cards are dealt by giving each player their five ‘side pile’ cards, placing three cards face down in the centre, and dealing the extra cards evenly as draw piles.
## What is the order of cards in speed?
The order of the cards is the standard Ace-10, Jack, Queen, King, where the values loop from highest back to lowest (i.e. aces placed on kings, kings placed on aces).
## How can I win at speed?
When the player runs out of cards they slap the table or hit both stacks and say “Speed!” to officially win. If a player fails to do whatever has been agreed on beforehand, he must take one of the central stacks as their draw pile and resume playing.
## Can you put a king on an ace in Speed?
When playing Speed you can either put an Ace or a Queen on top of a King, because the Ace is starting over from the top since the King is the highest card.
## Can you put the same number down in speed?
Two cards can be put down at once. You can not put down more than 2 at once. In Spit each player has a row of stock piles, usually five, with the top card face up, so all cards in play are visible to both players.
## How do you spit the card game?
Begin the game by having each player flip a card from his spit pile into the center of the game. Each player should also say “spit!” as he or she does this. These cards will begin the spite pile, where each player will try to place cards of ascending or descending value from his or her stockpile.
## What is a speed card game?
The Speed card game, also commonly known as Spit, is a game that moves very quickly. The object of Speed is to eliminate your hand of cards as quickly as possible while playing against one other opponent. Using older decks of cards is recommended, since they tend to offer better grip, and this card game can require quick card handling.
## What is the card game called Spit?
Spit (card game) Spit, also referred to as Slam or Speed, is a game of the shedding family of card games for two players.
## What is speed game?
Speed (card game) Speed is a game for two players or more of the shedding family of card games, in which each player tries to get rid of all of his or her cards first. | 606 | 2,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-21 | latest | en | 0.984943 |
http://mathhelpforum.com/calculus/212092-implict-differentiation-2cosysinx-1-a-print.html | 1,511,523,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807650.44/warc/CC-MAIN-20171124104142-20171124124142-00799.warc.gz | 188,789,418 | 2,796 | # Implict differentiation: 2cosysinx=1
• Jan 26th 2013, 07:27 PM
Furyan
Implict differentiation: 2cosysinx=1
Hello,
The question is to find the equation of the normal to the curve $2\cos y\sin x = 1$ at $(\tfrac{\pi}{4},\tfrac{\pi}{4})$.
I'm not confident with implicit differentiation, but using the product rule I got:
$2\sin x\dfrac{dy}{dx}(-\sin y) + 2\cos y\cos x = 0$
$\dfrac{dy}{dx}(2\sin x\sin y) - 2\cos y\cos x = 0$
$\dfrac{dy}{dx} = \dfrac{2\cos y\cos x}{2\sin x\sin y}$
$\dfrac{dy}{dx} = \cot x\cot y$
My question is, is it correct to write the derivative of $\cos y$ with respect to $x$ as $\dfrac{dy}{dx}(-\sin y)$ as I have done.
Thank you.
• Jan 26th 2013, 07:31 PM
chiro
Re: Implict differentiation: 2cosysinx=1
Hey Furyan.
Your working looks good for the implicit differentiation.
• Jan 26th 2013, 07:39 PM
Furyan
Re: Implict differentiation: 2cosysinx=1
Hey chiro,
That's a relief, thank you very much for looking it over :) | 353 | 954 | {"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-47 | longest | en | 0.8469 |
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CGAL::Point_d< Kernel > Member List
This is the complete list of members for CGAL::Point_d< Kernel >, including all inherited members.
cartesian(int i) CGAL::Point_d< Kernel > cartesian_begin() CGAL::Point_d< Kernel > Cartesian_const_iterator typedef CGAL::Point_d< Kernel > cartesian_end() CGAL::Point_d< Kernel > dimension() CGAL::Point_d< Kernel > homogeneous(int i) CGAL::Point_d< Kernel > homogeneous_begin() CGAL::Point_d< Kernel > Homogeneous_const_iterator typedef CGAL::Point_d< Kernel > homogeneous_end() CGAL::Point_d< Kernel > LA typedef CGAL::Point_d< Kernel > operator+(const Vector_d< Kernel > &v) CGAL::Point_d< Kernel > operator+=(const Vector_d< Kernel > &v) CGAL::Point_d< Kernel > operator-(const Origin &o) CGAL::Point_d< Kernel > operator-(const Point_d< Kernel > &q) CGAL::Point_d< Kernel > operator-(const Vector_d< Kernel > &v) CGAL::Point_d< Kernel > operator-=(const Vector_d< Kernel > &v) CGAL::Point_d< Kernel > operator<(const Point_d< Kernel > &q) CGAL::Point_d< Kernel > operator<=(const Point_d< Kernel > &q) CGAL::Point_d< Kernel > operator==(const Origin &) CGAL::Point_d< Kernel > operator>(const Point_d< Kernel > &q) CGAL::Point_d< Kernel > operator>=(const Point_d< Kernel > &q) CGAL::Point_d< Kernel > operator[](int i) CGAL::Point_d< Kernel > Point_d() CGAL::Point_d< Kernel > Point_d(int d, Origin) CGAL::Point_d< Kernel > Point_d(int d, InputIterator first, InputIterator last) CGAL::Point_d< Kernel > Point_d(int d, InputIterator first, InputIterator last, RT D) CGAL::Point_d< Kernel > Point_d(RT x, RT y, RT w=1) CGAL::Point_d< Kernel > Point_d(RT x, RT y, RT z, RT w) CGAL::Point_d< Kernel > transform(const Aff_transformation_d< Kernel > &t) CGAL::Point_d< Kernel > | 511 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.387104 |
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# How many different four-letter words can be formed (the
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How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(2!*2!*2!*3!)
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23 Feb 2005, 15:45
Could someone explain the soln..... Thanks.
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23 Feb 2005, 20:21
That's right. You have to divide off the repeating elements.
Think of it this way, how many ways are there to arrage 3 alphbets:
ABA
We can have ABA, BAA, and ABA again and finally AAB. But note that two of them are repeated due to the fact we have repeated letters, 'A'.
So the way to do this is to divide off the repeating elements. In my example, this would be 3!/2 = 3. (i.e. ABA, BAA, AAB)
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23 Feb 2005, 20:56
MEDITERRANEAN
total 13 letters
We want four letter words that the first letter is E and the last letter is R.
So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN
We have 8 different letters with E, A, N repeated twice.
So it would be P(8,2)+3=56+3=59
(A)
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24 Feb 2005, 08:54
I mean I understand that there are 11! possibilities for two spaces.
There is also 3 letters that are repeated twice (E,A, and N)
Now what I don't understand is where the 3! comes from.
currently I see this equation 11!/(2!*2!*2!)
I realize there is more, but how do we justify the *3! in the denominator?
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24 Feb 2005, 10:30
HongHu wrote:
MEDITERRANEAN
total 13 letters
We want four letter words that the first letter is E and the last letter is R.
So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN
We have 8 different letters with E, A, N repeated twice.
So it would be P(8,2)+3=56+3=59
(A)
Hong I followed the highlighted part , but then how come +3?????????
Could you plz explain that?
Last edited by Vijo on 25 Feb 2005, 03:54, edited 1 time in total.
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24 Feb 2005, 12:32
But I thought since the word starts with E and ends with R, technically we only have 2 E's that can be used. Can someone please explain?
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24 Feb 2005, 12:33
Christoph what's the OA on this one?
P(8,2)+3=56+3=59
HongHo, can you plz explain this a lil bit in details?
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25 Feb 2005, 00:47
HongHu seems closest - though I still dont completely understand his solution.
Nice qs bTw
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25 Feb 2005, 15:22
Ok my thought is this. For the middle two letters, you could have two different ones from the eight different letters, or you could have two same ones from the three repeated letters. So the first part is C(8,2), and the second part is 3 (EE, AA, NN).
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09 Mar 2005, 17:19
Christoph,
Let's get a resolution to this Q.
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01 May 2005, 13:16
christoph wrote:
How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(2!*2!*2!*3!)
how many four letter words if letters can be repeated?
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02 May 2005, 18:02
christoph wrote:
How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(2!*2!*2!*3!)
this is really a new concept in counting methods (permutation and combination), if hong's approach is the solution/answer.
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03 May 2005, 02:07
MA wrote:
christoph wrote:
How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(2!*2!*2!*3!)
this is really a new concept in counting methods (permutation and combination), if hong's approach is the solution/answer.
why ? we have a four-letter-word that begins with a E and ends with a R => E _ _ R. The two spaces in the middle can be occupied by 8 distinct letters, hence 8P2. but its also possible to have repeated letters such as EE, AA and NN. thats why we add 3. i think its just slightly different to a classical permutation-letter-problem.
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10 May 2005, 14:56
HongHu wrote:
MEDITERRANEAN
total 13 letters
We want four letter words that the first letter is E and the last letter is R.
So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN
We have 8 different letters with E, A, N repeated twice.
So it would be P(8,2)+3=56+3=59
(A)
You nailed it.
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Its not the fact its your attitude towards the fact
Re: PS - Combination [#permalink] 10 May 2005, 14:56
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# Calculating the point at which a piece of box section
### deflection calculator for rectangular tubing - PngLine
Jun 17,2017 Calculating the point at which a piece of box section#0183;Calculating the point at which a piece of box section steel will The next step is to refer to the following equation and do some However,this is where my knowledge falters and I would really appreciate some inputViews 1.2MCross Section Properties MechaniCalcNote that the first moment of the area is used when calculating the centroid of a cross section with respect to some origin (as discussed previously).The first moment is also used when calculating the value of shear stress at a particular point in the cross section.In this case,the first moment is calculated for an area that makes up a Tube Calculator - Rogue FabricationWhat You Should Use This Calculator For.You should use this calculator to compare materials,diameters,and wall thicknesses to find out how to make your designs safer.For example,lets say your local metal distributor has 1.75x.095 tube and 1.25 solid bar on sale from some huge bulk purchase that fell through with another customer.
### TORSIONAL SECTION PROPERTIES OF STEEL SHAPES
A calculation method is given by Galambos (1968).Monosymmetry Constant The monosymmetry constant, X,is used in calculating the buckling moment resistance of laterally unsupported monosymmetric beams loaded in the plane of symmetry (CSA 2000).In the case of a monosymmetric section that is symmetric about the vertical axis,the generalSquare Tubing Deflection Calculator Deflection of Hollow Online calculator to calculate deflection of hollow square tube.Hollow structural sections are made from metal with square or rectangular tube section.ASTM A500 is the common steel specification used for hollow structural tubes.This specification forms welded carbon steel tubing in round,square and rectangular shapes.Some results are removed in response to a notice of local law requirement.For more information,please see here.Previous123456Next10.6 Calculating Moments of Inertia - Physics LibreTextsAug 13,2020 Calculating the point at which a piece of box section#0183;In this example,we had two point masses and the sum was simple to calculate.However,to deal with objects that are not point-like,we need to think carefully about each of the terms in the equation.The equation asks us to sum over each piece of
### Some results are removed in response to a notice of local law requirement.For more information,please see here.12345NextArea Moment of Inertia - Typical Cross Sections I
Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I,is a property of shape that is used to predict deflection,bending and stress in beams..Area Moment of Inertia - Imperial units.inches 4; Area Moment of Inertia - Metric units.mm 4; cm 4; m 4; Converting between Units.1 cm 4 = 10-8 m 4 = 10 4 mm 4; 1 in 4 = 4.16x10 5 mm 4 = 41.6 cm 4 Some results are removed in response to a notice of local law requirement.For more information,please see here.Metal Weight Calculator - steel weight calculator Metal weight calculator online - free steel weight calculator.Has pre-entered densities for dozens of commonly-used metals and metal alloys like steel,aluminum,nickel,iron,copper,cadmium,gold,silver,etc.Calculate the weight of a steel beam,bar,tube,profiles,channels,or a simple metal sheet.
### How to determine the max load capcity of a steel square
Jul 21,2014 Calculating the point at which a piece of box section#0183;2.The next worst situation will be when a point load is applied to only one upper edge,near the middle of the tube.A dent in the edge will form.The same sort of failure as case 1 will then occur,as the loaded sidewall bends along its midline and so reduces the top to bottom separation.3.How to calculate maximum weight before a length of box Feb 09,2015 Calculating the point at which a piece of box section#0183;The box section must resist this torque with an arm distance of only the height of the box section,80 mm or 0.26 feet This results in a compressive force of 49,000 pounds or about 110,000 PSI on the cross section of the top of the box section.How to Determine the Area of Sectors and Segments of a use the formula to find the area of sector ACB..Area of a segment To compute the area of a segment like the one in the first figure,just subtract the area of the triangle from the area of the sector (by the way,theres no technical way to name segments,but you can call this one circle segment XZ) The following problem illustrates how to find arc length,sector area,and segment area:
### How to Calculate Miter Angles Hunker
Sep 25,2017 Calculating the point at which a piece of box section#0183;By far the easiest way to calculate miter angles on existing wall corners is using an electronic protractoran inexpensive,battery-powered tool that is a great addition to any home toolbox.The tool has adjustable arms that can be set to conform to wall angles,and a digital display that reveals the angles at which the adjoining surfaces meet.Help with angle iron calculation Engineers Edge Aug 31,2011 Calculating the point at which a piece of box section#0183;I think the solution is to attach a piece of angle iron across the bottom (underneath).Hopefully this will give the extra strength for support.What I don't know is what dimensions I'll need.I am also not certain if angle iron is the best solution,would a box section be stronger for instance? The width of the unit is 980mm.H Beam I Beam Weight Calculator Chart (Free to Use By using these two calculators,you can easily calculate the weight of H-beam and I-beam.Of course,for more calculations of different metal weights,you can refer to the following article. Calculating the point at which a piece of box sectiongt; Calculating the point at which a piece of box sectiongt; Theoretical Metal Weight Calculation Formula (30 Types of Metals) In addition,we also made a press brake tonnage calculator and press force calculator
### Formula for warping constant of box section - Structural
I have a box section with width and height 2 in and thickness 0.5 in.The warping constant that the software gives me is equal to 0.00730 in^6.Does anyone know if there is an explicit formula to calculate the warping constant of a box section?ENGINEERING Beam Deflection CalculatorsENGINEERING's Beam Deflection Calculators.Beam Deflection Calculators - Solid Rectangular Beams,Hollow Rectangular Beams,Solid Round BeamsDeflection of hollow rectangular beams - Beam - CalculatorBuilders use hollow rectangular beams in construction because such beams can withstand the forces of shearing and bending in both the x- and y-directions.They are also resistant to torsional forces,much more so than I-beams.This online Mechanical Engineering Calculator is to compute the deflection of hollow rectangular beams.
### Cross Section Properties MechaniCalc
Note that the first moment of the area is used when calculating the centroid of a cross section with respect to some origin (as discussed previously).The first moment is also used when calculating the value of shear stress at a particular point in the cross section.In this case,the first moment is calculated for an area that makes up a Cross Section Calculator - Engineering CalculatorCross Section Property Calculator Powered by WebStructural.This free cross section property tool calculates moment of inertia,polar moment of inertia and second moment of inertia for various shapes.Select a Shape.English (inches,kips,ksi) English (inches) Metric (mm) Calculate.Engineer's Calculator.Cross Section Calculator - Engineering CalculatorCross Section Property Calculator Powered by WebStructural.This free cross section property tool calculates moment of inertia,polar moment of inertia and second moment of inertia for various shapes.Select a Shape.English (inches,kips,ksi) English (inches) Metric (mm) Calculate.Engineer's Calculator.
### Chapter 5 Stresses In Beams - ncyu.edu.tw
cross section is dP = dA.Substituting = - (E/ )y,(a) Where y is the distance of dA from the neutral axis (NA).The resultant of the normal stress distribution over the cross section must be equal to the bending moment M acting about the neutral axis (z-axis).Figure 5.3 Calculating the resultant of the Normal stress acting on the crossCalculation the point at which a piece of box section Calculation the point at which a piece of box section steel will bend and not recover.Stephen Wood Senior Metalwork Technician,Wimbledon college of Art The following document is published using a creative commons license allowing anyone to alter and share the content for non commercial use.Calculating the point at which a piece of box section Apr 26,2010 Calculating the point at which a piece of box section#0183;This Work,Calculating the point at which a piece of box section steel will bend and not recover.,by swood is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported license.Post by swood
### Calculating the Statical or First Moment of Area of Beam
Consider the I-beam section shown below.In our previous tutorial we already found the centroid to be 216.29 mm from the bottom of the section.To calculate the statical moment of area relative to the horizontal x-axis,the section can be split into 4 segments as illustrated:Calculating the Statical or First Moment of Area of Beam Consider the I-beam section shown below.In our previous tutorial we already found the centroid to be 216.29 mm from the bottom of the section.To calculate the statical moment of area relative to the horizontal x-axis,the section can be split into 4 segments as illustrated:Calculating section factors - SteelConstructionfoThe section factor of a hot rolled open section,a hot finished/formed hollow section or fabricated girder is defined as the surface area of the member per unit length (A m) divided by the volume per unit length (V).It is measured in units of m-1.It is perhaps simpler to consider it as the heated perimeter of the exposed cross section (H p) divided by the total cross-sectional area (A).
### Calculating Torque With Examples - ThoughtCo
Oct 18,2018 Calculating the point at which a piece of box section#0183;To begin calculating the value of the torque,you have to realize that there's a slightly misleading point in the above set-up.(This is a common problem in these situations.) Note that the 15% mentioned above is the incline from the horizontal,but that's not the angle .The angle between r and F has to be calculated.Calculate Required Tube Size Using Structural Properties Calculator Results Disclaimer.The results provided herein were generated using recognized engineering principles and are for informational purposes only.All information is provided as is,exclusive of any warranty,including,without limitation,any implied warranty of merchantability,fitness for a particular purpose,or any other warranty Calculate Bending Stress of a Beam Section SkyCiv Cloud The general formula for bending or normal stress on the section is given by Given a particular beam section,it is obvious to see that the bending stress will be maximised by the distance from the neutral axis (y).Thus,the maximum bending stress will occur either at the TOP or the BOTTOM of the beam section depending on which distance is larger:
### Calculate Bending Stress of a Beam Section SkyCiv Cloud
The general formula for bending or normal stress on the section is given by Given a particular beam section,it is obvious to see that the bending stress will be maximised by the distance from the neutral axis (y).Thus,the maximum bending stress will occur either at the TOP or the BOTTOM of the beam section depending on which distance is larger:CALCULATING THE POINT AT WHICH A PIECE OF BOXCALCULATING THE POINT AT WHICH A PIECE OF BOX SECTION STEEL WILL BEND AND NOT RECOVER Its often the case in the metal workshop that the laws of physics are challenged in order to pursue a design aesthetic.Luckily,with some knowledge of physics and a little common sense we can employ these laws and prevent a lot of wasted time and effort.C channel beam loads - Structural engineering general Feb 10,2003 Calculating the point at which a piece of box section#0183;A = cross sectional area,from point where we wish to find the shear stress at,to an outer edge of the beam cross section (top or bottom) y' = distance from neutral axis to the centroid of the area A.I = moment of inertia for the beam cross section.b = width of the beam at the point we wish to determine the shear stress.
### Beam Stress Deflection MechaniCalc
The equation for shear stress at any point located a distance y 1 from the centroid of the cross section is given by where V is the shear force acting at the location of the cross section,I c is the centroidal moment of inertia of the cross section,and b is the width of the cross section.Basic Rigging Workbook - BNLCalculating Weight Exercises 1.Use weights of common materials table on page 4 of this workbook to calculate the weight of a steel plate 4 ft wide x 10 ft long x 1/2 inch thick.2.Use the weight table for pipe on page 32 to calculate the weight of a nominal 6-inch seamless steel pipe,Schedule 120,20 ft long.3.5 Ways to Calculate Center of Gravity - wikiHowJun 04,2020 Calculating the point at which a piece of box section#0183;To calculate the center of gravity of 2 objects on a see-saw,first identify the weight of each separate object.Choose a starting point,or datum,on one end of the see-saw and measure its distance from the center and each object.Find each objects moment by multiplying the distance by the objects weight,then add up the 3 moments.
### 17 BEAMS SUBJECTED TO TORSION AND BENDING -I
Methods of calculating the position of the shear centre of a cross section are found in standard textbooks on Strength of Materials.2.2 Classification of Torsion as Uniform and Non-uniform As explained above when torsion is applied to a structural member,its cross section may warp in17 BEAMS SUBJECTED TO TORSION AND BENDING -IMethods of calculating the position of the shear centre of a cross section are found in standard textbooks on Strength of Materials.2.2 Classification of Torsion as Uniform and Non-uniform As explained above when torsion is applied to a structural member,its cross section may warp in10.6 Calculating Moments of Inertia - Physics LibreTextsAug 13,2020 Calculating the point at which a piece of box section#0183;In this example,we had two point masses and the sum was simple to calculate.However,to deal with objects that are not point-like,we need to think carefully about each of the terms in the equation.The equation asks us to sum over each piece of
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Looking for steel stock and quoted price? | 3,257 | 15,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.889286 |
https://www.ocf.berkeley.edu/~shidi/cs61a/w/index.php?title=Linked_list&oldid=975 | 1,643,083,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304760.30/warc/CC-MAIN-20220125035839-20220125065839-00378.warc.gz | 936,564,157 | 8,667 | A linked list is a recursive data structure that represents a sequence of elements. It consists of a series of nodes. Each node contains a piece of data, as well as a pointer to the next node. The last element in the list is the empty linked list. The piece of data is called the "first" field of that linked list, and the pointer to the next node is called the "rest" field.
## Types
In a straight-forward linked list, a node's first field contains a value (string, number, etc.), while the second field will contain another linked list. Using this structure, a series of nested linked lists can form a list of values.
A deep linked list is slightly different. The first and second fields contain another linked list. A good way to visualize linked lists is to draw them out.
The ADT of a linked list is independent of its implementation. The functions are:
• `link(elem, list)` – returns a linked list with `elem` as the first item and `list` as the rest of the list
• `first(list)` – returns the first field of linked list `list`
• `rest(list)` – returns the rest field of linked list `list`
• `empty` – the empty linked list
The following are implementations of the ADT:
• with tuples:
```empty = lambda: 42
return (element, list)
def first(list):
return list[0]
def rest(list):
return list[1]```
• with `cons`:
```empty = lambda: 42
return cons(element, list)
def first(list):
return car(list)
def rest(list):
return cdr(list)```
• `Link(elem, list)` – returns a linked list with `elem` as the first item and `list` as the rest of the list
• `list.first` – returns the first field of linked list `list`
• `list.rest` – returns the rest field of linked list `list`
• `Link.empty` – the empty linked list
The basic code can be seen below. Note that `list.first` and `list.rest` are implemented as instance attributes, `Link.empty` as a class attribute, and `Link(first, rest)` via `__init__`:
```class Link:
"""A recursive list, with Python integration."""
empty = None
def __init__(self, first, rest=empty):
self.first = first
self.rest = rest
def __repr__(self):
rest = ''
else:
rest = ', ' + repr(self.rest)
def __str__(self):
rest = ''
else:
rest = ' ' + str(self.rest)[2:-2]
return '< {}{} >'.format(self.first, rest)```
## Operations
The recursive structure of a linked list suggests recursive algorithms for operations. Generally, the base case will be if the current node is the empty list. The recursive step will involve recursing on the rest of the list.
A linked list can also be operated on using iteration: a while loop that continues until the current node is the empty list and repeatedly updating a variable to the rest of the list.
## Examples
Here is a function that prints a linked list in a readable format:
```def print_linked_list(lst):
< 3 1 5 3 >
"""
s = '< '
while lst != empty:
s = s + repr(first(lst)) + ' '
lst = rest(lst)
print(s[:-1] + ' >')```
Here is a function that searches a linked list for a given item:
```def contains(lst, elem):
"""Returns True if linked list LST contains ELEM.
>>> contains(lst, 5)
True
>>> contains(lst, 4)
False
"""
if lst == empty:
return False
return first(lst) == elem or contains(rest(lst), elem)```
Here is a function that returns a new linked list that is the reverse of the one passed in:
```def reverse(lst):
"""Returns a new list that is the reverse of LST. | 828 | 3,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-05 | latest | en | 0.822545 |
https://www.geeksforgeeks.org/scala-trait-traversable-set-4/ | 1,580,253,217,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00279.warc.gz | 839,624,194 | 27,170 | # Scala Trait Traversable | Set-4
Prerequisite :-
Scala Trait Traversable | Set-1
Scala Trait Traversable | Set-2
Scala Trait Traversable | Set-3
It is recommended to view Set-1, Set2 and Set-3 before this one.
The operations are as follows:
• Folds:
The operations here are reduceRight, reduceLeft, /: or foldLeft, :\ or foldRight. This methods apply binary operation to the successive elements of the Traversable collection.
Example :
`// Scala program for operations ` `// of Folds ` ` ` `// Creating object ` `object` `Folds ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``8``, ``6``, ``4``, ``2``) ` ` ` ` ``// Applying method of ` ` ``// Folds ` ` ``// val y = x.foldLeft(0)(_ - _) or ` ` ``val` `y ``=` `(``0` `/``:` `x)(``_` `- ``_``) ` ` ` ` ``// Evaluates the Expression ` ` ``// given below and displays it ` ` ``// ((((0 - 8) - 6) - 4) - 2) ` ` ``println(y) ` ` ` ` ``} ` `} `
Output:
```-20
```
Here, foldLeft(0) is the desired operation where zero is the ‘initial value’.
This method applies binary operation between successive elements of the given Traversable collection, moving from left to right and starting with the initial value.
Example :
`// Scala program for operations ` `// of Folds ` ` ` `// Creating object ` `object` `Folds ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``8``, ``6``, ``4``, ``2``) ` ` ` ` ``// Applying method of ` ` ``// Folds ` ` ``// val y = x.foldRight(0)(_ - _) or ` ` ``val` `y ``=` `(x ``:``\ ``0``)(``_` `- ``_``) ` ` ` ` ``// Evaluates the Expression ` ` ``// given below and displays it ` ` ``// (8 - (6 - (4 - (2 - 0)))) ` ` ``println(y) ` ` ``} ` `} `
Output:
```4
```
Here, foldRight will perform binary operation between successive elements of the collection, moving from left to right and ending with initial value.
Example :
`// Scala program for operations ` `// of Folds ` ` ` `// Creating object ` `object` `Folds ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``8``, ``6``, ``4``, ``2``) ` ` ` ` ``// Applying method of folds ` ` ``val` `y ``=` `x.reduceLeft(``_` `- ``_``) ` ` ` ` ``// Evaluates the Expression ` ` ``// given below and displays it ` ` ``// (((8 - 6) - 4) -2) ` ` ``println(y) ` ` ``} ` `} `
Output:
```-4
```
Here, reduceLeft performs the binary operation between the successive elements of the collection like foldLeft but considering first element of the given collection as initial value.
Example :
`// Scala program for operations ` `// of Folds ` ` ` `// Creating object ` `object` `Folds ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``8``, ``6``, ``4``, ``2``) ` ` ` ` ``// Applying method of folds ` ` ``val` `y ``=` `x.reduceRight(``_` `- ``_``) ` ` ` ` ``// Evaluates the Expression ` ` ``// given below and displays it ` ` ``// (8 - (6 - (4 - 2))) ` ` ``println(y) ` ` ``} ` `} `
Output:
```4
```
Here, reduceRight will perform binary operation like foldRight but considering last element of the collection as initial value.
• Specific folds:
The operations here are min, max, product, and sum. These operations operate on distinct kind of collections of Traversable.
Example :
`// Scala program for operations ` `// of Specific folds ` ` ` `// Creating object ` `object` `Specificfold ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``3``, ``4``, ``7``, ``10``) ` ` ` ` ``// Applying method of ` ` ``// Specific folds ` ` ``val` `y ``=` `x.sum ` ` ` ` ``// Displays sum of all the ` ` ``// elements in the list ` ` ``println(y) ` ` ``} ` `} `
Output:
```24
```
Here, sum will return sum of all the elements in the collection given.
Example :
`// Scala program for operations ` `// of Specific folds ` ` ` `// Creating object ` `object` `Specificfold ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``2``, ``5``, ``10``, ``12``) ` ` ` ` ``//Applying method of ` ` ``//Specific folds ` ` ``val` `y ``=` `x.product ` ` ` ` ``//Displays product of all ` ` ``//the elements in the list ` ` ``println(y) ` ` ``} ` `} `
Output:
```1200
```
Here, product will return product of all the elements in the collection given.
Example :
`// Scala program for operations ` `// of Specific folds ` ` ` `// Creating object ` `object` `Specificfold ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``21``, ``15``, ``27``, ``22``) ` ` ` ` ``// Applying method of ` ` ``// Specific folds ` ` ``val` `y ``=` `x.max ` ` ` ` ``// Displays largest element ` ` ``// of the list ` ` ``println(y) ` ` ``} ` `} `
Output:
```27
```
Here, max will return largest element of all the elements in the collection given.
Example :
`// Scala program for operations ` `// of Specific folds ` ` ` `// Creating object ` `object` `Specificfold ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``21``, ``15``, ``27``, ``22``) ` ` ` ` ``// Applying method of ` ` ``// Specific folds ` ` ``val` `y ``=` `x.min ` ` ` ` ``// Displays smallest element ` ` ``// of the list ` ` ``println(y) ` ` ``} ` `} `
Output:
```15
```
Here, min will return smallest element of all the elements in the collection given.
• String operations :
The operations here are addString, mkString, stringPrefix. This operations are utilized to provide another methods for converting a given collection of Traversables to a String.
Example :
`// Scala program for operations ` `// of Strings ` ` ` `// Creating object ` `object` `Strings ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``7``, ``8``, ``9``, ``10``, ``11``) ` ` ``val` `v ``=` `List(``21``, ``19``, ``17``, ``15``) ` ` ` ` ``// Applying Strings operations ` ` ``val` `y ``=` `x.mkString(``" < "``) ` ` ``val` `w ``=` `v.mkString(``" > "``) ` ` ` ` ``// Displays all the elements ` ` ``// separated by (<) ` ` ``println(y) ` ` ``println(``"\n"``) ` ` ` ` ``// Displays all the elements ` ` ``// separated by (>) ` ` ``println(w) ` ` ``} ` `} `
Output:
```7 < 8 < 9 < 10 < 11
21 > 19 > 17 > 15
```
Here, mkString(“<") is the desired operation where, “<" is the separator.
This operation is utilized to return the given collection of elements separated by the given separator.
Example :
`// Scala program for operations ` `// of Strings ` ` ` `// Creating object ` `object` `Strings ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating List of numbers ` ` ``val` `x ``=` `List(``7``, ``8``, ``9``, ``10``, ``11``) ` ` ` ` ``// Creating StringBuilder ` ` ``val` `B ``=` `new` `StringBuilder(``"The numbers are: "``) ` ` ` ` ``// Applying Strings operation ` ` ``val` `y ``=` `x.addString(B, ``", "``) ` ` ` ` ``// Displays all the elements ` ` ``// of the list in the String ` ` ``// builder separated by (,) ` ` ``println(y) ` ` ``} ` `} `
Output:
```The numbers are: 7, 8, 9, 10, 11
```
Here, addString(B, “, “) is the desired operation where, “B” is a String builder and “, ” is a separator.
This operation Adds the given collection of the elements in the StringBuilder separated by the stated separator.
Example :
`// Scala program for operations ` `// of Strings ` ` ` `// Creating object ` `object` `Strings ` `{ ` ` ` ` ``// Main method ` ` ``def` `main(args``:` `Array[String]) ` ` ``{ ` ` ` ` ``// Creating a Set of numbers ` ` ``val` `x ``=` `Set(``7``, ``8``, ``9``, ``10``, ``11``) ` ` ` ` ``// Applying Strings operation ` ` ``val` `y ``=` `x.stringPrefix ` ` ` ` ``// Displays the name of the ` ` ``// collection used ` ` ``println(y) ` ` ``} ` `} `
Output:
```Set
```
Here, stringPrefix will return the name of the collection used.
• Copying operations:
The two Copying operations are copyToArray and copyToBuffer. These operations are utilized to Copy the elements of the Traversable Collection to an Array or to a buffer.
• View operations:
The operations here are view, view(from, to). These operations are utilized to produce a view over the collection of Traversable given.
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 3,161 | 10,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.547913 |
https://chestofbooks.com/reference/American-Cyclopaedia-5/Gunnery.html | 1,555,806,892,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530100.28/warc/CC-MAIN-20190421000555-20190421022555-00553.warc.gz | 389,454,006 | 12,550 | Gunnery , the art of using guns, gunpowder, and projectiles. The forces which are of moment in gunnery as affecting the course of projectiles are terrestrial gravitation and the resistance of the air. The former is so nearly uniform, both in amount and direction, that it may be so regarded. But the difficulties which appear when we investigate the resistance of air are so formidable, that hitherto mathematicians have utterly failed to find a general formula, and have been obliged to resort to purely empirical methods. The first quantity to be sought is a unit of resistance with which all other degrees of resistance may be compared; and this is usually taken as the resistance offered by the air to a body having a front 1 foot square, moving 1 foot in 1 second. This quantity cannot bo determined theoretically, but it is found by careful trial that the value of this unit depends upon the form of the front, as well as its area. It is also considerably influenced by the shape of the rear. Hutton has given the following ratios between the values of the resistance:
Hemisphere, convex side foremost................. 119
Sphere........................................... 124
Cone, point foremost, with a vertical angle of 25° 42'. 126
Disk............................................. 285
Hemisphere, flat surface foremost.................. 289
Cone, base foremost............................... 291
In these ratios it appears that the resistance to the cone is about the same as that to a sphere, notwithstanding the sharp point of the former. From recent experiments by Prof. Bashforth of Woolwich, it also appears that the resistance to an elongated shot with a hemispherical front is less than that to a spherical shot of equal diameter, in the ratio of 1.845 to 1.531. Newton, in his Principia, gives as the front of least resistance a figure having nearly the section of a pointed Gothic arch. In practice it has been found that the "pointed ogive" or pointed Gothic arch gives less resistance than any other front hitherto experimented with. Investigators have therefore been compelled to determine the values of the unit independently for every kind of projectile in use. The dependence of resistance of air upon velocity is also determined experimentally. The latest and most trustworthy researches, by Prof. Francis Bashforth of the Woolwich artillery school, show that for velocities ranging from 1,400 to 1,700 ft. a second the resistance varies nearly as the square of the velocity; for those between 1,100 and 1,400 ft. it varies more nearly as the cube of the velocity; while for still lower velocities the ratio is in some power higher than the cube.
Thus a 15-inch shot, moving 1,500 ft. a second, encounters a resistance amounting to nearly a ton and a half, while a 10-inch shot encounters about three fourths of a ton at the same velocity. The amount of resistance offered by the air, and many other important data in gunnery, are ascertained by measuring the velocity of a projectile in different parts of its path. This is accomplished by means of an electro-veloci-meter. The projectile is made to break a series of electric circuits at several points, separated by equal intervals. The electric circuit passes through a machine, which contains a cylinder revolving at a known rate, and by appropriate devices the ruptures of the circuit make visible marks upon this cylinder. By measuring the distance between these marks, and multiplying it by the rate of revolution, the time which elapsed between any two instants of rupture becomes known. - Besides retardation, projectiles moving in air are subject to deviations resulting from their rotary motions about their axes. Spherical shot are always made of smaller diameter than the bore of the gun from which they are fired; the difference in the two diameters being termed windage.
One of its consequences is, that spherical shot are subject to a series of rebounds from side to side or from top to bottom of the bore, which is called balloting, and which causes them to leave the bore with a rotary motion. Let us suppose, for instance, that at the last ballot (rebound) the shot strikes the right side of the bore, as in fig. 1, receiving a rotary motion in the direction indicated by the arrows. This motion, combined with the motion of translation, tends to augment the pressure of the opposing air in the direction A I, and to diminish it in the direction A r; and the result is the deflection of the path of the shot to the right. Hence the effect of the last ballot in the case supposed is, first, to throw the shot to the left, while the unequal pressure of the air gradually deflects it back again to the right. If the final ballot were on the left side, the deflections would be reversed; if upon the top, the range would be slightly increased; and if upon the bottom, the range would be diminished. These effects were investigated, and the results demonstrated experimentally, by Magnus. They are much aggravated when, by reason of irregular density, the centre of gravity of a ball does not coincide with its centre of figure.
They are greater in small than in large projectiles for three reasons: 1. The actual amount of windage is very nearly the same for all calibres, and hence is relatively less for larger calibres than for small ones; therefore the balloting and consequent rotation will be less. 2. Large projectiles can be made more nearly isotropic than small ones, and the centres of figure and of gravity are more nearly coincident. 3. The effects of resistance of air are very nearly proportional to the surface exposed, i. e., to the square of the calibre; while the inertia of the shot and its consequent power to resist these effects is proportional to its mass, i. e., to the cube of the calibre. No projectiles have less lateral deviation than the largest round shot, whether the range be long or short; but the deviations of small spherical shot are notoriously great. In using elongated projectiles, the purpose is to reduce the total resistance encountered in passing through the air and through the target. This is attained by reducing the area of resistance, while the mass is not reduced. Less velocity is lost by them in consequence of the smaller front they offer to atmospheric resistance, as compared with spherical shot of equal weight.
After reaching the target they are required, in order to penetrate it, to make smaller holes than spherical projectiles of equal weight, and hence, with an equal striking velocity, will penetrate further. To secure these advantages, the elongated shot must always move with its axis as nearly as possible tangent to its path. But there are several causes which tend to make it rotate about its shortest axis, or tumble. To prevent this, and to give stability to the position of the long axis, a rotary motion about this axis is given to the projectile. This motion is totally distinct from the rotation of spherical projectiles just described, and the resulting effect of resistance of air is altogether peculiar. By reference to fig. 2 it will be seen that if the axis of the projectile were always parallel to its initial position, the curvature of the path would cause the resistance of air to act more and more upon the lower side, while the air upon the upper side would be rarefied in the wake of the projectile. The rotation upon the condensed air beneath causes it to roll to the right or left, according to the original direction of rotation; to which deflection the name drift is given. But in reality the axis does not continue parallel to its initial position.
It describes very slowly a conical surface, the apex of which is the centre of gravity of the shot; and what is most singular, the direction of this axial motion in pointed projectiles is opposite to that of flat-fronted projectiles. The conical rotation (or precession) of the axis causes an increased drift, the amount of which is even greater than the rolling drift already described. With pointed shot this deviation is to the right, but with flat-fronted shot to the left. The point of the former also droops, turning obliquely downward and to the right; the flat front turns obliquely upward and to the left. During the flight the former is more nearly tangent to the path than the latter. For uniform projectiles, the drift at moderate ranges is tolerably constant, and may be allowed for in sighting; but for round shot it is hopelessly irregular, sometimes to the right and sometimes to the left. At long ranges the drift of the elongated shot also becomes irregular, and often excessive, amounting sometimes to 200 or 300 yards to the right of the object sighted.
There are also vertical deviations, causing over-or under-shooting. In many cases these errors are more serious than lateral drift; for instance, against a battalion of troops, the hull of a vessel, the crest of a parapet, or the body of a deer, where the object is more extended laterally than vertically, and is more liable to be missed by vertical than by lateral error. There is another kind of error which may be called longitudinal deviation, or variation in range. A series of projectiles fired under conditions as nearly alike as practicable will differ in range; partly because no two charges of powder can be made to give exactly the same initial velocity, and partly because slight differences in the forms of the projectiles occasion marked differences in the amount of vertical drift. Hence the form of the trajectory is of great importance. To avoid vertical errors as much as practicable, it is desirable to give a high velocity to the shot; since the swifter its motion, the less curvature will gravitation produce in its path. It is evident that a low or flat trajectory is more dangerous to an enemy than a high one; but the former requires a higher velocity in the projectile than the latter.
The trajectories of spherical shot are at first less curved than those of elongated shot; but in the latter part of the flight, at considerable ranges, this relation is reversed. This is because the initial velocity of round shot is almost always greater, and the terminal velocity less, than that of elongated shot; the curvature everywhere being very nearly proportional to the velocity. The so-called "dangerous space" is that part of a projectile's path which is not higher above the earth than 5 ft. 10 in., or the stature of a man. The dangerous space is evidently greater at short than at long ranges, since it depends upon the angle which the descending branch of the trajectory makes with the earth, being greater the less the angle of descent; and the longer the range the greater is the angle of descent; for, to obtain the longer range, the muzzle of the gun must be more elevated, and the descending branch, owing to the resistance of the air, always makes a larger angle with the earth than the ascending branch. - The force of a projectile is measured by the product of its mass into the square of its velocity. The force, although a prime factor in the efficiency of a projectile, is not the only one; for cases may arise in which the energy is too great.
Thus in firing at a wooden vessel, a shot with a slow motion, making a large irregular hole, and hurling splinters, will be more destructive than a swift shot, cutting cleanly through, with comparatively little injury. In curved fire from mortars and howitzers, a low velocity is not only necessary, but desirable. Most of the effects of projectiles are accomplished by penetrating the objects against which they are directed, and their work will generally be most effectively accomplished when their energy is moderately in excess of that required for complete penetration. In this connection the penetration of iron vessels becomes of great interest and importance. - The most systematic experiments to ascertain the effect of shot on iron targets have been summarized by Capt. W. H. Noble of the English artillery, who deduces the following rules: 1. If two shot, having the same diameter and form of head, strike with equal energy, the penetration will be the same, though one may be a light round shot, striking with a high velocity, and the other a long heavy shot, with a low velocity. 2. A plate will be equally penetrated by shot of different diameters, provided this energy on striking is proportional to the diameter.
Thus, a 12-inch shot must have twice as much energy as a 6-inch shot, in order to penetrate the same plate. 3. The resistance of plates to penetration varies as the square of the thickness. These rules are subject to certain qualifications, depending upon the shape of the head of the shot. A hemispherical head is disadvantageous, because it tends to bulge laterally, and the same is partially true of a flat-fronted shot. The best form is the pointed ogive, which passes through without materially bulging, and makes a hole no larger than its true diameter. The flatfronted shot usually rips out a piece, called a button, in the shape of the frustum of a cone, the larger base being detached from the back of the plate. This is carried into the wooden backing, giving an increased resistance as compared with the ogive. Spherical projectiles are liable to flatten against the target and break in pieces. It is apparent that when flattening occurs the increased diameter involves the necessity of making a larger hole in order to penetrate. The striking velocity may be so great that the projectile will be dashed to pieces by its impact, and its energy partially absorbed in its own destruction, instead of that of the target.
This is especially true of spherical shot, fired with heavy charges at short range against thick plates. In comparing the effects of spherical and elongated projectiles against iron plates, many quantities must be considered, some favoring one form and some the other; but the final result is strongly in favor of the elongated form. For penetrating earth and battering masonry, similar considerations are applicable. - Concerning the effective range of guns, there is much popular misapprehension. To the scientific gunner the maximum range is of so little moment that its extent for common infantry bullets or for the heaviest seacoast projectiles is unknown. The longest range known to us was attained by one of Sir Joseph Whitworth's projectiles, viz., about 11,100 yards, not quite 6 1/2 miles. The efficiency is greatest near the muzzle, and diminishes as the range increases. A range may be considered effective at which there is a reasonable probability of doing injury. For bullets the effective range will depend upon the way in which the enemy's troops are deployed. Against a skirmish line it cannot much exceed 500 yards, but against massed troops it may be as great as 1,500 yards.
With field projectiles an enemy may be harassed at 2,500 yards, or even 3,000. In the bombardment of cities the extreme range is sometimes resorted to, on the assumption that a projectile falling anywhere within the line of fortification may work damage. Effective range turns upon the higher question of probabilities of fire. - Thus far we have discussed projectiles only, since their properties constitute the basis of gunnery. Gunpowder is merely the agent for giving them energy, and the gun for giving them direction. When we examine the relations among the three elements, the problem is highly complicated. We have two forces: the inertia of the shot, and the elastic force of the gases evolved by the powder. It is supposed that the metal contained in a given projectile is cast into a solid cylinder, having the diameter of the bore of the gun. Its length is called the column of metal of the projectile, and constitutes a measure of its inertia. Equal velocities will be imparted to different projectiles when the mean intensity of the forces acting upon them during a given time is proportional to their respective columns of metal.
But the intensity of the force of gunpowder is highly variable at different portions of the path along the bore, being very great near the seat of the shot, and rapidly declining toward the muzzle; hence equal, velocities will be imparted only when, at different points in the path along the bore, the respective intensities are proportional to the columns of metal. A complete analysis of the relations existing between the force of gunpowder and the motion of the shot in the gun, in terms of time, space, and mass, has never been attempted; it is a very formidable problem, and its chief difficulty is our ignorance of the rate at which the gases are developed and the quantity of heat evolved. But the greater the resistance opposed to the expansion of the gases of gunpowder, the more rapidly will the powder burn and develop gases, and the higher will be their temperature. Such an increased resistance is offered by an increased column of metal; and hence the conclusion that a longer column of metal carries with it the power of developing more force from a given quantity of powder than a shorter one.
On the other hand, the shorter column of metal will still have the higher velocity, though the longer will have the greater energy (mass multiplied into square of velocity); the difference in energy in favor of the latter being due to its greater mass, which more than compensates for its lower velocity. But if the quantity of powder is proportional to the column of metal, a larger charge will develop at every moment more gas than a smaller charge, and give a more intense force. But a larger charge occupies more space in the bore, and robs the projectile of a part of its travel, and hence of a part of the time in which it can receive acceleration. Increasing the charge will increase velocity up to a certain point, but beyond that point will diminish it. In small cannon the maximum pressure is probably reached before the shot has travelled three inches, and in large guns before it has travelled a foot. The time occupied by the shot in traversing the bore probably ranges from 1/90 to 1/200 of a second, and depends mainly upon the length of the bore and the quantity of powder.
A bold attempt was made by Rodman in 1858 to measure the distribution of the forces of gunpowder, by placing pressure gauges along the bore to register the pressure at different points; and to measure the time of passing over different parts of the bore, by a series of ruptures of electric circuits. (For a description of the pressure gauge, and the electric velocimeter, see Gunpowder, and Velocimeter.) It is obvious that an increase either in the column of metal or in the charge involves an increase in the intensity of the pressure of the gases, and hence an increased strain upon the gun. As the strength of a gun is limited, both the column of metal and the charge must be regulated accordingly. It is the maximum pressure which is dangerous. In large guns this difficulty is serious. Not only is a higher pressure produced by the longer column of metal, but the pressure is distributed over a larger area of bore, and the bursting tendency is in the ratio of the product of these two quantities. The greater thickness of walls gives increased resistance, but this increase is in a lower ratio than that of the bursting tendency, and hence large guns are relatively weaker than small ones.
To compensate for this difficulty, constructors have resorted to metals of greater strength, and especially have modified the action of the powder, so that the maximum pressures have been reduced, and the subsequent lower pressures have been increased. Thus the total effort of the powder upon the shot is undiminished. (See Gunpowder.) The column of metal of a spherical shot is two thirds its calibre; that of an elongated shot is usually between one and three fourths and twice its calibre. The latter limit has been found to be about as great as the strength of the gun will permit in large calibres. It is sometimes exceeded with very little advantage in the smaller and intermediate calibres. The charge of powder varies from one fourth to one tenth the weight of the projectile. With round shot it is sometimes as high as one third; but it is found that the velocity is not much increased when the charge is greater than one fourth. The velocities imparted to round shot vary from 1,400 to 1,750 ft. per second, and those of elongated shot from 1,150 to 1,500 ft. - For a good introduction to the science of gunnery, see "Ordnance and Gunnery," by Major J. G. Benton, U. S. A., and "Treatise on Artillery," by Lt. Col. C. II. Owen, R. A.
FIG. 1.
Fig. 2. | 4,256 | 20,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-18 | longest | en | 0.950697 |
http://tailieu.vn/doc/ebc-fun-with-pencils-vui-voi-but-chi-phan-3-258827.html | 1,506,163,272,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689615.28/warc/CC-MAIN-20170923085617-20170923105617-00645.warc.gz | 335,548,061 | 20,763 | # EBC - Fun with pencils - Vui với bút chì - Phần 3
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## EBC - Fun with pencils - Vui với bút chì - Phần 3
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## Nội dung Text: EBC - Fun with pencils - Vui với bút chì - Phần 3
1. PART THREE A WORLD FOR YOUR FIGURES TO LIVE IN 97
2. PERSPECTIVE 98
3. HOW TO ESTABLISH FIGURES ON THE GROUND 99
4. PERSPECTIVE IN THE FIGURE 100
5. COMMON FAULTS 101
6. FURNITURE 102
7. HOW TO PROJECT FURNITURE ONTO THE GROUND PLANE Here is an excellent method for building furniture and figures on a ground plane. It is simpler 103
8. BUILDING AN INTERIOR FROM A GROUND PLAN-I 104
9. BUILDING AN INTERIOR FROM A GROUND PLAN-II 105
10. BUILDING AN INTERIOR FROM A GROUND PLAN-III 106
11. BUILDING AN INTERIOR FROM A GROUND PLAN-IV BEDTIME And here is the finished drawing. It’s fun to try inking in some of your pencil drawings. Get a bottle of waterproof black drawing ink. You can get a box of school water colors, also, and get still more fun out of it. Knowing just what is the correct perspective helps so much to give that solid, finished, and professional look. This procedure opens up a whole world for the little figures you have learned to draw. It is worth while to see what you can do with this method. It offers a possi- bility of setting some work, besides the thrill of doing it. Now we shall take up a new subject. 107
12. LIGHT AND SHADOW: THE PRINCIPLE Rays of light travel in straight lines. From any spot, the middle ray, the “per- pendicular to source,” would meet the earth and pass through its center. At the point directly under the source we establish the point DL, meaning “direction of light.” S will mean “source” at the top of the perpendicular, From the farthest limit of the shadow to DL, then up to the source and back to the shadow, forms a triangle. The third corner of the triangle will be called At, mean- ing “angle of light.” DL may be the vanishing point of the shadow or the base from which it proceeds outward. 108
13. A SIMPLIFIED METHOD FOR GROUND SHADOWS 109 | 618 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | longest | en | 0.734049 |
http://list.seqfan.eu/pipermail/seqfan/2016-December/068775.html | 1,695,670,807,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00715.warc.gz | 26,761,676 | 6,604 | # [seqfan] Re: A new sequence in Dutch magazine "Pythagoras"
Richard J. Mathar mathar at mpia-hd.mpg.de
Fri Dec 2 14:59:57 CET 2016
```In response to http://list.seqfan.eu/pipermail/seqfan/2016-December/017115.html
njas> By the way, what about the other new sequences in the same Pythagoras
njas> article? See "Andere Rijen" - can someone explain what they are? (And then
njas> they too should be added to the OEIS.)
njas> (Here is the link again: http://www.pyth.eu/jaargangen/Pyth55-6.pdf See
njas> page 21)
Two of the series on page 21 are just a notice of the editor that some
submitters forgot to include some members; after correction
they are actually A016105 and A175746.
To understand the remaining sequence on page 21 of PYTHAGORAS_JG55_No6.pdf
one probably needs to read the article starting on page 26 of the earlier issue
of the PYTHAGOROS_JG55_no5.pdf, which contains 21, 33, 57,.. as a table
read down columns: Here is a rough translation:
(by Arnout Jaspers)
Creating mega-periodic Fractions
If you write 1/7 as a decimal number, you get 0.bar(142857).
The bar(..) above the 6 digits means that the group
is repeated indefinitely. We say: the period of 1/7 has a period 6.
In the article "The period of a decimal number" in the January issue
we stated (without proof)
Take two prime numbers a and b. If their inverses 1/a and 1/b, written
as decimals, have periods p_a and p_b, the product 1/(a*b)
has period lcm(p_a,p_b). Here lcm means the least common multiple.
With the aid of 9-numbers of the previous article the proof is easy. We
show it by example, that is a=7 and b=17. We observed that 1/7 has period 6.
Also 1/17= 0.bar(0588235294117647) has period 16. Then the fraction 1/(7*17)
=1/119 should have, written as decimals, have period lcm(6,16)=48.
Earlier we showed that each periodic fraction may be written as a fraction
with a 9-number in the denominator:
0.bar(142857) = 142'857/999'999
and
0.bar(0.588235294117647)=588'235'294'117'647/9'999'999'999'999'999.
You see: the number of 9's in the denominator equals the period of the decimal
number.
1/7=142857/999'999
and
1/17 = 588'235'294'117'647/9'999'999'999'999'999
and this fits in, because 7*1542857 =999'999 and
17*588'235'294'117'647=9'999'999'999'999'999.
In fact the period of a fraction (written as a decimal): 999'999 is
the smallest 9-number with a factor of 7, and the smallest 9-number with
a factor of 17 is 9'999'999'999'999'999.
The writeup of a periodic fraction as a simple fraction with a 9-number
in the denominator now demands:
(*) Vor 1/(7*17), the product of 1/7 by 1/17, we need a 9-number in
the denominator with a factor of 7 as well as a factor of 17, so we may
write 1/119 = 1/(7*17)=(some integer number)/(9-number with factor 7 and factor 17).
Now we use two properties of 9-numbers that we already proved:
- a 9-number of P-1 digits has at least one factor P (if P is neither 2 nor 5)
- if the 9-number with N digits is the smallest 9-number with P as factor,
then all the 9-numbers N, 2N, 3N.. have one or more factors P.
Applied to our example this means we already know that 999'999 must contain
a factor 7 and 9'999'999'999'999'999 a factor 17. How do we construct a 9-number
that has a factor 7 as well as a factor 17? Easy: because of the second property
all 9-numbers with 6, 12, 18... digits contain a factor 7, and all 9-numbers with
16, 32,.. digits contain a factor 17. The smallest 9-numbers that is in both
sequences is the 9-number with lcm(6,16)=48 digits. By the second property these
two sequences are also the *only* 9-number with the factors 7 and 17, and
therefore the 9-number with 48 digits is indeed the smallest possible one.
This means the faction 1/119 is written as (a integer with 48 digits)/(9-number with 48 digits).
In decimal writeup this is a fraction with period 48, i.e.
1/19=0.bar(0084033613....521)
If a=b this rule cannot hold, as lcm(a,a)=a.
We return to the requiremnt writen in (*), but in view of the case a=b,
with a=7 as example:
(*) Vor 1/(7*7) we need a 9-number in the denmoinator with two factors 7,
so we can write 1/49=1/(7*7) = (an integer)/(9-number with two factors 7).
We already know that 1/7 has period 6, so the 9-numbers with 6, 12, 18...
digits all have a factor 7. But how does another factor 7 join? In general:
if the smallest 9-number with a factor P with N digits is 10^N-1, what is
the smallest 9-numbers with a factor P^2?
As 10^N-1 has a factor P, it is a multiple V of P, so we can write
10^N-1=V*P
whence
10^N=V*P+1
and
10^(P*N)=(10^N)^P=(V*P+1)^P.
Write this up with Newton's binomial expansion, (a+b)^M.
(VP+1)^P=(VP)^P+P*(VP)^(P-1)+...+P(P-1)*(VP)^2+P(VP)+1.
We now have
10^(PN)-1=(VP)^P+P*(VP)^(P-1)+...+P(P-1)*(VP)^2+P*(VP).
All terms on the right hand side have a factor P^2, so 10^(PN)-1 is a 9-number
with P*N digits that is divisible by P^2, precisely what we looked for.
But: is 10^(PN)-1 the *smallest* 9-number with factor P^2? Assume that there
is a smaller 9-number with factor P^2. This 9-number then has the form 10^(A*N)-1
(because all these 9-numbers have one or more factors) with A<P. In the same
manner as in the previous article happened for prime factor P one may proove
that only the 9-numbers 10^(A*N)-1, 10^(2*A*N)-1, 10^(3*A*N)-1 have one or more
factors P^2. In general: each 9-number 10^(Q*A*N)-1, with Q=1,2,3... has a factor P^2.
But we have not shown that in each case 10^(P*N)-1 has a factor P^2, therefore
that must show up in the sequence of 9-numbers. Is this possible?
This may only occur as P*N=Q*A*N, therefore with P=Q*A. But here that states
that P is the product of two integers, where P is a prime number. This cannot
be, therefore there is no smaller 9-number with factor P^2 than 10^(P*N)-1.
Back to the example 1/7. This fraction has period 6, so the denominator is
the 9-number with 6 digit, 10^6-1. We like to add a factor 7, so P is 7, thus
the first 9-number with two factors 7 is 10^(6*7)-1=10^42-1. The period of 1/(7*7)
should be 42. Indeed
1/49 = 0.(020408...51).
Base fractions with more than two prime factors.
Vor small prime factors a the period fo 1/a is already the maximum, a-1.
So 1/7 has period 6 and 1/17 period 16. That does not need to be always correct:
1/13 has period 6. Vor a larger prime G one cannot predict whether the inverse 1/G
has the maximum period length G-1; this becomes more unlikely as G increases.
This happens only as G appears in the (G-1)st 9-numbers as a prime factor.
Look for example at G of approximately 1 million. Then a prime factor G
has approximately a 1-millionth higher change to be earlier in the
sequence of 9-numbers. The larger the 9-numbers get, the larger the
probability that the number of prime factors (on the average) increases
as the 9-number increases.
So if you wish to construct a fraction with small numbers in the
denominator, but a period as long as possible, you need to choose a small
prime number P with longest possible period P-1. The fraction 1/P^k
then has a period (P-1)*P^(k-1).
All the prvious remarks are also valid for fractions 1/(a*b*c....) with more
than two distinct prime factors. Then you need to find the smallest 9-number
with factors a,b,c.., and this has lcm(p_a,p_b,p_c,...) digits.
So periods may increase swiftly. For example 1/(17*29*47) = 1/23171 is a frctio
with period lcm(16,28,46) =2576.
You may check yourself the calculation of the fraction with online calculators
for large numbers, for example https://defuls.ca/big-number-calculator.htm .
This - and other calculators - don't show digits after the dot; so even
a fraction 1/7 is shown as 0. But you may patch this simply: for the fraction
1/(17*29*47) key in:
10^10000/(17*29*47)
431 573 950 196 366 ...
You may check that the first 2576 digits are indeed the period, by search
with 'control-F' (Windows) or 'cmd-F' (Mac) for the first digits, then for
'431 573 950' (dont forget the blanks). You find the block again after 2576,
5152 and 7728 digits, but nowhere alse.
Product of arbitrary periodic fractions..
Finally we look at an more general case: take two arbitrary periodic
fractions, as 0.bar(523) and 0.bar(3054478), and multiply these. What is
the period length of the product? We note
0.bar(523)*0.bar(3054478) = 523/999 * 3054478/9999999
and search for a 9-number such that
523/9999 * 3054.378/9999999 = (an integer)/(a 9-number). (**)
Note: the product of two 9-numbers is not itself a 9-number (Observe:
(10^N-1)(10^M-1) = 10^(N+M)-10^N-10^M+1, which may not be written as 10^R-1
for some R. Otherwise we could have simply calculated 999*9999999, put
this 9-number in the denominator, 523*3054478, and that kills it.
To find the 9-number in the denominator, we decompose both 9-numbers
left of the equal sin in factors:
999=3^3*57 and 9999999=3^2*239*4649.
To write the product as a fraction with a 9-number in the denominator
we need a 9-number with the factors 3^5, 37, 239 and 4649. In the list
of factorization of 9-numbers (see pages 25-26 in the January issue)
we see that 239 and 4649 occur in the 9-numbers ith 7 and 14 digits,
thus also in those with 21, 28...digits. 37 appears in the 9-numbers
with 3, 6,9.. digits. Because lcm(3,7)=21, 239, 4649 and 37 occur in the
9-numbers with 21, 42, 63,...digits.
With the factor 3 we have not to deal differently because it occurs
in both denominators, comparable to what happens if you multiply
two identical reduced fractions. We said that if a 9-number with
N digits has a factor P, the 9-number with N*P ciffers has factor P^2.
This concludes that a 9-number with N*P^2 ciffers has a factor P^3,
and so forth.
All 9-numbers are divisible by 9, so they have a factor 3^2. So the
9-numbers with 3, 6,9,... digits have a factor 3^3, the 9-numbes with
9,18,27.. digits have a factor 3^4, and the 9-numbers with 27, 54, 81,... digits
have a fctor 3^5. The smallest 9-number with all the factors needed,
3^5, 37, 239 and 4649, therefore has lcm(3,7,27)=189 digits.
Now look again at the expression (**). The denominator, the 9-number
with 189 digits, has the factors 3^5, 37, 239 and 4649 and others, which
we symbolically replace by [others]. The numerator is the
number 523 *3054748 * [others]. Therefore
523/999 * 3054478/9999999 = 523*3054*478*others / (3^5*37*239*4649*others)
[others] carries the minimum of factors that is needed to pump up the
factors 3^5, 37, 239 and 4649 to a 9-number. Conclusion: the product
of a decimal fraction with period 3 and a decimal fraction with period 7
has at most period 189.
Note: We write 'at most' because with an arbitrary fraction with 999
in the denominator we cannot exclude that this may be reducable to
a smaller fraction, and likewise for a fraction with 9999999 in the
denominator. Example: each 9-number is divisible throuh 9 (therefore
also by 3), so if you can divide the numerator through 9 (or 3), such
a fraction can be reduced. All 9-numbers with an even number of digits
are divisible through 11, so if you can divide the numerator through 11,
the fraction can also be simplified. In that case the 9-number in the
denominator of the product of the two periodic fractions must have a smaller
number of factors, and may be replaced by a smaller 9-number. If you
chose for both numerators prime numbers, this is exculded and you
get a maximum period.
This way you may step by step compute the maximum period of fractions
with length 2,3,4.. This is summarized in the table on the bottom of the page.
This table shows that the product of two periodic fractions with the
same period delivers much longer periods than a product of two
periodic fractions with different periods.
This means for example, that if you chose two prime numbers of just 3
digits, say 359 and 617, the product 0.bar(359)*0.bar(167) has period 2997.
For two periodic fractions with period 5 you already arrive at a maximum
period of 499995, close to half a million! Prime numbers with 5 digits
are plentiful; lists of these are on the internet. The computer mentioned
earlier may handle numbers to half a million digits, so you can create
fractions with these enormous periods by yourself.
At last: the table suggests that the maximum length for two fractions
with period N equals N*(10^N-1). Whether this is true we leave as
an open problem to the reader...
1 2 3 4 5 6 7
----------------------------------
1 9 18 27 36 45 54 63
2 198 54 396 90 594 126
3 2997 108 135 5994 189
4 39996 180 3564 252
5 499995 270 351
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Documentation Center
# Distorted Mix Method for constructing copulas with tail dependence
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<rdf:Description>
<dc:creator>Li, Lujun</dc:creator>
<dc:date>2014-07-07</dc:date>
<dc:description xml:lang="es">Sumario: This paper introduces a method for constructing copula functions by combining the ideas of distortion and convex sum, named Distorted Mix Method. The method mixes different copulas with distorted margins to construct new copula functions, and it enables us to model the dependence structure of risks by handling the central and tail parts separately. By applying the method we can modify the tail dependence of a given copula to any desired level measured by tail dependence function and tail dependence coefficients of marginal distributions. As an application, a tight bound for asymptotic Value-at-Risk of order statistics is obtained by using the method. An empirical study shows that copulas constructed by this method fit the empirical data of SPX 500 Index and FTSE 100 Index very well in both central and tail parts.</dc:description>
<dc:identifier>https://documentacion.fundacionmapfre.org/documentacion/publico/es/bib/148453.do</dc:identifier>
<dc:language>spa</dc:language>
<dc:rights xml:lang="es">InC - http://rightsstatements.org/vocab/InC/1.0/</dc:rights>
<dc:type xml:lang="es">Artículos y capítulos</dc:type>
<dc:title xml:lang="es">Distorted Mix Method for constructing copulas with tail dependence</dc:title>
<dc:relation xml:lang="es">En: Insurance : mathematics and economics. - Oxford : Elsevier, 1990- = ISSN 0167-6687. - 07/07/2014 Volumen 57 Número 1 - julio 2014 </dc:relation>
</rdf:Description>
</rdf:RDF>
`````` | 501 | 1,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.574198 |
http://iea-etsap.org/forum/showthread.php?tid=165&pid=729&mode=threaded | 1,714,011,664,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00804.warc.gz | 17,667,950 | 7,712 | Calculation of INVCOST for technology with negative ILED earlier than the base year Antti-L Super Moderator Posts: 401 Threads: 18 Joined: May 2010 14-08-2019, 10:38 PM (This post was last modified: 14-08-2019, 10:54 PM by Antti-L.) pauldodds Wrote:For 2020, I can almost recreate the TIMES calculation (I'd be interested to know the cause of the 0.05% error, but am not worrying about it).The cause of the 0.05% error is the use of time-dependent general discount rates (G_DRATE). I think one of the main ideas behind the annualization of the investment costs is that, assuming zero risk premium (no NCAP_DRATE specified) and zero IDC (no NCAP_ILED specified), the lump-sum present value of the investment payments should be equal to the original investment cost (NCAP_COST), when the annual payments are discounted back to the beginning of the commissioning year. However, your calculation is inconsistent with this equivalence principle. For some extreme examples, consider buying an electric car in 2012, which has a price £30,000 and ELIFE=10. Assume first that G_DRATE is constant, with the values 1%, 10% and 100%. With all these discount rates, the lump-sum investment cost obtained by discounting the annual investment payments (£3700, £4439, £15015) back to the beginning of the commissioning year is still in all cases exactly £30,000. But now, assume that G_DRATE is 10% in 2012, but then either A) decreases to 1% or B) increases to 100% starting from 2013. Using your calculation method, in the first case A), the lump-sum investment cost would have a value of £42,459, and in case B) it would have a value of only £8,868! So, in case B) the electric car, which is bought in 2012 with the price £30,000, would effectively cost only £8,868, just because of the annualization of the investment cost. I think such a price reduction would be inconsistent, but perhaps you disagree? To avoid such artificial alteration in prices when G_DRATE is changing, TIMES currently does not follow your calculation, but discounts the payments back to the commissioning year by the G_DRATE of that year, and therefore in TIMES the lump-sum payment is consistent with NCAP_COST. Would you prefer that your approach is adopted instead? « Next Oldest | Next Newest »
Messages In This Thread Calculation of INVCOST for technology with negative ILED earlier than the base year - by pauldodds - 13-08-2019, 01:06 AM RE: Calculation of INVCOST for technology with negative ILED earlier than base year - by Antti-L - 13-08-2019, 04:24 AM RE: Calculation of INVCOST for technology with negative ILED earlier than the base year - by pauldodds - 14-08-2019, 03:09 AM RE: Calculation of INVCOST for technology with negative ILED earlier than base year - by Antti-L - 14-08-2019, 10:38 PM RE: Calculation of INVCOST for technology with negative ILED earlier than the base year - by pauldodds - 02-09-2019, 05:51 PM RE: Calculation of INVCOST for technology with negative ILED earlier than base year - by Antti-L - 03-09-2019, 09:28 PM
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## Presentation on theme: "Chapter 5 – Trigonometric Functions: Unit Circle Approach 5.4 - More Trigonometric Graphs."— Presentation transcript:
Chapter 5 – Trigonometric Functions: Unit Circle Approach 5.4 - More Trigonometric Graphs
Cosecant
Graphing y = Acsc(Bx - C) +D Graph the sine function with dotted lines. The max point of the sine function is the MINIMUM point of the cosecant function. The min point of the sine function is the MAXIMUM point of the cosecant function. Where the sine function and y = D intersect are the vertical asymptotes of the cosecant function.
Cosecant Example Graph the following equation:
Secant
Graphing y = Asec(Bx - C) + D Graph the cosine function with dotted lines. The max point of the cosine function is the MINIMUM point of the secant function. The min point of the cosine function is the MAXIMUM point of the secant function. Where the cosine function and y = D intersect are the vertical asymptotes of the secant function.
Secant Example Graph the following equation:
Tangent
Graphing y = Atan(Bx - C) + D Find two consecutive asymptotes A pair of consecutive asymptotes occur at Find the point midway between the asymptotes (this is the x-intercept if there is no vertical shift; the y-value is the D). Find the points on the graph that are ¼ and ¾ of the way between the asymptotes. These points will have the y-values of D+A and D-A respectively.
Tangent Example Graph the following equation:
Cotangent
Graphing y = Acot(Bx - C) + D Find two consecutive asymptotes A pair of consecutive asymptotes occur at Find the point midway between the asymptotes (this is the x-intercept if there is no vertical shift; the y-value is the D). Find the points on the graph that are ¼ and ¾ of the way between the asymptotes. These points will have the y-values of –A+ D and A+D respectively.
Cotangent Example Graph the following equation:
Graphing Summary
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Get Full Access to University Physics, Volume 3 - 17 Edition - Chapter 5 - Problem 17
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ISBN: 9781938168185 2032
Solution for problem 17 Chapter 5
University Physics, Volume 3 | 17th Edition
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Problem 17
Relativistic Energy
What happens to the mass of water in a pot when it cools, assuming no molecules escape or are added? Is this observable in practice? Explain.
Step-by-Step Solution:
Step 1 of 3
• pH and Acid/Base Balance • What is pH – Can be thought of as the power of Hydrogen concentration in a solution – Mathematical definition- negative logarithmic value of the hydrogen ion (H+) concentration, or • pH = -log [H+] – Negative Logarithm- a negative scale based on ten • As the hydrogen ion concentration goes up, the scale goes down • Example: 1 is ten times more that 2 » 3 is a hundred time less than 1 (10 x 10 = 100) • What is the pH for an Acid Neutral A base • pH of Common Products • Physiological pH • Normal body pH is between 7.35-7.45 or ~ 7.4 • So, normal body pH is slightly basic • What pH range is compatible with life – pH 6.8 to pH 8 • Acid Disassociation Constant • Given a weak acid “HA”, its disassociation in water is subject to the following equilibrium : + – – HA + H+O ↔ H–O + A or HA ↔ H + A • Ka is the disassociation constant for acid – The stronger the acid, the higher the Ka • Ka=1, almost completely disassociated • Ka=0, almost completely associated pKa- the acidity constant is often represented by the inverse of its logarithm • Types of acids 1. Volatile - carbonic acid (can leave solution and enter the atmosphere) 2. Fixed acids - sulfuric and phosphoric (doesn'
Step 2 of 3
Step 3 of 3
Related chapters
Unlock Textbook Solution | 524 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-21 | longest | en | 0.844522 |
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summary:
Complex numbers are constrained to move with their basis vectors, unable to explore all of spacetime.
description:
Complex numbers have some freedom to move in space. The 3 straight lines indicate a choice of Cartesian basis vectors, but other basis vectors could have been chosen. Complex numbers are used extensively in quantum mechanics. Yet the obvious limitations in the animations suggest we should rebuild the foundations of complex-valued quantum mechanics. Sounds like a lot of work!
command:
q_graph -out complex -dir vp -loop 0 -box 5 -command 'q_add_n -3 4 0 0 .002 -0.008 0 0 1000' -color yellow -command 'q_add_n 4 0 4 0 -.006 0 -0.008 0 1000' -color blue -command 'q_add_n -6 0 0 -6 .012 0 0 .012 1000' -color green
math
equation:
$(2 q + q^* + (i q i)^*) / 2 = (t, x, 0, 0)$
$(2 q + q^* + (j q j)^*) / 2 = (t, 0, y, 0)$
$(2 q + q^* + (k q k)^*) / 2 = (t, 0, 0, z)$
numbers:
tags
Math Tag:
real numbers
complex numbers
Programming Tag:
command line quaternions | 324 | 1,023 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 3, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-22 | latest | en | 0.726411 |
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Answered by: Raouf Rahiche Date created: May 22, 2022
## What is the pay rate for a 14 year old at McDonalds?
Asked By: Koushik Roy Date created: May 15, 2022
Typically a 14 year old person would get paid a rate of \$10 or a little more during the working week. In addition to this, the older you turn the more your rate increases either by a few cents or dollars. A 14 year old would average around \$150 for a 12 hour week.
Answered By: Hasan Abbasi Date created: May 16, 2022
## What is the pay rate for a 17 year old?
Asked By: Tom Date created: May 24, 2022
The NMW for 18- to 20-year-olds will increase from £5.60 to £5.90 per hour; The NMW for 16- and 17-year-olds will increase from £4.05 to £4.20 per hour; and.
Answered By: vestland Date created: May 25, 2022
## What is the hourly rate for part time?
Asked By: techraf Date created: May 08, 2022
To get the hourly equivalent rate for the job, divide \$40,000 by 2,080 hours (2,080 equals 40 hours per week times 52 weeks in a year). That equals \$19.23 per hour. Or, if you want help with the math, using an hourly to salary calculator may work better for you.
Answered By: Yuck Date created: May 08, 2022
## What is the going rate for pocket money for a 12 year old?
Asked By: kakarukeys Date created: May 17, 2022
How much?AgePocket Money4-6\$ 7.177-9\$ 7.0410-12\$11.3713-15\$14.11
Answered By: strager Date created: May 19, 2022
## What is the hourly rate at Amazon?
Asked By: JayDev Date created: May 15, 2022
Amazon.com Inc Jobs by Hourly RateJob TitleRangeAverageJob Title:Warehouse AssociateRange:\$13 - \$17Average:\$15Warehouse WorkerRange:\$12 - \$17Average:\$15Delivery DriverRange:\$12 - \$21Average:\$16Human Resources (HR) AssistantRange:\$16 - \$27Average:\$213 more rows•May 5, 2021
Answered By: kojow7 Date created: May 17, 2022
## What is the hourly rate for 20 000 a year?
Asked By: Cory Klein Date created: May 20, 2022
Yearly / Monthly / Weekly / Hourly Converter If you make £20,000 per year, your hourly salary would be £9.62. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 40 hours a week.
Answered By: Frank Farmer Date created: May 21, 2022
## What is the hourly rate for salary?
Asked By: Xenolion Date created: May 20, 2022
To determine your hourly wage, divide your annual salary by 2,080. If you make \$75,000 a year, your hourly wage is \$75,000/2080, or \$36.06. If you work 37.5 hours a week, divide your annual salary by 1,950 (37.5 x 52). At \$75,000, you hourly wage is \$75,000/1,950, or \$38.46.
Answered By: m4n0 Date created: May 23, 2022
## What is the pay rate for 21 year old?
Asked By: Dhanu K Date created: May 08, 2022
The NMW for 21- to 24-year-olds will increase from £7.05 to £7.38 per hour; The NMW for 18- to 20-year-olds will increase from £5.60 to £5.90 per hour; The NMW for 16- and 17-year-olds will increase from £4.05 to £4.20 per hour; and.
Answered By: cnst Date created: May 08, 2022
## What is the hourly rate for 26 000 a year?
Asked By: Jon Cage Date created: May 19, 2022
Converting £26,000 a year in another time unitConversionUnitBiweekly salary£26,000 a year is £1,000 per 2 weeksWeekly salary£26,000 a year is £500 per weekDaily salary£26,000 a year is £100 per dayHourly salary£26,000 a year is £12.50 per hour1 more row
Answered By: Tarun Lalwani Date created: May 20, 2022
## What is the maximum heart rate for 16 year old?
Asked By: Pekka Date created: May 17, 2022
What is a Typical Pulse?Age GroupNormal Heart Rate at RestChildren (ages 3-4 years)70 - 100 beats per minuteChildren (ages 5-7 years)65 - 100 beats per minuteChildren (ages 8-15 years)60 - 100 beats per minuteAdults (ages 16-18+ years)60 - 100 beats per minute
Answered By: Nabin Date created: May 19, 2022
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User
### What is a good resting heart rate for a 54 year old woman?
Resting Heart RateWOMENATHLETE54-6054-59EXCEL'T61-6560-64GOOD66-6965-68ABOVE AV70-7369-724 more rows
User
### What is a good resting heart rate for a 62 year old woman?
Your resting heart rate For most healthy adult women and men, resting heart rates range from 60 to 100 beats per minute.
User
### What is a normal resting heart rate for a 60 year old woman?
A resting heart rate is normal between 60-100 beats per minute. A resting heart rate is fast (i.e. tachycardic) at greater than 100 beats per minute.
User
### What is the average hourly rate for a gunsmith?
\$22 an hourSalary Recap The average pay for a Gunsmith is \$45,062 a year and \$22 an hour in the United States.
User
### What is the hourly rate for \$30000 a year?
\$15.38If you make \$30,000 per year, your hourly salary would be \$15.38. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 37.5 hours a week.
Guest
### What should my resting heart rate be at 30?
For a 30-year-old person, for example: 220 – 30 = 190. The target zone for a 30-year-old person would be between 50 and 85 percent of his or her maximum heart rate: 50 percent: 190 x 0.50 = 95 bpm. 85 percent: 190 x 0.85 = 162 bpm.
Guest
### Is 124 heart rate bad?
Moderate intensity: 50-69 percent of your maximum heart rate. 90 to 124 beats per minute for a 40-year-old. High intensity: 70-90 percent of maximum heart rate. 125 to 162 beats per minute for a 40-year-old.
Guest
### What is the average hourly rate for a carer?
Average London Care hourly pay ranges from approximately £9.62 per hour for Support Worker to £10.50 per hour for Care Assistant. The average London Care salary ranges from approximately £16,000 per year for Home Care Worker to £17,500 per year for Support Worker.
Guest
### What is the average hourly rate for a home carer?
How much does London Care in the United Kingdom pay? Average London Care hourly pay ranges from approximately £9.54 per hour for Support Worker to £10.50 per hour for Care Assistant. The average London Care salary ranges from approximately £16,000 per year for Home Care Worker to £17,500 per year for Support Worker.
Guest
### What is the best mortgage rate today?
Mortgage rate trendsProductRateLast week30-year fixed3.10%3.09%15-year fixed2.38%2.37%30-year jumbo mortgage rate3.15%3.14%30-year mortgage refinance rate3.14%3.13%16 hours ago
Guest
### What is the minimum hourly rate for a 15 year old?
\$7.25 per hourHow much should I be paid? An employer must pay you at least the federal minimum wage of \$7.25 per hour for all the hours that you work, except under certain circumstances.
Guest
### What is the hourly rate for a 21 year old?
The minimum wage is the base rate of pay that must be paid to workers who are permanent employees and aged 21 years or older. The current national minimum wage which takes effect on July 1st 2019 is \$19.49 per hour. An employer cannot pay you less than the minimum wage even if you agree to it.
Guest
### What is the hourly rate for KFC?
Average KFC hourly pay ranges from approximately \$21.84 per hour for Team Member to \$25.00 per hour for Waiter/Waitress. The average KFC salary ranges from approximately \$44,833 per year for Shift Manager to \$62,913 per year for Retail Manager.
Guest
### What is the minimum rate of pay for a 16-17 year old in the UK?
NLW and NMW rates from 1 April 2020Previous rateCurrent rate from 1 April 2020National Living Wage£8.21£8.7221-24 Year Old Rate£7.70£8.2018-20 Year Old Rate£6.15£6.4516-17 Year Old Rate£4.35£4.552 more rows•Apr 1, 2020.
Guest
### What is your desired hourly rate or annual salary GameStop?
Average GameStop hourly pay ranges from approximately \$8.21 per hour for Associate General Counsel to \$27.39 per hour for District Manager. The average GameStop salary ranges from approximately \$17,279 per year for Retail Manager to \$136,015 per year for Hadoop Developer.
Guest
### What is a dangerous heart rate for an infant?
Infants 1 to 11 months old: 80 to 160 beats per minute. Children 1 to 2 years old: 80 to 130 beats per minute. Children 3 to 4 years old: 80 to 120 beats per minute. Children 5 to 6 years old: 75 to 115 beats per minute.
Guest
### What is the average resting heart rate for a 60 year old?
A normal heart rate is between 60 and 100 bpm while you're resting.
Guest
### What is the hourly rate in Spain?
The average hourly wage (pay per hour) in Spain is 16 EUR. This means that the average person in Spain earns approximately 16 EUR for every worked hour.
Guest
### What is the hourly rate for 80000?
Yearly / Monthly / Weekly / Hourly Converter If you make \$80,000 per year, your hourly salary would be \$40.49. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 38 hours a week.
Guest
### What is the hourly rate of 18000?
Yearly / Monthly / Weekly / Hourly Converter If you make £18,000 per year, your hourly salary would be £8.65. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 40 hours a week.
Guest
### What is the hourly rate for 25000 a year?
\$12.50 per hourIn this case, you can quickly compute the hourly wage by dividing the annual salary by 2000. Your yearly salary of \$25,000 is then equivalent to an average hourly wage of \$12.50 per hour.
Guest
### What is a normal heart rate for a sick child?
More than 160 beats per minute in a child less than 12 months old. More than 150 beats per minute in a child 12-24 months old. More than 140 beats per minute in a child 2-4 years old.
Guest
### What is the average hourly rate for a dental assistant?
The average hourly rate for Dental Assistant ranges from \$16 to \$20 with the average hourly pay of \$18.
Guest
### What is the going rate for pocket money for a 13 year old?
Pocket money by ageAgeAverage pocket money12 years old£6.60 per week13 years old£6.68 per week14 years old£7.61 per week15 years old£8.38 per week5 more rows•Aug 17, 2011
Guest
### What is the success rate of minoxidil?
In a one-year observational study, 62 percent of the 984 men using 5 percent minoxidil reported a reduction in hair loss. As for hair regrowth, the drug was rated as “very effective” in 16 percent of participants, “effective” in 48 percent, “moderately effective” in 21 percent, and “ineffective” in 16 percent.
Guest
### What is the hourly rate for 35000?
\$17.95If you make \$35,000 per year, your hourly salary would be \$17.95. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 37.5 hours a week.
Guest
### What is a normal pulse rate for a 90 year old?
A normal resting heart rate for an adult is between 60 and 100 beats per minute. Both tachycardia and bradycardia can be indicators of other health conditions.
Guest
### What is the hourly rate for 20000 a year UK?
£20,000 a year is how much per hour? If you make £20,000 per year, your hourly salary would be £9.62. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 40 hours a week.
Guest
### What is a good resting heart rate for my age?
3 to 4 years old: 80 to 120 bpm. 5 to 6 years old: 75 to 115 bpm. 7 to 9 years old: 70 to 100 bpm. 10 years and older: 60 to 100 bpm.
Guest
### What is a normal heart rate for a 90 year old woman?
A normal resting heart rate for an adult is between 60 and 100 beats per minute.
Guest
### What is a good heart rate for a 70 year old?
60 years: 80 to 136 beats per minute. 65 years: 78 to 132 beats per minute. 70 years: 75 to 128 beats per minute.
Guest
### What is a normal pulse rate for a 80 year old woman?
As expected, the peak heart rate tended to decline with age. It averaged 175 beats per minute for 35-year-old women and about 135 beats per minute for 80-year-old women (Circulation, July 13, 2010).
Guest
### What hourly rate is 25k?
£25,000 a year is how much per hour? If you make £25,000 per year, your hourly salary would be £12.02. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 40 hours a week.
Guest
### What is a good heart rate for 50 year old?
So for a 50-year-old, maximum heart rate is 220 minus 50, or 170 beats per minute. At a 50 percent exertion level, your target would be 50 percent of that maximum, or 85 beats per minute. At an 85 percent level of exertion, your target would be 145 beats per minute.
Guest
### What is hourly rate for 125k?
\$64.10 per hourConverting \$125,000 a year in another time unitConversionUnitBiweekly salary\$125,000 a year is \$4,808 per 2 weeksWeekly salary\$125,000 a year is \$2,404 per weekDaily salary\$125,000 a year is \$481 per dayHourly salary\$125,000 a year is \$64.10 per hour1 more row
Guest
### What is the hourly rate for 72800 a year?
Your annual salary of \$72,500 would end up being about \$34.86 per hour.
Guest
### What is the hourly rate for 65000 a year?
about \$31.25 per hourBut if you get paid for 2 extra weeks of vacation (at your regular hourly rate), or you actually work for those 2 extra weeks, then your total year now consists of 52 weeks. Assuming 40 hours a week, that equals 2,080 hours in a year. Your annual salary of \$65,000 would end up being about \$31.25 per hour.
Guest
### What is a normal heart rate for a 3 month old baby?
Normal Results Newborns 0 to 1 month old: 70 to 190 beats per minute. Infants 1 to 11 months old: 80 to 160 beats per minute. Children 1 to 2 years old: 80 to 130 beats per minute. Children 3 to 4 years old: 80 to 120 beats per minute.
Guest
### What should my hourly rate be?
A common approach to figuring out an hourly rate is to divide the salary you want by the number of hours worked each year: 40 hours/week × 52 weeks/year = 2,080 hours.
Guest
### What is a good hourly rate in New Zealand?
The median for hourly earnings for wage and salary workers reached \$25.50 in the June 2019 quarter, up from \$25.00 last year (2.0 percent). | 4,831 | 17,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.941806 |
https://mystudywriters.com/maria-hernandez-case/ | 1,722,643,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00145.warc.gz | 327,422,356 | 31,083 | # Maria Hernandez Case
## Maria Hernandez Van Ness team 10
The Story: Maria Hernandez & Associates is a company that started its business with a cash deposit. On June 20,2004 Maria Hernandez transferred all her savings of \$30,000 into a new Bank account under her company’s name, two days later she transferred another \$20,000 which she had borrowed from her father on a 6% p. a. interest rate. Thus, with an amount of \$50,000 in its bank account Maria Hernandez and Associates was ready to start its life in the Webpage designing sector.
After the Bank account transactions, Maria Hernandez quickly took care of the initial expenses that included pre-paid rent for the new office, giving a security deposit for the same, buying used computers and software from her previous employers and also ordering and acquiring office stationary. On July 2,2004 Maria Hernandez & Associates opened its doors for business. The substance of our report covers the first two months of the company’s operations. At the start of the operations i. e.
July 2nd, 2004 the amount in the company bank account was \$12,000; however on August 31st, 2004 (roughly two months of operations) the amount had declined to \$6,600. We are therefore left with two key questions to answer. 1. How would we report on the operations of Maria Hernandez & Associates through August 31, 2004? Had the company made a profit as Maria Hernandez believed? If so, how can we explain the decline of cash in the bank? 2. What can we say about the status of the business on August 31, 2004?
To answer these questions we analyzed the company’s income statement and balance sheet for the months of July-August, 2004 and have come up with the following analysis and suggestions: Financial Ratios: Through the analysis of the Income Statement and Balance Sheet, we were able to extrapolate the following Ratios, which gave us an insight into the workings of Maria Hernandez & Associates |Financial Ratio |Figures | |Current Ratio |4. 17 | |Return on Equity |13% | Return in Assets |7% | |Profit Margin |9. 8% | |Debt to Equity |0. 74 | By and large, the ratio’s displayed are lower than ideal. However, given the fact that the operation is only 2 months old, the figures are very promising; especially since there was an increase in workload of the company in early August with four new clients by way of referrals. Considering all the ratios in more details we would like to start our analysis with ROE ratio that measures a company’s profitability.
We have 13% what means that the company is making 13 cents out of every dollar invested. This figure is relatively low, but for a start-up company it is rather satisfactory, because it indicates a growth opportunity with increasing operations. ROA ratio shows us how many dollars the company makes in relation to its assets, thus 7 cents per 1 dollar. The ratio is deceptive because by definition a lower ratio denotes inefficient use of assets. But considering a start-up that operates for only 2 months, there is a scope for improvement since the number of operations has been increasing.
In addition, this ratio can vary depending on the industry in which the company operates. This is why our suggestion to Maria Hernandez is to compare ROA every month in order to be able to realize how productive or unproductive the business is. Profit margin represents the percentage of revenue that a company keeps as profit after accounting for fixed and variable costs. In other words, it is company’s health indicator. The company is keeping 9. 8 cents of sales as earnings for every dollar that the company earns.
It is a good sign because the company was able to recoup the initial fixed costs and also showed a profit in the books within 2 months, on the other hand the usual trend for web-page design companies to show a profit is 1-2 years. Debt to equity ratio indicates extend to which the business relies on debt financing. As we know, Maria Hernandez borrowed \$20,000 from her father at 6% interest rate and invested \$30,000 cash from her own savings. In addition, the company made revenue of \$40,000 in cash that helped to cover all the expenses and operational purchases.
So, we can conclude that the company is growing on cash mainly and in the tech industry this ratio is bound to go down, because once the assets [computers and software] are acquired there is no need to take on debt to grow the company, as the growth can come from the revenue itself. On an average computer companies have a Debt to Equity ratio of under 0. 5 Current ratio that shows the ability of the company to pay off its liabilities at a given period of time is the only point of concern. As a rule the acceptable figure is between 1 and 2, in our case we have 4. 7, what means that Maria Hernandez can pay off her loan with interest however, she has some excessive cash on hand what indicates inefficient management of funds. Suggestions: We would first like to address the matter of treating the Interest and Depreciation. The interest is accumulating and since the interest has to be paid at the end of the year, the amount at the moment is incomplete. Therefore, the interest payable should be accounted in the Balance Sheet, and interest expense in the Income Statement. In case of the equipment, accumulated depreciation is to be taken into consideration.
The depreciation per month is \$750, thus the accumulated depreciation is \$1500 after 2 months of operation. As the expected life of the equipment is 3 years Maria Hernandez should credit the accumulated depreciation for 1/3 of the value of the assets, subtract accumulated depreciation from the equipment in the Balance Sheet and include depreciation expense in the Income Statement. An analysis of the Expense to Income ratio showed that currently 86% of the income is being used to write off expenses such as rent and salaries, which explains the decline in the bank balance as on August 31st.
We recommend reducing such expenses by keeping fewer full-time staff and hiring interns or keeping staff on a part-time basis at least for the initial period of the companies life. Conclusion: In conclusion we would like to say that Maria Hernandez & Associates is doing rather well as a Start-up company. The numbers are mostly in its favor, and are bound to get better as the life of the company progresses. The only flaw in the design is by way of the expenses incurred in form of Salaries, which can easily be fixed.
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