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http://chemistry.about.com/od/workedchemistryproblems/a/celsius-to-kelvin-temperature.htm | 1,406,092,673,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997874283.19/warc/CC-MAIN-20140722025754-00157-ip-10-33-131-23.ec2.internal.warc.gz | 28,648,318 | 13,364 | • Share
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# Celsius to Kelvin Temperature Conversion Example
## Example Problem Converting Celsius to Kelvin
Add 273 to a Celsius temperature to get Kelvin.
Gary S Chapman, Getty Images
The formula to convert Celsius to Kelvin is:
K = °C + 273
Here is an example problem that explains how to convert a temperature from degrees on the Celsius scale to Kelvin.
### Celsius to Kelvin Problem #1
Convert 27° C to Kelvin.
### Celsius to Kelvin Solution #1
K = °C + 273
K = 27 + 273
K = 300
300 K
Note that the answer is 300 K. Kelvin is not expressed in degrees.
### Celsius to Kelvin Problem #2
Convert 77° C to Kelvin.
### Celsius to Kelvin Solution #2
K = °C + 273
K = 77 + 273
K = 350
350 K
Anne Marie Helmenstine, Ph.D.
Explore Chemistry
By Category
2. Education
3. Chemistry
4. Chemistry Homework Help
5. Worked Chemistry Problems
6. Celsius to Kelvin Temperature Conversion Example | 273 | 1,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2014-23 | latest | en | 0.783452 |
https://www.usefullinks.org/cat/Combinatorial_algorithms.html | 1,723,159,192,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00196.warc.gz | 809,455,319 | 4,737 | # Category: Combinatorial algorithms
Garsia–Wachs algorithm
The Garsia–Wachs algorithm is an efficient method for computers to construct optimal binary search trees and alphabetic Huffman codes, in linearithmic time. It is named after Adriano Garsia and Michel
Bit-reversal permutation
In applied mathematics, a bit-reversal permutation is a permutation of a sequence of items, where is a power of two. It is defined by indexing the elements of the sequence by the numbers from to , rep
Fisher–Yates shuffle
The Fisher–Yates shuffle is an algorithm for generating a random permutation of a finite sequence—in plain terms, the algorithm shuffles the sequence. The algorithm effectively puts all the elements i
Heap's algorithm
Heap's algorithm generates all possible permutations of n objects. It was first proposed by B. R. Heap in 1963. The algorithm minimizes movement: it generates each permutation from the previous one by
Bender–Knuth involution
In algebraic combinatorics, a Bender–Knuth involution is an involution on the set of semistandard tableaux, introduced by , pp. 46–47) in their study of plane partitions.
Boltzmann sampler
A Boltzmann sampler is an algorithm intended for random sampling of combinatorial structures. If the object size is viewed as its energy, and the argument of the corresponding generating function is i
Kernighan–Lin algorithm
The Kernighan–Lin algorithm is a heuristic algorithm for finding partitions of graphs.The algorithm has important practical application in the layout of digital circuits and components in electronic d
Greedy algorithm
A greedy algorithm is any algorithm that follows the problem-solving heuristic of making the locally optimal choice at each stage. In many problems, a greedy strategy does not produce an optimal solut
Additive combinatorics is an area of combinatorics in mathematics. One major area of study in additive combinatorics are inverse problems: given the size of the sumset A + B is small, what can we say
Lemke–Howson algorithm
The Lemke–Howson algorithm is an algorithm that computes a Nash equilibrium of a bimatrix game, named after its inventors, Carlton E. Lemke and .It is said to be "the best known among the combinatoria
Lin–Kernighan heuristic
In combinatorial optimization, Lin–Kernighan is one of the best heuristics for solving the symmetric travelling salesman problem. It belongs to the class of local search algorithms, which take a tour
Robinson–Schensted–Knuth correspondence
In mathematics, the Robinson–Schensted–Knuth correspondence, also referred to as the RSK correspondence or RSK algorithm, is a combinatorial bijection between matrices A with non-negative integer entr
Jeu de taquin
In the mathematical field of combinatorics, jeu de taquin is a construction due to Marcel-Paul Schützenberger which defines an equivalence relation on the set of skew standard Young tableaux. A jeu de
Picture (mathematics)
In combinatorial mathematics, a picture is a bijection between skew diagrams satisfying certain properties, introduced by in a generalization of the Robinson–Schensted correspondence and the Littlewoo
Criss-cross algorithm
In mathematical optimization, the criss-cross algorithm is any of a family of algorithms for linear programming. Variants of the criss-cross algorithm also solve more general problems with linear ineq
SMAWK algorithm
The SMAWK algorithm is an algorithm for finding the minimum value in each row of an implicitly-defined totally monotone matrix. It is named after the initials of its five inventors, Peter Shor, Shlomo
Tompkins–Paige algorithm
The Tompkins–Paige algorithm is a computer algorithm for generating all permutations of a finite set of objects.
Loopless algorithm
In computational combinatorics, a loopless algorithm or loopless imperative algorithm is an imperative algorithm that generates successive combinatorial objects, such as partitions, permutations, and
Cycle detection
In computer science, cycle detection or cycle finding is the algorithmic problem of finding a cycle in a sequence of iterated function values. For any function f that maps a finite set S to itself, an
Reverse-search algorithm
Reverse-search algorithms are a class of algorithms for generating all objects of a given size, from certain classes of combinatorial objects. In many cases, these methods allow the objects to be gene
Steinhaus–Johnson–Trotter algorithm
The Steinhaus–Johnson–Trotter algorithm or Johnson–Trotter algorithm, also called plain changes, is an algorithm named after Hugo Steinhaus, Selmer M. Johnson and Hale F. Trotter that generates all of
Robinson–Schensted correspondence
In mathematics, the Robinson–Schensted correspondence is a bijective correspondence between permutations and pairs of standard Young tableaux of the same shape. It has various descriptions, all of whi
Map folding
In the mathematics of paper folding, map folding and stamp folding are two problems of counting the number of ways that a piece of paper can be folded. In the stamp folding problem, the paper is a str | 1,086 | 5,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.899953 |
https://www.cquestions.com/2008/01/c-program-for-find-power-of-number_03.html?showComment=1413432920920 | 1,726,504,000,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00797.warc.gz | 658,439,495 | 19,147 | ## INDEX
### FIND POWER OF A NUMBER USING C PROGRAM
How to calculate power of a number in c
How to write power in c
#include<stdio.h>
int main(){
int pow,num,i=1;
long int sum=1;
printf("\nEnter a number: ");
scanf("%d",&num);
printf("\nEnter power: ");
scanf("%d",&pow);
while(i<=pow){
sum=sum*num;
i++;
}
printf("\n%d to the power %d is: %ld",num,pow,sum);
return 0;
}
#### 31 comments:
Anubhav Shrivastava said...
Helpful post...thank you
Anonymous said...
this program shows segmentation fault in linux...
Shahid said...
Thanks alot my brother for helping us....................
Diwakar Mishra said...
thanx
Anuj said...
It is realy very helpfull for students
Anonymous said...
questns bagunayi boss!!!
Anonymous said...
Thanks
Anonymous said...
This comment has been removed by the author.
Anonymous said...
help me to calculate the power by using for loop
Unknown said...
#include
using namespace std;
int main()
{
int n,p,temp,i;
int total=0;
cout<<"Enter the number";
cin>>n;
cout<<"power of ";
cin>>p;
temp = n;
for(i=1;i<p;i++)
{
total = temp*n;
temp = total;
}
cout<<temp;
system("PAUSE");
cin.get();
return 0;
}
Anonymous said...
very helpful
Unknown said...
wap to c language
1
22
333
4444
55555
Anand said...
#include
main()
{
int i, j, k = 0;
for(i = 1; i <= 5; i++)
{
++k;
for(j = 1; j <= i; j++)
printf("%d",k);
printf("\n");
}
}
Rocky said...
#include
void main()
{
int n,i,j=1,k;
printf("enter n:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(k=1;k<=i;k++)
{
printf("%d",j);
}
printf("\n");
j++;
}
}
Unknown said...
#include
int main()
{
int num;
int pow;
int sum =1;
printf("\n enter the number \n");
scanf("%d",&num);
printf("\n enter the power of \n");
scanf("%d",&pow);
fo(;pow;pow--)
sum = sum *num;
printf("\n sum = %d \n", sum);
}
Nikhil said...
Program to enter the numbers till the user wants and at the end it should display the count of positive, negative and zeros entered.
Anonymous said...
/****Program to input 10numbers from the user and display whether it is even or odd****/
#include
#include
int main(void)
{
int number_of_terms=10;//Total number of input from user
int number[10];//Array to store the numbers given from users
int i;
printf("Enter 10 integers\n");
for(i=1;i<=10;i++)
scanf("%d",&number[i]);
i=1;
while(i!=(number_of_terms+1))
{
if(number[i]%2==0)
{
printf("%d is Even\n",number[i]);
}
else
{
printf("%d is Odd\n",number[i]);
}
i++;
}
getch();
return 0;
}
Unknown said...
Can anyone tell how to return the square of any given number in c...without using math.h,looping mechanisms and * sign..and ya it is possible..not kidding
Anonymous said...
x^2/m+x^5/m^3+x^8/m^5+x^11/m^7.............
please help........
RavindrA said...
I=j=1:
while(i<=1)
{
while(j<=i)
{
printf("%d",i);
}
printf("\n");
I++;
}
Anonymous said...
WAP in c to display follwing o/p....plz help
*
* *
* * *
* * * *
* * * * *
StudyWard Education said...
/*
WAP in c to display follwing o/p....plz help
*
* *
* * *
* * * *
* * * * *
*/
#include
void main()
{
int row,col,i,j;
printf("enter the no. of rows and columns\n");
scanf("%d \n %d",&row,&col);
for(i=0;i0)
printf(" ");
else
printf("*");
}
printf("\n");
}
}
StudyWard Education said...
This comment has been removed by the author.
StudyWard Education said...
#include
void main()
{
int row,col,i,j;
printf("enter the no. of rows and columns\n");
scanf("%d \n %d",&row,&col);
for(i=0;i0)
printf(" ");
else
printf("%d",i+1);
}
printf("\n");
}
}
computerFlashes said...
#include
int main()
{
int x,i,k,j;
printf("Enter any number \n ");
scanf("%d",&x);
for (i=1;i<=x;i++){
for(k=1;k<=i;k++)
printf("*");
for(j=1;j<=k*2;j++)
printf(" ");
printf("\n");}
return 0;
}
Unknown said...
for C# how to calculate the power of num
{
int num;
Console.WriteLine("enter the value");
num = Convert.ToInt32(Console.ReadLine());
num = num * num;
Console.WriteLine("power of given no = {0}",num);
Console.ReadLine();
}
Unknown said...
A program using for loop to find out X to the power N to the power M
#include
int main(){
int pow1,pow2,num,i;
long int sum1=1,sum2=1;
printf(" Enter a number: ");
scanf("%d",&num);
printf("\n Enter power 1: ");
scanf("%d",&pow1);
printf("\n Enter power 2: ");
scanf("%d",&pow2);
for(i=1;i<=pow1;i++)
{
sum1=sum1*num;
}
for(i=1;i<=pow2;i++)
{
sum2=sum2*sum1;
}
printf("\n%d to the power %d to the power %d is: %ld",num,pow1,pow2,sum2);
return 0;
}
Unknown said...
This comment has been removed by the author.
Unknown said...
Why you all are taking extra variable?
#include
int main()
{
int i; // you can take this value using scanf also.
for(i = 1; i <= 5; i++)
{
for(int j = 1; j <= i; j++)
{
printf("%d ",i); // print i here
}
printf("\n");
}
return 0;
}
Unknown said...
#include
int main()
{
int num,power,i,ans=1;
scanf("%d",&num);
scanf("%d",&power);
for(i=0;i<power;i++);
{
ans=num*ans;
}
printf("%d",ans);
return (0);
}
ans is not coming right
Unknown said...
Why does in for loop the formula is power= power * base? What does this mean? Help pls. | 1,508 | 5,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.618333 |
https://wlmb.github.io/2022/04/11/PWC160/ | 1,718,508,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00393.warc.gz | 552,646,869 | 5,177 | # Perl Weekly Challenge 160.
My solutions (task 1 and task 2 ) to the The Weekly Challenge - 160.
# Task 1: Four Is Magic
``````Submitted by: Mohammad S Anwar
You are given a positive number, \$n < 10.
Write a script to generate english text sequence starting with
the English cardinal representation of the given number, the
word ‘is’ and then the English cardinal representation of the
count of characters that made up the first word, followed by a
comma. Continue until you reach four.
Example 1:
Input: \$n = 5
Output: Five is four, four is magic.
Example 2:
Input: \$n = 7
Output: Seven is five, five is four, four is magic.
Example 3:
Input: \$n = 6
Output: Six is three, three is five, five is four, four is
magic.
``````
This is simple, as four is indeed magic (see below). We initialize an array with the digit names and use their length as an index to the next name, starting from the supplied argument. This fits into a oneliner.
``````perl -E '@e=qw(zero one two three four five six seven eight nine); for \$n(@ARGV){
while(\$n!=4){print "\$e[\$n] is ", \$e[\$n=length \$e[\$n]], ", "} say "four is magic"};
' 0 1 2 3 4 5 6 7 8 9
``````
Results:
``````zero is four, four is magic
one is three, three is five, five is four, four is magic
two is three, three is five, five is four, four is magic
three is five, five is four, four is magic
four is magic
five is four, four is magic
six is three, three is five, five is four, four is magic
seven is five, five is four, four is magic
eight is five, five is four, four is magic
nine is four, four is magic
``````
The corresponding full program would be:
`````` 1 # Perl weekly challenge 160
2 # Task 1: Four is magic
3 #
5 use v5.12;
6 use warnings;
7 die "Usage: ./ch-1.pl N1 [N2... ]\n".
8 "to find all trajectories between number names starting at index N1, N2..."
9 unless @ARGV;
10 my @names=qw(zero one two three four five six seven eight nine);
11 for my \$n(@ARGV){
12 say("Wrong input: \$n"), next if \$n>=@names;
13 while(\$n!=4){
14 print "\$names[\$n] is ", \$names[\$n=length \$names[\$n]], ", ";
15 }
16 say "four is magic"
17 };
``````
Example:
``````./ch-1.pl 0 1 2 3 4 5 6 7 8 9 10
``````
Results:
``````zero is four, four is magic
one is three, three is five, five is four, four is magic
two is three, three is five, five is four, four is magic
three is five, five is four, four is magic
four is magic
five is four, four is magic
six is three, three is five, five is four, four is magic
seven is five, five is four, four is magic
eight is five, five is four, four is magic
nine is four, four is magic
Wrong input: 10
``````
The reason four is magic is that it has four letters, and that the sequence started from any number converges to four. Thus, instead of hard-coding the magic four, we may test for a fixed point of the iteration that maps a `\$name` to the `\$name[length \$name]`. This allows the problem to be generalized to an arbitrary array and an arbitrary mapping of the array onto itself. The problem then is that there could be cycles as well as fixed points, which would lead to infinite loops if they are not detected. Thus, the following more general and more robust program. I still use the length of the string as a mapping function, but it could be changed.
`````` 1 # Perl weekly challenge 160
2 # Task 1: Four is magic
3 #
5 use v5.12;
6 use warnings;
7 die <<'END' unless @ARGV >= 2;
8 Usage: ./ch-1a.pl "S1 [S2... ]" N1 [N2...]\n
9 to find all trajectories between strings S1 S2...
10 starting from position N1, N2...
11 The mapping function uses the length of a string as an index
12 to the next string
13 END
14 my @strings=split " ", shift;
15 foreach my \$index(@ARGV){
16 my @seen;
17 while(1){
18 say("\$strings[\$index] is magic"), last if (my \$next=next_index(\$index))==\$index;
19 say("\$strings[\$index] is magic loop"), last if \$seen[\$index];
20 say("\$index->nothing"), last unless defined \$next;
21 \$seen[\$index]++;
22 print "\$strings[\$index] is \$strings[\$next], ";
23 \$index=\$next;
24 }
25 }
26 sub next_index {
27 my \$current=shift;
28 my \$next=length \$strings[\$current]; # Could use other mappings
29 return undef unless defined \$strings[\$next];
30 return \$next;
31 }
``````
Example:
``````./ch-1a.pl "zero one two three four five six seven eight nine" 0 1 2 3 4 5 6 7 8 9
``````
Results:
``````zero is four, four is magic
one is three, three is five, five is four, four is magic
two is three, three is five, five is four, four is magic
three is five, five is four, four is magic
four is magic
five is four, four is magic
six is three, three is five, five is four, four is magic
seven is five, five is four, four is magic
eight is five, five is four, four is magic
nine is four, four is magic
``````
Another example, with different strings,
``````./ch-1a.pl "red orange yellow green blue indigo violet" 0 1 2 3 4 5 6 7
``````
Results:
``````red is green, green is indigo, indigo is violet, violet is magic
orange is violet, violet is magic
yellow is violet, violet is magic
green is indigo, indigo is violet, violet is magic
blue is magic
indigo is violet, violet is magic
violet is magic
7->nothing
``````
In this case, there are two magic color names, as blue occupies position 4 (starting from 0) and it has four letters, while violet occupies position 6 and has six letters.
``````Submitted by: Mohammad S Anwar
You are give an array of integers, @n.
Write a script to find out the Equilibrium Index of the given
array, if found.
For an array A consisting n elements, index i is an
equilibrium index if the sum of elements of subarray A[0…i-1]
is equal to the sum of elements of subarray A[i+1…n-1].
Example 1:
Input: @n = (1, 3, 5, 7, 9)
Output: 3
Example 2:
Input: @n = (1, 2, 3, 4, 5)
Output: -1 as no Equilibrium Index found.
Example 3:
Input: @n = (2, 4, 2)
Output: 1
``````
To solve this task I can sum the leading and trailing subarrays for all possible cuts. I use PDL’s ability to slice arrays and operate on its elements in the following one-liners.
``````perl -MPDL -MPDL::NiceSlice -E 'say "Input: ", \$x=pdl(@ARGV); \$i="-1 No eq. found";
for(1..\$x->nelem-2){\$i=\$_ if \$x(0:\$_-1)->sumover==\$x(\$_+1:-1)->sumover} say "Output: \$i"
' 1 3 5 7 9
perl -MPDL -MPDL::NiceSlice -E 'say "Input: ", \$x=pdl(@ARGV); \$i="-1 No eq. found";
for(1..\$x->nelem-2){\$i=\$_ if \$x(0:\$_-1)->sumover==\$x(\$_+1:-1)->sumover} say "Output: \$i"
' 1 2 3 4 5
perl -MPDL -MPDL::NiceSlice -E 'say "Input: ", \$x=pdl(@ARGV); \$i="-1 No eq. found";
for(1..\$x->nelem-2){\$i=\$_ if \$x(0:\$_-1)->sumover==\$x(\$_+1:-1)->sumover} say "Output: \$i"
' 2 4 2
``````
Results:
``````Input: [1 3 5 7 9]
Output: 3
Input: [1 2 3 4 5]
Output: -1 No eq. found
Input: [2 4 2]
Output: 1
``````
The full version follows.
`````` 1 # Perl weekly challenge 160
2 # Task 2: Equilibrium index
3 #
5 use v5.12;
6 use warnings;
7 use PDL;
8 use PDL::NiceSlice;
9 die "Usage: ./ch-2.pl N1 [N2...] to find equilibrium index of an array of numbers"
10 unless @ARGV;
11 my \$input=pdl(@ARGV);
12 my \$result="-1 as no equilibrium index was found"; # default output
13 say "Input: ", \$input;
14 for(1..\$input->nelem-2){ #for all internal indices
15 \$result="\$_ as sum(".\$input(0:\$_-1).")==sum(".\$input(\$_+1:-1).")", last
16 if \$input(0:\$_-1)->sumover==\$input(\$_+1:-1)->sumover; # Found equilibrium
17 }
18 say "Output: \$result";
``````
Examples:
``````./ch-2.pl 1 3 5 7 9
./ch-2.pl 1 2 3 4 5
./ch-2.pl 2 4 2
``````
Results:
``````Input: [1 3 5 7 9]
Output: 3 as sum([1 3 5])==sum([9])
Input: [1 2 3 4 5]
Output: -1 as no equilibrium index was found
Input: [2 4 2]
Output: 1 as sum([2])==sum([2])
``````
Notice that the equilibrium point as defined above is not the center of mass of the array, as the array elements are summed disregarding their index with respect to the equilibrium index, accounting only for their sign (left or right). Thus the simpler calculation of the center of mass (the first moment of the array) wouldn’t work.
Written on April 11, 2022 | 2,631 | 8,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.935429 |
http://www.ux1.eiu.edu/~cfadd/1360/23EFields/Ch23Hmwk/23.star2.html | 1,516,354,973,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887849.3/warc/CC-MAIN-20180119085553-20180119105553-00330.warc.gz | 583,842,597 | 1,952 | Ch23 Homework Solutions
23.**. On a dry winter day, if you scuff your feet across a carpet, you build up a charge and get a shock when you touch a metal doorknob. In a dark room you can actually see a spark about 2.0 cm long. Air breaks down at a field strength of 3.0 x 106 N/C. Assume that just before the spark occurs, all the charge is in your finger, drawn there by induction due to the proximity of the doorknob. Approximate your fingertip as a sphere of diameter 1.5 cm, and assume that there is an equal amount of charge on the doorknob 2.0 cm away.
(a) How much charge have you built up?
(b) How many electrons does this correspond to?
Ooops. It is the diameter that is to be 1.5 cm, not the radius. I used Serway's assumption for your thumb but I did not include a charge on the doorknob. Since these are only approximations, I am going to leave this "as it is".
Q = 4.08 x 10 - 7 C
Q = 4.08 x 10 - 7 C [ 1 e / 1.6 x 10 - 19 C ]
Q = 2.55 x 1012 e
As we discussed in class, these are approximations and the E-field could break down the air any place along the 2 cm between your thumb and the door knob. | 319 | 1,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.935663 |
http://www.mathisfunforum.com/viewtopic.php?id=17216&p=8 | 1,397,937,293,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00151-ip-10-147-4-33.ec2.internal.warc.gz | 541,991,066 | 7,500 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #176 2012-03-04 11:42:00
anonimnystefy
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### Re: Whats my line?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #177 2012-03-04 11:46:34
bobbym
Online
### Re: Whats my line?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #178 2012-03-04 11:50:19
anonimnystefy
Real Member
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### Re: Whats my line?
Hi bobbym
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #179 2012-03-04 11:52:45
bobbym
Online
### Re: Whats my line?
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #180 2012-03-04 11:57:29
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Hi bobbym
That's because of the similarities in latin and cyrillic letters.
What you are referring to its MOCKBA.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #181 2012-03-04 12:01:47
bobbym
Online
### Re: Whats my line?
Okay, yes here we say Moscow. But americans are infamous for their lack of caring about other culture's pronunciations. Many people here do not have their correct name because it could not be pronounced by the idiot behind the desk and was therefore changed.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #182 2012-03-04 12:05:33
anonimnystefy
Real Member
Offline
### Re: Whats my line?
I know that.I especially hate the fuss about 'ć' and 'č'.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #183 2012-03-04 12:08:53
bobbym
Online
### Re: Whats my line?
My brother understands some Russian and can write using the Cyrillic alphabet, I cannot.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #184 2012-03-04 12:16:03
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Does he speak any Serbian?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #185 2012-03-04 12:18:12
bobbym
Online
### Re: Whats my line?
I do not think so. At the time he was learning it because he wanted to read Russian chess books. Back then there was only Yugoslavia.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #186 2012-03-04 12:26:44
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Okay.Which languages do you speak? I mean besides English.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #187 2012-03-04 12:29:43
bobbym
Online
### Re: Whats my line?
Smatterings of languages from chess and literature. I used to speak a little Japanese,German and Latin ( Roman Catholic Background ). I slowly forgot all of it and never learned my native tongue. So I would say zilch!
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #188 2012-03-04 20:19:37
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Hi bobbym
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #189 2012-03-04 20:26:56
bob bundy
Moderator
Offline
### Re: Whats my line?
hi bobbym and Stefy,
I 'accidentally' clicked on that hidden box, so now I know too!
I've just reviewed all the posts. I can't believe we scouted around the edge of the answer without ever suggesting it!
I guess that ends this round with a score of
bobbym 1 us 0
At least that means we can start a new "What's My Line?"
And, bobbym, by the rules of I Spy you get to post the next one.
As for Moscva, I'm afraid we Brits have been getting foreign names wrong since before you (USA) were born so I think you probably got it from us.
That's why we called Beijing Peking and pronounce Paris as Paris when the natives say Pari.
And our biggest error was to call the Aborigine Americans, Red Indians for which I humbly apologise.
Now that everyone is speaking 'English' perhaps we'll listen more carefully in future!
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #190 2012-03-04 20:31:48
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Hi bob
In Serbia we call Beijing Peking as well and for Vienna we use Beč, which is totally kaboobly doo.
As for the game, I am just waiting for a new one.They are fun and build teamwork.
Last edited by anonimnystefy (2012-03-04 20:33:07)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #191 2012-03-04 20:33:54
bob bundy
Moderator
Offline
### Re: Whats my line?
Hi Stefy,
Yes, the list of "Getting the Name Wrong" is very long!
That's why we English call it 'tea'. Most people here think we invented it too!
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #192 2012-03-04 20:38:06
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Well I don't know how it should be called, but if I'm not mistaken it originates from India.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #193 2012-03-04 20:43:43
bob bundy
Moderator
Offline
### Re: Whats my line?
hi
According to Wiki the most likely origin is China rather than India and the Mandarin word is pronounced 'cha'.
Still I must stop apologising for my ancestors and get on. I've got to spend the morning setting up a sound system for school so I'll say goodbye for now.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #194 2012-03-04 20:48:25
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Hi bob
Ok, see you.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #195 2012-03-04 22:00:52
bobbym
Online
### Re: Whats my line?
Hi Bob;
We have done the same with Paris and Beijing. But it has progressed way passed that. My last name is the name of the town where my grandparents were born! This is because the guards at Ellis island were unable to pronounce it. So they changed it to what they thought they could pronounce.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #196 2012-03-05 22:24:07
bobbym
Online
### Re: Whats my line?
What's my line?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #197 2012-03-05 22:32:36
bob bundy
Moderator
Offline
### Re: Whats my line?
hi bobbym,
Hope you are well today.
(1) Is he alive today as far as you know, obviously? (to Stefy: as it's black and white we might as well establish whether he is still around now or deceased.)
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #198 2012-03-05 22:36:01
bobbym
Online
### Re: Whats my line?
Hi Bob;
Howdy! Yes he is alive today.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #199 2012-03-05 23:06:36
anonimnystefy
Real Member
Offline
### Re: Whats my line?
Hi bob
Yes I understand that.And since he is alive we must take into account that he is now old.
To bobbym
1.Does he live in the USA?
Last edited by anonimnystefy (2012-03-05 23:07:14)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #200 2012-03-05 23:11:47
bobbym
Online
### Re: Whats my line?
He may have lived in many countries but he is not an American by birth.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 3,084 | 11,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2014-15 | longest | en | 0.906004 |
https://diydrones.com/members/mikiuse/content?type=DiscussionEntry&context=all | 1,709,562,463,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00823.warc.gz | 205,745,711 | 9,225 | Sort by
### What do the values of roll pitch and yaw mean?
Hi
I am struggling to figure out what the Roll = -0.02581, Pitch = -0.03569 and Yaw = -1.95976 values actually mean?
Is it in degrees/rad?
Is it relative to magnetic north? of to what? | 72 | 245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.913244 |
https://space.stackexchange.com/questions/21920/has-adjustable-yo-yo-de-spin-been-tested-or-demonstrated | 1,702,277,347,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00017.warc.gz | 584,715,256 | 43,460 | # Has adjustable Yo-Yo de-spin been tested or demonstrated?
The video in the interesting question How did the GoFast 2014 Rocket de-spin? shows a de-spin that went well. An initial very fast spin (for stabilization during propulsion) is almost perfectly de-spun during a maneuver that is so fast that it is hard to imagine it could be anything else but a Yo-Yo de-spin.
The mathematics behind this is summarized in this answer along with links to several original sources. Briefly, $r$ is the ratio of final to initial angular velocity, and will be zero at perfect de-spin but can still be positive or negative for undershoot or overshoot conditions. The remaining terms are described in detail in the link.
$$\frac{m(d+a)^2}{I}=\frac{1-r}{1+r}$$
It's a balance. So if the final angular speed is much much lower than the initial (near perfect cancellation), it suggests that both the initial spin rate and the de-spin were well controlled.
Another example of careful spin and de-spin cancellation (but not with a Yo-Yo) is discussed in this answer and in this answer about NASA's LSDS test.
But what if the initial spin had some uncertainty that could not be controlled?
Question: Has adjustable Yo-Yo de-spin been tested or demonstrated? Reading about a serious but untested proposal would also be interesting.
I can think of all kinds of possible ways to do this, last-second laser trimming of the wires, or just a clever break-away release, vernier yo-yos (say 10kg plus 160g, 80g, 40g, 20g, 10g independently addressable "fine yo-yos") but this kind of complexity really defeats the elegance and reliability of the Yo-Yo technique with fixed mass weights and premeasured wires, so maybe backup thrusters to re-adjust the spin before Yo-Yo would be a better way to stay with known technologies.
Still, have variable Yo-Yo's been tested or demonstrated?
• The stretch yo-yo despin was devised to account for some uncertainty and successfully tested in 1962. Is that what you are looking for? Jun 12, 2017 at 13:41
• @called2voyage wow! cool! yes!
– uhoh
Jun 12, 2017 at 14:16
In the 1960s the Goddard Space Flight Center devoted some study to the topic of error in yo-yo de-spin. It was acknowledged that spin-up errors and uncertainty of the spin axis moment of inertia were sources of de-spin error. Since de-spin error can be a source of many problems, which may even impact the success of the mission, the stretch yo-yo concept was devised to reduce de-spin error.
A device that greatly reduces spin-up errors and errors due to variations in spin moment of inertia is the stretch yo-yo (the stretch concept was first suggested by Henry Cornille of Goddard Space Flight Center (Reference 3). It can consist of a weight, a spring, a wire, and end fittings (Figure 1), or simply a weight, spring, and end fittings (Figure 2). The purpose of the spring is to compensate for errors in initial spin-up. For example, in a given stretch yo-yo application there will be a certain amount of stretching or elongation of the spring during normal operation. If the initial spin is greater than the nominal value, the spring will elongate more than normal during operation and reduce the spin to the desired value. If the initial spin is less than the nominal initial spin, the spring will elongate less than normal during functioning and correct for the under spin. The stretch yo-yo is a simple example of an adaptive control system: it senses the spin environment it is in and corrects accordingly. Also, it will be shown that the stretch yo-yo is relatively insensitive to variations (uncertainties) of the spin moment of inertia.
After analysis and design, it was decided that the stretch yo-yo despin should be tested on the Ariel I spacecraft, and in April 1962, it was successfully flown ("Analysis of the Dynamic Tests of the Stretch Yo-Yo De-Spin System" by William R. Mentzer):
Following dynamic analysis and development of design equations for the stretch yo-yo, the decision was made to fabricate and test stretch yo-yos on the Explorer XII (1961 v1) and Ariell (1962 01) type payloads. Three series of tests were conducted beginning with feasibility tests on the two payloads and concluding with flight qualification tests for the Ariel I payload. As a result of these tests, a stretch yo-yo was flown on the successful Ariel I spacecraft in April 1962.
In the record of the experimental results, the effects of the several variables involved in stretch yo-yo design can be readily noted. These variables include optimum spring constant, preload, initial spin rate, spin moment of inertia, and material strength.
You can read the full analysis of the tests in that document, but here's one last brief note on the effectiveness of the stretch yo-yo de-spin:
The Explorer XII type optimum spring scale tests clearly demonstrated the ability of the stretch yo-yo to compensate for errors in initial spin rate. In this series of tests, spin-up errors of +/-20 percent were reduced to within +/-1.5 percent of the design final spin rate.
• Page 6, Discussion of Test Result: "The Explorer XII type optimum spring scale tests clearly demonstrated the ability of the stretch yo-yo to compensate for errors in initial spin rate. In this series of tests, spin-up errors of +/-20 percent were reduced to within +/-1.5 percent of the design final spin rate." Incredible - thank you for this!
– uhoh
Jun 12, 2017 at 14:45
• @uhoh You're welcome! I wrapped that quote into the answer, because it really does show quite clearly how effective the technique is. Jun 12, 2017 at 14:48 | 1,256 | 5,591 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | longest | en | 0.951941 |
https://www.sparknotes.com/math/calcbc2/taylorseries/problems/ | 1,718,736,518,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00115.warc.gz | 897,288,032 | 56,381 | Problem : Compute the Taylor series for f (x) = 1/(1 + x).
The first few derivatives of the function are
f'(x) = f''(x) = f(3)(x) =
so f (0) = 1, f'(0) = - 1, f''(0) = 2, f(3)(0) = - 6. The general case is clearly that f(n)(0) = (- 1)nn!. Hence the Taylor series for f (x) is
p∞(x) = xn = (- 1)nxn = 1 - x + x2 - x3 + ...
Problem : What is the Taylor series of a polynomial p(x) = anxn + an-1xn-1 + ... + a0?
It is easy to check that the Taylor series of a polynomial is the polynomial itself! (All the coefficients of higher order terms are equal to 0.)
Problem : Find the Taylor series for the function g(x) = 1/ about x = 1.
The first couple derivatives of the function are
g'(x) = x-3/2 g''(x) = x-5/2 g(3)(x) = x-7/2
so g(1) = 1, g'(1) = - 1/2, g''(1) = (- 1/2)(- 3/2). We deduce that
g(n)(1) = (1)(3) ... (2n - 1) =
Hence the Taylor series for g(x) is
xn | 342 | 878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.849498 |
https://testbook.com/question-answer/ifleft-overrightarrow-a-overrightarro--602b58c6cfd3da0d46ad5020 | 1,632,813,439,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00381.warc.gz | 592,225,078 | 31,499 | # If $$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = 1$$, then the value of $$\frac{{\overrightarrow a \cdot \overrightarrow b \times \overrightarrow c }}{{\overrightarrow c \times \overrightarrow a \cdot \overrightarrow b }} + \frac{{\overrightarrow b \cdot \overrightarrow c \times \overrightarrow a }}{{\overrightarrow a \times \overrightarrow b \cdot \overrightarrow c }} + \frac{{\overrightarrow c \cdot \overrightarrow a \times \overrightarrow b }}{{\overrightarrow b \times \overrightarrow c \cdot \overrightarrow a }}$$ is:
This question was previously asked in
UP TGT Mathematics 2013 Official Paper
View all UP TGT Papers >
1. 1
2. -1
3. 2
4. 3
Option 4 : 3
Free
CT 1: हिन्दी (आदिकाल)
5057
10 Questions 40 Marks 10 Mins
## Detailed Solution
Given,
$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = 1$$
take L.H.S $$\dfrac{{\overrightarrow a \cdot \overrightarrow b \times \overrightarrow c }}{{\overrightarrow c \times \overrightarrow a \cdot \overrightarrow b }} + \dfrac{{\overrightarrow b \cdot \overrightarrow c \times \overrightarrow a }}{{\overrightarrow a \times \overrightarrow b \cdot \overrightarrow c }} + \dfrac{{\overrightarrow c \cdot \overrightarrow a \times \overrightarrow b }}{{\overrightarrow b \times \overrightarrow c \cdot \overrightarrow a }}$$
$$= \dfrac{[\vec a\; \vec b \;\vec c]}{[\vec c \; \vec a \;\vec b]} + \dfrac{[\vec b\; \vec c \;\vec a]}{[\vec a \; \vec b \;\vec c]} + \dfrac{[\vec c\; \vec a \;\vec b]}{[\vec b \; \vec c \;\vec a]}$$
= 1 + 1 + 1
= 3 | 505 | 1,558 | {"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} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.174559 |
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author nipkow Fri Jun 10 18:36:47 2005 +0200 (2005-06-10) changeset 16359 af7239e3054d parent 16306 8117e2037d3b child 16523 f8a734dc0fbc permissions -rw-r--r--
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wenzelm@12668 1 \chapter{The Basics} nipkow@8743 2 nipkow@8743 3 \section{Introduction} nipkow@8743 4 paulson@11405 5 This book is a tutorial on how to use the theorem prover Isabelle/HOL as a paulson@11405 6 specification and verification system. Isabelle is a generic system for paulson@11405 7 implementing logical formalisms, and Isabelle/HOL is the specialization paulson@11405 8 of Isabelle for HOL, which abbreviates Higher-Order Logic. We introduce paulson@11405 9 HOL step by step following the equation nipkow@8743 10 $\mbox{HOL} = \mbox{Functional Programming} + \mbox{Logic}.$ paulson@11456 11 We do not assume that you are familiar with mathematical logic. paulson@11456 12 However, we do assume that paulson@11456 13 you are used to logical and set theoretic notation, as covered paulson@11456 14 in a good discrete mathematics course~\cite{Rosen-DMA}, and paulson@11450 15 that you are familiar with the basic concepts of functional nipkow@11209 16 programming~\cite{Bird-Haskell,Hudak-Haskell,paulson-ml2,Thompson-Haskell}. nipkow@11209 17 Although this tutorial initially concentrates on functional programming, do nipkow@11209 18 not be misled: HOL can express most mathematical concepts, and functional nipkow@11209 19 programming is just one particularly simple and ubiquitous instance. nipkow@8743 20 nipkow@11205 21 Isabelle~\cite{paulson-isa-book} is implemented in ML~\cite{SML}. This has nipkow@11205 22 influenced some of Isabelle/HOL's concrete syntax but is otherwise irrelevant paulson@11450 23 for us: this tutorial is based on nipkow@11213 24 Isabelle/Isar~\cite{isabelle-isar-ref}, an extension of Isabelle which hides nipkow@11213 25 the implementation language almost completely. Thus the full name of the nipkow@11213 26 system should be Isabelle/Isar/HOL, but that is a bit of a mouthful. nipkow@11213 27 nipkow@11213 28 There are other implementations of HOL, in particular the one by Mike Gordon paulson@11450 29 \index{Gordon, Mike}% nipkow@11213 30 \emph{et al.}, which is usually referred to as the HOL system'' nipkow@11213 31 \cite{mgordon-hol}. For us, HOL refers to the logical system, and sometimes paulson@11450 32 its incarnation Isabelle/HOL\@. nipkow@8743 33 nipkow@8743 34 A tutorial is by definition incomplete. Currently the tutorial only nipkow@8743 35 introduces the rudiments of Isar's proof language. To fully exploit the power nipkow@11213 36 of Isar, in particular the ability to write readable and structured proofs, nipkow@15429 37 you should start with Nipkow's overview~\cite{Nipkow-TYPES02} and consult nipkow@15429 38 the Isabelle/Isar Reference Manual~\cite{isabelle-isar-ref} and Wenzel's nipkow@15429 39 PhD thesis~\cite{Wenzel-PhD} (which discusses many proof patterns) nipkow@15429 40 for further details. If you want to use Isabelle's ML level nipkow@8743 41 directly (for example for writing your own proof procedures) see the Isabelle nipkow@8743 42 Reference Manual~\cite{isabelle-ref}; for details relating to HOL see the nipkow@8743 43 Isabelle/HOL manual~\cite{isabelle-HOL}. All manuals have a comprehensive nipkow@8743 44 index. nipkow@8743 45 nipkow@8743 46 \section{Theories} nipkow@8743 47 \label{sec:Basic:Theories} nipkow@8743 48 paulson@11428 49 \index{theories|(}% nipkow@8743 50 Working with Isabelle means creating theories. Roughly speaking, a paulson@11428 51 \textbf{theory} is a named collection of types, functions, and theorems, nipkow@8743 52 much like a module in a programming language or a specification in a nipkow@8743 53 specification language. In fact, theories in HOL can be either. The general nipkow@8743 54 format of a theory \texttt{T} is nipkow@8743 55 \begin{ttbox} nipkow@15136 56 theory T nipkow@15141 57 imports B$$@1$$ $$\ldots$$ B$$@n$$ nipkow@15136 58 begin paulson@11450 59 {\rmfamily\textit{declarations, definitions, and proofs}} nipkow@8743 60 end nipkow@15358 61 \end{ttbox}\cmmdx{theory}\cmmdx{imports} nipkow@15136 62 where \texttt{B}$@1$ \dots\ \texttt{B}$@n$ are the names of existing paulson@11450 63 theories that \texttt{T} is based on and \textit{declarations, paulson@11450 64 definitions, and proofs} represents the newly introduced concepts nipkow@8771 65 (types, functions etc.) and proofs about them. The \texttt{B}$@i$ are the paulson@11450 66 direct \textbf{parent theories}\indexbold{parent theories} of~\texttt{T}\@. paulson@11450 67 Everything defined in the parent theories (and their parents, recursively) is nipkow@8743 68 automatically visible. To avoid name clashes, identifiers can be paulson@11450 69 \textbf{qualified}\indexbold{identifiers!qualified} paulson@11450 70 by theory names as in \texttt{T.f} and~\texttt{B.f}. paulson@11450 71 Each theory \texttt{T} must paulson@11428 72 reside in a \textbf{theory file}\index{theory files} named \texttt{T.thy}. nipkow@8743 73 nipkow@8743 74 This tutorial is concerned with introducing you to the different linguistic paulson@11450 75 constructs that can fill the \textit{declarations, definitions, and paulson@11450 76 proofs} above. A complete grammar of the basic nipkow@12327 77 constructs is found in the Isabelle/Isar Reference nipkow@12327 78 Manual~\cite{isabelle-isar-ref}. nipkow@8743 79 nipkow@8743 80 \begin{warn} paulson@11428 81 HOL contains a theory \thydx{Main}, the union of all the basic paulson@10885 82 predefined theories like arithmetic, lists, sets, etc. paulson@10885 83 Unless you know what you are doing, always include \isa{Main} nipkow@10971 84 as a direct or indirect parent of all your theories. nipkow@12332 85 \end{warn} nipkow@16306 86 HOL's theory collection is available online at nipkow@16306 87 \begin{center}\small nipkow@16306 88 \url{http://isabelle.in.tum.de/library/HOL/} nipkow@16306 89 \end{center} nipkow@16359 90 and is recommended browsing. In subdirectory \texttt{Library} you find nipkow@16359 91 a growing library of useful theories that are not part of \isa{Main} nipkow@16359 92 but can be included among the parents of a theory and will then be nipkow@16359 93 loaded automatically. nipkow@16306 94 nipkow@16306 95 For the more adventurous, there is the \emph{Archive of Formal Proofs}, nipkow@16306 96 a journal-like collection of more advanced Isabelle theories: nipkow@16306 97 \begin{center}\small nipkow@16306 98 \url{http://afp.sourceforge.net/} nipkow@16306 99 \end{center} nipkow@16306 100 We hope that you will contribute to it yourself one day.% paulson@11428 101 \index{theories|)} nipkow@8743 102 nipkow@8743 103 paulson@10885 104 \section{Types, Terms and Formulae} nipkow@8743 105 \label{sec:TypesTermsForms} nipkow@8743 106 paulson@10795 107 Embedded in a theory are the types, terms and formulae of HOL\@. HOL is a typed nipkow@8771 108 logic whose type system resembles that of functional programming languages paulson@11450 109 like ML or Haskell. Thus there are paulson@11450 110 \index{types|(} nipkow@8743 111 \begin{description} paulson@11450 112 \item[base types,] paulson@11450 113 in particular \tydx{bool}, the type of truth values, paulson@11428 114 and \tydx{nat}, the type of natural numbers. paulson@11450 115 \item[type constructors,]\index{type constructors} paulson@11450 116 in particular \tydx{list}, the type of paulson@11428 117 lists, and \tydx{set}, the type of sets. Type constructors are written nipkow@8771 118 postfix, e.g.\ \isa{(nat)list} is the type of lists whose elements are nipkow@8743 119 natural numbers. Parentheses around single arguments can be dropped (as in nipkow@8771 120 \isa{nat list}), multiple arguments are separated by commas (as in nipkow@8771 121 \isa{(bool,nat)ty}). paulson@11450 122 \item[function types,]\index{function types} paulson@11450 123 denoted by \isasymFun\indexbold{$IsaFun@\isasymFun}. nipkow@8771 124 In HOL \isasymFun\ represents \emph{total} functions only. As is customary, nipkow@8771 125 \isa{$\tau@1$\isasymFun~$\tau@2$\isasymFun~$\tau@3$} means nipkow@8771 126 \isa{$\tau@1$\isasymFun~($\tau@2$\isasymFun~$\tau@3$)}. Isabelle also nipkow@8771 127 supports the notation \isa{[$\tau@1,\dots,\tau@n$] \isasymFun~$\tau$} nipkow@8771 128 which abbreviates \isa{$\tau@1$\isasymFun~$\cdots$\isasymFun~$\tau@n$nipkow@8743 129 \isasymFun~$\tau$}. paulson@11450 130 \item[type variables,]\index{type variables}\index{variables!type} paulson@10795 131 denoted by \ttindexboldpos{'a}{$Isatype}, \isa{'b} etc., just like in ML\@. They give rise nipkow@8771 132 to polymorphic types like \isa{'a \isasymFun~'a}, the type of the identity nipkow@8771 133 function. nipkow@8743 134 \end{description} nipkow@8743 135 \begin{warn} nipkow@8743 136 Types are extremely important because they prevent us from writing nipkow@16359 137 nonsense. Isabelle insists that all terms and formulae must be nipkow@16359 138 well-typed and will print an error message if a type mismatch is nipkow@16359 139 encountered. To reduce the amount of explicit type information that nipkow@16359 140 needs to be provided by the user, Isabelle infers the type of all nipkow@16359 141 variables automatically (this is called \bfindex{type inference}) nipkow@16359 142 and keeps quiet about it. Occasionally this may lead to nipkow@16359 143 misunderstandings between you and the system. If anything strange nipkow@16359 144 happens, we recommend that you ask Isabelle to display all type nipkow@16359 145 information via the Proof General menu item \textsf{Isabelle} $>$ nipkow@16359 146 \textsf{Settings} $>$ \textsf{Show Types} (see \S\ref{sec:interface} nipkow@16359 147 for details). paulson@11450 148 \end{warn}% paulson@11450 149 \index{types|)} nipkow@8743 150 nipkow@8743 151 paulson@11450 152 \index{terms|(} paulson@11450 153 \textbf{Terms} are formed as in functional programming by nipkow@8771 154 applying functions to arguments. If \isa{f} is a function of type nipkow@8771 155 \isa{$\tau@1$ \isasymFun~$\tau@2$} and \isa{t} is a term of type nipkow@8771 156 $\tau@1$ then \isa{f~t} is a term of type $\tau@2$. HOL also supports nipkow@8771 157 infix functions like \isa{+} and some basic constructs from functional paulson@11428 158 programming, such as conditional expressions: nipkow@8743 159 \begin{description} paulson@11450 160 \item[\isa{if $b$ then $t@1$ else $t@2$}]\index{*if expressions} paulson@11428 161 Here $b$ is of type \isa{bool} and $t@1$ and $t@2$ are of the same type. paulson@11450 162 \item[\isa{let $x$ = $t$ in $u$}]\index{*let expressions} nipkow@13814 163 is equivalent to $u$ where all free occurrences of $x$ have been replaced by nipkow@8743 164 $t$. For example, nipkow@8771 165 \isa{let x = 0 in x+x} is equivalent to \isa{0+0}. Multiple bindings are separated nipkow@13814 166 by semicolons: \isa{let $x@1$ = $t@1$;\dots; $x@n$ = $t@n$ in $u$}. nipkow@8771 167 \item[\isa{case $e$ of $c@1$ \isasymFun~$e@1$ |~\dots~| $c@n$ \isasymFun~$e@n$}] paulson@11450 168 \index{*case expressions} nipkow@8771 169 evaluates to $e@i$ if $e$ is of the form $c@i$. nipkow@8743 170 \end{description} nipkow@8743 171 nipkow@8743 172 Terms may also contain paulson@11450 173 \isasymlambda-abstractions.\index{lambda@$\lambda$ expressions} paulson@11450 174 For example, nipkow@8771 175 \isa{\isasymlambda{}x.~x+1} is the function that takes an argument \isa{x} and nipkow@8771 176 returns \isa{x+1}. Instead of nipkow@8771 177 \isa{\isasymlambda{}x.\isasymlambda{}y.\isasymlambda{}z.~$t$} we can write paulson@11450 178 \isa{\isasymlambda{}x~y~z.~$t$}.% paulson@11450 179 \index{terms|)} nipkow@8743 180 paulson@11450 181 \index{formulae|(}% paulson@11450 182 \textbf{Formulae} are terms of type \tydx{bool}. paulson@11428 183 There are the basic constants \cdx{True} and \cdx{False} and nipkow@8771 184 the usual logical connectives (in decreasing order of priority): paulson@11420 185 \indexboldpos{\protect\isasymnot}{$HOL0not}, \indexboldpos{\protect\isasymand}{$HOL0and}, paulson@11420 186 \indexboldpos{\protect\isasymor}{$HOL0or}, and \indexboldpos{\protect\isasymimp}{$HOL0imp}, nipkow@8743 187 all of which (except the unary \isasymnot) associate to the right. In nipkow@8771 188 particular \isa{A \isasymimp~B \isasymimp~C} means \isa{A \isasymimp~(B nipkow@8771 189 \isasymimp~C)} and is thus logically equivalent to \isa{A \isasymand~B nipkow@8771 190 \isasymimp~C} (which is \isa{(A \isasymand~B) \isasymimp~C}). nipkow@8743 191 paulson@11450 192 Equality\index{equality} is available in the form of the infix function paulson@11450 193 \isa{=} of type \isa{'a \isasymFun~'a nipkow@8771 194 \isasymFun~bool}. Thus \isa{$t@1$ = $t@2$} is a formula provided $t@1$ paulson@11450 195 and $t@2$ are terms of the same type. If $t@1$ and $t@2$ are of type paulson@11450 196 \isa{bool} then \isa{=} acts as \rmindex{if-and-only-if}. paulson@11450 197 The formula nipkow@8771 198 \isa{$t@1$~\isasymnoteq~$t@2$} is merely an abbreviation for nipkow@8771 199 \isa{\isasymnot($t@1$ = $t@2$)}. nipkow@8743 200 paulson@11450 201 Quantifiers\index{quantifiers} are written as paulson@11450 202 \isa{\isasymforall{}x.~$P$} and \isa{\isasymexists{}x.~$P$}. paulson@11420 203 There is even paulson@11450 204 \isa{\isasymuniqex{}x.~$P$}, which paulson@11420 205 means that there exists exactly one \isa{x} that satisfies \isa{$P$}. paulson@11420 206 Nested quantifications can be abbreviated: paulson@11420 207 \isa{\isasymforall{}x~y~z.~$P$} means paulson@11450 208 \isa{\isasymforall{}x.\isasymforall{}y.\isasymforall{}z.~$P$}.% paulson@11450 209 \index{formulae|)} nipkow@8743 210 nipkow@8743 211 Despite type inference, it is sometimes necessary to attach explicit paulson@11428 212 \bfindex{type constraints} to a term. The syntax is nipkow@8771 213 \isa{$t$::$\tau$} as in \isa{x < (y::nat)}. Note that nipkow@10538 214 \ttindexboldpos{::}{$Isatype} binds weakly and should therefore be enclosed paulson@11450 215 in parentheses. For instance, paulson@11450 216 \isa{x < y::nat} is ill-typed because it is interpreted as paulson@11450 217 \isa{(x < y)::nat}. Type constraints may be needed to disambiguate paulson@11450 218 expressions paulson@11450 219 involving overloaded functions such as~\isa{+}, paulson@11450 220 \isa{*} and~\isa{<}. Section~\ref{sec:overloading} paulson@11450 221 discusses overloading, while Table~\ref{tab:overloading} presents the most nipkow@10695 222 important overloaded function symbols. nipkow@8743 223 paulson@11450 224 In general, HOL's concrete \rmindex{syntax} tries to follow the conventions of paulson@11450 225 functional programming and mathematics. Here are the main rules that you paulson@11450 226 should be familiar with to avoid certain syntactic traps: nipkow@8743 227 \begin{itemize} nipkow@8743 228 \item nipkow@8771 229 Remember that \isa{f t u} means \isa{(f t) u} and not \isa{f(t u)}! nipkow@8743 230 \item nipkow@8771 231 Isabelle allows infix functions like \isa{+}. The prefix form of function nipkow@8771 232 application binds more strongly than anything else and hence \isa{f~x + y} nipkow@8771 233 means \isa{(f~x)~+~y} and not \isa{f(x+y)}. nipkow@8743 234 \item Remember that in HOL if-and-only-if is expressed using equality. But nipkow@8743 235 equality has a high priority, as befitting a relation, while if-and-only-if nipkow@8771 236 typically has the lowest priority. Thus, \isa{\isasymnot~\isasymnot~P = nipkow@8771 237 P} means \isa{\isasymnot\isasymnot(P = P)} and not nipkow@8771 238 \isa{(\isasymnot\isasymnot P) = P}. When using \isa{=} to mean nipkow@8771 239 logical equivalence, enclose both operands in parentheses, as in \isa{(A nipkow@8743 240 \isasymand~B) = (B \isasymand~A)}. nipkow@8743 241 \item nipkow@8743 242 Constructs with an opening but without a closing delimiter bind very weakly nipkow@8743 243 and should therefore be enclosed in parentheses if they appear in subterms, as paulson@11450 244 in \isa{(\isasymlambda{}x.~x) = f}. This includes paulson@11450 245 \isa{if},\index{*if expressions} paulson@11450 246 \isa{let},\index{*let expressions} paulson@11450 247 \isa{case},\index{*case expressions} paulson@11450 248 \isa{\isasymlambda}, and quantifiers. nipkow@8743 249 \item nipkow@8771 250 Never write \isa{\isasymlambda{}x.x} or \isa{\isasymforall{}x.x=x} nipkow@12327 251 because \isa{x.x} is always taken as a single qualified identifier. Write nipkow@8771 252 \isa{\isasymlambda{}x.~x} and \isa{\isasymforall{}x.~x=x} instead. paulson@11450 253 \item Identifiers\indexbold{identifiers} may contain the characters \isa{_} nipkow@12327 254 and~\isa{'}, except at the beginning. nipkow@8743 255 \end{itemize} nipkow@8743 256 paulson@11450 257 For the sake of readability, we use the usual mathematical symbols throughout nipkow@10983 258 the tutorial. Their \textsc{ascii}-equivalents are shown in table~\ref{tab:ascii} in nipkow@8771 259 the appendix. nipkow@8771 260 paulson@11450 261 \begin{warn} nipkow@16359 262 A particular problem for novices can be the priority of operators. If nipkow@16359 263 you are unsure, use additional parentheses. In those cases where nipkow@16359 264 Isabelle echoes your input, you can see which parentheses are dropped nipkow@16359 265 --- they were superfluous. If you are unsure how to interpret nipkow@16359 266 Isabelle's output because you don't know where the (dropped) nipkow@16359 267 parentheses go, set the Proof General flag \textsf{Isabelle}$>$nipkow@16359 268 \textsf{Settings}$>$\textsf{Show Brackets} (see \S\ref{sec:interface}). paulson@11450 269 \end{warn} paulson@11450 270 nipkow@8743 271 nipkow@8743 272 \section{Variables} nipkow@8743 273 \label{sec:variables} paulson@11450 274 \index{variables|(} nipkow@8743 275 paulson@11450 276 Isabelle distinguishes free and bound variables, as is customary. Bound nipkow@8743 277 variables are automatically renamed to avoid clashes with free variables. In paulson@11428 278 addition, Isabelle has a third kind of variable, called a \textbf{schematic paulson@11428 279 variable}\index{variables!schematic} or \textbf{unknown}\index{unknowns}, nipkow@13439 280 which must have a~\isa{?} as its first character. paulson@11428 281 Logically, an unknown is a free variable. But it may be nipkow@8743 282 instantiated by another term during the proof process. For example, the nipkow@8771 283 mathematical theorem$x = x$is represented in Isabelle as \isa{?x = ?x}, nipkow@8743 284 which means that Isabelle can instantiate it arbitrarily. This is in contrast nipkow@8743 285 to ordinary variables, which remain fixed. The programming language Prolog nipkow@8743 286 calls unknowns {\em logical\/} variables. nipkow@8743 287 nipkow@8743 288 Most of the time you can and should ignore unknowns and work with ordinary nipkow@8743 289 variables. Just don't be surprised that after you have finished the proof of paulson@11450 290 a theorem, Isabelle will turn your free variables into unknowns. It nipkow@8743 291 indicates that Isabelle will automatically instantiate those unknowns nipkow@8743 292 suitably when the theorem is used in some other proof. nipkow@9689 293 Note that for readability we often drop the \isa{?}s when displaying a theorem. nipkow@8743 294 \begin{warn} paulson@11450 295 For historical reasons, Isabelle accepts \isa{?} as an ASCII representation paulson@11450 296 of the $$\exists$$ symbol. However, the \isa{?} character must then be followed paulson@11450 297 by a space, as in \isa{?~x. f(x) = 0}. Otherwise, \isa{?x} is paulson@11450 298 interpreted as a schematic variable. The preferred ASCII representation of paulson@11450 299 the $$\exists$$ symbol is \isa{EX}\@. paulson@11450 300 \end{warn}% paulson@11450 301 \index{variables|)} nipkow@8743 302 paulson@10885 303 \section{Interaction and Interfaces} nipkow@16306 304 \label{sec:interface} nipkow@8771 305 nipkow@16359 306 The recommended interface for Isabelle/Isar is the (X)Emacs-based nipkow@16359 307 \bfindex{Proof General}~\cite{proofgeneral,Aspinall:TACAS:2000}. nipkow@16359 308 Interaction with Isabelle at the shell level, although possible, nipkow@16359 309 should be avoided. Most of the tutorial is independent of the nipkow@16359 310 interface and is phrased in a neutral language. For example, the nipkow@16359 311 phrase to abandon a proof'' corresponds to the obvious nipkow@16359 312 action of clicking on the \textsf{Undo} symbol in Proof General. nipkow@16359 313 Proof General specific information is often displayed in paragraphs nipkow@16359 314 identified by a miniature Proof General icon. Here are two examples: nipkow@16359 315 \begin{pgnote} nipkow@16359 316 Proof General supports a special font with mathematical symbols known nipkow@16359 317 as x-symbols''. All symbols have \textsc{ascii}-equivalents: for nipkow@16359 318 example, you can enter either \verb!&! or \verb!\! to obtain nipkow@16359 319$\land$. For a list of the most frequent symbols see table~\ref{tab:ascii} nipkow@16359 320 in the appendix. nipkow@8771 321 nipkow@16359 322 Note that by default x-symbols are not enabled. You have to switch nipkow@16359 323 them on via the menu item \textsf{Proof-General}$>$\textsf{Options}$>$nipkow@16359 324 \textsf{X-Symbols} (and save the option via the top-level nipkow@16359 325 \textsf{Options} menu). nipkow@16306 326 \end{pgnote} nipkow@8771 327 nipkow@16306 328 \begin{pgnote} nipkow@16359 329 Proof General offers the \textsf{Isabelle} menu for displaying nipkow@16359 330 information and setting flags. A particularly useful flag is nipkow@16359 331 \textsf{Isabelle}$>$\textsf{Settings}$>$\textsf{Show Types} which nipkow@16359 332 causes Isabelle to output the type information that is usually nipkow@16306 333 suppressed. This is indispensible in case of errors of all kinds nipkow@16359 334 because often the types reveal the source of the problem. Once you nipkow@16359 335 have diagnosed the problem you may no longer want to see the types nipkow@16359 336 because they clutter all output. Simply reset the flag. nipkow@16306 337 \end{pgnote} nipkow@8771 338 paulson@10885 339 \section{Getting Started} nipkow@8743 340 nipkow@16359 341 Assuming you have installed Isabelle and Proof General, you start it by typing nipkow@16359 342 \texttt{Isabelle} in a shell window. This launches a Proof General window. nipkow@16359 343 By default, you are in HOL\footnote{This is controlled by the nipkow@16359 344 \texttt{ISABELLE_LOGIC} setting, see \emph{The Isabelle System Manual} nipkow@16359 345 for more details.}. nipkow@16359 346 nipkow@16359 347 \begin{pgnote} nipkow@16359 348 You can choose a different logic via the \textsf{Isabelle}$>\$ nipkow@16359 349 \textsf{Logics} menu. For example, you may want to work in the real nipkow@16359 350 numbers, an extension of HOL (see \S\ref{sec:real}). nipkow@16359 351 % This is just excess baggage: nipkow@16359 352 %(You have to restart Proof General if you only compile the new logic nipkow@16359 353 %after having launching Proof General already). nipkow@16359 354 \end{pgnote} | 7,361 | 22,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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} | 2.90625 | 3 | CC-MAIN-2019-43 | latest | en | 0.728661 |
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# math237_a7 - SOLUTIONS FOR ASSIGNMENT 7 Problem Set 5 A6 a...
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SOLUTIONS FOR ASSIGNMENT 7 Problem Set 5: A6: a). Use Lagrange multipliers to find the greatest and least distance of the curve 6 x 2 + 4 xy + 3 y 2 = 14 from the origin. b). Illustrate the result graphically by drawing the constraint curve g ( x, y ) = 0 , the level curves f ( x, y ) = C , and the gradient vectors f and g . Clearly indicate the relation between the level curves of f and the contraint curve at the maximum and minimum. a). We look for the maximum and minimum values of the square of the distance of a point ( x, y ) on the curve from the origin, f ( x, y ) = x 2 + y 2 , subject to the constraint g ( x, y ) = 14 , where g ( x, y ) = 6 x 2 + 4 xy + 3 y 2 . We solve the system: f = λ g, g ( x, y ) = 14 , which is equivalent to: (2 x, 2 y ) = λ (12 x + 4 y, 4 x + 6 y ) g ( x, y ) = 14 , or: x = λ (6 x + 2 y ) y = λ (2 x + 3 y ) g ( x, y ) = 14 . (1) Since λ negationslash = 0 ( λ = 0 implies from the first two equations of the system (1) that x = y = 0 which is not a solution of the system since g (0 , 0 = 0 negationslash = 14 ), we can elliminate λ between the first two equations of the system (1) and get: x (2 x + 3 y ) = y (6 x + 2 y ) ( y + 2 x )(2 y x ) = 0 y = 2 x or y = x/ 2 . If y = 2 x then the constraint equation g ( x, 2 x ) = 14 gives x 2 = 7 / 5 , and hence ( ± radicalBig 7 5 , 2 radicalBig 7 5 ) are two solutions of the system (1). If y = x/ 2 then the constraint equation g ( x, x/ 2) = 14 gives x 2 = 8 / 5 , and thus ( ± 2 radicalBig 2 5 , ± radicalBig 2 5 ) are other two solutions of the system (1).
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{[ snackBarMessage ]} | 653 | 1,949 | {"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} | 4.40625 | 4 | CC-MAIN-2018-17 | latest | en | 0.786088 |
https://stat.ethz.ch/pipermail/r-help/2014-April/373351.html | 1,585,855,070,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370507738.45/warc/CC-MAIN-20200402173940-20200402203940-00204.warc.gz | 702,241,079 | 2,378 | # [R] Generate Binary Matrix
David Carlson dcarlson at tamu.edu
Wed Apr 9 23:33:10 CEST 2014
```You could randomly assign 1 to a single column in each row and
then use binomial draws on the remaining 0's:
> set.seed(42)
> dimMat <- matrix(0, 1000, 4)
> dimMat[cbind(1:1000, sample.int(4, 1000, replace=TRUE))] <- 1
> dimMat[dimMat<1] <- sample(0:1, 3000, replace=TRUE, prob=c(.6,
.4))
> table(rowSums(dimMat))
1 2 3 4
217 402 310 71
> colSums(dimMat)
[1] 551 586 533 565
> sum(dimMat)
[1] 2235
-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-----Original Message-----
From: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org] On Behalf Of Doran, Harold
Sent: Wednesday, April 9, 2014 3:56 PM
To: r-help at r-project.org
Subject: [R] Generate Binary Matrix
I am trying to generate a binary matrix where every row in the
matrix is guaranteed to have at least one 1. Ideally, I would
like most rowSums to be equal to 2 or 3 with some 1s and some
4s. But, rowSums cannot be equal to 0.
I can tinker with the vector of probability weights, but in
doing so (in the way I am doing it) this causes for more rowSums
to be equal to 4 than I ideally would like, but this never helps
to guarantee a rowSum will not be equal to 0. I could post-hoc
tinker with any rows that are all 0, but seems like that may be
just inefficient.
Below is sample code, any ideas on how to best tackle this?
Harold
dimMat <- matrix(0, 1000, 4)
for(i in 1:1000){
dimMat[i, ] <- sample(c(0,1), 4, replace = TRUE,
prob = c(.3, .7))
}
table(rowSums(dimMat))
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https://stat.ethz.ch/mailman/listinfo/r-help | 566 | 1,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.802278 |
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# Phase transformation
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### Phase transformation
1. 1. Phase Transformation
2. 2. Phase transformationPhase transformation – Formation of a new phase having adistinct physical/chemical character and/or a different structurethan the parent phase.It involves two phenomena – Nucleation and Growth Nucleation – formation of a nucleus or tiny particles of thenew phase.A nucleus is formed when the Gibbs free energy, G, of thesystem decreases i.e. G becomes negative.Two types of nucleation – Heterogeneous and HomogeneousGrowth – Increase in size of the nucleus at the expense of theparent phase.
3. 3. Homogeneous nucleationIn homogeneous nucleation the probability of nucleation issame throughout the volume of the parent phase.The simplest example of nucleation is solidification of a metal.Above melting point Tm, liquid free energy, Gl < Gs (Solid freeenergy) and free energy change for solidification G > 0.Below Tm, G < 0 and nuclei of the solid phase form.
4. 4. Homogeneous nucleationThere are two contributions to free energy change, volumefree energy Gv and surface free energy, due to creation of anew surface.Taking the nucleus as a spherical particle of radius r G = 4/3r3Gv + 4r2 ------------------ (1)The tiny particle of the solid that forms first will be stable onlywhen it achieves a critical radius (r*). Below the critical radius itis unstable and is called embryo.Since this happens at the maximum of the G vs. r curvedG /dr = 4r2Gv + 8r = 0 2 * 16 3This yields r and G * Gv 3( Gv )2
5. 5. Heterogeneous nucleationHere, the probability of nucleation is much higher at certainpreferred sites such as mold wall, inclusions, grain boundaries,compared to rest of the parent phase.Example - Solidification of a liquid on an inclusion surface IL = IS + SL cos 2 * With a similar approach it can be shown that r S L and Gv 4 3 G *Het (2 3 cos cos3 ) G *Hom S( ) 3( Gv )2The small value of ensures that the energy barrier (G) iseffectively lowered in heterogeneous nucleation.
6. 6. Nucleation and Growth KineticsOnce the embryo exceeds the critical size r*, the growth ofthe nucleus starts. Nucleation continues simultaneously.Nucleation and growth rates are function of temp. Nucleationrate increases with cooling rate and degree of undercooling (T= Tm – T). High nucleation rate and low growth – Finer grain size.The over all transformation rate is the product of nucleationand growth rates.
7. 7. Fe-C Phase diagram
8. 8. Phases in Fe-C system-ferrite – Interstitial solid solution of C in BCC iron. Maxsolubility of C is 0.025%. Exists from 273C to 910C.Austenite () - Interstitial solid solution of C in FCC iron. Maxsolubility of C is 2.1%. Exists from 910C - 1394C. -ferrite (BCC) exists over the temp range of 1394C to1539C. Max solubility of C is 0.09%.Cementite, Fe3C - is an intermetallic compound. C content inFe3C is 6.67%.Graphite, the free form of C, also exists in the Fe-C system.Bainite (B) is another phase which forms in steels at highercooling rates.The hard phase martensite (M) forms below the bainitictemperature range at high cooling rates.
9. 9. Critical temperatures in Fe-C systemThe eutectoid temperature (727C) during heating andcooling is Ac1 and Ar1 respectively. A for arrêt (arrest), c forchauffage (heating) and r for refroidissement (cooling).At normal rates of heating or cooling Ac1 > Ar1.A2 at 768 C is the currie temp above which Fe turnsparamagnetic while heating.The temperatures corresponding to ( +)/ and ( +Fe3C)/phase boundaries are function of carbon content and arerepresented as A3 and Acm respectively.The eutectic temperature is 1146 C.The peritectic temperature is at 1495 C.
10. 10. Phase transformation in Fe-C systemPeritectic reaction at 1495 CL (0.53% C) + (0.09% C) (0.17% C)Eutectic reaction at 1146CL (4.3% C) (2.1 % C) + Fe3C (6.67% C). The eutecticmixture of austenite () and cementite (Fe3C) is calledLedeburite. Compositions right and left of 4.3% are called hyperand hypoeutectic steels (Cast iron) respectively. Eutectoid reaction at 727C (0.8 % C) (0.025% C) + Fe3C (6.67% C). The eutectoidmixture of ferrite () and cementite (Fe3C) is called Pearlite.Compositions right and left of 0.8% are called hyper andhypoeutectoid steels respectively.Compositions up to 2.1% C are steels and beyond this it isconsidered as cast iron.
11. 11. MicrostructuresA eutectoid steel (0.8% C) will have 100% pearlite (p) at roomtemperature (RT). The pearlite formed under equilibriumconditions consists of alternate lamellas of ferrite and Fe3C. Schematic of Pearlite – White areas are Hypoeutectoid steels – + p; hypereutectoid – Fe3C + p.Hypoeutectic cast irons consist of + ledeburite (Le) belowthe eutectic temp and p + Fe3C + Le at RT as the transformsto Fe3C and p at the eutectoid temp. Similarly hyper eutecticcast irons will have a structure of Fe3C + Le.
12. 12. T-T-T diagramThe relation between temperature and time for the formationof a phase is given by T-T-T or temp – time – transformationdiagrams also known as isothermal-transformation diagram.A typical T-T-T diagram is shown below. The phases formedon isothermal holding at a given temp for a certain period oftime are indicated.
13. 13. T-T-T diagramAt normal cooling rates pearlite (P) forms, higher coolingrates generates bainite (B). The size of pearlite or bainitedepends on the transformation temp.Martensite (M) forms when the steel is cooled below thematernsite start (Ms) temp at much higher cooling rate so thatthe nose of the T-T-T curve (shown dotted) is avoided (the longblue arrow) .Diffusion rates below Ms is so low that M transformationis a diffusionless process (the C content remains same).However, the crystal structure changes from FCC () to bodycentered tetragonal (BCT).
14. 14. C-C-T diagramIn actual practice a steel is generally cooled continuously.Continuous-cooling-transformation (C-C-T) diagrams depict thissituation.The C-C-T curve (Blue) is shifted to the right of the T-T-T(dashed) curve as continuous cooling transformation occurs atlower temperature and longer time compared isothermal holding.
15. 15. C-C-T diagramBainite generally does not form in steels during continuouscooling and hence the C-C-T curve ceases just below thenose.The microstructure (fine or coarse) depends on the coolingrate. Higher the cooling rate finer the microstructure is.Finer size pearlite is called sorbite and very fine sizepearlite is called troostite.The critical cooling rate is the one at which the coolingcurve just touches the nose of the C-C-T curve.A cooling rate higher than the critical rate is needed to formmartensite.
16. 16. ExamplesEx.1. A eutectoid steel is slowly cooled from 750 C to atemperature just below 727 C . Calculate the percentage offerrite and cementite.Solution: Eutectoid composition – 0.8% C, Ferrite composition- 0.025% C and cementite – 6.67% C.Apply the lever rule to get the percentages as% Ferrite = 100* (6.67 – 0.80)/(6.67 – 0.025) = 88.3%%Cementite = 100* (0.80 – 0.025)/(6.67 – 0.025) = 11.7%Ex.2. A carbon steel cooled from austenitic region contains9.1% ferrite. What is the C content in the steel?Let c be C content. Apply the lever rule0.091 = (6.67 – c)/(6.67 – 0.025)c = 0.1% C
17. 17. Referenceshttp://www.ce.berkeley.edu/~paulmont/CE60New/heat_treatment.pdfhttp://www.ce.berkeley.edu/~paulmont/CE60New/transformation.pdfhttp://www.synl.ac.cn/org/non/zu1/knowledge/phase.pdfhttp://www.youtube.com/watch?v=3xP1U_oDnfUKey words. Phase transformation; Nucleation; Homogeneousand heterogeneous nucleation; Growth; Fe-C phase diagram;Eutectoid reaction; T-T-T diagram; C-C-T diagram
18. 18. Quiz1. What are the different stages of phase transformation?2. What are homogeneous and heterogeneous nucleation?3. Derive the expression for critical radius of the nucleus?4. What are the different phases present in the Fe-C system?5. How many invariants reactions are present in the Fe-C system and what are those?6. What are microstructure of eutectoid, hypoeutectoid and hypereuctectoid steels obtained under equilibrium conditions?7. What are T-T-T and C-C-T diagrams? What is the fundamental difference between them?8. What should be the conditions for forming martensite in steels?9. Why is the martensitic transformation in steels a diffusionless process?
19. 19. Quiz10. What are sorbite and troostite?11. A plain-carbon steel contains 93 wt% ferrite and 7 % Fe3C. What is the average carbon content in the steel?12. A 0.9% C steel is slowly cooled from 900 C to a temperature just below 727 C . Calculate the percentages of proeutectoid cementite and eutectoid ferrite?13. A 0.4% C steel is slowly cooled from 940 C to (A) just above 727 C (B) just below 727 C. Calculate the amount of austenite and proeutectoid ferrite for case (A). Calculate the amount of proeutectoid ferrite and eutectoid ferrite and cementite for case (B). | 2,970 | 9,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36 | latest | en | 0.818737 |
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# Map Skills - PowerPoint PPT Presentation
Map Skills. Let’s learn the terms!. Map Terms. Map Terms. Compass Rose – used to find cardinal and intermediate directions on a map. It looks like this… Map Key – Used to show the symbols on a map. For Example:. Symbols. Map Terms.
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## PowerPoint Slideshow about 'Map Skills' - may
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### Map Skills
Let’s learn the terms!
Map Terms
• Compass Rose – used to find cardinal and intermediate directions on a map. It looks like this…
• Map Key – Used to show the symbols on a map. For Example:
Symbols
Map Terms
• Cardinal Directions are North, South, East and West.
Map Terms
• Intermediate Directions are Northeast, Southeast, Southwest, Northwest.
Map Terms
• Map Scale – the unit of length on a map and the corresponding length on the ground. There are several kind of map scales.
Map Terms
• Types of Map Scales:
• Verbal scale – Uses words to express the relationship between actual size of an area and its size as shown on a map.
Map Terms
• Types of Map Scales:
• Graphic scale – Uses a graphic image to express the relationship between the actual size of an area and its size as shown on a map.
Map Terms
• Types of Map Scales:
• Representative Fraction Scale – uses a fraction to express the relationship between the actual size of an area and its size as shown on a map.
Map Terms
• Axis - The Earth’s axis is an imaginary line that goes through the center of the Earth from the north pole to the south pole.
• Axis – The powerful earthquake that unleashed a devastating tsunami Friday appears to have moved the main island of Japan by 8 feet (2.4 meters) and shifted the Earth on its axis.
• "At this point, we know that one GPS station moved (8 feet), and we have seen a map from GSI (Geospatial Information Authority) in Japan showing the pattern of shift over a large area is consistent with about that much shift of the land mass," said Kenneth Hudnut, a geophysicist with the U.S. Geological Survey (USGS).
Images released by NASA show Japan's northeast coast before, left, and after flooding from the quake-induced tsunami.
Map Terms
• Locator Map - sometimes referred to simply as a locator, is typically a simple map used to show the location of a particular geographic area on a larger scale.
Map Terms
• Continents:
• Asia, Africa, North America, South America, and Antarctica, up to Europe then Australia, now you now the Seven Continents.
Map Terms
• Oceans –
• Pacific Ocean
• Atlantic Ocean
• Indian Ocean
• Arctic Ocean
Map Terms
• Hemispheres -
• Northern and Southern Hemisphere (separated by the equator)
• Eastern and Western Hemisphere (separated by the Prime Meridian)
• ***There is no such thing as the Northwestern Hemisphere, Southeastern Hemisphere, etc.) You are always in two hemispheres.***
Map Terms
• Line of Latitude - Reference lines that run parallel to the Equators plane.
• Line of Longitude – Reference lines that run from the North Pole to the South Pole.
Climate Zones
• Polar - the part of the Earth's surface forming a cap over a pole; characterized by frigid climate.
Climate Zones
• Temperate – the regions of Earth that lies either between the Tropic of Cancer and the Arctic Circle or the Tropic of Capricorn and the Antarctic Circle.
Climate Zones
• Tropical Zone – also known as the tropics is a region of the Earth surrounding the Equator. It is limited in latitude by the Tropic of Cancer in the northern hemisphere at approximately 23° 26′ 16″ (or 23.4378°) N and the Tropic of Capricorn in the southern hemisphere at 23° 26′ 16″ (or 23.4378°) S.
Named Lines of Latitude
• Line of Latitude - Reference lines that run parallel to the Equator’s plane.
Named Line of Latitude
• North Pole – SANTA LIVES THERE!!!
But Seriously…….
I LOVE SANTA!!
No.. Really it is this!!!
Named Line of Latitude
• North Pole - The northern end of the Earth's axis of rotation, located at 90° north latitude at a point in the Arctic Ocean.
Named Line of Latitude
• Arctic Circle - The parallel of latitude approximately 66°33 north. It forms the boundary between the North Temperate and North Frigid zones.
Named Lines of Latitude
• Tropic of Cancer - also referred to as the Northern tropic, is the circle of latitude on the Earth that marks the most northerly position at which the Sun may appear directly overhead at its zenith.
Named Lines of Latitude
• Equator - Marks the point of 0 degrees latitude and divides the Northern and Southern Hemispheres
Lines of Latitude
• Tropic of Capricorn - or Southern tropic, marks the most southerly latitude on the Earth at which the Sun can be directly overhead.
Named Line of Latitude
• Antarctic Circle - The parallel of latitude approximately 66°33 south. It forms the boundary between the South Temperate and South Frigid zones.
Named Line of Latitude
• South Pole - The southern end of the Earth's axis of rotation, located at 90° south latitude at a point in Antarctica.
Lines of Longitude or Meridians of Longitude
• Lines of Longitude – Reference lines that run from the North Pole to the South Pole.
• Prime Meridian – Marks the point of 0 degrees longitude and divides the Eastern and Western Hemispheres.
Kinds of Maps
• Physical Map - maps that shows countries of the world.
.
Kinds of Maps
• Political maps - are designed to show governmental boundaries of countries, states, and counties, the location of major cities, and they usually include significant bodies of water. Like the sample above, bright colors are often used to help the user find the borders.
Kinds of Maps
• Historical Map - shows locations, events, or points of interest for a specific time in history. They may also focus on a geographical location.
Kinds of Maps
• Distribution map – a map that shows where things are distributed or located.
• Climate
• Resources
• Products
• Vegetation
Kinds of Maps
• Climate Map – shows weather patterns in a given area.
Kinds of Maps
• Resource map – Shows areas that contain resources that are rich in that area.
A map of McDonald’s, Burger King, Pizza Hut, Taco Bell, Wendy’s, KFC, Jack in the Box, Hardee’s, Carl’s Jr, and In-N-Out
Kinds of Maps
• Products map - it is a map that shows products on in a particular state or area.
Kinds of Maps
• Vegetation Map - a map that shows particular types of vegetation that are grown in an area.
### The End
Be sure to study for your test!!!! | 1,602 | 6,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.847136 |
https://forum.math.toronto.edu/index.php?PHPSESSID=6606e9fcfgkaac837q091eqml5&topic=2509.msg7395 | 1,712,969,730,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00410.warc.gz | 255,290,261 | 5,351 | ### Author Topic: Quiz4-problemSixE (Read 1854 times)
#### Xuefen luo
• Full Member
• Posts: 18
• Karma: 0
##### Quiz4-problemSixE
« on: October 24, 2020, 02:50:58 AM »
Problem: Find a "Closed form" for the given power series:
\begin{align*} \sum_{n=1}^{\infty} z^{3n}.\end{align*}
Therefore, the closed form for $\sum_{n=1}^{\infty} z^{3n}$ is $\frac{1}{1-z^3} - 1$. | 152 | 371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-18 | latest | en | 0.575701 |
https://pro.arcgis.com/en/pro-app/latest/tool-reference/image-analyst/how-raster-calculator-works.htm | 1,695,780,809,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510238.65/warc/CC-MAIN-20230927003313-20230927033313-00169.warc.gz | 508,651,239 | 6,582 | # How Raster Calculator works
The Raster Calculator tool allows you to create and execute Map Algebra expressions in a tool. Like other geoprocessing tools, the Raster Calculator tool can be used in ModelBuilder, allowing you to integrate the power of Map Algebra into your workflows.
##### Note:
The Raster Calculator tool is not intended to be used in scripting environments and is not available in the standard Spatial Analyst ArcPy module. To learn more about scripting and Map Algebra, see What is Map Algebra in the Spatial Analyst extension help.
The Raster Calculator tool is designed to offer the following benefits:
• Implement single-line algebraic expressions.
• Support the use of variables in Map Algebra when in ModelBuilder.
• Apply Image Analyst operators on three or more inputs in a single expression.
• Use multiple Image Analyst tools in a single expression.
Raster Calculator is designed to execute a single-line algebraic expression using multiple tools and operators. When multiple tools or operators are used in one expression, the performance of this equation will generally be faster than executing each of the operators or tools individually.
## Use the Raster Calculator tool
There are three main areas on the tool dialog box that are used to create a Map Algebra expression: Rasters, Tools, and the expression box.
### Rasters
The input Rasters list identifies the input that can be used in the Map Algebra expression. The raster list contains the layers in the contents and datasets added with the Add raster button. When the tool is used in ModelBuilder, the raster list also contains Raster model variables.
### Tools
The Tools list is a selection of tools that can be used in the Map Algebra expression. When you click a tool in the list, the tool name and open and close parentheses () are placed in the expression where the pointer is currently positioned. The remaining input required by the tool must then be entered. A tool can be placed anywhere in the expression, but it should be placed in a position that produces valid Map Algebra syntax.
The operators in the list allow you to enter mathematical (addition, division, and so on) and logical (greater than, equal to, and so on) operators in the expression. When you select these symbols, that operator is entered in the expression where the pointer is currently positioned.
### Expression
The expression is the Map Algebra expression to be run. The expression must be entered with valid syntax. See the section below to learn more about the syntax rules for Map Algebra.
## Map Algebra language
Map Algebra is a simple and powerful algebra with which you can execute Image Analyst tools, operators, and functions to perform geographic analysis.
The Map Algebra used in Raster Calculator has a syntax, or a set of rules, that must be followed to create a valid expression. If these rules are not adhered to, the expression may be invalid and will not execute, or you may get results you did not expect.
Not only can you access Map Algebra from the Raster Calculator tool, but you can also access it in Python scripting using the Image Analyst ArcPy module. The Map Algebra syntax used in this tool is the same, with the following exceptions:
• You do not need to put the output raster name or the equal sign (=) in the expression, since the output name is specified in the Output raster parameter.
• You do not need to cast input data as a Raster object when using operators. | 686 | 3,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | longest | en | 0.851034 |
https://dsp.stackexchange.com/questions/69053/calculating-normalized-edge-frequencies-for-filter-design-in-matlab | 1,718,325,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00672.warc.gz | 196,739,383 | 39,193 | Calculating normalized edge frequencies for filter design in Matlab
What is the proper formula for converting ordinary frequency to normalized edge frequency in matlab for use in filter design commands like cheby2 etc The wikipedia page shows three different formulas for normalized frequency,i higlighted all those in attached snapshot And Matlab magnitude plot has written (x pi rad/sample) below plot,and i highlighted that also.
So which one of three formulas of wikipedia ,should we use in Matlab?
This is from the mathworks documentation on cheby2
For digital filters, the stopband edge frequencies must lie between 0 and 1, where 1 corresponds to the Nyquist rate—half the sample rate or $$\pi$$ rad/sample.
For discrete-time signals, we use the normalized frequency $$\omega$$ in radians (per sample), defined as
$$\omega=\frac{2\pi f}{f_s}\tag{1}$$
where $$f$$ is the actual frequency in Hertz, and $$f_s$$ is the sampling frequency.
Matlab uses $$(1)$$ normalized by $$\pi$$, i.e., edge frequencies etc. are defined by W$$=2f/f_s$$. The value W$$=1$$ corresponds to $$f=f_s/2$$, which is the Nyquist frequency.
Example: The cut-off frequency is $$f_c=500$$ Hz and the sampling frequency is $$f_s=2000$$ Hz (samples/second). According to $$(1)$$, the corresponding normalized frequency in radians (per sample) is
$$\omega_c=\frac{2\pi f_c}{f_s}=\frac{\pi}{2}\tag{2}$$
Consequently, when calling a Matlab routine you would use Wc$$=\omega_c/\pi=\frac12$$.
• In last line of your answer, Wc(upper case W) and wc(lower case w) are not exactly equal?why this difference? Commented Jul 13, 2020 at 7:30
• @engr: You're right, Wc and $\omega_c$ are different by a factor of $\pi$. That's just the way Matlab uses normalized frequency. Commented Jul 13, 2020 at 7:43 | 467 | 1,780 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-26 | latest | en | 0.881872 |
https://www.hometheaterforum.com/community/search/5127187/ | 1,611,286,248,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529080.43/warc/CC-MAIN-20210122020254-20210122050254-00411.warc.gz | 825,752,570 | 21,025 | Forum list
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# Search results
1. ### What's the next number...
1 10 11 22 14 45 Edit: I screwed this one up. The series should read: 1 10 11 12 14 24 I leave the orginal to show you how dumb I am. :b --Anders
2. ### PDA OS problem
A friend of mine has an IPAQ 3630, which he wanted to upgrade to Pocket PC 2002. Unfortunately, the installation was interrupted, and now the IPAQ wont start (it starts, but the screen just shows red vertical stripes). He can't restart the installation, because ActiveSync wont recognize the IPAQ...
3. ### Another bloody addictive game
Gridlock Currently, I'm stuck on level 25... --Anders
4. ### Famous House Whisky?
A guy mentioned to me that he was looking for a whisky called Famous House, which I didn't recognize. At first, I though he was refering to Famous Grosse (not sure about the spelling), but he assured me that he was not. I can't find anything about this and my current though is that he read...
5. ### 1+1=1?
An oldie, but fun: a = 1 b = 1 This gives us: a = b Now multiply both sides by a: aa = ab a^2 = ab Subtract both sides by b^2: a^2 - b^2 = ab - b^2 The conjugate rule now gives us: (a + b)(a - b) = ab - b^2 Extract b on the right side: (a + b)(a - b) = b(a - b) Divide...
6. ### What to get
Having promised my wife not to buy that many DVDs this year, I'm a bit limited to what I can buy. This also mean that some of the things I want have really piled up. I'm thinking I can probably get away with buying one or two of these this month, so what do you think: Die another Day Star...
7. ### Tom Lehrer concert?
Does anyone know if any filmed concert material with Tom Lehrer is available (at all), and who should be pushed into releasing it? I find his lyrics quite hilarious, and since much of the avialable material is live, one could always hope. Of course, it's so old that nothing may be available...
8. ### So, I'm thinking about getting a new tattoo...
My current thought is this across my back. Whattcha think? --Anders
9. ### Totally irrelevant car question
How many of you actually keep gloves in the glove compartment? --Anders
10. ### Jesus Christ Vampire Hunter
Well, what can you say? The story was awful. The dialogue was totally pointless, and poorly dubbed. The fighting scenes made Steven Segal seem like a talent, and there wasn't the faintest trace of any acting talent. In other words, I loved this movie. The title alone make sure there is no way...
11. ### Another what's-this-movie
It's a comedy, probably 80's, about three fat guys hired as nurses for some old guy. The old guys son wants him dead, so he hires the fat guys as nurses, since he figures they're so bad at what they do, they'll probably kill him. Ring any bells? I remember it being funny as hell when I was...
12. ### The origins of Gollum...
Over The Hedge --Anders
13. ### What's this series
I remember watching it a couple of years back on TV. It was about a vampire. Not really an evil one, as far as can remember. There was also this female ghost which apparently haunted him, but I can't remeber what it wanted with him. Ring a bell to anyone? Is this available on shiny discs...
14. ### Just finished TNG
What a rush. Having watched the whole series during a period of 7 months or so gives a great overview of it as a whole. I truly love it, and it makes me even more eager for DS9. Now, I do have a tiny question as well: Patrick Stewart is described as a "Shakespearean actor", a term which I've...
15. ### Quicktime trailers
I've been thinking about something: Why is it just about every trailer is only available in Quicktime format? Is there some conspiracy to make us (me) install yet another software that will be used for one purpose only? Or is there an actual reason for this? --Anders
16. ### What's this movie?
It's a childrens movie about a robotic dog, built to fight burglaries. It was along time ago, so it's hard to remember many details, but I do remember that you gave commands to the dog by saying numbers. Particulary, the number 100 meant that he would attack. So, what's the name of this...
17. ### Pac-Man the Movie?
I just read that a comapny called Crystal Sky have aquired the movie rights for Pac-Man. Apparently no actors, directors or writers have signed up yet. Hollywood Reporter What do you guys make of this? --Anders
18. ### Does canadian Jason X have bilingual cover?
Guess the topic covers it. --Anders
19. ### Whaaaa, what's this movie?
The only thing I remmeber is a guy (possibly a monster in disguise), sitting in an armchair, dipping worms and eating them. --Anders
20. ### I think I'm falling in love...
... with my wife's best friend. Yea, it sounds like Ricki Lake, but I'm serious. And I feel pretty bad about it. My wife and I don't have a great relationship. I still love her, but the feeling is not mutual. We also have two children. We don't fight all the time, we actually get along much...
21. ### Over the Hedge
I happened to run into this at IMDB today. Apparently there will be an animated movie based on the comic strip Over the Hedge. According to the Tree-That-Knows-Stuff, Jim Carrey will be doing the voice of RJ. I can't wait, though it looks like I have to. Does anyone know something more about...
22. ### My mind wanders
Last night, I finished my shower, got out and put on my robe. When I grabbed a towel to dry my hair, I suddenly realized that I'd forgotten to rinse the conditioner out. I just don't know what I'm thinking sometimes. --Anders
23. ### Is this wrong?
Being that everyone was very willing to discuss the morals of exchanging DVDs in another store than it was originally bought, here's another moral dilemma: My wife feels I buy far too many DVDs (I probably do). As you're all aware ST:TNG season 3 was released a couple of weeks ago. Is it...
24. ### The art of movies
Ok, this is probably not all that interesting to people in the US (or anyone outside Sweden), but some might find it interesting. Swedish filmmaker Vilgot Sjöman has filed charges against the swedish channel TV4. Why? Because his movie Alfred (about Alfred Nobel) was shown... with breaks for...
25. ### Beta testers wanted!
Been working for a while now on a new version of MoCaSys (possibly a competitor to DVD Profiler, but probably not a threat to them :) ). It holds info about all kinds of movie releases, (DVD, LD, VHS, VCD, etc). Well, I'm gonna release version 3.3 now, and it'd be most helpful if a bunch of you...
26. ### Oppinion on MoCaSys...
Yea, I keep promoting my own (movie) tracking software, MoCaSys, but noone seems to care. :) I must be doing something wrong.... Anyway, I'd like your oppinion on the layout of the online version. The information will be presented in the same way as in the downloadable version, so any...
27. ### The "correct" way to face a box set?
Ok, I know this is a bit anal, but... a well, you all are: :) Which is the "correct" way to face a box set in the shelf? I've personally always prefered the spines of the discs out, but moving towards changing that... What do you say? -Anders
28. ### Buffy Season 3 R4 trade
I currently have Buffy season 3 R4, which I would like to trade for the R2 UK version. Any takers? OR: I might also be willing to trade a sealed Army of Darkness (2 disc LE) for the UK Buffy S3. --Anders
29. ### Hellraiser III aspect?
I recently picked up Hellraiser III on DVD (canadian), which unfortunately turned out to be 1.33:1. My question is, what's the orginal aspect? And is the 1.33:1 an open matte or a pan & scan? --Anders
30. ### Region code scans
Need your help again, guys: I'm looking for scans of the region code ions. You know, the globe with the number, for all regions. Does anyone know where to get these? --Anders | 1,956 | 7,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.958779 |
https://stats.stackexchange.com/questions/489326/why-functions-sampled-from-a-linear-kernel-gaussian-process-are-guaranteed-to-be | 1,718,395,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00003.warc.gz | 488,964,747 | 42,782 | # Why functions sampled from a linear kernel Gaussian Process are guaranteed to be a linear function?
It's well known that a linear kernel Gaussian Process regression is equivalent to Bayesian Linear Regression, because the functions sampled from a linear kernel GP is bound to be a linear function. However, I have trouble to understand why this is the case. Just in case you are not sure, you can see it for your self with this nice visualization tool. Just choose the kernel type to be linear.
My confusion can be further boiled down to the following question.
Let's suppose a simple covariance matrix as below, which is derived from a linear kernel.
[[ 0. 0. 0.]
[ 0. 25. 50.]
[ 0. 50. 100.]]
If you draw a sample Y from a 3D multivariate Gaussian with this covariance matrix, and plot the vector Y against X=[1,2,3], the 3 points will always be on a straight line. I just can't figure out why this is the case. I hope someone can help give an intuitive explanation.
Observe that your covariance is degenerate, as it has rank 1. Sampling from this distribution does not give you as much "randomness" as expected as it can be written as the push-forward of a random variable on a lower dimensional space. Let's make this explicit! Notice that we can write your covariance matrix as $$\boldsymbol{\Sigma} = \begin{pmatrix}0 & 0 & 0 \\ 0 & 25 & 50 \\ 0 & 50 & 100 \end{pmatrix} = \begin{pmatrix}0 & 5 & 10 \end{pmatrix}\begin{pmatrix} 0 \\ 5 \\ 10 \end{pmatrix} = \boldsymbol{v}\boldsymbol{v}^{T}$$ which reveals the rank 1 structure of $$\boldsymbol{\Sigma}=\boldsymbol{v}\boldsymbol{v}^{T}$$. Notice that this structure directly comes from the linearity of the kernel! Denote the samples as $$\boldsymbol{y} \sim \mathcal{N}\left(\boldsymbol{0}, \boldsymbol{\Sigma}\right) \in \mathbb{R}^{3}$$. The trick is now rewriting this as $$\boldsymbol{y} \stackrel{(d)}{=} z\boldsymbol{v}$$ where $$z \sim \mathcal{N}(0,1)$$ (one dimensional!). This is due to the fact that scaling $$z$$ with $$v_i$$ in each component is still Gaussian and we can check the statistics $$\mathbb{E}[z\boldsymbol{v}] = 0 \hspace{3mm} \text{ and } \hspace{3mm} \text{cov}(z\boldsymbol{v}) = \boldsymbol{v} \text{ cov}(z)\boldsymbol{v}^{T} = \boldsymbol{v}\boldsymbol{v}^{T} = \boldsymbol{\Sigma}$$ which shows $$\boldsymbol{y} \stackrel{(d)}{=} z\boldsymbol{v}$$. But this means we can also sample $$\boldsymbol{y}$$ as $$z\boldsymbol{v}$$ which always is a random multiple of $$\boldsymbol{v}$$ (again, $$z$$ is one dimensional) and hence reveals that the three points will always lie on the line given by $$\boldsymbol{v}$$!
Essentially, having a low rank covariance implies that we only will access low-dimensional randomness although the random variable itself seems to live in a higher dimensional space. This is why all your points end up on a line!
Another, maybe helpful approach looks at the diagonalization of your covariance: $$\boldsymbol{\Sigma} = \boldsymbol{U}\boldsymbol{\Lambda}\boldsymbol{U}^{T}$$ Due to rank 1, $$\Lambda_{11} \not = 0$$ while $$\Lambda_{22}=\Lambda_{33}=0$$. Since Gaussians are invariant under rotations, we can essentially use this new rotated coordinate system where your covariance $$\boldsymbol{\Sigma}$$ becomes the diagonal covariance $$\boldsymbol{\Lambda}$$. But here you see that the second and third entry of your random vector are fixed since they have no variance ($$\Lambda_{22}=\Lambda_{33}=0$$)!
You can check out this answer for a more general explanation of this effect.
• Thanks for your answer. While I can see your line of reasoning proves sampling from the family of z*v will result in a covariance matrix as specified, I still have difficulty to understand why it is the only possible type of vector which can be drawn from a multivariate with such covariance matrix. To put it another way, why it is not possible for a sample from a multivariate with such covariance matrix is not on a linear line? Commented Nov 27, 2020 at 5:44
• You're welcome! I edited my answer and added another short paragraph. I hope this explains it more intuitively. A rank 1 covariance matrix essentially means that you can find a rotation in which the rotated system has a diagonal covariance but only the first eigenvalue is non-zero. This means that all other entries are deterministic (they have zero variance) and hence you land in a 1d subspace, given by a line. Commented Nov 27, 2020 at 8:29
Refer to Drawing values from the distribution section in Wikipedia. Let me use that to prove the linearity of functions with basic linear algebra.
We will use the following fact,
If we can write the covariance matrix $$\Sigma$$ of a multivariate normal distribution as $$\Sigma=AA^T$$ for any real $$A$$, then $$\mathbf{y} = \boldsymbol{\mu} + A \mathbf{z}$$ is a valid function drawn from $$\mathcal{N}(\boldsymbol{\mu},\Sigma)$$, where $$\mathbf{z}$$ is a function drawn from $$\mathcal{N}(\mathbf{o},I)$$ (standard multivariate normal distribution).
We want to show that any $$\mathbf{y}$$ follows $$m\mathbf{x}+b$$ (linear) form, where $$m$$ and $$b$$ are slope and offset respectively.
According to the distill article you refer to in the question (also in general), the linear kernel is given as follows, $$K(x,x') = \sigma^2(x-c)(x'-c) + \sigma_b^2$$ Writing it in covariance matrix form, $$\Sigma = K(\mathbf{x},\mathbf{x}) = \begin{bmatrix} \sigma^2(x_1-c)^2+\sigma_b^2 & \sigma^2(x_1-c)(x_2-c)+\sigma_b^2 & \cdots\\ \sigma^2(x_2-c)(x_1-c)+\sigma_b^2 & \sigma^2(x_2-c)^2+\sigma_b^2 & \cdots\\ \cdots& \cdots & \cdots \end{bmatrix}$$
If we want to write it in $$\Sigma = AA^T$$ form, $$A$$ would be, $$A= \begin{bmatrix} \sigma(x_1-c) & \sigma_b & 0 & \cdots\\ \sigma(x_2-c) & \sigma_b & 0 & \cdots\\ \cdots& \cdots & \cdots & \cdots\\ \sigma(x_n-c)& \sigma_b & 0 & \cdots \end{bmatrix}$$ Now, for GP, we take $$\boldsymbol{\mu}=\mathbf{o}$$, so, $$\mathbf{y}=A\mathbf{z}$$ is a valid function from $$\mathcal{N}(\mathbf{o},\Sigma)$$. $$\begin{bmatrix} y_1\\y_2\\\cdots\\y_n \end{bmatrix} =A\mathbf{z}= \begin{bmatrix} \sigma(x_1-c) & \sigma_b & 0 & \cdots\\ \sigma(x_2-c) & \sigma_b & 0 & \cdots\\ \cdots& \cdots & \cdots & \cdots\\ \sigma(x_n-c)& \sigma_b & 0 & \cdots \end{bmatrix} \begin{bmatrix} z_1\\z_2\\\cdots\\z_n \end{bmatrix}=z_1\sigma \begin{bmatrix} x_1\\x_2\\\cdots\\x_n \end{bmatrix}-(z_1\sigma c - z_2\sigma_b)$$
As you can see, $$\mathbf{y}$$ follows $$m\mathbf{x}+b$$ form, where slope $$m=z_1\sigma$$ and offset $$b=z_2\sigma_b - z_1\sigma c$$.
Hence proved :)
(let me know in comments if any step requires more clarification). | 1,977 | 6,599 | {"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": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-26 | latest | en | 0.89815 |
https://brainly.com/question/230747 | 1,485,273,290,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00099-ip-10-171-10-70.ec2.internal.warc.gz | 797,868,011 | 9,048 | 2014-12-17T20:13:12-05:00
### This Is a Certified Answer
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We have such equation
-3y-7=2 /+7 (add 7 both sides)
-3y-7+7=2+7
-3y-0=9
-3y=9 /:3 (divide both sides by 3)
-y=3 /*(-1) (multiply both sides by -1)
y=-3 - its the result
• Brainly User
2014-12-17T20:15:04-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
-3y-7=2
-3y-7+7=2+7
-3y=9
(-3y)/(-3)=9/(-3)
y=-3 | 285 | 950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-04 | latest | en | 0.902208 |
https://amelyachtowners.groups.io/g/main/message/43093?p=,,,20,0,0,0::created,0,,1,2,737,43093 | 1,623,494,857,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00625.warc.gz | 115,598,085 | 8,245 | #### Re: Power draw at anchor
Jim Anderson
Mark, Thanks you for the correction. My math was for 12V rather than 24. So hopefully the following is correct:
If your SM is equipped with 8 12V, 110Ah batteries and you are drawing 0.1 Amps continuously at 24V then theoretically it would take 73 days, 8 hours to run your batteries down to 60% charged.
If you were drawing 0.2A at 24V then it would be half of that, 36 days, 16 hours hours.
For the 12 battery "comfort pak" model it would be 110 days and 55 days, respectively.
8 X 110 = 880 Total Ah at 12V
880/2 = 440Ah at 24V
440 X 40% = 176 Usable Ah
176/0.1 = 1760 hours available at 0.1A draw
1760/24 = 73.33 Days
Hope that helps, or is at least mildly interesting trivia,
Jim
SM384 Sirena Azul
Join main@AmelYachtOwners.groups.io to automatically receive all group messages. | 256 | 843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.929368 |
https://study.com/academy/lesson/x-variable-definition-lesson-quiz.html | 1,563,698,054,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526940.0/warc/CC-MAIN-20190721082354-20190721104354-00142.warc.gz | 556,974,220 | 34,953 | # X Variable: Definition & Concept
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
In ways, the little x variable is the most popular item in algebra. Learn the many uses of this variable. Also find out why the x variable is not always shown in all algebraic equations and expressions.
## What is a Variable?
A variable is used to represent something unknown. In math, you encounter many problems that ask you to find a certain value - like finding the total number of items if there are two piles of the same item. If one pile contains two widgets and the other pile contains five widgets, we can use a variable to represent our total.
In some cases, the variable is easy to figure out. In other cases, the variable depends on other variables, too.
But, in all cases, the variable stands for an unknown quantity. Sometimes, you can solve it and other times, you can't solve it and would leave it in equation form.
## What's So Special About the x Variable?
The x variable seems ubiquitous in math. You see it everywhere. The variable x appears more often than any other variable. Why? This is one of those mathematical things that have always been around and everyone is okay with it. The letter x just seems like a good candidate for it because it stands out as different from any of the numbers. There is the phrase, 'x marks the spot,' that tells you the location of a particular destination. Similarly, in math, the x tells you the location of a point on a graph.
## Using the x Variable
The x variable is used in algebra as well as in calculus and higher math. Any math that has unknowns will usually involve the x variable. Using it requires following certain steps that remain the same no matter what level of math you are in.
Step one is that of locating the x variable that you want to solve for. If given a word problem, you will need to define your x variable and set it equal to the item you want to solve for. Say, for example, you get a problem that is asking you to find the area in between two squares - the smaller square being inside the larger square. You are given the areas of the squares. What will you set your x variable equal to? Yes, you will set the x variable equal to the area between the two squares.
Step two is to write down a formula or equation you can use to solve the problem. In our problem with the two squares, think of a way to calculate the area between the two squares. Difference is usually solved by subtraction. So, which area would you subtract from the other? You would subtract the smaller area from the larger area to find the difference.
The A1 is the area of the larger square and A2 is the area of the smaller square. By setting up your problem like this, you have created an equation with your x variable that you can solve.
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Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 751 | 3,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-30 | longest | en | 0.947976 |
https://astronomy.stackexchange.com/questions/45199/how-slow-would-you-age-on-a-double-gravity-planet | 1,716,638,630,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00032.warc.gz | 84,691,592 | 37,049 | # How slow would you age on a double gravity planet?
How slow would you age on a double gravity planet?
Gravity of the planet A is 10 units. If you are taken from planet A to planet B where gravity is double that of A, i.e. 20 units. How slowly would you age as compared to being on planet A, and how?
• BTW, gravitational time dilation on a planet is quite small. Eg, the difference in the time rate between the Earth's surface and its centre is $\approx 3.48×10^{-10}$, so over a century a clock at the centre of the Earth would be slow by ~1.1 seconds compared to a clock on the surface. Aug 8, 2021 at 0:03
A simple way to calculate the gravitational time dilation on the surface of a planet comes by assuming that
• the planet is spherically symmetric
• other sources of gravitational potential (a star, a moon) are negligible on the surface of the planet
In that case, the time experienced by the person on planet 1 is related to the time experienced by the person on planet 2 by
$${\Delta \tau_1 \over \Delta \tau_2} = \frac{\sqrt{1-\frac{2GM_1}{c^2R_1}}}{\sqrt{1-\frac{2GM_2}{c^2R_2}}}$$
On a planet, the factor $${2GM \over c^2R}$$ is usually very small, so the formula can be approximated to
$${\Delta \tau_1 \over \Delta \tau_2} = (1-\frac{GM_1}{c^2R_1})(1+\frac{GM_2}{c^2R_2})$$
This is valid if the two planets are still with respect to each other. If they are moving at a substantial speed also SR time dilation should be considered. In any case, the effect is really small.
• Just to extend this a little, if the density of the planets is the same, then surface acceleration is proportional to R, (and mass is proportional to R^3). Thus if the second planet has double the gravity, it would have double the radius, and eight times the mass. Plugging in the values for Earth mass and radius gives a ratio of 1.00000000209 times slower on the planet with more gravity. Aug 8, 2021 at 19:38
Gravitational time dilation depends on the gravitational potential, not the gravitational acceleration. It can't depend on the gravitational acceleration, because by the equivalence principle that can have any value you like, including zero for a free-falling observer. | 574 | 2,184 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-22 | latest | en | 0.915646 |
https://www.convertunits.com/from/arsheen/to/braccio | 1,623,872,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00022.warc.gz | 655,461,164 | 12,834 | ## ››Convert arsheen [Russia] to braccio [Italy]
arsheen braccio
How many arsheen in 1 braccio? The answer is 0.98425196850394.
We assume you are converting between arsheen [Russia] and braccio [Italy].
You can view more details on each measurement unit:
arsheen or braccio
The SI base unit for length is the metre.
1 metre is equal to 1.4060742407199 arsheen, or 1.4285714285714 braccio.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between arsheen [Russia] and braccio [Italy].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of arsheen to braccio
1 arsheen to braccio = 1.016 braccio
5 arsheen to braccio = 5.08 braccio
10 arsheen to braccio = 10.16 braccio
15 arsheen to braccio = 15.24 braccio
20 arsheen to braccio = 20.32 braccio
25 arsheen to braccio = 25.4 braccio
30 arsheen to braccio = 30.48 braccio
40 arsheen to braccio = 40.64 braccio
50 arsheen to braccio = 50.8 braccio
## ››Want other units?
You can do the reverse unit conversion from braccio to arsheen, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 498 | 1,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.725483 |
https://www.vedantu.com/maths/division-of-two-digit-numbers | 1,723,617,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641104812.87/warc/CC-MAIN-20240814061604-20240814091604-00079.warc.gz | 819,593,508 | 36,668 | Courses
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# Division of Two-Digit Numbers for Kids
Reviewed by:
Last updated date: 14th Aug 2024
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Views today: 4.10k
## Learning Maths with Two-digit Divisions
Division is the process of dividing a group of things into equal parts. There are 4 things to know about while dividing, i.e. dividend, divisor, quotient, and remainder. The two-digit division is a very small calculation and does not involve too many steps.
Dividend is the number that is being divided. Divisor is the number by which the dividend is being divided. Quotient is the result of division. Remainder is the amount that is left after the division. Division is the tool used to separately distribute objects among different people in equal numbers. Let’s see the division of 2-digit numbers in more detail.
## Process of Division
The following discusses the process of division in a step-by-step manner.
• Let's take the first digits of the dividend which should be equal to the number of digits that the divisor has.
• The number that is taken from the dividend is smaller than the divisor, we take the next digit of the dividend.
• We begin by dividing the first number of the dividend by the first digit of the divisor.
• The result of that division has to be written in the space of the quotient.
• The digit of the quotient is multiplied by the divisor, and the result is written below the dividend and subtracted.
• But in case the dividend is a smaller number we cannot do so. Therefore, we will need to choose a smaller number in the quotient till the time we can subtract it.
• Once we perform the subtraction, use the next digit of the dividend and repeat the process of the division until the point where there are no more remaining numbers present in the dividend.
Three Aspects in a Division
## Two-digit Division Questions: Solved Examples
In the below examples, we will solve some problems based on two-digit division questions.
Example 1. Divide 94 by 12.
Ans: The number 12 is multiplied by the number 7 so that it equals 84.
94 - 84 = 10
Quotient = 7
Remainder = 10
Example 2. Divide 96 by 16.
Ans: The number 16 is multiplied by 6, so that equals 96.
Quotient = 6
Remainder = 0
Example 3. Divide 88 by 17
Ans: The number 17 is multiplied by 5, so that equals 85.
Quotient = 5
Remainder = 3
Example 4. Divide 192 by 24
Ans: The number 24 is multiplied by 8, so that equals 192.
Quotient = 8
Remainder = 0
Example 5. Divide 51 by 32
Ans: The number 32 is multiplied by 1, so that equals 32.
Now, 51 - 32 = 19
So, 190 is left which is divided now.
Now, again, 32 is multiplied by 5, so that equals 160.
Now, 190 - 160 = 30
Quotient = 15
Remainder = 30
6. Divide 275 by 24
Ans: The number 24 is multiplied by 1, so that equals 24. Now, 27 - 24 = 3
Therefore, 35 is now the dividend. Now, again, 24 is multiplied by 1, so that equals 24
35-24 = 11
Quotient = 11
Remainder = 11
7. Divide 803 by 70
Ans: The number 70 is multiplied by 1, so that equals 70. Now, 80 - 70 = 10
Therefore, 103 is now the dividend. Now, again 70 is multiplied by 1, so that equals 70.
103 - 70 = 33
Quotient = 11
Remainder = 33
8. Divide 345 by 49
Ans: The number 49 is multiplied by 7 so that equals 343. Now, 345 - 343= 2
Therefore, 2 cannot be divided further as the divisor is much larger than the dividend.
Quotient = 7
Remainder = 2
9. Divide 4963 by 14
Ans: The number 14 is multiplied by 3 so that equals 42. Now, 49 - 42 = 7
Therefore, 763 is now the dividend but we use only 76. Now, again 14 is multiplied by 5 so that equals 70.
76 - 70 = 6
Now, 63 is the dividend and we multiply 14 with 4 so that equals 56
63 - 56 = 7
Quotient = 354
Remainder = 7
## Practice Problems
Q1. Divide 300 by 15
Ans: 20
Q2. Divide 155 by 25
Ans: Quotient = 6, Remainder = 5
Q3. Divide 267 by 30
Ans: Quotient = 8, Remainder =27
Two-Digit Division in Steps
## Summary
Division is an important tool for calculation in Mathematics. The four components of division are divisor, dividend, remainder, and quotient. This process of division involves subtraction as well, and it allows us to complete the division by using up all the values of the dividend. Some sums of two digit division questions are discussed in detail and then in practice problems sums are given for practising. We hope now you will no longer struggle with division and enjoy Maths even more.
## FAQs on Division of Two-Digit Numbers for Kids
1. What is the importance of division?
We can divide something into parts but whenever the dividing parts are more and the dividend value is large, we need to use the division rules and follow the process to find the right amount. Division leads to the quotient which is basically the number of parts a dividend can be divided into. If the value of dividend is not divisible by the divisor then a remainder can be found at the end of all steps. This value is what cannot be divided equally.
2. What is the use of division in everyday life?
The process of division involves separately grouping a lot. The application of division is in the equal separation of objects into various groups. It is used to share equally between people, to divide goods, etc.
A very regular example can be the sweets distributed on Independence Day. If there are 200 students and there are 400 sweets, then we can say that 2 sweets each will be received by every student.
3. In a two-digit division, what is the use of a divisor?
The divisor is the two-digit number in a two-digit division that is used to divide the dividend or the number which we need to divide in equal parts. The division can be stated as the total number of parts the dividend has to be divided into. For example, if we need to divide a cake among 5 kids, we shall cut the cake into 5 pieces here the divisor or the number of pieces is 5. | 1,544 | 5,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-33 | latest | en | 0.917233 |
https://www.univerkov.com/an-ice-block-weighing-15-kg-cooled-by-10-degrees-c-overnight-the-specific-heat-capacity-of-ice/ | 1,638,517,991,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362605.52/warc/CC-MAIN-20211203060849-20211203090849-00586.warc.gz | 1,087,341,012 | 6,292 | # An ice block weighing 15 kg cooled by 10 degrees C overnight. The specific heat capacity of ice
An ice block weighing 15 kg cooled by 10 degrees C overnight. The specific heat capacity of ice is 2100 J / kg * C. How much energy is required to heat the ice to its original temperature?
The amount of heat required to heat the ice to its original temperature can be calculated using the formula:
Q = C * m * ∆t, where Q is the amount of heat expended on ice heating (J), C is the specific heat capacity of ice (C = 2100 J / (kg * K)), m is the mass of the ice block (m = 15 kg ), ∆t is the change in the temperature of the ice block (∆t = 10 ºС).
Let’s calculate the required amount of heat:
Q = C * m * ∆t = 2100 * 15 * 10 = 315000 J = 315 kJ.
Answer: To heat an ice block, 315 kJ of heat is needed.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 283 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.937029 |
https://www.esaral.com/q/if-a-3-6-9-12-15-18-21-b-4-8-12-16-20-50081 | 1,712,920,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00429.warc.gz | 661,909,764 | 11,812 | # If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},
Question:
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},
C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii) D – C
Solution:
(i) A – B = {3, 6, 9, 15, 18, 21}
(ii) A – C = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 16, 20}
(v) C – A = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 20}
(vii) B – C = {20}
(viii) B – D = {4, 8, 12, 16}
(ix) C – B = {2, 6, 10, 14}
(x) D – B = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C = {5, 15, 20} | 440 | 713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-18 | longest | en | 0.630707 |
http://dangerousprototypes.com/docs/index.php?title=Mixed_voltage_interfacing_with_the_Bus_Pirate&diff=prev&oldid=1802 | 1,582,779,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146647.82/warc/CC-MAIN-20200227033058-20200227063058-00286.warc.gz | 36,587,362 | 6,983 | # Mixed voltage interfacing with the Bus Pirate
(Difference between revisions)
Revision as of 14:22, 20 August 2010 (view source)Ian (Talk | contribs) (→Examples)← Older edit Revision as of 15:26, 20 August 2010 (view source)Ian (Talk | contribs) (→3.3volts)Newer edit → Line 16: Line 16: ===3.3volts=== ===3.3volts=== + Select output type: + 1. Open drain (H=Hi-Z, L=GND) + 2. Normal (H=3.3V, L=GND) + + (1)> 2 + Ready * Interface directly with normal pin outputs * Interface directly with normal pin outputs * No pull-ups or Vpullup pin connection used * No pull-ups or Vpullup pin connection used
## Revision as of 15:26, 20 August 2010
This will be an updated guide to using the Bus Pirate pull-up resistors to interface stuff at different voltages.
## Contents
The Bus Pirate's on-board pull-up resistors can be used to interface electronics above and below the Bus Pirate's 3.3volt supply.
Read how the pull-up resistors on the Bus Pirate work.
Read our guide to pull-up resistors and high impedance pins.
## Examples
Here's some examples of how to interface different voltages using the Bus Pirate.
These examples assume you use the Bus Pirate on-board pullup resistors. If you use the Bus Pirate's on-board pull-up resistors, remember to connect the Vpullup pin to the circuit supply voltage. The Bus Pirate pull-up resistors are powered through the Vpullup pin.
```Some modes (1-Wire, I2C) only use high impedance pin outputs
```
### 3.3volts
Select output type:
1. Open drain (H=Hi-Z, L=GND)
2. Normal (H=3.3V, L=GND)
(1)> 2
• Interface directly with normal pin outputs
• No pull-ups or Vpullup pin connection used
### 5.0volts
• Choose Hi-Z output type
• Connect the Vpullup pin to the power supply of the 5volt circuit
• The 5volts can come from the Bus Pirate or an external supply
### 2.5volts
• Choose Hi-Z output type
• Connect the Vpullup pin to the power supply of the 2.5volt circuit
### Any voltage 1-Wire, I2C
• The Bus Pirate can interface with 1-Wire or I2C at 5.5volts or less
• Open collector buses like 1-Wire and I2C always use pull-up resistors
• These modes have no output configuration, outputs are always HiZ | 610 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.785042 |
https://blog.csdn.net/Lerbon23James/article/details/79951886 | 1,544,723,506,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825029.40/warc/CC-MAIN-20181213171808-20181213193308-00359.warc.gz | 559,670,254 | 33,213 | # 【状压dp】UVA1252二十个问题
#include<iostream>
#include<cstdio>
#include<cstring>
#define INF (20010)
using namespace std;
int m,n,A[2020],cnt[(1<<11)+5][(1<<11)+5],dp[(1<<11)+5][(1<<11)+5];
char s[2020];
bool vis[(1<<11)+5][(1<<11)+5];
int Dfs(int s1,int s2){
int &ans=dp[s1][s2];
if (vis[s1][s2]) return ans;
vis[s1][s2]=1;
if (cnt[s1][s2]<=1) return ans=0;
ans=INF;
int i;
for (i=1;i<=m;i++)
if (!((1<<(i-1))&s1))
ans=min(ans,max(Dfs(s1|(1<<(i-1)),s2|(1<<(i-1))),Dfs(s1|(1<<(i-1)),s2))+1);
return ans;
}
void Work(){
int i,j,len;
memset(A,0,sizeof(A));
memset(cnt,0,sizeof(cnt));
memset(vis,0,sizeof(vis));
for (i=1;i<=n;i++){
scanf("%s",s);
len=strlen(s);
for (j=0;j<len;j++) if (s[j]-'0') A[i]|=(1<<j);
}
for (i=0;i<=(1<<m)-1;i++)
for (j=1;j<=n;j++)
cnt[i][i&A[j]]++;
printf("%d\n",Dfs(0,0));
}
int main(){
while (cin>>m>>n){
if (!m&&!n) break;
Work();
}
} | 367 | 855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.179653 |
https://www.toppstodin.is/how-to-calculate-maintenance-calories-bodybuilding/ | 1,638,217,732,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00244.warc.gz | 1,171,661,390 | 17,459 | October 10, 2020
# How To Calculate Maintenance Calories Bodybuilding
By
This “go find out yourself” method to calculate maintenance calories takes more time and effort but is certainly more reliable than the starting values given by calorie calculators or any other method. Adjust as needed to maintain your weight.
Pin on Gym Training Guides and Workout Plans
Table of Contents
### In other words, how many calories you need if you did no physical activity.
How to calculate maintenance calories bodybuilding. If you are sedentary, multiply your bmr (1745) by 1.2 = 2094. Bmr= lean body mass (lbs) x 13.8 calories. I hope this post on how to.
You can obtain your lean body mass from body fat measurements. Operate at your new calorie intake level for 2 weeks, then check your overall body weight. First, the calculator will estimate your daily calorie maintenance level, which is the number of calories you require to maintain your current weight.
And cutting more than that is almost never the place to start. Maintenance calories are the total amount of calories required on a daily basis to maintain your body weight with no gains or losses in fat and/or muscle tissue. Here’s an example… let’s say you burn 2500 calories per day from a combination of exercise (weight training, cardio, etc.), normal daily activities (getting dressed, showering, driving, etc.), and your body doing the things it needs to do to sustain.
This is the rate at which a persons body burns energy at rest. The first is bmr, which stands for basal metabolic rate. It doesn't matter how hard you train in the gym.
Together, these three parts are called tdee (total daily energy expenditure).how to calculate maintenance calories: Also, how would you calculate your daily maintenance without an online calculator?. Most men require more calories because they are naturally more muscular than women, but maintenance depends on several factors, ie your body composition and age.
Read How To Create A Cryptocurrency Portfolio
How to calculate maintenance calories. The food options you choose each day will impact total calories consumed, but this difference varies day by day. In the united states (u.s), people consume more than 11% of their daily caloric intake from.
“3500 calorie rule” is not exactly accurate, but it serves the purpose for the course of 2 weeks. Every individual requires different amounts of energy per day as it depends on gender, age, size, and activity level; And they represent the number of calories you burn during the day.
Second, it will add or subtract the appropriate number of calories from your maintenance level to create either a calorie surplus for muscle growth or a calorie deficit for fat loss. Meaning, when calories in = calories out, you’ve reached your calorie maintenance level. To stay at your current weight, stick to the number of calories resulted from step #2.
You'd then either add or subtract 10 to 20% of that number depending on whether your goal is to drop body fat or build muscle. Calculate your maintenance calorie level with a formula. Say your daily calorie goal is 2,500 calories, and you hit your macros with 110 calories remaining.
That may not sound like a big deficit, but it can be enough! If you work out regularly, this number will allow you to slowly decrease fat and increase muscle mass. So, cutting an extra 250 calories from the aforementioned maintenance caloric intake would be about 1869 calories per day.
For activity recall, simply track all hikes, walks, and general activity outside of weightlifting for the weeks you are recording. As illustrated above, there are three parts that go into your maintenance calories. The most popular bodybuilding message boards!
This is based on a your age, weight, gender, and height. How to calculate maintenance calories: Resting metabolism, daily activity, and exercise.
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Key points calculate your maintenance calorie level with a formula. The most precise way to calculate your maintenance calories is to do diet and activity recall by recording the calories you eat and all the activity you do on a daily basis. Put together with the 250 calories expended via exercise, this equates to a 500 calorie per day deficit, or 3,500 calories per week, which would equal about one pound magically lost!
If you use an calculator, have you found that it is pretty accurate? How do you calculate your daily maintenance calorie needs? Calculate lean muscle mass vs.
The number one downfall for anyone who isn't achieving their fitness goals is nutrition. This calorie calculator estimates the number of calories needed each day to maintain, lose, or gain weight. The human body requires energy every day to maintain bodily functions and body weight.
Calories are energy that is essential for human health, and even the key is consuming the right amount of it; Learn more about different kinds of calories and their effects, and explore many other free calculators addressing the topics of finance, math, health, and fitness, among others. Also in my opinion, if you really are going to put fitness into your lifestyle, 2.
This is the best way to calculate calories for people who have a healthy weight, but want lower body fat levels. If your nutrition plan isn't on par with that, you're not going to get the best results. This is the total number of calories you need in order to maintain your current weight.
Hi,i like to know if there are any ways i can calculate a bmr for a teenager who is 14 years old, 6ft 2 and about 230lbs. Resting metabolism, daily activity, and exercise. Maintenance calories are based on two factors.
Calculating daily maintenance calories how do you calculate your daily maintenance. Together, these three parts are called tdee (total daily energy expenditure). It also generates a perfect customized meal for you with proper portion size to help you achieve your goal.
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Don't worry about filling in those additional calories. The reason is because he is overweight and is the son of my friend. Maintenance depends on your activity level, while your basal metabolic rate tells you how many calories you need to function properly (even if you nap all day).
(emma leigh strongly does not recommend using the stickies to calculate a teenager's bmr). How to calculate maintenance calories for teenagers? Focus on the macros, and the calories will even out over time.
Now, it's commonly believed that the calorie equivalent to 1 pound of fat is 3,500 calories.[1]
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Pin on Workout Nutrition & Diet Plans For Men And Women | 1,543 | 7,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.92417 |
http://2000clicks.com/MathHelp/Factoring3d.aspx | 1,555,786,831,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00376.warc.gz | 1,197,780 | 2,848 | # document.write (document.title)
Math Help > Basic Algebra > Factoring > Factoring Trinomials > Simplified AC Method
This is factoring page 3d. It explains the simplified AC Method
In the first page, I presented the perfect square trinomial and the difference of two squares.
In the second page, I presented a method of factoring four-term expressions with pairs of factors.
Pages 3a, 3b, 3c, and 3d give three methods of factoring trinomials:
Page 3a works by listing the factors of the "a" and "c" coefficients.
Page 3b completes the square.
Page 3c is the "AC Method", which has you multiply the "a" and "c" coefficients, and list all the factors of that product.
Some people say this is even easier than the AC Method.
## How to Factor Using the Simplified AC Method
First, consider the general form of the quadratic expression:
ax� + bx + c
where a, b, and c have no factors in common.
(If they have any factors in common, then "pull out" that factor separately, as in this example:
(2)(ax� + bx + c), if the common factor was 2.
Now look again at EXAMPLE 5:
6x² + 19x - 20
It has no factors in common. The "a" coefficient is 6, and the "c" coefficient is 20.
Next, consider what happens to this polynomial when we let y=ax, and so x=y/a:
ax� + bx + c
y�/a + by/a + ac/a
(1/a)(y� + by + ac)
This new polynomial is easier to factor, because the high-order coefficient is 1. To get ac in our example, we multiply the 6 and the -20 together to get -120. This gives us
(1/6)(y� + 19y - 120)
Next, factor the new polynomial. In our example, it is
(1/6)(y-5)(y+24)
Now, substitute the value of ax (6x, in our example) in place of y:
(1/6)(6x-5)(6x+24)
(6x-5)(x+4)
### Summary
You can factor a quadratic this way:
6x² + 19x - 20
(1/6)(y² + 19y - 120)
(1/6)(y-5)(y+24)
(1/6)(6x-5)(6x+24)
(6x-5)(x+4)
If you weren't able to factor a quadratic equation using any of the methods in this section, then either it can't be factored, or else the factors are not rational numbers. In either case, you'll find out what's going on using the QUADRATIC FORMULA -- that's the topic of the next factoring page. | 642 | 2,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2019-18 | latest | en | 0.862967 |
https://johnmjennings.com/is-there-intelligent-life-elsewhere-in-the-universe/ | 1,702,008,209,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00766.warc.gz | 380,678,147 | 57,877 | Select Page
Is There Intelligent Life Elsewhere in the Universe?
by | Mar 20, 2017
“Either we are alone, or we are not; either way is mind-boggling”Lee DuBridge (Astrophysicist and Science Advisor to Pres. Eisenhower).
Is there life elsewhere in the universe? In our galaxy? In 1960 Dr. Frank Drake (Prof. at U. of Calf. Santa Cruz) developed the equation now known as the Drake Equation. This equation attempts to quantify the number of sentient civilizations in our galaxy with whom we might come in contact. The Drake Equation is not supposed to be an actual scientific prediction of alien life, but rather a mechanism to raise discussion around how much intelligent alien life there might be. Here is the equation:
where:
N is the number of civilizations in our galaxy with which we might hope to be able to communicate;
and
R* is the number of stars in our galaxy (actually, the true formula is based on the rate of star formation)
fp is the fraction of those stars that have planets
ne is the average number of planets that can potentially support life per star that has planets
fl is the fraction of the above that actually go on to develop life at some point
fi is the fraction of the above that actually go on to develop intelligent life
fc is the fraction of civilizations that develop a technology that releases detectable signs of their existence into space
L is the length of time such civilizations release detectable signals into space.
Most plausible inputs to the equation generate numbers greater than one (meaning that there should be at least one other civilization out there with intelligence and ability to transmit communications).
Okay, let’s run this baby. Assume that the Milky Way has 200 billion stars, that 30% of those stars have planets and that there is about one planet per star (that has planets) that lie in the habitable zone and are rocky – meaning that they can potentially support life. Further, assume that a mere 1% of those planets go on to develop life, and then 10% of those planets with any sort of life develop intelligent life. Then 10% of those planets with intelligent life develop a technology that releases detectable signs of life into space and they do so for 100,000 years. The Drake Equation says that with those assumptions there should be about 60 communicating civilizations in our galaxy.
All this raises the question – where is everybody? Carl Sagan proposed that the real problem with finding any other intelligent life out there is the last factor – the length of time that such civilizations release detectable signals into space. Consider that we have only sent signals into space with the first radio broadcast in 1906 – so we have only been broadcasting 101 years. So, our transmitting of signals have only reached a mere 101 light-years away. Our galaxy is 100,000 light-years in diameter and we are in one of the outer arms. The radio signals from another civilization broadcasting for thousands or tens of thousands of years from the other side of the galaxy would not have reached us yet unless they started broadcasting many thousands of years before.
Here is a link to a site that will allow you to plug in your own numbers into the Drake Equation: http://www.activemind.com/Mysterious/Topics/SETI/drake_equation.html
1 Comment
1. A compelling argument for the existence of life, but an equally compelling argument that communication between us and other civilizations based on our knowledge of signal propagation is pretty remote. Also, consider if we detect a communication form another civilization that was sent several 1000 years ago, how relevant is that communication.
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Get the Interesting Fact of the Day delivered twice a week. Plus, sign up today and get Chapter 2 of John's book The Uncertainty Solution to not only Think Better, but Live Better. Don't miss a single post! | 817 | 3,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.941219 |
https://curriculum.illustrativemathematics.org/MS/teachers/2/6/7/practice.html | 1,624,261,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00501.warc.gz | 180,423,601 | 16,806 | # Lesson 7
Reasoning about Solving Equations (Part 1)
### Problem 1
There is a proportional relationship between the volume of a sample of helium in liters and the mass of that sample in grams. If the mass of a sample is 5 grams, its volume is 28 liters. (5, 28) is shown on the graph below.
1. What is the constant of proportionality in this relationship?
2. In this situation, what is the meaning of the number you found in part a?
3. Add at least three more points to the graph above, and label with their coordinates.
4. Write an equation that shows the relationship between the mass of a sample of helium and its volume. Use $$m$$ for mass and $$v$$ for volume.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 11.)
### Problem 2
Explain how the parts of the balanced hanger compare to the parts of the equation.
$$7=2x+3$$
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
For the hanger below:
1. Write an equation to represent the hanger.
2. Draw more hangers to show each step you would take to find $$x$$. Explain your reasoning.
3. Write an equation to describe each hanger you drew. Describe how each equation matches its hanger.
### Solution
For access, consult one of our IM Certified Partners. | 314 | 1,295 | {"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} | 3.875 | 4 | CC-MAIN-2021-25 | longest | en | 0.909163 |
https://proofwiki.org/wiki/Definition:Compact_Linear_Operator | 1,716,580,967,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00737.warc.gz | 418,426,957 | 10,631 | Definition:Compact Linear Operator
Definition
Normed Vector Space
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $T : X \to X$ be a compact linear transformation.
We say that $T$ is a compact linear operator.
Inner Product Space
Let $\struct {X, \innerprod \cdot \cdot}$ be an inner product space.
Let $T : X \to X$ be a compact linear transformation.
We say that $T$ is a compact linear operator. | 110 | 420 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.647934 |
https://support.nag.com/numeric/nl/nagdoc_27cpp/clhtml/d05/d05bac.html | 1,702,138,273,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00637.warc.gz | 597,431,716 | 9,558 | # NAG CL Interfaced05bac (volterra2)
## 1Purpose
d05bac computes the solution of a nonlinear convolution Volterra integral equation of the second kind using a reducible linear multi-step method.
## 2Specification
#include
void d05bac (
double (*ck)(double t, Nag_Comm *comm),
double (*cg)(double s, double y, Nag_Comm *comm),
double (*cf)(double t, Nag_Comm *comm),
Nag_ODEMethod method, Integer iorder, double alim, double tlim, double tol, Integer nmesh, double thresh, double work[], Integer lwk, double yn[], double errest[], Nag_Comm *comm, NagError *fail)
The function may be called by the names: d05bac or nag_inteq_volterra2.
## 3Description
d05bac computes the numerical solution of the nonlinear convolution Volterra integral equation of the second kind
$yt=ft+∫atkt-sgs,ysds, a≤t≤T.$ (1)
It is assumed that the functions involved in (1) are sufficiently smooth. The function uses a reducible linear multi-step formula selected by you to generate a family of quadrature rules. The reducible formulae available in d05bac are the Adams–Moulton formulae of orders $3$ to $6$, and the backward differentiation formulae (BDF) of orders $2$ to $5$. For more information about the behaviour and the construction of these rules we refer to Lubich (1983) and Wolkenfelt (1982).
The algorithm is based on computing the solution in a step-by-step fashion on a mesh of equispaced points. The initial step size which is given by $\left(T-a\right)/N$, $N$ being the number of points at which the solution is sought, is halved and another approximation to the solution is computed. This extrapolation procedure is repeated until successive approximations satisfy a user-specified error requirement.
The above methods require some starting values. For the Adams' formula of order greater than $3$ and the BDF of order greater than $2$ we employ an explicit Dormand–Prince–Shampine Runge–Kutta method (see Shampine (1986)). The above scheme avoids the calculation of the kernel, $k\left(t\right)$, on the negative real line.
## 4References
Lubich Ch (1983) On the stability of linear multi-step methods for Volterra convolution equations IMA J. Numer. Anal. 3 439–465
Shampine L F (1986) Some practical Runge–Kutta formulas Math. Comput. 46(173) 135–150
Wolkenfelt P H M (1982) The construction of reducible quadrature rules for Volterra integral and integro-differential equations IMA J. Numer. Anal. 2 131–152
## 5Arguments
1: $\mathbf{ck}$function, supplied by the user External Function
ck must evaluate the kernel $k\left(t\right)$ of the integral equation (1).
The specification of ck is:
double ck (double t, Nag_Comm *comm)
1: $\mathbf{t}$double Input
On entry: $t$, the value of the independent variable.
2: $\mathbf{comm}$Nag_Comm *
Pointer to structure of type Nag_Comm; the following members are relevant to ck.
userdouble *
iuserInteger *
pPointer
The type Pointer will be void *. Before calling d05bac you may allocate memory and initialize these pointers with various quantities for use by ck when called from d05bac (see Section 3.1.1 in the Introduction to the NAG Library CL Interface).
Note: ck should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d05bac. If your code inadvertently does return any NaNs or infinities, d05bac is likely to produce unexpected results.
2: $\mathbf{cg}$function, supplied by the user External Function
cg must evaluate the function $g\left(s,y\left(s\right)\right)$ in (1).
The specification of cg is:
double cg (double s, double y, Nag_Comm *comm)
1: $\mathbf{s}$double Input
On entry: $s$, the value of the independent variable.
2: $\mathbf{y}$double Input
On entry: the value of the solution $y$ at the point s.
3: $\mathbf{comm}$Nag_Comm *
Pointer to structure of type Nag_Comm; the following members are relevant to cg.
userdouble *
iuserInteger *
pPointer
The type Pointer will be void *. Before calling d05bac you may allocate memory and initialize these pointers with various quantities for use by cg when called from d05bac (see Section 3.1.1 in the Introduction to the NAG Library CL Interface).
Note: cg should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d05bac. If your code inadvertently does return any NaNs or infinities, d05bac is likely to produce unexpected results.
3: $\mathbf{cf}$function, supplied by the user External Function
cf must evaluate the function $f\left(t\right)$ in (1).
The specification of cf is:
double cf (double t, Nag_Comm *comm)
1: $\mathbf{t}$double Input
On entry: $t$, the value of the independent variable.
2: $\mathbf{comm}$Nag_Comm *
Pointer to structure of type Nag_Comm; the following members are relevant to cf.
userdouble *
iuserInteger *
pPointer
The type Pointer will be void *. Before calling d05bac you may allocate memory and initialize these pointers with various quantities for use by cf when called from d05bac (see Section 3.1.1 in the Introduction to the NAG Library CL Interface).
Note: cf should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d05bac. If your code inadvertently does return any NaNs or infinities, d05bac is likely to produce unexpected results.
4: $\mathbf{method}$Nag_ODEMethod Input
On entry: the type of method which you wish to employ.
${\mathbf{method}}=\mathrm{Nag_Adams}$
${\mathbf{method}}=\mathrm{Nag_BDF}$
For backward differentiation formulae.
Constraint: ${\mathbf{method}}=\mathrm{Nag_Adams}$ or $\mathrm{Nag_BDF}$.
5: $\mathbf{iorder}$Integer Input
On entry: the order of the method to be used.
Constraints:
• if ${\mathbf{method}}=\mathrm{Nag_Adams}$, $3\le {\mathbf{iorder}}\le 6$;
• if ${\mathbf{method}}=\mathrm{Nag_BDF}$, $2\le {\mathbf{iorder}}\le 5$.
6: $\mathbf{alim}$double Input
On entry: $a$, the lower limit of the integration interval.
Constraint: ${\mathbf{alim}}\ge 0.0$.
7: $\mathbf{tlim}$double Input
On entry: the final point of the integration interval, $T$.
Constraint: ${\mathbf{tlim}}>{\mathbf{alim}}$.
8: $\mathbf{tol}$double Input
On entry: the relative accuracy required in the computed values of the solution.
Constraint: $\sqrt{\epsilon }\le {\mathbf{tol}}\le 1.0$, where $\epsilon$ is the machine precision.
9: $\mathbf{nmesh}$Integer Input
On entry: the number of equidistant points at which the solution is sought.
Constraints:
• if ${\mathbf{method}}=\mathrm{Nag_Adams}$, ${\mathbf{nmesh}}\ge {\mathbf{iorder}}-1$;
• if ${\mathbf{method}}=\mathrm{Nag_BDF}$, ${\mathbf{nmesh}}\ge {\mathbf{iorder}}$.
10: $\mathbf{thresh}$double Input
On entry: the threshold value for use in the evaluation of the estimated relative errors. For two successive meshes the following condition must hold at each point of the coarser mesh
$Y1-Y2 maxY1,Y2,thresh ≤tol,$
where ${Y}_{1}$ is the computed solution on the coarser mesh and ${Y}_{2}$ is the computed solution at the corresponding point in the finer mesh. If this condition is not satisfied then the step size is halved and the solution is recomputed.
Note: thresh can be used to effect a relative, absolute or mixed error test. If ${\mathbf{thresh}}=0.0$ then pure relative error is measured and, if the computed solution is small and ${\mathbf{thresh}}=1.0$, absolute error is measured.
11: $\mathbf{work}\left[{\mathbf{lwk}}\right]$double Output
12: $\mathbf{lwk}$Integer Input
On entry: the dimension of the array work.
Constraint: ${\mathbf{lwk}}\ge 10×{\mathbf{nmesh}}+6$.
Note: the above value of lwk is sufficient for d05bac to perform only one extrapolation on the initial mesh as defined by nmesh. In general much more workspace is required and in the case when a large step size is supplied (i.e., nmesh is small), you must provide a considerably larger workspace.
On exit: if ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NW_OUT_OF_WORKSPACE, ${\mathbf{work}}\left[0\right]$ contains the size of lwk required for the algorithm to proceed further.
13: $\mathbf{yn}\left[{\mathbf{nmesh}}\right]$double Output
On exit: ${\mathbf{yn}}\left[\mathit{i}-1\right]$ contains the most recent approximation of the true solution $y\left(t\right)$ at the specified point $t={\mathbf{alim}}+\mathit{i}×H$, for $\mathit{i}=1,2,\dots ,{\mathbf{nmesh}}$, where $H=\left({\mathbf{tlim}}-{\mathbf{alim}}\right)/{\mathbf{nmesh}}$.
14: $\mathbf{errest}\left[{\mathbf{nmesh}}\right]$double Output
On exit: ${\mathbf{errest}}\left[\mathit{i}-1\right]$ contains the most recent approximation of the relative error in the computed solution at the point $t={\mathbf{alim}}+\mathit{i}×H$, for $\mathit{i}=1,2,\dots ,{\mathbf{nmesh}}$, where $H=\left({\mathbf{tlim}}-{\mathbf{alim}}\right)/{\mathbf{nmesh}}$.
15: $\mathbf{comm}$Nag_Comm *
The NAG communication argument (see Section 3.1.1 in the Introduction to the NAG Library CL Interface).
16: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
## 6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_CONVERGENCE
The solution is not converging. See Section 9.
NE_ENUM_INT
On entry, ${\mathbf{method}}=\mathrm{Nag_Adams}$ and ${\mathbf{iorder}}=2$.
Constraint: if ${\mathbf{method}}=\mathrm{Nag_Adams}$, $3\le {\mathbf{iorder}}\le 6$.
On entry, ${\mathbf{method}}=\mathrm{Nag_BDF}$ and ${\mathbf{iorder}}=6$.
Constraint: if ${\mathbf{method}}=\mathrm{Nag_BDF}$, $2\le {\mathbf{iorder}}\le 5$.
NE_ENUM_INT_2
On entry, ${\mathbf{method}}=\mathrm{Nag_Adams}$, ${\mathbf{iorder}}=〈\mathit{\text{value}}〉$ and ${\mathbf{nmesh}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{method}}=\mathrm{Nag_Adams}$, ${\mathbf{nmesh}}\ge {\mathbf{iorder}}-1$.
On entry, ${\mathbf{method}}=\mathrm{Nag_BDF}$, ${\mathbf{iorder}}=〈\mathit{\text{value}}〉$ and ${\mathbf{nmesh}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{method}}=\mathrm{Nag_BDF}$, ${\mathbf{nmesh}}\ge {\mathbf{iorder}}$.
NE_INT
On entry, ${\mathbf{iorder}}=〈\mathit{\text{value}}〉$.
Constraint: $2\le {\mathbf{iorder}}\le 6$.
On entry, ${\mathbf{lwk}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{lwk}}\ge 10×{\mathbf{nmesh}}+6$; that is, $〈\mathit{\text{value}}〉$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_REAL
On entry, ${\mathbf{alim}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{alim}}\ge 0.0$.
On entry, ${\mathbf{tol}}=〈\mathit{\text{value}}〉$.
Constraint: .
NE_REAL_2
On entry, ${\mathbf{alim}}=〈\mathit{\text{value}}〉$ and ${\mathbf{tlim}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{tlim}}>{\mathbf{alim}}$.
NW_OUT_OF_WORKSPACE
The workspace which has been supplied is too small for the required accuracy. The number of extrapolations, so far, is $〈\mathit{\text{value}}〉$. If you require one more extrapolation extend the size of workspace to: ${\mathbf{lwk}}=〈\mathit{\text{value}}〉$.
## 7Accuracy
The accuracy depends on tol, the theoretical behaviour of the solution of the integral equation, the interval of integration and on the method being used. It can be controlled by varying tol and thresh; you are recommended to choose a smaller value for tol, the larger the value of iorder.
You are warned not to supply a very small tol, because the required accuracy may never be achieved. This will usually force an error exit with ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NW_OUT_OF_WORKSPACE.
In general, the higher the order of the method, the faster the required accuracy is achieved with less workspace. For non-stiff problems (see Section 9) you are recommended to use the Adams' method (${\mathbf{method}}=\mathrm{Nag_Adams}$) of order greater than $4$ (${\mathbf{iorder}}>4$).
## 8Parallelism and Performance
d05bac makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
When solving (1), the solution of a nonlinear equation of the form
$Yn-αgtn,Yn-Ψn=0,$ (2)
is required, where ${\Psi }_{n}$ and $\alpha$ are constants. d05bac calls c05avc to find an interval for the zero of this equation followed by c05azc to find its zero.
There is an initial phase of the algorithm where the solution is computed only for the first few points of the mesh. The exact number of these points depends on iorder and method. The step size is halved until the accuracy requirements are satisfied on these points and only then the solution on the whole mesh is computed. During this initial phase, if lwk is too small, d05bac will exit with ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NW_OUT_OF_WORKSPACE.
In the case ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_CONVERGENCE or NW_OUT_OF_WORKSPACE, you may be dealing with a ‘stiff’ equation; an equation where the Lipschitz constant $L$ of the function $g\left(t,y\right)$ in (1) with respect to its second argument is large, viz,
$gt,u-gt,v≤Lu-v.$ (3)
In this case, if a BDF method (${\mathbf{method}}=\mathrm{Nag_BDF}$) has been used, you are recommended to choose a smaller step size by increasing the value of nmesh, or provide a larger workspace. But, if an Adams' method (${\mathbf{method}}=\mathrm{Nag_Adams}$) has been selected, you are recommended to switch to a BDF method instead.
In the case ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NW_OUT_OF_WORKSPACE, then if ${\mathbf{fail}}\mathbf{.}\mathbf{errnum}=6$, the specified accuracy has not been attained but yn and errest contain the most recent approximation to the computed solution and the corresponding error estimate. In this case, the error message informs you of the number of extrapolations performed and the size of lwk required for the algorithm to proceed further. The latter quantity will also be available in ${\mathbf{work}}\left[0\right]$.
## 10Example
Consider the following integral equation
$yt=e-t+∫0te-t-sys+e-ysds, 0≤t≤20$ (4)
with the solution $y\left(t\right)=\mathrm{ln}\left(t+e\right)$. In this example, the Adams' method of order $6$ is used to solve this equation with ${\mathbf{tol}}=\text{1.e−4}$.
### 10.1Program Text
Program Text (d05bace.c)
None.
### 10.3Program Results
Program Results (d05bace.r) | 4,304 | 14,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 128, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.770181 |
https://www.aqua-calc.com/convert/area/acre-to-square-fathom | 1,503,208,364,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00330.warc.gz | 904,045,864 | 9,545 | # Area conversion table
## Convert acre to square fathom
### (ac to ftm²)
#### Foods, Nutrients and Calories
Pumpkin, flowers, cooked, boiled, drained, with salt weigh(s) 141.6 gram per (metric cup) or 4.73 ounce per (US cup), and contain(s) 15 calories per 100 grams or ≈3.527 ounces [ calories | weight to volume | volume to weight | price | density ]
#### Gravels and Substrates
CaribSea, Freshwater, Instant Aquarium, Kon Tiki density is equal to 1601.85 kg/m³ or 100 lb/ft³ with specific gravity of 1.60185 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
#### Materials and Substances
Fasertonerde (Al2O3) weigh(s) 3.95 gram per (cubic centimeter) or 2.283 ounce per (cubic inch) [ weight to volume | volume to weight | price | density ]
#### Weight/Volume at Temperature
Grapeseed oil weigh(s) 0.915 gram per (cubic centimeter) or 0.529 ounce per (cubic inch) at 27°C or 80.6°F [ weight to volume | volume to weight | price ]
#### What is microgram per metric tablespoon?
The microgram per metric tablespoon density measurement unit is used to measure volume in metric tablespoons in order to estimate weight or mass in micrograms
#### What is kinematic viscosity measurement?
The kinematic viscosity (ν) is the dynamic viscosity (μ) divided by the density of the fluid (ρ), i.e. ν = μρ, where:
μ = Pa ⋅ s = N × s ⁄ m² = kg ⁄ m ⁄ s.
ρ = kg ⁄ m³.
ν = kg ⁄ m ⁄ s ⁄ (kg ⁄ m³) = m² ⁄ s. | 451 | 1,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.730539 |
http://math.stackexchange.com/questions/85540/many-one-reductions-vs-turing-reductions-and-ph?answertab=votes | 1,462,467,017,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127870.1/warc/CC-MAIN-20160428161527-00078-ip-10-239-7-51.ec2.internal.warc.gz | 186,863,014 | 16,627 | # Many-One Reductions vs Turing-Reductions and PH
One definition of $\mathsf{PH}$ uses Oracles and in this definition both $\mathsf{NP}$ and $\mathsf{coNP}$ are contained in P^NP which equals $\mathsf{P^{coNP}}$. It is believed that $\mathsf{NP}$ does not equal $\mathsf{coNP}$, in other words they are not many-one reducible to each other. If indeed this is proved, then doesn't it also hold that $\mathsf{P^{NP}}$ doesn't equal $\mathsf{P^{coNP}}$ since many-one reducibility imply Turing reducibility?
-
## migrated from cstheory.stackexchange.comNov 25 '11 at 15:24
This question came from our site for theoretical computer scientists and researchers in related fields.
## 1 Answer
The fact that many-one reductions are weaker than Turing reductions only means that NP = coNP implies P^NP = P^coNP, but not the converse. To illustrate this, think about the following: many-one polytime reductions are weaker than many-one exponential time reductions. But even though P and EXP are not equal by the time hierarchy, they are exp-time reducible to each other.
-
Your answer is totally the opposite of the answer to this question... cstheory.stackexchange.com/questions/8759/… – Tayfun Pay Nov 25 '11 at 2:48
So if A is many-one reducible to B then it means A is Turing reducible to B. If A is not many-one reducible to B then it doesn't necessarily mean that A is not Turing reducible to B. OKAY!!!! Phew! – Tayfun Pay Nov 25 '11 at 3:01 | 383 | 1,445 | {"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} | 3.03125 | 3 | CC-MAIN-2016-18 | latest | en | 0.909067 |
http://www.playonline-roulette.com/page/72/ | 1,585,515,881,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00222.warc.gz | 292,753,783 | 8,253 | ## The Martingales Probability and Casino Strategy
There are a few articles about the Martingale System on this website, however here I wanted to cover a little bit more information on Martingales Probability and the flaws of using this method in an online casino.
But before we start a word on randomness, at the heart of the martingale strategy is a requirement for randomness, it is a system that will fail very badly in the light of any bias. So be sure your casino is fair and random and beware leaving your luck in the hands of a random number generator – if only for your own piece of mind.
The concept of the martingale strategy is quite straight forward, you play consistently on a single bet type – usually 50/50 one like odd/even or red/black. You are relying on the low probabilities of repetition and on a mathematical inequality. The amount – S is put on this single event, say on red being spun, if the bet does not win then the Stake (S) is doubled to S2 on the second spin. The third spin will require 4S (double the last bet) if it loses and so on, doubling the bet until the win.
When the winning bet does occur then the final amount will exceed the losses from all the previous games, a fact which is actually algebraically provable.
Here’s some sample odds from American Roulette (the odds for European are slightly different but not greatly)
Some Martingales Probability
For example, the probability of black occurring
• 2 times in a row – 22.43%
• 3 times in a row – 10.62%
• 4 times in a row – 5.03 %
• 5 times in a row – 2.38%
When we look at this system, the mathematical certainty of doubling the bet bringing a win eventually, and the low probability of consecutive repetition of the same color – it looks extremely tempting and some will say infallible.
There are problems though, as mentioned it suffers badly if there is any bias although it could be argued that any strategy would. The reliance on randomness though makes playing ordinary online casinos with their computer generated spins less than appealing. It should only be used in a real life casino or where you can play online roulette here – DublinBet which is a real online casino from dublin played in real time.
There are however some problems which actually exist outside the realms of probability and mathematical certainty of the Martingale system. The system actually carries a major risk of consuming all your cash before paying out, it can be surprising how many times you’ll see long consecutive streaks of the same color for instance. Be careful how you try this if you do play online roulette
Even with sufficient funds you could find yourself in the position of placing an enormous bet to make a very small profit. If your nerve fails you, the house blocks your bet or you do not continue doubling your stake for any reason – then you lose. These are very real reasons combined with the fact that wins are small but regular, that make the Martingale system great in the land of probability but more flawed in the real world.
## How do you Bet in Roulette
Although when you see someone throwing hundreds of chips on a roulette table, it can look a bit daunting. It’s actually not that difficult, here’s some of the main number bets you can make in roulette –
Single Number – Quite simply placing your chip on a single number, e.g 1, 5, 0 or 34 etc.
Number (Split)
– Putting your chip on a line adjoining two numbers, you are betting on either of those numbers coming up
Number(Street)
– Betting on three adjoining numbers – to place this bet put your chips on the left line of the first number of the series. For 16,17 and 18 you’d place your chip on the left line of the box around the number 16
Number (Corner)
– A bet on four numbers whose position on the table make a square. To make this bet you put your chips on the line in the centre of the square.
Number(Line)
– This is a bet on six numbers, made up of two rows of three. Here you place your chips on the line to the left of the first number in the series and between the two rows.
Number (Dozen)
– You have actually got some alternative to making this bet depending on which 12 numbers you want. You can either bet on the first, second or third twelve which has it’s own section. So the first group includes the numbers 1,2,3 -12, the next 13-24 and so on. Remember though none of these groups include the zero
Number (Column)
– There are also three distinct column bets, to place on these numbers you simply put your chips on the first column (e.g. 1,4,7,10,13,16,19,22,25,28,31 or 34), or the 2nd, 3rd column. The bet is placed using the square at the bottom of each column)
These bets are classified as either ‘inside bets’ or ‘outside bets’ depending on which section they are in on the roulette table. I’ll put all the odds up in my next post on roulette betting. But remember if you play online roulette, then al the decent casinos have free games where you can test these for yourself without risking any money. One word of warning though, if you seem to be exceptionally ‘lucky’ on any roulette free game, be very cautious about using their real money game they maybe manipulating the odds deliberately. Personally I’d never trust any roulette game that did this.
There are many other bets | 1,178 | 5,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-16 | latest | en | 0.933314 |
https://www.easyelimu.com/kenya-secondary-schools-pastpapers/mocks/2017/kitui-mocks/item/825-mathematics-paper-2?tmpl=component&print=1 | 1,568,674,824,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00024.warc.gz | 850,166,756 | 9,772 | ## MATHEMATICS PAPER 2 - 2017 KITUI MOCK EXAMINATION
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Click the link below to download the full 2017 KITUI MOCK EXAMINATIONS pdf document, with all the topics.
SECTION I (50 MARKS)
Answer ALL the questions from this section.
1. Use logarithms to evaluate, (4 marks)
1. Make P the subject of the formula (3 marks)
1. Find the circle centre and radius whose equation is
3x2 + 3y2 + 18x – 6y + 18 = 0 (3 marks)
1. The volumes of two similar cylindrical containers are 27cm3 and 64cm3 Given that the height of the smaller container is 12cm, find the height of the larger container. (2 marks)
2. 3cm3 of water is added to 2cm3 of a certain medicine which costs sh.12 per cm3. The chemist sells the diluted medicine at sh.6 per cm3. Calculate the percentage profit.
(3 marks)
3. Given that 4y = 3 sin 2/5 for 0 < θ< 360o. Determine
1. The amplitude of the curve (1 mark)
2. The period of the curve. (1 mark)
1. Find the length BC of the following triangle if AC = 3.7cm, AB = 4cm and angle ABC = 63o. (3 marks)
1. Solve for x in the equation
27x - 1 3x + 1 = 729 (3 marks)
1. In the figure below ABCD is a cyclic quadrilateral. Point O is the centre of the circle. <ABO = 30o and <BCD = 110o.
Calculate the size of angle ADB. (2 marks)
1. Three people Mutua, Wanza and Kiilu contributed money to start a business. Mutua contributed a quarter of the money and Wanza two fifths of the reminder. Kiilu’s contribution was one and a half times that of Mutua. They borrowed the rest of the money from a bank which was sh.60,000 less than Kiilu’s contribution. Find the total amount required to start the business. (4 marks)
1. Simplify (3 marks)
1. Expand (2-1/4x)5 and use the first three terms to find the value of 1.9755 to four significant figures. (4 marks)
1. The radius of a spherical ball is measured as 7cm correct to the nearest centimeter. Determine to 2 decimal places, the percentage error in calculating the surface area of the ball. (3 marks)
1. Given that tan θ= 1/√5 where θ is an acute angle, find without using tables or calculator sin (90−θ) leaving your answer in the simplified surd form. (4 marks)
1. Given that a = 1.2, b = 0.02 and c = 0.2, express ac b in the form where m and n are integers. (3 marks)
1. The diagram below shows sector AOB of a circle centre O. <AOB = 1.5C and arc AB is of length 12cm.
1. Determine the radius OA of the circle. (1 mark)
2. Calculate the area of the shaded region. Give your answer correct to 3 s.f. (3 marks)
SECTION II (50 MARKS)
Answer any FIVE questions from this section
1. The table below shows the taxation rates.
Income (£ per month) Rate (%) 0-382 10 383-754 15 755-1126 20 1127-1498 25 1499-1870 30 1871-2242 35 over 2242 40
Mueni is housed by her employer but pays a nominal rent of sh.1200 per month. She is entitled to a personal relief of sh.950 per month. If her monthly P.A.Y.E is sh.7024,
1. Calculate her gross income. (5 marks)
2. In addition to the tax the following monthly deductions are also made
Sacco shares Ksh. 1200
Coop loan Ksh.1500
Union dues Ksh.300
Calculate
1. Her monthly salary. (3 marks)
2. Net monthly salary (2 marks)
1. Use a ruler and compasses only for all construction in this question.
1. Construct a triangle ABC in which AB = 8cm, BC = 7.5cm and <ABC = 1121/20 . (3 marks)
2. Measure the length of AC. (1 mark)
3. By shading the unwanted region show the locus of P within the triangle ABC such that AP BP, AP > 3cm. Mark the required region as P. (3 marks)
4. Construct a normal from C to meet AB produced at D. (1 mark)
5. Locate the locus of R in the same diagram such that the arc of triangle ARB is the arc of the triangle ABC. (2 marks)
1. In the triangle PQR below, L and M are points on PQ and QR respectively such that PL : LQ = 1 : 3 and Qm : mR = 1 : 2. Pm and RL intersect at X. Given that PQ = b and PR= c
1. Express the following vectors in terms of b and c.
1. QR (1 mark)
2. Pm (1 mark)
3. RL (1 mark)
2. By taking PX = hPm and RX = kRL where h and k are constants. Find two expressions of PX in terms of h, k, b and c. Hence determine the values of the constants h and k. (6 marks)
3. Determine the ratio LX : XR. (1 mark)
1. OABC is a parallelogram with vertices O(0, 0), A(2, 0), B(3, 2) and C(1, 2). OIAIBICI is the image of OABC under a transformation matrix.
1.
1. Find the coordinates of OIAIBICI (2 marks)
2. On the graph provided, draw OABC and OIAIBICI (2 marks)
2.
1. Find OIIAIIBIICII, the image of OIAIBICI under the transformation matrix. (2 marks)
2. On the same grid, draw OIIAIIBIICII. (1 mark)
3. Find the single matrix that maps OIIAIIBIICII onto OABC. (3 marks)
1. An aircraft leaves town P (30oS, 17oE) and moves directly towards Q (60oN, 17oE). It then moved at an average speed of 300 knots for 8 hours Westwards to town R. Determine
1. The distance PQ in nautical miles. (2 marks)
2. The position of town R. (4 marks)
3. The local time at R if local time at Q is 3.12p.m (2 marks)
4. The total distance moved from P to R in kilometers. (Take 1nm = 1.853km) (2 marks)
1. The figure below is a sketch of a curve whose equation is y = x2 + x + 5. It cuts the line y = 11 at points P and Q.
1. Find the area bounded by the curve y = x2 + x + 5 and the line y = 11 using the trapezium rule with 5 strips. (5 marks)
2. Calculate the difference in the area if the mid-ordinate rule with 5 ordinates was used instead of the trapezium rule. (5 marks)
1. The figure below represents a rectangular based pyramid VABCD. AB = 12cm and AD = 16cm. Point O is vertically below V and VA = 26cm.
Calculate:
1. The height, VO, of the pyramid. (4 marks)
2. The angle between the edge VA and the plane ABCD. (3 marks)
3. The angle between the planes VAB and ABCD. (3 marks)
1. The distances S metres from a fixed point O, covered by a particle after t seconds is given by equation
S = t3 – 6t2 + 9t + 5
1. Calculate the gradient to the curve at t = 0.5 seconds. (3 marks)
2. Determine the values of S at the maximum and minimum turning points of the curve. (4 marks)
3. On the space provided, sketch the curve of S = t3 – 6t2 + 9t + 5. (3 marks) | 1,915 | 6,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-39 | longest | en | 0.804283 |
http://mathgpselaboration.blogspot.com/2011/01/mathematical-practice.html | 1,532,229,418,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00537.warc.gz | 234,263,088 | 10,727 | ## Saturday, January 8, 2011
### Mathematical Practice
Mathematical Practice
As I mentioned previously, the State of Georgia has adopted the mathematics standards developed by the Common Core State Standards Initiative. These Standards will become the new state standards starting in the school year 2012-13. So, in this blog, I will try to discuss the specific CCSS standards and compare/contrast with the current GPS.
In the GPS, there are two sets of standards: content standards and process standards. The process standards are the five standards that are discussed at the end of each grade and relate directly to the process standards discussed in the NCTM Standards - Problem Solving, Reasoning, Connection, Communication, and Representation. The CCSS mathematics standards, in contrast, include a set of standards on mathematical practice. According to the CCSS, mathematical practice is a variety of "expertise that mathematics educators at all levels should seek to develop in their students," and the eight expertise are:
1. make sense of problems and persevere in solving them
2. reason abstractly and quantitatively
3. construct viable arguments and critique reasoning of others
4. model with mathematics
5. use appropriate tools strategically
6. attend to precision
7. look for and make use of structures
8. look for and express regularity in repeated reasoning
Some of the items in this list sound very similar to the current GPS process standards while others appear to be new and different. For example, the idea of persevering to solve problems is not explicitly stated in the current GPS, but if students were to learn from problem solving, it is essential that students persevere. On the other hands, some of the current GPS process standards are much more obviously related to the eight expertise while others may appear to be forgotten. However, a more detailed look at the mathematical practice does suggest that even those standards are still important. For example, the connection standards seem to be absent from the list of mathematical practice. However, the description of "modeling with mathematics" include the following:
Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
These descriptions of "mathematically proficient" students clearly suggest students must be able to connect their understanding of mathematics to things both within and outside of mathematics, and both within and outside of classrooms.
One of the main concern as we move forward with the CCSS is that these standards on mathematical proficiency will receive less attention just as the process standards of the current GPS do. In some ways, it is understandable as it is rather difficult to imagine these mathematical practice standards in action. Moreover, it is not quite clear how these standards will be assessed. Thus, it is natural for some teachers to focus on things that will be assessed. The authors of the CCSS, however, offers a suggestion that can guide us as we grapple with the content standards:
Expectations that begin with the word “understand” are often especially good opportunities to connect the practices to the content. Students who lack understanding of a topic may rely on procedures too heavily. ... In short, a lack of understanding effectively prevents a student from engaging in the mathematical practices.
In this respect, those content standards which set an expectation of understanding are potential “points of intersection” between the Standards for Mathematical Content and the Standards for Mathematical Practice. These points of intersection are intended to be weighted toward central and generative concepts in the school mathematics curriculum that most merit the time, resources, innovative energies, and focus necessary to qualitatively improve the curriculum, instruction, assessment, professional development, and student achievement in mathematics.
As I continue discussing the specific standards, I will try to keep this suggestion in mind. I would also like to encourage you to keep thinking about the mathematical practice standards as we go through this time of transition. | 869 | 4,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.95052 |
https://www.dca-cc.com/dca/databroker-dao | 1,669,516,657,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00879.warc.gz | 776,831,347 | 46,722 | # Dollar-cost averaging (DCA) calculator forDaTa eXchange DTX (DTX)backtesting
Visualise and calculate historical returns of investing \$100 in DTX every 7 days from Aug 2022 until now
#1090 DTX Today Price
\$0.02
10.08%
24H Range
\$0.02
avg \$0.02
\$0.02
Value in FIAT
\$1,548.2
\$148.2
+10.59%
in 3 months
1. Investment
2. Earnings
DTX selling price
\$0.02
1st order \$0.02
Average price
Selling price
\$0.02
\$0.02
Total Investment
\$1,400
Over 14 instalments of \$100, every 7 days
Earnings over time
Estimate the development of your earnings over time
DTX price over time
Price development vs. average cost
#### Fact
Investing \$100 in DTX from Aug 2022 to Nov 2022 every 7 days (\$1,400 in total) would result in \$1,548.2 of value! Average price of \$0.02 per 1DTX. +10.59%!
Profit/Loss every 7 days
Based on your purchase interval you would make profit 64% of the time
What is DaTa eXchange DTX?
As a decentralised marketplace for IoT sensor data using Blockchain technology, Databroker DAO enables sensor owners to turn generated data into revenue streams. This will open up a wealth of opportunities for various industries. Data will be used and become more effective.
## What is a DCA-CC Calculator and How to Use it?
If you want to test out your investment strategy, you'll need to understand how it works and what you're hoping to achieve. This is where the DCA-CC calculator comes in - it can help you see if your strategy will generate the return you want.
The calculator is separated into two modes: the dollar cost average calculator and the lump sum investing calculator. You can use either one to budget for your investments on a regular basis, or to invest all your money at once.
To use the DCA-CC, start by entering a DCA or lump sum investment amount. Then, select the time period, interval, and investment you want to use. The calculator will show you how your strategy would perform under those conditions. You can also experiment with different parameters to see how they affect your results.
And that's not all! The DCA-CC also lets you see how your investment would fare if you used the lump sum strategy. So if you're not sure which approach is right for you, this calculator can help you compare and make the best decision.
### What is DCA (Dollar Cost Averaging)?
DCA is like buying a little bit of your favorite cryptocurrency each week or month regardless of the price. By buying equal dollar amounts at regular intervals, you're helping to smooth out the bumps of a volatile market.
Think of it as when buying a house. When you want to buy a house, you don't just fork over all the cash upfront. You make a down payment, and then you pay the mortgage every month. Over time, the house is yours.
DCA is like that, but with investments. You spread your investment out over time, so you're less likely to buy when the market is high. And just like with a house, you eventually own more and more of your investment.
By buying a little bit of your favorite cryptocurrency each day, week or month, you're making small, regular payments that will help you get the coin you want without waiting for a price dip.
Of course, there is always the risk that the price of the coin could continue to fall. However, this risk can be mitigated by using DCA when the market is trending upwards.
### How to use DCA-CC to backtest your dollar cost average strategy?
Do you want to know how effective your dollar cost averaging strategy would have been in the past? The DCA-CC calculator can tell you for sure!
This tool is designed to help you backtest your investment strategy, so you can compare it against other strategies and decide which one is best for you.
When you first use the tool, we'll make some assumptions about your potential investment. For example, we'll assume you're investing \$10 in bitcoin every week for the past three years.
Of course, you can change the parameters at any time to get more accurate results. So why wait? Use the DCA-CC calculator tool now and find out how your investment strategy would have fared in the past.
## DCA Widgets
### Top 3 cards
Value in FIAT, BTC selling price and Total investment cards are the easiest to understand. However, we'll give a little more explanation:
#### What is DCA Value in FIAT card?
The Value in FIAT card is a great way to see the value of your investments after a dollar cost averaging period. This card can help you understand how DCA affects the value of your investment over time.
The scale on the lower part of the widget displays the investment to interest ratio. In other words: it shows how much of your investment is lost or how much was added to your investment due to the earnings.
#### What does the DCA Value in FIAT card show?
This card lets you know how much your cryptocurrency is worth in Fiat currency at the end of your investment period. In other words, the price you sell it at.
The card also shows you the price of your first order, so you can see how the market volatility affects your investment over time.
Lastly, the card shows the ratio of the selling price to the average price. This is helpful in determining the value of your investment strategy and how it impacts the selling price.
#### What is a DCA Total Investment card?
The total investment card calculates how much money you would have invested, given an initial investment and an investment interval, over a specified period of time.
For example, if you invest \$100 every month for 3 years, the total investment card would show you how much money you would have invested at the end of those 3 years.
### Charts
We are presenting two charts here: a chart of earnings over time, and a chart of price over time. These charts can help provide context and perspective, and allow you to see what would be different if you entered or exited the market at a different time.
#### Earnings over time
This chart shows how much money you've made over time from your investments. It includes your balance in FIAT (the dollar equivalent of cryptocurrencies) as well as your total investment up to that day.
#### BTC price over time
This chart shows the price of a given cryptocurrency over time, as well as the average cost of a cryptocurrency on any given day.
This chart can help you understand the value of dollar cost averaging as a strategy, and how it may impact your earnings.
### What is a Fact card?
The Fact card is an automated message that summarises all the information from all the charts in a short, sharable sentence.
### What is Profit/Loss card?
The Profit/Loss card is a tool that can help you to better manage your risks by understanding how often you might be making a profit. As with everyhing else here, this card can provide guidance and clarity in your decision-making process.
### What is Purchase history?
What you'll find here is a table of purchase data, which includes information like how much cryptocurrency you could buy with the money invested on a given day, or how much you would have profited or lost at a given point in history.
### What is Lump Sum Investing strategy?
Lump sum investing strategy is a method of investing where you invest a fixed sum of money all at once. This is in contrast to dollar-cost averaging, where you spread your investment into several installments over a period of time.
Lump sum investing has its pros and cons. On the plus side, you only have to make the investment decision once. And, if you’re investing in a volatile asset like cryptocurrency, you may benefit from buying when prices are low and selling when prices are high. On the downside, you could end up buying at the top of a market bubble – and we all know how those end.
So, should you go with lump sum investing or dollar-cost averaging? That depends on your investment goals and your personal risk tolerance. If you’re the type of person who can stomach the ups and downs of the market, and you believe in the long-term potential of the asset you’re investing in, then lump sum investing may be the way to go.
## How to use DCA-CC to backtest your lump sum investment strategy?
DCA-CC is a powerful, easy to use backtesting tool that can be used to test and optimise your investment strategy.
The DCA-CC calculator will help you calculate the performance of your investment strategy across different market conditions.
You can also compare your performance against other strategies such as dollar cost averaging.
It's easy to use and only takes a few seconds to set up.
Here's how it works:
1. Enter the amount of money you want to invest.
2. Choose the cryptocurrency you want to invest in.
3. Choose a time range
4. Press the "Calculate" button.
The DCA-CC calculator will then show you how much money you would have made if you had invested that money in the cryptocurrency at that price.
## Lump Sum Widgets
Both lump sum investment and DCA widgets are very similar, yet the strategy we employ is different. Some data is presented differently, so we will mention only these differences here.
### Top 3 cards
As with the DCA strategy, we have three key cards here: Value in FIAT, BTC selling price and Total investment.
#### What is a Lump sum selling price card?
This card displays the value of the chosen cryptocurrency - the current selling price.
It also displays the ratio of the current selling price to the buying price. This information is helpful in showcase the value of your investment strategy and its impact on your selling price over time.
#### What is a Lump sum Total Investment card?
The amount on the total investment card will always be different from the investment parameter you entered into the calculator. This is because we calculate the total investment using the DCA strategy. Doing this allows us to show you the impact of investing the same amount of money using two different strategies.
As a reminder, in the DCA strategy, the total investment card takes your initial investment and your investment interval and multiplies it over the time period provided. | 2,201 | 10,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.886232 |
http://perplexus.info/show.php?pid=3648&cid=27862 | 1,591,453,376,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00423.warc.gz | 89,599,221 | 4,465 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
The omitted age (Posted on 2005-10-20)
When completed, the cross-number below will have one digit (from 0-9) in each cell, and no zeros in row (a) and in column (a).
``` (a) (b) (c)
+---+---+
(a) | | |
+---+---+---+
(b) | | | |
+---+---+---+
(c) | | |
+---+---+
ACROSS : (a) Abigail´s age.
(b) Sum of Abigail´s age,
Blanche´s age,
Cynthia´s age, and
Darlene´s age.
(c) Blanche´s age.
DOWN : (a) Darlene´s age
(b) Sum of three of the ages in (b) across.
(c) Cynthia´s age.```
Whose age was omitted from (b) down, and what are the ages of the 4 women ?
Note: this can be solved by hand. Those who will use the computer, give the others some time before posting the solution obtained this way. Tk you.
See The Solution Submitted by pcbouhid Rating: 3.0000 (3 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(8): Re: Solution by hand*****remark | Comment 11 of 21 |
(In reply to re(7): Re: Solution by hand by pcbouhid)
If you ever post hints- you should notify everybody by placing a word "hint" or "spoiler" in the subject-
I would prefer to solve a puzzle like this without any guidance
stay well
Posted by Ady TZIDON on 2005-10-22 04:00:18
Search: Search body:
Forums (5) | 412 | 1,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-24 | latest | en | 0.866838 |
http://practicalphysics.org/inelastic-collision-trolleys.html | 1,558,297,770,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00029.warc.gz | 152,501,293 | 6,182 | # Inelastic collision of trolleys
##### Class practical
Collisions when objects 'stick together'.
#### Apparatus and materials
Per student pair
Dynamics trolleys, 2
Runway, friction-compensated
Ticker-tape vibrator
Cork
Needle or large pin
#### Health & Safety and Technical notes
The runways are heavy and long. They need to be handled with care to avoid damage to students or nearby equipment. Runways are best lifted into position by 2 people. Place a barrier to ensure the trolleys do not roll off the end of the bench.
The method by which the cork and pins are attached to the trolley will depend on the particular make used. With most, it is easy to wedge a pin so that it sticks out from the trolley. It may be easier to fix a pin on both trolleys and stick a cork on one of them rather than to fix the cork directly on the trolley.
Another way to get the trolleys to stick together after collision is to fix a block of Plasticine onto one trolley with strong insulating tape at the edges. For the other trolley, another piece of Plasticine can be used or a series of drawing pins can be poked through a strip of insulating tape, before that too is attached to the trolley.
Yet another method is to attach double sided adhesive pads to the trolleys.
#### Procedure
a Fix a needle on the front of the moving trolley and a cork on the back of the second trolley, so that the trolleys stick together on collision. They will then move on as one unit after collision.
b Use a single ticker-tape attached to the back of the first trolley and pulled through a ticker-tape vibrator, to record the motion before and after collision.
c Make a moving trolley collide with an identical trolley at rest.
d Make a moving trolley collide with a trolley double the mass (one trolley stacked on top of another).
e Analyze the tapes to see if momentum is conserved.
#### Teaching notes
1 In these experiments the colliding bodies have constant speeds before collision and constant speeds (of different size) after collision. Make it clear beforehand that the students are to measure speeds, not accelerations, before and after impact. They then calculate momentum. A lot of kinetic energy will be lost in this inelastic collision, up to 50%, but hopefully it will demonstrate that momentum is conserved.
2 You could say:
'Elastic' has a special meaning in science. A spring is termed 'elastic' if, when stretched and released the spring goes back in exactly the same way and to exactly the same length as it was originally. It shows no sign of fatigue or of permanent stretch. Springs of good, hardened steel are elastic over a large range of stretches. Rubber cords are not perfectly elastic. They show a little fatigue and after-effects so it is unfortunate that we call them 'elastic'.
In a collision, the objects approach each other, losing kinetic energy, come to rest momentarily then move apart. If they end up with the same total kinetic energy as they started with, then we say that the collision has been 'perfectly elastic'.
3 The collision is inelastic if the total kinetic energy is much less after collision than before. Some of the original energy has been transferred elsewhere; often to warm up the colliders. If two lumps of sticky clay are thrown at each other so that they stick together then the collision is completely inelastic. If the two masses were equal and the velocities were equal, the combined lump would be at rest. All the kinetic energy will have been transferred, probably to thermal energy. However, momentum is a vector quantity (energy is not) and the two equal and opposite lots of momentum before the collision add up to zero before and after the collision. Conservation of momentum always holds. | 807 | 3,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-22 | latest | en | 0.912717 |
https://a4accounting.com.au/handling-plurals-with-the-if-function/ | 1,702,219,033,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00724.warc.gz | 104,878,088 | 18,227 | # Handling plurals with the IF function
### Singular or plural?
Sometimes when creating text you need to handle plurals correctly. The IF function makes it easy.
When creating flexible text you can use things like “there were 2 car(s)” which handles the singular and plural together.
If you need to be a bit more accurate then you can use the IF function.
In the image below there are formulas in columns B and C that determine whether to use the singular or the plural.
The formula in cell B2 is
`="Car"&IF(A2<>1,"s","")`
This formula starts with the singular word and then the IF function is used to determine if we need to add the letter s or not.
A2<>1 means A2 is not equal to 1. In general when creating an IF function you try to have the most likely situation as the TRUE argument. In this case it is most likely that the number won’t be 1, so we handle that first. If the number isn’t 1 we add an s. If the number is 1 then we add nothing. To add nothing you use “” which is a blank in Excel.
When the plural is not just adding an s on the end you need to use the part of the word that doesn’t change. in the case of cell C2 the formula is
`="Compan"&IF(A2<>1,"ies","y")`
This IF functions add the necessary ending text to get the correct singular and plural.
These formulas are a bit shorter than using
`=IF(A2<>1,"Companies","Company")`
This IF function includes the whole words, the result is the same but the formula is slightly longer. You would use this for words like goose and geese.
Please note: I reserve the right to delete comments that are offensive or off-topic.
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## 2 thoughts on “Handling plurals with the IF function”
1. David says:
Thanks. Exactly what I needed. | 423 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | longest | en | 0.900152 |
https://puzzling.stackexchange.com/questions/7934/determine-the-next-image-in-the-sequence/56641 | 1,619,184,290,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00456.warc.gz | 570,454,163 | 39,080 | # Determine the next image in the sequence
This is a standardized test practice question asking for the next image in the sequence. Can anyone identify the pattern?
Here are two different methods that give the same solution which is:
Based on sequential logic:
From the gray diamonds, the fifth image could also be a combination of image one and three (as the fourth image could be a combination of image one and two).
The sequence for the arrow is up, down, left, right. If repeating, next should be up.
The color of the dots could be 0=white, 1=gray, 2=black (a base 3 system). So adding the dots for image one and two gives white-gray-black-gray which is the pattern for image 4 (starting at top left and going clockwise). It follows that the dots for image 5 should be white-gray-black-white.
The keyhole at the edge has a sequence of left, top, right, top. If repeating, the next should be left.
The square around the dot has a sequence of none, top left, none, bottom left. If repeating, the next should be none.
Putting this together fits the pattern for answer a.
Based on incorrect answers being formulated in regions surrounding the correct answer:
The most common elements in the five answers are arrow=up, dot colors=white-gray-black-white, keyhole=left, square around dot = none.
These most common elements coincide with answer a.
Option A
I've chosen this option because
I've noticed that when there are two grey circles in the big square, then the grey circle opposite to the direction that the arrow points to has a small square around it.
Also
I am assuming that there will not be more than one black circle in the big square as there is just one in each image.
Moreover,
The arrows in each image point at directions up,down,left and right and the "door" at left,up,right,up. So, the next arrow will probably point up and the "door" at left.
So, by elimination, I got the answer.
By eliminating,I think the answer is
A
Clue 1
The four dots in upper big square,must be 2 white,1 grey,1 black.The 2nd and 4th are not,but one grey dot is in a square.So I guess square turns a white dot to grey(or turns grey to black).
Looking at the choices,with the same rule,C and D fails,they have 2 grey dots without square surrounding,E fails too,it has square,but imagine it without square,it includes 2 grey dots too. The rest are A or B
Except for clue 1,I don't really have confidence to know which and why that is the answer,but here's what information I get at least.
Rest of Clues
Clue 2. As clue 1 said, the answer is either A or B,they have two same dots,white and black down there.So we have these two dots to find the pattern.
Clue 3. the three diamond below should be independent,they affects how the pattern above to be formed.That's why it gives us the 3 diamonds for the 5th one.
Clue 4. In clock wise,the dot next to the arrow is usually grey,except for the 4th one.Considering this,A would be the answer.
Clue 5. If A is the answer,comparing it with the 1st,they are similar,share the same arrow and mini-square.But here's what stops me,I can't go further.
I know it is not the solution you are looking for, but with finding similarities in options and elimination I choose...
C
Explanation:
4 of options has arrow up. b is eliminated
None of four leads (upper images) has two black circles. e is eliminated
Locations of gray, white and black circle in c and d is exactly the same. a is eliminated
Finally, between c and d, I choose c because it is more similar to the last eliminated option (a) than d.
I also think it's A. I simply realised that the big squares were only present in alternate boxes i.e, there was no big square on the first box , it was present in the second box and so on.Thus I cancelled out B and E since the answer should not have the big square.Then I saw that there was two white circles in the first box,only one in the second box , two again in the third box and so on. With this I managed to cancel out C and D as the answer required two white circles. This left me with the answer A. | 957 | 4,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.945352 |
http://convertit.com/Go/SalvageSale/Measurement/Converter.ASP?From=Greek+fathom&To=length | 1,657,080,029,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00082.warc.gz | 10,588,769 | 3,440 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Greek fathom = 1.851152 length (length) ``` Related Measurements: Try converting from "Greek fathom" to archin (Russian archin), arpentlin, chain (surveyors chain), cloth quarter, fathom, French, gradus (Roman gradus), Greek palm, Israeli cubit, league, light yr (light year), m (meter), marathon, mil, parsec, Roman mile, stadium (Roman stadium), survey foot, UK mile (British mile), verst (Russian verst), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: Greek fathom = 18,511,520,000 angstrom, 7,288 caliber (gun barrel caliber), 99.84 digitus (Roman digitus), 4.82E-09 earth to moon (mean distance earth to moon), 1.01 fathom, 4 Greek cubit, 18.22 hand, 3.34 Israeli cubit, .87386091 ken (Japanese ken), .00287155 li (Chinese li), 9.2 link (surveyors link), .00004387 marathon, 2.43 pace, .00047141 ri (Japanese ri), 6.25 Roman foot, .00125094 Roman mile, .02024444 soccer field, .010022 stadium (Roman stadium), 2.21 vara (Mexican vara), .00173524 verst (Russian verst).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 458 | 1,674 | {"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} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.694701 |
https://whatswho.com/nature-of-rectifier-output-ripple-factor-comparison-of-rectifiers/ | 1,611,249,878,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00439.warc.gz | 629,551,000 | 25,868 | ## Nature of Rectifier Output
Nature of Rectifier:- It has already been discussed that the output of a rectifier is pulsating d.c. as shown in Fig. 6.37. In fact, if such a waveform is carefully analyzed, it will be found that it contains a d.c. component and an a.c. component. The a.c. the component is responsible for the *pulsations in the wave. The reader may wonder how a pulsating d.c. voltage can have an a.c. component when the voltage never becomes negative. The answer is that any wave which varies in a regular manner has an a.c. component.
The fact that a pulsating d.c. contains both d.c. and a.c. components can be beautifully illustrated by referring to Fig. 6.38. Fig. 6.38(i) shows a pure d.c. component, whereas Fig. 6.38 (ii) shows the *a.c. component. If these two waves are added together, the resulting wave will be as shown in Fig. 6.38 (iii). It is clear that the wave is shown in Fig. 6.38 (iii) never becomes negative, although it contains both a.c. and d.c. components. The striking resemblance between the rectifier output wave shown in Fig. 6.37 and the wave is shown in Fig. 6.38 (iii) may be noted. It follows, therefore, that a pulsating output of a rectifier contains a d.c. component and an a.c. component.
## What isRipple Factor
The output of a rectifier consists of a d.c. component and an a.c. component (also known as ripple). The a.c. component is undesirable and accounts for the pulsations in the rectifier output. The effectiveness of a rectifier depends upon the magnitude of a.c. component in the output; the smaller this component, the more effective is the rectifier.
The ratio of r.m.s. value of a.c. component to the d.c. component in the rectifier output is known
as ripple factor i.e.
Therefore, the ripple factor is very important in deciding the effectiveness of a rectifier. The smaller the ripple factor, the lesser the effective a.c. component and hence more effective is the rectifier. Mathematical analysis. The output current of a rectifier contains d.c. as well as a.c. component. The undesired a.c. the component has a frequency of 100 Hz (i.e. double the supply frequency 50 Hz) and is called the ripple (See Fig. 6.39). It is a fluctuation superimposed on the d.c. component.
By definition, the effective (i.e. r.m.s.) value of total load current is given by :
It is clear that a.c. component exceeds the d.c. component in the output of a half-wave rectifier.
This results in greater pulsations in the output. Therefore, half-wave rectifier is ineffective for conversion of a.c. into d.c.
### For full-wave rectification
In full-wave rectification,
This shows that in the output of a full-wave rectifier, the d.c. component is more than the a.c.
component. Consequently, the pulsations in the output will be less than in half-wave rectifier. For
this reason, full-wave rectification is invariably used for conversion of a.c. into d.c.
Example 6.22. A power supply A delivers 10 V dc with a ripple of 0.5 V r.m.s. while the power
supply B delivers 25 V dc with a ripple of 1 mV r.m.s. Which is a better power supply?
Solution. The lower the ripple factor of a power supply, the better it is.
## Comparison of Rectifiers
A comparison among the three rectifier circuits must be made very judiciously. Although bridge
circuit has some disadvantages
, it is the best circuit from the viewpoint of overall performance. When the cost of the transformer is the main consideration in a rectifier assembly, we invariably use the bridge circuit. This is particularly true for large rectifiers which have a low-voltage and a high-current rating.
Gas Filled Tubes | Conduction in a Gas Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
Half Wave rectifier | Properties | Frequency| Ripples Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
Diode | Types | Properties | Applications Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
Chapter Review Topics |Problems |Discussion Questions Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
MCQ’s| Electrons|Atomic | Voltage |Thevenin’s Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
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1. CLICK HERE TO READ RELATED ARTICLES Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
2). CLICK HERE TO READ RELATED ARTICLES Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
3) CLICK HERE TO READ RELATED ARTICLES Nature of Rectifier Output | Ripple Factor | Comparison of Rectifiers
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https://cracku.in/3-in-the-following-question-select-the-related-numbe-x-ssc-cgl-16-aug-shift-3 | 1,660,730,607,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00021.warc.gz | 211,250,737 | 20,864 | Question 3
In the following question, select the related number from the given alternatives.7 : 56 : : 11 : ?
Solution
Expression : 7 : 56 : : 11 : ?
The pattern followed = $$n:n^2+n$$
Eg = $$7:7^2+7=7:56$$
Similarly, $$11^2+11=121+11=132$$
=> Ans - (C) | 98 | 262 | {"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} | 3.515625 | 4 | CC-MAIN-2022-33 | latest | en | 0.430104 |
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# In the adjoining figure D is a point on BC such that, ∠ABD = ∠CAD. If AB = 5cm, AD = 4cm and AC = 3cm. Find A (∆ACD) : A (∆BCA).
Last updated date: 18th Jul 2024
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Hint: Visualizing the given data in the question draws an appropriate figure. According to the properties of triangles, observe similarities between the two triangles in order to establish a relation between their angles or sides.
Given data,
∠ABD = ∠CAD. AB = 5cm, AD = 4cm and AC = 3cm.
Let us compare ∆ABC and ∆CAD
⟹ $\dfrac{{{\text{A}}\left( {\Delta {\text{ACD}}} \right)}}{{{\text{A}}\left( {\Delta {\text{BCA}}} \right)}} = {\text{ }}\dfrac{{{\text{A}}{{\text{D}}^2}}}{{{\text{A}}{{\text{B}}^2}}}$
⟹ $\dfrac{{{\text{A}}\left( {\Delta {\text{ACD}}} \right)}}{{{\text{A}}\left( {\Delta {\text{BCA}}} \right)}} = {\text{ }}\dfrac{{{4^2}}}{{{5^2}}}$
⟹ $\dfrac{{{\text{A}}\left( {\Delta {\text{ACD}}} \right)}}{{{\text{A}}\left( {\Delta {\text{BCA}}} \right)}} = {\text{ }}\dfrac{{16}}{{25}}$ | 399 | 1,096 | {"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} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.706205 |
https://www.teachoo.com/4046/1163/Misc-5---Show-f(x)--x3-is-injective---Chapter-1-Class-12-CBSE/category/To-prove-injective--surjective--bijective-(one-one---onto)/ | 1,695,707,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510149.21/warc/CC-MAIN-20230926043538-20230926073538-00592.warc.gz | 1,122,371,552 | 29,922 | To prove one-one & onto (injective, surjective, bijective)
Chapter 1 Class 12 Relation and Functions
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Misc 2 Show that the function f: R → R given by f(x) = x3 is injective. f(x) = x3 We need to check injective (one-one) f (x1) = (x1)3 f (x2) = (x2)3 Putting f (x1) = f (x2) (x1)3 = (x2)3 x1 = x2 Since if f (x1) = f (x2) , then x1 = x2 ∴ It is one-one (injective) Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 | 215 | 580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | longest | en | 0.885837 |
https://www.compadre.org/Physlets/electromagnetism/illustration24_1.cfm | 1,716,721,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00482.warc.gz | 638,753,723 | 5,897 | ## Illustration 24.1: Flux and Gaussian Surfaces
The bar graph shows the flux, Φ, through four Gaussian surfaces: green, red, orange and blue (position is given in meters, electric field strength is given in newtons/coulomb, and flux is given in N·m2/C). Restart. Note that this animation shows only two dimensions of a three-dimensional world. You will need to imagine that the circles you see are spheres and that the squares you see are actually boxes. Flux, Φ, is a measure of the amount of electric field through a surface. Gauss's law relates the flux to the charge enclosed (qenclosed) in a Gaussian surface:
Φ = qenclosed0 and Φ = ∫ E · dA = ∫ E cosθ dA
where ε0 is the permittivity of free space (8.85 x 10-12 C2/N·m2), E is the electric field, dA is the unit area normal to the surface, and θ is the angle between the electric field vector and the surface normal.
Begin by moving the green Gaussian surface around. What is the flux when the surface encloses the point charge? What is the flux when the point charge is not inside the surface? What about the red surface? Since the flux is the electric field times the surface area, why doesn't the size of the surface matter? As long as the point charge is enclosed, the flux is the same and is equal to qenclosed0. When the charge is not enclosed, the flux is always zero. Notice that both the green and red Gaussian surfaces can be moved to either enclose or not enclose the charge. Therefore, the two fluxes should, and do, agree. However, only when these surfaces are centered on the charge can you use them to determine the electric field.
The orange surface has a different symmetry from the point charge (and its electric field). With the orange surface, why doesn't the shape matter in finding the flux? Again, what matters is whether the charge is enclosed or not. Move the surface to a point where the flux is zero. Is the electric field zero at the surface of the box? If the electric field is not zero, why is the flux zero? If you think about flux as a flow of electric field through an area (a bit like fluid flow), which was the early analogy for electric field and flux, then when there is no charge inside, the electric field that comes into the box must also leave. There is no source of electric field inside the box. However, the cubical box no longer has the same symmetry (a spherical symmetry) of the point charge. While the flux is zero for these scenarios, the value of the flux cannot be used to determine the electric field. The integral ∫ E cosθ dA is not equal to the integral E ∫ cosθ dA because E is not uniform across the Gaussian surface.
Finally, try two charges using the blue surface. What happens when the blue surface encloses just one charge? What happens when it encloses both charges?
Illustration authored by Anne J. Cox.
Script authored by Mario Belloni.
Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College. | 682 | 3,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-22 | latest | en | 0.923733 |
http://www.r-bloggers.com/principal-component-analysis-pca-vs-ordinary-least-squares-ols-a-visual-explanation/ | 1,469,653,622,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827079.61/warc/CC-MAIN-20160723071027-00133-ip-10-185-27-174.ec2.internal.warc.gz | 660,573,942 | 16,940 | # Principal Component Analysis (PCA) vs Ordinary Least Squares (OLS): A Visual Explanation
September 16, 2010
By
(This article was first published on Cerebral Mastication » R, and kindly contributed to R-bloggers)
Over at stats.stackexchange.com recently, a really interesting question was raised about principal component analysis (PCA). The gist was “Thanks to my college class I can do the math, but what does it MEAN?”
I felt like this a number of times in my life. Many of my classes were focused on the technical implementations they kinda missed the section titled “Why I give a shit.” A perfect example was my Mathematics Principles of Economics class which taught me how to manually calculate a bordered Hessian but, for the life of me, I have no idea why I would ever want to calculate such a monster. OK, that’s a lie. Later in life I learned that bordered Hessian matrices are a second derivative test used in some optimizations. Not that I would EVER do that shit by hand. I’d use some R package and blindly trust that it was coded properly.
So back to PCA: as I was reading the aforementioned stats question I was reminded of a recent presentation that Paul Teetor gave at a August Chicago R User Group. In his presentation on spread trading with R he showed a graphic that illustrated the difference between OLS and PCA. I took some notes and went home and made sure I could recreate the same thing. If you have wondered what makes OLS and PCA different, open up an R session and play along.
The first observation to make is that regressing x ~ y is not the same as y ~ x even in a simple univariate regression. You can illustrate this by doing the following:
set.seed(2)
x <- 1:100
y <- 20 + 3 * x
e <- rnorm(100, 0, 60)
y <- 20 + 3 * x + e
plot(x,y)
yx.lm <- lm(y ~ x)
lines(x, predict(yx.lm), col=”red”)
xy.lm <- lm(x ~ y)
lines(predict(xy.lm), y, col=”blue”)
You should get something that looks like this:
So it’s obvious they give different lines. But why? Well, OLS minimizes the error between the dependent and the model. Two of these errors are illustrated for the y ~ x case in the following picture:
But when we flip the model around and regress x ~ y then OLS minimizes these errors:
Well let’s draw the first principal component the old school way:
#normalize means and cbind together
xyNorm <- cbind(x=x-mean(x), y=y-mean(y))
plot(xyNorm)
#covariance
xyCov <- cov(xyNorm)
eigenValues <- eigen(xyCov)\$values
eigenVectors <- eigen(xyCov)\$vectors
plot(xyNorm, ylim=c(-200,200), xlim=c(-200,200))
lines(xyNorm[x], eigenVectors[2,1]/eigenVectors[1,1] * xyNorm[x])
lines(xyNorm[x], eigenVectors[2,2]/eigenVectors[1,2] * xyNorm[x])
# the largest eigenValue is the first one
# so that’s our principal component.
# but the principal component is in normalized terms (mean=0)
# and we want it back in real terms like our starting data
# so let’s denormalize it
plot(xy)
lines(x, (eigenVectors[2,1]/eigenVectors[1,1] * xyNorm[x]) + mean(y))
# that looks right. line through the middle as expected
# what if we bring back our other two regressions?
lines(x, predict(yx.lm), col=”red”)
lines(predict(xy.lm), y, col=”blue”)
PCA minimizes the error orthogonal (perpendicular) to the model line. So first principal component looks like this:
The two yellow lines, as in the previous images, examples of two of the errors which the routine minimizes.
So if we plot all three lines on the same scatter plot we can see the differences:
The x ~ y OLS and the first principal component are pretty close, but click on the image to get a better view and you will see they are not exactly the same.
All the code from the above examples can be found in a gist over at GitHub.com. It’s best to copy and past from the github as sometimes WordPress molests my quotes and breaks the codez.
The best introduction to PCA which I have read is the one I link to on Stats.StackExchange.com. It’s titled “A Tutorial on Principal Components Analysis” by Lindsay I Smith.
R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more... | 1,068 | 4,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.962396 |
https://www.thehindubusinessline.com/portfolio/commodity-analysis/how-to-choose-between-a-long-call-and-a-short-put/article34800314.ece | 1,628,082,125,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154805.72/warc/CC-MAIN-20210804111738-20210804141738-00595.warc.gz | 1,015,194,388 | 53,975 | How to choose between a long call and a short put
| Updated on June 12, 2021
It is important to estimate the speed at which the underlying is likely to reach its target
In the past two weeks, we discussed the risk and the returns associated with short call and short put positions. The question is: How do you decide whether to set up a long or a short position? This week, we discuss issues relating to choosing between a long call and a short put.
Long call vs Short put
Suppose you have a view that the underlying is likely to move from 15730 to 15820. Further suppose you have to choose between a long position on the 15700 call or a short position on the 15700 put. Note that even though the 15700 call carries intrinsic value, we will consider that as an at-the-money option (ATM) in line with our rule that an ATM option is the one that is closest to the current spot price (ignoring the 50s strike for the Nifty Index). By the same argument, the 15700 put is also ATM. The next-week expiry 15700 call trades for 152 points and the same expiry 15700 put for 118 points.
Now, suppose the underlying value is 15820 at option expiry. The 15700 call would be worth 120 points, suffering losses despite a 90-point increase in the underlying. On the other hand, the 15700 put would have expired worthless, thereby generating 118-point profit. Even if the underlying trades at 15820 earlier in the expiry week, the short put would have provided higher gains than the long call. Why?
The reason has to do with time decay. Note that time decay lowers gains or increases losses on long positions but contributes to gains on the short positions. The 15700 call has 122 points of time value (option price less intrinsic value). The 90-point gain in the underlying was not enough to counter the erosion in time value because of passage of time.
So, does this mean shorting puts is typically profitable? The answer depends on several factors. One, the estimated upside price movement in underlying. Two, the premium on the short put and the time value of the long call. And three, when do you expect the underlying to reach its target price. The longer the time for the underlying to reach its price target, higher the likelihood that the short put will be more profitable than the long call.
Suppose the underlying moves from 15730 to 15818 (strike plus put premium) the day after you buy the 15700 call. The long call will generate higher profits. The reason is because the increase in intrinsic value and time value capture on the long call is greater than the gains from time decay on the short put. However, if the underlying were to reach 15818 a day later, the short put will be profitable. The pattern is similar if the underlying trades at 15852 (15700 plus call premium).
The above discussion brings to the fore a key issue with options trading- estimating the speed at which the underlying is likely to reach its target. Technical analysis can help you fix a price target. But you may be unable to determine when the target will be reached. You can improve the chances of winning by having a view on implied volatility. The point is that deciding between a long call and a short put may not be easy.
You should be mindful of the following before shorting a put. First, time decay accelerates as the option nears expiry. So, shorting options during expiry week is better, especially if the ATM call and put premiums are similar and you expect the underlying to move sideways or move marginally higher. Second, a short put is an obligation to buy the underlying. You will have to, therefore, maintain margins on your short position. Finally, shorting options exposes you to large infrequent losses, which can wipe out your frequent small gains from such short positions.
The author offers training programmes for individuals to manage their personal investments
Published on June 12, 2021 | 840 | 3,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-31 | latest | en | 0.951021 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-1-foundations-for-algebra-cumulative-test-prep-multiple-choice-page-74/2 | 1,582,094,080,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00154.warc.gz | 742,975,805 | 13,811 | ## Algebra 1
Substitute in the correct value for d and evaluate. $7d+7=14$ $7(1)+7\overset?=14$ $7+7\overset?=14$ $14=14\checkmark$ | 55 | 132 | {"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} | 3.921875 | 4 | CC-MAIN-2020-10 | latest | en | 0.271909 |
https://www.lido.app/calculations/cost-of-goods-sold-postgresql | 1,679,693,806,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00648.warc.gz | 989,434,475 | 8,985 | # How to Calculate Cost of Goods Sold in PostgreSQL
Learn how to calculate Cost of Goods Sold (COGS) with your inventory and purchases. Discover industry averages and related platforms like Stripe and Shopify.
Optimizing the Cost Of Goods (COGS) sold is one of the best ways to determine the total cost of your production that will be useful in calculating your gross profit. By definition, the Cost of Goods Sold is the direct cost of producing the goods sold by a company (thanks, Investopedia!). While the metric seems simple on the surface, it gives valuable insights into areas of future growth and improvement. We'll take you through how to calculate this metric, how to understand it in the context of your industry, and how to apply it to the platforms you're using now.
## How to Calculate the Cost of Goods Sold
You can calculate COGS by adding the beginning inventory and purchases made during the period, then subtracting the ending inventory from it. COGS can be assessed for various time-periods (i.e. day, week, month, year) and, if your purchases, beginning inventory, and ending inventory are tied to certain products or campaigns, you can assign a specific COGS for each.
If you’re confused about how to define the beginning inventory, purchases, and ending inventory, we’ve included some nifty definitions and examples below:
• Beginning inventory is the leftover inventory from the last year or month that was not sold. This includes the products or merchandise that were not sold in the previous month or year.
• Purchase during the period is the additional purchases or productions that the manufacturers or company has made. This includes the raw materials and cost of labor for the additional products made.
• Ending inventory is the number of products that are not sold at the end of the month or year. This will be used as the beginning inventory for your next month/year’s COGS calculation.
You may be wondering, how do I know which value to attribute to each unit of inventory if I bought materials at different prices? There are three methods to answer this question: Weighted Average Cost (WAC); First in, First out (FIFO); and Last in, First out (LIFO). Each presents a general rule to make calculating COGS consistent over time and across businesses. We recommend reading about each method’s pros and cons and deciding which works best for your business.
When choosing how to calculate COGS, remember: consistency is key! That means, if you decide to use LIFO for the month, you should do that for the next month’s COGS as well. Having consistent metrics will allow you to better identify the causes of good or bad performance.
## How to Calculate Metric in PostgreSQL
It can be difficult to calculate this metric in PostgreSQL, but we have a quick and easy solution, Lido.app. Read below to learn more and get started today.
## What is Lido?
Lido allows you to connect, analyze, and visualize all of your data in a single spreadsheet. Don't wait for engineers to build analysis dashboards! Lido provides a simple and easy solution to importing data from numerous platforms. Automatically import data from your favorite platforms such as Shopify, Facebook, Google Analytics, and many more and apply Lido's software to extract meaningful metrics from them. After applying Lido software to your data, you will be left with sleek, attractive dashboards to share with your stakeholders, rather than confusing and jumbled raw data. Furthermore, the dashboards are easily editable to focus on specific data or metrics. | 718 | 3,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-14 | latest | en | 0.941153 |
http://intermath.coe.uga.edu/topics/nmcncept/ratios/a04.htm | 1,505,872,755,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686117.24/warc/CC-MAIN-20170920014637-20170920034637-00077.warc.gz | 165,167,467 | 2,845 | Home | Number Concept | Ratio | Additional Investigations | Trigonometry Ratios
Trigonometry Ratios
Sine, cosine, and tangent are functions which compare an angle in a right triangle to the ratio of two of the sides. On a calculator they are marked sin, cos, and tan, respectively.
Using GSP, create and measure several right triangles (angles and sides). Find the sin(<CAB), cos(<CAB), and tan(<CAB), and determine which side ratios they are each equal to.
Repeat the process with <ACB and explain a general rule for determining the sine, cosine, and tangent of an angle in a right triangle.
Extensions
The slope of line AC is equivalent to which of these functions?
Submit your idea for an investigation to InterMath. | 162 | 727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-39 | latest | en | 0.873819 |
https://socratic.org/questions/how-do-you-find-the-limit-of-x-lnx-as-x-approaches-infinity-1 | 1,723,282,546,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00395.warc.gz | 425,247,090 | 5,906 | # How do you find the Limit of (x/lnx) as x approaches infinity?
$\infty$
${\lim}_{x \setminus \to \setminus \infty} \frac{x}{\ln} x$
this is in indeterminate form ie $\setminus \frac{\infty}{\infty}$ so you can use L'Hopitals Rule
$\setminus \implies {\lim}_{x \setminus \to \setminus \infty} \frac{x}{\ln} x = {\lim}_{x \setminus \to \setminus \infty} \frac{1}{\frac{1}{x}} = {\lim}_{x \setminus \to \setminus \infty} x = \setminus \infty$ | 165 | 442 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.337734 |
https://freeeducationweb.com/physics-of-electrostatics-ap-physics-iit-jee-neet/ | 1,701,201,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00505.warc.gz | 308,705,777 | 22,392 | Description
This course includes –
Coulombs law
1. Charge and electrostatic force
2. Coulomb’s law
3. Coulomb’s law and Newton’s force
4. Coulomb’s law and charge distribution
Electric fields
5. Electric field
6. Dipoles and dipole moment
7. Electric field due to ring
8. Electric field charged disc
9. Potential energy of an electric dipole.
Gauss Law
10. Electric flux
11. Measuring electric flux
12. Gauss law
13. Application of Gauss Law
14. Charge inside a cavity
How I make my courses:
When I create content for lessons,I think deeply around the areas where students struggle and feel confused. My lessons tackle these parts in depth. Also, I believe visual representation of various ideas in physics makes a lot of impact. The lessons have visuals and animations that are thought through for faster learning and absorption of the subject
And most importantly, I make myself available to answer questions of students enrolled in my course
My students (some are wonderful teachers too) wrote this to me
Bobbie Smith: Amazing explanations, I really learned a lot. Thank you.
Satyam Jha: amazing!! i could not understand vector physics in my class but here it is very easy to understand Thanks!!
Csaba (teacher): I learned new ideas. I’m looking to try them in my professional practice as a physics teacher. Thanks! 🙂
Fernando P. Radaza: It help me a lot to understand better about physics of Work, Power & Energy.
Chamara Dilshan: it’s good, explaining every small thing,it’s good to start physics beginners
Onofrio : The lessons given by the teacher are very interesting! Excellent course!
Simaran: Very deep understanding of the subject
Shiva: Very knowledgeable and sounds very nice and helpful
Gallina: Excellent the lessons held by the teacher with exhaustive explanations and well illustrated. Well done course!
Smith: Great course.The presentation is very clear. Thank you.
Pawan Kumar: The way to teaching us is amazing with all diagrams
Samit This course has a lot of good content and very well presented. Thank you
Dani (teacher): It was concise and consequent. The exercises were good exposed and explained. Simply excellent. I promise, that i will use some ideas in my every day practice in my classroom. I’m also teaching physics, but in HungarianI finished this course to improve my skills, first of all in interesting approaches, and foreign language skills as well. This course was exactly what I expected!
Who this course is for:
• Students preparing for IIT JEE or NEET
• AP Physics students
• IGCSE Students
• First year (freshman) engineering students
Requirements
• Students need to have finished class 10 physics
What you’ll learn
• What is electric charge
• Conductors and insulators
• Coulomb’s law
• Electric fields
• What is electric flux
• Gauss Law
• Application of Gauss Law (sphere, wire, parallel plates)
• How to measure electric flux | 722 | 3,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-50 | latest | en | 0.766266 |
http://www.eg.bucknell.edu/~phys211/fa2017/cal/lec17.html | 1,513,395,091,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581053.56/warc/CC-MAIN-20171216030243-20171216052243-00734.warc.gz | 339,115,727 | 3,041 | ## Lecture 17: Liquids, Gases, and Phase Transitions
October 26, 2017
• Study: Fig 7.1; Ex 1; Eq 7.23; Exs 4, 6; Eq 7.38; Exs 7, 8.
### Objectives
• (Continuing objective) Be able to relate concepts of thermodynamics and statistical mechanics to “everyday” situations and to discuss various applications of the concepts to practical problems in various fields of science, medicine and engineering.
• Give a molecular description of a solid, liquid and a gas; i.e., be able to explain how the motion of individual molecules and the patterns that they form differ between solids, liquids and gases.
• Describe qualitatively why melting and vaporization phase transitions occur, and what their associated latent heats are. Use latent heat of fusion or vaporization to calculate the heat released or absorbed during a phase transition.
• Relate thermal kinetic energy, thermal speed, and temperature for a solid, liquid, or gas.
• Describe the properties of monatomic and diatomic ideal gases, including their molar specific heat and speed of sound.
• Relate pressure, volume, temperature, and number of molecules or moles, or changes in the quantities, using the ideal gas law.
### Homework
• Friday's Assigned Problems: A47, A76; Supp CH 7: 1, 2, 3, 4, 5, 12, 13, 15
• Monday's Hand-In Problems: A48; Supp CH 6: 12, 13, 14, 16, 18; Supp CH 7: 8, 9, 10, 14
### Videos of example problems
To see the problem statement, click on the link below. To play the video example, click on the underlined words "Video Demonstration" near the top of the page with the problem statement.
• Video Example #1: Estimate melting temperature and latent heat for nickel.
• Video Example #2: Another example with Ideal Gas Law. Note that in the final case, the water is boiling. ans: 1.71
### Drinking bird reference of the day
"This is the greatest invention in the world!" -- Homer Simpson
### Pre-Class Entertainment
• Natural Blues, by Moby
• Norwegian Wood, by the Beatles
• Not Fade Away, by Buddy Holly and the Crickets
• No Rain, by Blind Melon
• No Roots, by Alice Merton
• Out in the Street, by Bruce Springsteen and the E Street Band
• Outa Space, by Billy Preston | 555 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.842201 |
https://gemseo.readthedocs.io/en/stable/_modules/gemseo.problems.scalable.parametric.core.problem.html | 1,656,316,163,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00050.warc.gz | 326,459,041 | 7,937 | gemseo / problems / scalable / parametric / core
# problem module¶
## Scalable problem - Problem¶
Classes:
TMProblem([n_shared, n_local, n_coupling, ...]) The scalable problem from Tedford and Martins, 2010, builds a list of strongly coupled models (TMSubModel) completed by a main model (TMMainModel) computing the objective function and the constraints.
Functions:
npseed seed(self, seed=None) rand(d0, d1, ..., dn) Random values in a given shape.
class gemseo.problems.scalable.parametric.core.problem.TMProblem(n_shared=1, n_local=None, n_coupling=None, full_coupling=True, noised_coupling=False, seed=1)[source]
Bases: object
The scalable problem from Tedford and Martins, 2010, builds a list of strongly coupled models (TMSubModel) completed by a main model (TMMainModel) computing the objective function and the constraints.
These disciplines are defined on an unit design space whose parameters comprised in [0, 1] (TMDesignSpace). This problem is defined by the number of shared design parameters, the number of local design parameters per discipline and the number of output coupling variables per discipline. The strongly coupled disciplines can be either fully coupled (one discipline depends on all the others) or circularly coupled (one discipline depends only on the previous one and the first discipline depends only on the last one).
Constructor.
Parameters
• n_shared (int) –
size of the shared design parameters. Default: 1.
By default it is set to 1.
• n_local (list(int)) –
sizes of the local design parameters for the different disciplines. Same length as n_coupling. If None, use [1, 1]. Default: None.
By default it is set to None.
• n_coupling (list(int)) –
sizes of the coupling parameters for the different disciplines. Same length as n_local. If None, use [1, 1]. Default: None.
By default it is set to None.
• full_coupling (bool) –
if True, fully couple the disciplines. Otherwise, use circular coupling. Default: True.
By default it is set to True.
• noised_coupling (bool) –
if True, add a noise component u_local_i on the i-th discipline output.
By default it is set to False.
• seed (int) –
seed for replicability.
By default it is set to 1.
Methods:
get_default_inputs([names]) Get default input values. Get the TM design space. Reset the TM design space.
get_default_inputs(names=None)[source]
Get default input values.
Parameters
names (list(str)) –
names of the inputs.
By default it is set to None.
Returns
name and values of the inputs.
Return type
dict
get_design_space()[source]
Get the TM design space.
Returns
instance of the design space
Return type
TMDesignSpace
reset_design_space()[source]
Reset the TM design space.
gemseo.problems.scalable.parametric.core.problem.npseed()
seed(self, seed=None)
Reseed a legacy MT19937 BitGenerator
Notes
This is a convenience, legacy function.
The best practice is to not reseed a BitGenerator, rather to recreate a new one. This method is here for legacy reasons. This example demonstrates best practice.
>>> from numpy.random import MT19937
>>> from numpy.random import RandomState, SeedSequence
>>> rs = RandomState(MT19937(SeedSequence(123456789)))
# Later, you want to restart the stream
>>> rs = RandomState(MT19937(SeedSequence(987654321)))
gemseo.problems.scalable.parametric.core.problem.rand(d0, d1, ..., dn)
Random values in a given shape.
Note
This is a convenience function for users porting code from Matlab, and wraps random_sample. That function takes a tuple to specify the size of the output, which is consistent with other NumPy functions like numpy.zeros and numpy.ones.
Create an array of the given shape and populate it with random samples from a uniform distribution over [0, 1).
Parameters
• d0 (int, optional) – The dimensions of the returned array, must be non-negative. If no argument is given a single Python float is returned.
• d1 (int, optional) – The dimensions of the returned array, must be non-negative. If no argument is given a single Python float is returned.
• ... (int, optional) – The dimensions of the returned array, must be non-negative. If no argument is given a single Python float is returned.
• dn (int, optional) – The dimensions of the returned array, must be non-negative. If no argument is given a single Python float is returned.
Returns
out – Random values.
Return type
ndarray, shape (d0, d1, ..., dn) | 1,024 | 4,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | latest | en | 0.705733 |
http://modsurski.com/csci161/topic16.html | 1,643,304,390,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00696.warc.gz | 46,098,248 | 3,822 | # Topic #16 – Models of Computation¶
## Finite State Machines/Automata (FSM/FSA)¶
• Simple models of computation.
• We can define a very very very very simple computer in terms of
• Input
• Output
• States
• And then we can geek out over how much work this little computer can do
• Like, what types of problems this computer can solve.
• FSM’s are VERY simple computers.
• But, just because they’re simple, doesn’t mean we can’t use them
• In fact, these little guys are powerful enough to solve A LOT problems we deal with every day.
• Vending Machines
• Traffic Lights
• Elevators
• Locks/Safe
• Regular Expressions
• Let’s say we want a lock that has the combination 7, 7, 3.
• Here’s a FSM for that lock
• Or, we can go the other way and say, given this machine, what input can it accept.
Activity
What input strings can this machine accept? What I mean is, what strings will get this machine to it’s final state.
• Think about it this way. Think of how simple the computational system is that is powerful enough to unlock a lock
• Computer Scientists like to think about what else can a computer this powerful do?
## Pushdown Automata (PDA) and Context-Free Grammars¶
• FSM’s are cool and all, but they can only solve certain types of problems. If we want to solve more complex problems (for lack of a better term), we need a more powerful machine.
• PDA’s are more powerful computational models
• They can solve strictly more problems than FSMs.
• Anything a FSM can do, a PDA can do
• There are a lot of things a PDA can do, that a FSM cannot
• PDA’s accept what we call Context-Free Grammars.
• Here’s an example of doing the same thing as the 2nd FSM.
S –> aA
S –> bB
A –> aA
A –> b
B –> bB
B –> a
• Here, the upper case letters are like special symbols that mean you can replace them in a string with whatever is on the right hand side.
• The lower case letters are just letters.
• String: ‘S’
• We have two options, let’s go with the first
• String: ‘aA’
• We now have an A, so let’s go with the first options
• String: ‘aaA’
• Let’s do it again
• String: ‘aaaA’
• Let’s do it 4 more times
• String: ‘aaaaaaaA’
• Now let’s go with the second option
• String: ‘aaaaaaab’
• No more upper case letters, so we’re done.
• String: ‘aaaaaaab’
• Here’s another one
S –> aSa
S –> bSb
S –> epsilon/ ‘’
(epsilon means empty string)
Activity
What strings can this system create? Try to generate a few and see if you can see the big picture on what it’s doing.
• Another one
S –> SS
S –> (S)
S –> ()
S –> epsilon/ ‘’
Activity
What strings can this system create? Try to generate a few and see if you can see the big picture on what it’s doing.
## Context-Sensitive Grammars¶
• Context matters now
S –> abc
S –> aAbc
Ab –> bA
Ac –> Bbcc
bB -> Bb
aB –> aa
aB –> aaA
## Turing Machines¶
• Turing Machines are even more powerful models of computation
• Basically, the computers we use today are kinda’ like these
• They’re not built like these, but they are as powerful
• Can solve the same problems.
• Also, we often say that our brains are at least as powerful as a Turing Machine.
• At least? | 852 | 3,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-05 | latest | en | 0.879356 |
http://slideplayer.com/slide/1456069/ | 1,526,858,728,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00089.warc.gz | 270,205,785 | 24,279 | # Energy & Work. Work involves a change in a system. changing an objects position heating or cooling a building, generating a image on the TV screen, moving.
## Presentation on theme: "Energy & Work. Work involves a change in a system. changing an objects position heating or cooling a building, generating a image on the TV screen, moving."— Presentation transcript:
Energy & Work
Work involves a change in a system. changing an objects position heating or cooling a building, generating a image on the TV screen, moving a speaker cone to make sound Since different tasks require different amounts of work, some things require more energy than others.
Work is… F = force in Newtons d = displacement in meters The angle between F & d Joule is the unit of WORK
Work is… Work- A quantity that measures the effects of a force acting over a distance. Work is a result of motion in the direction of the force. There is no work without motion (d=0). Distance-means distance in the direction of the force. If a force is vertical and motion is horizontal, No work is done.
Work is… MAXIMUM when = 0º MAXIMUM when Force // Displacement MINIMUM when = 90º MINIMUM when Force Displacement
Question #2 If you push vigorously against a brick wall, how much work do you do on the wall? a)A lot b)None c)Without numbers, how can we know? d)No idea
Answer #2: (b) None There is no work done on the wall as there is no displacement of the wall.
What are the units of WORK? Work is measured in Newton-meters (Nm) or foot-pounds (ftlb) A Newton-meter is called a JOULE (sounds like jewel) – Named after James Prescott Joule (1818-1889) – British physicist who established the mechanical equivalence of heat and discovered the first law of thermodynamics.
Find the work done by gravity when a 2.0 kg rock falls 1.5 m. w = F (d) cos( ) What is the formula for Force (F) F = m (g) or F = m (9.8 m/s 2 ) w = (m · g) (d) cos( ) w = (2.0kg · 9.8m/s 2 )(1.5m) cos(0°) w = 29 J
Negative Work... = 0° Cos(0°) = 1 = 180° Cos(180°) = -1
How much work is done when a man pushes a car with an 800 N constant force over a distance of 20 m? Question #3 a)0 J b)40 J c)800 J d)16000 J e)Im lost…
How much work is done when a man pushes a car with an 800 N constant force over a distance of 20 m? Answer #3: (d) 16000 J w = F (d) cos( ) w = (800 N)(20 m) cos(0°) w = 16 000 J
How much work is done by a woman pulling a loaded dolly 100 ft with a force of 150 lb at an angle of 45°? Question #4 a)0 ft-lb b)7879.8 ft-lb c)10606.6 ft-lb d)15000 ft-lb e)Im lost…
How much work is done by a woman pulling a loaded dolly 100 ft with a force of 150 lb at an angle of 45°? Answer #4: (c) 10606.6 ft-lb w = F (d) cos( ) w = (150 lb)(100 ft) cos(45°) w = 10606.6 ft-lb
Power & Work Work can be done at different rates. Since work involves the transfer of energy, the faster work is done, the quicker energy needs to be transferred. Power i s the measure of how fast work can be done. In other words, power is the rate at which energy is transferred.
Power is… W = work in Joules t = time in seconds WATT is the unit of power
Question #5 A woman exerts 100 N of force pushing a grocery cart 5 meters in 2.5 seconds. How much power did she exert? a)0 watt b)40 watt c)200 watt d)1250 watt e)Im lost…
Answer #5: (c) 200 watt A woman exerts 100 N of force pushing a grocery cart 5 meters in 2.5 seconds. How much power did she exert?
Horsepower (hp) is a commonly used unit of power. 1hp = 746 watts(W)
For example… Let's carry a box of books up a set of stairs. From experience, we know that running the books up the stairs takes more energy than walking the same distance (you would be more tired if you ran). But the amount of work done is the same since the books weighed the same and moved the same distance each trip. However, the work is done much faster if we run, so energy must be converted faster. Therefore, more power is required.
For example #2… Think of a racecar versus an economy car. They both can travel the same distance, but the race car does it much faster since it is capable of expending much more energy in much less time. This is because the more powerful car can convert energy quicker.
For example #3… Think of an 18-wheeler versus an economy car. They both can travel the same distance, but the economy car does it much faster since it is capable of expending much more energy in much less time. BUT… The truck can carry more weight (exert a greater force) and is more powerful…
Electrical Power Electrical Power is defined the same way. Work must be done to move electrons through the electrical devices (i.e.,resistance). More resistance means more work must be done to allow the device to operate. More electrical power means more energy is being converted. This electrical energy is supplied by the source of the electrical current, like a battery or generator.
Energy The ability to do work. An object has energy if it is able to produce change in itself or its surroundings.
Energy lets us do work Energy is the ability to do Work Energy is important to all living things in order to maintain life functions. Humans use energy to modify their environment and perform work. Energy is measured by the amount of work it is able to do. The units of energy are joules (J).
Energy exists in different forms 1. Mechanical energy (moving objects and their positions) 2. Radiant energy (light and solar energy) 3. Chemical energy (including the food you eat and fuels we burn) 4. Thermal or heat energy (molecules moving faster means more heat) 5. Electrical energy (electrons moving through a wire) 6. Nuclear energy (energy locked in the nucleus of an atom)
Energy can be transferred… Fossil fuels like coal and oil can be burned to heat water that boils into steam that turns a turbine to generate electricity that you use to operate a stereo. Chemical energy Thermal energy Thermal energy Kinetic energy Kinetic energy Electrical energy
Energy cannot be created or destroyed. In the example of riding a bicycle down a steep hill, you begin with a lot of potential energy at the top of the hill and gain kinetic energy as you coast down the hill. If you are not making the kinetic energy (movement down the hill), where does it come from? The answer is simple: your potential energy at the top is transformed into kinetic energy as you speed along.
Mechanical Energy Kinetic & Potential Kinetic is the energy of moving objects. Potential Energy is stored energy. Gravitational PE is energy due to position.
Mechanical Energy - II As you speed down a steep hill on a bicycle, you are moving and therefore have kinetic energy. But where did this energy come from? You probably already know that it came from your position at the top of the hill. At the top of the hill, you had the ability to do work (move the bicycle) purely because of where you were. You had the potential to perform the work of moving the bicycle. Whenever you work with mechanical energy, you probably are dealing with both forms together in the same system.
Potential Energy Energy that is a result of an objects position or condition. All potential energy is Stored Energy. – Pull back on a bow string and bend the bow. The object then possesses potential energy.
A rock on a table top has more potential energy than when it is on the ground due to its position. This is a form of gravitational potential energy. Fuel is an example of chemical potential energy, due to its ability to burn. Potential Energy
Gravitational Potential Energy Depends on mass and height. GPE = m(g)h m = mass g = acceleration due to gravity h = height -What are the Units of GPE?
SI units? m = kg g = m/s 2 h = m PE = (kg · m/s 2 ) * m = N*m = J
Question #6 A man lifts a 2 kilogram book from the floor to the top of a 1.25 meter tall table. What is the change in the books gravitational potential energy? a)0 joules b)+2.50 joules c)-2.50 joules d)+24.525 joules e)-24.525 joules
Answer #6: (d) +24.525 J A man lifts a 2 kilogram book from the floor to the top of a 1.25 meter tall table. What is the change in the books gravitational potential energy?
Question #7 A mouse now pushes a book (2 kg) off the table (1.25m). What is the change in the books gravitational potential energy? a)0 joules b)+2.50 joules c)-2.50 joules d)+24.525 joules e)-24.525 joules
Answer #7: (e) -24.525 J A man lifts a 2 kilogram book from the floor to the top of a 1.25 meter tall table. What is the change in the books gravitational potential energy?
Kinetic Energy Energy that appears in the form of motion. Depends on the mass and speed of the object in motion.
Kinetic Energy KE = (1/2)mv 2 m = mass v = velocity Unit for energy is Joule (J) it is defined as a Newton Meter.
SI units?
Kinetic Energy Energy due to motion. A brick falling at the same speed as a ping pong ball will do more damage. KE is dependent on mass. KE also depends on speed (v)
Which would affect the kinetic energy of an object more, doubling its mass or its velocity? doubling the mass would result in a doubling of the KE. doubling the velocity would quadruple the KE. Kinetic Energy
Question #8 What is the KE of a 1140 kg (2513 lb) car driving at 8.95 m/s (20 mph)? a)0 joules b)5101.5 joules c)10203 joules d)4.57x10 4 joules e)Im lost…
Answer #8: (d) 45658.4 J What is the KE of a 1140 kg (2513 lb) car driving at 8.95 m/s (20 mph)?
Question #9 What is the KE of 11 pound rabbit running at 302.6 mph? Note: 1 ton = 907.185 kg and 1 mph = 0.447 m/s. a)0 joules b)5101.5 joules c)10203 joules d)4.57x10 4 joules e)Im lost…
Answer #9 What is the KE of 11 pound rabbit running at 302.6 mph? Note: 1 pound = 0.454 kg and 1 mph = 0.447 m/s.
Recall, Law of Conservation of Energy Energy can not be created nor destroyed. Energy can change from one form to another. The total energy in the universe is constant.
Conservation of Energy In a roller coaster all of the energy for the entire ride comes from the conveyor belt that takes the cars up the first hill.
Examples A 400 kg roller coaster car sits at the top of the first hill of the Magnum XL200. If the hill is 151 ft (46 m) tall, what is the potential energy of the cart? What is the speed of the cart at the bottom (what do you need to ignore?) How much KE and PE does the car have half way down the hill?
Answers Energy at Top = Energy at Bottom Ignoring friction – assume 100% energy conversion Energy at Top – GPE = 400 kg x 9.8 m/s 2 x 46 m = 180 320 Joules – KE = 0 Energy at Bottom – GPE = 0 – KE = 1/2 x 400 kg x v 2 180 320 = 200 v 2 Velocity at Bottom = 30 m/s = 67 mph
Answers Energy at Top = Energy at Bottom Energy at Halfway point? – 1/2 PE = 90 160 J – 1/2 KE = 90 160 J Speed at Halfway point = ? – 1/2 of 30 m/s = 15 m/s = 33.5 mph NO !!!!!! – 1/2 mv 2 = 90 160 – Velocity = 21.2 m/s = 47.5 mph
The End
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Equality Management I have a series of slides, one of which shows the difference the average price of each line by itself (in meters) versus the average price of all the other lines (in centimeters). When they are in actual value, their difference will drop. But the price is proportional, which means that the average price should be 0. Imagine you are putting in 200 meters of water with this technology and you’re getting a new bottle, but the average price of the water is 250. That’s as close as you can get. You will get a lower average price of the same amount of water to put in, which is in fact higher. It looks complicated, but does it really change go price or price at the footer, or the top;???????? But maybe when you put a price right in the footer, they will have a lower average price, or they will be buying more value given the cost of the line you’re putting in. But this is very simplified. When you put the price of 200 meters of water on the footer, it’s 0.150 – 0.200 = 0.230. But when you put the same price in the footer it’s the same difference. But the difference is 0.230 / 0.230 = 0.180. This is done basically when the legs of your meter are pointing right to right. In the footer you are putting meters which have a lower price than the lower average price. But that’s not a problem in the footer because the leg of the meter has the lowest price.
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This is especially true in smaller containers as small as one meter. In the feet side of the meter, the lower average price is less than the higher price. In containers you can see the farr and the nearr. In the footers you can see the farr, but the nearr is the lower average price. In both feet you can see the distances. The price when you put 20 meters by foot is exactly zero. The opposite is the price when you take hundreds of meters by foot, just 0.200. They don’t change. It is very close to zero when you put a 100 meter by foot in the feet as it is a 20 meter by foot in the feet. This may look like a price fluctuation, but it doesn’t really have to deal with that problem because none of the other items on the price will change. But before you get a more detailed explanation, I wanted to express in real life what I think the amount of water consumed by different machines in different countries is. I used a model at ZDNet (in Microsoft Visual C#) with a model of carbon dioxide stored in a magnetic stirrer. Once everything was in the process, I replaced the fuel cells of the machine with a carbon dioxide in situ. The water had every step in my water cycle and the carbon dioxide released was the least, and my carbon dioxide in the stirrer was the top. The Modeling Comparison!!! (Note: As the model was based on a simple mathematical model, I just don’t like the model or even the appearance of it. Overall, what made it run the way I wanted to run it was its size. I cannot imagine how small the modelEquality Management – Creating Public Awareness and Performing Public Policies! By The Editor __________ In the past 3 months, our team has been on the road touring companies and now our annual review update! From companies like Uber and San Francisco to public policy research firms like Google, we are spending a huge amount of time writing literature on public policies and how the media is an important source of knowledge. We published hundreds of articles on issues related to public policy, and we have written numerous papers on issues that have relevance to our field and other publications. Some of our articles will be given away when we review these articles. | 802 | 3,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.956484 |
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# Aeronautical engineering and airplane design online
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gliding angle 1 in 7,
and the average width
assumed depending on
the form and strength
of the section).
Then, since gliding
W
angle = ^- , every /
pounds of strut
\vright will give rise to 1 pound of resistance, in addition to
the aerodynamic resistance of the struts.f We can, therefore,
W
write T = 1- R, where T = thrust due to the struts, and
It is their aerodynamic resistance. Simplifying, we have
G = W + 7R, but, since this expression has a maximum value
for the least efficient strut, the reciprocal is here employed,
14300
and multiplied by the constant 14300, giving C = = jp i 7^
The best strut under the conditions above specified is then
the one showing the highest value for C. The reason for
choosing this particular value for the multiplier is that it
makes C = 100 for the best strut of the first and largest series
which we shall consider.
If the speed of the machine for which the struts are being
selected is greater than 60 miles an hour, the resistance be-
comes of greater importance as compared with the weight,
and the merit factors for those sections which, although heavy,
offer very low resistances are relatively improved. If the glid-
ing angle is flatter than 1 in 7, a similar effect ensues.
On the other hand, if it becomes necessary to use struts
having a diameter of more than 1 inch or thereabouts, the ad-
OGILVIE'S SECTIONS
vantage inclines toward the sections which have the greatest
strength for their weight, and the relative importance of re-
sistance is diminished, since, in similar sections, weight varies
as the square of the breadth and resistance only as the first
power. These effects are, however, of slight importance, and
would not be likely to change the merit factors enough to have
serious influence on the choice of a section in any given case.
The question of strength will be taken up more fully in
another section of the course. It will suffice to say here that
the strengths of two struts have been considered to be equal
when their moments of inertia about their longitudinal axes
are equal.
Strut Sections Developed by Ogilvie
We may now proceed to the examination of definite data for
a number of series of struts, tested at various times and places.
The following figures are the result of experiments performed
at the N. P. L. at the suggestion of Alec Ogilvie, the sections
being illustrated in Fig. 1.
/ = moment of inertia for the section in question about its
longitudinal axis (inches* for a strut 1 inch wide).
R = resistance in pounds of 100 feet of strut 1 inch wide at
60 miles per hour.
W = weight in pounds of 100 feet of spruce strut 1 inch
wide.
b = width of strut whose strength will be equal to that of a
strut of section a, and 1 inch wide.
W = weight of 100 feet of spruce strut of width b.
C M = merit factor at 60 miles per hour.
No.
W
W
a
.167
104.4
41.6
1.00
41.8
19
b
.049
81.9
16.4
1.36
30.3
18
c
.090
59.2
30.4
1.17
41.6
27
d
.124
36.9
34.8
1.08
40.6
45
e
.074
63.0
33.4
1.23
50.6
24
1
.134
28.6
37.7
1.06
42.4
56
.094
54.9
30.0
1.15
39.7
30
h
.119
12.8
39.7
1.09
47.1
99
i
427
12.8
41.0
1.07
47.0
100
;
.119
13.5
39.7
1.09
47.1
96
i
.111
13.5
38.0
1.11
46.8
94
I
.106
29.9
36.4
1.12
45.6
51
m
.106
45.9
36.6
1.12
45.9
35
n
.171
14.2
51.9
0.99
509
97
o
.146
13.5
47.0
1.03
49.9
97
p
.128
18.7
44.1
1.07
50.5
75
q
.245
15.1
71.0
0.91
58.9
93
r
.227
16.4
67.2
0.93
58.1
87
s
.194
13.5
62.0
0.96
57.2
97
t
.209
13.5
66.1
0.95
59.7
95
.115
24.6
42.5
1.10
51.4
59
t Relationships between weight and resistance on a glide will be
fully considered in Section 12.
Many very interesting conclusions can be drawn from this
table. In the first place, it is evidently of the utmost im-
portance to avoid rapid changes in curvature. Several sec-
tions, notably, e and I, although they appear to have a very
smooth outline, oppose a large resistance simply because the
transition from the entrance to the run is so abrupt that the
air-flow cannot follow its contour, and violent eddy-making
ensues.
8fi
AERODYNAMICAL THEORY AND DATA
The good performance of several sections so formed indi-
cates that it may be wise actually to run the sides of the strut
parallel for some little distance, as illustrated by q and t.
This is counteracted, however, by the fact that skin-friction
increases in proportion to the " wetted surface " of the strut.
It is for this reason that the very longest sections did not give
such low resistances as those of more moderate form. This
matter of the ratio of length of section to width will be dis-
cussed more fully somewhat later, in connection with another
series of tests.
It will be seen, too, that the resistance is little affected by
the chopping off of a portion of the tail in such a manner as
to leave it straight across. Examples of this are furnished
by n, ( and i. This is due to the fact that it has not been pos-
sible in any strut yet designed to totally eliminate the region
of deadwater behind the strut. As will be evident from any
section of air-flow about a fair-shaped section, the lines of
flow always leave the contour of the strut some distance short
of the extreme rear. Since no changes made in the contour
within this region will have any decided effect on the re-
sistance, it avails nothing to go to the trouble and expense
involved in the attempt to construct a wooden strut running
out to a sharp point at the back.
Another Series of Struts Tested at the N. P. L.
at the same laboratory, the sections being those actually em-
Dt Haulllant
Beta
B.F.34
c
Baby
FIG. 2. STRUT SECTIONS
TESTED AT N. P. L.
G>
PIG. 3. N. P. L.
STRUTS
ployed in machines then existing. The outlines of the sections
tested are shown in Fig. 2, and the characteristics are given
below.
Blerlot A 070
Bleriot B 107
Fmrma n 074
De Harilland. .052
Baby 110
B.F. 34 279
B.F. 35 238
188
K
51.0
52.7
49.3
54.9
17.0
i: -..r.
13.5
14.8
34.9
25.2
20.5
41.6
93.2
si! 7
61.7
b
1.24
1.11:
1.22
1 .:;
1.11
-v
0.92
0.97
40.0
43.8
36.8
r,i.:.
n.9
70.0
58.0
Om
80
31
26
7K
90
of the symbols being the same as in the tables already given,
except that n = the ratio of the length to width of section.
n
2.
2.5
3.
3.5
4.
4.5
5.
.094
.117
.141
.104
.iss
.211
.235
R
24.8
13.7
13.4
11.4
11.2
11.7
12.1
W
32.0
40.0
48.1
50.1
04.1
72.1
80.1
b
1.15
1.0!)
1.04
1.00
0.97
O.94
0.92
W
42.3
47.5
52.1
M.1
)i'J.'J
67.8
73.7
59
94
9(i
in:,
LOS
99
94
Tims it is apparent that the best of these sections are inatc-
rinlly superior to the best of the sections tested by Ogilvie.
both in resistance and in merit factor. In Fig. 4 resistance of
RESISTANCE
OP
R.A.K STRUTS
FIG. 4.
UESISTAVCE AND FACTORS OP MEKIT FOR R. A.
STRUTS
100 feet of strut at 60 miles per hour, and merit factor at 60
miles per hour, are plotted against ratio of length to width.
As this ratio diminishes, the air-flow about the strut takes on
a very uncertain character, and the values when n is less than
2 are rather doubtful. Such extremely short sections as this
are also undesirable from the standpoint of lateral stability.
as will be shown in another section of the Course. On the
other hand, n may be considerably pi-eater than the absolute
optimum value without any great disadvantage, so it will be
well in general to employ a ratio of four, or even a slightly
higher figure. The photographs of Mow aliout strut sections.
reproduced in Fig. ~>. show clearly why such a procedure c-.-m
It will be seen that these figures simply supplement and
confirm the conclusions already deduced from the more exten-
sive and systematic investigations directed by Mr. Ogilvie.
TeU on Struts, Length to Width Varied
As a result of these and other tests. series of struts em-
bodying the best features of those already tried, and varying
only in the ratio of length of section to width, was made and
tested at the National Physical laboratory. Three rep re
live members of tin- series are shown in Fig. .'!. The table
below gives the characteristics of these struts, the meaning
Two Kiffcl Struts
Two struts of somewhat the same section as those just di-
eiisscd have recently l.een I, '-led by Kifl'el. and show remark-
ably low resistances. Their outlines arc uivcn in Fiir. (i. For
Xo. 1, having n equal to :!.L'.'I, If equals !i.7 pounds, while for
No. 2. with a somewhat sharper entry, i is L'.lKi and R is only
8.7 pounds. I'arl of this improvement oxer the best of the
Knglish tests, hoxvexei. i- undoubtedly due to the higher wind
speed which is secured in Kiffel's laboratory, the resistance
coefficient having a tendency to rise as the speed of test i^.
decreased.
AERODYNAMICAL THEORY AND DATA
67
Effect of Length of Struts
We now turn our attention to the effect of the length of
the strut. While this point is less important than was gen-
erally supposed a few years ago, and while its effects are
largely determined by the nature of the surfaces in which the
strut terminates, the experimental results bearing on the mat-
ter should nevertheless be studied. For this data we are
indebted to Mr. Thurston, who has described his results in
the series of articles already cited. As the result of a great
nel would be exceedingly difficult to devise. The matter might
well be investigated in an outdoor, full-scale plant such as
that at St. Cyr.
Resistance of Inclined Struts
The only point which remains to be studied is the resistance
of struts which are not normal to the line of flight. Some
much more recent tests by Mr. Thurston have covered this
point, and show very surprising results. Struts of square,
rectangular, circular, and stream-line section were tested at
angles from to 90 degrees, and the effects of the ends of the
strut offering a direct resistance when inclined were overcome
by the use of the method of differences: that is, tests were
made first on a strut 34 inches long, and then on one 16
inches long, the difference of the figures obtained being equal
to the resistance of an 18-inch section of an infinite strut.
The ratio of the resistance of a strut inclined at various
BETA
FIG. 5.
DE HAVILLAND
ILLUSTRATING FLOW AROUND STRUTS
many experiments on manifold different types of strut, he
came to the conclusion that resistance for a strut with free
ends could best be expressed by the formula B = KltV 1 -
.0073fF 2 , where R is the resistance in pounds, I and *, re-
spectively, the length and thickness of the strut in feet, K a
constant, and V the speed in miles per hour.
It is evident from this equation that, even with the lowest
values of K yet obtained, the effects of length will be prac-
Fio. 6. Two EIFFEL STRUTS
tically negligible when the length is more than 50 times the
thickness, as it generally is. Since, in addition, the case of a
strut with free ends is one which never occurs in practise,
resistance may be considered as independent of length-thick-
ness ratio for all the purposes of design.
The form of air-flow about the wing may have very decided
effects on the resistance of interplane struts, but we have no
means of knowing how great these aro. and experiments cover-
ing this point and susceptible of performance in a wind tun-
INCMMATIOM Of BAH TO WIMO
FIG. 7. DATA FOR INCLINED STRUTS
angles to the resistance of a normal strut of like section and
equal projected length is plotted in Fig. 7. It will be seen
that the resistance at 30 degrees to the wind is less than one-
third of that at 90 degrees, and this large difference is by no
means accounted for by the difference in length of section
parallel to the wind. When a circular strut is placed at an
angle of 30 degrees to the wind, the section parallel thereto
is an ellipse having a length of twice its width, and the resist-
ance of an elliptical strut such as this, when placed normal.
is only 36 per cent less than that for a circular section.
About 45 per cent of the reduction due to inclination thus
remains unaccounted for.
Since, however, the curve of reduction is substantially a
sine curve, and is therefore very flat at the ends, there is
very little advantage to be gained from inclining a stream-
line strut unless it is inclined at least 30 degrees to the nor-
mal. This reduced resistance should, however, be kept in
mind as a point in favor of the staggered biplane. Eiffel
also made a few tests on struts inclined 30 degrees from the
normal, the results cheeking very well with Mr. Thurston's.
The Effect of Changing the DV Product for Struts
As was shown in Chapter 10, the resistance coefficient is
not an absolute constant, but is a function of VI), when-
V Is the speed and I) the diameter of the strut. The coeffi-
cient tends to decrease as VI) increases, but the change for
values of I'D (in foot/second units) above 6 is extremely small,
as Eiffel lias demonstrated. The tests made at thp National
Physical Laboratory have been made with a value of VD
equal to only 2.5. whereas, in an actual machine, this quantity
would never be likely to fall below 5, and is generally from
7 to 10.
i is AERODYNAMICAL THEORY AND DATA
We can therefore deduce from Kitlel's experiments that it References for Part I. Chapter 11
is safe to reduce the values for resistance here given (for the
N. P. L. tests) by about 25 per cent in applying them to a " strut*." FHoht. June is. 1012.
design. This indicates that, as was hinted above, the superior- -Aerodynamic RMMUM of struts. Bare, and wires." by A. v.
ity of Eiffel's strut sections is more apparent than real, and Thurston ; Aeronautical Journal, April and July, 191::.
that the best sections yet available are the N. P. L. sections Technical Reports of the British Advisory Committee on Aeronautics.
having fineness ratios of from 3.5 to 4.5. The correction 1911-12. 1912-13.
given here should be applied only to Struts of fairly good "The Resistance of Inclined Struts In a Uniform Air Curri'iit," by A.
section, as the value of VD has much less effect on those sec- i^ 1 " 1 "" " " Dd "' Tonnsteln ' Aeron <"" lcal """' Janu "5'-
tions for which the resistance is relativelv IUL-II. and in which
" Nouvelles Recbercues sur In Resistance de 1'Alr et 1'Avlatlon." by G.
there is more effect due to turbulence than to skin friction. Eiffel. (1914 edition.)
Chapter XII
Resistance and Performance
Nomenclature
It may be useful to restate the symbols which we employ
in considering performance curves, ascent and descent.
IT" = weight of the machine;
A = area of the wings.
i = angle of incidence of the wings.
L = lift.
K u = lift coefficient.
D = drag of wings.
K., = drag coefficient.
11 = resultant of lift and drag on the wings.
P = parasite or structural resistance of a machine.
Dt = total resistance or drag = T) -\- P.
R t = total resultant air force on a machine.
// = -propeller thrust.
6 = angle of flight path with the horizontal.
Structural and Wing Resistance for the British B.E.2
In Chapter 4, a problem was worked out on the sustentation
and resistance of wing surfaces, which in spite of some rough
50 00
MILES PER HOUR
FIG. 1. PERFORMANCE CURVES FOR THE B.E.2
assumptions, illustrated the main performance curves and cal-
culations employed. In Fig. 1 are shown curves for the Brit-
ish B. E. 2. It is not a particularly modern machine, but has
been worked out so thoroughly that it deserves particularly
careful study.
The body or parasite resistance which includes the resis-
tance of the wing bracing, chassis, etc.. as well as the resistance
of the body proper, is taken as varying as T'" 2 and allowance
has been made for propeller slip stream velocity. The body
resistance is seen to play an unimportant part at low speeds.
But at about 53 miles per hour it becomes greater than the
plane or wing resistance, and at high speeds it. is almost twice
as great as the wing resistance. This emphasizes the imppr-
tance of minimizing the resistance for a high-speed machine.
However good a wing section itself may be, high structural
resistance will make high speeds impossible.
The plane resistance curve has a minimum value at about
65 miles per hour and increases on either side of this speed.
It is interesting to follow out how this increase in resist-
ance on either side occurs. At high speeds, the angles of
incidence and the drift coefficients are small but the speeds
are very great, and the increase in wing resistance is obvious.
At small speeds on the other hand the airplane is flying at
large angles of incidence to give the necessary sustentation and
the drift coefficients are large. The shape of the total re-
sistance curve follows from the summation of the two.
Theoretical Laws for Minimum Thrust and
Minimum Horsepower
From a theoretical treatment of the question, the following
interesting law has been derived :
Minimum thrust is required to overcome the resistance of an
airplane when Hie parasite resistance is equal to the drag of
the wings.
For a proof of this law, reference to Chasseriaud and
Espitallier is appended. In the case we have selected, illus-
trated in Fig. 1, the structural air resistance and the wing
drag are equal at a speed of 53 miles an hour, while the
minimum resistance is at 49 miles per hour. The law does not
seem to be borne out by practice, though it may be occasion-
ally useful as a rough check.
The minimum horsepower required generally occurs at a
low speed, but not at the minimum speed; and its position
will vary for every machine. Another theoretically deduced
law states that:
Minimum horsepower is required irln'ii Hie machine is mov-
ing at a speed at which the wing resistance is three times the
body resistance.
" This law is often highly inaccurate, but may be useful.
Effective or Propeller Horsepower Available Curve
Typical curves for these are also illustrated in Fig. 1, and
are of the greatest interest to the designer. In establishing
such curves it is generally assumed that the engine is running
at the rated revolutions per minute and that in designing the
propeller the efficiency for this revolution per minute at every
airplane speed is known. Thus assuming an engine which
delivers 140 horsepower at an ail-plane speed of 80 miles an
hour, the propeller having an efficiency of 75 per cent at this
speed, the available horsepower will be
140 X 75
100
= 105 horsepower.
69
70
AERODYNAMICAL THEORY AND DATA
Since the power of a propeller is given by the product of
us thrust into the speed and the speed of the propeller is the
speed of the airplane, it follows that when the propeller is
delivering sufficient power, it is also delivering sufficient
thrust. Hence propeller horsepower available is sufficient for
all practical consideration, and propeller thrust curves need
not be included in a performance chart.
Minimum and Maximum Speed; Maximum Excess
Power; Best Climb; Descent
The maximum and minimum speeds of an airplane are gen-
erally given by the two points of intersection of the propeller
horsepower available and the total horsepower required. If
the machine is highly- powered, and the propeller efficient, the
two curves may not intersect at the speed at which the lift
becomes insufficient, and the airplane would climb at stalling
angle, unless the engine is considerably throttled down. The
climb decreases the angle of incidence, and checks stalling.
It is thus a decided advantage to have excess available power
at high angles.
It is a simple matter to deduce the speed of climb from the
excess power. This is absorbed in raising the machine.
Ex
power
Total weight X climb per second
551)
The maximum excess power does not occur at the lowest
speed. To find it, we must measure the maximum ordinate
between the available propeller horsepower and the total re-
quired horsepower. In Fig. 1 this is to be found at 48 miles
per hour. The excess is 21 horsepower and the weight of the
machine is 1650 pounds.
Climb =
21 X 550
= 7 feet per second or 420 feet per min-
ute. This is, however, only the initial rate of climb. As the
machine rises, the density of the air, the power of the engine,
In practice, the pilot need not know the change of in-
cidence that he produces to climb, although for a given ma-
chine it is an easy matter to calculate the correct angle from
the performance curves. In Dr. Hunsaker's words, " a care-
ful man moves his elevator slowly until he has placed him-
self on the desired trajectory." Part of the art of aviation is
to do this without exceeding safe limits, for obviously there is
a limit to the rate of climb the engine can handle. If the
machine is put on a climb too steep for the power of the ma-
chine, the speed is suddenly lost, the controls become ineffec-
tive, and the machine has stalled.
In descent, very analogous considerations obtain. The
pilot decreases his angle of incidence to a negative value. At
this angle the speed required for sustentation is beyond that
of the maximum, and the propeller horsepower is insufficient.
If D = deficiency in horsepower,
n Total weight X velocity of descent.
Mt
The machine descends and gains the required speed under tin-
action of gravity.
The Two Regions of Control. Control by I limiilin-
Consider the performance curves of the same machine, the
Hnti-h I'..K.'_' shown in Fig. 1. Suppose the machine to be
flown iit in degrees at the point J/ with the engine throttled.
so that there is equilibrium, and the power curve is as shown.
26 horsepower. The pilot wishing t<> rise will naturally in
crease his angle of incidence to say 12 degrees. He will thru
require 30 horsepower while the throttled engine will deliver
even less than the 2(j horsepower through the propeller. In-
stead of rising the machine will fall.
Suppose now that flying at the same point and under the
same conditions he wishes to descend, and decreases his angle
to 8 degrees. He will now have an excess of power of 3
horsepower as can be seen from the curves and will ascend
instead of descend. There is therefore a region of reverse
controls, known to French authors as the regime lent.
At the point M . when the pilot wishes to rise and in-
creases his angle of incidence, he does indeed obtain excess
power and rises. Here the controls are normal and the region
is known as regime rapide. For an inexperienced pilot the
regime lent is dangerous. Even if he knows the angle of in-
cidence at which he is working, he is likely to get into diffi-
culties.
With a flexible engine, an expert pilot can operate an air-
plane in the slow speed region by manipulation of the throttle
40 M 60 TO SO 90 100
MILES PEU HOUR
Fiu. 2. VARYING SPEED RANGE WITH ENGINE THROTTLED
alone. In Fig. 2 the propeller horsepower available is shown
with the engine throttled down to various speeds for a design
taken i'rom Dr. Hunsaker's pamphlet, to which reference is
appended. For each speed of the engine there is a different
maximum and minimum speed of the airplane, and a different
speed range. If the airplane is living at the minimum speed
in the regime lent region at a certain revolution per minute,
the pilot can by unthrottling his engine pass to a larger speed
!.-iii'_ r e, obtain excess power and climb without changing his
forward speed or angle of incidence. When an engine is
throttled the danger of reversed controls is still greater, lie-
cause the speed range becomes so very small. Kven the best
of pilots may mistake his position on the curve.
In French airplane contests, a premium has been placed on
low speeds, and the regime lent with throttling has been
largely and successfully used. Such operation does not seem | 6,268 | 23,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-22 | latest | en | 0.930829 |
https://www.physicsforums.com/threads/i-dont-understand-an-approximation-in-an-expression-in-stat-mech.774600/ | 1,508,708,140,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825464.60/warc/CC-MAIN-20171022203758-20171022223758-00591.warc.gz | 973,963,434 | 16,937 | # I don't understand an approximation in an expression in stat. mech.
1. Oct 5, 2014
### fluidistic
1. The problem statement, all variables and given/known data
Hello guys, I fail to understand a mathematical approximation I see in a solved exercise.
The guy reached a partition function of $Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]$ and he wants to analyze the case $T>> \omicron$.
He states that with the change of variables $x=l(l+1)\frac{\omicron}{T}$, $Z\approx \frac{T}{\omicron} \int _0^\infty e^{-x}dx$.
I really don't understand this last step.
2. Relevant equations
Sum becomes integral. The change of variables.
3. The attempt at a solution
Making the change of variables, I understand that the sum transforms into an integral and I also understand why the limits of the integral are 0 and infinity (because l goes from 0 to infinity and thus x too).
I am unable to perform the change of variables and get rid of l's outside the exponential.
I'd appreciate if someone shed some light. Thanks.
2. Oct 5, 2014
### vela
Staff Emeritus
I think the l's get sucked into what becomes dx since the derivative of l(l+1) is 2l+1.
3. Oct 5, 2014
### RUber
Notice that once you insert the substitution, $Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]$, $=\sum_{l=0}^\infty (2l+1) \exp [-x]$.
Then see that $dx=(2l+1)\frac{\omicron}{T}$.
So, $2l+1=dx\frac{T}{\omicron}$.
Since $T>>o$ the integral is a good approximation, since it is the limit as $l(l+1)o/T \to 0$.
Putting this all together, $Z = \int_{l=0}^\infty \frac{T}{\omicron} \exp [-x] dx$.
4. Oct 6, 2014
### fluidistic
Thank you guys. I missed the dx part indeed. | 537 | 1,699 | {"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} | 3.78125 | 4 | CC-MAIN-2017-43 | longest | en | 0.854429 |
https://www.tristansonlinephysicsnotes.com/springs-and-hooke-s-law | 1,696,409,493,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00258.warc.gz | 1,129,907,590 | 90,701 | top of page
Springs and Hooke's Law
Everyone deals with springs all the time. Go find one and play with it, a shoelace works enough. Pull on one end and you'll feel the same amount force on your other hand and see the spring stretch outward. Using Newton's 1st law, the force the spring produces is opposite what your hand produces, and the other end has the same situation as shown below.
It might be a bit confusing, but there are two ways to draw the free body diagram (FBD) for this situation. In a statics problem, the forces are balanced on every object in the system and therefore are balanced at every point (often called a node) in the system. The FBD on the objects is simple, and it tells us that all the forces we can see have the same magnitude, but nothing about the spring force, it only sees normal forces where it contacts the hands. The other diagram looks at the atoms at the ends of the springs, one side is pushing on the hand atoms and the other side is pulling on other spring atoms. The point of this was to show that the applied force on a spring is going to be equal and opposite to the internal forces, the middleman of normal forces can be ignored.
Looking at the point on the right, the applied force would be considered positive because it points in the positive x direction and the spring would be under tension. The material would stretch to the right as shown below.
As shown above, the restoring force points in the opposite direction of the applied force and displacement, so a positive displacement has a negative spring force (restoring force is the same thing). This gives the relationship discovered by Robert Hooke, referred to as Hooke's law. The variable "k" is called a spring constant, and is a function of material properties, shape, mechanical advantages, temperature, and more. Determining spring constant for basic shapes like solid bars is covered on the solid mechanics page (not written yet), but shapes like coil springs are far more complicated and not covered in basic physics courses, so empirical (meaning experimental) data will be used in your classes.
bottom of page | 444 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-40 | longest | en | 0.945739 |
https://notebook.community/BrentDorsey/pipeline/gpu.ml/notebooks/07_Optimize_Model | 1,722,850,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00238.warc.gz | 343,481,545 | 7,125 | # Optimize Trained Models for Inference
## Graph Transform Tool
Great Blog Post by Pete Warden from Google
## Types of Optimizations
• Remove training-only operations (checkpoint saving, drop out)
• Strip out unused nodes
• Remove debug operations
• Fold batch normalization ops into weights (super cool)
• Quantize weights
## Original Model (CPU)
### File Size
WARNING: If this directory doesn't exist, you need to train the model in an earlier notebook!!
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu/
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb
``````
``````
In [ ]:
from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
import re
from tensorflow.core.framework import graph_pb2
def convert_graph_to_dot(input_graph, output_dot, is_input_graph_binary):
graph = graph_pb2.GraphDef()
with open(input_graph, "rb") as fh:
if is_input_graph_binary:
else:
with open(output_dot, "wt") as fh:
print("digraph graphname {", file=fh)
for node in graph.node:
output_name = node.name
print(" \"" + output_name + "\" [label=\"" + node.op + "\"];", file=fh)
for input_full_name in node.input:
parts = input_full_name.split(":")
input_name = re.sub(r"^\^", "", parts[0])
print(" \"" + input_name + "\" -> \"" + output_name + "\";", file=fh)
print("}", file=fh)
print("Created dot file '%s' for graph '%s'." % (output_dot, input_graph))
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb'
output_dot='/root/notebooks/unoptimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/unoptimized_cpu.dot \
-o /root/notebooks/unoptimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/unoptimized_cpu.png', width=1024, height=768)
``````
## Strip Unused Nodes
``````
In [ ]:
%%bash
transform_graph \
--in_graph=/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb \
--out_graph=/root/models/optimize_me/linear/cpu/strip_unused_optimized_cpu.pb \
--inputs='x_observed' \
--transforms='
strip_unused_nodes'
``````
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/strip_unused_optimized_cpu.pb
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/strip_unused_optimized_cpu.pb'
output_dot='/root/notebooks/strip_unused_optimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/strip_unused_optimized_cpu.dot \
-o /root/notebooks/strip_unused_optimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/strip_unused_optimized_cpu.png')
``````
## Remove Nodes
Remove pesky `Identity` and `CheckNumerics`
``````
In [ ]:
%%bash
transform_graph \
--in_graph=/root/models/optimize_me/linear/cpu/strip_unused_optimized_cpu.pb \
--out_graph=/root/models/optimize_me/linear/cpu/remove_nodes_optimized_cpu.pb \
--inputs='x_observed' \
--transforms='
strip_unused_nodes
remove_nodes(op=Identity, op=CheckNumerics)'
``````
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/remove_nodes_optimized_cpu.pb
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/remove_nodes_optimized_cpu.pb'
output_dot='/root/notebooks/remove_nodes_optimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/remove_nodes_optimized_cpu.dot \
-o /root/notebooks/remove_nodes_optimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/remove_nodes_optimized_cpu.png')
``````
## Fold Constants
``````
In [ ]:
%%bash
transform_graph \
--in_graph=/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb \
--out_graph=/root/models/optimize_me/linear/cpu/fold_constants_optimized_cpu.pb \
--inputs='x_observed' \
--transforms='
strip_unused_nodes
remove_nodes(op=Identity, op=CheckNumerics)
fold_constants(ignore_errors=true)'
``````
### File Size
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/fold_constants_optimized_cpu.pb
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/fold_constants_optimized_cpu.pb'
output_dot='/root/notebooks/fold_constants_optimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/fold_constants_optimized_cpu.dot \
-o /root/notebooks/fold_constants_optimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/fold_constants_optimized_cpu.png')
``````
## Fold Batch Normalizations
Prereq: `fold_constants`
``````
In [ ]:
%%bash
transform_graph \
--in_graph=/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb \
--out_graph=/root/models/optimize_me/linear/cpu/fold_batch_norms_optimized_cpu.pb \
--inputs='x_observed' \
--transforms='
strip_unused_nodes
remove_nodes(op=Identity, op=CheckNumerics)
fold_constants(ignore_errors=true)
fold_batch_norms
fold_old_batch_norms'
``````
### File Size
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/fold_batch_norms_optimized_cpu.pb
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/fold_batch_norms_optimized_cpu.pb'
output_dot='/root/notebooks/fold_batch_norms_optimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/fold_batch_norms_optimized_cpu.dot \
-o /root/notebooks/fold_batch_norms_optimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/fold_batch_norms_optimized_cpu.png')
``````
## Quantize Weights
Prereq: `fold_batch_norms`
``````
In [ ]:
%%bash
transform_graph \
--in_graph=/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb \
--out_graph=/root/models/optimize_me/linear/cpu/quantize_weights_optimized_cpu.pb \
--inputs='x_observed' \
--transforms='
strip_unused_nodes
remove_nodes(op=Identity, op=CheckNumerics)
fold_constants(ignore_errors=true)
fold_batch_norms
fold_old_batch_norms
quantize_weights'
``````
### File Size
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu/
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/quantize_weights_optimized_cpu.pb
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/quantize_weights_optimized_cpu.pb'
output_dot='/root/notebooks/quantize_weights_optimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/quantize_weights_optimized_cpu.dot \
-o /root/notebooks/quantize_weights_optimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/quantize_weights_optimized_cpu.png')
``````
## Combine All Optimizations
``````
In [ ]:
%%bash
transform_graph \
--in_graph=/root/models/optimize_me/linear/cpu/unoptimized_cpu.pb \
--out_graph=/root/models/optimize_me/linear/cpu/fully_optimized_cpu.pb \
--inputs='x_observed' \
--transforms='
strip_unused_nodes
remove_nodes(op=Identity, op=CheckNumerics)
fold_constants(ignore_errors=true)
fold_batch_norms
fold_old_batch_norms
quantize_weights'
``````
### File Size
``````
In [ ]:
%%bash
ls -l /root/models/optimize_me/linear/cpu/
``````
### Graph
``````
In [ ]:
%%bash
summarize_graph --in_graph=/root/models/optimize_me/linear/cpu/fully_optimized_cpu.pb
``````
``````
In [ ]:
input_graph='/root/models/optimize_me/linear/cpu/fully_optimized_cpu.pb'
output_dot='/root/notebooks/fully_optimized_cpu.dot'
convert_graph_to_dot(input_graph=input_graph, output_dot=output_dot, is_input_graph_binary=True)
``````
``````
In [ ]:
%%bash
dot -T png /root/notebooks/fully_optimized_cpu.dot \
-o /root/notebooks/fully_optimized_cpu.png > /tmp/a.out
``````
``````
In [ ]:
from IPython.display import Image
Image('/root/notebooks/fully_optimized_cpu.png')
`````` | 2,518 | 8,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.567158 |
http://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt | 1,464,161,979,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274191.57/warc/CC-MAIN-20160524002114-00190-ip-10-185-217-139.ec2.internal.warc.gz | 188,558,626 | 19,483 | # Calculating :$((\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))…((\sqrt{3}+\tan(29^\circ))$
What is the easiest way to calculate :
$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ))$$
-
This is Problem J39 from Volume 4 of Mathematical Reflections. – Byron Schmuland Aug 30 '12 at 15:21
$\sqrt3+\tan(30^\circ-x)=\sqrt3+\frac{\frac{1}{\sqrt3}-\tan x}{1+ \frac{1}{\sqrt3}\tan x}$ $=\sqrt 3+\frac{1-\sqrt 3 \tan x}{\sqrt 3+\tan x}=\frac{4}{\sqrt 3+\tan x}$
$$\Longrightarrow(\sqrt 3+\tan(30^\circ-x))(\sqrt 3+\tan x)=4$$
Put $x=1,2,....,14^\circ$.
To complete, $x=15^\circ$, $(\sqrt 3+\tan(30^\circ-15^\circ))(\sqrt 3+\tan 15^\circ)=4$
$\implies (\sqrt 3+\tan 15^\circ)^2=4\implies \sqrt 3+\tan 15^\circ=2$ as $\tan 15^\circ>0$.
So, the answer should be $4^{14}\cdot 2=2^{29}$
A little generalization : Assuming $A≠n\frac{\pi}{2}$(where $n$ is any integer) so that $\cot A$ and $\tan A$ are non-zero finite,
$\cot A+ \tan(A-y)= \cot A+ \frac{\tan A-\tan y}{1+\tan A\tan y}=\frac{\cot A + \tan A}{1+\tan A\tan y}=\frac{\csc^2A}{\cot A + \tan y}$ (multiplying the numerator & the denominator by $\cot A$)
$\implies (\cot A+ \tan(A-y))(\cot A + \tan y)=\csc^2A$ , here $A=30^\circ$.
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+1 Very nice, concise, clear. – DonAntonio Aug 30 '12 at 9:28
How does one come up with this kind of stuff?? – NikolajK Aug 30 '12 at 9:41
@Nick Kidman: it's simple. You need some practice and the rest will come. – user 1618033 Aug 30 '12 at 10:14
Note that $\sqrt{3} = \tan(60^{\circ})$, so your product is $$(\tan(60^{\circ}) + \tan (1^\circ))\cdot((\tan(60^{\circ}) +\tan(2^\circ))\cdots((\tan(60^{\circ})+\tan(29^\circ)).$$ Now use the identity (which is very simple to prove by just writing $\tan$ in terms of $\sin$ and $\cos$ and adding the resulting fractions) $$\tan(A) + \tan(B) = \frac{\sin(A+B)}{\cos(A)\cos(B)}.$$ For your product, set $A = 60^{\circ}$, so we have $$(\tan(60^{\circ}) + \tan (1^\circ))\cdot((\tan(60^{\circ}) +\tan(2^\circ))\cdots((\tan(60^{\circ})+\tan(29^\circ)) \\ = \frac{\sin(61^{\circ})\sin(62^{\circ})\cdots\sin(89^{\circ})}{(\cos(60^{\circ}))^{29}\cos(1^{\circ})\cos(2^{\circ})\cdots\cos(29^{\circ})}.$$ And using $\cos(x)=\sin(90^{\circ}-x)$, this reduces to $$\frac{1}{(\cos(60^{\circ}))^{29}}=2^{29}.$$
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This is great. I think it's even more clear if you do the $\cos(x)=\sin(90°-x)$ step first: You show that $\sqrt{3} + \tan(x)=2\frac{\cos(30-x)}{\cos(x)}$, and so when the product makes $x$ run from $1°$ to $29°$, all the $\cos$'s cancel. – NikolajK Aug 30 '12 at 9:54
@Nick I see what you mean, the way you suggest I think is a better way to set it out in the end, but if you haven't solved the question yet and this is the first time you're seeing a solution to it, I think this is a clearer way to see just how the sin's and cos's cancel. Though, in the end, it's just a matter of personal preference I suppose. – Monty Gill Aug 30 '12 at 14:46
Maybe you wanna go this way:
$$P=(\sqrt{3} + \tan (1^\circ))((\sqrt{3} +\tan(2^\circ))\cdots((\sqrt{3}+\tan(29^\circ))$$ $$\frac{P}{2^{29}}=\frac{(\sqrt{3} \cos(1^\circ) + \sin (1^\circ))}{2\cos(1^\circ)}\frac{(\sqrt{3} \cos(2^\circ) + \sin (2^\circ))}{2\cos(2^\circ)}\cdots\frac{(\sqrt{3} \cos(29^\circ) + \sin (29^\circ))}{2\cos(29^\circ)}$$
Above I used the fact that $$\sin(60^\circ+\alpha^\circ)=\frac{\sqrt3}{2}\cos \alpha^\circ+\frac{1}{2}\sin \alpha^\circ$$ then $$\frac{P}{2^{29}}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\cos1^\circ\cos2^\circ\cdots\cos29^\circ}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\sin89^\circ\sin88^\circ\cdots\sin61^\circ}=1$$ $$P={2^{29}}.$$
Q.E.D.
- | 1,457 | 3,640 | {"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} | 4.34375 | 4 | CC-MAIN-2016-22 | longest | en | 0.585108 |
https://semisignal.com/tag/stipple-patterns/ | 1,725,810,193,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00849.warc.gz | 493,784,894 | 10,061 | ## Posts Tagged ‘stipple patterns’
### Circular stipple patterns on an HTML5 canvas
In my previous post on stipple patterns, I presented code to draw a few simple stipple patterns based on drawing single pixels at fixed locations. In this post, I’ll present something just a bit more complex: drawing circles to create a circular stipple patten, again writing a shader that makes use of the GraphicsCore and FXController classes.
`Shader.circleStippleShader = function (imageData, bufWrite, index, x, y, r, g, b, a, passNum, frameNum, maxFrames){ var alpha = 1.0; var r1 = r / 255.0; var rF = Math.floor((alpha * r1 + (1.0 - alpha)) * 255.0); var circleMaxDiam = 12; // circle at every 12th pixel, also defines max diameter of circle, var circleMaxRadius = circleMaxDiam / 2; // maximum radius of circle // figure out the x, y indices of the circle we're within // x,y need to be shifted by the circle radius b/c circleMaxDiam defined the offset of the circle center // ... e.g. going along the x-axis, we are within the next circle not at x/circleMaxDiam, but at (x+6)/circleMaxDiam var iX = Math.floor((x + circleMaxRadius) / circleMaxDiam); var iY = Math.floor((y + circleMaxRadius) / circleMaxDiam); // multiply the circle indices by the diameter to get the actual coordinates of the circle's center var targetX = iX * circleMaxDiam; var targetY = iY * circleMaxDiam; // calculate squared distance to the circle we are within var dist = (targetX - x) * (targetX - x) + (targetY - y) * (targetY - y); if (dist < 25) { GraphicsCore.setPixel(bufWrite, index, rF, 0, 0, 255); } else { GraphicsCore.setPixel(bufWrite, index, r, g, b, 255); }}Shader.circleStippleShader.numPassesRequired = 1; `
Conceptually, we define the center of a circle at every 12th pixel (both along the x and y axis). At every pixel (x,y) we figure out which circle we are within, and calculate the distance to the center. If the distance is less than our threshold (25), we change the color of the pixel (use the red channel only).
### Stipple patterns on an HTML5 canvas
I wanted to play around a bit with stipple patterns after seeing stippling done with photos on the LinkedIn news feed. However, what I’m going to present is not what LinkedIn does. LinkedIn applies the stipple pattern as a background-image on a DOM element above a <img> element with a (fairly low resolution) JPEG – the stippling may help to alleviate the negative visual impact of the low-resolution image. What I’m going to show is how to do stippling on an HTML5 canvas, which allows for a much greater degree of freedom in terms of what’s possible, but is also slower and requires a modern browser.
I’m going to make use of the GraphicsCore and FXController classes in a previous post, Gaussian blur on an HTML5 canvas. In that post I presented the concept of writing shaders as plug-in to the FXController class to apply different per-pixel effects. What I’m going to present are shaders for a few simple stipple patterns. Applying the shader is simply a matter of passing it into the constructor for the FXController class, e.g.
`var theShader = Shader.crossStippleShader;var fxCtrlr = new FXController(ctxSource, ctxDest, theShader, width, height, 100, 1);fxCtrlr.init();`
### Checkerboard Stipple
This shader has the effect of creating a checkerboard pattern.
The source pixel is preserved if (x+y) % 2 == 0, otherwise the pixel’s alpha is reduced to 66.
`Shader.checkerboardStippleShader = function (imageData, bufWrite, index, x, y, r, g, b, a, passNum, frameNum, maxFrames){ if( (x+y)%2 == 0) { GraphicsCore.setPixel(bufWrite, index, r, g, b, 255); } else { GraphicsCore.setPixel(bufWrite, index, r, g, b, 66); } }Shader.checkerboardStippleShader.numPassesRequired = 1; `
### Dot Stipple
This shader blends a white pixel into the source image where x%2 == 0 && y%2 == 0, in effect creating a dotted grid pattern.
The alpha blending code is a straightforward implementation of alpha compositing, but since we’re blending with white (where r=1.0, g=1.0, and b=1.0) the equation is simplified and there is no second color value; we’re just biasing the source color by the alpha value. Also note that this is different than simply changing the alpha of the source pixel (as was done in the Checkerboard Stipple shader), here we’re always blending with white, in the previous shader we’re blending with whatever is the background of the DOM element.
`Shader.dotStippleShader = function (imageData, bufWrite, index, x, y, r, g, b, a, passNum, frameNum, maxFrames){ var alpha = 0.8; var r1 = r / 255.0; var rF = Math.floor((alpha*r1 + (1.0-alpha)) * 255.0); var g1 = g / 255.0; var gF = Math.floor((alpha*g1 + (1.0-alpha)) * 255.0); var b1 = b / 255.0; var bF = Math.floor((alpha*b1 + (1.0-alpha)) * 255.0); if( x%2 == 0 && y%2 == 0) { GraphicsCore.setPixel(bufWrite, index, rF, gF, bF, 255); } else { GraphicsCore.setPixel(bufWrite, index, r, g, b, 255); } }Shader.dotStippleShader.numPassesRequired = 1;`
### Quincunx Stipple
With this shader we blend in a white pixel at every 4 pixels (x%4 == 0 && y%4 == 0, the target pixel) and also at the 4 orthogonally adjacent pixels around the target, creating a quincunx pattern.
`Shader.quincunxStippleShader = function (imageData, bufWrite, index, x, y, r, g, b, a, passNum, frameNum, maxFrames){ var alpha = 0.78; var r1 = r / 255.0; var rF = Math.floor((alpha*r1 + (1.0-alpha)) * 255.0); var g1 = g / 255.0; var gF = Math.floor((alpha*g1 + (1.0-alpha)) * 255.0); var b1 = b / 255.0; var bF = Math.floor((alpha*b1 + (1.0-alpha)) * 255.0); if( (x%4 == 0 && y%4 == 0) || ((x+1)%4 == 0 && y%4 == 0) || ((x-1)%4 == 0 && y%4 == 0) || (x%4 == 0 && (y+1)%4 == 0) || (x%4 == 0 && (y-1)%4 == 0) ) { GraphicsCore.setPixel(bufWrite, index, rF, gF, bF, 255); } else { GraphicsCore.setPixel(bufWrite, index, r, g, b, 255); } }Shader.quincunxStippleShader.numPassesRequired = 1;`
### Cross Stipple
Similar to the quincunx stipple, but we blend in a white pixel at every 6 pixels (x%6 == 0 && y%6 == 0, the target pixel), the 4 orthogonally adjacent pixels around the target, and 4 additional pixels extending beyond the orthogonals, creating a cross (“+”) pattern.
`Shader.crossStippleShader = function (imageData, bufWrite, index, x, y, r, g, b, a, passNum, frameNum, maxFrames){ var alpha = 0.78; var r1 = r / 255.0; var rF = Math.floor((alpha*r1 + (1.0-alpha)) * 255.0); var g1 = g / 255.0; var gF = Math.floor((alpha*g1 + (1.0-alpha)) * 255.0); var b1 = b / 255.0; var bF = Math.floor((alpha*b1 + (1.0-alpha)) * 255.0); if( (x%6 == 0 && y%6 == 0) || ((x+1)%6 == 0 && y%6 == 0) || ((x-1)%6 == 0 && y%6 == 0) || ((x+2)%6 == 0 && y%6 == 0) || ((x-2)%6 == 0 && y%6 == 0) || (x%6 == 0 && (y+1)%6 == 0) || (x%6 == 0 && (y-1)%6 == 0) || (x%6 == 0 && (y+2)%6 == 0) || (x%6 == 0 && (y-2)%6 == 0) ) { GraphicsCore.setPixel(bufWrite, index, rF, gF, bF, 255); } else { GraphicsCore.setPixel(bufWrite, index, r, g, b, 255); } }Shader.crossStippleShader.numPassesRequired = 1; `
That’s all for now. There’s tons of variations possible with only minor code changes to alter blending, color, and the shape of the stipple patten. | 2,358 | 7,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.683554 |
http://ac-immacolata.it/turr/cylinder-cone-and-sphere-formulas.html | 1,596,897,036,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00357.warc.gz | 3,328,119 | 17,361 | ### Cylinder Cone And Sphere Formulas
Specifically, the cylinder's volume formula is and the cone's volume formula is. The volume of a cylinder. Volume of a partial sphere. Enter one value and choose the number of decimal places. Find the total surface area (including the base) and volume of the tank. Area of a sphere: 4 x 3. The circumference of a circle is always taken as the important concept in Geometry and Trigonometry. As it is in circular shape so it has diameter and radius. How to use the volume formulas to calculate the volume. In the definition of a prism, the bases were forced to be congruent polygons. Pyramids, prisms, cylinders and cones - YouTube. A round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point (the centre). Anil Surface Area of a Cone Formula. Since the diameter of a sphere equals two radii or d = 2r, you can use another form of the equation which is: A = 4 * π * (d / 2)² = π * d². Surface Area of Cylinders. Pyramids, prisms, cylinders and cones - YouTube. EXAMPLE 1 Finding the Volume of a Cylinder and a Cone Topic 2 Volume Volume of a Cylinder Words The volume V of a cylinder is the product of the area of the base and the height of the cylinder. The radius of each cylinder is the same at an intersection point so an appropriate sphere still fills the gaps. Enter one value and choose the number of decimal places. Each penny is 0. Male or Female ? Volume of a partial right cylinder. Volume Formulas Volume Formulas : We know that solid occupies some region in space and the magnitude of this region is called the Volume of the solid. How to Find the Surface Area and Volume of Prisms - From the TCS high school mathematics 'How-to Library'. Each shape will have a unique formula, but in addition to the formula, you'll need measurements of the height (h) and radius (r). See more ideas about Geometry, Teaching math and 8th grade math. The latter is π r³, making the volume of the sphere 4/3 π r³. Watch this video and write down the formulas: Circle: Cylinder: Cone: Sphere:. A visual explanation of the volume of a cone, a sphere and a cylinder. Practice: Volume of cylinders, spheres, and cones word problems Volume formulas review Math · Basic geometry · Volume and surface area · Volume of cones, cylinders, and spheres. Find the volumes of the shapres below. Cylinder Volume Calculator The measurements that are inserted by the user will be used by the online Cylinder calculator to compute electronically the volume of the cylinder. Visualizing the Volume of a Sphere Formula | Deriving the Algebraic Formula With Animations - Duration: (solid shapes song- including sphere, cylinder, cube, cone, and pyramid) - Duration: 3:49. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. For any point S on the diameter AC of the sphere, suppose we look at a cross section of the three solids obtained by slicing the three solids with a plane containing point S and parallel to the base of the cylinder. So we want a cone area formula for just the curved surface of the cone. Then imagine cutting both the sphere and cylinder at any. We like to find the volume of the cylinder which is outside the cone (yellow portion). 14)(3)²(6) ∴ The volume of the cylinder = 169. Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. OP is the radius of the sphere. Volume of a Sphere and/or Cone Interactive. Since a cylinder's volume formula is V = Bh, then the volume of a cone is one-third that formula, or V = Bh/3. The reference area for a cylinder is also the same as that used for the disk except the radius is that of the cylinder's circular cross-section. What Archimedes discovered was that if the cross-sections of the cone and sphere are moved to H (where. But why is a sphere's surface area four times its shadow? How many Liters are there in one meter cube Solve Volume of Cylinder - Duration: 6:42. Make sense of problems and persevere in solving them. Surface Area of Sphere and Hemisphere Here we will discuss surface area of sphere and hemisphere Sphere is nothing but a any type of ball. So, the volume of a cone is equal to the volume of the first disk times the sum of all the cylinders, which we can get using the summation of squares formula. To end the Mini-Lesson, I show the students the formula for finding the volume of a sphere. Without him, we can not talk about Picasso or Matisse because he was the one to have moved on from impressionism to post-impressionism and broke classic styles of art forms to lead the next. The diagram shows a solid cone. We get: (Vcone + Vsphere)r = (Vcylinder) r 2. Find the total surface area (including the base) and volume of the tank. Let's fit a cylinder around a cone. The only information needed is the length of the interior wall that's left in the sphere after the core has been drilled out and one (or more) common geometric formula(s). The results we provide are accurate, but. Use the formulas below to find the volume of many different geometric shapes. Geometry - Volume and surface area of spheres. Diameter : A line segment through the center of a sphere and with the end points on its boundary is called its diameter. The radio station is giving the winner of this challenge a prize pack that includes tickets to see his or her favorite band in concert. The formula for calculating the surface area of sphere is given by: SA = 4 × π × r2. Cylinder The height is 8 inches and the radius is 2 inches. ICSE Class 10 Maths Mensuration: Cylinder, Cone and Sphere. Standard(s): 8. 2) An ice cream package is cone shaped. A sphere has no faces, a cone has one circular face, and a cylinder has two circular faces. 1415, and r is the radius of the circle. Weight Density formula is given by The Density is expressed in Kg/L. The volume of a cylinder: V=π*r^2*h Volume of a cone: V=π*r^2*(h/3) Volume of a sphere: V=(4/3)π*r^3. know the volume formulas for cones and cylinders. Perimeter of an ellipse: 4 exact series and a dozen approximate formulas! Vertical fall against fluid resistance (including viscous and quadratic drag). Some other standard units of Volume are mm 3, dm 3, m 3 and km 3. 4 Volumes of Prisms and Cylinders. Use the formulas for the volumes of cylinders, cones, and spheres to solve a variety of real-world problems. Right, acute, and obtuse. Spherical Zone. Write the letter of the answer that matches the problem. Circle Formulas. Archimedes showed that not only are these total areas equal, but the areas cut off by any planes perpendicular to the. Practice: Volume of cylinders, spheres, and cones word problems Volume formulas review Math · Basic geometry · Volume and surface area · Volume of cones, cylinders, and spheres. The volume of a cube = side times side times side. + 2B = 2?r·h + 2·?r2 The variable “r” stands for the RADIUS OF THE BASE. Surface area is calculated by adding up all external sides of the cylinder. Cone: A solid with a circular base and a curved side that ends in one point; It has one vertex. What is the radius of the sphere? Ex: Esther and Jasmine each bought ice cream from Chilly’s Ice Cream Parlor. As students complete this lesson, a firmer grasp of the big ideas should be seen, and. As it is in circular shape so it has diameter and radius. Volume and Surface Area of a Sphere. How Many Cones Does It Take To Fill a Sphere? In this 3 act math task, the teacher will show short video clips to help students understand where the Volume of a Sphere formula comes from. If the sphere is flattened, it will fill. Sphere Surface area of a sphere: S=4πr2 Where S is the surface area of a sphere and r is the radius. 61 cm3 _____ 2. In this example, r and h are identical, so the volumes are πr 3 and 1 ⁄ 3 π r 3. When we measure liquids, the unit that we use commonly is liter. Over 70 formulas included. Over the past year, I have been on a mission to try and make some of the formulas we use in the intermediate math courses in Ontario (Middle School for our friends in the U. A cone is ⅓ the volume of a cylinder, or 1⁄3πr²h. Use this volume calculator to easily calculate the volume of common bodies like a cube, rectangular box, cylinder, sphere, cone, and triangular prism. The cross-sectional area of a thin layer with a vertical distance a (same as figure 1) from the center of the base consists of two concentric circles. Sphere Formulas Where r is the radius of the sphere Cone Formulas Where r is the radius of cone, h is the height of cone and s is the slant height of the cone. What is the relationship between the volume of a cone and cylinder when they both have the same radius and height?. Thus, the sphere had diameter , the cylinder had diameter and height , and the cone had base diameter and height (see Figure). Find the volumes of the shapres below. 264 Lesson 26 U nerstan Volume of Cylinders, Cones, and Spheres ©urriculum ssociats opyig is ot prmitt Solve. Lines that never, ever cross. We like to find the volume of the cylinder which is outside the cone (yellow portion). Annular ring on cylinder base or top to inside of right circular cylinder. Firstly, determine the volume of a cone and volume of sphere using formula, volume of sphere = 4/3 πr 3 and volume of cone = 1/3 πr 2 h Further, equate volume of a cone and volume of sphere; so that we get the radius of sphere. Volume of a partial hemisphere. We discuss parts of the formula and how it relates to the area of a circle. 61 cm3 _____ 2. I know the formulas for finding the volume of cylinders, cones, and spheres; I can find the volume of a cylinder. The formula for volume is 4/3 pi r cubed. You will get the conical cylinder when the edge of the cylinder is cut off. In the figure above, drag the orange dot to change the radius of the sphere and note how the formula is used to calculate the volume. Thus, a rheometer can be considered as a special type of viscometer. 2) (A) a cone and a cylinder (B) a hemisphere and a cone (C) frustum of a cone and a cylinder (D) sphere and cylinder 4. know the volume formulas for cones and cylinders. Complete the equations to find the volumes of the sphere and the cylinder in terms of p. I can find the volume of a cone. Drag into world. Title: Volume of a Cylinder, Cone, and Sphere 1 Volume of a Cylinder, Cone, and Sphere. Pyramids, prisms, cylinders and cones - YouTube. Use this calculator to easily calculate the surface area of common bodies like a cube, rectangular box, cylinder, sphere, cone, and triangular prism. By (date), when given (5) real-life problems on calculating the volume of a solid, examples on finding the volumes of cones, cylinders, and spheres (e. Anil Surface Area of a Cone Formula. Volume of Cylinders, Cones, and Spheres Essential Question: How do you find the volume of a cylinder, cone, and sphere? Learning Goal: Students will know the formula for the volume of cylinders, cones, spheres and use them to solve real-world and mathematical problems. S Formula for surface area of a sphere= 4πr2 324π = 4πr2 Substitute 324π for S. You will need to refer to the reference sheet as you do the problems. The radius of a sphere is half of its diameter. Volumes of Pyramids and Cones. Volume of a right cone: V= 1 3 Bh or V= 1 3 πr2h where B is the area of the base and h is the height of the cone. Then click Calculate. Comparing Spheres, Cones and Cylinders Video. Archimedes Discovers the Volume of a Sphere. The formula for the Volume of Preview this quiz on Quizizz. A bucket of height 8cm and made up of copper sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as 3cm and 9m respectively. Using this calculator, we will understand methods of how to find the surface area and volume of the hemisphere. First section (Green) has cylinders in. This can be calculated by adding the volume of cylinder and cone. 9: Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. If you do not know it, you can find the side length (s) using the radius (r) and the cone's height (h). Second section (Amber) has a cone and sphere in. Some other standard units of Volume are mm 3, dm 3, m 3 and km 3. volume of the hollow cylinder πr2 2h — πr2 1h = π(r2 2 – r2 1)h. OP is the radius of the sphere. Volume formulas. Equations The drag force equation used for the calculation on this page is (Blevins, 2003 and Munson et al. Volume of the frustum. Volume of a sphere. Your job is to accurately construct the box to package the speakers. Find a missing measurement (height, radius, or. Watch this video and write down the formulas: Circle: Cylinder: Cone: Sphere:. The formula for the volume of a cone is 1/3*pi*r^2*h. Schaum's Outline series in Mathematics. Find the Volume Cone 6. 1 To find the surface area of a prism and a cylinder NAT: CC G. All of the dimensions shown in blue are equal. Volume of Core: 1/3 π r² 2r. Please practice hand-washing and social distancing, and check out our resources for adapting to these times. Find the Volume Pyramid 12m rr2h Cylinder 5. Second section (Amber) has a cone and sphere in. First section (Green) has cylinders in. Calculate area of Cylinder from radius. A cone is a solid three-dimensional geographical figure with a flat circular base (or roughly circular base) from which it tapers smoothly to a point known as the vertex or apex. Processing. Volumes of Cones, Cylinders, and Spheres - Matching Worksheet Match the word problems to their answers. GCSE Maths Formula Sheet Rules of Indices Rule 1: When you multiply indices of the same number you add the powers. 9: Know the formulas for the volume of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. Calculations at a sphere. The 'base' of the cone will be at the top of the cylinder, and the point at the bottom will be at the center of the hemisphere. Graphing Complex Solutions. Cylinder, Cone, and Sphere Volume - Duration: 2:32. Find a missing measurement (height, radius, or diameter) for a cylinder, cone, or sphere given the volume. But why is a sphere's surface area four times its shadow? How many Liters are there in one meter cube Solve Volume of Cylinder - Duration: 6:42. Formulas for SA and Volume of Cylinder,Cone, and Sphere. The formula for the volume of a sphere is 4⁄3πr³. Â Let us begin with figures like Cylinders, Cones, and Spheres. Physical and chemical changes Essay Challenge #1 — The Penny Problem: The first challenge to complete is the Penny Problem. Find the flux of F = zi +xj +yk outward through the portion of the cylinder x2 +y2 = a2 in the first octant and below the plane z = h. To start off the challenge, the radio station has placed pennies in a cylindrical glass jar. Math · Basic geometry · Volume and surface area · Volume of cones, cylinders, and spheres. Volume of an ellipsoid. Whether it's a sphere or a circle, a rectangle or a cube, a pyramid or a triangle, each shape has specific formulas that you must follow to get the correct measurements. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. have to creat a geosolid class that has 3 sub class i choose sphere cylinder and cone, i got the first two to work but my cone class works except when it calculates the volume it keep getting 0 no matter what number is use for the hieght and radius. Surface area of cylinders, spheres, and cones. mensuration formulas In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like cylinder, cone and sphere. Archimedes specified that the density of the cone is four times the density of the cylinder and the sphere. This foldable is the perfect way to introduce calculating the volume of cylinder, cones, and spheres. Volume of a Cylinder Volume of a Cylinder. Simple worksheet showing formulas for spheres and cones. A cone is one third of the volume of a cylinder. Class Activity: Volume of a Cone Worksheet (you need Volume of a Sphere and/or Cone Interactive to do this) MORE PRACTICE. Processing. Finally, we looked at spheres, like a basketball and the planets. 29) Volume and surface area of spheres (Eighth grade - Q. List of 3D inertia tensors. I can find the volume of cylinders, cones, and spheres in real world problems. Visualizing the Volume of a Sphere Formula | Deriving the Algebraic Formula With Animations - Duration: (solid shapes song- including sphere, cylinder, cube, cone, and pyramid) - Duration: 3:49. Find a mix of spheres and hemispheres in these 3D shapes worksheets. Free download of step by step solutions for class 10 mathematics chapter 20 - Cylinder, Cone and Sphere (Surface Area and Volume) of ICSE Board (Concise - Selina Publishers). Volume of a hollow cylinder Volume of an elliptic truncated cone. Rewrite as. Students then have three sheets green, yellow and red which they can attempt. In the good old days the students didn't have to memorize the formulas, but those days are gone. Volume of a hemisphere = 2 / 3 πr3. The cross-sections are all circles with radii SR, SP, and SN, respectively. First of all, Imagine how a cylinder is made. Some of the worksheets for this concept are Spheres date period, Volumes of solids, Infinite pre algebra, Volume, Volume, Volume, Volume, Lesson 48 pyramids cones and spheres. The formula for the volume of a cone is: Since the base area is a circle, again we can substitute the area formula for a circle into the volume formula, in place of the base area. have to creat a geosolid class that has 3 sub class i choose sphere cylinder and cone, i got the first two to work but my cone class works except when it calculates the volume it keep getting 0 no matter what number is use for the hieght and radius. 0 Equation Volume of a Cylinder, Cone, and Sphere Volume Cylinder Previous Formulas Learned Area and Circumference of a Circle Cylinder Volume of a Cylinder Volume of a Cylinder Volume of Cylinders Class Practice Volume of Cylinders. Example: The volume of a sphere with a radius of length 3 inches: Sphere Volume = 4/3 x 3. Cone volume= 1/3Bh or 1/3( πr2)h 5. On the standard cone there is an edge between the nose and the cylinder which forms the body of the rocket. For example, to calculate the volume of a cone with a radius of 5cm and a height of 10cm: The area within a circle = πr2 (where π (pi) is approximately 3. Sphere: A solid figure where all points are an equal distance from the center point. 436 cubic inches and the height is 40 inches? Formula of the Volume of a Sphere: _____ 6in 8in 16 m 360in. asked by Maya on March 22, 2012; Algebra 1--How Am I Supposed to Solve This?! Suppose a cone and a cylinder have the same radius and that the slant height l of the cone is the same as the height h of the. By (date), when given (5) real-life problems on calculating the volume of a solid, examples on finding the volumes of cones, cylinders, and spheres (e. A great question to investigate, especially for honors sections: Why doesn't the formula "sphere volume= 2/3Bh" work? (does a sphere have a base or height—it only has a diameter). The measure of space that an object or material occupies is called as volume. Learn with flashcards, games, and more — for free. (Try to imagine 3 cones fitting inside a cylinder, if you can!). Cylinders and Cones. The density of the material is derived from its specific gravity. The circumference of a Sphere Formula in mathematics could be given as – Where, π=3. Geometric figure with volume and surface formulas – stock illustration Vector illustration, line design — Vector by kovalto1. The formula for the volume of a sphere is 4⁄3πr³. (A) a sphere and a cylinder (B) a hemisphere and a cylinder (C) two hemispheres (D) a cylinder and a cone. Multiply the area of the base times the height. Roughly 1/4 of the space is empty from the optimal packing. They notice that the circle has the same. ” is a famous saying Paul Cézanne wrote for the next generation (Wikipedia, Paul Cézanne 2015). Circumference formula. Surface area of a sphere The surface area formula for a sphere is 4 x π x (diameter / 2) 2, where (diameter / 2) is the radius of the sphere (d = 2 x r), so another way to write it is 4 x π x radius 2. Volume calculator will determine the volume of the most common geometric solids. 3) I have a snow globe with a diameter of 3 cm. 9 - Volume and 3D Shapes - Solve real-world and mathematical problems involving volume of cylinders, cones and spheres. Typically these formula are written as V = Bh (prism or cylinder), V =(1/3) Bh (pyramid or cone), or V =(4/3) r 3 (sphere). The reference area for a cylinder is also the same as that used for the disk except the radius is that of the cylinder's circular cross-section. It's a ball. 9 cm; Possible explanation: The paperweights have the same radius, so a cone with a Sphere Cylinder Cone V 5 4 p ··3 r3 VBh V 5 5 1 Bh. Annular ring on cylinder base or top to inside of right circular cylinder. Flux through a cylinder and sphere. Notice how, if we add up subtract multiply the volume of the cone and the sphere, we get exactly the volume of the cylinder! Surface Area of a Sphere. (SHOW YOUR WORK and don't forget the appropriate units. A neat relationship between the volume of a sphere and a cylinder. Geometry - Surface Area - Prisms, Cylinders, Cones, Pyramids and Spheres. com's basic geometry & shapes calculators, formulas & examples to deal with length, area, surface, volume, points, lines, dimensions, angles & curves calculations of 2 or 3 dimensional (2D or 3D) geometric shapes. Projection of a Sphere - Central Projection. 2 Surface Area and Volume of Spheres. Cylinder Volume Calculator The measurements that are inserted by the user will be used by the online Cylinder calculator to compute electronically the volume of the cylinder. Find a missing measurement (height, radius, or. cylinder = b h = r 2 h pyramid = (1/3) b h cone = (1/3) b h = 1/3 r 2 h sphere = (4/3) r 3. A capsule is a three-dimensional geometric shape comprised of a cylinder and two hemispherical ends, where a hemisphere is half a sphere. Round to the nearest hundredth. Then compare the volumes. Using this calculator, we will understand methods of how to find the surface area and volume of the sphere. mensuration formulas In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like cylinder, cone and sphere. Volume formulas for cone, cube, sphere, cylinder - name it and you will get the information in the articles below. V = 1 3π ⋅ r2 ⋅ h. Plus, if we write the ratio as 2 to 3, that will help teachers recognize that this question is an opportunity to equate two cones with the volume of a sphere. Volume of Pyramids, Cones and Spheres Square pyramid (begin{array}{rl} text{Volume}& = cfrac{1}{3}times text{area of base} times \ &. But why is a sphere's surface area four times its shadow? How many Liters are there in one meter cube Solve Volume of Cylinder - Duration: 6:42. Finally, we looked at spheres, like a basketball and the planets. To improve this 'Volume of a circular truncated cone Calculator', please fill in questionnaire. The volume of a cube = side times side times side. A cone is ⅓ the volume of a cylinder, or 1⁄3πr²h. Learn with flashcards, games, and more — for free. 7 A cylinder and a cone have the same radius The volume of the cone is twice the volume of the cylinder How many times greater is the height of the cone than the height of the cylinder? Explain your reasoning 270 M M M C pr2h 8pr2h 8 p ··3 r2h 2 3 pr2h Solid D, solid A, solid B, solid C; Possible explanation: All of the volumes have the term. *Use volume formulas to solve real-world problems involving spheres, cones and cylinders Volume: The amount of space an object takes up *Can always find volume by doing Area of the Base (B) times the height Units: units³ Formulas: ***CANNOT convert 𝟏 𝟑 to a decimal · Cylinder – V = πr²h r = radius, h = height · Sphere – 𝑉=4 3. Surface Area of a Cylinder Right circular cylinder and its radius , height. Volume Cones Spheres And Cylinders Answer Key. Second section (Amber) has a cone and sphere in. Presentation Summary : Volume of Pyramids, Cones, and Sphere. Drag coefficient of blunt nose and rounded nose cylinders versus fineness ratio l/d. In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. STANDARDS OF MATHEMATICAL CONTENT: Solve real ‐ world and mathematical problems involving volume of cylinders, cones, and spheres. Geometric figure with volume and surface formulas – stock illustration Vector illustration, line design — Vector by kovalto1. Animation-The Cylinder Answers Use the same technique as with the Sphere to have the Cylinder bob up and down before answering the Sphere. Physical and chemical changes Essay Challenge #1 — The Penny Problem: The first challenge to complete is the Penny Problem. Sphere with a diameter of 6 inches. Spherical Cap. We have created a simple visualization using which you can remember the formula of these figures. I have not asked them to memorize the volume formula of a cylinder, but after day 1, they all know it. Frustum Pyramid. Verify the answer using the formulas for the volume of a sphere, and for the volume of a cone, In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. 3X3 System Possibilities. The top half of the cone can be written as. , mensuration is a heavily tested topic. The problem can be generalized to other cones and n-sided pyramids but for the moment consider the right circular cone. We assume you know the volume of this cylinder: volume is area of the base multiplied by height. Volume and sufrace formulas for all geometric figures - formula, cube, parallelepiped, pyramid, cone, sphere, cylinder. Cone is geometric body which is formed by a set of rays from apex (vertex) point and cross any flat surface, and at the intersection form basis of the cone. Online calculators and formulas for a surface area and other geometry problems. This volume calculator used to calculate the various simple shapes of volume such as cone, cube, ball, cylinder and rectangular tank using the known values. Cylinder, Cone, and Sphere Volume - Duration: 2:32. Volume of a sphere. Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Find the volume of the ice cream cone. First section (Green) has cylinders in. Circle Formulas. Tags: Question 19 answer choices. VOLUME OF A CONE:_____ Practice problems: 18mm What is the radius of the circular base if the volume of the cone is 5068. The formula for calculating the surface area of sphere is given by: SA = 4 × π × r2. 3D Problem Solving - Sphere resting inside of a Cone. This free volume calculator can compute the volumes of common shapes, including that of a sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. 416 km² 19) A sphere with a diameter of 20 yd. The sphere is a space figure having all its points an equal distance from the center point. Mathematical Handbook of Formulas and Tables. Volume of Sphere Volume of Cylinder V 5 4 3 ··3 pr V 5 pr2h. I know the formulas for finding the volume of cylinders, cones, and spheres; I can find the volume of a cylinder. The volume of a cone is ¹/₃ × π × r² × l. 4 Volumes of Prisms and Cylinders. Volume of Cylinder: π r² h. Now, when my students practice using the formulas for volume of cylinders, cones, and spheres, we do a lot of practice focused on understanding the formulas. Example Question #2 : Cylinders, Pyramids, Cones, And Spheres Volume Formulas: Ccss. 3) I have a snow globe with a diameter of 3 cm. Surface Area of Pyramids and Cones. Then click Calculate. Watch this video and write down the formulas: Circle: Cylinder: Cone: Sphere:. The volume is the amount of total space on the interior of the solid. Male or Female ? Volume of a partial right cylinder. In this article, the lateral surface area of different figures including cuboid, cube, cylinder, cone, and sphere. Cylinder, Cone, and Sphere Volume - Duration: 2:32. They both have just one base and they converge to a point, the vertex. Volume Cones Spheres And Cylinders Answer Key - Displaying top 8 worksheets found for this concept. (Try to imagine 3 cones fitting inside a cylinder, if you can!) Now let's fit a cylinder around a sphere. The formulas for calculating Surface Area and Volume are. Volume of a sphere. First section (Green) has cylinders in. Your spheres are each V_s= 4/3*pi*d^3. A cylinder shaped jar has a radius of 2 cm and a height of 6 cm. Thus, a rheometer can be considered as a special type of viscometer. cube = a 3. 28) Volume of pyramids and cones (Eighth grade - Q. The principal formulae derived in On the Sphere and Cylinder are those mentioned above: the surface area of the sphere, the volume of the contained sphere, and surface area and volume of the cylinder. Single dimensions and two dimensions shapes like straight line or square, circle, triangle have zero volume in three dimensional space. For a right circular cone calculator click here. Make sense of problems and persevere in solving them. There are proven benefits of this cross-lateral brain activity: - new learning - relaxation. Okay, a cylinder neatly encloses a sphere, so that the curve of the cylinder touches the sphere in a circle at its widest, and the top and bottom faces of the cylinder are tangent to the sphere. Ex: The volume of a sphere is 113. cylinder covered by a half sphere as shown. For example: 54x 53= 54+3= 57 Rule 2: When you divide indices of the same number you subtract the powers. Volume Of Pyramids, Cones, And Sphere PPT. Volume Formulas Volume Formulas : We know that solid occupies some region in space and the magnitude of this region is called the Volume of the solid. All of the dimensions shown in blue are equal. 9 Math Scale: Notes: Volume of a Cylinder:. A cone, sphere and cylinder of radius r and height h The above formulas can be used to show that the volumes of a cone , sphere and cylinder of the same radius and height are in the ratio 1 : 2 : 3 , as follows. Substituting in the frustum volume formula and simplifying gives: Now, use the similar triangle relationship to solve for H and subsitute. Algebra V = Bh = π r 2h Volume of a Cone. From the above figure, you can see that the curved surface area of cylinder is equal to the area of rectangle taken intially. know the volume formulas for cones and cylinders. One of the standard unit of volume is cubic centimeter (cm 3). Surface area of a cone - derivation. Review the formulas for the volume of prisms, cylinders, pyramids, cones, and spheres. Bellwork Students will turn in homework. If the volume of the solid is #2750 pi#, what is the area of the base of the cylinder?. Note – This is not actual derivation! This is just to memorize it. I can find the volume of a cone. If the volume of a cylinder is four-thirds pi r squared, what fraction of the cylinder does the sphere occupy? Okay, well, let's think about this. Some other standard units of Volume are mm 3, dm 3, m 3 and km 3. This can be calculated by adding the volume of cylinder and cone. By using this website, you agree to our Cookie Policy. I have not asked them to memorize the volume formula of a cylinder, but after day 1, they all know it. A cylinder is similar to a prism, but its two bases are circles, not polygons. Volume of a Cylinder Volume of a Cylinder. Volume of a sphere formula. Use the given formulas to find the volume of the given figures. Volume of a Cone vs Cylinder. Teacher's Page Title: A Cure for the Volume of a Cone, Cylinder and Sphere Common Core Standards: 8th Grade Geometry (8. We must now make the cylinder's height 2r so the sphere fits perfectly inside. Right, acute, and obtuse. 1) 7 km 22 km 2) 9 mi 10 mi 3) 7 mi 4 mi 4) 7 km 6 km Find the volume of each figure. First section (Green) has cylinders in. 264 Lesson 26 Unerstan Volume of Cylinders, Cones, and Spheres ©urriculum ssociats opyig is ot prmitt. Cone is geometric body which is formed by a set of rays from apex (vertex) point and cross any flat surface, and at the intersection form basis of the cone. For example, if the radius of a sphere is 6 units, then the volume would be $$\frac{4}{3}\pi (6)^{3}=288\pi$$ or approximately $$904$$ cubic units. uk Volume of Prisms, Cones, Pyramids & Spheres (H) - Version 2 January 2016 Find a formula for h in terms of x Give your answer in its simplest form. have to creat a geosolid class that has 3 sub class i choose sphere cylinder and cone, i got the first two to work but my cone class works except when it calculates the volume it keep getting 0 no matter what number is use for the hieght and radius. Angles and degrees. To see why the areas are the same, first imagine the sphere sitting snugly inside the cylinder. Find the Volume Cone 6. cube = 6 a 2. FORMULAS FOR PERIMETER, AREA, SURFACE, VOLUME Cylinder Volume = r2 X formulas for different shaped bases. In an organized way… and label each “cylinder,” “cone,” and “sphere”Use pi button for pi, NOT 3. Cone Cylinder h— Sphere Volume Formulas 3. I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. On the 3D side, note the flat circular disk in shape #7. Circumference formula. 2) (A) a cone and a cylinder (B) a hemisphere and a cone (C) frustum of a cone and a cylinder (D) sphere and cylinder 4. Megan Millan 106,614 views. But why is a sphere's surface area four times its shadow? How many Liters are there in one meter cube Solve Volume of Cylinder - Duration: 6:42. The final result is the ration of the volume of the cylinder to the volume of the sphere. Since each side of a square is the same, it can simply be the length of one side cubed. Verify the answer using the formulas for the volume of a sphere, $$V = \frac{4}{3}\pi r^3$$, and for the volume of a cone, $$V = \frac{1}{3} \pi r^2 h$$. Now, when my students practice using the formulas for volume of cylinders, cones, and spheres, we do a lot of practice focused on understanding the formulas. 262 Lesson 26 Understand Volume of Cylinders, Cones, and Spheres ©urriculum ssociats opyig is ot prmitt Solve. Over 70 formulas included. The figure is made up of a cylinder and a half sphere. Learn with flashcards, games, and more — for free. All that is left to do is normalize this by dividing it the volume of the sphere. Hexagonal Prism. Surface area of Cylinder• Look at the net of the. 8WB8-73 ª2014 University of Utah Middle School Math Project in partnership with the. Surface area is the measure of the area of the surface of a 3-dimensional geometric shape or object and is measured in square units, such as square inches or feet. The purpose of the article is to help you learn basics of 3-D geometry and encapsulate some of the important formulae and tricks. Finally, we looked at spheres, like a basketball and the planets. Student Objectives I can apply formulas to calculate the volume of cones, cylinders, and spheres. A cone is ⅓ the volume of a cylinder, or 1⁄3πr²h. The formula for calculating the surface area of sphere is given by: SA = 4 × π × r2. The latter is π r³, making the volume of the sphere 4/3 π r³. Given a sphere with radius r, a cone with radius r and height 2r, and a cylinder with radius r and height 2r, the sum of the volume of the cone and sphere is equal to the volume of the cylinder. To determine the surface area of the whole cube multiple the surface area of one side by six (the total number of equal surfaces on the cube). Know the formulas for the volumes of cones, cylinders, and spheres and use them to. SArect = 2 lw + 2 wh + 2 lh. 4 530 prenumeranter. The problem can be generalized to other cones and n-sided pyramids but for the moment consider the right circular cone. Find a missing measurement (height, radius, or. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real world and mathematical problems Volume of Cylinders, Cones and Spheres The NYS Grade 8 Reference Sheet contains many of the volume formulas you need to do the problems below. Spherical Cap. Megan Millan 105,933 views. For instance, the general formula for the surface area of a cone is: However, a normal cone has both the curved part and a circular base. Angles and degrees. Here is the cone with a “seam”: Here V = 2, E = 2, F = 2, so V – E + F = 2 – 2 + 2 = 2. Then compare the volumes. V = 1 3π ⋅ r2 ⋅ h. The cross-sectional area of a thin layer with a vertical distance a (same as figure 1) from the center of the base consists of two concentric circles. Cylinders and Cones (Math Journal 2, p. Let the radius be r and the height be h (which is 2r for the sphere), then the volume of cone is. Surface Area of Sphere and Hemisphere Here we will discuss surface area of sphere and hemisphere Sphere is nothing but a any type of ball. have to creat a geosolid class that has 3 sub class i choose sphere cylinder and cone, i got the first two to work but my cone class works except when it calculates the volume it keep getting 0 no matter what number is use for the hieght and radius. For instance, the general formula for the surface area of a cone is: However, a normal cone has both the curved part and a circular base. 14159) times the diameter d squared times the height h divided by twelve; V = pi * d^2 * h / 12 A parabolic cone has a smooth curved surface and a sharp pointed nose. Visualizing the Volume of a Sphere Formula | Deriving the Algebraic Formula With Animations - Duration: 3:12. Cylinder, Cone, and Sphere Volume - Duration: 2:32. Surface Area Formulas for Geometric Figures. The sphere is a space figure, which all its points at an equal distance from the center point. PROBLEM 1 : Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum. , 1998 and others): F = 0. }\) You need to learn this formula - or at least understand where it comes from. OP is the radius of the sphere. Further, equate volume of a cone and volume of sphere; so that we get the radius of sphere. Find the volume of a prism that has the base 5 and the height 3. Verify that your result is a maximum or minimum value using the first or second derivative test for extrema. 80 ft Concrete To complete a construction job, a contractor needs 78 cubic yards of concrete. Let our calculator do the complex formulas for you to ensure an accurate reading. In this example, r and h are identical, so the volumes are πr 3 and 1 ⁄ 3 π r 3. Each shape will have a unique formula, but in addition to the formula, you'll need measurements of the height (h) and radius (r). Volume formulas. Cylinders may be right or oblique. The volume of a cone is one third of the volume of a cylinder. The problem can be generalized to other cones and n-sided pyramids but for the moment consider the right circular cone. Complete the equations to find the volumes of the sphere and the cylinder in terms of p. Cylinder: A solid with two parallel circular bases. Spherical Sector. A sphere is a perfectly round geometrical 3D object. Lines that cross, forming right angles. Practice: Volume of cylinders, spheres, and cones word problems Volume formulas review Math · Basic geometry · Volume and surface area · Volume of cones, cylinders, and spheres. The sphere is a space figure, which all its points at an equal distance from the center point. Multiply the area of the base times the height. Volume of Cylinder: 2π r^3. Weight Density formula is given by The Density is expressed in Kg/L. 3X3 System Possibilities. Find the Volume Rectangular Prism: V = I w h 14 cm 3. Area of Cube, Rectangular prism, Cylinder, Cone, Sphere. plane, sphere, cylinder, cone, torus), where the simpler surface would b e chosen first. 0 Equation Volume of a Cylinder, Cone, and Sphere Volume Cylinder Previous Formulas Learned Area and Circumference of a Circle Cylinder Volume of a Cylinder Volume of a Cylinder Volume of Cylinders Class Practice Volume of Cylinders. Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more. On the 3D side, note the flat circular disk in shape #7. MATHS(NCERT) CLASS-9 SURFACE AREAS AND VOLUMES CHAPTER-13 FORMULAS OF CUBOID,CUBE, CYLINDER,CONE. Surface Area of Sphere and Hemisphere Here we will discuss surface area of sphere and hemisphere Sphere is nothing but a any type of ball. Purpose/Big Idea: Conceptually understand the formula for volume of a cone/sphere. Surface Area of Pyramids and Cones. This is not a necessarily a project to print on a 3D printer. a more detailed explanation (examples and solutions) of each volume formula. Cylinder, Partially Filled Calculate the volume, length, radius, or diameter of a partially filled laying down. Volume is measured in "cubic" units. Know the formulas for the volume of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. First section (Green) has cylinders in. I've sketched a cylinder with a circular base in Figure 21. Example Question #2 : Cylinders, Pyramids, Cones, And Spheres Volume Formulas: Ccss. GCSE Maths Formula Sheet Rules of Indices Rule 1: When you multiply indices of the same number you add the powers. The sphere and cylinder have the same radius and the height of the cylinder is equal to the diameter of the sphere. I can find the volume of a sphere. Volumes of Pyramids and Cones. Learn the names for angles of all sizes. have to creat a geosolid class that has 3 sub class i choose sphere cylinder and cone, i got the first two to work but my cone class works except when it calculates the volume it keep getting 0 no matter what number is use for the hieght and radius. Purpose/Big Idea: Conceptually understand the formula for volume of a cone/sphere. Find volumes of cones, cylinders, and spheres (8. The volume is the amount of total space on the interior of the solid. The formula for the volume of a cone is 1/3*pi*r^2*h. Algebra V = Bh = π r 2h Volume of a Cone. Students should have an understanding of why the formula works and how the formula relates to the measure and the figure. Looking at this in reverse, each cone is one-third the volume of a cylinder. 2 Surface Area and Volume of Spheres. OP is the radius of the sphere. Find a mix of spheres and hemispheres in these 3D shapes worksheets. of spheres, cylinders, cones and tori to three-dimensional data. The height of the cylinder is 50 feet and its diameter is 80 feet. 5cm³ (cubic centimeters) to 1 d. Standard(s): 8. Find the area of the triangle if sides are 12 cm. Loading Click Here To See Circle, Sphere and Cylinder Formulas. Suppose a sphere with radius r is placed inside a cylinder whose height and radius both equal the diameter of the sphere. 8 cm and a height of 15 cm. Each end is a circle so the surface area of each end is π * r 2, where r is the radius of the end. Hyperboloid of One Sheet. Teaching the volume of cylinders, cones, and spheres is all about the formulas. To improve this 'Volume of a circular truncated cone Calculator', please fill in questionnaire. The only things that are required to fill in are the radius of the base and the height of the cylinder for which the user needs to calculate the volume. Volume formulas for cone, cube, sphere, cylinder - name it and you will get the information in the articles below. l Triangular Prism: V = 4. The volume V of a cylinder is easy: it's the area of the end (which is just the area of the circle) times the height h: The surface area SA is the area of the ends (which are just circles), plus the area of the side, which is a. Learn about lines, rays, and line segments. Given a sphere with radius r, a cone with radius r and height 2r, and a cylinder with radius r and height 2r, the sum of the volume of the cone and sphere is equal to the volume of the cylinder. Visualizing the Volume of a Sphere Formula | Deriving the Algebraic Formula With Animations - Duration: 3:12. The sphere is a space figure having all its points an equal distance from the center point. The latter is π r³, making the volume of the sphere 4/3 π r³. have to creat a geosolid class that has 3 sub class i choose sphere cylinder and cone, i got the first two to work but my cone class works except when it calculates the volume it keep getting 0 no matter what number is use for the hieght and radius. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. Calculate volume of geometric solids. Note – This is not actual derivation! This is just to memorize it. Teaching the volume of cylinders, cones, and spheres is all about the formulas. 5 + l) = 217. MATHS(NCERT) CLASS-9 SURFACE AREAS AND VOLUMES CHAPTER-13 FORMULAS OF CUBOID,CUBE, CYLINDER,CONE. Formulas for SA and Volume of Cylinder,Cone, and Sphere. Volume of the tip cone. Volume of a sphere. You need to divide 40 cm by 2 to. Hyperboloid of One Sheet. Notice that the surface area around the circular walls of a cylinder of radius r and height 2r (i. Volume of a torus. = πrl is a formula for the Lateral Area of a cone, where r is the radius of the base and l the lateral height (slant height) of the cone. We must now make the cylinder's height 2r so the sphere fits perfectly inside. By (date), when given (5) real-life problems on calculating the volume of a solid, examples on finding the volumes of cones, cylinders, and spheres (e. As a first step, you’ll need to review earlier learning about the solids, surface areas and use the formulas for volume. Find the area and circumference of the circles below. CCSS Math: HSG. The Volumes of cones, cylinders, and spheres exercise appears under the 8th grade (U. Therefore we have, π x r x (r + l) = 217, which is same as π x 3. Archimedes imagined taking a circular slice out of all three solids. Purpose/Big Idea: Conceptually understand the formula for volume of a cone/sphere. one outside the sphere (circumscribed) so its volume was greater than the sphere's, and one inside the sphere (inscribed) so its volume was less than the sphere's. Remembering the volume formulas for all of these shapes can be extremely difficult. Student Objectives I can apply formulas to calculate the volume of cones, cylinders, and spheres. 14 × 5 × 5 = 78. Cylinder, Cone, and Sphere Volume - Duration: 2:32. Second section (Amber) has a cone and sphere in. Students will use the formula for the volume of a pyramid, cone, and sphere to solve problems. Explanation: The surface area of cone = π x r x r + π x r x l where l is the slant height and r is the radius of the base. Calculator online on how to calculate volume of capsule, cone, conical frustum, cube, cylinder, hemisphere, pyramid, rectangular prism, triangular prism and sphere. The volume of a cylinder: V=π*r^2*h Volume of a cone: V=π*r^2*(h/3) Volume of a sphere: V=(4/3)π*r^3. Discussion of Volume Calculation This web page is designed to compute volumes of storage tanks for engineers and scientists; however, it may be useful to anyone who needs to. Presentation Summary : Volume of Pyramids, Cones, and Sphere. The volume of a cylinder with hemispherical ends is calculated by entering the known values for radius and length. Geometry formulas vector illustration - Buy this stock vector and explore similar vectors at Adobe Stock Sales: 888-649-2990. It is given by the formula, πr 2 h, where r is the radius of the circular base and h is the height of the cylinder. A cone has one circular base and a vertex that is not on the base. The volume enclosed by a sphere is given by the formula. This song's hook makes these formulas easy to remember. We now show how to calculate the flux integral, beginning with two surfaces where n and dS are easy to calculate — the cylinder and the sphere. 14)(3)²(6) ∴ The volume of the cylinder = 169. On the 3D side, note the flat circular disk in shape #7. The general formula represents the most basic conceptual understanding of the moment of inertia. The cylinder’s HEIGHT is labeled “h. Use the formulas below to find the volume of many different geometric shapes. Cylinders (which are like tubes, but with caps on the ends) also come up occasionally. Rewrite as. The ice cream cone we’ve got doesn’t have a circular base – that’s where the sphere goes. my first method a Cylinder say "I know, I know!" Drag in a Cylinder say "The volume formula for a sphere is 4/3 times pi times radius cubed. As it is in circular shape so it has diameter and radius. MathPlanetVideos. V = Bh Write the formula. The area is the sum of these two areas. Frustum cone. The formulas for calculating Surface Area and Volume are. Cylinder Calculator revised August, 2014 If you know two variables (area, volume, radius, height) it solves for the other two. Many subjects are also addressed but will come under one of the above headings. Parallelepiped. 264 Lesson 26 Unerstan Volume of Cylinders, Cones, and Spheres ©urriculum ssociats opyig is ot prmitt. The drag coefficient for this shape is given as 1. The song also gives examples of objects of these shapes. Cylinders and Cones. If the given cylinder is of radius r, then t. Apart from these, some software also feature options to calculate diagonal, height, length, breadth, etc. 8 cm and a height of 15 cm. Students will complete "Volume of 3D Solids Ticket 2. Cone Cylinder h— Sphere Volume Formulas 3. Volume and surface formulas. Students must be able to solve problems and determine the solutions. Find the Volume Sphere 3 cm. The results we provide are accurate, but. Each formula has calculator. Formulas and explanation below. cylinder = b h = r 2 h pyramid = (1/3) b h cone = (1/3) b h = 1/3 r 2 h sphere = (4/3) r 3. Archimedes imagined taking a circular slice out of all three solids. Sphere: A solid figure where all points are an equal distance from the center point. In general, the moment of inertia is a tensor, see below. Students can use clay to model a cone and a cylinder to help them see the relationship (MP4). Basically, for any rotating object, the moment of inertia can be calculated by taking the distance of each particle from the axis of rotation (r in the equation), squaring that value (that's the r 2 term), and multiplying it times the mass of that particle. So, to find the volume of a hollow sphere, we must subtract the volume of the hollow region from the volume of the overall sphere. 9) Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. Formulas*)Perimeter,)Circumference,)Area,)Surface)Area,)Volume)*)of2D)and)3D)Shapes) ♦! 2D)–geometric)shape:square,rectangle,triangle,circle,trapezoid. Scientific Calculator. Class Activity: Volume of a Cone Worksheet (you need Volume of a Sphere and/or Cone Interactive to do this) MORE PRACTICE. Volume of a right cone: V= 1 3 Bh or V= 1 3 πr2h where B is the area of the base and h is the height of the cone. 8 cm and a height of 15 cm. Second section (Amber) has a cone and sphere in. Practice: Volume of cylinders, spheres, and cones word problems Volume formulas review Math · Basic geometry · Volume and surface area · Volume of cones, cylinders, and spheres. Volume of a torus. The cross-sections are all circles with radii SR, SP, and SN, respectively. Find formulas of the surface area of cylinder, cone, and sphere 8. A cone, sphere and cylinder of radius r and height h The above formulas can be used to show that the volumes of a cone , sphere and cylinder of the same radius and height are in the ratio 1 : 2 : 3 , as follows. Surface Area of Sphere and Hemisphere Here we will discuss surface area of sphere and hemisphere Sphere is nothing but a any type of ball. 3 Surface Area of Prisms and Cylinders. Then click Calculate. Volume of a spheroidal cap. Cylinder Surface area Calculation. A right cone is a cone with its vertex above the center of its base.
7e0bftn6x6mdcow, v57slx699b72, e3o3ck1kvmr, kqp557c8i23z, 4lo0b6k2zgsfv5, zl0pp6wguqo, 32mtez9m2zx5, c51yf9olp9k7i, 7m4y86aiqjvve, 3c1s1ihutlje, lsoc3hsl3w5arar, d9oy4zcuj74h, 0z1hsn6qe5yg, 7cyje9nrv1u7, 13g71mdcde, a66nei1dype, doaa4r5f1ve08, h1fxb6cpk49shz, jtpa7vi75sel9, hx2zaqedz53j, ja8ksrxf08, 8sved0lns7805a5, y86mqp6apoc, ijwl96dv443v, gybcclarkwqd, 70t9wimd33056, h6qwp3zly18m7rc, 3juxnfiy5n4b3, f6h44xptamsi, 7tgybydxo5tfnjo | 12,889 | 52,819 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-34 | latest | en | 0.89312 |
https://www.printablemultiplication.com/multiplication-place-value-chart/ | 1,680,218,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00516.warc.gz | 1,029,535,255 | 15,984 | Multiplication Place Value Chart
Learning multiplication following counting, addition, as well as subtraction is good. Kids understand arithmetic via a organic progression. This advancement of understanding arithmetic is usually the subsequent: counting, addition, subtraction, multiplication, and ultimately department. This statement brings about the query why discover arithmetic with this series? Most importantly, why understand multiplication after counting, addition, and subtraction before division?
1. Youngsters find out counting first by associating graphic items making use of their hands. A perceptible illustration: How many apples exist from the basket? More abstract example is when aged have you been?
2. From counting figures, the following plausible phase is addition combined with subtraction. Addition and subtraction tables can be very valuable teaching tools for youngsters because they are visual resources creating the transition from counting much easier.
3. Which will be discovered up coming, multiplication or division? Multiplication is shorthand for addition. At this point, youngsters have a company knowledge of addition. Consequently, multiplication is definitely the following rational type of arithmetic to discover.
Overview basic principles of multiplication. Also, evaluate the basic principles utilizing a multiplication table.
We will evaluation a multiplication example. Using a Multiplication Table, flourish 4 times three and acquire a solution 12: 4 by 3 = 12. The intersection of row a few and column four of your Multiplication Table is twelve; twelve is definitely the response. For children starting to understand multiplication, this is certainly simple. They can use addition to eliminate the issue therefore affirming that multiplication is shorthand for addition. Example: 4 by 3 = 4 4 4 = 12. It is really an superb overview of the Multiplication Table. The added benefit, the Multiplication Table is visible and demonstrates returning to discovering addition.
In which will we begin studying multiplication using the Multiplication Table?
1. Very first, get acquainted with the table.
2. Get started with multiplying by one particular. Start off at row number one. Go on to column # 1. The intersection of row 1 and column one is the solution: one.
3. Recurring these actions for multiplying by one. Multiply row a single by columns a single through 12. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Replicate these techniques for multiplying by two. Increase row two by posts one particular by way of 5 various. The answers are 2, 4, 6, 8, and 10 respectively.
5. Let us leap ahead of time. Repeat these steps for multiplying by 5. Increase row 5 by posts one by means of 12. The replies are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now let us raise the quantity of problems. Repeat these methods for multiplying by a few. Increase row 3 by columns one by way of twelve. The solutions are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are confident with multiplication to date, consider using a examination. Resolve the next multiplication problems in your head and after that compare your responses on the Multiplication Table: flourish half a dozen and 2, multiply nine and 3, grow a single and 11, grow several and 4, and multiply several as well as 2. The issue answers are 12, 27, 11, 16, and 14 respectively.
Should you got 4 out of 5 difficulties correct, design your own multiplication exams. Estimate the solutions in your mind, and check them utilizing the Multiplication Table. | 807 | 3,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-14 | latest | en | 0.936167 |
https://m.basicmusictheory.com/b-flat-lydian-mode | 1,561,495,762,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999946.25/warc/CC-MAIN-20190625192953-20190625214953-00183.warc.gz | 528,237,580 | 7,808 | # B-flat lydian mode
The Solution below shows the B-flat lydian mode notes on the piano, treble clef and bass clef.
The Lesson steps then explain how to identify the mode note interval positions, choose note names and scale degree names.
For a quick summary of this topic, have a look at Mode.
All Keys On 1 page
## Solution - 2 parts
### 1. B-flat lydian mode
This step shows the ascending B-flat lydian mode on the piano, treble clef and bass clef. It also shows the scale degree chart for all 8 notes.
The B-flat lydian mode has 1 flat.
B-flat lydian mode note names
Note no.Note intervalNote name
1tonicThe 1st note of the B-flat lydian mode is Bb
2Bb-maj-2ndThe 2nd note of the B-flat lydian mode is C
3Bb-maj-3rdThe 3rd note of the B-flat lydian mode is D
4Bb-aug-4thThe 4th note of the B-flat lydian mode is E
5Bb-perf-5thThe 5th note of the B-flat lydian mode is F
6Bb-maj-6thThe 6th note of the B-flat lydian mode is G
7Bb-maj-7thThe 7th note of the B-flat lydian mode is A
8Bb-perf-8thThe 8th note of the B-flat lydian mode is Bb
Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram.
These note names are shown below on the treble clef followed by the bass clef.
B-flat lydian mode degrees
Note no.Degree name
1Bb is the tonic of the B-flat lydian mode
2C is the supertonic of the B-flat lydian mode
3D is the mediant of the B-flat lydian mode
4E is the subdominant of the B-flat lydian mode
5F is the dominant of the B-flat lydian mode
6G is the submediant of the B-flat lydian mode
7A is the leading tone of the B-flat lydian mode
8Bb is the octave of the B-flat lydian mode
### 2. B-flat lydian mode descending
This step shows the descending B-flat lydian mode on the piano, treble clef and bass clef.
No. Note 1 2 3 4 5 6 7 A G F E D C Bb
## Lesson steps
### 1. Piano key note names
This step shows the white and black note names on a piano keyboard so that the note names are familiar for later steps, and to show that the note names start repeating themselves after 12 notes.
The white keys are named using the alphabetic letters A, B, C, D, E, F, and G, which is a pattern that repeats up the piano keyboard.
Every white or black key could have a flat(b) or sharp(#) accidental name, depending on how that note is used. In a later step, if sharp or flat notes are used, the exact accidental names will be chosen.
The audio files below play every note shown on the piano above, so middle C (marked with an orange line at the bottom) is the 2nd note heard.
### 2. B-flat lydian mode tonic note and one octave of notes
This step shows an octave of notes in the B-flat lydian mode to identify the start and end notes of the mode.
The numbered notes are those that might be used when building this mode.
The B-flat lydian mode starts on note B-flat.
Since this mode begins with note Bb, it is certain that notes 1 and 13 will be used in this mode.
Note 1 is the tonic note - the starting note - Bb, and note 13 is the same note name but one octave higher.
No. Note 1 2 3 4 5 6 7 8 9 10 11 12 13 Bb B C C# / Db D D# / Eb E F F# / Gb G G# / Ab A Bb
### 3. B-flat lydian mode note interval positions
This step applies the B-flat lydian mode note positions to so that the correct piano keys and note pitches can be identified.
In their simplest / untransposed form, modes do not contain any sharp or flat notes.
This can be seen by looking at the Mode table showing all mode names with only white / natural notes used.
The lydian mode uses the W-W-W-H-W-W-H note counting rule to identify the note positions of 7 natural white notes starting from note F.
The B-flat lydian mode re-uses this mode counting pattern, but starts from note Bb instead.
To count up a Whole tone, count up by two physical piano keys, either white or black.
To count up a Half-tone (semitone), count up from the last note up by one physical piano key, either white or black.
The tonic note (shown as *) is the starting point and is always the 1st note in the mode.
Again, the final 8th note is the octave note, having the same name as the tonic note.
No. Note 1 2 3 4 5 6 7 8 Bb C D E F G A Bb
One or more note in this mode has a sharp or flat, which means that this mode has been transposed to another key.
### 4. B-flat lydian mode notes
This step tries to assign note names to the piano keys identified in the previous step, so that they can be written on a note staff in the Solution section.
The 7 unique notes in a mode need to be named such that each letter from A to G is used once only - and so each note name is either a natural white name(A..G) , a sharp(eg. F-sharp) or a flat(eg. G-flat).
The rule ensures that every position of a staff is used once and once only - whether that position be a note in a space, or a note on a line.
This is needed to ensure that when it comes to writing the mode notes on a musical staff (eg. a treble or bass clef), there is no possibility of having 2 G-type notes, for example, with one of the notes needing an accidental next to it on the staff (a sharp, flat or natural symbol).
Applying the rule below ensures that when accidental adjustment symbols are added next to staff notes as part of composing music based on that mode, these accidentals will indicate that the adjusted note is not in that mode.
To apply this rule, firstly list the white key names starting from the tonic, which are shown the white column below.
Then list the 7 notes in the mode so far, shown in the next column.
For each of the 7 notes, look across and try to find the white note name in the mode note name.
If the natural white note can be found in the mode note, the mode note is written in the Match? column.
The 8th note - the octave note, will have the same name as the first note, the tonic note.
B-flat lydian mode
No.WhiteMode NoteMatch?
1BBbBb
2CCC
3DDD
4EEE
5FFF
6GGG
7AAA
8BBbBb
For this mode, all notes have a match, and so the Match? column shows the mode note names.
### 5. B-flat lydian mode descending
This step shows the notes when descending the B-flat lydian mode, going from the highest note sound back to the starting note.
For all modes, the notes names when descending are just the reverse of the ascending names.
So assuming octave note 8 has been played in the step above, the notes now descend back to the tonic.
No. Note 1 2 3 4 5 6 7 A G F E D C Bb
### 6. B-flat lydian mode degrees
This step shows the B-flat scale degrees - Tonic, supertonic, mediant, subdominant, dominant, submediant, etc.
Each of the notes in this mode has a degree name, which describes the relationship of that note to the tonic(1st) note.
Scale degree names 1,2,3,4,5,6, and 8 below are always the same for all modes (ie. 1st note is always tonic, 2nd is supertonic etc.) , but obviously the note names will be different for each mode / key combination.
In this mode, the 7th note is called the leading note or leading tone because the sound of the 7th note feels like it wants to resolve and finish at the octave note, when all mode notes are played in sequence.
It does this because in this mode, the 7th note is only 1 semitone / half-tone away from the 8th note - the octave note. The lydian mode shares the same property - it only has one semitone / half-tone between the 7th and 8th notes.
In contrast, all other modes, including for example the phrygian mode, have a whole tone (two semitones, two notes on the piano keyboard) between the 7th and 8th notes, and the 7th note does not lean towards the 8th note in the same way. For these other modes, the 7th note is called the subtonic.
B-flat lydian mode degrees
Note no.Degree name
1Bb is the tonic of the B-flat lydian mode
2C is the supertonic of the B-flat lydian mode
3D is the mediant of the B-flat lydian mode
4E is the subdominant of the B-flat lydian mode
5F is the dominant of the B-flat lydian mode
6G is the submediant of the B-flat lydian mode
7A is the leading tone of the B-flat lydian mode
8Bb is the octave of the B-flat lydian mode | 2,227 | 8,029 | {"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} | 3.140625 | 3 | CC-MAIN-2019-26 | longest | en | 0.824639 |
https://enigmaticcode.wordpress.com/2018/09/19/tantalizer-445-key-problem/ | 1,569,125,619,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575076.30/warc/CC-MAIN-20190922032904-20190922054904-00125.warc.gz | 456,943,885 | 21,027 | # Enigmatic Code
Programming Enigma Puzzles
## Tantalizer 445: Key problem
From New Scientist #996, 15th April 1976 [link]
Uncle Tom’s bungalow is well endowed with doors, as you can see. He locks them each night, to keep out things that go bump. He would dearly love to do it by passing through each door just once and locking it behind him, so as to finish safely locked up in his bedroom. Alas, it cannot be done. Ah, but wait a minute. It could be done, if he had one of the doors bricked up. By Jove, that’s it.
Precisely which door would he be better without?
[tantalizer445]
### One response to “Tantalizer 445: Key problem”
1. Jim Randell 19 September 2018 at 8:01 am
This Python 3 program runs in 284ms.
Run: [ @repl.it ]
```from collections import defaultdict
from enigma import join, printf
# the doors and the rooms they link
doors = (
'A13', 'B14', 'C12', 'D24', 'E02', 'F34', 'G34',
'H46', 'I35', 'J04', 'K05', 'L05', 'M06', 'N06',
)
# construct the graph: room -> door -> room
graph = defaultdict(dict)
for (d, r1, r2) in doors:
graph[r1][d] = r2
graph[r2][d] = r1
# from room r go through (and lock) k more doors
def solve(r, k, ds=[]):
# are we done?
if k == 0:
yield (ds, r)
else:
# choose an unlocked door
for (d, t) in graph[r].items():
if d in ds: continue
yield from solve(t, k - 1, ds + [d])
# choose a starting room (inside the house)
for r in '123456':
# try to lock 13 of the doors
for (ds, t) in solve(r, 13):
# which door is missing
missing = list(d[0] for d in doors if d[0] not in ds)[0]
printf("missing = {missing} [{r} -> {ds} -> {t}]", ds=join(ds))
break
```
Solution: He would be better without door “C”.
Labelling the rooms 1, 2, 3, 4, 5, 6 and “outside” 0 we can construct a graph with the rooms as nodes and the doors as edges.
A Eulerian path (a path that visits all the nodes using each edge once), can only exist if the start and end node have odd degree and the remaining nodes have even degree.
The nodes in our graph have degrees (= number of doors):
room 0: 6 doors
room 1: 3 doors
room 2: 3 doors
room 3: 4 doors
room 4: 6 doors
room 5: 3 doors
room 6: 3 doors
So we have too many nodes with odd degree. Removing an edge will reduce the degree of the connected nodes by 1, and we can only remove one edge. So we need to remove the edge from two nodes of odd degree that are connected together.
The only two nodes of odd degree that are connected together are 1 and 2. The door between rooms 1 and 2 is C. So eliminating door C gives us a graph where rooms 5 and 6 have odd degree and the remaining rooms have even degree, so we can make a path between room 5 and room 6 that passes through each of the remaining doors. So if one of these rooms is Tom’s bedroom he can start in the other one and finish in his bedroom with all the doors locked.
Here is a diagram of path between rooms 5 and 6 that passes through each door (except C) once.
It might be easier just to keep door C permanently locked, rather than brick it up.
Note that to lock all the doors he will have to leave and re-enter the house 3 times through different doors.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 918 | 3,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.90939 |
https://www.jiskha.com/display.cgi?id=1299088744 | 1,511,360,542,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806609.33/warc/CC-MAIN-20171122141600-20171122161600-00703.warc.gz | 798,517,008 | 3,785 | # Statistics
posted by .
In a single toss of a fair die, what is the probability of obtaining an even number divisible by three?
• Statistics -
What even number is divisible by 3 that is found on a die?
Each number has P = 1/6
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More Similar Questions | 513 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-47 | latest | en | 0.89674 |
https://reviews.llvm.org/D145230 | 1,716,381,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00714.warc.gz | 419,595,839 | 28,771 | # [ScalarEvolution] Apply loop guards against min/max for its argumentsClosedPublic
Authored by dmakogon on Mar 3 2023, 4:28 AM.
# Details
Reviewers
mkazantsev nikic fhahn aleksandr.popov
Commits
rGb60758374b7b: [SCEV] Apply loop guards against min/max for its arguments
Summary
This replaces several rewriting rules in ScalarEvolution::applyLoopGuards that are applied to min/max expressions with the equivalent ones but applied to its arguments.
So currently given we have a loop guard min(a, b) >= c, the min expression gets rewritten as max(c, min(a, b)).
With such approach, we're unable to apply the rewrite if min operands are zext for example (min(zext(a), zext(b))), however it's equivalent to the expression zext(min(a, b)) for which we can apply the rewrite.
We can rewrite the min operands instead with these expressions:
• a --> max(c, a) and
• b --> max(c, b)
and this would allow us to apply the loop guard in this and similar cases: min(zext(a), zext(b)) would get rewritten as min(zext(max(c, a)), zext(max(c, b))) instead of just being skipped.
The table of replaced rules (omitting predicates signedness for simplicity):
Old rule New rules min(a, b) >= c --> max(c, min(a, b)) a --> max(a, c) and b --> max(b, c) min(a, b) > c --> max(c + 1, min(a, b)) a --> max(a, c + 1) and b --> max(b, c + 1) max(a, b) <= c --> min(c, max(a, b)) a --> min(a, c) and b --> min(b, c) max(a, b) < c --> min(c - 1, max(a, b)) a --> min(a, c - 1) and b --> min(b, c - 1)
# Diff Detail
### Event Timeline
dmakogon created this revision.Mar 3 2023, 4:28 AM
Herald added a project: Restricted Project. Mar 3 2023, 4:28 AM
dmakogon requested review of this revision.Mar 3 2023, 4:28 AM
Herald added a project: Restricted Project. Mar 3 2023, 4:28 AM
dmakogon updated this revision to Diff 502100.EditedMar 3 2023, 4:42 AM
There is one regression I didn't notice initially (trip-count-minmax.ll). Posted fix for review - https://reviews.llvm.org/D145231
dmakogon edited the summary of this revision. (Show Details)Mar 3 2023, 11:59 AM
dmakogon edited the summary of this revision. (Show Details)Mar 3 2023, 12:24 PM
dmakogon updated this revision to Diff 502225.Mar 3 2023, 12:39 PM
Cleanup, got rid of copy-paste code
mkazantsev requested changes to this revision.Mar 5 2023, 8:41 PM
llvm/lib/Analysis/ScalarEvolution.cpp
15075
This is not correct (c + 1 may overflow). Why didn't you just keep it as
`// 'min(a, b) > c' -> '(a > c) && (b > c)';`
?
15077
Same
15100
Bug when RHS = SINT_MIN.
15119
Why do you think that eq/ne cannot come here?
This revision now requires changes to proceed.Mar 5 2023, 8:41 PM
This comment was removed by mkazantsev.
mkazantsev added a comment.EditedMar 5 2023, 8:50 PM
Okay, seems that same bug is in underlying code. I suggest to fix it first. It seems to be turning predicates that are trivially true into things that may be false .
llvm/lib/Analysis/ScalarEvolution.cpp
15119
Question withdrawn, I see.
llvm/lib/Analysis/ScalarEvolution.cpp
15072–15079
This comment is out of place. You are not constructing c + 1 or c - 1 here, it is done in the caller code. Here enough to say that
```min(a, b) > c -> (a > c) && (b > c); // same for >=
max(a, b) < c -> (a < c) && (b < c); // same for <=```
llvm/lib/Analysis/ScalarEvolution.cpp
15092–15093
I propose refactoring like:
```case CmpInst::ICMP_UGT:
RHS = RHS - 1;
LLVM_FALLTHROUGH;
case CmpInst::ICMP_UGE:
RewrittenRHS = getUMaxExpr(RewrittenLHS, RHS);```
mkazantsev added a comment.EditedMar 5 2023, 9:20 PM
Generally, you are just replicating the bug and not introducing a new one. I don't insist that fix is blocking this one, you can make them in any order you like more. Please address restructuring comments. Also fix the bug with max/min having another number of operands than 2.
llvm/lib/Analysis/ScalarEvolution.cpp
15091–15094
There can be more (or less?) than 2 operands.
llvm/lib/Analysis/ScalarEvolution.cpp
15121–15122
What if the operands are also min/max? Maybe use a DFS traversal instead to collect all leaves?
llvm/lib/Analysis/ScalarEvolution.cpp
15121–15122
I mean, non-recursive traversal using stack.
llvm/lib/Analysis/ScalarEvolution.cpp
15102
You should set RewrittenRHS if it happens.
15104
Can we just always call AddMinMaxRewrites and, if it's not min or max, just to the trivial thing?
mkazantsev added a comment.EditedMar 5 2023, 9:40 PM
So far plans look follwing:
1. Refactor switch cases into fallthroughs;
2. Then, make an NFC that calls a function. Instead of
`RewrittenRHS = getUMaxExpr(RewrittenLHS, RHS);`
smth like
`RewrittenRHS = AddRewrites(RewrittenLHS, RHS);`
The handling of predicate can be moved inside this function. You don't need to have 2 predicate switches for calling, you just need one. Maybe 2nd (small) switch is only for checks and adding + - 1.
1. Into this function, add handling of min/max (non-recursive stack-based traversal). The returned value should be the rewrite of original RewrittenLHS no matter if it's a min or not.
As a side thing - add checks that, whenever you compute RHS + - 1, it is not a MAX/MIN int respectively. In this case we should not rewrite anything.
dmakogon updated this revision to Diff 502950.EditedMar 7 2023, 1:11 AM
Thanks for the suggestions, Max.
1. Now the switch uses fallthroughs.
2. The expressions are now processed in a worklist, for min/max we enqueue their operands (this also means we support any number of operands).
3. Suggest to fix +/- 1 bug in a follow-up. The bug was there before, this patch doesn't expose it anyway.
And also there was no use in adding a new function, the traversal is done in the same function as before.
dmakogon marked 10 inline comments as done.Mar 7 2023, 1:14 AM
dmakogon marked an inline comment as done.
Underlying code seems buggy to me. I'd prefer it fixed in follow-ups.
Your change mostly LG, if possible, do variable renaming as a separate patch.
Please wait couple of days in case if someone else has concerns before landing this.
llvm/lib/Analysis/ScalarEvolution.cpp
15085–15099
I know it's not your checks, but this can be changed further.
• Why pointers are checked only for ult? ugt will also make us computations with it, and why not check it for signed?
• Does this whole thing even make any sense for pointers?
• Why we do umax(..., UMIN + 1) correction for unsigned but not smax(..., SMIN + 1) for signed? Why don't we do the same correction for umin(RHS, UMAX - 1) when we are about to add one? Looks like an attempt to hide some bugs.
15114
Can variable renaming be done separately?
mkazantsev accepted this revision.Mar 7 2023, 2:00 AM
This revision is now accepted and ready to land.Mar 7 2023, 2:00 AM
This revision was landed with ongoing or failed builds.Mar 13 2023, 10:06 AM
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https://math.stackexchange.com/questions/2542415/standard-way-of-using-sine-and-cosine-to-find-tangent | 1,713,826,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00739.warc.gz | 339,542,413 | 35,313 | Standard way of using sine and cosine to find tangent
There was question I came upon, and I was stumped. The question was: evaluate the sine ,cosine, and tangent of the angle without using a calculator.
I was given $-\pi/6$. I know that sine is $-1/2$ and cosine is $\sqrt{3}/2$.
Normally I know that $\tan = \sin/\cos$. And doing so gives $-1/\sqrt{3}$. The problem was that the actual answer is $-\sqrt{3}/3$, and I can't seem to figure out how that can be possible
• $-\frac 1{\sqrt 3} = -\frac {\sqrt 3}{3}$. Those are both the same answer. Nov 29, 2017 at 7:09
$-\frac 1{\sqrt{3}} = -\frac 1{\sqrt{3}} * \frac {\sqrt{3}}{\sqrt{3}} = - \frac {\sqrt 3}{3}$.
$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt 3}$. To see this, rationalise the denominator of the latter by multiplying top and bottom by $\sqrt 3$. | 265 | 806 | {"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-2024-18 | latest | en | 0.90486 |
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Online Practice Quiz in Op-Amp Applications part 2 from the book Electronic Devices and Circuit Theory 10th Edition by Robert L. Boylestad.
This is the Online Practice Quiz in Op-Amp Applications Part 2 from the book, Electronic Devices and Circuit Theory 10th Edition by Robert L. Boylestad. If you are looking for a reviewer in Electronics Engineering this will definitely help. I can assure you that this will be a great help in reviewing the book in preparation for your Board Exam. Make sure to familiarize each and every questions to increase the chance of passing the ECE Board Exam.
### Continue Part II of the Online Practice Quiz
#### Quiz in Op-Amp Applications
Question 11. Determine the output voltage for this circuit with a sinusoidal input of 2.5 mV.
A. –0.25 V
B. –0.125 V
C. 0.25 V
D. 0.125 V
Question 12. Calculate the cutoff frequencies of a bandpass filter with R1 = R2 = 5 kΩ and C1 = C2 = 0.1 μF.
A. fOL = 318.3 Hz, fOH = 318.3 Hz
B. fOL = 636.6 Hz, fOH = 636.6 Hz
C. fOL = 318.3 Hz, fOH = 636.6 Hz
D. fOL = 636.6 Hz, fOH = 318.3 Hz
Question 13. Determine the output voltage when V1 = –V2 = 1 V.
A. 0 V
B. –2 V
C. 1 V
D. 2 V
Question 14. Calculate the output voltage.
A. –6.00 mV
B. 6.0 mV
C. 6.12 mV
D. –6.12 mV
Question 15. Determine the output voltage when V1 = –V2 = –1 V.
A. 0 V
B. –2 V
C. 1 V
D. 2 V
Question 16. Calculate the output voltage if V1 = –0.2 V and V2 = 0 V.
A. 0 V
B. –6.6 V
C. –4 V
D. 2 V
Question 17. Calculate IL for this circuit.
A. 3 mA
B. 4 mA
C. 5 mA
D. 6 mA
Question 18. Calculate the output voltage.
A. 3.02 V
B. 2.03 V
C. 1.78 V
D. 1.50 V
Question 19. Calculate the output voltage if V1 = V2 = 0.15 V.
A. 0 V
B. 4.65 V
C. 6.45 V
D. –6.45 V
Question 20. Calculate the output voltage if V1 = 33 mV and V2 = 02 mV.
A. 0 V
B. –6.6 V
C. –4 V
D. 2 V
### More Practice Quiz in Op-Amp Applications
Practice Quiz Part 1
Practice Quiz Part 2
Practice Quiz Part 3
Practice Quiz Part 4
Practice Quiz Part 5
Practice Quiz Part 6
### See: Complete List of Practice Quizzes
Note: After taking this particular quiz, you can proceed to check all the topics.
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## A Book of Abstract Algebra, Second Edition (1982)
### Chapter 25. FACTORING POLYNOMIALS
Just as every integer can be factored into primes, so every polynomial can be factored into “irreducible” polynomials which cannot be factored further. As a matter of fact, polynomials behave very much like integers when it comes to factoring them. This is especially true when the polynomials have all their coefficients in a field.
Throughout this chapter, we let F represent some field and we consider polynomials over F. It will be found that F[x] has a considerable number of properties in common with . To begin with, all the ideals of F[x] are principal ideals, which was also the case for the ideals of .
Note carefully that in F[x], the principal ideal generated by a polynomial a(x) consists of all the products a(x)s(x) as a(x) remains fixed and s(x) ranges over all the members of F[x].
Theorem 1 Every ideal of F[x] is principal.
PROOF: Let J be any ideal of F[x]. If J contains nothing but the zero polynomial, J is the principal ideal generated by 0. If there are nonzero polynomials in J, let b(x) be any polynomial of lowest degree in J. We will show that J=⟨(b(x)⟩, which is to say that every element of J is a polynomial multiple b(x)q(x) of b(x).
Indeed, if a(x) is any element of J, we may use the division algorithm to write a(x) = b(x)q(x) + r(x), where r(x) = 0 or deg r(x) < deg b(x). Now, r(x) = a{x) −b(x)q(x); but a(x) was chosen in J, and b(x) ∈ J; hence b(x)q(x) ∈ J. It follows that r(x) is in J.
If r(x) ≠ 0, its degree is less than the degree of b(x). But this is impossible because b(x) is a polynomial of lowest degree in J. Therefore, of necessity, r(x) = 0.
Thus, finally, a(x) = b(x)q(x); so every member of J is a multiple of b(x), as claimed. ■
It follows that every ideal J of F[x] is principal. In fact, as the proof above indicates, J is generated by any one of its members of lowest degree.
Throughout the discussion which follows, remember that we are considering polynomials in a fixed domain F[x] where F is a field.
Let a(x) and b(x) be in F[x]. We say that b(x) is a multiple of a(x) if
b(x) = a(x)s(x)
for some polynomial s(x) in F[x]. If b(x) is a multiple of a(x), we also say that a(x) is a factor of b(x), or that a(x) divides b(x). In symbols, we write
a(x)∣b(x)
Every nonzero constant polynomial divides every polynomial. For if c ≠ 0 is constant and a(x) = a0+…+anxn, then
hence ca(x). A polynomial a(x) is invertible iff it is a divisor of the unity polynomial 1. But if a(x)b(x) = 1, this means that a(x) and b(x) both have degree 0, that is, are constant polynomials: a(x) = a, b(x) = b, and ab = 1. Thus,
the invertible elements of F[x] are all the nonzero constant polynomials.
A pair of nonzero polynomials a(x) and b(x) are called associates if they divide one another: a(x)∣b(x) and b(x)∣a(x). That is to say,
a(x) = b(x)c(x) and b(x) = a(x)d(x)
for some c(x) and d(x). If this happens to be the case, then
a(x) = b(x)c(x) = a(x)d(x)c(x)
hence d(x)c(x) = 1 because F[x] is an integral domain. But then c(x) and d(x) are constant polynomials, and therefore a(x) and b(x) are constant multiples of each other. Thus, in F[x],
a(x) and b(x) are associates iff they are constant multiples of each other.
If a(x) = a0 + ⋯ + anxn, the associates of a(x) are all its nonzero constant multiples. Among these multiples is the polynomial
which is equal to (1/an)a(x), and which has 1 as its leading coefficient. Any polynomial whose leading coefficient is equal to 1 is called monk. Thus, every nonzero polynomial a(x) has a unique monic associate. For example, the monic associate of 3 + 4x + 2x3 is .
A polynomial d(x) is called a greatest common divisor of a(x) and b(x) if d(x) divides a(x) and b(x), and is a multiple of any other common divisor of a(x) and b(x); in other words,
(i)d(x)∣a(x) and d(x)∣b(x), and
(ii)For any u(x) in F[x], if u(x)∣a(x) and u(x)∣b(x), then u(x)∣d(x). According to this definition, two different gcd’s of a(x) and b(x) divide each other, that is, are associates. Of all the possible gcd’s of a(x) and b(x), we select the monic one, call it the gcd of a(x) and b(x), and denote it by gcd[a(x), b(x)].
It is important to know that any pair of polynomials always has a greatest common divisor.
Theorem 2 Any two nonzero polynomials a(x) and b(x) in F[x] have a gcd d(x).Furthermore, d(x) can be expressed as alinear combination
d(x) = r(x)a(x) + s(x)b(x)
where r(x) and s(x) are in F[x].
PROOF: The proof is analogous to the proof of the corresponding theorem for integers. If J is the set of all the linear combinations
u(x)a(x) + υ(x)b(x)
as u(x) and υ(x) range over F[x], then J is an ideal of F[x], say the ideal ⟨d(x)⟩ generated by d(x). Now a(x) = la(x) + 0b(x) and b(x) = 0a(x) + 1b(x), so a(x) and b(x) are in J. But every element of 7 is a multiple of d(x), so
d(x)∣a(x) and d(x)∣b(x)
If k(x) is any common divisor of a(x) and b(x), this means there are polynomials f(x) and g(x) such that a(x) = k(x)f(x) and b(x) = k(x)g(x). Now, d(x) ∈ J, so d(x) can be written as a linear combination
hence k(x)∣d(x). This confirms that d(x) is the gcd of a(x) and b(x). ■
Polynomials a(x) and b(x) in F[x] are said to be relatively prime if their gcd is equal to 1. (This is equivalent to saying that their only common factors are constants in F.)
A polynomial a(x) of positive degree is said to be reducible over F if there are polynomials b(x) and c(x) in F[x], both of positive degree, such that
a(x) = b(x)c(x)
Because b(x) and c(x) both have positive degrees, and the sum of their degrees is deg a(x), each has degree less than deg a(x).
A polynomial p(x) of positive degree in F[x] is said to be irreducible over F if it cannot be expressed as the product of two polynomials of positive degree in F[x]. Thus, p(x) is irreducible iff it is not reducible.
When we say that a polynomial p(x) is irreducible, it is important that we specify irreducible over the field F. A polynomial may be irreducible over F, yet reducible over a larger field E. For example, p(x) = x2 + 1 is irreducible over ; but over it has factors (x + i)(xi).
We next state the analogs for polynomials of Euclid’s lemma and its corollaries. The proofs are almost identical to their counterparts in ; therefore they are left as exercises.
Euclid’s lemma for polynomials Let p(x) be irreducible. If p(x)∣ a(x)b(x), then p(x)∣a(x) or p(x)∣b(x).
Corollary 1 Let p(x) be irreducible. If p(x) ∣a1(x)a2(x) ⋯ an(x), then p(x) ∣ ai(x) for one of the factors ai(x) among a1(x),. . ., an(x).
Corollary 2 Let q1(x), …, qr(x) and p(x) be monic irreducible polynomials. If p(x) ∣ q1 (x) … qr(x), then p(x) is equal to one of the factors q1(x),...,qr(x).
Theorem 3: Factorization into irreducible polynomials Every polynomial a(x) of positive degree in F[x] can be written as a product
a(x) = kp1(x)p2(x) … pr(x)
where k is a constant in F and p1(x), …, pr(x) are monic irreducible polynomials of F[x].
If this were not true, we could choose a polynomial a(x) of lowest degree among those which cannot be factored into irreducibles. Then a(x) is reducible, so a(x) = b(x)c(x) where b(x) and c(x) have lower degree than a(x). But this means that b(x) and c(x) can be factored into irreducibles, and therefore a(x) can also.
Theorem 4: Unique factorization If a(x) can be written in two ways as a product of monic irreducibles, say
a(x) = kp1(x) ⋯ pr(x) = lq1(x) ⋯ qs(x)
then k = l, r = s, and each pi(x) is equal to a qJ(x).
The proof is the same, in all major respects, as the corresponding proof for ; it is left as an exercise.
In the next chapter we will be able to improve somewhat on the last two results in the special cases of [x] and [x]. Also, we will learn more about factoring polynomials into irreducibles.
EXERCISES
A. Examples of Factoring into Irreducible Factors
1 Factor x4 − 4 into irreducible factors over , over , and over .
2 Factor x6 − 16 into irreducible factors over , over , and over .
3 Find all the irreducible polynomials of degree ≤ 4 in 2[x].
# 4 Show that x2 + 2 is irreducible in 5[x]. Then factor x4 − 4 into irreducible factors in 5[x]. (By Theorem 3, it is sufficient to search for monic factors.)
5 Factor 2x3 + 4x + 1 in 5[x]. (Factor it as in Theorem 3.)
6 In 6[x], factor each of the following into two polynomials of degree 1 : x, x + 2, x + 3. Why is this possible?
B. Short Questions Relating to Irreducible Polynomials
Let F be a field. Explain why each of the following is true in F[x]:
1 Every polynomial of degree 1 is irreducible.
2 If a(x) and b(x) are distinct monic polynomials, they cannot be associates.
3 Any two distinct irreducible polynomials are relatively prime.
4 If a(x) is irreducible, any associate of a(x) is irreducible.
5 If a(x)≠, a(x) cannot be an associate of 0.
6 In p[x], every nonzero polynomial has exactly p − 1 associates.
7 x2 + 1 is reducible in p[x] iff p = a + b where ab ≡ 1 (mod p).
C. Number of Irreducible Quadratics over a Finite Field
1 Without finding them, determine how many reducible monic quadratics there are in 5[x]. [HINT: Every reducible monic quadratic can be uniquely factored as (x + a)(x + b).]
2 How many reducible quadratics are there in 5[x]? How many irreducible quadratics?
3 Generalize: How many irreducible quadratics are there over a finite field of n elements?
4 How many irreducible cubics are there over a field of n elements?
D. Ideals in Domains of Polynomials
Let F be a field, and let J designate any ideal of F[x]. Prove parts 1−4.
1 Any two generators of J are associates.
2 J has a unique monic generator m(x). An arbitrary polynomial a(x) ∈ F[x] is in J iff m(x) ∣ a(x).
3 J is a prime ideal iff it has an irreducible generator.
# 4 If p(x) is irreducible, then ⟨p(x)⟩ is a maximal ideal of F[x]. (See Chapter 18, Exercise H5.)
5 Let S be the set of all polynomials a0 + a1x + ⋯ + anxn in F[x] which satisfy a0 + a1 + ⋯ + an = 0. It has been shown (Chapter 24, Exercise E5) that S is an ideal of F[x]. Prove that x − 1 ∈ S, and explain why it follows that S =⟨− 1⟩.
6 Conclude from part 5 that F[x]/⟨x − 1⟩ ≅ F. (See Chapter 24, Exercise F4.)
7 Let F[x, y] denote the domain of all the polynomials Σ aijxiyj in two letters x and y, with coefficients in F. Let J be the ideal of F[x, y] which contains all the polynomials whose constant coefficient in zero. Prove that J is not a principal ideal. Conclude that Theorem 1 is not true in F[x, y].
E. Proof of the Unique Factorization Theorem
1 Prove Euclid’s lemma for polynomials.
2 Prove the two corollaries of Euclid’s lemma.
3 Prove the unique factorization theorem for polynomials.
F. A Method for Computing the gcd
Let a(x) and b(x) be polynomials of positive degree. By the division algorithm, we may divide a(x) by b(x):
a(x) = b(x)ql(x) + r1,(x)
1 Prove that every common divisor of a(x) and b(x) is a common divisor of b(x) and r1(x).
It follows from part 1 that the gcd of a(x) and b(x) is the same as the gcd of b(x) and r1(x). This procedure can now be repeated on b(x) and r1(x); divide b(x) by r1(x):
b(x) = r1(x)q2(x)r2(x)
Next
r1(x) = r2(x)q3(x) + r3(x)
Finally,
rn1(x) = rn (x)qn+1(x) + 0
In other words, we continue to divide each remainder by the succeeding remainder. Since the remainders continually decrease in degree, there must ultimately be a zero remainder. But we have seen that
gcd[a(x),b(x)] = gcd[b(x), r1(x)] = ⋯ = gcd[rn1(x), rn(x)]
Since rn(x) is a divisor of rn1(x), it must be the gcd of rn(x) and rn.1. Thus,
rn(x) = gcd[a(x), b(x)]
This method is called the euclidean algorithm for finding the gcd.
# 2 Find the gcd of x3 + 1 and x4 + x3 + 2x2 + x − 1. Express this gcd as a linear combination of the two polynomials.
3 Do the same for x24 — 1 and x15 − 1.
4 Find the gcd of x3 + x2 + x + 1 and x4 + x3 + 2x2 + 2x in 3[x].
G. A Transformation of F[x]
Let G be the subset of F[x] consisting of all polynomials whose constant term is nonzero. Let h : GG be defined by
h(a0 + a1x + ⋯ + anxn) = an + an1x + ⋯ + a0xn
Prove parts 1−3:
1 h preserves multiplication, that is, h[a(x)b(x)] = h[a(x)]h[b(x)].
2 h is injective and surjective and hh = ε.
3 a0 + a1x + ⋯ + anxn is irreducible iff an + anlx + ⋯ + a0xn is irreducible.
4 Let a0 + a1x + ⋯ +anxn = (b0 + ⋯ +bmxm)(c0+ ⋯ +cqxq). Factor
an + an1x + ⋯ + a0xn
5 Let a(x) = a0 + a1x + ⋯ + anxn and â(x) = an + an 1x + ⋯ + a0xn. If cF, prove that a(c) = 0 iff â(1/c) = 0.
| 3,866 | 12,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-05 | longest | en | 0.946625 |
https://www.xloypaypa.pub/codeforces-round-706-lets-go-hiking-game/ | 1,726,278,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00723.warc.gz | 989,152,276 | 7,291 | ## Solution
Not hard, But need consider carefully. Still can be kind of game problem.
Let define the concepts below:
1. Top of mountainBottom of mountain。Top of mountain means the point $i$ which $p_i > p_{i\pm{1}}$. Bottom of mountain means the $i$ which $p_i<p_{i\pm{1}}$.
2. Left/Right side of mountain. Left/Right the side between two nearby top of mountain and bottom of mountain.
3. Length of side. Of course the length of the side of mountain. The length of side can apply for left side and right side.
conclusion:
Let's say the maximal value of Length of side is max.
If and only if there is only one top of mountain $i$, both the lenght of left side of $i$ and the lenght of right side of $i$ are max and the max is odd. Then this $i$ is the only solution.
Otherwise, just print 0.
Prove case by case:
1. If Qingshan not choose the top of mountain which length is not max. Then Dainel only need start from the bottom of mountain which length is max.
2. If there are no less than two side's length is max, and the these two side is not for the same top/bottom of mountain. Then whatever Qingshan choose (of course one of top of mountain), Daniel only need choose another one. So no possible two win. Just print 0.
3. If the two side which length is max is two side for a bottom of mountain. Then Daniel only need choose the bottom of mountain. Then Qingshan will lose. Just print 0.
4. If only one side which length is max. Then Dainel only need choose the point in the side. Which distance between that point and the top of mountain is even, and this distance longer than length of another side. Then Qingshang will always lose. Because if Qingshan go to Daniel side, because it's even, it will lose. If Qingshan go to another side, then the length is lesser than Daniel. So just print 0.
5. If both side of top of mountain is max. But the max is even. Then of cours, Qingshan will chose the top and Dainel will chose the bottom. If Qingshan go with Dainel side, then because of even, Qingshan will lose. If Qingshan go with another side, then because of same length， Qingshan will lose. Just print 0.
6. If not all the case above. the case is the length of both side of top is same, and this lenght is max, and this max is odd. Then Qingshan chose the top, Daniel chose bottom can win. Print 1.
## Code
Forgot case #5 when I was solving this problem at beginning. Got WA answer before I realized that. | 614 | 2,419 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.882095 |
https://concrete-calculator.org/lumber-calculator/ | 1,723,427,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00722.warc.gz | 131,925,032 | 9,795 | # Lumber Calculator
Lumber Calculator Length (feet): Width (inches): Height (inches): Quantity: Unit Price: Calculate
Are you planning a woodworking or construction project and need to figure out the amount of lumber required? Calculating lumber requirements can be tricky, but with the right tools and knowledge, you can save time and avoid costly mistakes.
That’s where a lumber calculator comes in handy. Here’s how you can use it to estimate your lumber n
### How To Use Lumber Calculator
Before you start calculating, you need to know the project requirements. Consider the type of project, the size of the structure, and the type of lumber you plan to use.
Enter the project requirements into the lumber calculator. This includes the length, width, and thickness of each board, as well as the number of boards required.
Once you enter the required information, the lumber calculator will estimate the amount of lumber required for your project. It will also calculate the total cost of the lumber based on the current market prices.
If you want to make changes to your project or adjust the lumber requirements, you can go back and adjust the calculation in the lumber calculator.
Using a lumber calculator can save you time and help you avoid costly mistakes when planning your woodworking or construction project. Try out our lumber calculator, along with our other construction calculators, to make your next project a success.
## FAQs
### How Do I Calculate How Much Lumber I Need?
You can calculate how much lumber you need by using a lumber calculator. Enter the project requirements such as the length, width, and thickness of each board, as well as the number of boards required, and the calculator will estimate the amount of lumber needed.
### How Is Project Lumber Calculated?
Project lumber is calculated by determining the project requirements such as the type of project, the size of the structure, and the type of lumber you plan to use. Then, use a lumber calculator to estimate the amount of lumber required based on the project requirements.
### How Do You Calculate The Area of a Lumber?
To calculate the area of a lumber, multiply the length by the width of the board. For example, if the board is 8 feet long and 4 inches wide, the area would be calculated as (8 feet x 4 inches) / 12 inches per foot = 2.67 square feet. | 483 | 2,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.900571 |
https://www.wise-geek.com/how-do-i-choose-the-best-bond-yield.htm | 1,713,249,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00092.warc.gz | 996,353,795 | 13,671 | Finance
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# How do I Choose the Best Bond Yield?
Kristie Lorette
Kristie Lorette
A bond yield is the amount of return on your investment or amount of interest earned on a bond that you invest your money in. High yield bonds return a higher return on the investment than those with lower interest rates or yields. Part of choosing the best bond yield is to calculate the yield of the bond, which considers the price paid to buy the bond, the face value of the bond and the interest payment of the bond.
To calculate the yield on a bond, divide the face value or coupon rate of the bond by the price you pay to buy the bond. For example, a bond issued at face value of \$1,000 US Dollars (USD) that has a 10% coupon rate yields \$100 USD in interest. When the same bond, with a face value of \$1,000 USD, sells at a discount rate of \$800 USD, then the yield on the bond increases to 12.5%.
Adversely, a bond with a face value of \$1,000 USD that sells at a premium of \$1,200 USD pushes the bond yield down to 8.33%. So, a major part of choosing the best bond yield is to evaluate whether you are paying face value, a discount or premium price to buy the bind in the first place.
The second major calculation is to calculate the yield to maturity (YTM). Unfortunately, this calculation is not a simple equation. Calculating this figure requires you to use a bond yield to maturity calculator, which you can find online, or use a financial calculator to input the figures. The answer to this calculation, however, tells you how much money you will earn on your investment in the bond if you hold it until the maturity date on the bond.
In short, however, if you understand how bonds behave, you can choose the best bond yield. Essentially, as bond prices go up, the bond yield goes down. When bond prices go down, then the bond yield goes up. This means that if you have the opportunity to buy a bond at a discount price, then you will yield more in the long-run than if you pay a premium or face value.
If you buy the bond at a premium price, then you will earn less of a yield than if you purchase it at face value or at a discount. If you buy the bond at face value, then you fall right in the middle — not earning as much as if you buy the bond at a discount, but earning more than buying the bond at a premium. | 577 | 2,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.912819 |
https://gmatclub.com/forum/for-which-of-the-following-functions-f-is-f-x-f-1-x-for-all-x-85751-20.html | 1,553,110,683,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202450.86/warc/CC-MAIN-20190320190324-20190320212038-00000.warc.gz | 498,068,213 | 150,958 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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For which of the following functions f is f(x) = f(1-x) for all x?
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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27 Mar 2014, 10:47
rajatsp wrote:
Please do let me know if there are more similar questions besides the ones you mentioned.
Thank you
Rajat
Here are several more:
for-which-of-the-following-functions-does-f-x-f-2-x-155813.html
for-which-of-the-following-does-f-a-f-b-f-a-b-164979.html
for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
Hope this helps.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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02 Sep 2014, 07:35
Bunuel, would you generally recommend picking numbers as a strategy for such questions (especially if the functions are for ALL X)? Or rather the algebraic approach?
Thanks!
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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02 Sep 2014, 08:13
bgpower wrote:
Bunuel, would you generally recommend picking numbers as a strategy for such questions (especially if the functions are for ALL X)? Or rather the algebraic approach?
Thanks!
Usually number plugging works just fine for such questions.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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03 Sep 2014, 01:14
VeritasPrepKarishma wrote:
metallicafan wrote:
Is there a faster way to solve this question rather than replacing each "x" by (1-x)? Thanks!
For which of the following functions $$f$$ is $$f(x) = f(1-x)$$ for all x?
A. $$f(x) = 1-x$$
B. $$f(x) = 1-x^2$$
C. $$f(x) = x^2 - (1-x)^2$$
D. $$f(x) = (x^2)(1-x)^2$$
E. $$f(x) = x / (1-x)$$
Tip: Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. I will intuitively check D first since it involves multiplication of the terms.
Agreed; did in the same way
Option D returns the same term by substituting x with (1-x)
In the OA, had they expanded option D then calculation would be required
$$x^2 (1-x)^2 = [x(1-x)]^2 = (x - x^2)^2 = x^2 - 2x^3 + x^4$$
Now replacing x with (1-x) would had required serious calculation.....
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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15 May 2015, 08:46
I basically substituted and checked each option. Somehow I get this intuition that in such questions, one should actually start from option E and work backwards, rather than starting from option A and moving forward. The reason being that GMAT would not make it so easy that option A or B itself would satisfy:). Any observations? Has anyone come across such questions (on functions), where the correct answer is option A/B?
A. f(x) = 1 - x
=> f(1-x) = 1 - (1-x) = x
So, f(x) <> f(1-x)
B. f(x) = 1 - x^2
=> f(1-x) = 1 - (1-x)^2 = 1 - (1+x^2 - 2x) = 2x - x^2
So, f(x) <> f(1-x)
C. f(x) = x^2 - (1 - x)^2
=> f(1-x) = (1-x)^2 - [(1-(1-x)]^2 = (1-x)^2 - x^2
So, f(x) <> f(1-x)
D. f(x) = x^2*(1 - x)^2
=> f(1-x) = (1-x)^2 * [(1-(1-x)]^2 = (1-x)^2 * x^2
So, f(x) = f(1-x)
E. f(x) = x/(1 - x)
=> f(1-x) = (1-x)/[(1-(1-x)] = (1-x)/x
So, f(x) <> f(1-x)
Hence, D
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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26 Dec 2015, 16:20
Big algebra is good but not prefer in these type if question. Plugin is best approach:
Assume x = 2.
$$f(2) = 2^2(1-2)^2$$
$$= 4(-1)^2 = 4*1= 4$$
$$f(1-2) or f(-1) = -1^2(1+1)^2$$
$$=1*2^2= 1*4=4$$
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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11 Jan 2016, 14:21
Abhii46 wrote:
You need to check for every option.
Either you can replace x in every option with 1-x or you can take x = 1 and then check for which option f(1) = f(1-1) = f(0).
In A f(1) = 1-1 = 0. f(0) = 1-0 = 1.
In B f(1) = 1 - 1^2 = 0. f(0) = 1 - 0 = 1
In C f(1) = 1^2 - (1 - 1 )^2 = 1. f(0) = 0 - (1-0)^2 = 0 - 1 = -1
in D f(1) = 1(1-1)^2 = 0. f(0) = 0.(1-0)^2 = 0. ( Right answer )
And for option E you can take x = 2 because if you take x = 1 in denominator the denominator becomes zero, which makes the f(x) undefined.
f(2) = 2/(1 - 2) = -2. f(1-2) = f(-1) = -1/(1-(-1)) = -1/2.
Please give a kudo if you like my explanation.
This explanation is much more clear than the other one; now I have understood the question!!! :D
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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11 Apr 2016, 17:36
I honestly don't get this one. Please elaborate, stared at it for 2 and half hours. lol
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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11 Apr 2016, 18:30
g3lo18 wrote:
I honestly don't get this one. Please elaborate, stared at it for 2 and half hours. lol
Have you gone through @Bunuel's solution at for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html#p644387 ? It is a well detailed and comprehensive solution.
As for the starting point, f(x) stands for a way to mention a relation in 'x'.
When I say, y=f(x), then there is a relation between y and x (it could be a linear relation in x or quadratic relation in x or cubic relation in x etc). In standard terms, we say that y is a function of x.
Now, in a similar fashion, if I mention y=f(1-x) ---> means that y is a function of (1-x).
The question is asking for the given 5 options of f(x), which of them will follow the relation f(x)=f(1-x) ---> simply check all the given options by substituting 1-x for x.
Example, option (a): f(x)=1-x ---> substitute 1-x for x ---> f(1-x)=1-(1-x) = 1-1+x = x $$\neq$$ f(x). Keep working like this for other options and stop when you find the one that works.
Alternately, you can plug in x=0 and see which of the options gives you f(x)=f(1-x) ---> f(0) = f(1-0)=f(1).
Hope this helps.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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12 Jun 2016, 22:33
Given: f(x)= f(1-x)
This means if we replace x by 1 - x in the function, still the result should be same.
Checking each option:
A f(x)=1-x
f(1 - x) = 1 - (1-x) = x. Not equal to f(x)
INCORRECT
B f(x)=1-x^2
f(1-x) = 1- (1-x)^2 = 1 - (1 +x^2 - 2x). Not equal to f(x)
INCORRECT
C f(x)=x^2-(1-x)^2
f(1-x) = (1-x)^2 - (1 - 1 +x)^2 = (1-x)^2 -x^2. Not equal to f(x)
INCORRECT
D f(x)=x^2(1-x)^2
f(1-x) = (1-x)^2(1 - 1 + x)^2 = (1-x)^2*x^2.
This is equal to f(x)
CORRECT
E f(x)= x/(1-x)
f(1-x) = 1-x/1-x + x = (1-x)/x. Not equal to f(x)
INCORRECT
Correct Option: D
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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30 Mar 2017, 16:24
1
study wrote:
For which of the following functions f is f(x) = f(1-x) for all x?
A. f(x) = 1 - x
B. f(x) = 1 - x^2
C. f(x) = x^2 - (1 - x)^2
D. f(x) = x^2*(1 - x)^2
E. f(x) = x/(1 - x)
Since we are not given any restrictions on the value of x, let’s let x = 1. Thus, we are determining for which of the following functions is f(1) = f(1-1), i.e., f(1) = f(0). Next, we can test each answer choice using our value x = 1.
A. f(x) = 1 - x
f(1) = 1 - 1 = 0
f(0) = 1 - 0 = 1
Since 0 does not equal 1, A is not correct.
B. f(x) = 1 - x^2
f(1) = 1 - 1^2 = 1 - 1 = 0
f(0) = 1 - 0^2 = 1 - 0 = 1
Since 0 does not equal 1, B is not correct.
C. f(x) = x^2 - (1 - x)^2
f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1
f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1
Since 1 does not equal -1, C is not correct.
D. f(x) = x^2*(1 - x)^2
f(1) = 1^2*(1 - 1)^2 = 1(0)= 0
f(0) = 0^2*(1 - 0)^2 = 0(2) = 0
Since 0 equals 0, D is correct.
Alternate Solution:
Let’s test each answer choice using x and 1 - x.
A. f(x) = 1 - x
f(x) = 1 - x
f(1 - x) = 1 - (1 - x) = x
Since 1 - x does not equal x, A is not correct.
B. f(x) = 1 - x^2
f(x) = 1 - x^2
f(1 - x) = 1 - (1 - x)^2 = 1 - (1 + x^2 -2x) = 2x - x^2
Since 1 - x^2 does not equal 2x - x^2, B is not correct.
C. f(x) = x^2 - (1 - x)^2
f(x) = x^2 - (1 - x)^2 = x^2 - (1 + x^2 - 2x) = 2x - 1
f(1 - x) = (1 - x)^2 - (1 - (1 - x))^2 = 1 + x^2 - 2x - x^2 = 1 - 2x
Since 2x - 1 does not equal 1 - 2x, C is not correct.
D. f(x) = x^2*(1 - x)^2
f(x) = x^2*(1 - x)^2
f(1 - x) = (1 - x)^2*(1 - (1 - x))^2 = (1 - x)^2*x^2
Since x^2*(1 - x)^2 equals (1 - x)^2*x^2, D is correct.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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18 Jul 2017, 17:37
Most people seem to suck at explaining these function related questions. Here is a good video explaining some function skills that helps on other questions. https://www.youtube.com/watch?v=T6-Zdr5w_bE
The key to these questions is understanding that f(x) is the function. Meaning that f(3) would mean everytime you see an x you sub in a 3. Or in this case we sub in an X-3. So for instance the first question is F(x) = 1-x............. Imagine the X is a blank or a blank parentheses waiting to be filled in by a number. so F(3) = 1-3 or F(x) = 1 - X or F (blank)= 1-(blank) or F(1-x) = 1-(1-x)
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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23 Mar 2018, 09:21
Wow, Bunuel. Thank you for that explanation. After so much studying I have never seen such a simple explanation. Definitely one of those eye-opening moments. Plugging in x and x-1 to the expression must yield the same result.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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06 May 2018, 18:28
can i choose non integers in this question ? can someone solve the D option using 1/4 as an example ?
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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07 May 2018, 00:42
pranavpal wrote:
can i choose non integers in this question ? can someone solve the D option using 1/4 as an example ?
Yes, you can plug 1/4 for x but the question is WHY work with fractions when you can use easier numbers?
D. $$f(x)=f(\frac{1}{4}) = (\frac{1}{4})^2*(1 - \frac{1}{4})^2=\frac{9}{256}$$
$$f(1-x) =f(\frac{3}{4})=(\frac{3}{4})^2*(1 - \frac{3}{4})^2=\frac{9}{256}$$
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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18 Jun 2018, 08:52
For which of the following functions f(x)f(x) is the relation f(f(x))=f(f(f(f(x))))f(f(x))=f(f(f(f(x)))) NOT true for at least some values of xx not equal to zero?
f(x)=−|x|f(x)=−|x|f(x)=2−xf(x)=2−xf(x)=3xf(x)=3xf(x)=4xf(x)=4xf(x)=5f(x)=5
How to arrive solution to this question
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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13 Nov 2018, 16:51
This question was difficult for me to understand. The following video cleared it up for me in about 30 seconds. It's actually very simple once you understand what it's asking:
Thanks everyone
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For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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24 Nov 2018, 17:24
Think of functions in terms of inputs and outputs. We want a function for which the input of x will lead to exactly the same result as the input of (1-x). Below is a video explanation
For a function question, it is almost always best to pick numbers, and to choose numbers that are small and manageable. Lets chose 1 for x. The quantity (1-x) would therefore equal 0.
If you put each of those inputs into the function in answer choice A you'll see very quicky that the two outputs are not equal to each other. You'll see very quickly that answer B is also incorrect. Answer C takes a bit more time. The correct answer is D.
The question if easiest to answer if you create separate columns for f(x) and f(1-x). Once again, a video explanation can be found here:
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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29 Dec 2018, 09:53
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study wrote:
For which of the following functions f is f(x) = f(1-x) for all x?
A. $$f(x) = 1 - x$$
B. $$f(x) = 1 - x^2$$
C. $$f(x) = x^2 - (1 - x)^2$$
D. $$f(x) = x^2*(1 - x)^2$$
E. $$f(x) = \frac{x}{(1 - x)}$$
Let's try plugging in an easy value for x. How about x = 0.
So, we can reword the question as, For which of the following functions is f(0)=f(1-0)
In other words, we're looking for a function such that f(0) = f(1)
A) f(x)=1-x
f(0)=1-0 = 1
f(1)=1-1 = 0
Since f(0) doesn't equal f(1), eliminate A
B) f(x) = 1 - x^2
f(0) = 1 - 0^2 = 1
f(1) = 1 - 1^2 = 0
Since f(0) doesn't equal f(1), eliminate B
C) f(x) = x^2 - (1-x)^2
f(0) = 0^2 - (1-0)^2 = -1
f(1) = 1^2 - (1-1)^2 = 1
Since f(0) doesn't equal f(1), eliminate C
D) f(x) = x^2(1-x)^2
f(0) = 0^2(1-0)^2 = 0
f(1) = 1^2(1-1)^2 = 0
Since f(0) equals f(1), keep D for now
E) f(x) = x/(1-x)
f(0) = 0/(1-0) = 0
f(1) = 1/(1-1) = undefined
Since f(0) doesn't equal f(1), eliminate E
Since only D satisfies the condition that f(x)=f(1-x) when x=0, the correct answer is D
Cheers,
Brent
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] 29 Dec 2018, 09:53
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Display posts from previous: Sort by | 5,785 | 16,345 | {"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} | 3.53125 | 4 | CC-MAIN-2019-13 | latest | en | 0.884936 |
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Price: - Paid
## Overview
The husband and wife team producing these apps have created a series of lovely tools that will really help kids to become more “naturally” mathematical. The learning outcomes are set out very clearly in each app so that parents/teachers can guide kids with the mathematical concepts as they use the apps. They are fantastic resources and we have given them an EAS Certification of 5 Stars and EAS Recommended Status.
## Teacher Review
Balancing Act is an amazing game made up of three levels that challenges children to add, subtract, multiply and divide in order to create the number at the end of the equation. On the left hand side of the equation, the learner is given a series of numbers and he/she must make sure that the correct numbers and operations are chosen in order to come to the right conclusion; in other words, so that the equation balances out!
The first level only looks at addition and subtraction; the second level looks at addition, subtraction and multiplication, the third level include division into the mix. The app teachers kids to link into the order of operations so as to get to the right answer correctly.
If the learner gets the answer wrong, he/she is given the chance to try again.
The best way to get a lot out of this app is to follow along with your child so that you can help and explain any of the concepts. Teachers can also use it in the classroom in order to promote discussion about the order of operations and the numbers that will balance the equation out.
The illustrations within the app are very well done. As a suggestion, we would love to see the girl move about, wobble and balance out as the learner progresses with the work, it would make the whole app rather more interactive and entertaining. However, once again, works due to its simplicity and ability to really drill on a basic yet important concept.
## From the Developer
How to play
The Balancing Act game challenges the player has to make an arithmetic equation balance by choosing where to put the addition, subtraction, multiplication or division signs. When the operations have been placed, tap the Check button to see if the expression is correct and the equation balances.
Well done if the equation balances! But don’t worry if the expression is wrong. Have another go and get it right next time.
Play at three levels
Balancing Act can be played at three levels. At Level 1, only the operations of + and – are needed to complete the expression. At Level 2 there is a choice of +, – or ×. At Level 3 any of the four operations, +, –, × or ÷, might be needed to complete the expression.
Order of Operations
The player needs to know the order of operations when choosing where to put them. Multiplication and division are done first, so 5 + 3 × 4 = 17 and not 32 because the multiplication 3 × 4 = 12 is done before the addition 5 + 3. | 617 | 2,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-05 | longest | en | 0.964757 |
http://seaintarchive.org/mailarchive/1996a/msg00471.html | 1,493,494,692,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123560.51/warc/CC-MAIN-20170423031203-00529-ip-10-145-167-34.ec2.internal.warc.gz | 336,376,434 | 4,077 | Need a book? Engineering books recommendations...
# [SEAOC] Concrete Dilema
• To: seaoc(--nospam--at)seaoc.org
• Subject: [SEAOC] Concrete Dilema
• From: "Eric J. Scott" <sac39899(--nospam--at)saclink1.csus.edu>
• Date: Fri, 12 Apr 1996 18:50:26 -800
• Priority: normal
```I am studying to take the PE exam on the 19th, and I have run into a
snag on determining where the neutral axis is in a RC beam. The
method that I have used in the past is to simply use the expression:
a=As*fy/.85*f'c*b and a=B1*c
However, in flipping through my concrete book and Lindburg's PE
review manual I found another method which uses a factor multiplied
with the depth, or:
c=k*d where k=sqrt(2*p*n+(p*n)^2)-p*n
p= the steel ratio (since rho is not part of my keyboard)
n= the modular ratio of E (steel)/ E (conc.)
I know that one expression is based on Ultimate Strength and the
other is based on Working Stress, but since they both assume a
cracked section shouldn't they give similar values?
The example that I used was a concrete beam with the following
properties:
b=18" d=37" f'c=3000 psi fy=60,000 psi As=6.0 in^2
Ec=3,122,000 psi Es=29,000,000 psi n=9.29 k=0.333
The location I get for the neutral axis using the Ultimate Strength
method is 9.23". The value I get using the Working Stress Method is
12.34".
I have less than a week till the exam and I would really appreciate
it if anyone can let me know which method is the correct one.
Thanks,
==============================
Eric J. Scott
escott(--nospam--at)csus.edu
ASCE-YMF Membership Chair, Sacramento
Cline, Agee, and Swedin
Engineers and Architects
==============================
...
``` | 489 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-17 | latest | en | 0.892955 |
https://www.physicsforums.com/threads/does-the-quantum-space-of-states-have-countable-or-uncountable-basis.1004990/ | 1,627,922,674,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154321.31/warc/CC-MAIN-20210802141221-20210802171221-00286.warc.gz | 992,692,486 | 25,933 | # Does the quantum space of states have countable or uncountable basis?
• A
Gold Member
Summary:
Hard to express, but smth like "an apparent problem in choosing between a "big" Hilbert space of quantum states with non-countable basis and a "small" Hilbert space generated by a countable set of basis vectors".
It's probably more kind of math question.
I consider a wave function of a harmonic oscillator, i.e. a particle in a parabolic well of potential. We know that the Hamiltonian is a Hermitian operator, and so its eigenstates constitute a full basis in the Hilbert space of the wave function states. We also know that this basis is countable.
On the other side, the arbitrary state may also be considered as a weighted integral of delta-functions, and such delta-functions are obviously a non-countable basis which creates a "bigger" space than what the eigenstates of the original Hamiltonian could generate.
I then wonder what should be called a "true" Hilbert space for such a case? A continuous spectrum of eigenfunctions is common in QM, it is generated for instance by a "free" (no potential part) Hamiltonian, so it's not a completely weird idea that the space of the function should have uncountable basis for practically ANY potential, including parabolic - after all, the wave function at a single moment may be arbitrary. Also a question may be asked, how a delta-wave-function would evolve in a parabolic well potential, but such a function obviously lies out of the subspace generated by the countable set of eigenstates of the Hamiltonian of the parabolic potential, so how such a problem should be approached, while we are used to a countable basis for the Harmonic oscillator?
Delta2 and Demystifier
atyy
There is the rigged Hilbert space formalism, which allows delta functions to be considered "generalized" basis functions (although these are not permitted in the Hilbert space formalism). However, we still do not consider a delta function to be an allowed quantum state (which must be square integrable), although we allow it to be a basis vector for mathematical operations.
http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf
Quantum Mechanics in Rigged Hilbert Space Language
https://arxiv.org/abs/quant-ph/0502053
The role of the rigged Hilbert space in Quantum Mechanics
https://arxiv.org/abs/1411.3263
Quantum Physics and Signal Processing in Rigged Hilbert Spaces by means of Special Functions, Lie Algebras and Fourier and Fourier-like Transforms
Enrico Celeghini, Mariano A. del Olmo
sysprog, Delta2, vanhees71 and 3 others
Gold Member
There is the rigged Hilbert space formalism, which allows delta functions to be considered "generalized" basis functions (although these are not permitted in the Hilbert space formalism). However, we still do not consider a delta function to be an allowed quantum state (which must be square integrable), although we allow it to be a basis vector for mathematical operations.
Thanks, atyy. Having said that, could you clarify what space should be considered as "full", "true" etc. for the case of harmonic oscillator, and if it's a rigged space, what does it practically/theoretically mean that this space cannot be generated by the harmonic oscillator Hamiltonian? Like, in the end, does the Hamiltonian produce a full basis via its eigenstates or not? And whether some state from the rigged space (a nontrivial one i.e. laying out of the space with countable basis) may be used as an initial condition for a harmonic oscillator Schrodinger equation? I think the questions I am asking may be somehow incorrect as well, so some clarifications of the validity of such questions may be helpful also. I should read the books you provided by 'immediate help' may be of value now...
Last edited:
Not rigid space, it is called rigged Hilbert space (in the sense of "upgraded" Hilbert space).
vanhees71 and MichPod
hilbert2
Gold Member
The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.
Gold Member
Not rigid space, it is called rigged Hilbert space (in the sense of "upgraded" Hilbert space).
oops. my poor English. sorry. fixing this.
Gold Member
The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.
But, speaking formally, why would the Hamiltonian define what is the Hilbert space for a particular arrangement, but not, say, a coordinate or momentum operator for which the spectrum is (always) continuous?
stevendaryl
Staff Emeritus
The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.
I would clarify that statement a little bit. What you really mean is that space of all eigenstates of the Hamiltonian. There might be countably many independent eigenstates, or there might be continuum many. But the Hilbert space is not forced on you by the Hamiltonian. The Hilbert space is (as I understand it) the collection of all square-integrable functions of 3D space. The Hilbert space is the same whether the Hamiltonian is the harmonic oscillator or the Coulomb potential.
PeterDonis
Mentor
2020 Award
The Hilbert space is (as I understand it) the collection of all square-integrable functions of 3D space. The Hilbert space is the same whether the Hamiltonian is the harmonic oscillator or the Coulomb potential.
Not quite. The Hilbert space for a particular system is the set of square-integrable functions that are solutions of the Schrodinger Equation with the applicable Hamiltonian. This is generally a proper subset of all square integrable functions (i.e., some functions are not possible solutions and are excluded from the Hilbert space). As long as that proper subset also satisfies all the axioms of a Hilbert space, there is no issue.
bhobba and Delta2
stevendaryl
Staff Emeritus
Not quite. The Hilbert space for a particular system is the set of square-integrable functions that are solutions of the Schrodinger Equation with the applicable Hamiltonian.
That's a little bit apples-to-oranges, because the Hilbert space contains functions of space, while solutions to Schrodinger's equation are functions of space and time. Do you mean the space of all functions ##f(x)## such that there exists a solution ##\psi(x,t)## to the Schrodinger equation, and a time ##t_0## such that ##f(x) = \psi(x,t_0)##?
PeterDonis
Mentor
2020 Award
Do you mean the space of all functions ##f(x)## such that there exists a solution ##\psi(x,t)## to the Schrodinger equation, and a time ##t_0## such that ##f(x) = \psi(x,t_0)##?
Yes.
stevendaryl
Staff Emeritus
Yes.
Okay, so what's an example of a square-integrable function that can't be a solution to Schrodinger's equation?
Oh, I guess an infinite square well gives an example: No function that is nonzero outside the well is a possible solution.
atyy
Thanks, atyy. Having said that, could you clarify what space should be considered as "full", "true" etc. for the case of harmonic oscillator, and if it's a rigged space, what does it practically/theoretically mean that this space cannot be generated by the harmonic oscillator Hamiltonian? Like, in the end, does the Hamiltonian produce a full basis via its eigenstates or not? And whether some state from the rigged space (a nontrivial one i.e. laying out of the space with countable basis) may be used as an initial condition for a harmonic oscillator Schrodinger equation? I think the questions I am asking may be somehow incorrect as well, so some clarifications of the validity of such questions may be helpful also. I should read the books you provided by 'immediate help' may be of value now...
The eigenstates of the Hamiltonian is a basis for all the physical states. This is a countably infinite basis. Each Hamiltonian eigenstate is an allowed physical state.
The eigenstates of the position and momentum operators are not bases for all physical states, but they are generalized bases (normally, we aren't so particular, and we don't bother to use the term "generalized" and we simply call them "bases"). They are uncountably infinite bases. In the Hilbert space formulation, all bases are countable (the Hilbert spaces of quantum mechanics are separable Hilbert spaces). In the rigged Hilbert space formulation, when we allow generalized bases, those can be countable (eg. Hamiltonian eigenstates) or uncountable (position eigenstates). While each Hamiltonian eigenstates is an allowed physical state, each position eigenstate is not an allowed physical state.
The rigged Hilbert space formalism roughly corresponds to what physicists do with Dirac notation. However, the first rigorous mathematical formulation of quantum mechanics was the Hilbert space formalism, which meant that the Dirac notation was somewhat "quick and dirty". Later it was found that many of the "quick and dirty" practices could be included in a rigorous formulation, if one extended the Hilbert space formalism to rigged Hilbert spaces.
bhobba and Demystifier
atyy
The reference by Celeghini and del Olmo says:
"lt seems trivial, but ##\{|n \rangle \}## has the cardinality of the natural numbers ##\aleph_{0}## and, as all bases in a Hilbert space have the same cardinality, the structure we have constructed (the quantum space on the line ##\mathbb{R}##) is not an Hilbert space but a Rigged Hilbert space."
dextercioby
PeterDonis
Mentor
2020 Award
Oh, I guess an infinite square well gives an example: No function that is nonzero outside the well is a possible solution.
More generally, any Hamiltonian that includes a potential which is only finite in a bounded region will have a restricted set of possible solutions.
Even more generally, any Hamiltonian that includes a potential which is not bounded at infinity (e.g., the harmonic oscillator) will, I think, have a restricted set of possible solutions.
Gold Member
In the rigged Hilbert space formulation, when we allow generalized bases, those can be countable (eg. Hamiltonian eigenstates) or uncountable (position eigenstates). While each Hamiltonian eigenstates is an allowed physical state, each position eigenstate is not an allowed physical state.
Do I understand you correctly that the same rigged Hilbert space is somehow considered as having both countable and uncountable basis?
Also, it looks like a plain wave solution of a free Hamiltonian cannot be considered as an "allowed physical state" with this approach. Not arguing, just thinking...
stevendaryl
Staff Emeritus
Even more generally, any Hamiltonian that includes a potential which is not bounded at infinity (e.g., the harmonic oscillator) will, I think, have a restricted set of possible solutions.
So the basis states for Harmonic Oscillator states are not complete (in the sense that there are square-integrable functions that are not expressible as linear combinations?)
PeterDonis
Mentor
2020 Award
So the basis states for Harmonic Oscillator states are not complete
They're complete for the Hilbert Space of possible solutions of the harmonic oscillator Schrodinger Equation. I don't think they're complete for the Hilbert space of all square integrable functions on the real numbers, but that's not the relevant Hilbert space for the harmonic oscillator.
Keith_McClary
Gold Member
they're complete for the Hilbert space of all square integrable functions on the real numbers
Proof of completeness.
atyy
stevendaryl
Staff Emeritus
They're complete for the Hilbert Space of possible solutions of the harmonic oscillator Schrodinger Equation. I don't think they're complete for the Hilbert space of all square integrable functions on the real numbers, but that's not the relevant Hilbert space for the harmonic oscillator.
At this point, I'm just curious about the mathematical issue:
Some definitions:
If ##f(x)## and ##g(x)## are square-integrable functions, say that ##f \approx g## if
##\int_{-\infty}^{+\infty} |f(x) - g(x)|^2 dx = 0##.
Let ##\psi_n(x)## be the ##n^{th}## Harmonic oscillator basis state. If ##f(x)## is a complex-valued function on ##(-\infty, +\infty)##, let's say that ##f(x)## is representable by HO basis states if there is a sequence of complex numbers ##C_n## such that ##\sum_n C_n \psi_n(x)## converges to some function ##g(x)## where ##f \approx g##.
So is every square-integrable function ##f(x)## representable by HO basis states? If not, what is a counter-example?
Keith_McClary
Gold Member
So is every square-integrable function representable by HO basis states? If not, what is a counter-example?
Yes. You probably didn't see my post about completeness a minute before.
atyy
stevendaryl
Staff Emeritus
There's a step that is omitted in the proof. Maybe it's because it's obvious? There are two different claims: (1) that ##f(x)## has a nonzero overlap with at least one basis state.
(2) that ##f(x)## can be represented as a (possibly infinite) sum of basis states.
The article you pointed to proved the first statement. Does the second follow obviously? I suppose you could just keep taking out the part of ##f(x)## that is not in the overlap, and then apply the theorem again. If that process converges, then you get a representation?
A second observation is that the proof is for functions that are square-integrable when multiplied by ##e^{-x^2}##.
atyy
Do I understand you correctly that the same rigged Hilbert space is somehow considered as having both countable and uncountable basis?
Also, it looks like a plain wave solution of a free Hamiltonian cannot be considered as an "allowed physical state" with this approach. Not arguing, just thinking...
In the Hilbert space formalism, the Hilbert space of quantum mechanics is always separable and has a countable basis.
In the rigged Hilbert space formalism, the generalized formalism allows both countable and uncountable bases in a generalized sense.
Yes, a plain wave solution cannot be considered an allowed physical state in both the Hilbert space and the rigged Hilbert space formalisms.
Both Hilbert space and rigged Hilbert space formalisms are the same with respect to physical content, so the Hilbert space formalism is enough. The rigged Hilbert space formalism makes the elegant but apparently mathematically dirty Dirac formalism closer to something mathematically respectable.
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vanhees71, mattt, MichPod and 2 others
Keith_McClary
Gold Member
A second observation is that the proof is for functions that are square-integrable when multiplied by .
That's a bit confusing:
An equivalent formulation of the fact that Hermite polynomials are an orthogonal basis for L2(R, w(x) dx) consists in introducing Hermite functions (see below), and in saying that the Hermite functions are an orthonormal basis for L2(R). - Wikipedia
atyy
A. Neumaier
The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.
All Hilbert spaces used in quantum mechanics (and most in QFT) are separable, i.e., have a countable basis. Most are also infinite-dimensional - then they also have continuous representations.
For example, the state space of a single particle can be expressed in the position or momentum representation, where each state is a superposition of uncountably many improper states, or in an angular momentum representation, where the basis is countable, given by the spherical harmonics (n,s,p,d,f,....)
This is independent of whether particles can escape to infinity. The latter is a property of the particular Hamiltonian used, not of the Hilbert space.
Okay, so what's an example of a square-integrable function that can't be a solution to Schrodinger's equation?
Oh, I guess an infinite square well gives an example: No function that is nonzero outside the well is a possible solution.
The Hamiltonian of the infinite square well does not act on the space of square integrable functions on ##R##, only on the space of square integrable functions on the support of the well. Thus the latter is the correct Hilbert space for this potential, and this is not a counterexample. Indeed, there are no counterexamples.
They're complete for the Hilbert Space of possible solutions of the harmonic oscillator Schrodinger Equation. I don't think they're complete for the Hilbert space of all square integrable functions on the real numbers, but that's not the relevant Hilbert space for the harmonic oscillator.
No. The harmonic oscillator basis is complete for that space. More generally, the collection of orthonormal eigenstates of any self-adjoint Hamiltonian on ##:^2(R)## with discrete spectrum is complete in this space. Eigenstates of self-adjoint Hamiltoniand with continuous spectrum need the rigged Hilbert space extension, which allows for superpositions of uncountably many basis states.
a planr wave solution of a free Hamiltonian cannot be considered as an "allowed physical state" with this approach.
No, because it is not square integrable.
Last edited:
vanhees71, strangerep and gentzen | 4,009 | 17,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.931349 |
https://www.aqua-calc.com/calculate/weight-to-volume/substance/perfluoromethyl-blank-iodide | 1,611,002,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00522.warc.gz | 663,785,846 | 7,566 | # Volume of Perfluoromethyl iodide
## perfluoromethyl iodide: convert weight to volume
### Volume of 100 grams of Perfluoromethyl iodide
centimeter³ 55.56 milliliter 55.56 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.23 inch³ 3.39 US fluid ounce 1.88 liter 0.06 US gallon 0.01 meter³ 5.56 × 10-5 US pint 0.12 metric cup 0.22 US quart 0.06 metric tablespoon 3.7 US tablespoon 3.76 metric teaspoon 11.11 US teaspoon 11.27
### The entered weight of Perfluoromethyl iodide in various units of weight
carat 500 ounce 3.53 gram 100 pound 0.22 kilogram 0.1 tonne 0 milligram 100 000
#### How many moles in 100 grams of Perfluoromethyl iodide?
There are 510.44 millimoles in 100 grams of Perfluoromethyl iodide
#### Foods, Nutrients and Calories
LOBSTER and CHEESE BITES, UPC: 085239064634 contain(s) 286 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Torpedo Beach weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Ytterbium [Yb] weighs 6 965 kg/m³ (434.81075 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 15W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
A troy pound is a troy weight measurement unit
The fuel consumption or fuel economy measurement is used to estimate gas mileage and associated fuel cost for a specific vehicle.
long tn/pt to mg/pt conversion table, long tn/pt to mg/pt unit converter or convert between all units of density measurement.
#### Calculators
Calculate gas pressure at different temperatures | 562 | 1,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.698029 |
https://uk.mathworks.com/matlabcentral/cody/players/2789428-richard-zapor/solved | 1,601,434,023,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00753.warc.gz | 593,646,734 | 21,883 | Cody
# Richard Zapor
Rank
Score
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#### Problem 1123. Rubik's Cube: 30 Moves or Less : Minimum Moves
Created by: Richard Zapor
#### Problem 937. Rubik's Mini Cube: Solve Randomized Cube in 11 Moves or Less; Score is by Time (msec)
Created by: Richard Zapor
#### Problem 886. Rubik's Cube: Solve Randomized Cube - Score : Moves
Created by: Richard Zapor
#### Problem 879. Perform Rubik's Cube Moves
Created by: Richard Zapor
#### Problem 931. Rubik's Mini-Cube: Solve Randomized Mini-Cube - Score : Moves
Created by: Richard Zapor
#### Problem 1139. Rubik's Cube: 30 Moves or Less : Contest Scoring (Time/Size/Moves)
Created by: Richard Zapor
#### Problem 1137. Rubik's Cube: 30 Moves or Less : Cody Size
Created by: Richard Zapor
#### Problem 44943. Calculate Amount of Cake Frosting
Created by: Pooja Lalan
#### Problem 2265. 2048 Next Move
Created by: Ned Gulley
Tags game, 2048
#### Problem 43269. Get all prime factors
Created by: Jamil Kasan
#### Problem 52. What is the next step in Conway's Life?
Created by: Cody Team
Tags life, automata
#### Problem 42651. Vector creation
Created by: ruta bhat
Tags easy, vector
#### Problem 3076. Create a vector
Created by: Carlton
Tags vector, uab
#### Problem 2631. Flip the vector from right to left
Created by: Pritesh Shah
#### Problem 1257. PONG 001: Player vs Wall, 4 Lives, Interactive download
Created by: Richard Zapor
#### Problem 46618. Kaggle 2020 Drone Delivery Contest
Created by: Richard Zapor
#### Problem 3047. Scrabble Scores - 2
Created by: goc3
Tags game, board, word
#### Problem 682. Image Processing 002 : Fix Vignetting in a Visible Sensor
Created by: Richard Zapor
#### Problem 882. Solve Rubik's Cube - One Rotation
Created by: Richard Zapor
Tags game, rubik
#### Problem 44240. Area of a Square
Created by: Jerry Gu
#### Problem 44092. Find the minimal value in N*N Matrix
Created by: Said BOUREZG
Tags matrix, basics, f
#### Problem 44235. Shift elements of vector left
Created by: Yash Ratanpara
Tags easy, vector, basics
#### Problem 44232. Relation between functions "dec2bin" & "dec2binvec"
Created by: Said BOUREZG
Tags char, flip, minus
#### Problem 44227. Divisible by 21
Created by: Said BOUREZG
#### Problem 44103. Find the maxmum value of N*N Matrix
Created by: Said BOUREZG
Tags max
#### Problem 44096. Matrix game: Winner takes all
Created by: Dongdong Yue
#### Problem 43562. Mastermind
Created by: Bert
Tags fun, difficult, hard
#### Problem 44239. Mastermind IV: Optimal Solver - max of 5 guesses
Created by: Richard Zapor
#### Problem 44238. Mastermind III: Solve in 1
Created by: Richard Zapor
#### Problem 44237. Mastermind II: Solve in 8 or less
Created by: Richard Zapor
#### Problem 44236. Mastermind I: Solve all 1296 cases
Created by: Richard Zapor
#### Problem 1512. Clock Solitaire
Created by: Nicholas Howe
#### Problem 44079. GJam 2017 Kickstart: Leader (Large)
Created by: Richard Zapor
#### Problem 44078. GJam 2017 Kickstart: Leader (Small)
Created by: Richard Zapor
#### Problem 44077. GJam 2017 Kickstart: Vote (Large)
Created by: Richard Zapor
#### Problem 44076. GJam 2017 Kickstart: Vote (Small)
Created by: Richard Zapor
Tags gjam, 2017, vote
#### Problem 44075. GJam 2017 Kickstart: Parentheses (Large)
Created by: Richard Zapor
#### Problem 44074. GJam 2017 Kickstart: Parentheses (Small)
Created by: Richard Zapor
#### Problem 1138. Rubik's Cube: 30 Moves or Less: Minimum Avg Time
Created by: Richard Zapor
#### Problem 42914. Counting the Grand Primes
Created by: John D'Errico
#### Problem 42892. Transpose of matrix
Created by: Pritesh Shah
#### Problem 42856. Block average
Created by: Peng Liu
#### Problem 42854. Crunch that matrix!
Created by: James
Created by: HH
#### Problem 42853. Fifteen Parity Check
Created by: Richard Zapor
Tags bug, mod, fifteen
#### Problem 42844. Detect pair of equal values in a Matrix
Created by: Pramit Biswas
Tags basic, matlab
#### Problem 42845. generate number in particular way
Created by: Pramit Biswas
Tags basic, matlab
#### Problem 42846. Wien's displacement law
Created by: Daniel Pereira
#### Problem 42847. Isothermal Expansion
Created by: Daniel Pereira
1 – 50 of 1,100 | 1,220 | 4,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-40 | latest | en | 0.677416 |
http://mathhelpforum.com/pre-calculus/49645-hyperbola.html | 1,481,107,598,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542060.60/warc/CC-MAIN-20161202170902-00172-ip-10-31-129-80.ec2.internal.warc.gz | 180,445,325 | 10,478 | # Thread: Hyperbola
1. ## Hyperbola
Hyperbola is at the center of coordinate system. Its foci (?) are on abscissa. There are two points on the hyperbola; these are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{9})$. What is hyperbola's equation?
(I'm having trouble with solving two equations that follow.)
2. Originally Posted by courteous
Hyperbola is at the center of coordinate system. Its foci (?) are on abscissa. There are two points on the hyperbola; these are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{9})$. What is hyperbola's equation?
(I'm having trouble with solving two equations that follow.)
Since its foci are on the abscissa [x-axis] and the hyperbola is centered at the origin, we know that the hyperbola has the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a>b$
Now plug the two points into this equation, and you will generate two new equations:
At $\left(4,\frac{4\sqrt{7}}{9}\right)$, we get $\frac{16}{a^2}-\frac{112}{81b^2}=1$
At $\left(\frac{3\sqrt{5}}{2},2\right)$, we get $\frac{45}{4a^2}-\frac{4}{b^2}=1$
After solving the system of equations, I got $a=\sqrt{\frac{981}{53}}$
But I'm getting $b^2=-\frac{1744}{171}\implies b\text{ is complex.}$
Are you sure that you wrote down the coordinates correctly?
--Chris
3. Originally Posted by Chris L T521
Are you sure that you wrote down the coordinates correctly?
Sorry!!! No!!! Sorry, Chris! The points are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{3})$. The $T_2$ y-coordinate's denominator is 3 (not 9).
I've done it so many times that I've automatically restarted with partially already calculated number. | 525 | 1,628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.85066 |
https://howigotjob.com/articles/how-long-is-a-decade-how-long-is-a-century/ | 1,713,137,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00416.warc.gz | 287,328,745 | 21,695 | Inspiring Job Search Stories
# HOW LONG IS A DECADE? HOW LONG IS A CENTURY?
HOW LONG IS A DECADE? HOW LONG IS A CENTURY?
• 1 CENTURY IS EQUIVALENT TO 10 DECADE
• CENTURY IS EQUIVALENT TO A PERIOD OF 100 YEARS
• 1 DECADE IS EQUIVALENT TO 0.1 CENTURY
• DECADE IS EQUIVALENT TO A PERIOD OF 10 YEARS
LET US EXPAND MORE!
• ONE MINUTE MAKES SIXTY SECONDS
• ONE HOUR MAKES 60 MINUTES
• 60 MINUTES x 60 SECONDS = 3,600 SECONDS
• 24 HOURS IN A DAY
• 12 hours am to noon
• 12 hours pm to midnight = 24 hours
• 1 hour= 60 minutes
• 12 hours x 60 minutes = 720 minutes (am)
• 12 hours x 60 minutes = 720 minutes (pm)
• 24 hours x 60 minutes = 1,440 minutes in a day
• 1,440 minutes x 60 seconds = 86, 400 seconds in a day
28 days for the month of February
1 day = 1, 440 minutes
1,440 minutes x 60 seconds = 86, 400 seconds in a day
28 days x 1, 440 minutes = 40, 320 minutes in a month
40, 320 minutes for the month of February x 60 seconds = 2, 419, 200 seconds for 28days
How about the months with 30 days?
30 days for the month of April, June, September, November
1 day = 1, 440 minutes
1,440 minutes x 60 seconds = 86, 400 seconds in a day
30 days x 1, 440 minutes =43, 200 minutes in a month
April, June, September, November = 120 days
120 days x 1,440 minutes = 172, 800 minutes in 4 months
43, 200 minutes in a month x 4 = 172, 800
172, 800 minutes for 4 months x 60 seconds = 10, 368, 000 seconds for 120 days
How about the months with 31 days?
31 days for the month of January, March, May, July, August, October, December
1 day = 1, 440 minutes
1,440 minutes x 60 seconds = 86, 400 seconds in a day
31 days x 1,440 minutes = 44, 640 minutes in a month
January, March, May, July, August, October, December
217 days x 1, 440 minutes = 312, 480 minutes in 7 months
44,640 minutes in a month x 7 = 312, 480
312, 480 minutes for 7 months x 60 seconds = 18, 748, 800 seconds for 217 days
Let us shorten it. In a Year it has 365 days (12 months)
365 days x 1440 minutes = 525, 600 minutes
525, 600 minutes x 60 seconds = 31, 536, 000
Very long, isn’t it? But it should not be boring but be thankful every day and have fun and value every moment. Let’s continue.
How to know how many years equals a decade and how long
Decade = years divided by 10
1 year is equivalent to 0.1000664383561 Decade
0.1000664383561 multiply by 10 = 1 Decade
1 year has 525, 600 minutes x 10 years or 1 Decade = 5, 256, 000 minutes
5, 256, 000 minutes x 60 = 315, 360, 000 seconds for a decade
Think about how long those moments were.
You just have to be thankful if you went through those seconds, minutes, days, years and decades. Always remember every second is important because it is a great blessing to be awake every morning.
Let us continue…
Centuries = years divided by 100
Century is equivalent to 100 years
If a year has 365 days or 12 months
365 days x 1440 minutes = 525, 600 minutes
525, 600 minutes x 60 seconds = 31, 536, 000
And a decade is equivalent to 10 years
1 year is equivalent to 0.1000664383561 Decade
0.1000664383561 multiply by 10 = 1 Decade
1 year has 525, 600 minutes x 10 years or 1 Decade = 5, 256, 000 minutes
5, 256, 000 minutes x 60 = 315, 360, 000 seconds for a decade
1 decade is equivalent to 0.1 centuries it simply means we need to multiply to achieve 100 years or a century.
10 years (decade) multiply by 10 = 100 years or a century
It takes ten decades to form a century or a hundred years
If a year has 365 days or 12 months
365 days x 1440 minutes = 525, 600 minutes
525, 600 minutes x 60 seconds = 31, 536, 000
And a decade is equivalent to 10 years
1 year is equivalent to 0.1000664383561 Decade
0.1000664383561 multiply by 10 = 1 Decade
1 year has 525, 600 minutes x 10 years or 1 Decade = 5, 256, 000 minutes
5, 256, 000 minutes x 60 = 315, 360, 000 seconds for a decade
5,256,000 minutes of a decade multiply by 10 = 52, 560, 000 minutes for a century
52, 560, 000 minutes x 60 = 3, 153, 600, 000 seconds for a century
Just simple as in a year it has 525,600 minutes x 100 years = 52, 560, 000 minutes
Years, Decades and centuries are used to measure time.
Perhaps many are blessed to reach a Century their age nowadays especially now that we are struggling in pandemic always eat nutritious foods to keep the body strong and healthy, avoid unhealthy foods such as junk foods, alcohol and other luxuries, avoid crowded place and wear face masks and bring alcohol every time. Take care of yourself and make it a habit to pray and give thanks for every day that you wake up and get up because life is a blessing. God bless!!!
HOW LONG IS A DECADE? HOW LONG IS A CENTURY?
Scroll to top | 1,453 | 4,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.755549 |
https://scicomp.stackexchange.com/questions/36355/are-spatial-boundary-conditions-required-for-pdes-discretized-with-method-of-lin/36356#36356 | 1,632,649,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00390.warc.gz | 537,227,595 | 39,806 | # Are spatial boundary conditions required for PDEs discretized with Method of Lines?
As far as I understand, you need to define boundary conditions in time and space to select a unique solution to a PDE and make it solvable. However, in ODEs I only need to specific the initial value in time which I think of as a boundary condition. If I now use the method of lines to transform a PDE into a system of ODEs, do I need the spatial boundary conditions? If yes, are they imposed by fixing some of the "lines" to given functions? If no, how can it be that they are first required but then somehow the need for them disappears?
I got this question because I did this with the 2D acoustic wave equation and the result looks reasonable, yet I did not impose any spatial boundary conditions. I have also read somewhere (that I sadly cannot find anymore) that there are implicit boundary conditions. Does that mean that me not specifying any conditions is equivalent to choosing some implicitly?
Let's take the example of the unsteady one-dimensional heat equation inside a solid on a domain $$x\in[0,1]$$: $$\partial_t u - D\partial_{xx} u = 0$$ with the initial profil $$u(0,x) = u_0(x)$$ at $$t=0$$, and Dirchlet boundary conditions enforcing that the wall at $$x=0$$ (respectively $$x=1$$) is at temperature $$u_{L}$$ (respectively $$u_{R}$$). If we use discretise our solid with a uniformely distributed set of points $$i \in [1,N]$$ at positions $$x_i = i \Delta x$$ with $$\Delta x = 1/(N+1)$$ and use second order centered finite differences for the diffusion term, we obtain the following semi-discrete set of ODEs: $$d_t u_i = D\frac{u_{i+1} - 2u_i + u_{i-1}}{\Delta x^2}$$ for $$i \in [2, N-1]$$. The boundary points $$i=1$$ and $$i=N$$ must be taken care of separately to ensure the boundary conditions are respected. Let's take the example of the boundary at $$x=0$$. The temperature $$u(x=0)$$ is enforced to be 0. Our first point is as $$x=1/(n+1)$$, therefore we can write: $$d_t u_1 = D\frac{u_{2} - 2u_1 + u_{L}}{\Delta x^2}$$
Similarly for our last point: $$d_t u_N = D\frac{u_{N-1} - 2u_N + u_{R}}{\Delta x^2}$$
We have to use these formulations, otherwise we would not even be able to compute the centered finite difference approximation of the diffusion terms at these outer points. An error would for example be to aproximate these terms with a decentered fintie difference formula, e.g for the point $$i=1$$: $$d_t u_1 = D\frac{ 2u_1 - 5u_2 + 4 u_3 - u_4}{\Delta x^2}$$ which uses a second-order forward finite difference expression of the diffusion term. But then, there is not guarantee that the temperature at $$x=0$$ will remain at $$u_L$$, i.e. your system will most likely have an unphysical behaviour.
Sometimes we rather use a Neumann boudary condition to specify a given heat flux at the boundary. Let's take the case of a Neumann BC at $$x=0$$, with a zero heat flux, i.e. $$\frac{\partial u}{\partial x}(x=0) = 0$$. This condition can be discretised as $$\frac{u_1-u_L}{\Delta x}=0$$, which means that $$u_L(t) = u_1(t)$$ at all times $$t$$. You can then inject this value of $$u_L$$ in the second equation of this answer.
Technically, a discretized PDE is integrated in time as follows:
• the time integration routines gives you the discrete solution profile $$u_i, i \in [1,N]$$ at time $$t_n$$
• we compute the boundary values $$u_L$$ and $$u_R$$ so as to enforce the proper boundary conditions
• we compute the time derivatives of our discrete solution values and return them to time integration routine.
I don't know how you've managed to integrate your system without specifying BCs at all times, maybe its just luck in your formulation or the particular equation you consider. But in the case of the discrete heat equation I've treated here, I see 3 potential ways of having this problem:
• you forget to specify $$d_t u_1$$ and then you actually enforce a Dirichlet boundary condition that reads $$u(t,x_1)= u_0(x_1)$$, i.e. the wall temperature remains equal to its initial value
• you forget the term $$u_L$$ in the third equation, which the means you implicitly assume $$u_L=0$$ .
• you used the forward finite difference expression, thus omitting the BC value. In this case, we can see what the implcitly asusmed of $$u_L$$ is by equating the 3rd and 5th equation and we obtain: $$u_L = -4u_1 + 6u_2 - 4u_3 + u_4$$, that is $$u_L$$ is relatively "unpredictable", and consequently so will be $$u_1$$, and $$u_2$$ and so on...
• Thank you very much for the detailed answer! One thing I did not mention in the question is that I discretized with finite elements in space with triangular elements. Then I get two the mass and stiffness matrices and the equation $Au_{tt} = Bu^{(t)}$ that can be solved for the second t-derivates which I then give to the ODE solver. And here I just used the $A$ and $B$ that came out of the computation. However, my suspicion is that the entries for the boundary points are wrong because they are not surrounded by triangles on all sides. They are on the boundary after all. Nov 23 '20 at 12:56
• I haven't done any finite elements for years, so I'd rather not answer on that :/ As far as I remember the boundary conditions are already taken into account when you write your weak formulation of the problem. Nov 23 '20 at 12:58
• From reading this sfepy.org/doc-devel/solving_pdes_by_fem.html, it looks like if you don't do anything for implementing BC in FEM you end up with Neumann zero flux BC (what they call "natural" BC); at least that's what it looks like in the example discussed there, not sure how general this is. Nov 23 '20 at 19:24 | 1,506 | 5,626 | {"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-39 | latest | en | 0.90642 |
http://www.jiskha.com/display.cgi?id=1352438343 | 1,498,135,542,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319265.41/warc/CC-MAIN-20170622114718-20170622134718-00368.warc.gz | 617,528,970 | 3,560 | # Science
posted by on .
What is the wavelength of a 600Hz sound wave ( assume the speed of sound is 340 m/s )?
• Science - ,
L = 340m/s * 1s/600c = 0.567 m. | 56 | 161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-26 | latest | en | 0.827683 |
https://www.kidzworld.com/article/863-the-pain-formula | 1,550,576,116,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489933.47/warc/CC-MAIN-20190219101953-20190219123953-00065.warc.gz | 863,373,855 | 49,433 | # The Pain Formula
December 27, 2006
## Quiz The Coach
What would hurt more - being hit by a fastball from one of baseball's best pitchers, being nailed by a slapshot from a hockey superstar or being smoked by a tennis ball served up by Andy Roddick? Kidzworld looks for the anwers with the Pain Formula.
## The Pain Formula - How Much Pain?
To figure out how much pain is caused by a baseball, tennis ball or hockey puck, the weight of each object is multiplied by the speed that it's travelling.
## The Pain Formula - Fastball Pain
Let's start with a fastball from one of baseball's top pitchers, which flys over the plate at 100 miles per hour (160 kms). A baseball weighs about five ounces (142 grams).
• 100 miles per hour x 5 ounces = 500 Units of Pain if you're not looking when the pitcher throws towards home plate. That's why batters wear batting helmets.
• ## The Pain Formula - Slapshot Pain
Dion Phaneuf of the Calgary Flames, who has the hardest shot in the NHL, slaps the puck towards a goalie at about 101 miles per hour (162 kms). A hockey puck weighs 6.1 ounces (173 grams.)
• 101 miles per hour x 6.1 ounces = 616 Units of Pain. Now you know why goalies wear so much padding.
• #### The Pain Formula - Tennis Serve Pain
A serve from Andy Roddick flies across the tennis court at 150 miles per hour (240 kms). A tennis ball weighs 1.07 ounces (30.4 grams).
• 150 miles per hour x 1.07 ounces = 160.5 Units of Pain.
• ##### The Pain Formula - Watch Out For The Slapshot
Kidzworld's Pain Formula has figured out that getting nailed by a slapshot would produce the most pain. But there are a few factors that haven't been considered, such as where the ball or puck hits you, how much bounce each object has and the angle that the puck or ball hits you at. For example, it hurts a lot more when you get smoked in the groin than it does getting nailed in the shoulder. Objects that are squishy and bouncy have "more give" so they hurt less than an object that is rock hard. That's why the tennis ball hurts less than the baseball and why your parents might let you shoot your younger brother with a Nerf ball but won't let you pelt him with rocks.
• If you've got a question about sports science or sports injuries, to Kidzworld.
• Related Stories:
• Human Growth Hormone
• World's Most Dangerous Sports
• Your Worst Sports Injuries
• More Fitness and Exercise Info!
• You May Like
Related Content | 584 | 2,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.936245 |
https://docs.rs/bbox/latest/bbox/ | 1,722,747,982,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00613.warc.gz | 171,898,324 | 8,963 | # [−][src]Crate bbox
bbox is crate for managing axis aligned 3d Bounding Boxes. Bounding Boxes can be created, dilated, transformed and joined with other Bounding Boxes using CSG operations. Finally you can test whether or not a Bounding Box contains some point and what approximate distance a Point has to the Box.
# Examples
Intersect two Bounding Boxes:
```use nalgebra as na;
let bbox1 = bbox::BoundingBox::<f64>::new(&na::Point3::new(0., 0., 0.),
&na::Point3::new(1., 2., 3.));
let bbox2 = bbox::BoundingBox::<f64>::new(&na::Point3::new(-1., -2., -3.),
&na::Point3::new(3., 2., 1.));
let intersection = bbox1.intersection(&bbox2);```
Rotate a Bounding Box:
```use nalgebra as na;
let rotation = na::Rotation::from_euler_angles(10., 11., 12.);
let bbox = bbox::BoundingBox::<f64>::new(&na::Point3::new(0., 0., 0.),
&na::Point3::new(1., 2., 3.));
let rotated_box = bbox.transform(&rotation.to_homogeneous());```
Is a point contained in the Box?
```use nalgebra as na;
let bbox = bbox::BoundingBox::<f64>::new(&na::Point3::new(0., 0., 0.),
&na::Point3::new(1., 2., 3.));
let result = bbox.contains(&na::Point3::new(1., 1., 1.));```
Calculate approximate distance of a point to the Box:
```use nalgebra as na;
let bbox = bbox::BoundingBox::<f64>::new(&na::Point3::new(0., 0., 0.),
&na::Point3::new(1., 2., 3.));
let distance = bbox.distance(&na::Point3::new(1., 1., 1.));```
## Structs
BoundingBox 3D Bounding Box - defined by two diagonally opposing points. | 477 | 1,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.528022 |
https://nrich.maths.org/public/leg.php?code=-333&cl=2&cldcmpid=10424 | 1,540,262,749,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00205.warc.gz | 746,055,417 | 9,659 | # Search by Topic
#### Resources tagged with Investigations similar to Multiply Multiples 2:
Filter by: Content type:
Age range:
Challenge level:
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### New House
##### Age 7 to 11 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
##### Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Tiles on a Patio
##### Age 7 to 11 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### Halloween Investigation
##### Age 7 to 11 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Tiling
##### Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Birthday Cake Candles
##### Age 7 to 11 Challenge Level:
This challenge involves calculating the number of candles needed on birthday cakes. It is an opportunity to explore numbers and discover new things.
##### Age 7 to 11 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### Room Doubling
##### Age 7 to 11 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Calcunos
##### Age 7 to 11 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Stairs
##### Age 5 to 11 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Polo Square
##### Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Bean Bags for Bernard's Bag
##### Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Round and Round the Circle
##### Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Newspapers
##### Age 7 to 11 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Making Boxes
##### Age 7 to 11 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### Cubes Here and There
##### Age 7 to 11 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### My New Patio
##### Age 7 to 11 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### More Plant Spaces
##### Age 7 to 14 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Magazines
##### Age 7 to 11 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### 3 Rings
##### Age 7 to 11 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Ice Cream
##### Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### More Children and Plants
##### Age 7 to 14 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Plants
##### Age 5 to 11 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### Crossing the Town Square
##### Age 7 to 11 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Building with Rods
##### Age 7 to 11 Challenge Level:
In how many ways can you stack these rods, following the rules?
### Division Rules
##### Age 7 to 11 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Train Carriages
##### Age 5 to 11 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### Street Party
##### Age 7 to 11 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Sets of Numbers
##### Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Magic Constants
##### Age 7 to 11 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
##### Age 7 to 11 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Polygonals
##### Age 7 to 11 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### More Transformations on a Pegboard
##### Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Cuboid-in-a-box
##### Age 7 to 11 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### Hexpentas
##### Age 5 to 11 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### Fencing Lambs
##### Age 7 to 11 Challenge Level:
A thoughtful shepherd used bales of straw to protect the area around his lambs. Explore how you can arrange the bales.
### Abundant Numbers
##### Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Two by One
##### Age 7 to 11 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Sticks and Triangles
##### Age 7 to 11 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### Two on Five
##### Age 5 to 11 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Ip Dip
##### Age 5 to 11 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### It Was 2010!
##### Age 5 to 11 Challenge Level:
If the answer's 2010, what could the question be?
### Three Sets of Cubes, Two Surfaces
##### Age 7 to 11 Challenge Level:
How many models can you find which obey these rules?
### Making Cuboids
##### Age 7 to 11 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Escher Tessellations
##### Age 7 to 11 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
### Tea Cups
##### Age 7 to 14 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour. | 2,202 | 9,654 | {"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} | 3.53125 | 4 | CC-MAIN-2018-43 | latest | en | 0.913438 |
http://stackoverflow.com/questions/10865089/comparing-multiple-arrays?answertab=oldest | 1,419,726,474,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447554236.61/warc/CC-MAIN-20141224185914-00097-ip-10-231-17-201.ec2.internal.warc.gz | 95,327,947 | 17,731 | # Comparing multiple arrays
I'm trying to figure out how to write a tie-breaker function. The following records are all tied for one reason or another, so to break the tie, we sort each result and then go through each line. The tie is broken at the first point there is a difference.
In the example below on the first pass \$result[c] is eliminated, but a and b are still tied. Then on the second pass a is eliminated because it is greater than b. So the result is b, a, c
``````\$result[a] = array(1, 3, 4, 5);
\$result[b] = array(1, 2, 3, 7);
\$result[c] = array(2, 3, 3, 5);
``````
And to make it even more complicated, I won't always have the same number of results to compare. It could be anything more than 2.
I really hope this makes sense.
-
I've added an answer that is 100% functional and produces the correct output. It will even work for an arbitrary number of competitors. – nickb Jun 2 '12 at 20:54
In php, you can actually compare arrays directly using relation operators
``````if (\$result['a'] > \$result['b']) {
}
``````
php will just loop through the sub arrays, comparing their elements. There's more detail if you want to read http://php.net/manual/en/language.operators.comparison.php
Anyway, you can take advantage of this and just sort it.
``````asort(\$result);
print_r(\$result);
``````
If you need a way to get the n'th place entry, do
``````asort(\$result);
print_r(\$result);
\$ranked = array_keys(\$result);
\$secondPlace = \$ranked[1]; // a
print_r(\$result[\$secondPlace]);
``````
and if you need the rank indice of a letter
``````\$letterRanks = array_flip(\$ranked);
echo \$letterRanks['a']; // 1, for second
``````
-
This nailed it. Once I have my results arrays sorted, which I had to do anyway the asort function did exactly what I needed with one line of code. – Mcg1978 Jun 3 '12 at 11:22
``````\$i = 0;
while (count(\$result) > 1) {
\$winners = array();
\$best = PHP_INT_MAX;
foreach (\$result as \$x) {
if (\$x[\$i] == \$best) {
\$winners[] = \$x;
} else if (\$x[\$i] < \$best) {
\$winners = array(\$x);
\$best = \$x[\$i];
}
}
\$i++;
\$result = \$winners;
}
``````
This is just a quick and dirty piece of code... it doesn't handle the situation where the arrays are different sizes. Mostly because I'm not sure which of array(1,2,3) or array(1,2) should "win". Also, it doesn't do any array bounds checking or handle the situation where more than one array is tied after all elements have been compared.
-
Thanks for the answer Bob. Luckily the arrays will always be the same size. But your example resets the \$result array after every iteration, surely that won't work? Oh, and if they're still tied after that process it falls to another process so we'd just need to know if it failed to resolve the ties. – Mcg1978 Jun 2 '12 at 19:29
Here is a working solution. Let me know if you need a better explanation of how it works. I've left the debugging statements in so you should be able to discern what it is doing. This solution will work with an arbitrary number of competitors, as long as they each have the same `\$num_elements` in each array.
``````\$result = array();
\$result['a'] = array(1, 3, 4, 5);
\$result['b'] = array(1, 2, 3, 7);
\$result['c'] = array(2, 3, 3, 5);
\$num_elements = 4; // In each array
\$num_competitors = count( \$result);
\$finish_order = array();
\$keys = \$winners = array_keys( \$result);
// \$i is the current index into each competitor's array
// \$j is the current index into the \$keys array (telling us the current competitor)
// \$k is the next index into the \$keys array (telling us the next competitor, i.e. the competitor to compare the current competitor with)
for( \$i = 0; \$i < \$num_elements; \$i++) {
// If we've eliminated all but one winner, we're done!
if( count( \$winners) == 1) {
\$finish_order[] = array_pop( \$winners);
break;
}
echo 'Element number ' . \$i . "\n";
for( \$j = 0; \$j < \$num_competitors; \$j++) {
// If we've already eliminated this competitor, continue;
if( !isset( \$winners[\$j])) continue;
for( \$k = \$j + 1; \$k < \$num_competitors; \$k++) {
// If we've already eliminated this competitor, continue;
if( !isset( \$winners[\$k])) continue;
echo "\t - Who wins: " . \$result[ \$keys[\$j] ][\$i] . ' from ' . \$keys[\$j] . ' or ' . \$result[ \$keys[\$k] ][\$i] . ' from ' . \$keys[\$k] . "?\n";
if( \$result[ \$keys[\$j] ][\$i] < \$result[ \$keys[\$k] ][\$i]) {
echo "\t\t *** " . \$keys[\$k] . ' is out!' . "\n";
\$finish_order[] = \$keys[\$k];
unset( \$winners[\$k]);
}
if( \$result[ \$keys[\$j] ][\$i] > \$result[ \$keys[\$k] ][\$i]) {
echo "\t\t *** " . \$keys[\$j] . ' is out!' . "\n";
\$finish_order[] = \$keys[\$j];
unset( \$winners[\$j]);
}
}
}
}
echo "Game over - Result order is: " . implode( ', ', array_reverse( \$finish_order));
``````
Output:
``````Element number 0
- Who wins: 1 from a or 1 from b?
- Who wins: 1 from a or 2 from c?
*** c is out!
Element number 1
- Who wins: 3 from a or 2 from b?
*** a is out!
Game over - Result order is: b, a, c
``````
Demo
- | 1,522 | 5,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2014-52 | latest | en | 0.861948 |
http://reference.wolfram.com/legacy/language/v10.4/ref/GeoRange.html | 1,508,686,595,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825308.77/warc/CC-MAIN-20171022150946-20171022170946-00824.warc.gz | 277,403,116 | 10,577 | # Wolfram Language & System 10.4 (2016)|Legacy Documentation
This is documentation for an earlier version of the Wolfram Language.
BUILT-IN WOLFRAM LANGUAGE SYMBOL
# GeoRange
GeoRange
is an option for GeoGraphics and GeoStyling that specifies the range of latitude and longitude to include.
## DetailsDetails
• GeoRange can be used in GeoGraphics and GeoStyling.
• The following settings can be used:
• All region corresponding to the entire world is included Automatic automatically select the region to display Full same as All when used in geo graphics {{lat1,lat2},{lon1,lon2}} explicit limits for latitude and longitude {All,{lon1,lon2}} complete latitude range and specified longitude range {{lat1,lat2},All} specified latitude range and complete longitude range r explicit radius assuming to be in meters quantity explicit radius around the specified or computed geo center entity region corresponding to the property of entity "Country" use bounding box of enclosing country "World" use bounding box of whole world
• All and Full both mean the complete range of the corresponding axis, namely for latitude and for longitude. They are therefore equivalent to when used on their own but can be also used as individual ranges, say {All,{-100,100}}.
• An explicit range is defined by latitude and longitude ranges in degrees.
• Specifying an entity uses the geo bounding box of its property.
• GeoRange->quantity represents a range enclosing a geo circle of radius quantity around the center specified by GeoCenter or by the geo primitives of GeoGraphics.
• For a longitude range , the upper boundary will be taken modulo 360 so that .
• GeoRangePadding can be used to extend the bounding box defined by GeoRange.
• AbsoluteOptions extracts the explicit form of GeoRange specifications when Automatic settings are given.
## ExamplesExamplesopen allclose all
### Basic Examples (4)Basic Examples (4)
Automatically select the region to plot:
In[1]:=
Out[1]=
Choose the geo range to show the surrounding country:
In[1]:=
Out[1]=
Set the geo range to an explicit distance:
In[1]:=
Out[1]=
Geo radii specified as pure numbers are assumed to be in meters:
In[2]:=
Out[2]=
In[3]:=
Out[3]=
In[4]:=
Out[4]=
In[5]:=
Out[5]=
Draw Brazil while showing the entire world:
In[1]:=
Out[1]=
Explicitly specify a geo range that includes the entire world:
In[2]:=
Out[2]=
Use the symbol All:
In[3]:=
Out[3]=
In[4]:=
Out[4]= | 575 | 2,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-43 | latest | en | 0.634114 |
https://www.geeksforgeeks.org/uhg-interview-experience-on-campus/?ref=rp | 1,695,513,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00714.warc.gz | 861,230,106 | 37,037 | Open In App
# UHG Interview Experience (On-Campus)
Round 1: The round 1 was an online test, on cocubes platform comprising of two sections.
Aptitude (30 questions in 30 minutes.): Again, it had two subsections.
• logical reasoning. 20 questions , 20 minutes. All the questions were paragraph based.
• Quantitative reasoning: 10 questions, 10 minutes. Again, paragraph based questions.
Coding Section: Two questions, very easy, 30 minutes.Every one got different sets. Some of the questions were:
Round 2: 35 people were shortlisted for this pen-paper-code round. People were divided in groups of nine and were being called group by group. Every group was given a question which was completely open-ended, and we had to write code in any language of our choice. Each individual was given 45 minutes. the four questions were:
• Problem Statemenrt: Given a graph implementation like , for eg, Facebook, find the minimum degree of separation betwen two given people. It is also given that the graph is implemented by using linked lists. The people who could see that the question demanded minimum path between two given nodes in a graph represented through adjacency list solved the problem by applying Djikstra algorithm after taking any of the given nodes as source.
• Problem Statement: Find the trending words on Twitter. Again, a very blunt, open-ended question that didn’t define much. They wanted to test the analytical, coding skills as well as thought process of the students. They were expecting features like, frequency-based sorting of the words giving priority to the timestamps(that’s what trending is, right!), remove the special characters like #,\$,@, etc, do not count the common words like is, am,are, the, preositiions and all. Two or more similar words, for example, #metooo and #meeeeeetoooouuuuuu.. be cosnidered same and many things else.
• Problem Statement: Maximise profits in stock given you can buy and sell stocks given k times. https://www.geeksforgeeks.org/maximum-profit-by-buying-and-selling-a-share-at-most-k-times/
• Problem Statement: Krithika has got a new prime subscription and have watched k movies. Help the prime people in suggesting her the (k+1)th movie. Basically, design a personalized recommendation engine. (w/o using ML libraries obviously! They wanted a C/C++/Java/Python code). This is the question I was asked. Again, a very open-ended question. They had come with plans to grill us and check our thought process. How I approached this problem was, I took both the conditions to filter out the remaining movies which are:
• 1. Item-based filtering: You have watched this movie, so you are more likely to watch movies of this type. For this, I had created a structure in C++ called movies, which contained different components like Genre, Rating, Studio, an array of Actors , release date, and an array of struct(Persons) which contained the list of people who had watched the movie.
• 2. User-based Filtering: People who have watched this movie have also watched thse movies. For the implementation of this feature, I created a structure person which contained name, age, nationality, profession and an array of struct movie that he has watched. Now, the process was, for each of the movie she has watched, I was comparing it first with all the features of every movie present in the database which is nothing but right now, an array of movies, and generating a score accoridng to an arbitrarily assigned priority right now. For the user-based filtering, for each of the movie that she has watched, there must be an array of persons who have also watched that movie, so again, traverse through that array and then inside that array, there must be an array of movies he would have watched, compare it with those and then generate the final maximum score from all of these. Finally, from all the scores generated the movie with the highest score is recommended.
Round 3(Technical Interview): After writing the code for the last round and submitting it, everyone was being called one by one and were being asked to explain the code and then discussion about the projects done and some SQL queries. In my case, there were three people togerther whom I had to explain my logic first, then, they started questioning.
• You have assigned the priority arabitrarily to generate the score. What if this priority is wrong for the person? If the recommendation is wrong, the person will obviously not watch the movie or leave it in between. So, let’s take a time-limit, and if until then, the movie is not watched, some other feature will be given priority and some other movie will be recommended and the different features are put into a circular queue. Apart from this, all the recommended and not watched movies should be kept in an array options and keep recommending them from time to time. Like giving them rest for now, and then comparing again, with the max score movies to go for recommendation.
• Although you have covered many great features, to do so, the time complexity is O(n^3), and for such a large database, dynamically calculating and showing this will impossible. How are you going to improve this? Well, for comparisions, I can partiiton the movies in different buckets so that the exhaustivity can be reduced, and put them in different sets which works in logn for searching. That’s how for two level of loops are converted to logn, thereby reducing the time complexity to n(logn)^2. | 1,183 | 5,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-40 | latest | en | 0.950356 |
https://allindiajobalerts.in/physics-quiz-7/ | 1,632,050,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00240.warc.gz | 152,880,882 | 30,779 | # Electronic Devices and Circuits-4
Physics Electronic Devices and Circuits-4 Multiple Choice Question Answers for all competitive exams Preparation for NEEET / JEE and KEA Preparation.
0
Created on By wefru
Electronic Devices and Circuits 4
1 / 10
Q31) What would happen, if the signal Xd passes through the feedback network?
2 / 10
Q32) Consider a single stage CE amplifier is estimated to possess the bandwidth of about 2MHz in addition to the resistive load of 500 ohm. What should be the value of source resistance in order to get the required bandwidth for the hybrid π equivalent circuit in accordance to the transistor assumptions given below? hfe = 100, gm = 30 mA , r'bb = 80Ω, Cc = 3pF, fT = 200MHz, Ce = 20pF, fH = 5MHz, r' be = 2kΩ
3 / 10
Q33) What should be the value of unity gain frequency for a short circuit CE transistor with gain of 30 at 4MHz and cut-off frequency of about 100 kHz?
4 / 10
Q34) Miller's theorem is applicable in a single stage CE hybrid π model in order to deal with ________
5 / 10
Q35) Which capacitors assists in preventing the loss of gain due to negative feedback without affecting the DC stability of R-C Coupled amplifier?
6 / 10
Q36) Which among the following is not an advantage of RC coupled amplifiers?
7 / 10
Q37) Which among the below mentioned circuits resemble its behaviour similar to that of an amplifier in high frequency region, as the response decreases with an increase in frequency?
8 / 10
Q38) Why is the Darlington configuration not suitable for more than two transistors?
9 / 10
Q39) What should be the value of resistance between collector and ground (R0) for below drawn schematic of transistor amplifier, comprising h-parameters as hie = 1.1 kΩ, hfe = 50, hre= hoe = 0 with short-circuit capacitors?
10 / 10
Q40) Consider the assertions given below.
A. Replacement of each coupling and bypass capacitors by a short circuit
B. Replacement of transistor by its hybrid equivalent model for further analysis
C. Replacement of DC voltage sources by a short circuit
Which is the correct sequential order of steps to be carried out for analysis of a transistor amplifier circuit? | 566 | 2,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | longest | en | 0.817625 |
https://www.goldsim.com/Courses/BasicGoldSim/Unit6/Lesson4/ | 1,719,106,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00851.warc.gz | 690,446,590 | 7,580 | # Lesson 4 - Specifying the Simulation Settings for Time
In the previous Lesson we discussed in general terms how we represent time in GoldSim by discretizing time into timesteps. In this Lesson we will see how we actually define the timestep in a GoldSim model.
Timestepping options are defined using the Simulation Settings dialog. You can open the Simulation Settings dialog by pressing F2, choosing Run | Simulation Settings from the main menu bar, or by pressing the Simulation Settings button in the toolbar:
Open the dialog now so we can examine it:
This dialog has four tabs: Time, Monte Carlo, Globals and Information. For now, we are going to focus on the Time tab (the default tab that is displayed).
The dialog has lots of options (since GoldSim’s timestepping features are powerful and flexible). However, we only need to focus on a few of these for now, and the options we will use for now are quite simple to understand. We will discuss some of the advanced features in subsequent Units.
The first option you should notice is the Time Basis. This has three options: “Elapsed Time”, “Calendar Time” and “Static Model”. In an Elapsed Time simulation (the default), you must specify a simulation Duration. The simulation is then tracked in terms of the elapsed time since the simulation began. In a Calendar Time simulation, you enter a Start Time and an End Time, and the simulation is tracked in terms of the calendar time. In a Static Model simulation, the model does not step through time at all; it does a static calculation (which can occasionally be of value when doing a Monte Carlo simulation, as we shall see in Unit 11). For the next several Units, we are going to carry out Elapsed Time simulations (and only use the basic timestepping options). We will discuss the other two types of simulations (and more advanced timestepping options) in subsequent Units.
If we select “Elapsed Time” for the Time Basis, we must then select a Duration. The Duration must be entered as a number followed by a valid unit abbreviation (it cannot be a link to another element). Typical unit abbreviations that you might use include sec, min, hr, day, and yr.
Note: If you are defining your Duration in terms of years (yr), it is important that you understand how the unit yr is actually defined. All units have fixed definitions. The definition for the unit yr in GoldSim is 365.25 days. Similarly, the unit mon (month) is defined as 30.4375 days. We will discuss the use of these units further (and the care that must be taken if you choose to use them) in Unit 10.
Once we select our Duration, we can select the timestep length (Basic Step). Like the Duration, the Basic Step must be entered as a number followed by a valid unit abbreviation (it cannot be a link to another element).
Note: The Basic Step does not need to divide evenly into the Duration. If it does not, the final step is simply adjusted (shortened) accordingly.
For the simple Exercises we will carry out in the next several Units, there is only one other option that we need to be aware of for now. At the top of the page, there is a field named Time Display Units. These define the default time units that are used in several parts of GoldSim. The most important of these is how the x-axis is labeled when plotting time history results (for Elapsed Time simulations).
GoldSim has many more advanced timestepping features that we have not discussed here. However, we’ve covered what you need to know about Time settings in GoldSim for now. We will discuss some of the advanced features in subsequent Units. You can also read more about timestepping in GoldSim Help. | 824 | 3,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.8267 |
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