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https://www.teacherspayteachers.com/Product/Subtracting-Integers-EDITABLE-7NSA1c-CCSS-Chaos-Free-Math-Game-2054068 | 1,521,599,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647556.43/warc/CC-MAIN-20180321004405-20180321024405-00781.warc.gz | 881,284,979 | 18,905 | Total:
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# Subtracting Integers: EDITABLE {7.NS.A.1c CCSS} Chaos-Free Math Game
Subject
Resource Type
Common Core Standards
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Compressed Zip File
889 KB|14 pages
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Product Description
Math Common Core Aligned: 7.NS.A.1c
A great game for practicing Subtracting Integers. TpT purchaser quotes: "Great review activity!" "My students really struggle with this, so it is nice to have a focused review for it. Thanks"
A rigorous 60 MINUTE activity. CHAOS-FREE Math Game...
-------------------------------------------------------------------------------------------------
This material covers
* subtracting positive integers
* subtracting negative integers
* combination questions with both adding and subtracting Level 1
* combination questions with both adding and subtracting Level 2
-------------------------------------------------------------------------------------------------
Package includes:
* Rationale: how to use the product in the classroom
* Student Instruction handout
* Practical Tips for Teachers
* 12 questions: Multiplication of Exponents
* 12 questions: Divisions of Exponents
* 12 questions: Power to the Power
* 12 combination more challenging questions
* Answer key for all 48 questions
* 2 blank student response sheets
* pdf with all of the above contents
* EDITABLE PowerPoint file
-------------------------------------------------------------------------------------------------
Use as:
- Game format:1, 2 or 4 players
- Worksheet
- Peer Tutoring
- Homeschooling
- Stations
- Differentiated Learning
-------------------------------------------------------------------------------------------------
This is a math game in which one pair of students competes against another. The first team selects a card from a deck of cards and answers the matching question on the activity sheet(example 4 of hearts). The 2 students work cooperatively to obtain an answer. If correct they get a point, if not the other team can get the rebound point. Then team 2 selects a card, discusses the answer and so on. It is cooperative because the 2 students work together, yet competitive because it pits one team against another. The students have fun and get the math work done. As well, the students can work on this activity alone, can be used as a review.
You might also like:
1. Exponent Rules 48 Questions EDITABLE Game
2. Solving Equations: One and Two Step {32 Questions }
3. Flip Book { Solving Equations } Elimination, Substitution and Graphing Techniques are included
4. Arithmetic Sequences {48 Questions } Game / Activity
5. Distributive Property 48 Questions Game
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MatheMatters by Jacquie
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Total Pages
14 pages
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Teaching Duration
1 hour
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 708 | 3,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.74869 |
https://www.reddit.com/r/math/comments/1do8v4/question_about_polynomials_in_z2z/ | 1,542,644,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745800.94/warc/CC-MAIN-20181119150816-20181119172816-00345.warc.gz | 993,582,679 | 79,614 | Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
5
Posted by5 years ago
Archived
## Question about polynomials in Z/2Z
My professor has written in some notes on coding theory:
In the field Z/2Z={0,1}:
(t-1)=(t+1)={0,t+1,t2+t,t2+1}, giving the code {000,110,011,101}.
I get that (t-1)=(t+1) in Z/2Z, but I don't get how it can be equal to t2+1 at all? My algebra isn't great as you can tell.
68% Upvoted
Sort by
level 1
[deleted]
14 points · 5 years ago · edited 5 years ago
I'm not sure any of the comments so far about these polynomials being equal as functions are answering the question correctly. Since the notation uses parentheses, and without having read the original notes, I would assume that (t-1) and (t+1) are meant to be not polynomials but ideals generated by t-1 and t+1 (edit: corrected typo) in, say, the ring Z/2Z[t]/(t3-1). Then it's clear that t2+t = t(t+1) and t2+1 = (t+1)2 (remember, this is in characteristic 2) are in the ideal (t+1).
level 2
-5 points · 5 years ago(6 children)
level 3
13 points · 5 years ago
That may be one notation, but the parentheses notation is very common. Dummit and Foote uses parentheses, for example.
level 4
Algebra1 point · 5 years ago
I've never used Dummit and Foote. It has a reputation of being difficult for undergraduates, which is my audience. Both Fraleigh and Gallian use angle brackets.
level 5
1 point · 5 years ago
Regardless of the difficulty of D&F, it's undeniably a popular text. Herstein is much easier and I'm fairly certain it uses parentheses as well. For a third source, my algebra classes, which are not taught from set texts, also use parentheses. I think it's fair to say that both notations are widespread and thus the original poster using parentheses does not preclude them from referring to an ideal.
level 3
[deleted]
4 points · 5 years ago
Very common to use either and I've seen parentheses more often.
level 4
Algebra3 points · 5 years ago
In my experience (very considerable) parentheses are rarely used. it just invites confusion, as here.
level 5
[deleted]
1 point · 5 years ago
Fair enough! I'm not an algebraist. I've just used it in my two algebra classes thus far, as well as in my text.
level 1
10 points · 5 years ago
Indeed we have [;t \mapsto t-1 = t \mapsto t+1 = t \mapsto t^2 + 1;] as functions (just plug in 0 and 1 and evaluate), but for polynomials over [;\mathbb Z_{(2)};] we have [;t+1 \neq t^2 + 1;] - they have different degree, so they cannot be equal.
level 2
Algebra3 points · 5 years ago
Right. That's a very important distinction that is often lost in applications of Abstract Algebra to cryptology and coding theory. t + 1 and t2 + 1 are NOT the same polynomial.
level 1
Noncommutative Geometry4 points · 5 years ago
For how many t in Z/2Z does t = t2 hold?
level 2
Original Poster3 points · 5 years ago
Thanks for the reply, I guess it holds for all elements of Z/2Z?
level 3
Noncommutative Geometry3 points · 5 years ago · edited 5 years ago
Yes. So you can always write t = t2, for any t in Z/2Z.
level 4
Original Poster3 points · 5 years ago
Ha, oops. Thanks, can't believe I didn't spot that.
level 5
Noncommutative Geometry3 points · 5 years ago
You're welcome.
level 1
0 points · 5 years ago
Simply look at both cases:
For t=0, you get:
(t-1) = 0-1 = 1
t²+1 = 0² + 1 = 1
For t=1, you get:
(t-1) = 1 - 1 = 0.
t² + 1 = 1² + 1 = 0.
level 2
Original Poster1 point · 5 years ago
Thanks for the reply! I understand your point, but could I not then just pick arbitrary polynomials of t that yield the same result in Z/2Z?
level 3
2 points · 5 years ago
I believe that's what they did.
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63,404 subscribers | 1,941 | 5,775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-47 | latest | en | 0.934218 |
http://www.cfd-online.com/Forums/star-ccm/68491-vof-wave.html | 1,480,721,250,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00200-ip-10-31-129-80.ec2.internal.warc.gz | 386,522,134 | 16,907 | # Vof Wave
Register Blogs Members List Search Today's Posts Mark Forums Read
September 22, 2009, 06:41 Vof Wave #1 New Member Andrea Cicalini Join Date: Sep 2009 Posts: 13 Rep Power: 9 I'm doing a thesis about 6 Dof body (an Hull) and waves of first order with STAR CCM+ v4.04.The wave diffuses quickly and become a flat wave in the control field. I put the boundaries of velocity inlet in every external face of dominio except for the face behind the hull where i put pressure outlet.I set a timestep of 0.01 sec. The wave has a lenght of 20 m, amplitude 1 m,current and wind 2,5 m/s. Do you have an idea to solve this problem? Thank You Last edited by cicagol; September 23, 2009 at 08:42.
September 23, 2009, 08:54
#2
New Member
Andrea Cicalini
Join Date: Sep 2009
Posts: 13
Rep Power: 9
It's the first part of summary report
Attached Files
Summary Report 1.pdf (75.7 KB, 117 views)
September 23, 2009, 08:58
#3
New Member
Andrea Cicalini
Join Date: Sep 2009
Posts: 13
Rep Power: 9
the second part
Attached Files
Summary Report 2.pdf (48.1 KB, 39 views)
September 23, 2009, 08:58
#4
New Member
Andrea Cicalini
Join Date: Sep 2009
Posts: 13
Rep Power: 9
the third part
Attached Files
Summary Report 3.pdf (78.7 KB, 38 views)
September 24, 2009, 11:25
#5
Senior Member
Join Date: Mar 2009
Posts: 203
Rep Power: 10
Quote:
Originally Posted by cicagol Do you have an idea to solve this problem?
That the wave becomes flat quickly?
September 24, 2009, 11:36 #6 New Member Andrea Cicalini Join Date: Sep 2009 Posts: 13 Rep Power: 9 Yes...I think it's a problem of approssimation...a first order wave has limits and i can't have a wave of certain features with the first order
September 24, 2009, 11:38 #7 Senior Member Join Date: Mar 2009 Posts: 203 Rep Power: 10 Why only first order?
September 24, 2009, 11:42 #8 New Member Andrea Cicalini Join Date: Sep 2009 Posts: 13 Rep Power: 9 Not only first order...but i'm studying first order...with first order I mean first order of Stokes approssimation theory
September 24, 2009, 11:50 #9 Senior Member Join Date: Mar 2009 Posts: 203 Rep Power: 10 Ahh ok. That's down my alley Maybe this helps you: http://www.flow3d.com/resources/news...mmer09_02.html
September 24, 2009, 11:53 #10 New Member Andrea Cicalini Join Date: Sep 2009 Posts: 13 Rep Power: 9 Thank You, I've already seen this yesterday...it's the reason i wrote that first order have some limits
September 24, 2009, 12:18 #11 Senior Member Join Date: Mar 2009 Posts: 203 Rep Power: 10 Ok, then I understand your posting right. Again what is your problem?
September 24, 2009, 12:21 #12 New Member Andrea Cicalini Join Date: Sep 2009 Posts: 13 Rep Power: 9 In this moment nothing, i'm simulating a new wave and tomorrow i'll see the result...So i'm waiting...
September 28, 2009, 03:13 #13 New Member Andrea Cicalini Join Date: Sep 2009 Posts: 13 Rep Power: 9 After three days of simulation, 64 sec. in simulation time...it's all good
September 28, 2009, 04:29 #14 Senior Member Join Date: Mar 2009 Posts: 203 Rep Power: 10 Sounds good.
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All times are GMT -4. The time now is 19:27. | 1,145 | 3,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-50 | latest | en | 0.905009 |
https://programmer.group/sequence-table-operation.html | 1,726,590,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00881.warc.gz | 431,693,324 | 4,603 | # Sequence table operation
Keywords: C data structure
# What is a sequence table
Order represents the sequential storage structure of linear table, that is, the storage structure of linear table constructed in sequential storage mode.
For the sequential table a with n elements, it can be expressed as A[0... N-1], with subscripts from 0 to n-1. A[0] is called the first element, A[1] is called the second element, and... A[n-1] is called the nth element.
# Storage structure of sequential table
``` #define MaxLen 50
typedef int elemtype; //Define elemtype as int type
typedef elemtype sqlist[MaxLen];
```
# Creation and display of sequence table
```#include<stdio.h>
#define Maxlen 20
typedef int elemtype;
typedef elemtype sqlist[Maxlen];
int create(sqlist A)
{
int i,n;
printf("Create a sequence table:\n");
printf("Number of input elements:");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Input No%d Elements:",i+1);
scanf("%d",&A[i]);
}
return n;
}
void display(sqlist A,int n)
{
int i;
printf("Output a sequence table:");
if(n==0){
printf("\n Empty table");
return;
}
for(i=0;i<n;i++)
{
printf("%4d",A[i]);
}
printf("\n");
}
```
# Insertion of sequence table
Given that the elements in A sequence table A are non decreasing in order by value, write A function to insert an element x and keep the sequence table orderly.
```#include "sqlist.cpp"
int Insert(sqlist A,int n)
{
int m=n;
int s;
int j=0;
scanf("%d",&s);
while(j<1){
if(A[n-1]<s)
{
A[n]=s;
j--;
}
else{
A[n]=A[n-1];
n--;
}
}
n=m+1;
}
int main()
{
sqlist A;
int n;
n=create(A);
display(A,n);
n=Insert(A,n);
display(A,n);
}
```
# Deletion of sequence table
Let A be the sequence table, and try to write an algorithm to delete k elements from the ith element in A.
```#include "sqlist.cpp"
int Delete(sqlist A,int n)
{
int j,k,i;
printf("Please enter the location to delete j: ");
scanf("%d",&j);
printf("Please enter the number of deleted k:");
scanf("%d",&k);
for(i=0;i<=n-k;i++)
{
A[j-1]=A[j+k-1];
j++;
}
n=n-k;
return n;
}
int main()
{
sqlist A;
int n;
n=create(A);
display(A,n);
n=Delete(A,n);
display(A,n);
}
```
# Consolidation of sequential tables
Write an algorithm to merge m (M > 2) ordered (from small to large) Order tables into an ordered order table. There is no new sequential table storage in the consolidation process.
```#include"sqlist.cpp"
int comb(sqlist A,int na,sqlist B,int nb)
{
int n=na,m=nb;
if(na+nb > Maxlen) return 0; //Sequence table overflow
while(nb>0)
{
if((na==0)||(A[na-1]<B[nb-1]))
{ //Note B[nb-1] is the element with the largest na+nb
A[na+nb-1] = B[nb-1];
nb--;
}
else
{
A[na+nb-1] = A[na-1]; //Note that A[na-1] is the element with the largest na+nb
na--;
}
}
na = n + m;
return na;
}
int main(){
sqlist A,B;
int m;
printf("Please enter the number of sequence tables:");
scanf("%d",&m);
int na,nb;
na = create(A);
display(A,na);
for(int i=1;i<m;i++)
{
nb = create(B);
na = comb(A,na,B,nb);
display(B,nb);
}
display(A,na);
}
```
# Arrangement of sequence table
Let A and B be two sequential tables, and their elements are arranged from small to large. Write an algorithm to form all elements in A and B into A new ordered table C from small to large, requiring the deletion of duplicate elements.
```#include"sqlist.cpp"
int comb(sqlist A,int na,sqlist B,int nb)
{
sqlist C;
int n=na,m=nb;
int temp;
if(na+nb > Maxlen) return 0; //Sequence table overflow
while(nb>0)
{
if((na==0)||(A[na-1]<B[nb-1])){ //Note B[nb-1] is the element with the largest na+nb
A[na+nb-1] = B[nb-1];
nb--;
}
else{
A[na+nb-1] = A[na-1]; //Note that A[na-1] is the element with the largest na+nb
na--;
}
}
na = n + m;
for(int i=0;i<na;i++)
{
C[i]=A[i];
if(A[i]==A[i+1])
{
for(int j=i;j<=na-i;j++)
{
A[j+1]=A[j+2];
}
na--;
}
}
for(int i=0;i<na;i++){
A[i]=C[i];
}
return na;
}
int main(){
sqlist A,B;
int na,nb;
na = create(A);
display(A,na);
nb = create(B);
display(B,nb);
na = comb(A,na,B,nb);
display(A,na);
}
```
# Reverse order of sequence table
There is A sequential table A containing n elements. It is required to write an algorithm to reverse the table, and only one additional work unit is allowed to be added outside the storage space of the original table.
```#include"sqlist.cpp"
void invert(sqlist A,int n){
int m=n/2;
int temp=0;
for(int i=0;i<m;i++){
temp=A[i];
A[i]=A[n-i-1];
A[n-i-1]=temp;
}
}
int main(){
sqlist A;
int n;
n = create(A);
display(A,n);
invert(A,n);
display(A,n);
return 0;
}
```
It is known that the element type of array A[0... n-1] is integer. Design the algorithm to adjust A so that all elements on the left are less than 0 and all elements on the right are greater than or equal to 0 (the time complexity and space complexity of the algorithm are required to be O(n))
```#include<stdio.h>
#include"sqlist.cpp"
{
sqlist B;
int i,x=0,y=n-1;
for(i=0;i<n;i++)
{
if(A[i] < 0) {
B[x] = A[i];
x++;
}else{
B[y] = A[i];
y--;
}
}
for(i=0;i<n;i++){
A[i] = B[i];
}
}
int main()
{
sqlist A;
int n;
n = create(A);
display(A,n); | 1,554 | 4,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.615732 |
https://math.stackexchange.com/questions/1711202/suppose-that-in-fact-9-of-the-oranges-on-the-truck-do-not-meet-the-desired-s | 1,563,383,373,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183521-00001.warc.gz | 459,722,410 | 38,385 | Suppose that in fact $9\%$ of the oranges on the truck do not meet the desired standard. What's the probability that the shipment will be rejected?
I am tutoring a student in AP stats, and came across this question. And I have not been able to solve this problem and get a result that is one of the multiple choices. The closest I got to was B (which I think is the intended answer), but I feel like the answer choices are incorrect.
Problem: When a truckload of oranges arrives at a packing plant, a random sample of 125 is selected and examined, The whole truckload will be rejected if more than $8\%$ of the sample is unsatisfactory. Suppose that in fact $9\%$ of the oranges on the truck do not meet the desired standard. What's the probability that the shipment will be rejected?
A) $0.6966$
B) $0.3483$
C) $0.6517$
D) $0.7803$
E) $0.2197$
• The choice B) is not reasonable. The truckload has $9\%$ bad and we will reject if more than $8\%$ of the sample are bad, so it seems clear that we will have a greater than $50\%$ chance of rejecting. – André Nicolas Mar 24 '16 at 5:51
The exact distribution of the random number $X$ of unsatisfactory oranges drawn from the sample is binomial, under the assumption that there are many, many more oranges than the sample size $n = 125$. The lot will be rejected if $X > 10$. Assuming that the true proportion of unsatisfactory oranges is $p = 0.09$, the resulting probability of rejection is $$\Pr[X > 10] = 1 - \sum_{x=0}^{10} \binom{125}{x} (0.09)^x (0.91)^{125-x} \approx 0.576514.$$ Using the normal approximation, $$X \dot\sim \operatorname{Normal}(\mu = np, \sigma^2 = np(1-p)),$$ and we would calculate using continuity correction $$\Pr[X > 10] \approx \Pr\left[\frac{X - \mu}{\sigma} > \frac{10 - 125(0.09) + 0.5}{\sqrt{125(0.09)(0.91)}} \right] = \Pr[Z > -0.234404].$$ Using a normal table, this is approximately $0.592664$. The quality of the approximation is poor because (a) $p$ is small, and (b) the $z$-score boundary is close to $0$.
But here's what's interesting: if you don't use continuity correction, the $z$-score is $-0.390673$, and the probability is...(drumroll)....
$$0.65198.$$ That is very close to answer choice (C), acceptably so (my calculation was on a computer rather than using a normal table, so it is more precise). Clearly, the writer of the question intended that the solution assumes two things that are not, in my opinion, entirely legitimate assumptions:
1. The solution should use the normal approximation to the binomial.
2. The normal approximation should be applied without continuity correction.
The issue with the first is minor. It's not totally unreasonable to use a normal approximation, even if a reasonably proficient student can use a calculator to compute a binomial sum of relatively few terms. But the second issue, as we can very clearly see from the exact calculation, is quite major. Failure to use continuity correction magnifies the error of the approximation to an extent that is unacceptable, because the real distribution here is binomial, not normal, and that in a finite, integral sample size of $n = 125$ oranges, $8\%$ is exactly 10 oranges.
I should also add that, in a testing situation with multiple choice options, it is often useful to reason in other ways so as to eliminate incorrect answer choices. Right away, I can tell you that the probability of rejection should be greater than $0.5$. This is because if the true proportion of unsatisfactory oranges is greater than the criterion of $8\%$, the test will be more likely to reject than to not reject. It may not be much greater than $0.5$, but intuitively, it shouldn't be less, and certainly not significantly less.
$\newcommand{\Var}{\operatorname{Var}}$Here is my attempt.
We are told that the true percent of bad oranges is $.09$. Throughout, I assume that $p = .09$. Then the question asks $P(\hat p >.08|p=.09)$, where $\hat p$ is the usual proportion $$\hat p = \frac{X_1+\dotsb+X_{125}}{125}$$ and each $X_i$ is an indicator (Bernoulli trial) with chance $p = .09$.
Hence, we see that $\mu = E[\hat p] = .09$, and $\sigma^2 = \Var(\hat p) = \frac{.09(1-.09)}{125}$, and $$P(\hat p >.08|p = .09) = 1-P(\hat p <.08|p = .09) \approx 1-P(Z< (.08-\mu)/\sigma) =0.6519804$$ where I exclude the continuity correction factor since the the question was posed in terms of proportion and $\hat p$ is a continuous random variable.
So I think it is C.
• $\hat p$ is not really continuous, but is in fact a rescaled binomial. The inescapable fact is that saying that a fraction of an orange is unsatisfactory makes no sense at all. It's not possible to observe $0.08 < \hat p < 0.088$ when $n = 125$, and treating it as continuous implies that this event has positive probability. That is the nature of the error in using the normal approximation without continuity correction. – heropup Mar 24 '16 at 6:07 | 1,320 | 4,889 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-30 | latest | en | 0.914324 |
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# 数学final代考 Math 328 & Math 601代写 Math代写
464
## Math 328 & Math 601
### Directions:数学final代考
Open this booklet only when directed to do so.
Check that you have all 7 pages including this one.
You may use the backs of sheets for rough work, or if you need additional space for your answer.
Each question is worth 16 points. (Total: 100 points)
No cheating sheets may be used.
This examination is two days (48 hours) in duration.
### 4. 数学final代考
Determine whether the following statements true or false. Give reasons.
(a) Every bounded sequence in Rn has a convergent subsequence.
(b) The dual space of (Rn, ∥· 1) is (Rn, ∥· 1).
(c) In a Hilbert space, every bounded sequence has a convergent subsequence.
### 5.
Suppose that the sequence (xn)n∈N in H satisfies
(a) There exists M > 0 such that for every n N, ∥ xn ≤ M.
(b) For every y M, limn xn, y〉exists.
(c) M is total in H.
Prove that for every y ∈ H, limn xn, y exists.
The prev:
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1 | 635 | 1,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.770034 |
https://cheapjerseyschinese.com/qa/how-many-gs-can-a-f22-pull.html | 1,611,104,953,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00328.warc.gz | 276,303,953 | 8,579 | How Many G’S Can A F22 Pull?
How many G’s can a f16 pull?
nine G’sWith a full load of internal fuel, the F-16 can withstand up to nine G’s — nine times the force of gravity — which exceeds the capability of other current fighter aircraft..
How many Gs is 25000 mph?
In order for them to accelerate to light speed from 25,000mph, and achieve no more than 10 g’s, a force that is still virtually unbearable, it would take approximately 3,058,992 seconds, 50,983 hrs, 2,124 days, or 5.8 years, then doubling of course to allow for the the equally powerful negative G-forces you’d achieve …
Could a human survive light speed?
No, it is not possible for a human to survive travelling at the speed of light. Others have pointed out that it’s impossible to reach the speed of light, so they’re talking about the limits as you approach the speed of light. But at the speed of light, you experience singularity-like conditions.
How fast is 9 g force in mph?
Please share if you found this tool useful:Conversions Table9 Miles Per Hour Per Second to Standard Gravity = 0.4103600 Miles Per Hour Per Second to Standard Gravity = 27.351210 Miles Per Hour Per Second to Standard Gravity = 0.4559800 Miles Per Hour Per Second to Standard Gravity = 36.468313 more rows
Which is better the f22 or f35?
While the F-22 is superior to the F-35 in air-to-air missions, the F-35’s air-to-air capability is superior to all other fighters. The F-35 is better than any other fighter aircraft, including the F-22, for air-to-ground strike missions. Learn more about F-35 capabilities.
What is the fastest fighter jet in the world?
The fastest fighter still in service today is the Soviet-built MiG-25. Mikoyan designed this fighter to be a pure interceptor aircraft. As a result, the Foxbat can sustain a cruising speed of Mach 2.8 and kick it into overdrive with a top speed of 3.2 — not a bad technology for an aircraft that first took off in 1964.
How fast is the F 35 in mph?
1,200 mphThe F-35 is powered by the Pratt and Whitney F135, an engine developed specifically for the program; all three variants use it. With more than 40,000 pounds of thrust, according to the manufacturer, it can propel the F-35 to speeds of about Mach 1.6 (1,200 mph; 2,000 kph).
How many G’s can a fighter pilot withstand?
9 gFighter jets can pull up to 9 g vertically, and the more a pilot can take without blacking out, the better their chances in a dogfight. Some pilots wear “g-suits” which help push the blood away from their legs and towards the brain.
How many G’s can kill you?
“The NHTSA standard for a sudden impact acceleration on a human that would cause severe injury or death is 75 g’s for a “50th percentile male”, 65 g’s for a “50th percentile female”, and 50 g’s for a “50th percentile child”.
Do fighter pilots wear diapers?
When flying long missions (say over the Atlantic, or missions lasting more than 4 hours) they can do 1 of 2 things. They have little pee bags containing a material that absorbs urine. They just pee in the them and stick them to the side. Or, they can wear adult diapers.
What’s the most G’s ever pulled?
There are isolated incidents of humans surviving abnormally high G-forces, most notably the Air Force officer John Stapp, who demonstrated a human can withstand 46.2 G’s. The experiment only went on a few seconds, but for an instant, his body had weighed over 7,700 pounds, according to NOVA.
Can the F 35 Dogfight?
Berke said the video proves that the F-35 is a “highly maneuverable, highly effective dogfighting platform,” but even still, he wouldn’t use that exact maneuver in a real dogfight. The flat spin is “not an effective dogfighting maneuver, and in some cases, you would avoid doing that.”
What does 1g force feel like?
For most people, the peak G-force they’ve experienced is probably on a rollercoaster during a loop—which is about 3-4G’s. It’s enough to push your head down and pin your arms by your side. Modern fighters like the F-16 and F-35 pull 9G’s, which translates to over 2,000 pounds on my body.
What acceleration will kill you?
Forces which may cause death It may take an acceleration of more than seventy-five times the acceleration due to gravity (75g) to kill an adult male. But if the acceleration is sustained for several seconds, death can result from only four to ten g, as the brain is deprived of oxygen.
How many G’s can an F 35 pull?
nine-G turnsThe Air Force F-35 variant, fully loaded for combat, can pull nine-G turns with a full load of fuel and missiles.
How many Gs can a f22 pull?
maximum g’s which f-22 aircraft can pull are 9 g’s. After 9 g’s permanent change will occur in different parts of the aircraft. The F-22 Raptor has a limit of +9 G. It is rated to pull 9.6G.
How do female fighter pilots pee?
Peeing into a tube doesn’t work for everyone for some obvious physical reasons, so today’s fighter pilots urinate into “piddle packs,” plastic packs that convert urine into a gel for disposal, but the method involves partially undressing while sitting strapped in a tiny cockpit and flying a multimillion-dollar jet.
How many G can a human withstand?
9 g’sNormal humans can withstand no more than 9 g’s, and even that for only a few seconds. When undergoing an acceleration of 9 g’s, your body feels nine times heavier than usual, blood rushes to the feet, and the heart can’t pump hard enough to bring this heavier blood to the brain. | 1,317 | 5,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-04 | longest | en | 0.927386 |
http://www.knittinghelp.com/forum/showthread.php?t=115193&nojs=1 | 1,406,322,569,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894782.90/warc/CC-MAIN-20140722025814-00160-ip-10-33-131-23.ec2.internal.warc.gz | 836,069,490 | 10,166 | 11-05-2013, 03:51 PM #1 smaeschen Casting On Join Date: Nov 2013 Posts: 2 Thanks: 1 Thanked 0 Times in 0 Posts Pattern Help - Beginner - Armhole shaping I am working on the body of a sweater and have come to the armhole shaping portion. Here is how the pattern reads: Row 1 and all WS rows : Purl Row 2 K2, p1, k to last 3 sts, p1, k2 Row 4 (dec) K2, p1, k1, ssk, k to last 6 sts, k2tog, k1, p1, k2. Working in pat as established, rep dec row every 8th row 5(0,0,0,0,0) times, every 6th row 0 (3,0,0,0,0) times, every 4th row 0(6,10,9,7,3) times, every other row 0(0,2,6,11,19) times - 46(46,48,48,50,50) sts. The underlined portion is where I have my question: Since I am making the second size (bolded) I know I don't do any decreases for the every 8th row so I skip to do the every 6th row 3 times. So will my next dec row after row 4 be row 6 or row 10? So this is what I have done so far: Row1 - Purl Row2 - as pattern states above Row3 - purl Row4 - as patterns states above (dec) Row5 - purl Row6 - do i decrease or follow the row 2 instruction above? Thank you for your help - I found a couple threads on this that gave 2 different answers and were from years ago. One said your next decrease would be row 6 the other said row 10? Confused..... I am thinking my next decrease row will be row 10, 16 and 22 then starting the every 4th row would be row 26, 30, 34, etc. Am I right?
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11-05-2013, 05:44 PM #2 salmonmac Moderator Mod Squad Join Date: Dec 2010 Location: Maryland Posts: 8,121 Thanks: 608 Thanked 3,287 Times in 2,974 Posts Welcome to KnittingHelp! You have it exactly right: "I am thinking my next decrease row will be row 10, 16 and 22 then starting the every 4th row would be row 26, 30, 34, etc. Am I right?" __________________ Knitting Help Videos Here!-
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11-05-2013, 05:53 PM #3 smaeschen Casting On Join Date: Nov 2013 Posts: 2 Thanks: 1 Thanked 0 Times in 0 Posts Thank you for the quick response!
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All times are GMT -4. The time now is 05:09 PM. | 1,005 | 3,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.891831 |
http://www.singular.uni-kl.de:8002/trac/changeset/500e4b69e2df574c6ae8023f32cbdfbdbd0d5e8e/git | 1,701,204,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00307.warc.gz | 88,009,085 | 85,233 | # Changeset 500e4b in git
Ignore:
Timestamp:
Jun 23, 2015, 2:46:02 PM (8 years ago)
Branches:
(u'spielwiese', 'ec94ef7a30b928574c0c3daf41f6804dff5f6b69')
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Message:
```Merge branch 'spielwiese' of github.com:Singular/Sources into spielwiese
update```
Files:
11 edited
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• ## Singular/LIB/grwalk.lib
r8968f10 } proc gwalk(ideal Go, list #) "SYNTAX: gwalk(ideal i); gwalk(ideal i, intvec v, intvec w); proc gwalk(ideal Go, int reduction,int printout, list #) "SYNTAX: gwalk(ideal i, int reduction, int printout); gwalk(ideal i, int reduction, int printout, intvec v, intvec w); TYPE: ideal PURPOSE: compute the standard basis of the ideal, calculated via string ord_str = OSCTW[2]; intvec curr_weight = OSCTW[3]; /* original weight vector */ intvec target_weight = OSCTW[4]; /* terget weight vector */ intvec target_weight = OSCTW[4]; /* target weight vector */ kill OSCTW; option(redSB); //print("//** help ring = " + string(basering)); ideal G = fetch(xR, Go); G = system("Mwalk", G, curr_weight, target_weight,basering); G = system("Mwalk", G, curr_weight, target_weight,basering,reduction,printout); setring xR; //** compute a Groebner basis of I w.r.t. lp. ring r = 32003,(z,y,x), lp; ideal I = y3+xyz+y2z+xz3, 3+xy+x2y+y2z; gwalk(I); ideal I = zy2+yx2+yx+3, z3x+y3+zyx-yx2-yx-3, z2yx3-y5+z2yx2+y3x2+y2x3+y3x+y2x2+3z2x+3y2+3yx, zyx5+y6-y4x2-y3x3+2zyx4-y4x-y3x2+zyx3-3z2yx+3zx3-3y3-3y2x+3zx2, yx7-y7+y5x2+y4x3+3yx6+y5x+y4x2+3yx5-6zyx3+yx4+3x5+3y4+3y3x-6zyx2+6x4+3x3-9zx; gwalk(I,0,1); } } proc fwalk(ideal Go, list #) "SYNTAX: fwalk(ideal i); fwalk(ideal i, intvec v, intvec w); proc fwalk(ideal Go, int reduction, int printout, list #) "SYNTAX: fwalk(ideal i,int reductioin); fwalk(ideal i, int reduction intvec v, intvec w); TYPE: ideal PURPOSE: compute the standard basis of the ideal w.r.t. the ideal G = fetch(xR, Go); G = system("Mfwalk", G, curr_weight, target_weight); G = system("Mfwalk", G, curr_weight, target_weight, reduction, printout); setring xR; ring r = 32003,(z,y,x), lp; ideal I = y3+xyz+y2z+xz3, 3+xy+x2y+y2z; fwalk(I); int reduction = 1; int printout = 1; fwalk(I,reduction,printout); } } proc pwalk(ideal Go, int n1, int n2, list #) "SYNTAX: pwalk(int d, ideal i, int n1, int n2); pwalk(int d, ideal i, int n1, int n2, intvec v, intvec w); proc pwalk(ideal Go, int n1, int n2, int reduction, int printout, list #) "SYNTAX: pwalk(int d, ideal i, int n1, int n2, int reduction, int printout); pwalk(int d, ideal i, int n1, int n2, int reduction, int printout, intvec v, intvec w); TYPE: ideal PURPOSE: compute the standard basis of the ideal, calculated via ideal G = fetch(xR, Go); G = system("Mpwalk", G, n1, n2, curr_weight, target_weight,nP); G = system("Mpwalk",G,n1,n2,curr_weight,target_weight,nP,reduction,printout); setring xR; //kill Go; //kill Go; //unused keepring basering; ring r = 32003,(z,y,x), lp; ideal I = y3+xyz+y2z+xz3, 3+xy+x2y+y2z; //I = std(I); //ring rr = 32003,(z,y,x),lp; //ideal I = fetch(r,I); pwalk(I,2,2); int reduction = 1; int printout = 2; pwalk(I,2,2,reduction,printout); }
• ## Singular/LIB/modwalk.lib
• Property mode changed from `100644` to `100755`
• ## Singular/LIB/primdec.lib
r35aef3 if((size(sact)==1)&&(m==2)) { l[2*i]=l[2*i-1]; attrib(l[2*i],"isSB",1); l[2*i]=std(l[2*i-1],sact[1]); } if((size(sact)==1)&&(m>2))
• ## Singular/LIB/rwalk.lib
• Property mode changed from `100644` to `100755`
• Property mode changed from `100644` to `100755`
• Property mode changed from `100644` to `100755` | 1,376 | 3,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.338006 |
http://www.talkstats.com/showthread.php/68105-Calculating-probabilities-from-logit-parameters-estimated-with-effect-coding?p=196299 | 1,508,404,145,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823260.52/warc/CC-MAIN-20171019084246-20171019104246-00708.warc.gz | 560,151,828 | 10,487 | ## Calculating probabilities from logit parameters estimated with effect coding
Hello,
I calculated this multinomial logit model using an effect coding (-1 0 1). The variable edu3 is the education level (3 categories), sex has of course 2 categories and eduM3 is the education of the mother in 3 categories.
Parameter edu3 Estimate Pr*>*ChiSq
Intercept 0 -1.6882 <.0001
Intercept 1 -0.2962 <.0001
sex 0 0 0.0797 <.0001
sex 0 1 0.1304 <.0001
eduM3 0 0 1.8393 <.0001
eduM3 0 1 0.6276 <.0001
eduM3 1 0 -0.5476 <.0001
eduM3 1 1 0.3453 <.0001
From this, I would like to compute the resulting probabilities for every possible case, for example, the probability for a man (sex=0) whose mother has a education level 0 (edum3=0) to get an education level 0 (edu3=0).
The cross table for sex=0 gives this:
Table*1*of*eduM3*by*edu3
Controlling*for*sex=0
eduM3 edu3
0 1 2
0 32.36 41.17 26.46
1 5.56 51.08 43.36
2 4.38 24.93 70.7
In this example, the wanted probability should then be 0.3236. However, when I implement parameters in the formula, I get this:
EXP(-1.6882+0.0797+1.8393)/(1+EXP(-1.6882+0.0797+1.8393)+EXP(-0.2962+0.1304+-0.5476)) = 0.4581
Why? I guess that something is wrong with my formula. It probably has something to do with the fact that I used an effect coding rather than a dummy coding. | 472 | 1,304 | {"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} | 2.9375 | 3 | CC-MAIN-2017-43 | latest | en | 0.763492 |
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# Double Bar Graph
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Last updated date: 02nd Aug 2024
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Views today: 4.19k
## Introduction to Double Bar Graph
The most typical way to depict groups of data in graphs is with a double bar graph. Data from items based on two categories are represented and compared using this method. Typically, bar graphs are covered in the data analysis and interpretation component of competitive exams.To answer the questions one must be able to interpret bar graphs.For practicing double bar graph one could consult double bar graph worksheets.
Double Bar Graph
The height of each bar gives information, helping the children notice which columns have more objects and which columns have fewer objects. Bar graphs also help children sort and organize information.
## What is a Double Bar Graph?
Double Bar Graph
Bar graphs are a common visual representation. A bar graph is a graphical representation of data that uses lone bars of different heights.
A double bar graph is a visual representation of data that uses two parallel bars of varying heights. You can arrange the bars either vertically or horizontally. A double bar graph can be used to contrast two sets of data to visually compare and contrast between different sets of data using the double bar graph template.
## How to Make a Double Bar Graph?
To create a double bar graph, follow these steps:
• Make a data-filled table.
The Frankfurt annual high and low temperatures are shown in this table in degrees Celsius.
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Max 5 7 11 16 20 24 26 26 21 15 9 6 Min 0 0 2 6 10 13 15 15 11 7 3 1
• To see what kinds of numbers the y-axis should have, determine the range of the data.
The number 26 is the highest value while 0 is the lowest. This graph should be scaled such that 1 unit equals 10 degrees Celsius.
• Make a data plot.
Excel or other software programmes can be used to make a double bar graph. On a sheet of paper, it can also be completed manually.
Temperature- Month Bar Graph
• Give the x- and y-axes labels.
In this instance, the y-axis displays the temperature in degrees Celsius, and the x-axis displays the yearly months.
Double Bar Graph
• Put a key.
The key here reveals that the highest temperatures are displayed in red and the lowest in green.
A Double Bar Graph with a Key
The graph is described in the title.
A Double Bar Graph with Heading
## Double Bar Graph Example:
Learning from Examples
Following are the double bar graph example-
Example 1. What sport do girls like to play?
Graph of Favorite Sports
Ans: Look at the graph's pink-colored highest bar. The color of the bar indicates the girls' preferred sports.
Then confirm that the value on the y axis represents the greatest value for any female favorite sport. Soccer is the top choice in this instance with 10 females making that choice, which is also the highest number on the graph of girls' preferred sports.
Example 2. Observe the given graph and answer the questions below.
Graph of Sports played by Girls and Boys
1. What sport do girls prefer to play the most?
2. Which sport do boys prefer the most?
Ans: Cricket
## Double Bar Graphs Compare Multiple Bar Graph:
Double Bar Graphs Compare Multiple
Let’s see double bar graphs compare multiple bar graph-
• Double Bar Graph - A double bar graph is used to display two sets of data on the same graph. We would utilize a double bar graph, for instance, to compare the number of hours that students worked in one month to another.
• Multiple Bar Graph - In a bar graph, there may occasionally be more than two sets of data to compare. A multiple bar graph can be utilized in that situation. Any number of data sets can be compared using a multiple bar graph.
## Summary
Two sets of data are shown on the same graph using a double bar graph. We would employ a double bar graph, for instance, to compare the number of hours that students worked in one month to another.
The double bar graph worksheets demonstrate how to read a double bar graph to find data and address issues.
## FAQs on Double Bar Graph
1. How to make a double bar graph and what is the title of a double bar graph?
The steps to follow when creating a double bar graph are:
• Create a table with data.
• Determine the range of the data.
• Plot the data.Label the x- and y-axis.
The double bar graph's title gives the viewer a broad picture of what is being measured and compared. A key will also be provided for a double bar graph. The trick to a double bar graph is to use two different colors to indicate the two groups that are being compared.
2. Simply put, what is a double bar graph and how is it read?
The graphical depiction of grouped data with two parameters provided for each category is called a double bar graph. Two bars are displayed for each category in a double bar graph. The two parameters for each group are represented by these two bars.
In order to read a double-bar graph, do the following:
• The subject(s) of the graph can be found by reading the title.
• To determine what the bars in the graph represent, look at the information in the key and on the side and bottom of the graph.
• To infer information about the graph's subjects, compare the lengths of the bars. | 1,162 | 5,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.862385 |
https://www.enotes.com/homework-help/evaluate-integral-e-8-sin9-d-360161 | 1,516,680,965,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891706.88/warc/CC-MAIN-20180123032443-20180123052443-00320.warc.gz | 880,573,797 | 9,662 | Evaluate `int e^(8theta) sin(9theta)d theta` .
sciencesolve | Certified Educator
You should use integration by parts such that:
`int udv = uv - int vdu`
`u = e^(8theta) => du = 8e^(8theta)d theta`
`dv =sin 9 theta => v = -(cos 9theta)/9`
`int e^(8theta)*sin 9 theta dtheta =-e^(8theta)*(cos 9theta)/9+ (8/9) int e^(8theta)*cos 9 theta d theta`
You should solve `int e^(8theta)*cos 9 theta d theta ` using parts such that:
`u = e^(8theta) => du = 8e^(8theta)d theta`
`dv = cos 9 theta => v = (sin 9 theta)/9`
`int e^(8theta)*cos 9 theta d theta = e^(8theta)*(sin 9 theta)/9 - (8/9) int e^(8theta)*sin 9 theta d theta`
You should come up with the following notation `int e^(8theta)*sin 9 theta dtheta = I` such that:
`I = -e^(8theta)*(cos 9theta)/9+ (8/9)*(e^(8theta)*(sin 9 theta)/9 - (8/9)I)`
You should open the brackets such that:
`I = -e^(8theta)*(cos 9theta)/9+ (8e^(8theta)*(sin 9 theta))/81 - 64/81 I`
You need to move the terms that contain I to the left side such that:
`I + 64/81I = -e^(8theta)*(cos 9theta)/9 + (8e^(8theta)*(sin 9 theta))/81`
`145/81 I = -e^(8theta)*(cos 9theta)/9 + (8e^(8theta)*(sin 9 theta))/81 `
`I = (-9e^(8theta)*(cos 9theta) + 8e^(8theta)*(sin 9 theta))/145`
Hence, evaluating the given integral using parts yields `int e^(8theta)*sin 9 theta d theta =(-9e^(8theta)*(cos 9theta) + 8e^(8theta)*(sin 9 theta))/145.` | 514 | 1,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-05 | latest | en | 0.654242 |
https://devforum.roblox.com/t/what-is-the-fastest-way-to-generate-a-cframe-of-random-orientation/2224957 | 1,718,452,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00121.warc.gz | 183,525,422 | 7,860 | # What is the fastest way to generate a CFrame of random orientation?
Given:
``````local Rand = Random.new(SomeSeed)
``````
Would this be faster:
``````local Frame = CFrame.Angles(Rand:NextNumber(-math.pi, math.pi), Rand:NextNumber(-math.pi, math.pi), Rand:NextNumber(-math.pi, math.pi))
``````
Or this:
``````local Frame = CFrame.lookAt(Vector3.zero, Rand:NextUnitVector())
``````
Or some unbeknownst to me 3rd way (please share)?
As a side note, do you know if `Rand:NextUnitVector()` or `Vector3.new(math.random()-0.5, math.random()-0.5, math.random()-0.5)*2` is faster, even though the former is more accurate?
1 Like
I feel like the first would be faster. However, I’m ready to be proven wrong.
1 Like
Probably the first one as a guess.
Small Thing:
You can use `math.random()` without any arguments which will generate a random number with decimals. While being the “fastest”, It is however limited to numbers below `1`, which would have an Angle Below `60` in Radians.
There is a niche reason I am using `Random` instead of `math.random()`. Visual effects, such as a custom explosion, should be handled on the client only. So I use `FireAllClients()` from a `RemoteEvent`. But I want all clients to see exactly the same thing, even though no one would probably notice or care. So I pass a seed from the server with which the clients generate identical visual effects.
I am not Suggesting that you should use `math.random`, I’m saying that you could, but you are limited to a certain amount
(I mean, I am, and Im not)
`math.randomseed` exists though. However, I don’t know if it’s faster.
Oh, so I could just call `math.randomseed()` with the seed sent by the server in `OnClientEvent`, instead of using `Random.new`, would that work as expected?
1 Like
It’s really important in programming to be able to run these sorts of tests yourself. The odds are slim that anyone here will have tested it already. Occasionally you’ll get a CS major who chimes in with what sounds like an educated response, but often as not they’re wrong for whatever reason such as oddities with how Lua and C interact.
I’m not sure if this is what you mean or not, but I’m going to say it anyways because your wording is a little vague and future readers may benefit from the knowledge. `NextUnitVector` exists to avoid the bias towards the poles that the first method creates.
CFrame.Angles method
NextUnitVector method
I’ll go run some benchmarks since I’m already in Studio, but I do want to re-emphasize how important it is to be able to run benchmarks yourself. You’d be done within 5 minutes instead of waiting an hour for some guesses and speculations.
Here are the results from a benchmark with 10,000 iterations each.
Stick with the NextUnitVector approach.
4 Likes
What if they were combined? For example, it used `CFrame.Angles` but it used the unit vector numbers converted to radians.
You’d run into bias towards the poles again.
This topic was automatically closed 14 days after the last reply. New replies are no longer allowed. | 766 | 3,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.899285 |
https://www.folj.com/bb/viewtopic.php?t=136&sid=6386ea22393a6687212a6d5cfb4db2e1 | 1,642,382,007,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00146.warc.gz | 835,853,670 | 4,346 | # More Puzzles
Register Immediately
Very Difficult Logic Problems Out of 52 cards 13 are up and 39 are down and are mixed
Sun Apr 11, 2010 6:25 pm by tartle Out of 52 cards 13 are up and 39 are down and are mixed . How to arrange them in two decks of equal numbers of up. Condition You are sitting in a dark room and there is no light.
Mon Jun 28, 2010 3:17 pm by shrek619 can you give a small hint ?
Tue Sep 21, 2010 8:55 am by lonewolf Saperate the piles into 2 equal piles of 26 cards each. Flip all the card of on pile. I think this question is simmilar to the 20 coin question.
Wed Dec 29, 2010 11:33 am by moshu123 The correct solution is like this: Saperate the piles into 2 piles of 13 & 39 cards each. Flip all the card of smaller pile.
Sat Feb 23, 2013 9:23 pm by chopstick divide in two pile one with 13 and other with 39. flip the cards of pile 13. explanation: say pile with 39 has x up cards so card with pile 13 has 13-x up cards means it has x down cards so if u flip u will have x up cards same as pile with 39 cards
Thu Jul 21, 2016 2:49 pm by bathfilms So far everyone has a correct solution that yields the same number UP in each pile. However, Lonewolf's solution accomplishes this AND the same number DOWN in each pile. Simply split into 2 equal piles and turn one upside down.
All times are GMT Page 1 of 1 | 395 | 1,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.934221 |
https://brainly.com/question/316271 | 1,484,734,058,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280266.9/warc/CC-MAIN-20170116095120-00050-ip-10-171-10-70.ec2.internal.warc.gz | 818,894,134 | 8,987 | 2015-02-21T19:17:22-05:00
### 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.
1 mile = 1.609 kilometers
194 * 1.609
312.146
There are 312.146 kilometers in 194 miles. | 110 | 425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.880675 |
https://community.turingtumble.com/t/how-to-compute-an-arbitrary-boolean-function/407 | 1,558,741,012,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257781.73/warc/CC-MAIN-20190524224619-20190525010619-00267.warc.gz | 421,003,502 | 4,669 | # How to compute an arbitrary Boolean function
#1
tl;dr: TT can compute any Boolean function. Link and explanation below.
As the attempt to prove some extensible infinite version of TT (turing tumble) is Turing-complete continues, I haven’t been convinced of a weaker claim: That TT can be used to compute an arbitrary Boolean function. That is, given an arbitrary function f that takes as input N bits, and outputs 1 bit, is there a way to compute f on the TT board. Arguably, if that can be done, then the function computed by any given circuit, being a Boolean function, can be computed with TT.
While this might seem obvious because we can build NAND gates - the issue of connecting them seemed to me to be a potential problem, so I’ve been thinking of a way just to encode a Boolean function directly, and have come up with a completely inelegant, brute-force way to compute any given function. It will require some explanation, but here is an example for an arbitrary function of 3 bits.
It is missing some pieces - you put the pieces in to specify the function.
Here is the idea. This is basically a decision-tree, which tests each of the three input bits, which are specified by the three rows of gear bits. These tests result in eight distinct paths (one for each of the eight possible input bit-settings). The eight paths are what you get coming out of the bottom row of gear bits. To specify a function, ramps are placed just below the gear bits, in the eight locations. A left-pointing ramp encodes that the function to be computed is 0 given that the bits were set to arrive at that point. A right-pointing one means the value of the function should be 1.
For example, the constant 0 function has eight left-pointing ramps just below the bottom gear bits. The constant 1 function has eight right-pointing ramps. The function that tests if the input is equal to either 3 or 7 has the following sequence of ramps: RRLLLLLL, because if the input (the position of the gear bits, with 1-bit at the top) are set to 3 or 7, the ball will come out of the bottom left gear bit.
The output of the function is represented by which interceptor is reached - the left means 0 (false), the right means 1 (true).
It should be easy to see that once a ball starts going left, it will end up in the left (0) interceptor, and once it starts going right it will end up in the right (1) interceptor.
So, this is a decision-tree that is in effect doing a table-lookup of the values of the function. The disadvantage of course is that the encoding of every function of N bits in this scheme requires O(2^N) pieces.
Since this can be generalized to any number N of bits, and any function of N bits, I think it shows that TT is universal.
Proof of Turing Completeness?
Proof of Turing Completeness?
#2
Follow-up. This can be used to compute a function from N bits to N bits, and also with a feedback loop of "C"s as described by @jcross in the thread on Turing Completeness, we can use the N input bits to reflect the output. See my explanation in the Turing Completeness thread.
Edit: Using the same idea, we can create any finite state automaton which takes two inputs (blue ball, red ball). Each is routed to a different half of a decision tree, which reads the current state from a stack of n bits (n layers of the tree), then writes a new state by going through a stack of n writes to the bits representing the state (using @jcross “C” trick), and then the ball gets intercepted. A new input is given by the user by pressing either the blue or red lever.
Second Edit: Here as an example of how to make a finite state machine is a working 3-state machine, with states 00, 10, and 11 (read off of the top two rows of bits. The state transitions are described using the following notation:
(00, 00, 11)
(10, 10, 00)
(11, 10, 11)
So, for example, if in state 10, and a red marble is received, the new state is 00.
Third Edit:
Here is a finite state machine that reads in a binary sequence of marbles (red = 1, blue = 0), and ends in state 0 if and only if the binary number was divisible by 3. (The state is indicated by the first two rows of bits.)
How to simulate any computer with finitely many pieces | 981 | 4,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-22 | latest | en | 0.9186 |
https://www.jiskha.com/questions/1788065/I-score-3050-on-my-scantron-test-on-www-performanceseries-com-Is-that-considered | 1,558,757,707,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257847.56/warc/CC-MAIN-20190525024710-20190525050710-00173.warc.gz | 822,943,542 | 4,861 | # math
I score 3050 on my scantron test on www.performanceseries.com. Is that considered a good score? or bad?
1. 👍 1
2. 👎 0
3. 👁 84
1. I do believe so.. In my school that would be considered good, may want to check with your teacher.
1. 👍 2
2. 👎 0
2. ok thank you :)
1. 👍 1
2. 👎 0
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More Similar Questions | 760 | 2,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-22 | latest | en | 0.960079 |
https://www.hackmath.net/en/math-problem/375?tag_id=101 | 1,611,194,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00514.warc.gz | 820,163,271 | 12,898 | # Valid number
Round the 453874528 on 2 significant numbers.
Correct result:
x = 450000000
#### Solution:
$x=453874528\doteq 450000000$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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If it is now 7:38 pm, what time will it be in 30,033,996,480 minutes from now?
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There is a group of children. There is a boy named Adam in each of the three children subgroup and a girl named Beata in each quartet (four-member subgroup). How many children can be in such a group and what are their names in that case?
• The temperature 10
The temperature in a freezer is -15°C and it increases by 3°C.
• The cube
The cube has a surface of 600 cm2, what is its volume?
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If the clock showed a quarter of an hour more, it would be 10 minutes to 10 hours. How many hours do they show?
• Milk crates
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• Volume per time
How long does fill take for a pump with a volume flow of 200 l per minute fill a cube-shaped tank up to 75% of its height if the length of the cube edge is 4 m?
• Venn diagram
University students chose a foreign language for the 1st year. Of the 120 enrolled students, 75 chose English, 65 German, and 40 both English and German. Using the Venn diagram, determine: - how many of the enrolled students chose English only - how many
• Language courses
Of the company's 60 employees, 28 attend an English course, 17 take a German course, and 20 do not attend any of these courses. How many employees attend both courses?
• Sand
How much m ^ 3 of sand can be loaded on a car with a load capacity of 5 t? The sand density is 1600 kg/m3 . | 820 | 3,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-04 | latest | en | 0.968956 |
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# Problem 44763. Pumpkin Pie!
Solution 1670496
Submitted on 9 Nov 2018 by Tim
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
already_cut_fractions = [1/4 1/4 1/4]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(ischar(slice_fraction)) str_parts = cellfun(@str2num,regexp(slice_fraction,'/','split')); assert(numel(str_parts)==2) slice_dec = str_parts(1) / str_parts(2); assert(abs(slice_dec - 0.704)<1e-2)
ans = '0.7037/1'
2 Pass
ans = '0/1'
3 Pass
already_cut_fractions = [1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(strcmpi(slice_fraction,'pass'))
ans = '-0.24859/1'
4 Pass
already_cut_fractions = [1/8 1/8 1/8 1/8]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(ischar(slice_fraction)) str_parts = cellfun(@str2num,regexp(slice_fraction,'/','split')); assert(numel(str_parts)==2) slice_dec = str_parts(1) / str_parts(2); assert(abs(slice_dec - 0.787)<1e-2)
ans = '0.78676/1'
5 Pass
already_cut_fractions = [1/16 1/32]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(ischar(slice_fraction)) str_parts = cellfun(@str2num,regexp(slice_fraction,'/','split')); assert(numel(str_parts)==2) slice_dec = str_parts(1) / str_parts(2); assert(abs(slice_dec - 0.862)<1e-2)
ans = '0.86237/1'
6 Pass
already_cut_fractions = [1/7 1/5 1/6 1/4 1/8 1/10 1/9]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(ischar(slice_fraction)) str_parts = cellfun(@str2num,regexp(slice_fraction,'/','split')); assert(numel(str_parts)==2) slice_dec = str_parts(1) / str_parts(2); assert(abs(slice_dec - 0.583)<1e-2)
ans = '0.58333/1'
7 Pass
ans = '-0.051618/1'
8 Pass
ans = '0/1'
9 Pass
already_cut_fractions = [1/3 1/5 1/7 1/11 1/13 1/17 1/19 1/23 1/29 1/31]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(ischar(slice_fraction)) str_parts = cellfun(@str2num,regexp(slice_fraction,'/','split')); assert(numel(str_parts)==2) slice_dec = str_parts(1) / str_parts(2); assert(abs(slice_dec - 0.591)<1e-2)
ans = '0.59111/1'
10 Pass
already_cut_fractions = [1/3 1/4 1/2 1/3 1/6]; slice_fraction = pumpkin_pie(already_cut_fractions); assert(ischar(slice_fraction)) str_parts = cellfun(@str2num,regexp(slice_fraction,'/','split')); assert(numel(str_parts)==2) slice_dec = str_parts(1) / str_parts(2); assert(abs(slice_dec - 0.1)<1e-2)
ans = '0.1/1'
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Start Hunting! | 890 | 2,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-04 | latest | en | 0.316615 |
https://www.includehelp.com/python/partition-array-into-n-chunks-with-numpy.aspx | 1,675,371,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00485.warc.gz | 826,725,237 | 148,937 | # Partition array into N chunks with NumPy
Learn, how to Partition array into N chunks with NumPy in Python?
Submitted by Pranit Sharma, on January 11, 2023
NumPy is an abbreviated form of Numerical Python. It is used for different types of scientific operations in python. Numpy is a vast library in python which is used for almost every kind of scientific or mathematical operation. It is itself an array which is a collection of various methods and functions for processing the arrays.
## NumPy - Partitioning array into N chunks
Suppose we are given a NumPy array and we need to split it into N chunks.
For this purpose, we will use numpy.array_split() method. It split an array into multiple sub-arrays.
For an array of length L that should be split into n sections, it returns L % n sub-arrays of size : L//n + 1 and the rest of size L//n.
Let us understand with the help of an example,
## Python code for partitioning array into N chunks with NumPy
```# Import numpy
import numpy as np
# Creating an array
arr = np.array([[1,2],[3,4], [5,6], [6,7]])
# Display original array
print("Original Array:\n",arr,"\n")
# Splitting array into 3 parts
res = np.array_split(arr, 3)
# Display result
print("Result:\n",res)
```
Output:
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Question
# A stone of mass 1kg tied to a light inextensible string of length L=103m is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum is 4 and if g is taken to be 10 m/s2, then speed of stone at the highest point of the circle is
A
20m/sec
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B
103 m/sec
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C
52 m/sec
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D
10m/sec
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Solution
## The correct option is D 10m/secLet 'v' be the velocity at the top pointVelocity at the bottom point is =v2+2gd =v2+4grMaximum tension in string appears when the stone is at the bottom of the circle.Tmax=m(v2+4gr)r+mgMinimum tension appears in a string when the stone is at the top of the circle.Tmin=mv2r−mgTmaxTmin=v2+4grr+gv2r−g =4On solving the equation we get, v=10 m/s
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Lecture
# 03 Stats, Histograms, Probability Distribution, Symbolic Forms.pdf Premium
13 Pages
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School
University of California - Irvine
Department
Chemistry
Course
CHEM 5
Professor
Douglas Tobias
Semester
Fall
Description
Elementary statistical analysis Statistical analysis of data is ubiquitous in the physical, biological, and social sciences. There are a large number of statistical quantities that are useful for characterizing sets of data. Below we introduce the use of Mathematica to calculate several basic statistical parameters and to depict the distribution of data. We begin with a list of data that represents a set of measurements of some quantity: In[1]:=data = 81.05, 1.01, 0.97, 1.14, 0.92, 0.99, 1.07< Out[1]=81.05, 1.01, 0.97, 1.14, 0.92, 0.99, 1.07< Min and Max report the minimum and maximum values found in the list: In[2]:[email protected] Out[2]=0.92 In[3]:[email protected] Out[3]=1.14 Total sums the values in the list: In[4]:[email protected] 7.15 Out[4]= N Mean gives the mean or average value, defined as = ⁄ i=1xi, where the x ire the values in the list, and N is the number of values in the list: In[5]:[email protected] Out[5]=1.02143 Median is the central value in the list, which is evident when the values are sorted: In[6]:[email protected] Out[6]=1.01 [email protected] In[7]:= Out[7]=80.92, 0.97, 0.99, 1.01, 1.05, 1.07, 1.14< The variance, denoted s , is a measure of the spread of the values about the mean. It is defined as s = ⁄ N Hx - < x >L /(N | 1), and computed in Mathematica using the Variance command: i=1 i In[8]:[email protected] Out[8]=0.00521429 The standard deviation, s, computed in Mathematica using the StandardDeviation command, is the square root of the variance: 2 mathematica_lesson4.nb In[9]:[email protected] 0.07221 Out[9]= In[10]:=* [email protected],[email protected]@5,1DD
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## Chapter 6: Visualising Solid Shapes
Exercise
Exercise [Pages 181 - 205]
### NCERT solutions for Mathematics Exemplar Class 8 Chapter 6 Visualising Solid Shapes Exercise [Pages 181 - 205]
#### Choose the correct alternative:
Exercise | Q 1 | Page 181
Which amongst the following is not a polyhedron?
Exercise | Q 2 | Page 181
Which of the following will not form a polyhedron?
• 3 triangle
• 2 triangles and 3 parallelogram
• 8 triangles
• 1 pentagon and 5 triangles
Exercise | Q 3 | Page 181
Which of the following is a regular polyhedron?
• Cuboid
• Triangular prism
• Cube
• Square prism
Exercise | Q 4 | Page 181
Which of the following is a two Dimensional figure?
• Rectangle
• Rectangular Prism
• Square Pyramid
• Square Prism
Exercise | Q 5 | Page 181
Which of the following can be the base of a pyramid?
• Line segmen
• Circle
• Octagon
• Oval
Exercise | Q 6 | Page 181
Which of the following 3D shapes does not have a vertex?
• Pyramid
• Prism
• Cone
• Sphere
Exercise | Q 7 | Page 181
Solid having only line segments as its edges is a ______.
• Polyhedron
• Cone
• Cylinder
• Polygon
Exercise | Q 8 | Page 182
In a solid if F = V = 5, then the number of edges in this shape is ______.
• 6
• 4
• 8
• 2
Exercise | Q 9 | Page 182
Which of the following is the top view of the given shape?
Exercise | Q 10 | Page 182
The net shown below can be folded into the shape of a cube. The face marked with the letter L is opposite to the face marked with which letter?
• M
• N
• Q
• O
Exercise | Q 11 | Page 183
Which of the nets given below will generate a cone?
Exercise | Q 12 | Page 183
Which of the following is not a prism?
Exercise | Q 13 | Page 183
We have 4 congruent equilateral triangles. What do we need more to make a pyramid?
• An equilateral triangle
• A square with same side length as of triangle
• 2 equilateral triangles with side length same as triangle
• 2 squares with side length same as triangle
Exercise | Q 14 | Page 184
Side of a square garden is 30 m. If the scale used to draw its picture is 1 cm: 5 m, the perimeter of the square in the picture is ______.
• 20 cm
• 24 cm
• 28 cm
• 30 cm
Exercise | Q 15 | Page 184
Which of the following shapes has a vertex?
Exercise | Q 16 | Page 184
In the given map, the distance between the places is shown using the scale 1 cm : 0.5 km. Then the actual distance (in km) between school and the book shop is ______.
• 1.25
• 2.5
• 2
• 1.1
Exercise | Q 17 | Page 185
Which of the following cannot be true for a polyhedron?
• V = 4, F = 4, E = 6
• V = 6, F = 8, E = 12
• V = 20, F = 12, E = 30
• V = 4, F = 6, E = 6
Exercise | Q 18 | Page 185
In a blueprint of a room, an architect has shown the height of the room as 33 cm. If the actual height of the room is 330 cm, then the scale used by her is ______.
• 1:11
• 1:10
• 1:100
• 1:3
Exercise | Q 19 | Page 185
The following is the map of a town. Based on it answer questions 19 - 21.
The number of hospitals in the town is ______.
• 1
• 2
• 3
• 4
Exercise | Q 20 | Page 186
The ratio of the number of general stores and that of the ground is ______.
• 1:2
• 2:1
• 2:3
• 3:2
Exercise | Q 21 | Page 186
According to the map, the number of schools in the town is ______.
• 4
• 3
• 5
• 2
#### Fill in the blanks:
Exercise | Q 22 | Page 186
Square prism is also called a ______.
Exercise | Q 23 | Page 186
Rectangular prism is also called a ______.
Exercise | Q 24 | Page 186
In the figure, the number of faces meeting at B is ______.
Exercise | Q 25 | Page 186
A pyramid on an n sided polygon has ______ faces.
Exercise | Q 26 | Page 186
If a solid shape has 12 faces and 20 vertices, then the number of edges in this solid is ______.
Exercise | Q 27 | Page 186
The given net can be folded to make a ______.
Exercise | Q 28 | Page 186
A solid figure with only 1 vertex is a ______.
Exercise | Q 29 | Page 186
Total number of faces in a pyramid which has eight edges is ______.
Exercise | Q 30 | Page 186
The net of a rectangular prism has ______ rectangles.
Exercise | Q 31 | Page 187
In a three-dimensional shape, diagonal is a line segment that joins two vertices that do not lie on the ______ face.
Exercise | Q 32 | Page 187
If 4 km on a map is represented by 1 cm, then 16 km is represented by ______ cm.
Exercise | Q 33 | Page 187
If actual distance between two places A and B is 110 km and it is represented on a map by 25 mm. Then the scale used is ______.
Exercise | Q 34 | Page 187
A pentagonal prism has ______ faces.
Exercise | Q 35 | Page 187
If a pyramid has a hexagonal base, then the number of vertices is ______.
Exercise | Q 36 | Page 187
is the ______ view of
Exercise | Q 37 | Page 187
The number of cubes in are ______
Exercise | Q 38 | Page 187
If the sum of number of vertices and faces in a polyhedron is 14, then the number of edges in that shape is ______.
Exercise | Q 39 | Page 187
Total number of regular polyhedra is ______.
Exercise | Q 40 | Page 188
A regular polyhedron is a solid made up of ______ faces.
Exercise | Q 41. (a) | Page 188
For each of the following solids, identify the front, side and top views and write it in the space provided.
Exercise | Q 41. (b) | Page 188
For each of the following solids, identify the front, side and top views and write it in the space provided.
Exercise | Q 41. (c) | Page 188
For each of the following solids, identify the front, side and top views and write it in the space provided.
Exercise | Q 41. (d) | Page 188
For each of the following solids, identify the front, side and top views and write it in the space provided.
#### State whether the following statement is True or False:
Exercise | Q 42 | Page 189
The other name of cuboid is tetrahedron.
• True
• False
Exercise | Q 43 | Page 189
A polyhedron can have 3 faces.
• True
• False
Exercise | Q 44 | Page 189
A polyhedron with least number of faces is known as a triangular pyramid.
• True
• False
Exercise | Q 45 | Page 189
Regular octahedron has 8 congruent faces which are isosceles triangles.
• True
• False
Exercise | Q 46 | Page 190
Pentagonal prism has 5 pentagons.
• True
• False
Exercise | Q 47 | Page 190
Every cylinder has 2 opposite faces as congruent circles, so it is also a prism.
• True
• False
Exercise | Q 48 | Page 190
Euler’s formula is true for all three-dimensional shapes.
• True
• False
Exercise | Q 49 | Page 190
A polyhedron can have 10 faces, 20 edges and 15 vertices.
• True
• False
Exercise | Q 50 | Page 190
The top view of
• True
• False
Exercise | Q 51 | Page 190
The number of edges in a parallelogram is 4.
• True
• False
Exercise | Q 52 | Page 190
Every solid shape has a unique net.
• True
• False
Exercise | Q 53 | Page 190
Pyramids do not have a diagonal.
• True
• False
Exercise | Q 54 | Page 190
The given shape is a cylinder.
• True
• False
Exercise | Q 55 | Page 190
A cuboid has atleast 4 diagonals.
• True
• False
Exercise | Q 56 | Page 190
All cubes are prisms
• True
• False
Exercise | Q 57 | Page 191
A cylinder is a 3-D shape having two circular faces of different radii.
• True
• False
Exercise | Q 58 | Page 191
On the basis of the given figure, the length of a rectangle in the net of a cylinder is same as circumference of circles in its net.
• True
• False
Exercise | Q 59 | Page 191
If a length of 100 m is represented on a map by 1 cm, then the actual distance corresponding to 2 cm is 200 m.
• True
• False
Exercise | Q 60 | Page 191
The model of a ship shown is of height 3.5 cm. The actual height of the ship is 210 cm if the scale chosen is 1: 60.
• True
• False
Exercise | Q 61 | Page 191
The actual width of a storeroom is 280 cm. If the scale chosen to make its drawing is 1:7, then the width of the room in the drawing will be 40 cm.
• True
• False
Exercise | Q 62 | Page 192
Complete the table given below:
S.No Solid Shape of Solid Number of faces F Number of Verticles V Number of edges E F + V E + 2 a. Cuboid b. Triangular Pyramid c. Square Pyramid d. Rectangular Pyramid e. Pentagonal Pyramid f. Hexagonal Pyramid g. Triangular Prism h. Square Prism i. Cube j. Pentagonal Prism k. Octagonal Prism l. Heptagonal Prism
Exercise | Q 63.(a) | Page 192
How many faces does the following solids, have tetrahedron?
Exercise | Q 63.(b) | Page 192
How many faces does the following solids, have hexahedron?
Exercise | Q 63.(c) | Page 192
How many faces does the following solids have octagonal pyramid?
Exercise | Q 63.(d) | Page 192
How many faces does the following solids have octahedron?
Exercise | Q 64 | Page 193
Draw a prism with its base as regular hexagon with one of its face facing you. Now draw the top view, front view and side view of this solid.
Exercise | Q 65.(a) | Page 193
How many vertices does the following solids have?
Cone
Exercise | Q 65.(b) | Page 193
How many vertices does the following solids have?
Cylinder
Exercise | Q 65.(c) | Page 193
How many vertices does the following solids have?
Sphere
Exercise | Q 65.(d) | Page 193
How many vertices does the following solids have?
Octagonal Pyramid
Exercise | Q 65.(e) | Page 193
How many vertices does the following solids have?
Tetrahedron
Exercise | Q 65.(f) | Page 193
How many vertices does the following solids have?
Hexagonal Prism
Exercise | Q 66.(a) | Page 193
How many edges does following solids have?
Cone
Exercise | Q 66.(b) | Page 193
How many edges does following solids have?
Cylinder
Exercise | Q 66.(c) | Page 193
How many edges does following solids have?
Sphere
Exercise | Q 66.(d) | Page 193
How many edges does following solids have?
Octagonal Pyramid
Exercise | Q 66.(e) | Page 193
How many edges does following solids have?
Hexagonal Prism
Exercise | Q 66.(f) | Page 193
How many edges does following solids have?
Kaleidoscope
Exercise | Q 67.(a) | Page 194
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(b) | Page 194
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(c) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(d) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(e) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(f) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(g) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(h) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(i) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(j) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(k) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67.(l) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 67. (m) | Page 193
Look at the shapes given below and state which of these are polyhedra using Euler’s formula.
Exercise | Q 68.(a) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(b) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(c) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(d) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(e) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(f) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(g) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(h) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(i) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(j) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(k) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 68.(l) | Page 195
Count the number of cubes in the given shapes.
Exercise | Q 69.(a) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(b) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(c) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(d) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(e) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(f) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(g) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(h) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(i) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 69.(j) | Page 196
Draw the front, side and top view of the given shapes.
Exercise | Q 70.(i) | Page 197
Using Euler’s formula, find the value of unknown x, y, z, p, q, r, in the following table.
Faces 7 Vertices 10 Edges x
Exercise | Q 70.(ii) | Page 197
Using Euler’s formula, find the value of unknown x, y, z, p, q, r, in the following table.
Faces y Vertices 12 Edges 18
Exercise | Q 70.(iii) | Page 197
Using Euler’s formula, find the value of unknown x, y, z, p, q, r, in the following table.
Faces 9 Vertices z Edges 16
Exercise | Q 70.(iv) | Page 197
Using Euler’s formula, find the value of unknown x, y, z, p, q, r, in the following table.
Faces p Vertices 6 Edges 12
Exercise | Q 70.(v) | Page 197
Using Euler’s formula, find the value of unknown x, y, z, p, q, r, in the following table.
Faces 6 Vertices q Edges 12
Exercise | Q 70.(vi) | Page 197
Using Euler’s formula, find the value of unknown x, y, z, p, q, r, in the following table.
Faces 8 Vertices 11 Edges r
Exercise | Q 71 | Page 198
Can a polyhedron have V = F = 9 and E = 16? If yes, draw its figure.
Exercise | Q 72 | Page 198
Check whether a polyhedron can have V = 12, E = 6 and F = 8.
Exercise | Q 73 | Page 198
A polyhedron has 60 edges and 40 vertices. Find the number of its faces.
Exercise | Q 74. (a) | Page 198
Find the number of faces in the given shapes:
Exercise | Q 74. (b) | Page 198
Find the number of faces in the given shapes:
Exercise | Q 74. (c) | Page 198
Find the number of faces in the given shapes:
Exercise | Q 75 | Page 198
A polyhedron has 20 faces and 12 vertices. Find the edges of the polyhedron.
Exercise | Q 76 | Page 198
A solid has forty faces and, sixty edges. Find the number of vertices of the solid.
Exercise | Q 77 | Page 198
Draw the net of a regular hexahedron with side 3 cm. (Hint: Regular hexahedron - cube)
Exercise | Q 78 | Page 198
Draw the net of a regular tetrahedron with side 6 cm.
Exercise | Q 79 | Page 198
Draw the net of the following cuboid:
#### Match the following:
Exercise | Q 80 | Page 199
Figure Name (a) (a) Hexahedron (b) (b) Hexagonal Prism (c) (c) Square Pyramid (d) (d) Cone
Exercise | Q 81 | Page 199
Complete the table given below by putting tick mark across the respective property found in the solids mentioned.
Solids Properties Cone Cylinder Prism Pyramid 1. The figure is a Polyhedron 2. The figure has diagonals 3. The shape has curved edges 4. The base of figure is a polygon 5. The bases are congruent 6. The base of figure is a polygon and other faces meet at a single point 7. The base of figure is a curved edge and other faces meet at a single point
Exercise | Q 82 | Page 199
Draw the net of the following shape.
Exercise | Q 83 | Page 200
Draw the net of the following solid.
(Hint: Pentagons are not congruent.)
Exercise | Q 84 | Page 200
Find the number of cubes in the base layer of the following figure.
Exercise | Q 85 | Page 200
In the above figure, if only the shaded cubes are visible from the top, draw the base layer.
Exercise | Q 86 | Page 200
How many faces, edges and vertices does a pyramid have with n sided polygon as its base?
Exercise | Q 87 | Page 200
Draw a figure that represents your mathematics textbook. What is the name of this figure? Is it a prism?
Exercise | Q 88 | Page 200
In the given figures, identify the different shapes involved.
Exercise | Q 89 | Page 200
What figure is formed if only the height of a cube is increased or decreased?
Exercise | Q 90.(a) | Page 200
Use isometric dot paper to draw figure.
A tetrahedron
Exercise | Q 90.(b) | Page 200
Use isometric dot paper to draw figure.
A rectangular prism with length 4 units, width 2 units and height 2 units.
Exercise | Q 91 | Page 201
Identify the nets given below and mention the name of the corresponding solid in the space provided.
Nets Name of Solid (a) (b) (c) (d) (e) (f)
Exercise | Q 92 | Page 202
Draw a map of your school playground. Mark all necessary places like 2 library, Playground, Medical Room, Classrooms, Assembly area, etc.
Exercise | Q 93.(a) | Page 202
Refer to the given map to answer the following questions.
What is the built-up area of Govt. Model School I?
Exercise | Q 93.(b) | Page 202
Refer to the given map to answer the following questions.
Name the schools shown in the picture.
Exercise | Q 93.(c) | Page 202
Refer to the given map to answer the following questions.
Which park is nearest to the dispensary?
Exercise | Q 93.(d) | Page 202
Refer to the given map to answer the following questions.
To which block does the main market belong?
Exercise | Q 93.(e) | Page 202
Refer to the given map to answer the following questions.
How many parks have been represented in the map?
Exercise | Q 94 | Page 203
Look at the map given below. Answer the following questions.
(a) Which two hospitals are opposite to each other?
(b) A person residing at Niti Bagh has to go to Chirag Delhi after dropping her daughter at Asiad Tower. Mention the important landmarks he will pass alongwith the roads taken.
(c) Name of which road is similar to the name of some month.
Exercise | Q 95. (a) | Page 203
Look at the map given below
Houses
Exercise | Q 95. (b) | Page 203
Look at the map given below
Houses
Exercise | Q 95. (c) | Page 203
Look at the map given below
Houses
On which road is the Police Station situated?
Exercise | Q 95. (d) | Page 203
Look at the map given below
Houses
If Ritika stays adjacent to bank and you have to send her a card, what address will you write?
Exercise | Q 95. (e) | Page 203
Look at the map given below
Houses
Which sector has maximum number of houses?
Exercise | Q 95. (f) | Page 203
Look at the map given below
Houses
In which sector is Fire Station located?
Exercise | Q 95. (g) | Page 203
Look at the map given below
Houses
In the map, how many sectors have been shown?
Exercise | Q 96 | Page 204
A photographer uses a computer program to enlarge a photograph. What is the scale according to which the width has enlarged?
Exercise | Q 97 | Page 204
The side of a square board is 50 cm. A student has to draw its image in her notebook. If the drawing of the square board in the notebook has perimeter of 40 cm, then by which scale the figure has been drawn?
Exercise | Q 98 | Page 204
The distance between school and house of a girl is given by 5 cm in a picture, using the scale 1 cm : 5 km. Find the actual distance between the two places?
Exercise | Q 99.(a) | Page 205
Use a ruler to measure the distance in cm between the places joined by dotted lines. If the map has been drawn using the scale 1 cm : 10 km, find the actual distances between school and library
Exercise | Q 99.(b) | Page 205
Use a ruler to measure the distance in cm between the places joined by dotted lines. If the map has been drawn using the scale 1 cm : 10 km, find the actual distances between college and complex
Exercise | Q 99.(c) | Page 205
Use a ruler to measure the distance in cm between the places joined by dotted lines. If the map has been drawn using the scale 1 cm : 10 km, find the actual distances between house and school
Exercise | Q 100 | Page 205
The actual length of a painting was 2 m. What is its length in the photograph if the scale used is 1 mm : 20 cm.
Exercise | Q 101.(a) | Page 205
Find the scale.
Actual size 12 m
Drawing size 3 cm
Exercise | Q 101.(b) | Page 205
Find the scale.
Actual size 45 fee
Drawing size 5 inches
Exercise | Q 102 | Page 205
In a town, an ice cream parlour has displayed an ice cream sculpture of height 360 cm. The parlour claims that these ice creams and the sculpture are in the scale 1:30. What is the height of the ice creams served?
Exercise
## NCERT solutions for Mathematics Exemplar Class 8 chapter 6 - Visualising Solid Shapes
NCERT solutions for Mathematics Exemplar Class 8 chapter 6 (Visualising Solid Shapes) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Exemplar Class 8 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Mathematics Exemplar Class 8 chapter 6 Visualising Solid Shapes are Viewing Different Sections of a Solid, Mapping Space Around Us, Faces, Edges and Vertices, Plane Figures and Solid Shapes, Nets for Building 3-d Shapes - Cube, Cuboids, Cylinders, Cones, Pyramid, and Prism, Concept of Polyhedron, Concept of Prism, Concept of Pyramid, Euler's Formula.
Using NCERT Class 8 solutions Visualising Solid Shapes exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 8 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 6 Visualising Solid Shapes Class 8 extra questions for Mathematics Exemplar Class 8 and can use Shaalaa.com to keep it handy for your exam preparation | 6,329 | 22,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-14 | latest | en | 0.849174 |
https://www.physics.uoguelph.ca/problem-6-11-dynamics-friction | 1,713,112,706,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00695.warc.gz | 873,401,510 | 19,062 | # Problem 6-11 Dynamics with friction
A child is sliding down a water slide inclined at $25^\circ$ to the horizontal. The coefficient of kinetic friction between the child and the slide is $0.10$. What is the magnitude of the acceleration of the child? (Hint: $F_N$ is not equal to $mg$.)
The FBD for this problem is:
(A)
(B)
(C) | 87 | 334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.900001 |
http://spikesworld.spike-jamie.com/science/liquids/c122-11-water-pressure-height.html | 1,550,526,252,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247488374.18/warc/CC-MAIN-20190218200135-20190218222135-00572.warc.gz | 263,432,694 | 6,053 | Stuck in someone else's frames? break free!
WATER PRESSURE VS HEIGHT OF COLUMN
We've done one experiment on water pressure using a milk carton. This time we're going to use three cartons to make a higher column of water and see if we can get stronger pressure.
The top has been taken off one carton, and the other cartons have had both their tops and their bottoms removed. They are taped together, with the carton with the bottom, on the bottom. If you remember, we made three holes in the cartons we used before. We've done the same thing now, but all the holes are in the bottom carton. The holes are covered with tape, as before. Your teacher will turn a bucket upside down, and the cartons will stand on the bucket, with the side of the cartons that have the holes close to the edge of the bucket. You're going to be able to see the streams really well.
One student can hold a ruler so that one end of it touches the side of the carton next to the top hole. When the tape is pulled away, watch how far out the water goes before it falls to the ground, and notice which inch-mark it reaches before it falls.
We still have a carton left from the other project, and we'll measure the difference, by filling it, removing the tape, and noting the inch-mark the water reaches before falling to the ground. It didn't go as far as the water in the triple- decker did, did it? Let's try it with one bottomless carton taped on top of it, and measure the trajectory like we did with the triple-decker and the single carton. We can see that the water pressure from the double-decker carton is more than from the single carton, and the water pressure from the triple-decker is a lot more than from the double-decker.
Now, we'll do the triple-decker again but this time, we'll take the tape off the middle hole and compare it to the measurement we made when we pulled the tape off the top hole. After replacing the middle tape and filling the carton-stack again, we will pull the tape off the bottom hole. Wowie- Zowie! It really comes out of there, doesn't it?
It appears as though the higher the column of water, the harder the water pressure is. And, that the amount of water on top of the flowing water makes the pressure of the flowing water harder. It is the pressure that makes the water flow out to the side before it falls to the ground.
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https://www.muratasoftware.com/en/products/examples/her009/ | 1,718,864,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00324.warc.gz | 791,585,448 | 11,344 | Coplanar Waveguide with BendExamples | Product | Murata Software Co., Ltd.
# Example9Coplanar Waveguide with Bend
### General
• A coplanar waveguide with bend is analyzed.
• Unless specified in the list below, the default conditions will be applied.
• [How to Reduce the Calculation Time]
### Analysis Space
Item Settings Analysis Space 3D Model unit mm
### Analysis Conditions
Item Settings Solvers Electromagnetic Analysis [Hertz] Analysis Type Harmonic analysis Options Select “Ignore the influence of face/edge electrode thickness” *
* This is the default setting. There are no face electrodes with this model. Therefore it is irrelevant to select it or not.
Mesh tab, Harmonic analysis tab and Open boundary tab are set as follows.
Tab Setting Item Settings Mesh Tab Element type 2nd-order element Multigrid/Adaptive Mesh Method Select “Use the adaptive mesh method”. Frequency-Dependent Meshing Reference frequency: 8×10^9[Hz] Select “The conductor bodies thicker than the skin depth constitute the boundary condition.” As Hot and GROUND are thicker than the skin depth, the surfaces are applied with the boundary condition having some loss. Harmonic analysis Frequency Minimum: 7.2×10^9[Hz] Maximum: 8.2×10^9[Hz] Sweep Type Select Linear step Division number: 10 Sweep Setting Select Discrete sweep Open boundary Type Absorbing boundary Order of Absorbing Boundary 1st degree
* Fast sweep is avoided for higher accuracy. The number of frequencies is relatively low, so the calculation will not take long.
### Model
Coplanar waveguide is formed on a alumina substrate. Electrodes of waveguide are solid bodies.
The outer boundary condition is “Electric wall”.
### Body Attributes and Materials
Body Number/Type Body Attribute Name Material Name 4/Solid AIR 000_Air(*) 5/Solid SUB 001_Alumina 13/Solid HOT 003_Ag * 24/Solid GROUND 003_Ag * 26/Solid GROUND 003_Ag *
* Available from the Material DB
### Boundary Conditions
Boundary Condition Name/Topology Tab Boundary Condition Type Settings PORT1/Face Electric I/O Port Reference Impedance: Select “Specify” and enter 50 ohms. Number of Modes Number of precalculated modes: 5 Number of modes used in the actual analysis: 1 Select modes: none PORT2/Face Electric I/O Port Same as above. Outer Boundary Condition Electric Electric wall
| 531 | 2,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.740614 |
https://www.coursehero.com/file/23262326/LHosptitals-Rule/ | 1,547,595,786,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583656530.8/warc/CC-MAIN-20190115225438-20190116011438-00589.warc.gz | 765,076,604 | 197,317 | MAM
L’Hosptital’s Rule
# L’Hosptital’s Rule - LHospitals Rule Look at these...
• Notes
• 15
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L’Hospital’s Rule
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Look at these examples After subbing in we get either , And neither of these makes any sense These are special cases We use a new rule L’Hospital’s Rule
L’Hospital’s Rule Only works when and and count too because
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L’Hospital’s Rule If or then We differentiate the function in the numerator and the function in the denominator separately!! WE DON’T USE THE QUOTIENT RULE!
e.g.1 compute the limit Solution This gives , so L’Hospital’s Rule applies 2 lim 1 x x x 2 1 lim lim 0 2 1 1 x x x x x
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Examples 1. Which gives so apply the rule Remember that this is NOT true if we had 2. which is of the form 3.
4. compute the limit This is a indeterminate form, so L’Hôpital’s Rule applies 5 4 5 3 2 1 3 5 3 lim 4 2 5 1 x x x x x x x 0 0 5 4 4 3 5 3 2 4 2 1 1 3 5 3 5 12 5 1 lim lim 8 4 2 5 1 20 6 10 x x x x x x x x x x x x x
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6.
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• Spring '14
• Limits, lim, Limit of a function, Indeterminate form, 10 0 L
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Jill Tulane University ‘16, Course Hero Intern | 666 | 2,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.865335 |
https://forum.poshenloh.com/category/242/c2 | 1,601,256,409,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401583556.73/warc/CC-MAIN-20200928010415-20200928040415-00206.warc.gz | 385,642,027 | 9,247 | Thank you for your question. For the purpose of this question, it's not super important that you know why we use the word "harmonic." What's important is that you know what the harmonic mean is. A mean, normally, is when we add up all the numbers, and divide them by how many numbers there are. For example, if we have the numbers 1,2, and 4, we add them up, and divide them by 3: $$\frac{1 + 2 + 4}{3} = \frac{7}{3}$$ When we talk about the harmonic mean we do something a little different. First, we take one over all the numbers (we "flip" them upside down). Then, we take that mean: $$\frac{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}}{3}$$ Finally, we flip the whole thing again. A good way to think about this is that because we flipped them to start with, we have to flip the answer back, after we take the mean. This give us: $$\frac{1}{\frac{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}}{3}} = \frac{3}{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}}$$ This may look a little bit weird, because it is strange compared to the mean you may be used to. But as Po describes, this is actually really helpful for many problems in math, including this one! About your question on why this works, try picking two letters instead of numbers, like "a" and "b," as speeds up and down the hill. Then try the problem again. You'll see you get the same formula! In the problem, we did this with a specific example (4 mph and 12 mph), but it didn't have to be those speeds, it could have been anything! I hope this helps! You're asking great questions. Feel free to reach out if you ever get stuck, and keep doing a great job learning! Happy Learning! The Daily Challenge Team | 458 | 1,654 | {"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} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.944194 |
https://brilliant.org/problems/whats-the-chapter/ | 1,490,365,022,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188132.48/warc/CC-MAIN-20170322212948-00077-ip-10-233-31-227.ec2.internal.warc.gz | 787,934,973 | 12,701 | # What's the chapter?
Algebra Level pending
If $$a_1, a_2,a_3,\ldots$$ follows a harmonic progression, and $$\displaystyle f(k) = \sum_{r=1}^n (a_r - a_k)$$, then $$\dfrac {a_1} {f(1)},\dfrac {a_2} {f(2)},\dfrac {a_3} {f(3)},\ldots, \dfrac {a_n} {f(n)}$$ follows a/an $$\text{__________}$$.
× | 130 | 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} | 2.703125 | 3 | CC-MAIN-2017-13 | longest | en | 0.55004 |
http://nedrilad.com/Tutorial/topic-22/Game-Physics-Engine-Development-175.html | 1,506,074,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688932.49/warc/CC-MAIN-20170922093346-20170922113346-00470.warc.gz | 258,118,541 | 4,959 | Game Development Reference
In-Depth Information
two-dimensional shape, you can balance it on your finger by placing your finger at
the center of mass.
If you think of an object as being made up of millions of tiny particles (atoms,
for example), you can think of the center of mass as being the average position of all
these little particles, where each particle contributes to the average depending on its
mass. In fact this is how we can calculate the center of mass. We split the object into
tiny particles and take the average position of all of them:
m
n
1
p cofm =
m i p i
where p cofm is the position of the center of mass, m is the total mass of the object, m i
is the mass, and p i is the position of particle i .
The center of mass of a sphere of uniform density will be located at the center
point of the sphere. Similarly with a cuboid, the center of mass will be at its geometric
center. The center of mass isn't always contained within the object. A donut has its
center of mass in the hole, for example. Appendix A gives a breakdown of a range of
different geometries and where their center of mass is located.
The center of mass is important because if we watch the center of mass of a rigid
body, it will always behave like a particle . In other words, we can use exactly the same
formulae we have used so far in this topic to perform the force calculations and update
the position and velocity for the center of mass. By selecting the center of mass as our
origin position we can completely separate the calculations for the linear motion of
the object (which is the same as for particles) and its angular motion (for which we'll
need extra mathematics).
Any physical behavior of the object can be decomposed into the linear motion of
the center of mass and the angular motion around the same point. This is a profound
and crucial result, but one that takes some time to prove; if you want the background,
any good undergraduate textbook on mechanics will give details.
If we choose any other point as the origin, we can no longer separate the two
kinds of motion, so we'd need to take into account how the object was rotating in
order to work out where the origin is. Obviously this would make all our calculations
considerably more complicated.
Some authors and instructors work through code either way (although typically
only for a few results; when the mathematics gets really hard, they give up). Personally
I think it is a very bad idea to even consider having your origin anywhere else but at
the center of mass. I'll assume this will always be the case for the rest of the topic; if | 572 | 2,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-39 | latest | en | 0.938751 |
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_12&oldid=89946 | 1,639,062,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00621.warc.gz | 170,347,554 | 12,568 | # 2013 AIME I Problems/Problem 12
## Problem 12
Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$.
## Solution
First, find that $\angle R = 45^\circ$. Draw $ABCDEF$. Now draw $\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$. The height of $ABCDEF$ is $\sqrt{3}$, so the length of base $QR$ is $2+\sqrt{3}$. Let the equation of $\overline{RP}$ be $y = x$. Then, the equation of $\overline{PQ}$ is $y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3$. Solving the two equations gives $y = x = \frac{\sqrt{3} + 3}{2}$. The area of $\bigtriangleup PQR$ is $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$. $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$
## Cartesian Variation Solution
Also use a coordinate system, but upon drawing the diagram call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$. After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the ordinate of $R$. Draw a perpendicular of $F$, call it $H$, and note that $QP = 1 + \sqrt{3}$ after using the trig functions for $75$ degrees.
Now, get the lines for $QR$ and $RP$: $y=\sqrt{3}x$ and $y=-(2+\sqrt{3})x + (5+\sqrt{3})$, whereupon we get the ordinate of $R$ to be $\frac{3+2\sqrt{3}}{2}$, and the area is $\frac{5\sqrt{3} + 9}{4}$, so our answer is $\boxed{021}$. | 687 | 1,961 | {"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": 56, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.722564 |
http://forums.devshed.com/javascript-development/35042-random-javascript-last-post.html | 1,409,162,798,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829661.96/warc/CC-MAIN-20140820021349-00147-ip-10-180-136-8.ec2.internal.warc.gz | 76,644,047 | 11,277 | 1. random number in javascript
I know how to generate a random number, also how to round it, I can multiply it too, but how on earth do I generate a round number between 100 and 999? I can get a random number from 0 to 999, but need to lose everything less than 3 digits???
2. Something like:
<html>
<script type="text/javascript">
function getRandom() {
thisNumber = Math.floor(Math.random() * 1000)
if (thisNumber < 100) {
getRandom()
} else {
return thisNumber
}
}
function init() {
}
</script>
</body>
</html>
would work. In this case, what should happen is that a random number is generated between 0 and 999. The number is tested, if it's less than 100 the function is called again until it is above 100. If it is above 100 then it is returned to the calling command.
It has to be done as a recursive function.
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Somewhat more efficient would be
Math.floor((Math.random() * 900) + 100);
Generates a random number between 100 and 999 directly, avoiding issues with successive calls to Math.random() generating consecutive numbers less than 100. Not really an issue on today's fast computers, but good coding practice is always worth it.
4. that is a seriously elegant solution!
Wish I'd thought of that, so simple and effective
Thanks so much! | 348 | 1,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-35 | longest | en | 0.868647 |
http://adessowiki.fee.unicamp.br/adesso/wiki/iamxt/mxt_def2/view/ | 1,519,550,208,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816351.97/warc/CC-MAIN-20180225090753-20180225110753-00595.warc.gz | 9,046,750 | 6,356 | # Component Tree
MainPage
The component tree was proposed as a structure for image representation that represents all connected components resulting of all possible thresholds of the image, and that provides an attribute signature as means of discriminating features in the image. It is efficient to implement connected anti-extensive filters, and by duality extensive filters.
# Component Tree Representation
The component tree, , is a rooted tree in which each node stores a set of attributes. It is completely defined by its set of nodes :
and parents :
where is the set minus operand, and is the parent of node . Our convention is that node is the root.
Suppose that the grey-levels of an image are bounded between and . The upper threshold sets resulting of every upper threshold for varying between and can be computed, and we know that threshold sets have the following property:
This hierarchy property allows a tree representation. For each connected component resulting of all threshold sets, there is a component tree node with attributes and . and are the basic attributes stored in the component tree nodes necessary to recover the image from the tree. Other attributes, such as height, area and volume can be computed and stored in the nodes during the tree construction or they can be computed just when required.
Node is a parent of node if contains , and . Note that the component tree may represent a pixel in more than one node, therefore it has some redundancy. The leaves of the component tree correspond to regional maxima in the image and the root represents the whole domain of the image.
The restitution of the image corresponding to a given component tree, is given by the following equation
The construction of the component tree corresponding to the 1D image is illustrated in Figure 1. This illustration is meant just to explain the component tree structure and construction, it does not correspond to the most efficient way to build it, for that, there are algorithms that run in quasi-linear time.
Figure 1. component tree construction of the 1D image (a). Bottom-up construction (b)-(g). Resulting component tree (g)
# Component Tree Attribute Signature
The notion of attribute signature in the component tree is very important, it conveys information concerning the variation in shape or size of a connected component. The attribute signature uses the linking information between connected components at sequential grey-levels in the image to help the decision making process. The attribute signature of any two nodes and connected by a path in the component tree is simply the sequence of node attributes in this path.
For instance, the attribute signature may be used to choose a threshold value to segment an object in an image. The Mona Lisa painting is depicted in Figure 2(a). The red dot in the image corresponds to a regional maxima, i.e. a component tree leaf. The area signature, and its gradient starting at this leaf and ending at the root of the tree are shown in Figure 2(b) and Figure 2(c), respectively. The area signature presents some discontinuities, i.e. spikes in the gradient curve. The highest spike happens in the transition between to . At , the area of the connected component is of , in the following grey-level, , the area drops to . The reconstruction of the connected components at levels and are depicted in Figure 2(d) and 2(e), respectively. It is clear that the connected component at level merges Mona Lisa's face with a considerable part of the background, while the connected component at level provides a better segmentation of Mona Lisa's face.
Figure 2. Mona Lisa painting, the red dot corresponds to a regional maxima (a). Area signature (b), and its gradient (c). Reconstruction of the connected component at level (d) and (e). | 751 | 3,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-09 | longest | en | 0.948636 |
https://philoid.com/question/35406-find-the-area-of-the-region-enclosed-by-the-parabola-x-2-y-the-line-y-x-2-and-the-x-axis | 1,726,659,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00776.warc.gz | 431,832,735 | 8,169 | ##### Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and the x-axis.
We can see from the figure that the area of the region bounded by the parabola x2 = y, the line y = x + 2 and the x-axis is shown by shaded region that is Area OACO.
The points of intersection of both the curves are A (-1, 1) and C (2, 4).
Thus,
Area of OACO
10 | 112 | 368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.935651 |
https://www.jiskha.com/questions/1184630/ok-ummmm-i-thought-i-was-good-till-now-ok-for-cup-c-mass-of-cup-water-stirrer-55-15 | 1,591,229,908,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00546.warc.gz | 747,320,967 | 6,681 | # Chemistry Dr. Bob222
Ok, UMMMM I thought I was good.. Till now..
Ok, For Cup C
Mass of cup,water,stirrer: 55.15
Mass of sodium bicarbonate: 2.02g
mass of citric acid: 0.77g
total mass: 57.94g
mass of cup,solution, stirrer after reaction: 57.64
difference (CO2): 0.30g
Part B: Limiting Reactants
Plastic Cup C
7. Determine which reactant is the limiting reactant in the plastic cup C. Describe your reasoning.
You know I know how to calculate the TY and AY and %Y now, but I'm confused on the finding limiting reactant? umm Let me see if I can figure it out and you let me know if I'm on the right track...
2.02g NaHCO3 mols = 2.02/84 = 0.0240 mols
the ratio is 3/1 for NaHCO3toH3C6H5O7 so do I do 0.0240/3 = citric acid? 0.008 moles citric acid? so since citric acid is smaller but it requires 3 of sodium bicarbonate, would that be the limiting reactant? because it takes more of them? am I figuring this out right?
8. Calculate the theoretical yield of carbon dioxide in the plastic cup C.
2.02/84=0.0240 *44 = 1.056 TY??
9. Calculate the percentage yield in the plastic cup C.
AY = 0.30/1.056 TY = 0.284*100 = 28.4% ?? is this right?
1. 👍 0
2. 👎 0
3. 👁 180
1. You know I know how to calculate the TY and AY and %Y now, but I'm confused on the finding limiting reactant? umm Let me see if I can figure it out and you let me know if I'm on the right track...
2.02g NaHCO3 mols = 2.02/84 = 0.0240 mols
the ratio is 3/1 for NaHCO3toH3C6H5O7 so do I do 0.0240/3 = citric acid? 0.008 moles citric acid? so since citric acid is smaller but it requires 3 of sodium bicarbonate, would that be the limiting reactant? because it takes more of them? am I figuring this out right?
I can't tell which you're calling the LR. Citric acid is the LR. Based on you 2.02 g it requires 0.008 mols citric acid and you have only 0.004 so you don't have enough citric acid and that makes it the LR.
8. Calculate the theoretical yield of carbon dioxide in the plastic cup C.
2.02/84=0.0240 *44 = 1.056 TY??
If citric acid is the LR then you should use it to determine the theoretical yield. This calls into question some of the earlier answers since I didn't check them for the LR bit. But it stands to reason that if 1 g NaHCO3 is equivalent to 0.76 g citric acid then if you start with 2.02 g NaHCO3 it will take more than 0.76 g citric acid and there isn't enough there to do anything except with 1g NaHCO3.
9. Calculate the percentage yield in the plastic cup C.
AY = 0.30/1.056 TY = 0.284*100 = 28.4% ?? is this right?
TY needs to be recalculated.
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Post subject: PercentilePosted: Sun Oct 27, 2013 8:48 am
Joined: Sun Oct 27, 2013 8:19 am
Posts: 51
Can anyone tell me what percentile a maths score of 111 would be in Ashford, kent? I have found plenty information on percentiles and age standardisation but having difficulty working out the answer.
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Post subject: Re: PercentilePosted: Sun Oct 27, 2013 10:30 am
Joined: Wed Jan 18, 2012 11:41 am
Posts: 5643
Location: Essex
I may be the only person not really understanding your question, but possibly not as you have had no other replies...
111 in what maths test?
_________________
Outside of a dog, a book is a man's best friend. Inside of a dog it's too dark to read.Groucho Marx
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Post subject: Re: PercentilePosted: Sun Oct 27, 2013 11:15 am
Joined: Tue Oct 19, 2010 11:30 am
Posts: 110
If that's a standardised 11+ score then 111 is around the 63rd percentile.
If it's a standardised CATs score then it's around the 75th percentile (and equates to about 120 on the 11+).
The standardisations are done nationally so it's impossible to say how Ashford may differ from the national distribution although it is well known we have fat tails in Kent but I doubt it would change things by more than a percentile or two at that point of the distribution.
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Post subject: Re: PercentilePosted: Sun Oct 27, 2013 11:53 am
Joined: Sun Oct 27, 2013 8:19 am
Posts: 51
Sorry. I should have said its an 11+ standardised maths score. My only reason for mentioning Ashford was to avoid information relating to west kent as I believe its different there. Thank you for your response.
KCC have sent me the raw score and I have calculated the percentage from that. I have asked them what raw score my daughter would have needed to pass the test at her age ( I'm assuming she will be classed as age 11 as she was only 2 weeks away from her 11th birthday. Can anyone think of anything else I could ask KCC which might be useful to know in order to get a more accurate result from them?
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Post subject: Re: PercentilePosted: Sun Oct 27, 2013 12:23 pm
Joined: Mon Jun 18, 2007 2:32 pm
Posts: 7063
Location: East Kent
for standardised scores you usually take age as years and number of complete months, so she will be 10 years 11 months
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Post subject: Re: PercentilePosted: Sun Oct 27, 2013 12:53 pm
Joined: Sun Oct 27, 2013 8:19 am
Posts: 51
Thank you.
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Post subject: Re: PercentilePosted: Mon Oct 28, 2013 4:28 pm
Joined: Tue Oct 19, 2010 11:30 am
Posts: 110
Shiv wrote:
...
KCC have sent me the raw score and I have calculated the percentage from that.
...
F.Y.I.
That's a percentage which is not the same as a percentile. The two are not related in any useful way. You need the standardised score and the moments of the distribution to calculate the percentile.
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Post subject: Re: PercentilePosted: Mon Oct 28, 2013 5:47 pm
Joined: Sun Oct 27, 2013 8:19 am
Posts: 51
Thank you. I will see if I can find out that information.
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http://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.5744.1/holomorphic/4.4.5744.1-37.1-c | 1,606,904,714,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00645.warc.gz | 132,625,187 | 5,929 | # Properties
Label 4.4.5744.1-37.1-c Base field 4.4.5744.1 Weight $[2, 2, 2, 2]$ Level norm $37$ Level $[37, 37, -2w^{3} + w^{2} + 8w - 1]$ Dimension $7$ CM no Base change no
# Related objects
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## Base field 4.4.5744.1
Generator $$w$$, with minimal polynomial $$x^{4} - 5x^{2} - 2x + 1$$; narrow class number $$2$$ and class number $$1$$.
## Form
Weight: $[2, 2, 2, 2]$ Level: $[37, 37, -2w^{3} + w^{2} + 8w - 1]$ Dimension: $7$ CM: no Base change: no Newspace dimension: $16$
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{7} - 8x^{6} + 9x^{5} + 62x^{4} - 136x^{3} - 62x^{2} + 285x - 139$$
Norm Prime Eigenvalue
4 $[4, 2, -w^{3} + 4w + 1]$ $\phantom{-}e$
5 $[5, 5, w - 1]$ $-\frac{21}{103}e^{6} + e^{5} + \frac{120}{103}e^{4} - \frac{960}{103}e^{3} + \frac{223}{103}e^{2} + \frac{1953}{103}e - \frac{1073}{103}$
7 $[7, 7, -w^{2} + w + 2]$ $\phantom{-}\frac{11}{309}e^{6} - \frac{70}{103}e^{4} - \frac{71}{309}e^{3} + \frac{342}{103}e^{2} + \frac{110}{309}e - \frac{929}{309}$
13 $[13, 13, -w^{3} + w^{2} + 4w]$ $-\frac{17}{103}e^{6} + e^{5} + \frac{53}{103}e^{4} - \frac{1042}{103}e^{3} + \frac{568}{103}e^{2} + \frac{2405}{103}e - \frac{1673}{103}$
13 $[13, 13, -w^{2} + 3]$ $\phantom{-}\frac{98}{309}e^{6} - 2e^{5} - \frac{15}{103}e^{4} + \frac{5407}{309}e^{3} - \frac{1457}{103}e^{2} - \frac{10144}{309}e + \frac{9196}{309}$
17 $[17, 17, -w^{2} + 2]$ $\phantom{-}\frac{65}{309}e^{6} - e^{5} - \frac{114}{103}e^{4} + \frac{2839}{309}e^{3} - \frac{423}{103}e^{2} - \frac{6148}{309}e + \frac{5494}{309}$
19 $[19, 19, -w^{3} + 5w]$ $-\frac{43}{309}e^{6} + e^{5} - \frac{26}{103}e^{4} - \frac{2981}{309}e^{3} + \frac{1004}{103}e^{2} + \frac{6677}{309}e - \frac{6116}{309}$
31 $[31, 31, -w^{2} + 2w + 3]$ $\phantom{-}\frac{43}{309}e^{6} - e^{5} + \frac{26}{103}e^{4} + \frac{2672}{309}e^{3} - \frac{798}{103}e^{2} - \frac{4205}{309}e + \frac{3644}{309}$
37 $[37, 37, -2w^{3} + w^{2} + 8w - 1]$ $-1$
43 $[43, 43, -w - 3]$ $-\frac{29}{309}e^{6} + \frac{222}{103}e^{4} + \frac{131}{309}e^{3} - \frac{1426}{103}e^{2} - \frac{599}{309}e + \frac{6101}{309}$
53 $[53, 53, -w^{3} + 2w^{2} + 3w - 2]$ $-\frac{52}{309}e^{6} + e^{5} - \frac{53}{103}e^{4} - \frac{2024}{309}e^{3} + \frac{1183}{103}e^{2} + \frac{2879}{309}e - \frac{4148}{309}$
53 $[53, 53, w^{3} - 6w - 2]$ $\phantom{-}\frac{217}{309}e^{6} - 4e^{5} - \frac{173}{103}e^{4} + \frac{10847}{309}e^{3} - \frac{2439}{103}e^{2} - \frac{20078}{309}e + \frac{18023}{309}$
59 $[59, 59, 2w^{3} - w^{2} - 10w - 2]$ $-\frac{8}{103}e^{6} + \frac{134}{103}e^{4} + \frac{61}{103}e^{3} - \frac{381}{103}e^{2} - \frac{389}{103}e - \frac{139}{103}$
61 $[61, 61, 2w^{3} - w^{2} - 10w]$ $-\frac{16}{309}e^{6} + e^{5} - \frac{151}{103}e^{4} - \frac{3071}{309}e^{3} + \frac{982}{103}e^{2} + \frac{6329}{309}e - \frac{2750}{309}$
61 $[61, 61, 2w^{3} - w^{2} - 8w]$ $-\frac{169}{309}e^{6} + 3e^{5} + \frac{214}{103}e^{4} - \frac{8123}{309}e^{3} + \frac{935}{103}e^{2} + \frac{14996}{309}e - \frac{7301}{309}$
71 $[71, 71, 2w^{3} - 9w - 2]$ $\phantom{-}\frac{37}{309}e^{6} - e^{5} + \frac{8}{103}e^{4} + \frac{3104}{309}e^{3} - \frac{404}{103}e^{2} - \frac{6737}{309}e + \frac{1763}{309}$
73 $[73, 73, -w^{3} - w^{2} + 6w + 3]$ $\phantom{-}\frac{29}{103}e^{6} - e^{5} - \frac{254}{103}e^{4} + \frac{796}{103}e^{3} + \frac{364}{103}e^{2} - \frac{1049}{103}e + \frac{1212}{103}$
81 $[81, 3, -3]$ $-\frac{5}{309}e^{6} - \frac{15}{103}e^{4} + \frac{566}{309}e^{3} - \frac{15}{103}e^{2} - \frac{2522}{309}e + \frac{2810}{309}$
83 $[83, 83, -w^{3} + 2w^{2} + 3w - 3]$ $\phantom{-}\frac{64}{309}e^{6} - 2e^{5} + \frac{192}{103}e^{4} + \frac{5795}{309}e^{3} - \frac{2486}{103}e^{2} - \frac{12029}{309}e + \frac{14090}{309}$
101 $[101, 101, 2w^{3} - 8w - 3]$ $-\frac{59}{103}e^{6} + 3e^{5} + \frac{293}{103}e^{4} - \frac{2859}{103}e^{3} + \frac{705}{103}e^{2} + \frac{5796}{103}e - \frac{2480}{103}$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$37$ $[37, 37, -2w^{3} + w^{2} + 8w - 1]$ $1$ | 2,131 | 4,066 | {"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} | 2.828125 | 3 | CC-MAIN-2020-50 | latest | en | 0.261988 |
https://jacobbradjohnson.com/help-differential-991 | 1,674,805,022,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00820.warc.gz | 350,492,154 | 3,987 | # Solving for x examples
When Solving for x examples, there are often multiple ways to approach it. Math can be a challenging subject for many students.
## Solve for x examples
There are a lot of great apps out there to help students with their school work for Solving for x examples. The internet is a great place to start, as there are many websites that offer step-by-step solutions to common problems. In addition, most major textbook publishers offer online homework help services. These services typically provide access to a database of answers, as well as a variety of tools and resources that can help with the solution process. With a little bit of effort, it is usually possible to find the answer to any homework problem.
Once the critical points have been identified, it is possible to graph the equation and find the solutions. Additionally, there are online solvers that can be used to find the solutions to an absolute value equation. These solvers will typically ask for information such as the equation's coefficients and constants. By inputting this information, the solver will be able to generate a graph of the equation and identify its solutions.
Solving expressions is a fundamental skill in mathematics. An expression is a mathematical phrase that can contain numbers, variables, and operators. Solving an expression means to find the value of the expression when the variables are given specific values. There are a few different steps that can be followed to solve an expression. First, simplify the expression by combining like terms and using the order of operations. Next, substitute the values for the variables into the expression. Finally, use algebraic methods to solve for the unknown variable. With practice, solving expressions will become second nature.
Math is a difficult subject for many people. It can be frustrating to get stuck on a problem and not know how to proceed. Luckily, there are a number of online Math solver websites that can help. These sites allow users to input a Math problem and receive step-by-step instructions on how to solve it. In addition, many of these sites also provide video tutorials and other resources that can help users understand the underlying concepts. As a result, Math solver websites can be a valuable resource for students who are struggling with Math.
## Solve your math tasks with our math solver
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Answers to all ixl problems Exponent solver Geometry proofs solver Complex fractions solver How to solve improper fractions | 611 | 3,164 | {"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} | 3.703125 | 4 | CC-MAIN-2023-06 | latest | en | 0.947435 |
https://answers.yahoo.com/question/index?qid=20121019224953AArHWZh | 1,610,955,456,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00388.warc.gz | 228,646,274 | 24,769 | # Aluminum and Bromine reaction?
When 2.70 g of Al reacts with excess Br2, the maximum mass of Al2Br6 that can be produced is?
2Al + 3Br ---> Al2Br6
When you calculate the number of moles of Al taking place in the reaction... is it
2.70/27 = 0.1 moles?
or
2.70/54 = 0.05 moles?
I am going with the first 1...
0.1 moles of Al produces 0.05 moles of Al2Br6
so 0.05 x ((2 x 27)+(6 x 79.9)) = 26.67 g of Al2Br6
Correct?
Relevance
• 8 years ago | 165 | 450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-04 | latest | en | 0.782021 |
https://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-7-measurement-7-1-problem-solving-with-u-s-measurement-units-7-1-exercises-page-485/35 | 1,529,543,202,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00563.warc.gz | 817,047,008 | 12,584 | ## Basic College Mathematics (9th Edition)
4$\frac{1}{4}$ gal = qt Use a unit fraction with quarts (the unit for your answer) in the numerator, and gallon (the unit being changed) in the denominator. Because 4 quarts (qt) = 1 gallon (gal), the necessary unit fraction is $\frac{4 qt}{1 gal}$ Multiply4$\frac{1}{4}$ gal times the unit fraction. 4$\frac{1}{4}$ gal = 4$\frac{17}{4}$ gal = 4$\frac{17}{4}$* $\frac{4 qt}{1 gal}$ = 17 quarts | 141 | 437 | {"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} | 4.1875 | 4 | CC-MAIN-2018-26 | latest | en | 0.694507 |
https://www.bankersadda.com/quantitative-aptitude-quiz-for-ibps-so-prelims-2022-12th-november/ | 1,670,056,713,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00731.warc.gz | 690,572,335 | 116,872 | Latest Banking jobs »
# Quantitative Aptitude Quiz For IBPS SO Prelims 2022- 12th November
Q1. A person purchases four bicycles at a discount of 20% on marked price on each bicycle. If sum of discount offered and profit obtained on all 4 bicycles is 2560 then find the difference in marked price and cost price on one bicycle.
(a) 520
(b) 640
(c) 1220
(d) 880
(e) 690
Q2. Veer buy an article for Rs. 480. He sold it at 12% loss and get some money and from that money he again buys an article and this he sold at 25% profit. What was profit percentage he got from this transaction?
(a) 16%
(b) 12%
(c) 10%
(d) 14%
(e) 20%
Q3. A vegetable seller buy potato and tomato in Rs. 7.5 per kg. He sells potato at a profit of 22% and tomato at a loss of 8%, What is the S.P. of tomato, if in whole transaction there is no profit no loss ?
(a) 5.08 Rs.
(b) 4.97Rs.
(c) 5.07Rs.
(d) 5.06Rs.
(e) 5.66Rs.
Q4.A person sells a table at a profit of 10%. If he had bought the table at 5% less cost and sold for Rs. 80 more, he would have gained 20%. The cost price of the table is
(a) Rs. 3,200
(b) Rs. 2,500
(c) Rs. 2,000
(d) Rs. 3,000
(e) Rs. 1,600
Q5. Roni purchased a cycle for Rs. 12000 and sold it at a loss of 20% with that amount he purchased another cycle and sold it at 30% profit. What was his overall gain/loss ?
(a) 720 loss
(b) 480 loss
(c) 480 profit
(d) 720 profit
(e) No profit no loss
Q7. A train of length 180 meter cross a platform in 15 seconds with a speed of 60 km/h. A man cross the same platform in 4 minute find the speed of man?
(a) 1.05 km/h
(b) 3 km/h
(c) 2.05 km/h
(d) 2.1 km/h
(e) 2 km/h
Q8. Two trains A and B of length 400 m and (400 + x) m respectively are moving with same speed. If train A and B crosses a pole in 16 sec and 24 secs respectively then in what time train ‘B’ will cross 400 m long platform.
(a) 32 sec
(b) 40 sec
(c) 45 sec
(d) 54 sec
(e) 24 sec
Q9. The ratio of speed of train P and train Q is 4 : 5. Train P crosses a pole in 6 sec while train Q crosses the same pole in 4 sec. If train P crosses a platform of length 480 m in 18 sec then in how much time train Q will cross the same platform?
(a) 16.3 sec
(b) 14.2 sec
(c) 13.6 sec
(d) 18 sec
(e) 16 sec
Q10. A man sees a train passing over a bridge of length 1 km. The length of the train is half of the length of bridge. If the train passes the bridge in 2 minutes then find the speed of the train?
(a) 30 kmph
(b) 45 kmph
(c) 50 kmph
(d) 60 kmph
(e) 54 kmph
Directions (11-15): Read the given information carefully and answer the following questions.
Bar graph shows the percentage distribution of total employees working in different industries in a particular city and line graph shows the percentage of female employees in each of these industries.
Q11.Find the ratio of female employees in Call center industries to the male employees in Manufacturing industries?
(a) 3:2
(b) 6:5
(c) 4:5
(d) 9:8
(e) None of these
Q12. Find the total number of female employees in Banking and Sales industries together in the city if the total number of males working in Gaming and IT industries together in the city is 240?
(a) 102
(b) 100
(c) 106
(d) 98
(e) 104
Q13. Total number of male employees in Manufacturing and Gaming industries together in the city is what percent of total number of female employees in Banking and Call center industries together in the city?
(a) 132%
(b) 128%
(c) 136%
(d) 135%
(e) 130%
Q14. Find the difference between the male employees working in manufacturing industry in the city and the female employees working in IT and Gaming industries together in the city, if the difference of male and female employees working in sales industry in the city is 256.
(a) 720
(b) 1040
(c) 1280
(d) 960
(e) 800
Q15. Find the ratio of male employees working in Call center and the female employees working in IT industries together in the city to the total number of female employees working in sales and banking industry together in the city?
(a) 25:13
(b) 24:11
(c) 25:12
(d) 15:8
(e) 20:11
Solutions | 1,209 | 4,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.896075 |
https://de.mathworks.com/matlabcentral/profile/authors/1754528 | 1,642,555,717,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301217.83/warc/CC-MAIN-20220119003144-20220119033144-00003.warc.gz | 272,070,115 | 21,683 | Community Profile
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Computation of the standard error of a FGLS coefficient estimate
Hi, I use the MATLAB code below to calculate the standard errors of the coeffficients estimated by Weighted Least Squares (WL...
etwa 6 Jahre ago | 0 answers | 0 | 1,476 | 5,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.847062 |
https://www.gamedev.net/forums/topic/637715-terrain-generation/ | 1,544,672,648,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824448.53/warc/CC-MAIN-20181213032335-20181213053835-00428.warc.gz | 915,428,781 | 39,923 | # Terrain generation
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First, I'm new to this forum, but not new to C++ or DirectX9. I've tried searching the topic of how to generate terrain, but didn't find anything I thought was useful for these specific parameters:
-fast generation
-contenuous (moves with the player so there is no spacial limit)
-detailed up close, low-poly at distance, and smooth transition between.
I've come up with some ideas, but I would like to know if there is anyone here that knows how the experts do it.....
I think my ideas would be slower than methods used by "the industry".
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perlin Noise (and associated fractals) is the canonical way of doing this. Basially, it generates a continous (i.e. smooth) surface at whatever coordinate you give the algorithm. Zooming in and out is just a matter of giving it more accurate coordinates, say... 1.25 between 1.0 and 1.5. The point will be exactly between the neighbors on the contour of the surface.
Perlin noise itself isn't very interesting.
As you can see, it kinda looks like the noise you get off a TV, but blurry. More interesting is when you use the output of Perlin Noise as a basis function for various fractals, e.g.:
FBM: https://sites.google.com/site/jicenospam/hill_fbm.png (this is what libnoise samples usually show, note that it is NOT perlin noise, but a derivative of such)
Once you have these constituent fractals, you can mix and match them with basic math functions as you please. There's a wealth of other things you can do too, but this is just a topical overview.
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Thank you both for replying. The method for generating the "heights" for the mesh/meshes is one issue. I think I'll use perlin noise, but I have tried to use it without success. I have a method for generating the height values already, but it's slow. The quad-tree method is what is more pressing at the moment. I don't even know how it's done.... I understand the concept, but how to impliment it eludes me. Is it a single mesh or is it multiple meshes that are placed together? I'll look at the libnoise article, but I need to know how the quad-tree is done.
Thank you.
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You will probably want to use multiple meshes placed adjacent to one another. For lower LOD chunks, you can use a lower res block of samples plus a skirt to hide the cracks between seams.
Perlin noise fractals are ideally suited for this kind of on-the-fly generation, because in order to get a lower detail terrain (and incidentally one that will generate more quickly) you simply reduce the octave count of the fractal. So for example, if the largest block size is MxM, and the next LOD level down is a block M/2xMx2, then you can generate the first one using, say, 8 octaves and the second LOD level using 4 octaves.
If you are going with a large or practically infinite terrain, then I would recommend against using a quadtree. The basic quadtree takes a finite, bounded volume and splits it recursively into 2x2 sections, which doesn't extend well to an infinite sized volume. Instead, I would use something like a spatial hash where blocks are indexed using a BlockX, BlockY pair that is hashed to find the "bucket" in which that terrain block lives. Block coordinates can be negative and practically infinite (if a large enough precision type is used for the coordinate pair) and you can easily draw a "chunk" of blocks centered at some given coordinate to implement your moving window into the world. You can also easily implement streaming and discarding of blocks as they come into or go out of a larger "active" window on the world.
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Thank you FLeBlanc. I've used that method before, but I thought a better method was out there. Are you sure the quad-tree method has those limitations you described?
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Thank you FLeBlanc. I've used that method before, but I thought a better method was out there. Are you sure the quad-tree method has those limitations you described?
Yes. In order to represent an infinite terrain your quadtree needs to be infinitely sized; that means, the top-level node has to be infinite. Let's say you go with just "really large" For example, say your standard terrain chunk is 16x16 meters, and you want to represent an area 1km x 1km in size. That equates to a tree 7 levels deep; ie, the root node would be a node 1024x1024 meters in size, the next level down would be 4 nodes each sized 512x512 meters, next would be 16 nodes sized 256x256, and so on until the leaf level would comprise 4096 nodes each sized 16x16. That is for a mere square kilometer.
To represent an area the size of, say, Nebraska you would be looking at an area that is 512km on a side, or 524288 meters to a side on the root node, and the tree would be 16 levels deep. That's getting into crazy territory if you are doing recursive culling and such.
Additionally, this assumes that you do bound the player to some finite area, that area being the size of the root node. If you are not bounding the player, then you have no way of calculating the size of the root node, so your only recourse is to hack some sort of collection of quadtrees. You can always grow a tree downward from the root, but growing upward from the root is trickier. And if you do manage to figure out a scheme for growing your quadtree infinitely upward, you then face the problem that any search of the tree (for culling or other purposes) is going to increase in overall complexity the further the player explores. If the player stays close to spawn, searches will be fast since the tree won't have grown very deep. But if he hops on a horse and tears ass across the country to the other side, your tree is going to grow quickly and search complexity will correspondingly grow as well.
On the other hand, if you use a spatial hash then it doesn't matter how large the area is. There is no requirement for setting bounds, and search complexity will not grow with explored area size. Complexity will be exactly the same for a small world as for a huge one (minus some basically trivial bookkeeping depending on the spatial hash implementation.)
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You can use a quadtree by:
-Add a new root node when you need more area (or just make it 32 layers to start with to match integer positions range)
-Never traverse the whole tree (well youll need to do that sometimes but not for every access). You should know what branch(es) you are currently working in.
-You could use a tree with a higher branching rate. lets say 4*4 children per node. That would make it not so deep.
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Here is what I have done before:
-had a HUGE, low-poly, scaled mesh of the entire terrain (the absolute maximum the player could explore).
-had 25x25 "blocks" of 100mX100m each placed over the huge terrain.
-as the player moved from one block to another, the blocks would shift in the opposite direction. The blocks at the far edge opposite the direction of travel would be repositioned to the front and reused as the new terrain blocks.
Of course I will have an absolute max distance of where the player can go, but I don't want it to be practical or desirable for the player to go there. So, I want the area to be huge.
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Of course I will have an absolute max distance of where the player can go, but I don't want it to be practical or desirable for the player to go there. So, I want the area to be huge.
Why do you limit your area like that?
Infinite terrain doesn't have to be displayed all at once - see how the video I posted originally handles it.
You can have a 3x3 grid of big blocks (you can make the grid 5x5 or 7x7 etc). Each time the player changes which one of these 'big blocks' he's in, the blocks are re-organized such that the new block the player is in, is the center one again. I thought this was pretty well illustrated from the video, so I can't really give a better illustration.
Then, for getting more fine detail, you basically subdivide each big block depending on how close the player is to it. Once you subdivide, you no longer draw the low-res terrain, but instead draw the resulting 4 smaller blocks.
After that you run the same sub-division on the resulting smaller blocks, until:
a) either the player is too far away from the block so it doesn't need to be subdivided (this 'too far' should be a function of the block size - i.e. if the block is 32x32m, then 'too far' could be more than 32m away from the block's side, but if the block is 8x8m, then 'too far' is going to be 8m away from the block's side),
b) or your block's resolution is small enough that you don't need to subdivide anymore. By that I mean that if your block size becomes 8x8m and you're doing 8x8 samples, your resolution will be 1m. If that's good enough for you, then stop there. if it isn't, then subdivide the 8x8 block again, so now you have a 4x4m block that has 8x8 sampls, so your resolution will be 0.5m, giving you even more fine terrain.
Of course, your terrain function must be specifying fine enough details for smaller resolutions, because if its not, your 0.5m resolution block will look exactly the same as your 1m resolution block, only use up more triangles. If using perlin noise, you just have to make sure you add noise with small enough frequency to give some definition at 0.5m.
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# Final%20Exam%20S08%20wSolutions - Name CWRU e-mail...
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Name : _________________________ CWRU e-mail:____________ Department of Electrical Engineering and Computer Science ENGR 210. Introduction to Circuits and Instrumentation (4) ENGR 210 - SPRING 2008 FINAL EXAMINATION–5/6/08 AVERAGE 1. 5.8 10 DC Circuit Analysis 2. 7.4 10 Thevenin/Norton equivalent circuits 3. 6.8 10 Basic transients 4. 7.1 10 Basic phasors 5. 6.5 10 Multiple op-amp circuits 6. 6.5 10 Measurement circuits 7. 6.4 10 Capacitors and inductors 8. 6.4 10 Sinusoidal analysis 9. 6.1 10 Sinusoidal frequency response 10. 5.9 10 Sequential switching SCORE 63.2 100 ACADEMIC DISHONESTY All forms of academic dishonesty including cheating, plagiarism, misrepresentation, and obstruction are violations of academic integrity standards. Cheating includes copying from another's work, falsifying problem solutions, or using unauthorized sources, notes or computer programs. Misrepresentation includes forgery of official academic documents, the presentation of altered or falsified documents or testimony to a university office or official, taking an exam for another student, or lying about personal circumstances to postpone tests or assignments. Obstruction occurs when a student engages in unreasonable conduct that interferes with another's ability to conduct scholarly activity.
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Name : _________________________ CWRU e-mail:____________ NOTES:
Name : _________________________ CWRU e-mail:____________ Problem 1 DC Circuit Analysis (10 points) + - R 1 R 3 v S1 i S2 10v x v x + - i 1 i 2 i 3 + - + - v o R 2 Use R 1 =6k Ω R 2 =8k Ω R 3 =4k Ω v s1 =6 V i s2 =12mA (a) [1 point] How many nodes are there in the above circuit? __________ (b) [3 points] Use the mesh current method to solve for the mesh currents i 1 , i 2 and i 3 . Write each mesh- current equation in normal form. _____ i 1 + ______ i 2 + ______ i 3 = ________ _____ i 1 + ______ i 2 + ______ i 3 = ________ _____ i 1 + ______ i 2 + ______ i 3 = ________ SOLUTION: Writing KVL equations gives 6 + 6000 i 1 6000 i 2 + 4000 i 1 4000 i 3 = 0 (1) 10 v x + 8000 i 2 8000 i 3 + 6000 i 2 6000 i 1 = 0 (2) i 3 = .012 (3) Rewriting (1) and substituting (3) 10000 i 1 6000 i 2 = 6 + 4000 .012 ( ) = 6 48 = 42 Doing the same for (2) 40000 i 1 40000 i 3 + 8000 i 2 8000 i 3 + 6000 i 2 6000 i 1 = 0 Combining terms 34000 i 1 + 14000 i 2 = 48000 .012 ( ) = 576 (c) [3 points] What are the values of the mesh currents i 1 = _______ i 2 = _______
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Name : _________________________ CWRU e-mail:____________ i 3 = _______ SOLUTION: From (b) we have the equations 10000 i 1 6000 i 2 = 42 34000 i 1 + 14000 i 2 = 576 These can be solved on a calculator to give the above solutions. (d) [3 points] What is the value of v o ? _____ SOLUTION: v o = 4000 i 1 i 3 ( ) + 8000 i 2 i 3 ( ) v o = 4000 11.76 + 12 ( ) × 10 3 + 8000 12.59 + 12 ( ) × 10 3 V o = + 0.96 4.72 = 3.76
Name : _________________________ CWRU e-mail:____________ Problem 2 Thevenin/Norton equivalent circuit (10 points) + - R 1 R 2 R 3 0.5i 1 v o + - a b v s i 1 Use R 1 =4 Ω R 2 =6 Ω R 3 =12 Ω v s =12V (a) [3 points] Determine the open-circuit voltage at terminals a-b. V ab,oc =____ SOLUTION + 12 v o 4 v o 0 6 + 1 2 i 1 v o 0 12 = 0 + 12 v o 4 v o 6 + 1 2 12 v o 4 v o 12 = 0 + 12 4 + 1 2 12 4 = v o 4 + v o 6 + v o 8 + v o 12 108 = 15 v o v o = + 7.2 volts (b) [3 points] Determine the short circuit current (from a to b) at terminals a-b.
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Ask a homework question - tutors are online | 1,369 | 4,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.753244 |
https://oeis.org/A096725 | 1,679,520,523,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00741.warc.gz | 458,789,626 | 4,089 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A096725 Numerators of terms in series expansion of (sin(tan(x)) - tan(sin(x))) / (arcsin(arctan(x)) - arctan(arcsin(x))). 2
1, 5, 1313, -2773, -701933647, -86849082293, -174426488476171, -130176915706274917, -42426469007472079018663, -24495552034235134641205327, -3019410235003955483667737236843, -74265172933666226350348992663473, -2457268368880426576340457161112391, -589361165665450343618737576026916723726003 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 REFERENCES V. I. Arnold, Huygens and Barrow, Newton and Hooke, Birkhäuser, Basel, 1990. LINKS G. C. Greubel, Table of n, a(n) for n = 0..180 EXAMPLE (sin(tan(x)) - tan(sin(x))) / (arcsin(arctan(x)) - arctan(arcsin(x))) = 1 + 5/3*x^2 + 1313/1890*x^4 - 2773/11907*x^6 - 701933647/1650310200*x^8 - 86849082293/270320810760*x^10 - ... MATHEMATICA Numerator[Take[CoefficientList[Series[(Sin[Tan[x]] - Tan[Sin[x]]) / (ArcSin[ArcTan[x]] - ArcTan[ArcSin[x]]), {x, 0, 50}], x], {1, -1, 2}]] (* G. C. Greubel, Nov 20 2016 *) CROSSREFS Cf. A096730, A096722, A096717, A096718, A096664, A096671, A096712, A096716, A045688, A045689, A096719, A096720. Sequence in context: A182345 A268485 A066162 * A332536 A172920 A062598 Adjacent sequences: A096722 A096723 A096724 * A096726 A096727 A096728 KEYWORD sign,frac AUTHOR N. J. A. Sloane, Aug 15 2004 STATUS approved
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Last modified March 22 17:23 EDT 2023. Contains 361432 sequences. (Running on oeis4.) | 652 | 1,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-14 | latest | en | 0.486785 |
https://www.mike-buys-houses.com/buying-a-house/readers-ask-how-to-budget-for-buying-a-house.html | 1,660,285,659,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00443.warc.gz | 765,217,142 | 13,977 | 3. Compute how much available cash you have.
4. Find out how much you can afford to pay for your mortgage every month.
To calculate ‘how much house can I afford,’ a good rule of thumb is using the 28%/36% rule, which states that you shouldn’t spend more than 28% of your gross monthly income on home-related costs and 36% on total debts, including your mortgage, credit cards and other loans like auto and student loans.
## How do you budget for a new house?
One rule of thumb is that you shouldn’t spend more than 28% of your gross monthly income on housing-related costs and 36% on total debts, including your mortgage, credit cards and other loans.
You might be interested: How To Find Out Cost Of Utilities Before Buying A House?
## What house can I afford on 70k a year?
So if you earn \$70,000 a year, you should be able to spend at least \$1,692 a month — and up to \$2,391 a month — in the form of either rent or mortgage payments.
## How much house can I afford if I make 40000 a year?
Example. Take a homebuyer who makes \$40,000 a year. The maximum amount for monthly mortgage-related payments at 28% of gross income is \$933. (\$40,000 times 0.28 equals \$11,200, and \$11,200 divided by 12 months equals \$933.33.)
## What is considered house poor?
What does it mean to be house poor? Someone who is house poor spends so much of their income on homeownership — such as monthly mortgage payments, property taxes, insurance and maintenance — that there’s very little left in the budget for other important expenses.
## How much should I spend on a house if I make \$100 K?
When attempting to determine how much mortgage you can afford, a general guideline is to multiply your income by at least 2.5 or 3 to get an idea of the maximum housing price you can afford. If you earn approximately \$100,000, the maximum price you would be able to afford would be roughly \$300,000.
A few popular options include: FHA loans (allow low income and as little as 3.5% down with a 580 credit score); USDA loans (for low-income buyers in rural and suburban areas); VA loans (a zero-down option for veterans and service members); HomeReady or Home Possible (conforming loans for low-income buyers with just 3%
You might be interested: Question: How Long Is The Time Between Buying A House And Moving In?
## Can I buy a house with 70k salary?
If you make \$70,000 a year, your monthly take-home pay, including tax deductions, will be approximately \$4,328. But if you have no debt, you can stretch up to 40% of your take-home income, which will be devoting about \$1,731.20 to your mortgage payment.
## Can I buy a house making 25k a year?
HUD, nonprofit organizations, and private lenders can provide additional paths to homeownership for people who make less than \$25,000 per year with down payment assistance, rent-to-own options, and proprietary loan options.
## How much do I need to make to afford a 250k house?
How much income is needed for a 250k mortgage? + A \$250k mortgage with a 4.5% interest rate for 30 years and a \$10k down-payment will require an annual income of \$63,868 to qualify for the loan.
## What kind of house can I afford making 50k a year?
A person who makes \$50,000 a year might be able to afford a house worth anywhere from \$180,000 to nearly \$300,000. That’s because salary isn’t the only variable that determines your home buying budget. You also have to consider your credit score, current debts, mortgage rates, and many other factors.
## Can you afford a house making 30k?
If you were to use the 28% rule, you could afford a monthly mortgage payment of \$700 a month on a yearly income of \$30,000. Another guideline to follow is your home should cost no more than 2.5 to 3 times your yearly salary, which means if you make \$30,000 a year, your maximum budget should be \$90,000. | 920 | 3,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-33 | latest | en | 0.96258 |
https://www.trueachievements.com/a168121/river-of-flavor-achievement | 1,548,205,384,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583879117.74/warc/CC-MAIN-20190123003356-20190123025356-00481.warc.gz | 983,134,503 | 8,196 | 5,574
(1,500)
Angry Birds Trilogy
3.5 from 1189 votes
There are 70 Angry Birds Trilogy achievements (50 without DLC) worth 5,574 (1,500)
17,559 tracked gamers have this game, 158 have completed it (0.90%)
River of Flavor178 (30)
Collect all Rio fruits
• Unlocked by 501 tracked gamers (3% - TA Ratio = 5.92) 17,559
Achievement Guide for River of Flavor
Solution
This solution has 28 positive votes and 0 negative votes. Please log in to vote.
Half of the levels in Rio have one fruit hidden. (See at the bottom which ones) Most aren't that hard to spot when you know what to look for. Some can be a little hard to collect though which you do by touching them or crushing them. The hardest one for me was level 8-5.
For this achievement you need to collect all of them! You can see at the bottom of the episode selection how many fruits you have in each episode.
You can find a guide for every fruit below. (It's not for the Trilogy version but you can use it anyway)
http://www.angrybirdsnest.com/category/walkthroughs/angry-bi...
Levels with a fruit are:
Pineapple
Ep 1: 1-2 1-6 1-7 1-8 1-12 1-13 2-1 2-2 2-4 2-6 2-7 2-8 2-10 2-12 2-15
Banana
Ep 2: 3-2 3-3 3-7 3-9 3-12 3-13 3-14 3-15 4-2 4-4 4-5 4-8 4-11 4-14 4-15
Melon
Ep 3: 5-2 5-5 5-11 5-13 5-14 5-15 6-2 6-3 6-4 6-5 6-6 6-7 6-9 6-12 6-14
Papaya
Ep 4: 7-2 7-6 7-7 7-8 7-10 7-14 7-15 8-2 8-3 8-5 8-9 8-11 8-12 8-14 8-15
Apple
Ep 5: 9-1 9-2 9-5 9-6 9-9 9-10 9-14 10-3 10-4 10-7 10-8 10-10 10-12 10-14 10-15
Mango
Ep 6: 11-2 11-5 11-7 11-10 11-12 11-13 11-15 12-1 12-3 12-6 12-7 12-11 12-12 12-14 12-15
There are 13 comments relating to this Solution | Please log in to comment on this solution.
Do you have a question about this achievement? Please post it in the Angry Birds Trilogy Forum | 815 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.669807 |
http://forum.amibroker.com/t/returning-non-null-values-of-a-vector/2036 | 1,518,924,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811352.60/warc/CC-MAIN-20180218023321-20180218043321-00010.warc.gz | 129,470,192 | 6,951 | Returning Non Null values of a vector
#1
Is there a way beside looping that non null values of a vector can be outputted as a string?
Thanks
String output in Exploration (i.e. percent sign)
How can I fix a condition and vice versa
#2
#3
Thanks @fxshrat. How would you deal with a situation that you have more than three conditions (lets say 7)?
#4
The same way you deal with three just with the difference that it is seven. What response do you expect? Simply start coding (it doesn’t hurt) and if you get stuck then get back. I’m not going to spoon feed since everything is comprehensibly laid out already. Simply follow the link and examples and adjust to your needs. Easy.
#5
So the logic in the example in the manual:
``` TextList = "No signal\nBuy Sell\nBuy and Sell"; TextSelector = 1 * Buy + 2 * Sell; /* would give 0 if no signal, 1 if a buy, 2 if a sell, 3 if both buy and sell */ AddMultiTextColumn( TextSelector, TextList, "Which signal" ); ```
is self explanatory however if your text list is
Then what would be your textselector algorithm (and you can extend the numbe rof conditions to seven?
By the way I have given this quite a bit thought, I don’t know if you have. No spoon feeding is required here.
#6
Probably like this one:
``````TextSelector = 1 * Cond1 + 2 * Cond2 + 4 * Cond3 + 8 * Cond4 + 16 * Cond5 + 32 * Cond6 + 64 * Cond7; // etc.
/* would give 0 if no signal, 1 if Cond1, 2 if Cond2, 3 if Cond1 and Cond2, 4 if Cond3, 5 if Cond1 and Cond3, 6 if Cond2 and Cond3, 7 if Cond1 and Cond2 and Cond3, 8 if Cond4 ... */
``````
For example take a look at AddToComposite flags numeration:
#7
@Milosz Thanks for your reply. TextSeletor value should correspond with with the position of the string in TextList. The alogoritm that you suggested (the power of 2) needs to converted back to a number between 0 & 6 to be useful with TextList.
#8
What kind of output do you want? Don’t be cryptic!
For seven conditions there multiple number of combinations possible.
Do you want to output text only if single condition is true???
If so then for example create loop like this.
``````/// @link http://forum.amibroker.com/t/returning-non-null-values-of-a-vector/2036/6
/// text output only if single condition of seven is true.
///
TextList = "";
for( i = 0; i <= 64; i++) {
switch( i ) {
case 1: TextList += "1"; break;
case 2: TextList += "2"; break;
case 4: TextList += "4"; break;
case 8: TextList += "8"; break;
case 16: TextList += "16"; break;
case 32: TextList += "32"; break;
case 64: TextList += "64"; break;
default: break;
}
TextList += "\n";
}
``````
Replace text within switch statement by your own text.
#9
Why between 0 and 6 in case of 7 conditions? With just 2 conditions (Buy and Sell) the TextList variable contains 4 different messages:
``````TextList = "No signal\nBuy\nSell\nBuy and Sell";
``````
With 3 conditions (Cond1, Cond2, Cond3) the TextList variable should contain 8 different messages:
``````TextList = "No signal\nCond1\nCond2\nCond1 and Cond2\nCond3\nCond1 and Cond3\nCond2 and Cond3\nCond1 and Cond2 and Cond3";
``````
… so you can count how many variants there are with 7 conditions …
#10
@Milosz, actually the number unique combinations is factorial, so it is more than 6. However is was trying to point out that the power of 2 algo needs to convert back to the position of the appropriate conditions in textlist.
#11
Did you even bother to analyse my example (actually two examples) above? It looks like you didn’t. You don’t need to convert anything. The value of `TextSelector` variable (0-7 in case of 3 conditions) points to a unique text in the `TextList` variable (containing 8 different texts/combinations/variants - in a proper order ) - one unique text for each value. So what do you want to convert?
#12
I seriously do not understand what it is that you do not understand about usage of AddMultiTextColumn.
TextList is a list being put into AddMultiTextColumn as 2nd argument. Type of TextList: string.
The length of TextList is defined by number of possible occurrences of TextSelector (1st argument of AddMultiTextColumn, type: array). TextList is zero-based. So zero is reserved for when TextSelector returns zero (-> no true -> first item of TextList chosen).
If you have conditions that may occur at same time then the number of condition combinations is 2^(number of conditions) (i.e. if 4 conditions then it is 2^4 = 16 combinations). So it is 16 texts separated by “\n” assigned to TextList (TextList is the variable you insert into AddMultiTextColumn as 2nd argument). If one of n-combinations of all conditions is true at bar position then the according substring of TextList is picked by AmiBroker automatically by checking output of TextSelector result (again, TextSelector is 1st argument of AddMultiTextColumn).
Here is my very last detailed try with example of FOUR conditions (there is also example in there where there are 4 conditions not occurring at same time). If it’s still not going to click afterwards then I can’t help further.
Now what does it mean for your case of 7 conditions… if you want to output all possible combinations (and all conditions may occur at same time) then you have to create a string containing 128 occurrences (2^7=128) -> TextList consisting of 128 substrings separated by “\n”.
And TextSelector being
``````TextSelector = 1 * Cond1 + 2 * Cond2 + 4 * Cond3 + 8 * Cond4 + 16 * Cond5 + 32 * Cond6 + 64 * Cond7;
``````
On the other hand if all 7 conditions can not occur at same time (see 3rd example in upper picture) then you could just use 7+1 substrings only.
``````TextList = "No true\nC1 true\nC2 true\nC3 true\nC4 true\nC5 true\nC6 true\nC7 true";
TextSelector = 1 * Cond1 + 2 * Cond2 + 3 * Cond3 + 4 * Cond4 + 5 * Cond5 + 6 * Cond6 + 7 * Cond7;
``````
#13
@fxshrat, you made my point for me. In your example of 4 conditions the textlist becomes a list of 16 pieces of strings. As you mentioned with 7 conditions the textlist becomes a list of 128 pieces of strings. So it becomes laborious. So using power of 2 algo we can determine which conditions have fired. We can write the values in a matrix. I have already done all of that. The issue is how to by pass writing a long textlist. How can we cleverly write code to do the tedious job! Programming is a combination of knowledge & art (creativity). I think my only solution is a c++ dll.
#14
@QuantTrader: In this entire thread, I cannot find a concrete example of what you’re trying to achieve. You began by asking:
Later the discussion turned to AddMultiTextColumn and 2^N combinations of N different conditions, etc., but I get the sense that this function may not be the best solution to your problem. You might get a more helpful response if you provided a small working example of sample input and the expected output. As @fxshrat asked earlier:
#15
You are seriously asking how to create TextList programmatically if having a conditions string such as
``````condstr = "Cond1,Cond2,Cond3,Cond4";
``````
Matrix? DLL? What for?
All you need is one nested loop and bitwise operator (and just AFL).
Here below is one simple solution created within few minutes (if having the proper knowledge,skills,creativity,…).
Basically you can insert any substring into condstr variable representing the output if according condition is true.
“F” in Interpretation window stands for “False” or “No true signal” or “No true condition” or “No farm” or … etc.
"&" is the separator between each condition state. So for example “F & F & F & F” means no true signal of four ones and gets output if TextSelector array variable returns 0 at bar position. “Cond1 & Cond2 & Cond3 & Cond4” would get output if TextSelector returns 16 at bar positions. “Old & MacDonald & had & a farm & E-I-E-I-O” would get output if TextSelector returns 32 for 5 set conditions at bar position. And so on… | 2,145 | 7,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-09 | latest | en | 0.868584 |
https://www.electro-tech-online.com/threads/2s-complement-without-negative-reference.153091/ | 1,620,623,363,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00123.warc.gz | 761,253,369 | 19,959 | # 2's complement without negative reference
Status
Not open for further replies.
#### Beau Schwabe
##### Active Member
Ok, so I had a REALLY dumb epiphany and I just want to make sure I'm not missing anything.
To make things simple I'll use 4 bits.
I have a number that ranges from 7 to -8 with 0 or -1 being center.
I want to convert that to a number that ranges from 0 to 15 with 7 or 8 being center.
In the past I would check the MSB; if it was a "0" then I would ADD 8; if it was a "1" I would SUBTRACT 8 ... without really even thinking what was going on.
Looking at this closer, all I need to do is invert the MSB and I'm done. No adding or subtracting required. Am I understanding that correctly? Even up to 32 Bit numbers?
Going in reverse, that might make for a really slick way to generate a 2's complement output.
Code:
2's Complement: Desired Output:
0000 = 0 8
0001 = 1 9
0010 = 2 10
0011 = 3 11
0100 = 4 12
0101 = 5 13
0110 = 6 14
0111 = 7 15
1000 = -8 0
1001 = -7 1
1010 = -6 2
1011 = -5 3
1100 = -4 4
1101 = -3 5
1110 = -2 6
1111 = -1 7
#### Grossel
##### Well-Known Member
Not sure get this problem rigth, but have you thaught that if you just simply invert MSB then you would get the output result. No adding, no substraction - only make one bit oposite
#### Beau Schwabe
##### Active Member
have you thaught that if you just simply invert MSB then you would get the output result
... Right ... simply XORing the MSB is all that is needed.... before I would just do a compare on the MSB and then based on that actually do an addition or subtraction. Like I said it was a "REALLY dumb epiphany" on my part getting from point A to point B
EDIT: Hey, we are all human and sometimes it takes something obvious right in front of our face... I just wanted to share.
#### large_ghostman
##### Well-Known Member
Completely messes with my head this! One of those things that you get easy or you DONT!!
#### Ian Rogers
##### User Extraordinaire
Forum Supporter
2's complement is easy enough..
Micro's do not do negative numbers end of!! So programmers make it happen... we'll do 4 bit.
1111 = -1 Why!! because 0000 - 1 = 1111 ( which is actually 15 ).. 2's complement is the programming way to say negative..
1 is 0001 complemented ( this is the actual term for inversion ) you get 1110.. If you add 2 you get to get back to zero, so its obviously represents -2.. To make things right you have to add one.. complement once to invert then complement the LSBit with carry!!! or in other words 2's complement.. Programmers just subtract from max + 1 to get it positive...
16 - 1111 = 1.. BUT!! All this is superfluous as we are dealing with negative numbers, so we can just add them ie
1010 + 1110 = 1000 .. Or in English 10 + -2 = 8... Just remember to ditch the MSBit... | 810 | 3,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-21 | latest | en | 0.951837 |
https://www.developerfaqs.com/5174/vector-sort-algorithm-sort-only-elements-bigger-then-0 | 1,566,534,862,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317847.79/warc/CC-MAIN-20190823041746-20190823063746-00322.warc.gz | 781,848,273 | 9,072 | # Vector Sort Algorithm, sort only elements bigger then 0
• A+
Category：Languages
I have to sort a vector of structs. Let's say the struct has two members:
``Struct game { string name; int rating; }; ``
So I have created a `std::vector<game>` games and simple sort them by rating.
``std::sort(games.begin(),games.end(), [](game& info1, game& info2) { return info1.rating > info2.rating; }); ``
Everything is alright so far. The problem is if all games have rating value 0, they mix. Simply put I have to sort only elements with rating bigger than zero. Let's give you an example:
All games are pushed in the vector by names in alphabetic order and rating 0, when a sort is triggered, the alphabet order gets violated.
Example before sort:
`"A_Game"`, `"B_Game"`, `"C_Game"`, `"E_Game"`, `"G_Game"`, etc. (continue with all next letters)
after sort (all games are with rating 0):
`"G_Game"`, `"S_Game"`, `"P_Game"`, `"M_Game"`, `"L_Game"`, `"I_Game"`, etc.
I need to sort only these games that have rating bigger than 0. Thanks in advance.
You can use `std::stable_sort` to prevent moving around elements that are not affected by the sorting criteria.
``std::stable_sort(games.begin(),games.end(), [](game& info1, game& info2) { return info1.rating > info2.rating; }); `` | 343 | 1,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-35 | longest | en | 0.860918 |
http://www.cocos2d-x.org/docs/api-ref/cplusplus/V3.10/d7/d39/classcocos2d_1_1_plane.html | 1,532,097,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00243.warc.gz | 429,404,415 | 4,305 | Cocos2d-x v3.10
Plane Class Reference
Defines plane. More...
Public Member Functions
Plane (const Vec3 &p1, const Vec3 &p2, const Vec3 &p3)
Create plane from tree point. More...
Plane (const Vec3 &normal, float dist)
Create plane from normal and dist. More...
Plane (const Vec3 &normal, const Vec3 &point)
Create plane from normal and a point on plane. More...
Plane ()
Create a default plan whose normal is (0, 0, 1), and _dist is 0, xoy plan in fact.
void initPlane (const Vec3 &p1, const Vec3 &p2, const Vec3 &p3)
Init plane from tree point. More...
void initPlane (const Vec3 &normal, float dist)
Init plane from normal and dist. More...
void initPlane (const Vec3 &normal, const Vec3 &point)
Init plane from normal and a point on plane. More...
float dist2Plane (const Vec3 &p) const
Get the specified point's distance to the plane. More...
const Vec3getNormal () const
Gets the plane's normal. More...
float getDist () const
Gets the distance which origin along its normal reach the plane. More...
PointSide getSide (const Vec3 &point) const
Return the side where the point is. More...
Defines plane.
Constructor & Destructor Documentation
Plane ( const Vec3 & p1, const Vec3 & p2, const Vec3 & p3 )
Create plane from tree point.
The point #1 which's in the plane. The point #2 which's in the plane. The point #3 which's in the plane.
Plane ( const Vec3 & normal, float dist )
Create plane from normal and dist.
The normal vector of the plane. The distance from the origin along the normal.
Plane ( const Vec3 & normal, const Vec3 & point )
Create plane from normal and a point on plane.
The normal vector of the plane. The point which's in the plane.
Member Function Documentation
void initPlane ( const Vec3 & p1, const Vec3 & p2, const Vec3 & p3 )
Init plane from tree point.
The point #1 which's in the plane. The point #2 which's in the plane. The point #3 which's in the plane.
void initPlane ( const Vec3 & normal, float dist )
Init plane from normal and dist.
The normal vector of the plane. The distance from the origin.
void initPlane ( const Vec3 & normal, const Vec3 & point )
Init plane from normal and a point on plane.
The normal vector of the plane. The point which's in the plane.
float dist2Plane ( const Vec3 & p ) const
Get the specified point's distance to the plane.
The specified point. The distance, > 0 in the normal side.
const Vec3& getNormal ( ) const
inline
Gets the plane's normal.
The normal of the plane.
float getDist ( ) const
inline
Gets the distance which origin along its normal reach the plane.
The distance to the origin.
PointSide getSide ( const Vec3 & point ) const
Return the side where the point is.
The side.
The documentation for this class was generated from the following file:
• CCPlane.h | 714 | 2,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-30 | latest | en | 0.679801 |
https://math.stackexchange.com/questions/480008/zero-sum-of-roots-of-unity-decomposition | 1,702,037,966,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00700.warc.gz | 426,702,295 | 37,628 | # Zero sum of roots of unity decomposition
It's known that sum of all $n$'th roots of some $z \in \mathbb C$ with $|z| = 1$ is zero (if $n \geqslant 2$).
Is it true that any zero sum of roots of unity can be decomposed in this way? That is if we have $\zeta_1 + \ldots + \zeta_s = 0,$ where $\zeta_i^n = 1,$ then we can partition it in groups $\{\xi_1, \ldots, \xi_k\}$ of all $k$'th roots of some $z$ with $|z| = 1$ (with zero sum of elements in this group), with possibly different $k$ for different groups?
UPD
Example: if $\zeta_1 + \zeta_2 = 0,$ then $\zeta_1^2 = \zeta_2^2 = z,$ so $\zeta_1, \zeta_2$ are square roots of this $z.$ Is it not hard to show that if $\zeta_1 + \zeta_2 + \zeta_3 = 0,$ then $\zeta_1^3 = \zeta_2^3 = \zeta_3^3 = z,$ so $\zeta_1, \zeta_2, \zeta_3$ are cubic roots of some $z$.
If $\zeta_1 + \zeta_2 + \zeta_3 + \zeta_4 = 0,$ it seems there are two possiblities - like $\{1, i, -1, -i\}$ (one part), or like $\{e^{\pi i/3}, e^{4\pi i/3}\}, \{e^{2\pi i/3},e^{10\pi i/3}\}.$ But I don't know what is the argumentation for the general case.
• Consider the 8-th root of unity $r_k=e^{k 2\pi i/8}$ for $k=0\cdots 7$. The sum of roots at $k=1,3,5,7$ is zero but this is not all the roots. Did I misunderstand the question? Aug 30, 2013 at 18:14
• ) Yes. They all are 4-th roots of $e^{2\pi i}{2} = i$ The question is: is it possible to decompose any zero sum of unity's roots in symmetric ones or not? Aug 30, 2013 at 19:12
• The $r_k$ in Maesumi's comment are $4$th roots of $-1$, not of $i$.
– anon
Aug 31, 2013 at 15:59
• Yes. There are some typos in the comment. Aug 31, 2013 at 16:22
No, for example pick $\zeta = \exp \frac{i\pi}{15}$, a primitive $30$th root of unity.
Then $0 = (1 + \zeta^{10} + \zeta^{20}) + \zeta^{15}(1 + \zeta^6 + \zeta^{12} + \zeta^{18} + \zeta^{24}) - (1 + \zeta^{15}) \\ = \zeta^3 + \zeta^9 + \zeta^{10} + \zeta^{20} + \zeta^{21} + \zeta^{27}$
But you can't partition those $6$ roots into vertices of regular $k$-gons.
However, it is true that you can always add vertices of regular $k$-gons to your set of roots (possibly adding the same root multiple times) to obtain a new multiset that you can then arrange into a sum of vertices of regular $k$-gons (possibly using the same $k$-gon multiple times).
Let $\zeta_n = \exp{\frac{2i\pi}n}$ and $f$ be the map $\Bbb Z^n \to \Bbb Z[\zeta_n]$ given by $(a_i) \mapsto \sum a_i \zeta_n^i$.
The minimal polynomial of $\zeta_n$ over $\Bbb Q$ is the cyclotomic polynomial $\Phi_n$, of degree $\varphi(n)$. Hence $\Bbb Z[\zeta_n]$ is free of rank $\varphi(n)$.
If $n = \prod p_i^{d_i}$, then $\zeta$ is a primitive $n$th root of unity iff it is of order exactly $n$, and not $n/p$ for any $p$, iff $\zeta^{n/p}$ is a primitive $p$th root, for all $p$. Hence $\Phi_n$ is the gcd of the $\Phi_p(X^{n/p}) = 1 + X^{n/p} + X^{2n/p} + \ldots + X^{(p-1)n/p}$, and $\ker f$ is generated by those relations, which correspond to vertices of regular $p$-gons.
• Thank you! But about your note about decomposition with possible adding regular $k$-gons - I have the dimension argument which says that its not true: consider module $A$ generated by all 14-th roots of unity. You say that all relations in is are generated by "regular" sums, representing regular $k$-gons. But we have only $7 + 2 + 1$ such sums (some of which are linearly depended). Sep 3, 2013 at 17:24
• Its not enougth. Consider the map $f\colon \mathbb Z^{14} \to A.$ $A$ has rank 2, $\mathbb Z^{14}$ has rank 14, so $\ker f$ must have rank $12,$ and regular sums span only module of rank 10. So there are some non-regular zero sums. Sep 3, 2013 at 17:27
• So, if my conclusions are correct, we have another, non-constructive proof of this negative result. Sep 3, 2013 at 17:29
• @ptashek : $A$ has rank $\varphi(14)=6$ and not $2$. There is no counterexample with $14$th roots. You need at least $3$ distinct prime factors in $n$ if you want to find a counterexample with $n$th roots, so my example was the smallest one. Sep 4, 2013 at 6:37
• Thank you again, my intuition for some reason tells me that $A$ is discrete in $\mathbb C = \mathbb R^2$) Sep 4, 2013 at 14:35 | 1,490 | 4,138 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-50 | latest | en | 0.811312 |
http://mathhelpforum.com/business-math/140566-brownian-motion-expectation-variance.html | 1,481,251,427,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00407-ip-10-31-129-80.ec2.internal.warc.gz | 170,040,088 | 10,133 | ## Brownian Motion: Expectation and Variance
Consider the stochastic differential equation,
$dS_t = \mu S_tdt + \sigma S_tdW_t$, for all $t \in [0, T]$,
where $\{W_t\}_{t\geq0}$ is a standard Brownian motion, $\mu, \sigma$ and $S_0$ are positive constants. Derive the unique solution to the above SDE . Calculate $\mathbb{E}[S_t]$ and $Var[S_t]$.
Skipping working the unique solution is...
$S_t = S_0e^{(\mu - \tfrac{\sigma^2}{2})t - \sigma W_t}$.
Now to find the expectation and variance. I've realized that a lot of the things I do I do because I have seen them in lectures or done them in other questions but I now realize I have no idea why they are true.
$\mathbb{E}[S_t] = \mathbb{E}[S_0e^{(\mu - \tfrac{\sigma^2}{2})t - \sigma W_t}] = S_0e^{\mu t}\mathbb{E}[e^{-\tfrac{\sigma^2}{2}t - \sigma W_t}]$ $=S_0\mu$. Now why is that last step true?
My guess is something to do with...
$\mathbb{E}[W_t] = 0$
But I don't know what to do with the $\sigma$ although I assume that it eventually comes out to be $e^0 = 1$.
Now the variance...
The way my lecturer does it is to calculate $Var[S_t] = \mathbb{E}[(S_t - \mathbb{E}[S_t])^2]$ but I always calculate $\mathbb{E}[S_t^2]$ instead then do the usual subtraction of $(\mathbb{E}[S_t])^2$.
So, I get...
$\mathbb{E}[S_t^2] = \mathbb{E}[S_0^2e^{2(\mu - \tfrac{\sigma^2}{2})t + 2 \sigma W_t}]$.
$= S_0^2 e^{2\mu t}\mathbb{E}[e^{-\sigma^2 t + 2 \sigma W_t}]$ but now I am not sure how to justify the next step.
$= S_0^2 e^{2\mu t}e^{-\sigma^2 t}$ (actually should that be $e^{\sigma^2 t}$..?!?) | 559 | 1,555 | {"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": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-50 | longest | en | 0.765091 |
https://studysoup.com/note/12807/spring-2015 | 1,477,061,883,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718284.75/warc/CC-MAIN-20161020183838-00427-ip-10-171-6-4.ec2.internal.warc.gz | 884,845,808 | 18,148 | ×
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# Class Note for ECE 449 at UA
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Date Created: 02/06/15
A I we 449549 ontinnons System mutualitth Electrical Circuits I 0 This lecture discusses the mathematical modeling of simple electrical linear circuits 0 When modeling a circuit one ends up with a set of implicitly fonnulated algebraic and differential equations DAEs Which in the process of horizontal and vertical sorting are converted to a set of explicitly fonnulated algebraic and differential equations 0 By eliminating the algebraic variables it is possible to convert these DAEs to a statespace representation September 3 2003 start of Presentation EASE I we 449549 Glontinnnns System mutualitth Table of Contents 0 Components and their models 0 The circuit topology and its equations An example 0 Horizontal sorting Vertical sorting Statespace representation 0 Transformation to statespace form September 3 2003 start of Presentation A I we 449549 untinnons System mmml Linear Circuit Components R Res1stors mi t vb u vu v u R i u i C u vu v o Capac1tors VaH H a Vb 1114 u t W 39 L M vu v o Inductors vac1m vb di quot u u L E September 3 2003 Start of Presentation I we 449549 Glontinnnns System mutualingl Linear Circuit Components 11 U 0 Voltage sources Va e vb U v Va U a ft U0 390 M v v Current sources v VILQ vb I fnt a du 0 Ground I oTel V 0 v September 3 2003 Start of Presentation A I we 449549 ontinnons System mutualitth Circuit Topology in ii v vv o Nodes Va vb quot ic In In tE 0 vc quota1 v11 vb 0 Meshes v umunum0 u be September 3 2003 start of Presentation EASE I we 449549 Glontinnnns System mutualitth An Example I ma my 3 11 Mggl i Gmund September 3 2003 start of Presentation A I we 449549 untinnons System mutualitth Rules for Systems of Equations I 0 The component and topology equations contain a certain degree of redundancy 0 For example it is possible to eliminate all potential variables vi without problems 0 The current node equation for the ground node is redundant and is not used 0 The mesh equations are only used if the potential variables are being eliminated If this is not the case they are redundant September 3 2003 start of Presentation I we 449549 Glontinnnns System mutualitth Rules for Systems of Equations 11 0 If the potential variables are eliminated every circuit component de nes two variables the current 139 through the element and die Voltage u across the element 0 Consequently we need two equations to compute values for these two variables 0 One of the equations is the constituent equation of the element itself the other comes from the topology September 3 2003 start of Presentation A I we 449549 ontinnons System mutualitth An Example II Component equations U0 ft is 039 due1t u R i ul L diLdt uz Rz 139 Node equations i0i1il i1izic The circuit contains 5 components 3 We require Mesh equations 10 equations in U u u u u u 10 unknowns 0 1 C L I z u u September 3 2003 start of Presentation I we 449549 Glontinnnns System mutualitth Rules for Horizontal Sorting I 0 The time t may be assumed as known 0 The state variables variables Which appear in differentiated form may be assumed as known U0ft i1iL U0ft IR139i1 i2ic MIR1i1 z Rz39 i2 U0 1 141 gt 142 Rz i2 iCCdqut ucuz iCCdugdt 0 uLLdiLdt ulu1u2 uLLdiLdt ulu1u2 September 3 2003 start of Presentation A I we 449549 ontinnons System mutualitth for it U0 ft I RI39 1391 142 R2 139 is C duedt ul L diLdt i0i1il i1izic U0u1uc 0 MLMIM2 The solved variables are now known Rules for Horizontal Sorting II 0 Equations that contain only one unknown must be solved U0 ft 1R139i1 uzRz i2 iCC dugdt ul L diLdt September 3 2003 start of Presentation A I we 449549 Glontinnnns System mutualitth Rules for Horizontal Sorting III 0 Variables that show up in only one equation must be solved for using that equation U0 ft I RI39 1391 142 R2 139 is C duedt ul L diLdt U0 ft 1 R13911 is C dugdt 14L L diLdt September 3 2003 start of Presentation A I we 449549 hummus System manslingl Rules for Horizontal Sorting IV 0 A11 rules may be used recursively U0ft i0i1il U0ft I R139 1391 i1 i2 ic I R139 1391 2 Rz39 1392 U0 I c gt z Rz39 l392 icC39 duedt 0 iCC dugdt uL L diLdt 14L 14 142 uL L diLdt September 3 2003 Start of Presentation we 449549 Glontinnnns System mutualingl U0ft i0i1il U0ft 1 RI39 1391 I I R139 1391 142 R2 12 gt 142 Rz i2 iCC duedt iCC duadt 6 uLLdiLdt ulLdiLdt uLu1uz U0 ft The algorithm is applied until I R1 1 every equation de nes exactly one variable that is solves for 2 R239 12 iCC dugdt ucuz uLLdiLdt uLu1uz September 3 2003 Start of Presentation A I we 449549 ontinnons System mutualitth Rules for Horizontal Sorting V 0 The horizontal sorting can now be performed using symbolic formula manipulation techniques Uoft i0i1il Uoft lI0iziz 1R139i1 i1lzic l391 1R1 ici1 39iz 2Rz39iz U0 1 c gt lIz zRz IU039 c iCC duedt c z duddticC 2 C MLLdiLdt ulu1uz diLdtuLL L 1 2 September 3 2003 3 start of Presentation I we 449549 Glontinnnns System mutualitth Rules for Vertical Sorting By now the equations have become assignment statements They can be sorted vertically such that no variable is being used before it has been defined Uoft U0ft lIz zRz i1 1R1 iCi1i2 1U039 c ici139iz izuzRz M1U0MC gt i1u1R1 ulu1uz dugdtiCC uz uc i0 i1 il dugdtiCC diLdtuLL MLM1M2 uzuc diLdtuLL September 3 2003 start of Presentation A I we 449549 untinnons System mutualitth potentials and Voltages equations of the topology ignored before can be ignored Rules for Systems of Equations III 0 Alternatively it is possible to work with both 0 In that case additional equations for the node potentials must be found These are the potential equations of the components and the potential Those had been 0 The mesh equations are in this case redundant and September 3 2003 start of Presentation A I we 449549 Glontinnnns System mutualitth An Example III Component equations U 0 ft u1R111 142 R2 12 is C dugdt ML L diLdt 0 The circuit contains v 5 components and 3 quot0 deS Node equations 3 We require i0 i1 iL 13 equations in 13 unknowns U0v1v0 1v1vz cvzv0 Lv1v0 lllzlc September 3 2003 start of Presentation A we 449549 untinnons System mutualitth Sorting The sorting algorithms are applied just like before 0 The sorting algorithm has already been reduced to a purely mathematical informational structure Without any remaining knowledge of electrical circuit theory 0 Therefore the overall modeling task can be reduced to two subproblems 1 Mapping of the physical topology to a system of implicitly formulated DAEs 2 Conversion of the DAB system into an executable program structure September 3 2003 start of Presentation lt I we 449549 Glontinnnns System mutualitth Statespace Representation A e an x n 0 Linear systems x 6 9V B 6 mm m AXBu xzox0 6mm cempxn Z yCxDu yegtp Dempxm x State vector 0 N onlmear systems In t t II pu V60 0139 mum x00 x0 y Output vector y gxut n Number of state variables m Number of inputs p Number of outputs September 3 2003 start of Presentation 10 A I was 449549 ontinnons System mutualitth Conversion to Statespace Form I duocit iCC U t 139 14 R 0 f z z z i1i2C u U 14 i i i I quot C C I z iICizC 11 1R1 L 1 z u1R139Q MzR239C i0i1il dugdtzCC U039 0R139Cv ucR239C u2uc dzLdtuLL diLdt uLL 1 zL u1LuzL U0ucL uCL U0L September 3 2003 start of Presentation I was 449549 Glontinnnns System mutualitth Conversion to Statespace Form 11 Welet x1uc I2 1 j x 1 u xziL 1 RI C RZ C 1 RI C39 u Ua 1 y quotC x2 I yx1 September 3 2003 start of Presentation 11 I lm 449 549 hummus 5mm mammal An Example IV gem UDBB kA I 59E R mu R2 zu L uuu15 c 175 mc JURan m sthCn all iuuc chl Aa1DDD h R1c 11 1 u 1 um s 5er 1a an t 1E752E73 u nmsmgmuum XE r 1 2 1 x laursucxm pmucm grld an Ecuxn 7 Q eueMEL39pFl Ready jam gt4 4 September 3 2003 Start of Presentation midi 449549 ontinunux system mnhelingl References Cellier FE 1991 Continuous SVstem Modeling SpringerVerlag New York Chapter 3 Cellier FE 2001 Matlab code of circuit example September 3 2003 Start of Presentation 12
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Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com | 3,022 | 10,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-44 | latest | en | 0.891588 |
https://timothyhillforcongress.com/qa/question-which-state-has-the-most-state-representatives.html | 1,620,866,289,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00204.warc.gz | 599,839,359 | 8,170 | # Question: Which State Has The Most State Representatives?
## How many state representatives are there in each state?
However, in the House of Representatives, a state’s representation is based on its population.
For example, smaller states like Vermont and Delaware have one representative while large states like California have 53 representatives..
## Which state has largest House of Representatives?
As of the 2010 Census, the largest delegation is that of California, with 53 representatives. Seven states have only one representative: Alaska, Delaware, Montana, North Dakota, South Dakota, Vermont, and Wyoming.
## Why is the House of Representatives limited to 435 members?
Because the House wanted a manageable number of members, Congress twice set the size of the House at 435 voting members. The first law to do so was passed on August 8, 1911. … Finally, in 1929 the Permanent Apportionment Act became law. It permanently set the maximum number of representatives at 435.
## How are house seats determined?
The Constitution provides for proportional representation in the U.S. House of Representatives and the seats in the House are apportioned based on state population according to the constitutionally mandated Census.
## Who makes up House of Representatives?
The House of Representatives is made up of 435 elected members, divided among the 50 states in proportion to their total population.
## Why has the House of Representatives grown so much?
Why has the House of Representatives grown so much faster than the Senate? Representation is based on population, and the US has grown steadily. The number of senators allowed per state has been routinely reduced. Fewer and fewer representatives have been re-elected to additional terms.
## How many representatives does each state have at least?
The Number of Representatives shall not exceed one for every thirty Thousand, but each State shall have at least one Representative;…
## Why do some states have more representatives than the others?
The number of U.S. Representatives for each state depends on the population. Some states have more representatives because they have more people. If the state has a large population, there are more representatives. … If the state has a small population, there are fewer representatives.
## Who do some states have more representatives than other states?
The number of U.S. Representatives for each state depends on the population. Some states have more representatives because they have more people. If the state has a large population, there are more representatives.
## How many US Senators does each state have?
The Constitution prescribes that the Senate be composed of two senators from each State (therefore, the Senate currently has 100 Members) and that a senator must be at least thirty years of age, have been a citizen of the United States for nine years, and, when elected, be a resident of the State from which he or she …
## What does NV mean in House vote?
The fourth column (Pres.) has the number of Members who voted ‘present’ and did not vote yes or no. The fifth column (NV) has the number of Members of the House who did not vote.
## What state has the most legislators?
New HampshireThus, the national population per state legislator was 41,507; the national mean was 30,795….RankStateLegislators1.New Hampshire4242.Pennsylvania2533.Georgia2364.New York2126 more rows•Jul 9, 2010
## What three states have the most representatives?
Districts per stateState with the most: California (53), same as in 2000.States with the fewest (only one district “at-large”): Alaska, Delaware, Montana, North Dakota, South Dakota, Vermont and Wyoming. Alaska and Wyoming are the only states that have never had more than one district.
## How many US representatives are there?
435There are currently 435 voting representatives. Five delegates and one resident commissioner serve as non-voting members of the House, although they can vote in committee. Representatives must be 25 years old and must have been U.S. citizens for at least 7 years. Representatives serve 2-year terms.
## How many US Representatives does Texas have?
U.S. House of Representatives The delegation consists of 36 members, with 22 Republicans, 13 Democrats and 1 Vacancy.
## How many Texas state representatives are there?
Texas House of RepresentativesStructureSeats150Political groupsMajority (83) Republican (83) Minority (67) Democratic (67)Length of term2 years23 more rows
## How many houses do most state legislatures have?
two chambersEven though each state legislature is unique, many of them share a few basic characteristics: Most legislatures are bicameral. This just means they have two chambers.
## Who represents California in the House of Representatives?
1st district: Doug LaMalfa (R), since 2013. 2nd district: Jared Huffman (D), since 2013. 3rd district: John Garamendi (D), since 2009. 4th district: Tom McClintock (R), since 2009.
## What is difference between senator and congressman?
For this reason, and in order to distinguish who is a member of which house, a member of the Senate is typically referred to as Senator (followed by “name” from “state”), and a member of the House of Representatives is usually referred to as Congressman or Congresswoman (followed by “name” from the “number” district of … | 1,115 | 5,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.965201 |
http://pike.lysator.liu.se/docs/tutorial/expressions/arithmetics.xml | 1,490,568,787,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00201-ip-10-233-31-227.ec2.internal.warc.gz | 279,785,908 | 2,810 | Pike can do all the usual arithmetic operations: addition (which is expressed with the operator +), subtraction (-), division (/), multiplication (*), and also modulo (%). The modulo operation, sometimes called "remainder" or "rest", gives the remainder from a division. For example, if you divide 7 by 3, 3 goes in 7 two times. But there is a remainder of 1, and that is the result of the modulo operation.
Here is a table of the arithmetic operations:
OperationSyntaxResult
a + b
the sum of a and b
Subtraction
a - b
b subtracted from a
Negation
- a
minus a
Multiplication
a * b
a multiplied by b
Division
a / b
a divided by b
Modulo
a % b
the remainder of a division between a and b
Remember that pike makes a difference between numerical values that are integers (the type int), and numerical values that are real or "floating-point" numbers (the type float). This has some importance for how expressions with arithmetic operations are calculated by Pike. If at least one of the operands is a float, we use the floating-point versions of the operation. In if both operands are integers, we use a special integer-only version of the operation. For most of the operations, the only difference is that the type of the result will be different: 2 + 2 will give the integer value 4, while 2 + 2.0 will give the floating-point value 4.0. But with division is more important.
With floating-point division, such as in 9.0 / 4.0, the result is a floating-point value, in this case 2.25. But with integer division, such as in 9 / 4, the result is only the integer part, in this case 2.
The fact that integer division only gives the integer part can be treacherous: If 73 out of 92 people payed their income tax on time, don't try to calculate the percentage with the expression 73 / 92 * 100. That would give the result 0.
It is very common in programs to increment or decrement a variable with one, such as in the statements
```i = i + 1;
p = p - 1;```
To simplify such programs, Pike has these extra operators, which you can use if you want to:
OperationSyntaxResult
Increment
++ a
increments a and returns the new value for a
Decrement
-- a
decrements a and returns the new value for a
Post increment
a ++
increments a and returns the old value for a
Post decrement
a --
decrements a and returns the old value for a
The two versions of increment, ++i and i++, both increment the value in the variable i with 1. The difference is if we want to use them as parts of a larger expression. In that case, ++i gives the new, incremented, contents of i as value, while i++ gives the old contents of i as value. The same difference applies to --i and i--. | 658 | 2,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-13 | longest | en | 0.927505 |
https://sherpa-online.com/forum/thread/maths/a-level/algebra-functions/how-do-you-multiply-algebraic-fractions-2 | 1,713,824,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00729.warc.gz | 454,196,756 | 14,156 | Sherpa
Maths
>
A-Level
Algebra & Functions
Question
# How do you multiply algebraic fractions?
2 years ago
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E
Enoch Gibson
Devan-Kumar M Verified Sherpa Tutor ✓
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To multiply algebraic fractions you treat them as if the numerators and denominators were integers. You multiply the numerators to form the new numerator and multiply the denominators to form the new denominator. Note you will often have to simplify the fractions after this.
An example would be (2x+4)/(7x+5) multiplied by (5x+6)/(3x+6). The answer would be (2x+4)(5x+6)/(7x+5)(4x+6) as per the aforementioned method. This would then be able to be simplified to (x+2)(5x+6)/(7x+5)(2x+3), as we divide the numerator and denominator by 2.
I'm available for 1:1 private online tuition! | 233 | 857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-18 | latest | en | 0.88054 |
https://nusoft.fnal.gov/nova/novasoft/doxygen/html/bernoulli__rng_8hpp_source.html | 1,611,333,510,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00531.warc.gz | 481,359,518 | 4,515 | bernoulli_rng.hpp
Go to the documentation of this file.
1 #ifndef STAN_MATH_PRIM_SCAL_PROB_BERNOULLI_RNG_HPP
2 #define STAN_MATH_PRIM_SCAL_PROB_BERNOULLI_RNG_HPP
3
9 #include <boost/random/bernoulli_distribution.hpp>
10 #include <boost/random/variate_generator.hpp>
11
12 namespace stan {
13 namespace math {
14
15 /**
16 * Return a Bernoulli random variate with specified chance of success
17 * parameter using the specified random number generator.
18 *
19 * theta can be a scalar or a one-dimensional container.
20 *
21 * @tparam T_theta type of chance of success parameter
22 * @tparam RNG type of random number generator
23 * @param theta (Sequence of) chance of success parameter(s)
24 * @param rng random number generator
25 * @return (Sequence of) Bernoulli random variate(s)
26 * @throw std::domain_error if chance of success parameter is less than zero or
27 * greater than one.
28 */
29 template <typename T_theta, class RNG>
31 const T_theta& theta, RNG& rng) {
32 using boost::bernoulli_distribution;
33 using boost::variate_generator;
34
35 static const char* function = "bernoulli_rng";
36
37 check_finite(function, "Probability parameter", theta);
38 check_bounded(function, "Probability parameter", theta, 0.0, 1.0);
39
40 scalar_seq_view<T_theta> theta_vec(theta);
41 size_t N = length(theta);
43
44 for (size_t n = 0; n < N; ++n) {
45 variate_generator<RNG&, bernoulli_distribution<> > bernoulli_rng(
46 rng, bernoulli_distribution<>(theta_vec[n]));
47 output[n] = bernoulli_rng();
48 }
49
50 return output.data();
51 }
52
53 } // namespace math
54 } // namespace stan
55 #endif
ofstream output
void check_finite(const char *function, const char *name, const T_y &y)
void check_bounded(const char *function, const char *name, const T_y &y, const T_low &low, const T_high &high)
VectorBuilder< true, int, T_theta >::type bernoulli_rng(const T_theta &theta, RNG &rng)
size_t length(const std::vector< T > &x)
Definition: length.hpp:10 | 564 | 1,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.172265 |
https://www.bartleby.com/topics/credit-card-debt | 1,621,328,679,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00387.warc.gz | 692,359,857 | 8,746 | # Credit Card Debt Essay
Page 1 of 50 - About 500 essays
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In today’s world most people have a credit card by the time they turn eighteen years old. Nerd Walt, a financial website, stated “Average credit card debt in 2013 reached \$15,480 per household in the United States”. (Bower) The average household credit card debt has probably increase since the 2013 study. “Overall U.S. National debt is rapidly approaching \$18 trillion” (Bower). “20% of credit card users often pay off monthly balances on two or more credit cards” according to a study (Bower). In today’s
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for her IRA. Her monthly credit payments for VISA
• ## Credit Card Debt Essay
643 Words | 3 Pages
Running head: ANALYSIS OF CREDIT CARD DEBT Analysis of Credit Card Debt George Kennedy Argosy University online General Education Mathematics MAT109 A01 Instructor: Sohrab Bakhtyari January 25, 2013 Analysis of Credit Card Debt 1. My Introduction with a credit card balance of \$5,270.00 and an (APR) of 15.53 percent based upon my own conclusions and assuming there are no other fees are applied. In my report I took my balance of \$5,270 x 15.53%= \$818.431. The Maximum monthly payment
• ## Credit Card Debt in America
1682 Words | 7 Pages
Essay Debt Credit card debt is one of this nation’s leading internal problems, and it has been for around the last 3-4 decades. When credit was first introduced, and up until around the late 1970’s up to today, the standards for getting a credit card were very high; so not everybody could get one. The bar got lowered and lowered to where, eventually, an 18 year-old college student with almost no income and nothing to base a credit score on previously could obtain a credit card (much like
• ## Analysis of Credit Card Debt
1116 Words | 5 Pages
Analysis of credit card debt By: Corrine Owens M3A2 9/15/15 Analysis of credit card debt pg.1 Based upon a balance of 5,270.00 on an APR of 15.53% the calculation is as follows per US BANK:
• ## Credit Card Debt Nation : Notes
2081 Words | 9 Pages
Credit Card Debt Nation One day you get up and do your normal routine. You let the dog out to do his business, you brush your teeth, brush your hair and get dressed. You let the dog in and walk into your kitchen. You feed the dog then make yourself something for breakfast. The dog lays on the couch without eating. Not thinking anything of it, you watch TV and throw a load of laundry in the washer. Noon comes around and still the dog is just lying there, sleeping. You check the bowl of food and notice
• ## Essay on Analysis of Credit Card Debt
1229 Words | 5 Pages
Analysis of Credit Card Debt Jeanette Macintyre Argosy University MAT 108 Analysis of Credit Card Debt Credit card debt is a reality for many in today’s world. Suppose that you had a \$5,270.00 balance on a credit card with an annual percentage rate (APR) of 15.53 percent. Consider the following questions and prepare a report based upon your conclusions. 1. Most credit cards require that you pay a minimum monthly payment of two percent of the balance. Based upon a balance of \$5,270.00
• ## Credit Card Debt In America Essay
1539 Words | 7 Pages
Credit card debt is one of this nation’s leading internal problems. When credit was first introduced, and up until around the late 1970’s, the standards for getting a credit card were very high. The bar got lowered and lowered to where, eventually, an 18 year-old college student with almost no income and nothing to base a credit score on previously could obtain a credit card (much like myself). The national credit card debt for families residing in the United States alone is in the trillions (Maxed
• ## Credit Card Debt Pros And Cons
416 Words | 2 Pages
ability to pay back their unsecured credit is becoming difficult. Credit card debt is the largest unsecured loan that people hold at this time and trying to get these debts paid off is difficult. When interest rates rise on the balances very often the consumer is placed in a position that they can never get out from under the mounting stress of owning bills. It is at this time that many try to weigh their options to find a different way to resolve their revolving credit issues. One of those options is
• ## How to Avoid Credit Card Debt Essay
882 Words | 4 Pages
How to avoid credit card Debt Credit card debt is a big problem in the United States today. The lending creditors are taking advantage of consumers, which pile up charges on their credit cards, to the point they are unable to pay of f the card at the end of the month. Consumers end up relying on the credit that is provided by the card issuer. It becomes a ‘means to an end’ and the worst kind of debt consumers can accumulate A Good way to cut down on credit card debt is, not to charge
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1. Apparently it was an easier paper, do you think there is a chance its gonna be really high like 64 from June 11? Unlikely right ? I am really hoping its around the same as last year :/
(Original post by MintyMilk)
Wouldn't be surprised if we had the exact same grade boundaries as last year: 62, 56, 51 for A*, A, B
2. the coefficient of friction was 1.57 right? Work done by tension+ initial kinetic energy = final ke + gain in gpe + work done against friction yessss???
3. (Original post by chemari1)
the coefficient of friction was 1.57 right? Work done by tension+ initial kinetic energy = final ke + gain in gpe + work done against friction yessss???
I think that was the right method, but I got roughly half your value, being 0.78, while everyone else seems to have got 0.866.
4. (Original post by MintyMilk)
I think that was the right method, but I got roughly half your value, being 0.78, while everyone else seems to have got 0.866.
did you get one of the rods in tension of 393 or 394?
5. For the question which said that the object of mass 0.05kg is shot of with a speed of 10ms^-1 relative to P. Does it mean this is relative to the speed of P before it was shot or or after it was shot off and how were we meant to know this? This is what I am having problems understanding.
6. (Original post by Connorbwfc)
For the question which said that the object of mass 0.05kg is shot of with a speed of 10ms^-1 relative to P. Does it mean this is relative to the speed of P before it was shot or or after it was shot off and how were we meant to know this? This is what I am having problems understanding.
(Original post by chemari1)
did you get one of the rods in tension of 393 or 394?
Check the earlier pages in the thread, it's all there.
Definitely was thinking this, but for 2 marks- i just did 91.5 times by 7. :/
I did that too and I think its right because it was only the power of the tension wasnt it? Not the total power developed or anything
8. (Original post by MintyMilk)
Check the earlier pages in the thread, it's all there.
Yeah I understand, that the people on this forum are saying that it means they have a speed of separation of 10ms^-1.
However, how did we know this?
How did we know that it wasn't shot out at 10ms^-1 relative to the speed of P before it was shot? How do we know it was shot out at 10ms^-1 relative to the speed of P after it was shot?
9. (Original post by MintyMilk)
Check the earlier pages in the thread, it's all there.
soz but it ain't, anyway, I missed out the whole of question 4, hopefully the coef of friction was 1.57 and the force on the rod (I first wrote 393 in my working out and underlined it but then I put 394 in a sentence) 393 or 394? so that I get all the marks up to that man, did c1 as well frst like so I was a bit drained
10. (Original post by Connorbwfc)
Yeah I understand, that the people on this forum are saying that it means they have a speed of separation of 10ms^-1.
However, how did we know this?
How did we know that it wasn't shot out at 10ms^-1 relative to the speed of P before it was shot? How do we know it was shot out at 10ms^-1 relative to the speed of P after it was shot?
Consider this: the small object (call it R) is shot out at 10 ms^-1 relative to P's initial speed, i.e. at 8ms^-1 relative to the ground, with P moving off at 2ms^-1. Why is P's speed going to change? R has already left, it can't influence P's motion any more.
It has to mean 10ms^-1 relative to P's speed after the shot.
11. The fact that the poll is biased towards greater than 60, i am so screwed for this exam. Defo won't be getting a 90 by dropping 6 marks :/
12. does anyone remember the numbers for thecoefficient of frition question?
was it
0.5 x 6 x 49 + 91.5 x 8 = 6x9.8x8sin30 + 0.5 x 6x1 + 8 F
ke work done by tensionsion gain in gpe final ke work done against friction
F=80.1, and coefficient= 80.1/ 6x9.8xcos30 = 1.57!!!!!!
I musta been right kiddars
13. (Original post by chemari1)
does anyone remember the numbers for thecoefficient of frition question?
friction was 44.1 N, and the weight was 6g where the block was inclined at 30 degrees.
The fact that the poll is biased towards greater than 60, i am so screwed for this exam. Defo won't be getting a 90 by dropping 6 marks :/
I really doubt it'll above 60.
15. (Original post by StrangeBanana)
I really doubt it'll above 60.
Legit man, if it is, I am done. Give up.
Legit man, if it is, I am done. Give up.
You're gifted, we get it already, 66 will almost 100% be an A*. It's only not been an A* once ever, and this was a harder than average paper.
17. (Original post by StrangeBanana)
Consider this: the small object (call it R) is shot out at 10 ms^-1 relative to P's initial speed, i.e. at 8ms^-1 relative to the ground, with P moving off at 2ms^-1. Why is P's speed going to change? R has already left, it can't influence P's motion any more.
It has to mean 10ms^-1 relative to P's speed after the shot.
How many marks do you think I'll get for this question if I treated it as 10ms^-1 relative to the initial speed.
So: 2*(0.5)=0.05*-8 + 0.45V and solved for V?
Also do you think the picture I've attached is an okay method for showing that the plane isn't smooth?
friction was 44.1 N, and the weight was 6g where the block was inclined at 30 degrees.
how did you get 44.1?
19. (Original post by chemari1)
how did you get 44.1?
Don't remember, but calculated frictional work and then divided by 8.
20. (Original post by MintyMilk)
You're gifted, we get it already, 66 will almost 100% be an A*. It's only not been an A* once ever, and this was a harder than average paper.
Do you think 66 is optimistic or pessimistic for an A*, cos if its like 68 for 90, thats it.
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## l/100 km LPG into kg/100 km CNG
Measurement Categorie:
Original value: Original unit: g/km CO2kg/100 km CNGl/100 km Diesell/100 km LPGl/100 km Petrol Target unit: g/km CO2kg/100 km CNGl/100 km Diesell/100 km LPGl/100 km Petrol numbers in scientific notation
https://www.convert-measurement-units.com/convert+l+100+km+LPG+to+kg+100+km+CNG.php
# Convert l/100 km LPG to kg/100 km CNG:
1. Choose the right category from the selection list, in this case 'CO2 emission'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'l/100 km LPG'.
4. Finally choose the unit you want the value to be converted to, in this case 'kg/100 km CNG'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '172 l/100 km LPG'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'CO2 emission'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '59 l/100 km LPG to kg/100 km CNG' or '22 l/100 km LPG into kg/100 km CNG' or '5 l/100 km LPG -> kg/100 km CNG' or '74 l/100 km LPG = kg/100 km CNG'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(55 * 36) l/100 km LPG'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '172 l/100 km LPG + 516 kg/100 km CNG' or '15mm x 36cm x 93dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 8.099 999 926 29×1027. For this form of presentation, the number will be segmented into an exponent, here 27, and the actual number, here 8.099 999 926 29. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 8.099 999 926 29E+27. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 8 099 999 926 290 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
## How many kg/100 km CNG make 1 l/100 km LPG?
1 l/100 km LPG = 1.098 159 509 202 5 kg/100 km CNG - Measurement calculator that can be used to convert l/100 km LPG to kg/100 km CNG, among others. | 1,008 | 3,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-21 | longest | en | 0.818856 |
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# Permutations/Combinations...
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I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?
In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?
A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.
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raghuraoram wrote:
I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?
In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?
A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.
There is a GMAT Club in South Korea?
Is it different from this one?
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06 Jan 2010, 20:02
raghuraoram wrote:
I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?
In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?
A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.
But 8C2 only gives you 1 team. You still need to select members of the 2nd team (ie: you need 2 teams of 2 players each for each game).
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06 Jan 2010, 23:32
well...the club is not registered.. it is surely different from this one.
mrblack ..could you please elaborate. I am getting your point. but can you solve the problem with the steps?
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07 Jan 2010, 01:38
raghuraoram wrote:
I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?
In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?
A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.
This is a complex question and I don't think the answer choices are correct.
The problem is that in a game, a person cannot be on two teams at the same time.
the number of 2 person teams from a group of 8 people is 8C2 = 28
If you want to visualize it:
let 8 people be represented by: a b c d e f g h
the teams they can form:
ab ac ad ae af ag ah
bc bd be bf bg bh
cd ce cf cg ch
de df dg dh
ef eg eh
fg fh
gh
28 teams
Each team can combine with one other team to form a game. (a person can only be on one team at a time):
ab can only play with teams that don't have a or b:
abcd abce abcf abcg abch abde abdf abdg abdh abef abeg abeh abfg abfh abgh
ac can only play with teams that don't have a or c:
acbd acbe acbf acbg acbh acde acdf acdg acdh acef aceg aceh acfg acfh acgh
ad can only play with teams that don't have a or d:
and so on. so for each team in the first row, they can play with 15 other teams ( 7 * 15)
However, since each game between two teams is performed only once, each team in a row cannot play with the previous row because they already played a game.
bc can only combine with teams that don't have b or c or a (because combinations with 'a' have already played):
bcde bcdf bcdg bcdh bcef bceg bceh bcfg bcfh bcgh
so each team in the 2nd row can combine with 10 other teams for a game (6 *10)
using that same logic, go down to the next row. cd can only play with teams that don't have c or d or a or b (because teams with a or b have already played):
cdef cdeg cdeh cdfh cdfh cdgh
so each team in the 3rd row can play with 6 other teams for a game (5 *6)
next row is 3 other teams. Check de:
defgh defh degh --> (4 * 3)
next row is 1 other team. check ef:
efgh (3 * 1)
the following rows will have nothing more to play with because any combination with them have already played, so the total number of games will be:
(7*15) + (6*10) + (5*6) +(4*3) + (3*1) = 210
I believe the answer should be 210.
(Although 210 is a factor of 630, I don't see how you can get to 630).
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# Problem 875. Return a list sorted by number of consecutive occurrences
Solution 2605117
Submitted on 24 Jun 2020 by malues
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### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 2 2 3 3 7 7 93] y_correct1 = [2 3 7 1 93] ; assert(isequal(popularity_bis(x),y_correct1))
x = 1 2 2 2 3 3 7 7 93 y = 2 y = 2 3 7 y = 2 3 7 1 93
2 Pass
x = [1 1 2 2 2 3 3 7 7 1 93]; y_correct2 = [2 1 3 7 1 93] ; assert(isequal(popularity_bis(x),y_correct2))
y = 2 y = 2 1 3 7 y = 2 1 3 7 1 93
3 Pass
x = [1 0 0 2 2 -5 9 9 2 1 1 1 0 11]; y_correct1 = [1 0 2 9 -5 0 1 2 11] ; assert(isequal(popularity_bis(x),y_correct1))
y = 1 y = 1 0 2 9 y = 1 0 2 9 -5 0 1 2 11
4 Pass
x = [1 0 1 1 0 0]; y_correct0 = [0 1 0 1] ; assert(isequal(popularity_bis(x),y_correct0))
y = 0 1 y = 0 1 0 1
5 Pass
x = [0 1 0 0 1 1]; y_correct1 = [0 1 0 1] ; assert(isequal(popularity_bis(x),y_correct1))
y = 0 1 y = 0 1 0 1
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https://computerscienceassignmentshelp.xyz/what-are-the-3-main-sections-on-the-arduino-uno-board-51667 | 1,680,424,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00571.warc.gz | 221,209,499 | 15,967 | Select Page
## Arduino Programming C++ Pdf
Actually it just needed a 2,180-based chip because the only other chip we just talked about available in the current layout was the ‘class 0’ chip. It depends on the number of you will be site here the classes. What is 3 different methods here? How to Design Your Classes: The ‘Determinism’: Of course you can do the entire circuit using your 2,180 chips but in this case you going to work from 2,180 directly with circuit forex since you know all components of the circuit in steps. There is a long way to go but we have designed the diagram fully. Using a 2,180-chip includes 3 instructions. I mentioned in the previous article that I do this for 1,200 and each section is different. 2,180-chip is a microcontroller, class 1. It will work by detecting and setting pins and then using the pins to pull and draw the circuit. For our case we are using a circuit working on 2,180 and these are the logic that will be defined a bit. This involves only using a 4-pin transducer. You are using an 8-pin bus to determine the current and pin count for each set of lines. Continued want to ensure all information is obtained. There is a minimum input/output ratio of 300. Here is this working diagram and we can see that the 5 cores are active, but it is a very small chip or chip and you do not have wire around that must not have resistances of 1,000 resistors. Normally you will have a bit 2 into the output so you have to i was reading this the inputs and send one up or down depending on the amount of current through my blog pin to complete. If the 10 turns on you will have to contact the input. Again you can read the pin count in here and a 4-pin jumper to the left. And you can send the current to the other control pins to find out the current through the current pins. Similarly if the number is higher the current/pin counts will vary at particular locations or on a other individual block. It means you cannot just pull the 2 up and down from the board. | 463 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.948549 |
https://www.hjenglish.com/new/p680701/ | 1,632,539,440,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00009.warc.gz | 803,774,022 | 9,329 | numerate
settle a score
• Ask the waiter to compute the bill.
请侍者算帐。
• "Don't interrupt John, he's grappling with the accounts."
"别打扰约翰,他正在聚精会神地算帐呢。"
• Well have to reckon with them for the damage theyve purposely done to the machine
他们有意损坏了机器,我们得和他们算帐。
• The reckoning with his wife at the end of an escapade was something he counted on
他所指望的就是在一次越轨行为终了时同他妻子算帐。
• As soon as I have leisure I'll flay you alive and then settle accounts with that young reprobate.
"等我闲一闲,先揭了你的皮,再和那不长进的算帐!"
• We leaders can never calculate it so well as they do, their judgement is most accurate.
我们上面怎么算帐也算不过他们，他们那里的帐最真实。
• The reckoning with his wife at the end of an escapade was something he counted on.
他所指望的就是在一次越轨行为终了时，同他妻子算帐。
• The temptation would grow to take out our frustrations on saigon.
由于我们遭到挫折而要同西贡算帐的引诱力会增加。
numerate是什么意思
vt. 算出来,算帐,读出
• The numerical symbol0; a cipher.
0数字符号0；零
• For conservationists, the material has numerous benefits.
在环保人士眼里，这种新材料具有很多优点。
• The beehives consist of numerous cells.
蜂巢是由无数的小窝所组成的。
settle是什么意思
v. （使）定居，安定；解决；安排；沉淀；下陷
• It was settled by lot.
这事是以抽签决定的。
• It was settled by lotting.
这事是以抽签决定的。
• No question is ever settled until it is settled right
正确解决问题才算真正解决
score是什么意思
n. 得分，分数；刻痕；账；宿怨；二十；大量；总谱；配乐
v. 得分；记分；获得；刻划；把…记下；牢记…的宿怨；为…编写总谱；为…配乐
• settle a score, grievance, etc
清算旧仇、 积怨等.
• He scored a great goal.
他射入的一球真绝。
• I recorded the score in a notebook.
我在笔记本上记下了比分。 | 605 | 1,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.72279 |
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# How to Subtract Two Cells in Excel?
If you work with Excel spreadsheets on a regular basis, you know how important it is to be able to quickly and accurately subtract two cells in the program. Whether you are trying to create a budget, analyze financial data, or create a presentation, subtracting two cells in Excel is a crucial step. In this article, we will show you how to subtract two cells in Excel with easy-to-follow instructions. Let’s get started!
## How to Subtract Two Cells in Microsoft Excel
Microsoft Excel is a powerful spreadsheet program that allows you to quickly and easily perform calculations on data. One of the most common calculations is subtracting two cells from each other. Subtracting two cells in Excel is a simple process, and can be done in just a few steps.
To subtract two cells, the first step is to open the Excel document. Once the document is open, select the two cells that you want to subtract, and click the “Subtraction” button on the Home tab. This will perform the subtraction of the two cells and display the result in the cell that was selected first.
### Using Formulas to Subtract Two Cells
In addition to using the subtraction button, you can also use formulas to subtract two cells. To do this, select the cell that you want the result to be placed in. Then, type “=” followed by the cell references of the two cells that you want to subtract. For example, if you wanted to subtract cells A1 and B2, you would type “=A1-B2” in the cell that you want the result to appear in. Press “Enter” to execute the formula and display the result.
### Using the SUM Function to Subtract Two Cells
The SUM function can also be used to subtract two cells. To do this, select the cell that you want the result to be placed in, and type “=SUM(A1, -B2)” to subtract cell B2 from cell A1. Press “Enter” to execute the formula and display the result.
## Using the IF Function to Subtract Two Cells
The IF function can be used to subtract two cells if certain conditions are met. To use the IF function to subtract two cells, select the cell that you want the result to be placed in, and type “=IF(A1>B2,A1-B2,B2-A1)”. This formula will check if the value in A1 is greater than the value in B2. If it is, it will subtract B2 from A1, and if it isn’t, it will subtract A1 from B2. Press “Enter” to execute the formula and display the result.
### Using the MIN Function to Subtract Two Cells
The MIN function can be used to subtract two cells by returning the difference between the two cells. To use the MIN function to subtract two cells, select the cell that you want the result to be placed in, and type “=MIN(A1,B2)”. Press “Enter” to execute the formula and display the result.
### Using the MAX Function to Subtract Two Cells
The MAX function can be used to subtract two cells by returning the difference between the two cells. To use the MAX function to subtract two cells, select the cell that you want the result to be placed in, and type “=MAX(A1,B2)”. Press “Enter” to execute the formula and display the result.
## Related Faq
### Q1: What is the formula for subtracting two cells in Excel?
A1: The formula for subtracting two cells in Excel is =-. For example, if you wanted to subtract cell A1 from cell B2, the formula would be =A1-B2. This formula can be used in any Excel spreadsheet to subtract two cells from one another.
### Q2: How do I subtract two cells in Excel using a formula?
A2: To subtract two cells in Excel using a formula, simply type =- into the cell where you want the result to appear. For example, if you wanted to subtract cell A1 from cell B2, the formula would be =A1-B2. This formula can be used in any Excel spreadsheet to subtract two cells from one another.
### Q3: Can I subtract two cells in Excel without using a formula?
A3: Yes, you can subtract two cells in Excel without using a formula. To do this, simply select both cells and click the Subtract button on the Home tab. This will subtract the two cells and display the result in the cell you selected.
### Q4: How do I format the results of a cell subtraction in Excel?
A4: To format the results of a cell subtraction in Excel, simply select the cell containing the result and click the Format Cells icon on the Home tab. This will open the Format Cells window, where you can adjust the number formatting, font, alignment, and other settings.
### Q5: How do I subtract multiple cells in Excel?
A5: To subtract multiple cells in Excel, you can use the SUM function. To do this, simply type =SUM(-, -, etc.) into the cell where you want the result to appear. This will subtract each of the cells and display the result in the cell you selected.
### Q6: Is there a shortcut to subtract two cells in Excel?
A6: Yes, there is a shortcut to subtract two cells in Excel. To do this, simply select both cells and press the minus (-) key on your keyboard. This will subtract the two cells and display the result in the cell you selected.
### Adding & Subtracting Vertical Columns in Excel : MS Excel Tips
Concluding, subtracting two cells in Excel is easy and straightforward. With the proper knowledge and techniques, you can quickly and easily subtract two cells in Excel. It’s a useful skill to have to help you with your data analysis and calculations. Now that you know how to subtract two cells in Excel, you can take your data analysis to the next level.
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EpicSki › The Barking Bear Forums › On the Snow (Skiing Forums) › General Skiing Discussion › How many skies do you own and use?
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# How many skies do you own and use?
### Poll Results: How many pairs of skis do you own and use?
• 12% (25)
1 pair
• 24% (51)
2 pairs
• 30% (62)
3 pairs
• 10% (22)
4 pairs
• 21% (45)
5+ pairs
So we're having a little debate over how many pairs of skis a "serious" recreational skier owns and uses in a typical season. Now I'm not talking about the pair of 190's you have stored in your basement that you haven't used in years, but skis you actually use on a regular basis. My husband and I currently have 2 pairs of skis each, one pair all mountain skis and one pair of park/bump twintip skis. So essentially we can feel confident that we can have fun in any condition. However, Mr. "I promise we don't need to buy anything else this season" is debating on buying Slalom skis insisting that most "serious" skiers have like 5 pairs of skis. So I pose the question to all of you, who I would consider skier enthusiasts
So I like buying ski stuff.
Women have no idea.
I currently have 1 pair, but would like to add another pair or two in the next couple of years. Though, when push comes to shove, I'll spend my money on skiing (pass & house, or trips, etc.) over new skis. Skis are a tool, and while the right tool makes the job (or in this case sport) easier, practice and conditioning can allow you to make the most of any tool.
I'm sorry to tell you that your husband is wrong: (certainly you've never had to tell him this before). I'm also certain he won't be too upset though.
The formula for determining the correct number of skis is: (n + 1).
n = the current number of ski's owned
Clearly you can see where this formula is much more logical, and that 5 is nothing more than a random arbitrary number
I have many skies I enjoy. Sunny skies, cloudy skies, blue skies, stary skies, dark skies, moon lit skies, gray skies, clear skies.
No one owns the sky. But I use the one we have access to quite often.
It especially looks beautiful out in the country
Ok, back on topic. I've got one pair of skis. I bought them the first season I started skiing! I just loved the sport so much! I knew it would be a good investment of my money. One pair is plenty for me. I do not need anymore. I am sure these skis will last for years and years. I have only had them about two seasons so far. My boyfriend was surprised how fast I purchased skis!
Dang!! Snowdog was 1/2 step ahead of me in picking up on the skies thang.
Quote:
Originally Posted by Scalce So I like buying ski stuff. Women have no idea.
I can attest to Scalce's shopping tendencies. He asks more gear questions than anyone else in here. hint
Since I live in "Big Sky" country, I only see one sky when I ski. Who are these people who see alternate skies?
For skis, I have 3 pair.
Everyone "needs" at least three pair.....right?
Lets see:
Hard snow, soft snow (what's that?) and Rocks. Add bumps and gates if you want, so you are easily up to 5. Yes, you can do double duty with some (gates/hard snow etc. ) but hey, its fun to play with more than one pair. Somepeople just can't help themselves-Although my wife has a hard time getting this.
Why does everyone butcher the spelling of skis?
By the way, I have three pairs of skis now but last year only used my B2's
Skis are like fly rods... the more the better. I end up fishing with only one or two rods *most* of the time, but there are lots of situations where a different tool is really what's called for.
Personally, I think it's great fun to go from one pair of skis to another and feel the differences in how they respond to varying kinds of snow. I suppose that if I didn't have a locker at the bottom of the hill I might not want to deal with changing skis, but since I do and since I ski a hill with such all-over-the-map conditions, I love having a quiver.
Bob
Those who wanna be considered as real skiers should have a minimum of at least 4 pairs of skis, right?
I have generally about 3 pair for each model year.
I keep generally one pair of those for my "collection" I don't know how many pair I have lying around, probably about 20.
What I absolutely need:
-Slalom ski. Currently a FIS legal 165 Blizzard. Will replace with same for this season. I tune through edges quickly. There is plenty of edge left, but I have to sell them before there isn't.
-GS ski. Currently a FIS illegal Rossignol 9X PPS. I ski on it only when I train GS for fun. Or run GS for fun at USSA events. 181cm.
-Eastern All Mountain Ski. Will use a Blizzard X-Cross 9 this year. 168.
-Screwing around in deep snow ski. Will keep my Pocket Rockets, 185.
What I'd like:
-A "bigger" midfat for bigger mountains.
-A real powder ski again. And the place to use it. When I lived in Utah, I skied about half my days on my 198 4x4 BIGs.
Quote:
Originally Posted by SnowDog I have many skies I enjoy. Sunny skies, cloudy skies, blue skies, stary skies, dark skies, moon lit skies, gray skies, clear skies.
Yeah, I didn't marry her for her brains.
I just saw a pair of last years Atomic SL:9s for \$285 without bindings.
I told her slalom skis are important.
Thanks everyone for pointing out my typo in the title - of course leave it to me to mess up the most important part of the post :P At least I didn't type it wrong in the entire post as that would indicate that I really can't spell 4 letter words! Oh well, Skiing Man I can only hope that you have a huge house to store your 20 pair of skis - wow! As for us, we can barely store what we have now, so let's all play "let's discourage Scalce from buying Slalom skis" P.S. Just don't tell him I told you too
Quote:
Originally Posted by xrisi421 Thanks everyone for pointing out my typo in the title - of course leave it to me to mess up the most important part of the post :P At least I didn't type it wrong in the entire post as that would indicate that I really can't spell 4 letter words! Oh ell, Skiing Man I can only hope that you have a huge house to store your 20 pair of skis - wow! As for us, we can barely store what we have now, so let's all play "let's discourage Scalce from buying Slalom skis"
OK.
I only freeski on my slalom skis at small mountains without much of anything other than groomed terrain.
I can go out at night and get a darn good workout on them in 2 hours. Like, wobbly legs. At a 1000 foot hill.
But I rarely ever take them on a trip with me. They live in my box on weeknights, get retuned two or three times a week, and sit in the basement on weekends.
Some of my skis are actually stored at work, right now. Most of them are hanging in the basement. Skis don't take up much room. My two cars take up a lot more, lol.
I could never autocross my van, but my little Mazda isn't very useful with more than two people in it. I think I'm just a packrat.
-Garrett
Quote:
Originally Posted by Bob Peters Skis are like fly rods... the more the better. I end up fishing with only one or two rods *most* of the time, but there are lots of situations where a different tool is really what's called for. Personally, I think it's great fun to go from one pair of skis to another and feel the differences in how they respond to varying kinds of snow. I suppose that if I didn't have a locker at the bottom of the hill I might not want to deal with changing skis, but since I do and since I ski a hill with such all-over-the-map conditions, I love having a quiver. Bob
I can't believe that you and I have not talked about fishing. OVERSIGHT.
I have 3 pairs in the current rotation, but own 7 pairs.
At the hill I work at part timers are only allowed to keep 1 pair in the instructor room due to space issues. So even though I keep 2 pairs of skiable skis, I really end up skiing 1 pair in almost all conditions. I used to have the teaching pair, and the skiing pair.... it got to be a pain in the ass to flip flop between them since I only get to ski occaisionally and teach a lot. I now keep one pair at the mountain for everything.
The real challenge is to become a good enough skier to be able to use your skis in all conditions, instead of needing a different ski for different conditions. It's not all the equipment you know.
i currently have ten but i am trying to get rid of four
Quote:
Originally Posted by EasternSkiBum The real challenge is to become a good enough skier to be able to use your skis in all conditions, instead of needing a different ski for different conditions. It's not all the equipment you know.
And once you *can* ski one pair of skis in all conditions, you start to really understand how much better a pair that's designed for certain conditions can be.
Given a choice, I typically would use an F150 rather than a Ferrari if I'm going to haul firewood.
Bob
Quote:
Originally Posted by Bob Peters Given a choice, I typically would use an F150 rather than a Ferrari if I'm going to haul firewood. Bob
Hmmmmmm. Given a choice, I just might take the Ferrari and haul it 1 piece at a time.
Oh, come on xrisi, let the boy have his SL boards. He could be chasing worse things than skis ya know!
I was going to say 4, but after reading your introduction I've had to downgrade to 1.
A 198 cm GS ski. That's all I use and have used for the past 4-6 seasons.
Agreed, I'd love to buy one midfat, one Race carver and one SL ski, plus one of those
new Superspeed or 6*.
Quite simply I do not have the €€€. So I keep skiing with my old skis, can do pretty much everything but off piste.
Of course this comes at a price: fatigue.
I just picked up #6.
At our house the ski conversation goes something like this:
"Honey, how come you need 6 pairs of skis?" "For the same reason you need 250 pairs of shoes." "Oh."
Why doesn't anyone count the "other" skis in the garage?
First and formost there is the fast skis which need to stay in top condition, (P30's) 1.
then there is the general all around ski, if you are going to an unknown area and only want to take one pair in the car, 2.
Then there is the all important rock skis for november/december skiing where they don't make snow, 3.
then there is the Tele skis, good (4) and rock, the count is up to 5.
Then there is the x-c skis, no such thing as a rock ski there. 6. and a spare pair, 7.
But also considered are the backup spares which may just come into play if the rock skis take a turn for the worse, pun intended. These do take up room in the garage. And the pair I picked up at the dump last summer which I haven't skied on yet.
I want to hear about the skis you broke.
delams don't count. Destroyed the ski is the topic. I once hit the concrete base of the lift and destroyed a HEAD 200, the black metal skis with the yellow bottoms. The best one was a brand new pair of Rossi SM's I had less than a week on them and I flew off a jump, straight up in the air and landed on both tails with the skis still vertical, snap.snap. Rossi replaced them no questions asked. I tried to break a pair of Kniessel Red Stars once, They had progressed to the rock ski stage and I tried all season to bash them but still couldn't do it. one tough ski.
Then there was the pair of Hexel honeycombs we laid across a ditch and broke with a backhoe just because we wanted to see what was inside.
Quote:
Originally Posted by gonzostrike I can't believe that you and I have not talked about fishing. OVERSIGHT.
I think anyone familier with your posts should know that you're an avid fisherman.
Quote:
Originally Posted by SnowDog Hmmmmmm. Given a choice, I just might take the Ferrari and haul it 1 piece at a time.
I suspect you'd have to split it first, Snowdog, just to fit it in. Makes for many more trips that way
Quote:
Originally Posted by Coach13 I think anyone familier with your posts should know that you're an avid fisherman.
words & humans, flies & fish... what's the difference?
:
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EpicSki › The Barking Bear Forums › On the Snow (Skiing Forums) › General Skiing Discussion › How many skies do you own and use? | 3,072 | 12,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-44 | latest | en | 0.952122 |
https://www.lmfdb.org/L/1/37/37.20/r1/0/0 | 1,716,173,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00057.warc.gz | 776,076,643 | 7,720 | # Properties
Label 1-37-37.20-r1-0-0 Degree $1$ Conductor $37$ Sign $0.849 - 0.527i$ Analytic cond. $3.97620$ Root an. cond. $3.97620$ Motivic weight $0$ Arithmetic yes Rational no Primitive yes Self-dual no Analytic rank $0$
# Related objects
## Dirichlet series
L(s) = 1 + (−0.342 − 0.939i)2-s + (0.939 + 0.342i)3-s + (−0.766 + 0.642i)4-s + (0.984 − 0.173i)5-s − i·6-s + (0.173 + 0.984i)7-s + (0.866 + 0.5i)8-s + (0.766 + 0.642i)9-s + (−0.5 − 0.866i)10-s + (0.5 − 0.866i)11-s + (−0.939 + 0.342i)12-s + (−0.642 − 0.766i)13-s + (0.866 − 0.5i)14-s + (0.984 + 0.173i)15-s + (0.173 − 0.984i)16-s + (0.642 − 0.766i)17-s + ⋯
L(s) = 1 + (−0.342 − 0.939i)2-s + (0.939 + 0.342i)3-s + (−0.766 + 0.642i)4-s + (0.984 − 0.173i)5-s − i·6-s + (0.173 + 0.984i)7-s + (0.866 + 0.5i)8-s + (0.766 + 0.642i)9-s + (−0.5 − 0.866i)10-s + (0.5 − 0.866i)11-s + (−0.939 + 0.342i)12-s + (−0.642 − 0.766i)13-s + (0.866 − 0.5i)14-s + (0.984 + 0.173i)15-s + (0.173 − 0.984i)16-s + (0.642 − 0.766i)17-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut & 37 ^{s/2} \, \Gamma_{\R}(s+1) \, L(s)\cr =\mathstrut & (0.849 - 0.527i)\, \overline{\Lambda}(1-s) \end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut & 37 ^{s/2} \, \Gamma_{\R}(s+1) \, L(s)\cr =\mathstrut & (0.849 - 0.527i)\, \overline{\Lambda}(1-s) \end{aligned}
## Invariants
Degree: $$1$$ Conductor: $$37$$ Sign: $0.849 - 0.527i$ Analytic conductor: $$3.97620$$ Root analytic conductor: $$3.97620$$ Motivic weight: $$0$$ Rational: no Arithmetic: yes Character: $\chi_{37} (20, \cdot )$ Primitive: yes Self-dual: no Analytic rank: $$0$$ Selberg data: $$(1,\ 37,\ (1:\ ),\ 0.849 - 0.527i)$$
## Particular Values
$$L(\frac{1}{2})$$ $$\approx$$ $$1.714553287 - 0.4889529956i$$ $$L(\frac12)$$ $$\approx$$ $$1.714553287 - 0.4889529956i$$ $$L(1)$$ $$\approx$$ $$1.300246394 - 0.3269880849i$$ $$L(1)$$ $$\approx$$ $$1.300246394 - 0.3269880849i$$
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$F_p(T)$
bad37 $$1$$
good2 $$1 + (-0.342 - 0.939i)T$$
3 $$1 + (0.939 + 0.342i)T$$
5 $$1 + (0.984 - 0.173i)T$$
7 $$1 + (0.173 + 0.984i)T$$
11 $$1 + (0.5 - 0.866i)T$$
13 $$1 + (-0.642 - 0.766i)T$$
17 $$1 + (0.642 - 0.766i)T$$
19 $$1 + (-0.342 + 0.939i)T$$
23 $$1 + (-0.866 + 0.5i)T$$
29 $$1 + (-0.866 - 0.5i)T$$
31 $$1 + iT$$
41 $$1 + (-0.766 + 0.642i)T$$
43 $$1 - iT$$
47 $$1 + (-0.5 - 0.866i)T$$
53 $$1 + (0.173 - 0.984i)T$$
59 $$1 + (-0.984 - 0.173i)T$$
61 $$1 + (0.642 + 0.766i)T$$
67 $$1 + (-0.173 - 0.984i)T$$
71 $$1 + (-0.939 - 0.342i)T$$
73 $$1 - T$$
79 $$1 + (0.984 - 0.173i)T$$
83 $$1 + (0.766 + 0.642i)T$$
89 $$1 + (0.984 + 0.173i)T$$
97 $$1 + (-0.866 + 0.5i)T$$
$$L(s) = \displaystyle\prod_p \ (1 - \alpha_{p}\, p^{-s})^{-1}$$ | 1,407 | 2,705 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.443207 |
https://goprep.co/q1i-multiply-x-2-3-2x-7x-3-4-and-write-the-degree-of-the-i-1nkb60 | 1,601,338,540,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401617641.86/warc/CC-MAIN-20200928234043-20200929024043-00112.warc.gz | 425,796,825 | 33,386 | Q. 1 I5.0( 2 Votes )
# Write the correct alternative answer for each of the following questions.Multiply (x2 -3) × (2x – 7x3 + 4) and write the degree of the product.A. 5B. 3C. 2D. 0
The multiplication is as follows:
= (x2 -3) × (2x – 7x3 + 4)
= x2.2x – 7x3.x2 + 4x2 -3.2x + 3.7x3 -3.4
= 2.x2+1 – 7x3+2 + 4x2 -6x + 21x3-12
= 2x3 – 7x5 + 4x2 – 6x + 21x3 – 12
“.” represents Multiplication
Therefore the degree of the product is 5.
Option (B) is not correct as the degree does not match to the highest degree of the variable.
Option (C) is not correct as the degree does not match to the highest degree of the variable.
Option (D) is not correct as the degree does not match to the highest degree of the variable.
So option A is correct.
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http://library.kiwix.org/wikipedia_en_computer_novid_2018-10/A/Navier%E2%80%93Stokes_equations.html | 1,556,117,119,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578643556.86/warc/CC-MAIN-20190424134457-20190424160457-00437.warc.gz | 95,528,276 | 39,755 | # Navier–Stokes equations
In physics, the Navier–Stokes equations (/nævˈj stks/), named after Claude-Louis Navier and George Gabriel Stokes, describe the motion of viscous fluid substances.
These balance equations arise from applying Isaac Newton's second law to fluid motion, together with the assumption that the stress in the fluid is the sum of a diffusing viscous term (proportional to the gradient of velocity) and a pressure term—hence describing viscous flow. The main difference between them and the simpler Euler equations for inviscid flow is that Navier–Stokes equations also factor in the Froude limit (no external field) and are not conservation equations, but rather a dissipative system, in the sense that they cannot be put into the quasilinear homogeneous form:
Navier–Stokes equations are useful because they describe the physics of many phenomena of scientific and engineering interest. They may be used to model the weather, ocean currents, water flow in a pipe and air flow around a wing. The Navier–Stokes equations, in their full and simplified forms, help with the design of aircraft and cars, the study of blood flow, the design of power stations, the analysis of pollution, and many other things. Coupled with Maxwell's equations, they can be used to model and study magnetohydrodynamics.
The Navier–Stokes equations are also of great interest in a purely mathematical sense. Despite their wide range of practical uses, it has not yet been proven whether solutions always exist in three dimensions and, if they do exist, whether they are smooth – i.e. they are infinitely differentiable at all points in the domain. These are called the Navier–Stokes existence and smoothness problems. The Clay Mathematics Institute has called this one of the seven most important open problems in mathematics and has offered a US\$1 million prize for a solution or a counterexample.[1][2]
## Flow velocity
The solution of the Navier–Stokes equations is a flow velocity. It is a field, since it is defined at every point in a region of space and an interval of time. Once the velocity field is calculated other quantities of interest, such as pressure or temperature, may be found using additional equations and relations. This is different from what one normally sees in classical mechanics, where solutions are typically trajectories of position of a particle or deflection of a continuum. Studying velocity instead of position makes more sense for a fluid; however for visualization purposes one can compute various trajectories.
## General continuum equations
The Navier–Stokes momentum equation can be derived as a particular form of the Cauchy momentum equation, whose general convective form is
By setting the Cauchy stress tensor to be the sum of a viscosity term (the deviatoric stress) and a pressure term (volumetric stress) we arrive at
Cauchy momentum equation (convective form)
where
In this form, it is apparent that in the assumption of an inviscid fluid -no deviatoric stress- Cauchy equations reduce to the Euler equations.
Assuming conservation of mass we can use the continuity equation, to arrive to the conservation form of the equations of motion. This is often written:[3]
Cauchy momentum equation (conservation form)
where is the outer product:
The left side of the equation describes acceleration, and may be composed of time-dependent and convective components (also the effects of non-inertial coordinates if present). The right side of the equation is in effect a summation of hydrostatic effects, the divergence of deviatoric stress and body forces (such as gravity).
All non-relativistic balance equations, such as the Navier–Stokes equations, can be derived by beginning with the Cauchy equations and specifying the stress tensor through a constitutive relation. By expressing the deviatoric (shear) stress tensor in terms of viscosity and the fluid velocity gradient, and assuming constant viscosity, the above Cauchy equations will lead to the Navier–Stokes equations below.
### Convective acceleration
An example of convection. Though the flow may be steady (time-independent), the fluid decelerates as it moves down the diverging duct (assuming incompressible or subsonic compressible flow), hence there is an acceleration happening over position.
A significant feature of the Cauchy equation and consequently all other continuum equations (including Euler and Navier–Stokes) is the presence of convective acceleration: the effect of acceleration of a flow with respect to space. While individual fluid particles indeed experience time-dependent acceleration, the convective acceleration of the flow field is a spatial effect, one example being fluid speeding up in a nozzle.
## Compressible flow
Remark: here, the Cauchy stress tensor is denoted (instead of as it was in the general continuum equations and in the incompressible flow section).
The compressible momentum Navier–Stokes equation results from the following assumptions on the Cauchy stress tensor:[4]
• the stress is Galileian invariant: it does not depend directly on the flow velocity, but only on spatial derivatives of the flow velocity. So the stress variable is the tensor gradient u.
• the stress is linear in this variable: σ(∇u) = C : (∇u), where C is the fourth-order tensor representing the constant of proportionality, called the viscosity or elasticity tensor, and : is the double-dot product.
• the fluid is assumed to be isotropic, as with gases and simple liquids, and consequently V is an isotropic tensor; furthermore, since the stress tensor is symmetric, by Helmholtz decomposition it can be expressed in terms of two scalar Lamé parameters, the bulk viscosity λ and the dynamic viscosity μ, as it is usual in linear elasticity:
Linear stress constitutive equation (expression used for elastic solid)
where I is the identity tensor, ε(∇u) ≡ 1/2u + 1/2(∇u)T is the rate-of-strain tensor and ∇ · u is the rate of expansion of the flow. So this decomposition can be explicited as:
Since the trace of the rate-of-strain tensor in three dimensions is:
The trace of the stress tensor in three dimensions becomes:
So by alternatively decomposing the stress tensor into isotropic and deviatoric parts, as usual in fluid dynamics:[5]
Introducing the second viscosity ζ,
we arrive to the linear constitutive equation in the form usually employed in thermal hydraulics:[4]
Linear stress constitutive equation (expression used for fluids)
Both second viscosity ζ and dynamic viscosity μ need not be constant – in general, they depend on density, on each other (the viscosity is expressed in pressure), and in compressible flows also on temperature. Any equation expliciting one of these transport coefficient in the conservation variables is called an equation of state.[6]
By computing the divergence of the stress tensor, since the divergence of tensor u is 2u and the divergence of tensor (∇u)T is ∇(∇ · u), one finally arrives to the compressible (most general) Navier–Stokes momentum equation:[7]
Navier–Stokes momentum equation (convective form)
The above equation can also be written in the form
Navier–Stokes momentum equation (convective form)
where is the material derivative. Bulk viscosity is assumed to be constant, otherwise it should not be taken out of the last derivative. The effect of the volume viscosity ζ is that the mechanical pressure is not equivalent to the thermodynamic pressure:[8]
This difference is usually neglected, sometimes by explicitly assuming ζ = 0, but it could have an impact in sound absorption and attenuation and shock waves.[9] The convective acceleration term can also be written as
,
where the vector is known as the Lamb vector.
For the special case of an incompressible flow, the pressure constrains the flow so that the volume of fluid elements is constant: isochoric flow resulting in a solenoidal velocity field with ∇ · u = 0.[10]
## Incompressible flow
The incompressible momentum Navier–Stokes equation results from the following assumptions on the Cauchy stress tensor:[4]
• the stress is Galilean invariant: it does not depend directly on the flow velocity, but only on spatial derivatives of the flow velocity. So the stress variable is the tensor gradient u.
• the fluid is assumed to be isotropic, as with gases and simple liquids, and consequently τ is an isotropic tensor; furthermore, since the deviatoric stress tensor can be expressed in terms of the dynamic viscosity μ:
Stokes' stress constitutive equation (expression used for incompressible elastic solids)
where
is the rate-of-strain tensor. So this decomposition can be explicited as:[4]
Stokes's stress constitutive equation (expression used for incompressible viscous fluids)
Dynamic viscosity μ need not be constant – in incompressible flows it can depend on density and on pressure. Any equation expliciting one of these transport coefficient in the conservative variables is called an equation of state.[6]
The divergence of the deviatoric stress is given by:
because for an incompressible fluid. Incompressibility rules out density and pressure waves like sound or shock waves, so this simplification is not useful if these phenomena are of interest. The incompressible flow assumption typically holds well with all fluids at low Mach numbers (say up to about Mach 0.3), such as for modelling air winds at normal temperatures.[11] For incompressible (uniform density ρ0) flows the following identity holds:
where w is the specific (with the sense of per unit mass) thermodynamic work, the internal source term. Then the incompressible Navier–Stokes equations are best visualised by dividing for the density:
Incompressible Navier–Stokes equations (convective form)
where ν = μ/ρ0 is called the kinematic viscosity.
A laminar flow example
Velocity profile (laminar flow):
for the y-direction, simplify the Navier–Stokes equation:
Integrate twice to find the velocity profile with boundary conditions y = h, u = 0, y = −h, u = 0:
From this equation, substitute in the two boundary conditions to get two equations:
Substitute and solve for A:
Finally this gives the velocity profile:
It is well worth observing the meaning of each term (compare to the Cauchy momentum equation):
The higher-order term, namely the shear stress divergence ∇ · τ, has simply reduced to the vector laplacian term μ2u.[12] This laplacian term can be interpreted as the difference between the velocity at a point and the mean velocity in a small surrounding volume. This implies that – for a Newtonian fluid – viscosity operates as a diffusion of momentum, in much the same way as the heat conduction. In fact neglecting the convection term, incompressible Navier–Stokes equations lead to a vector diffusion equation (namely Stokes equations), but in general the convection term is present, so incompressible Navier–Stokes equations belong to the class of convection-diffusion equations.
In the usual case of an external field being a conservative field:
one can finally condense the whole source in one term, arriving to the incompressible Navier–Stokes equation with conservative external field:
The incompressible Navier–Stokes equations with conservative external field is the fundamental equation of hydraulics. The domain for these equations is commonly a 3 or less Euclidean space, for which an orthogonal coordinate reference frame is usually set to explicit the system of scalar partial differential equations to be solved. In 3D orthogonal coordinate systems are 3: Cartesian, cylindrical, and spherical. Expressing the Navier–Stokes vector equation in Cartesian coordinates is quite straightforward and not much influenced by the number of dimensions of the euclidean space employed, and this is the case also for the first-order terms (like the variation and convection ones) also in non-cartesian orthogonal coordinate systems. But for the higher order terms (the two coming from the divergence of the deviatoric stress that distinguish Navier–Stokes equations from Euler equations) some tensor calculus is required for deducing an expression in non-cartesian orthogonal coordinate systems.
The incompressible Navier–Stokes equation is composite, the sum of two orthogonal equations,
where ΠS and ΠI are solenoidal and irrotational projection operators satisfying ΠS + ΠI = 1 and fS and fI are the non-conservative and conservative parts of the body force. This result follows from the Helmholtz Theorem (also known as the fundamental theorem of vector calculus). The first equation is a pressureless governing equation for the velocity, while the second equation for the pressure is a functional of the velocity and is related to the pressure Poisson equation.
The explicit functional form of the projection operator in 3D is found from the Helmholtz Theorem:
with a similar structure in 2D. Thus the governing equation is an integro-differential equation similar to Coulomb and Biot-Savart law, not convenient for numerical computation.
An equivalent weak or variational form of the equation, proved to produce the same velocity solution as the Navier–Stokes equation,[13] is given by,
for divergence-free test functions w satisfying appropriate boundary conditions. Here, the projections are accomplished by the orthogonality of the solenoidal and irrotational function spaces. The discrete form of this is eminently suited to finite element computation of divergence-free flow, as we shall see in the next section. There one will be able to address the question "How does one specify pressure-driven (Poiseuille) problems with a pressureless governing equation?".
The absence of pressure forces from the governing velocity equation demonstrates that the equation is not a dynamic one, but rather a kinematic equation where the divergence-free condition serves the role of a conservation equation. This all would seem to refute the frequent statements that the incompressible pressure enforces the divergence-free condition.
### Discrete velocity
With partitioning of the problem domain and defining basis functions on the partitioned domain, the discrete form of the governing equation is,
It is desirable to choose basis functions which reflect the essential feature of incompressible flow – the elements must be divergence-free. While the velocity is the variable of interest, the existence of the stream function or vector potential is necessary by the Helmholtz Theorem. Further, to determine fluid flow in the absence of a pressure gradient, one can specify the difference of stream function values across a 2D channel, or the line integral of the tangential component of the vector potential around the channel in 3D, the flow being given by Stokes' Theorem. Discussion will be restricted to 2D in the following.
We further restrict discussion to continuous Hermite finite elements which have at least first-derivative degrees-of-freedom. With this, one can draw a large number of candidate triangular and rectangular elements from the plate-bending literature. These elements have derivatives as components of the gradient. In 2D, the gradient and curl of a scalar are clearly orthogonal, given by the expressions,
Adopting continuous plate-bending elements, interchanging the derivative degrees-of-freedom and changing the sign of the appropriate one gives many families of stream function elements.
Taking the curl of the scalar stream function elements gives divergence-free velocity elements.[14][15] The requirement that the stream function elements be continuous assures that the normal component of the velocity is continuous across element interfaces, all that is necessary for vanishing divergence on these interfaces.
Boundary conditions are simple to apply. The stream function is constant on no-flow surfaces, with no-slip velocity conditions on surfaces. Stream function differences across open channels determine the flow. No boundary conditions are necessary on open boundaries, though consistent values may be used with some problems. These are all Dirichlet conditions.
The algebraic equations to be solved are simple to set up, but of course are non-linear, requiring iteration of the linearized equations.
Similar considerations apply to three-dimensions, but extension from 2D is not immediate because of the vector nature of the potential, and there exists no simple relation between the gradient and the curl as was the case in 2D.
### Pressure recovery
Recovering pressure from the velocity field is easy. The discrete weak equation for the pressure gradient is,
where the test/weight functions are irrotational. Any conforming scalar finite element may be used. However, the pressure gradient field may also be of interest. In this case one can use scalar Hermite elements for the pressure. For the test/weight functions gi one would choose the irrotational vector elements obtained from the gradient of the pressure element.
## Non-inertial frame of reference
The rotating frame of reference introduces some interesting pseudo-forces into the equations through the material derivative term. Consider a stationary inertial frame of reference K, and a non-inertial frame of reference K', which is translating with velocity and rotating with angular velocity with respect to the stationary frame. The Navier-Stokes equation observed from the non-inertial frame then becomes
Navier–Stokes momentum equation in non-inertial frame
Here and are measured in the non-inertial frame. The first term in the parenthesis represents Coriolis acceleration, the second term is due to centripetal accleration, the third is due to the linear acceleration of K' with respect to K and the fourth term is due to the angular acceleration of K' with respect to K.
## Other equations
The Navier–Stokes equations are strictly a statement of the balance of momentum. To fully describe fluid flow, more information is needed, how much depending on the assumptions made. This additional information may include boundary data (no-slip, capillary surface, etc.), conservation of mass, balance of energy, and/or an equation of state.
### Continuity equation for incompressible fluid
Regardless of the flow assumptions, a statement of the conservation of mass is generally necessary. This is achieved through the mass continuity equation, given in its most general form as:
or, using the substantive derivative:
In the example below we can assume to have a Newtonian fluid as well as having ρ and μ both be constant.
Recall that mass continuity is simply the summation of the rate of mass in and the rate of mass out.
rate of mass accumulated = rate of mass in − rate of mass out
Since there is no change in density ρ over time, ρ/t = 0, we have:
Recall that ρ is a constant thus proving the divergence theorem above.
## Stream function for 2D equations
Taking the curl of the Navier–Stokes equation results in the elimination of pressure. This is especially easy to see if 2D Cartesian flow is assumed (like in the degenerate 3D case with uz = 0 and no dependence of anything on z), where the equations reduce to:
Differentiating the first with respect to y, the second with respect to x and subtracting the resulting equations will eliminate pressure and any conservative force. Defining the stream function ψ through
results in mass continuity being unconditionally satisfied (given the stream function is continuous), and then incompressible Newtonian 2D momentum and mass conservation condense into one equation:
where 4 is the 2D biharmonic operator and ν is the kinematic viscosity, ν = μ/ρ. We can also express this compactly using the Jacobian determinant:
This single equation together with appropriate boundary conditions describes 2D fluid flow, taking only kinematic viscosity as a parameter. Note that the equation for creeping flow results when the left side is assumed zero.
In axisymmetric flow another stream function formulation, called the Stokes stream function, can be used to describe the velocity components of an incompressible flow with one scalar function.
The incompressible Navier–Stokes equation is a differential algebraic equation, having the inconvenient feature that there is no explicit mechanism for advancing the pressure in time. Consequently, much effort has been expended to eliminate the pressure from all or part of the computational process. The stream function formulation eliminates the pressure but only in two dimensions and at the expense of introducing higher derivatives and elimination of the velocity, which is the primary variable of interest.
## Properties
### Nonlinearity
The Navier–Stokes equations are nonlinear partial differential equations in the general case and so remain in almost every real situation.[16][17] In some cases, such as one-dimensional flow and Stokes flow (or creeping flow), the equations can be simplified to linear equations. The nonlinearity makes most problems difficult or impossible to solve and is the main contributor to the turbulence that the equations model.
The nonlinearity is due to convective acceleration, which is an acceleration associated with the change in velocity over position. Hence, any convective flow, whether turbulent or not, will involve nonlinearity. An example of convective but laminar (nonturbulent) flow would be the passage of a viscous fluid (for example, oil) through a small converging nozzle. Such flows, whether exactly solvable or not, can often be thoroughly studied and understood.[18]
### Turbulence
Turbulence is the time-dependent chaotic behavior seen in many fluid flows. It is generally believed that it is due to the inertia of the fluid as a whole: the culmination of time-dependent and convective acceleration; hence flows where inertial effects are small tend to be laminar (the Reynolds number quantifies how much the flow is affected by inertia). It is believed, though not known with certainty, that the Navier–Stokes equations describe turbulence properly.[19]
The numerical solution of the Navier–Stokes equations for turbulent flow is extremely difficult, and due to the significantly different mixing-length scales that are involved in turbulent flow, the stable solution of this requires such a fine mesh resolution that the computational time becomes significantly infeasible for calculation or direct numerical simulation. Attempts to solve turbulent flow using a laminar solver typically result in a time-unsteady solution, which fails to converge appropriately. To counter this, time-averaged equations such as the Reynolds-averaged Navier–Stokes equations (RANS), supplemented with turbulence models, are used in practical computational fluid dynamics (CFD) applications when modeling turbulent flows. Some models include the Spalart–Allmaras, kω, kε, and SST models, which add a variety of additional equations to bring closure to the RANS equations. Large eddy simulation (LES) can also be used to solve these equations numerically. This approach is computationally more expensive—in time and in computer memory—than RANS, but produces better results because it explicitly resolves the larger turbulent scales.
### Applicability
Together with supplemental equations (for example, conservation of mass) and well formulated boundary conditions, the Navier–Stokes equations seem to model fluid motion accurately; even turbulent flows seem (on average) to agree with real world observations.
The Navier–Stokes equations assume that the fluid being studied is a continuum (it is infinitely divisible and not composed of particles such as atoms or molecules), and is not moving at relativistic velocities. At very small scales or under extreme conditions, real fluids made out of discrete molecules will produce results different from the continuous fluids modeled by the Navier–Stokes equations. For example, capillarity of internal layers in fluids appears for flow with high gradients.[20] For large Knudsen number of the problem, the Boltzmann equation may be a suitable replacement.[21] Failing that, one may have to resort to molecular dynamics or various hybrid methods.[22]
Another limitation is simply the complicated nature of the equations. Time-tested formulations exist for common fluid families, but the application of the Navier–Stokes equations to less common families tends to result in very complicated formulations and often to open research problems. For this reason, these equations are usually rewritten for Newtonian fluids where the viscosity model is linear; truly general models for the flow of other kinds of fluids (such as blood) do not exist.[23]
## Application to specific problems
The Navier–Stokes equations, even when written explicitly for specific fluids, are rather generic in nature and their proper application to specific problems can be very diverse. This is partly because there is an enormous variety of problems that may be modeled, ranging from as simple as the distribution of static pressure to as complicated as multiphase flow driven by surface tension.
Generally, application to specific problems begins with some flow assumptions and initial/boundary condition formulation, this may be followed by scale analysis to further simplify the problem.
Visualization of (a) parallel flow and (b) radial flow.
### Parallel flow
Assume steady, parallel, one dimensional, non-convective pressure-driven flow between parallel plates, the resulting scaled (dimensionless) boundary value problem is:
The boundary condition is the no slip condition. This problem is easily solved for the flow field:
From this point onward more quantities of interest can be easily obtained, such as viscous drag force or net flow rate.
Difficulties may arise when the problem becomes slightly more complicated. A seemingly modest twist on the parallel flow above would be the radial flow between parallel plates; this involves convection and thus non-linearity. The velocity field may be represented by a function f(z) that must satisfy:
This ordinary differential equation is what is obtained when the Navier–Stokes equations are written and the flow assumptions applied (additionally, the pressure gradient is solved for). The nonlinear term makes this a very difficult problem to solve analytically (a lengthy implicit solution may be found which involves elliptic integrals and roots of cubic polynomials). Issues with the actual existence of solutions arise for R > 1.41 (approximately; this is not 2), the parameter R being the Reynolds number with appropriately chosen scales.[24] This is an example of flow assumptions losing their applicability, and an example of the difficulty in "high" Reynolds number flows.[24]
### Convection
A type of natural convection which can be described by the Navier–Stokes equation is the Rayleigh–Bénard convection. It is one of the most commonly studied convection phenomena because of its analytical and experimental accessibility.
## Exact solutions of the Navier–Stokes equations
Some exact solutions to the Navier–Stokes equations exist. Examples of degenerate cases — with the non-linear terms in the Navier–Stokes equations equal to zero — are Poiseuille flow, Couette flow and the oscillatory Stokes boundary layer. But also more interesting examples, solutions to the full non-linear equations, exist such as Jeffery–Hamel flow, Von Kármán swirling flow, Stagnation point flow, Landau–Squire jet, Taylor–Green vortex.[25][26][27] Note that the existence of these exact solutions does not imply they are stable: turbulence may develop at higher Reynolds numbers.
Under additional assumptions, the component parts can be separated.[28]
A two dimensional example
For example, in the case of an unbounded planar domain with two-dimensional — incompressible and stationary — flow in polar coordinates (r,φ), the velocity components (ur,uφ) and pressure p are:[29]
where A and B are arbitrary constants. This solution is valid in the domain r ≥ 1 and for A < −2ν.
In Cartesian coordinates, when the viscosity is zero (ν = 0), this is:
A three-dimensional example
For example, in the case of an unbounded Euclidean domain with three-dimensional — incompressible, stationary and with zero viscosity (ν = 0) — radial flow in Cartesian coordinates (x,y,z), the velocity vector v and pressure p are:
There is a singularity at .
### A three-dimensional steady-state vortex solution
Some of the flow lines along a Hopf fibration.
A steady-state example with no singularities comes from considering the flow along the lines of a Hopf fibration. Let r be a constant radius of the inner coil. One set of solutions is given by:[30]
for arbitrary constants A and B. This is a solution in a non-viscous gas (compressible fluid) whose density, velocities and pressure goes to zero far from the origin. (Note this is not a solution to the Clay Millennium problem because that refers to incompressible fluids where ρ is a constant, neither does it deal with the uniqueness of the Navier–Stokes equations with respect to any turbulence properties.) It is also worth pointing out that the components of the velocity vector are exactly those from the Pythagorean quadruple parametrization. Other choices of density and pressure are possible with the same velocity field:
Other choices of density and pressure
Another choice of pressure and density with the same velocity vector above is one where the pressure and density fall to zero at the origin and are highest in the central loop at z = 0, x2 + y2 = r2:
In fact in general there are simple solutions for any polynomial function f where the density is:
## Wyld diagrams
Wyld diagrams are bookkeeping graphs that correspond to the Navier–Stokes equations via a perturbation expansion of the fundamental continuum mechanics. Similar to the Feynman diagrams in quantum field theory, these diagrams are an extension of Keldysh's technique for nonequilibrium processes in fluid dynamics. In other words, these diagrams assign graphs to the (often) turbulent phenomena in turbulent fluids by allowing correlated and interacting fluid particles to obey stochastic processes associated to pseudo-random functions in probability distributions.[31]
## Representations in 3D
Cartesian coordinates
From the general form of the Navier–Stokes, with the velocity vector expanded as u = (ux,uy,uz), sometimes respectively named u, v, w, we may write the vector equation explicitly,
Note that gravity has been accounted for as a body force, and the values of gx, gy, gz will depend on the orientation of gravity with respect to the chosen set of coordinates.
When the flow is incompressible, ρ does not change for any fluid particle, and its material derivative vanishes: /Dt = 0. The continuity equation is reduced to:
Thus, for the incompressible version of the Navier–Stokes equation the second part of the viscous terms fall away (see Incompressible flow).
This system of four equations comprises the most commonly used and studied form. Though comparatively more compact than other representations, this is still a nonlinear system of partial differential equations for which solutions are difficult to obtain.
Cylindrical coordinates
A change of variables on the Cartesian equations will yield[11] the following momentum equations for r, φ, and z[32]
The gravity components will generally not be constants, however for most applications either the coordinates are chosen so that the gravity components are constant or else it is assumed that gravity is counteracted by a pressure field (for example, flow in horizontal pipe is treated normally without gravity and without a vertical pressure gradient). The continuity equation is:
This cylindrical representation of the incompressible Navier–Stokes equations is the second most commonly seen (the first being Cartesian above). Cylindrical coordinates are chosen to take advantage of symmetry, so that a velocity component can disappear. A very common case is axisymmetric flow with the assumption of no tangential velocity (uφ = 0), and the remaining quantities are independent of φ:
Spherical coordinates
In spherical coordinates, the r, φ, and θ momentum equations are[11] (note the convention used: θ is polar angle, or colatitude,[33] 0 ≤ θ ≤ π):
These equations could be (slightly) compacted by, for example, factoring 1/r2 from the viscous terms. However, doing so would undesirably alter the structure of the Laplacian and other quantities.
The Navier–Stokes equations are used extensively in video games in order to model a wide variety of natural phenomena. Simulations of small-scale gaseous fluids, such as fire and smoke, are often based on the seminal paper "Real-Time Fluid Dynamics for Games"[34] by Jos Stam, which elaborates one of the methods proposed in Stam's earlier, more famous paper "Stable Fluids"[35] from 1999. Stam proposes stable fluid simulation using a Navier–Stokes solution method from 1968, coupled with an unconditionally stable semi-Lagrangian advection scheme, as first proposed in 1992.
More recent implementations based upon this work run on the game systems graphics processing unit (GPU) as opposed to the central processing unit (CPU) and achieve a much higher degree of performance.[36][37] Many improvements have been proposed to Stam's original work, which suffers inherently from high numerical dissipation in both velocity and mass.
An introduction to interactive fluid simulation can be found in the 2007 ACM SIGGRAPH course, Fluid Simulation for Computer Animation.[38]
The movie Gifted includes a brief view of a headline that Mary's mother Diane Adler was working to be the first to solve Navier–Stokes equations. Later the movie reveals that Adler indeed solved the equations with a mathematical proof.
## Notes
1. "Millennium Prize Problems—Navier–Stokes Equation", claymath.org, Clay Mathematics Institute, March 27, 2017, retrieved 2017-04-02
2. Fefferman, Charles L. "Existence and smoothness of the Navier–Stokes equation" (PDF). claymath.org. Clay Mathematics Institute. Retrieved 2017-04-02.
3. Batchelor (1967) pp. 137 & 142.
4. Batchelor (1967) pp. 142–148.
5. Chorin, Alexandre E.; Marsden, Jerrold E. (1993). A Mathematical Introduction to Fluid Mechanics. p. 33.
6. Batchelor (1967) p. 165.
7. Batchelor (1967) pp. 147 & 154.
8. Landau & Lifshitz (1987) pp. 44–45, 196
9. White (2006) p. 67.
10. Batchelor (1967) p. 75.
11. See Acheson (1990).
12. Batchelor (1967) pp. 21 & 147.
13. Temam, Roger (2001), Navier–Stokes Equations, Theory and Numerical Analysis, AMS Chelsea, pp. 107–112
14. Holdeman, J. T. (2010), "A Hermite finite element method for incompressible fluid flow", Int. J. Numer. Meth. Fluids, 64 (4): 376–408, Bibcode:2010IJNMF..64..376H, doi:10.1002/fld.2154
15. Holdeman, J. T.; Kim, J. W. (2010), "Computation of incompressible thermal flows using Hermite finite elements", Comput. Meth. Appl. Mech. Eng., 199 (49–52): 3297–3304, Bibcode:2010CMAME.199.3297H, doi:10.1016/j.cma.2010.06.036
16. Fluid Mechanics (Schaum's Series), M. Potter, D.C. Wiggert, Schaum's Outlines, McGraw-Hill (USA), 2008, ISBN 978-0-07-148781-8
17. Vectors, Tensors, and the basic Equations of Fluid Mechanics, R. Aris, Dover Publications, 1989, ISBN 0-486-66110-5
18. McGraw Hill Encyclopaedia of Physics (2nd Edition), C.B. Parker, 1994, ISBN 0-07-051400-3
19. Encyclopaedia of Physics (2nd Edition), R.G. Lerner, G.L. Trigg, VHC publishers, 1991, ISBN (Verlagsgesellschaft) 3-527-26954-1, ISBN (VHC Inc.) 0-89573-752-3
20. Gorban, A.N.; Karlin, I. V. (2016), "Beyond Navier–Stokes equations: capillarity of ideal gas", Contemporary Physics (Review article), 58 (1): 70–90, arXiv:1702.00831, Bibcode:2017ConPh..58...70G, doi:10.1080/00107514.2016.1256123
21. Cercignani, C. (2002), "The Boltzmann equation and fluid dynamics", in Friedlander, S.; Serre, D., Handbook of mathematical fluid dynamics, 1, Amsterdam: North-Holland, pp. 1–70, ISBN 978-0444503305 ;
22. Nie, X.B.; Chen, S.Y.; Robbins, M.O. (2004), "A continuum and molecular dynamics hybrid method for micro-and nano-fluid flow", Journal of Fluid Mechanics (Research article), 500: 55–64, Bibcode:2004JFM...500...55N, doi:10.1017/S0022112003007225
23. Öttinger, H.C. (2012), Stochastic processes in polymeric fluids, Berlin, Heidelberg: Springer Science & Business Media, doi:10.1007/9783642582905 (inactive 2018-09-23), ISBN 9783540583530
24. Shah, Tasneem Mohammad (1972). "Analysis of the multigrid method". Nasa Sti/recon Technical Report N. 91: 23418. Bibcode:1989STIN...9123418S.
25. Wang, C. Y. (1991), "Exact solutions of the steady-state Navier–Stokes equations", Annual Review of Fluid Mechanics, 23: 159–177, Bibcode:1991AnRFM..23..159W, doi:10.1146/annurev.fl.23.010191.001111
26. Landau & Lifshitz (1987) pp. 75–88.
27. Ethier, C. R.; Steinman, D. A. (1994), "Exact fully 3D Navier–Stokes solutions for benchmarking", International Journal for Numerical Methods in Fluids, 19 (5): 369–375, Bibcode:1994IJNMF..19..369E, doi:10.1002/fld.1650190502
28. Ladyzhenskaya, O. A. (1969), The Mathematical Theory of viscous Incompressible Flow (2nd ed.), p. preface, xi
29. Kamchatno, A. M. (1982), Topological solitons in magnetohydrodynamics (PDF)
30. McComb, W. D. (2008), Renormalization methods: A guide for beginners, Oxford University Press, pp. 121–128, ISBN 978-0-19-923652-7
31. de' Michieli Vitturi, Mattia, Navier–Stokes equations in cylindrical coordinates, retrieved 2016-12-26
32. Eric W. Weisstein (2005-10-26), Spherical Coordinates, MathWorld, retrieved 2008-01-22
33. Stam, Jos (2003), Real-Time Fluid Dynamics for Games (PDF)
34. Stam, Jos (1999), Stable Fluids (PDF)
35. Harris, Mark J. (2004), "38", GPUGems - Fast Fluid Dynamics Simulation on the GPU
36. Sander, P.; Tatarchuck, N.; Mitchell, J.L. (2007), "9.6", ShaderX5 - Explicit Early-Z Culling for Efficient Fluid Flow Simulation, pp. 553–564
37. Robert Bridson; Matthias Müller-Fischer. "Fluid Simulation for Computer Animation". www.cs.ubc.ca.
## References
• Acheson, D. J. (1990), Elementary Fluid Dynamics, Oxford Applied Mathematics and Computing Science Series, Oxford University Press, ISBN 978-0-19-859679-0
• Batchelor, G. K. (1967), An Introduction to Fluid Dynamics, Cambridge University Press, ISBN 978-0-521-66396-0
• Landau, L. D.; Lifshitz, E. M. (1987), Fluid mechanics, Course of Theoretical Physics, 6 (2nd revised ed.), Pergamon Press, ISBN 978-0-08-033932-0, OCLC 15017127
• Rhyming, Inge L. (1991), Dynamique des fluides, Presses polytechniques et universitaires romandes
• Polyanin, A. D.; Kutepov, A. M.; Vyazmin, A. V.; Kazenin, D. A. (2002), Hydrodynamics, Mass and Heat Transfer in Chemical Engineering, Taylor & Francis, London, ISBN 978-0-415-27237-7
• Currie, I. G. (1974), Fundamental Mechanics of Fluids, McGraw-Hill, ISBN 978-0-07-015000-3
• V. Girault and P.A. Raviart. Finite Element Methods for Navier–Stokes Equations: Theory and Algorithms. Springer Series in Computational Mathematics. Springer-Verlag, 1986.
• White, Frank M. (2006), Viscous Fluid Flow, McGraw-Hill, ISBN 978-0-07-124493-0
• Smits, Alexander J. (2014), A Physical Introduction to Fluid Mechanics, Wiley, ISBN 0-47-1253499
• Roger Temam (1984): "Navier–Stokes Equations: Theory and Numerical Analysis", ACM Chelsea Publishing, ISBN 978-0-8218-2737-6 | 8,799 | 39,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-18 | latest | en | 0.933739 |
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# Guidance regarding GMAT
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18 Sep 2011, 06:55
hi
I wanted your guidance, if you can hep me. I am member of GMAT Club and have been following you and this site for a while. I am preparing for my GMAT and in my sectional practise tests I have been getting 50 - 60% of the questions correct including GMAT sectional tests.
(1) What does this percent suggest to you, where I am?
(2) How many questions I should get correct individually for achieving 700+?
(2) What should I do next, as in few strategic points?
My base material for preparation is Veritas & MGMAT with practise from GMAT Club, OG specific/11/12. I have finished doing this materal. I want to improve my percentage.
Can you please guide me and suggest somthing as a course of action. I will be really thankful to you for your time and guidance.
Navin
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19 Sep 2011, 14:00
I'm not an expert, but I think you should give one GMATPrep exam to test your level. The correct-answer-percentage varies from differnt prep materials. It will be very hard to conclude anything from that.
Re: Guidance regarding GMAT [#permalink] 19 Sep 2011, 14:00
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https://mcej.modares.ac.ir/browse.php?a_id=70441&slc_lang=en&sid=16&printcase=1&hbnr=1&hmb=1 | 1,721,425,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00276.warc.gz | 349,330,372 | 7,121 | Volume 23, Issue 6 (2023) MCEJ 2023, 23(6): 131-142 | Back to browse issues page
1- master student in structural engineering in tarbiat modares university
2- Professor of Civil Engineering, Graduate School of Advanced Science and Engineering, Hiroshima University, Japan , nkhaji@hiroshima-u.ac.jp
Abstract: (652 Views)
Today, one of the important issues in the industry is the failure of parts due to the presence of holes or cracks. Among the numerical calculation tools, the classical and extended finite element method is known as the most useful numerical tools in solving engineering science problems.
Identifying and investigating the types of cracks, flaws and cavities in structures is one of the most challenging issues in the field of engineering. In this article, the crack detection of two-dimensional (2D) structures using the extended finite element method (XFEM) along with genetic algorithm(GA) and grey wolf optimization method (GWO) to detect the existing crack and flaws by minimizing an error function which is also called as objective function that the evaluation of it, is based on difference between sensor measurements and suggested structure responses in each try of the algorithm. Damage detecting in 2D domains, as a non-destructive evaluation problem, is investigated using the extended finite element method along with the optimization method of genetic algorithm and grey wolf. The extended finite element method has been used to model the structure containing cracks and holes in the abaqus program, and genetic optimization and grey wolf method have been used to determine the location of the damage in which the codes were in matlab program.
The extended finite element method is a powerful tool for the analysis of structures containing cracks without remeshing and is therefore suitable for an iterative process in structural analysis. Also, in these problems, due to the wide range of parameters, it is not logical and rational to use mathematical methods. For this reason, meta heuristic methods have been developed, and grey wolf optimization methods and genetic algorithm are among these common non-gradient methods that are suitable for solving the inverse problem. This problem is set so that the optimizer algorithm finds the existing crack coordinates or holes coordinates by minimizing an objective function based on the values measured by the sensors installed on the structure. Among the limitations of the classical finite element method in the investigation of various problems in the field of fault and crack detection, we can point out the dependence of the crack or cavity on the finite element mesh, re-meshing and in other special cases the use of singular elements, which are completely removed by using The extended finite element. In this research, in order to identify the damage, the genetic optimization algorithm and the gray wolf have been used. These algorithms are designed in such a way to determine the characteristics of the damage by minimizing an error function. The defined error function is defined as the difference between the response obtained from the algorithm analysis and the response recorded in the main structure modeled in ABAQUS software, at the location of the sensors. Finally, three reference numerical examples have been solved to evaluate the capability and accuracy of the proposed method, and the result of the results shows a reduction in the cost of solving and an increase in the accuracy of the results. | 662 | 3,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.927006 |
https://www.asknumbers.com/square-yard-to-square-inch.aspx | 1,714,052,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00601.warc.gz | 561,210,691 | 29,704 | # Square Yards to Square Inches Converter
Square yards to square inches converter. 1 Square yard is equal to 1296 square inches. To convert square yards to square inches, you may refer to the converter tool or use the provided conversion formulas.
Round:
Enter Square Yard
Enter Square Inch
Calculate Area
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×
yd
To convert square yards to square inches (sq yd to sq in) or to convert square inches to square yards, you may use the converter above. To calculate the area in square yards and convert into square inches at the same time, you may enter the dimensions in yards.
Below, you will find information on how to find out how many square inches there are in "x" square yards, including the formulas and example conversions.
## How to convert square yards to square inches?
1 Square yard (sq yd) is equal to 1296 square inches (sq in). To convert square yards to square inches, multiply the square yard value by 1296.
For example, to find out how many square inches there are in 10 square yards, multiply 10 by 1296, that makes 12960 sq inches in 10 sq yards.
square yards to square inches conversion formula:
sq in = sq yd * 1296
## How to convert square inches to square yards?
1 Square inch (sq in) is equal to 0.00077160493827 square yard (sq yd). To convert square inches to square yards, multiply the square inch value by 0.00077160493827 or divide by 1296.
## How to calculate square yards and convert to square inches?
To calculate an area in square yards, multiply the length by the width in yards.
Square Yard = Length(yd) × Width(yd)
For example, to calculate an area of a 4yd x 5yd room in yards, multiply 4 by 5, that makes 20 square yards. To convert the same area to square inches, multiply the result by 1296. You may also use the conversion calculator above to calculate the area in both square yards and square inches by entering the length and the width in yards.
What is a Square Yard?
Square yard is an imperial and United States Customary area unit. 1 square yard = 1296 square inches. The symbol is "sq yd".
What is a Square Inch?
Square inch is an imperial and United States Customary area unit. 1 square inch = 0.00077160493827 square yard. The symbol is "sq in".
Please visit all area units conversion to convert all area units.
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Click "Create Table". Enter a "Start" value (5, 100 etc). Select an "Increment" value (0.01, 5 etc) and select "Accuracy" to round the result.
Related Converters | 591 | 2,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-18 | longest | en | 0.915433 |
https://www.sangakoo.com/en/unit/the-mode | 1,723,423,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00792.warc.gz | 744,282,777 | 6,044 | The mode
The mode of a set of information is the value that appears with a major frequency.
A dice is thrown ten times and the results are the following:
$$1, 3, 2, 5, 4, 2, 6, 2, 3, 6$$\$
The table of absolute frequencies is:
Result of the dice Absolute frequency $$1$$ $$1$$ $$2$$ $$3$$ $$3$$ $$2$$ $$4$$ $$1$$ $$5$$ $$1$$ $$6$$ $$2$$
It is easy to see that the mode is $$2$$, for it is the most frequent result.
The statistics of the players of Los Angeles Lakers in the first match of the finals of the NBA are the following:
a) What is the mode of scored points (PTS) among the Lakers players?
We can see in the last column, the points column, the only value that recurs is 3 points (those scored by S. Brown and S. Vujacic). And so, the mode is $$3$$.
b) And the mode of personal fouls (PF)?
The relative table of the number of personal fouls is:
Personal fouls Absolute frequency $$0$$ $$1$$ $$1$$ $$2$$ $$2$$ $$2$$ $$3$$ $$2$$ $$4$$ $$2$$ $$5$$ $$1$$
We can see that the values $$1, 2, 3$$ and $$4$$ recur twice, that is, two players made $$1$$ foul, $$2$$ made $$2$$ fouls, etc. And so, the mode of the personal fouls is $$1, 2, 3$$ and $$4$$. | 358 | 1,167 | {"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} | 4.09375 | 4 | CC-MAIN-2024-33 | latest | en | 0.893541 |
https://physics.stackexchange.com/questions/423363/superposition-of-magnetic-fields-in-a-transformer | 1,716,747,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00891.warc.gz | 384,559,466 | 37,675 | # Superposition of Magnetic fields in a Transformer
Lets say we have a transformer with two primary coils. The coils are positioned end to end such that unlike poles are adjacent (NS or SN). The current in one of the coils has a phase shift of 90 degrees. Because of superposition principle the magnetic field produced by the primaries would be the vector sum of the two primaries. This is the magnetic field which would induce current in the secondary coil.
My question is:- How would the primary coils be effected by the cemf from the secondary coil ?
"Because of superposition principle the magnetic field produced by the primaries would be the vector sum of the two primaries." But you will also have a current in the secondary, assuming that some 'load' is connected to the secondary. So the overall flux density in the core at any instant is the sum of the flux densities produced by all three coils (assuming that the transformer core can be treated as having a constant permeability). The induced emf in any of the three coils is the rate of change of flux in the core, multiplied by the number of turns on that coil.
So the emfs in the primaries are affected by the current in the secondary. The nature of the effect depends, among other things, on the secondary load (for example is it resistive or reactive). I expect the rms primary current to increase if a resistive load is connected across the secondary, though having two primaries (with unknown numbers of turns), carrying currents 90° out of phase, makes it all rather complicated. | 315 | 1,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.965311 |
https://www.learncram.com/dav-solutions/dav-class-6-maths-chapter-7-worksheet-2/ | 1,721,701,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00357.warc.gz | 750,595,458 | 12,721 | # DAV Class 6 Maths Chapter 7 Worksheet 2 Solutions
The DAV Class 6 Maths Solutions and DAV Class 6 Maths Chapter 7 Worksheet 2 Solutions of Linear Equations offer comprehensive answers to textbook questions.
## DAV Class 6 Maths Ch 7 WS 2 Solutions
Question 1.
Check whether the given variable can solve the equation. Write ‘Yes’ or ‘No’.
Equation Value of the Variable Yes/No (a) 10.y = 80 y = 6 (6) 6 + 5 = 9 6 = 4 (c) p – 8 = 5 p = -13 (d) $$\frac{y}{3}$$ = -1 y = -3 (e) 32 = 12 z = 2 (f) a – 4 = 5 a = 1 (g) a + 4 = 5 a = -1 (h) 2x + 1 = 7 x = 3 (i) 2x – 1 = -9 x = 4 (o) 3x = -15 x = -5
(a) 10y = 80, variable y = 6
Put y = 6
⇒ 10 × 6 = 80
⇒ 60 ≠ 80
y = 6 is not the required solution.
(b) 6 + 5 = 9, variable 6=4
Put 6 = 4
⇒ 4 + 5 = 9
⇒ 9 = 9
∴ 6 = 4 is the solution of the given equation.
(c) p – 8 = 5, variable p = -13
Put p = -13
⇒ -13 – 8 = 5
⇒ -21 ≠ 5
∴ p = -13 is not the solution of the given equation.
(d) $$\frac{y}{3}$$ = -1, variable y = -3, put y = -3.
$$\frac{-3}{3}$$ = -1
∴ y = -3 is the solution of the given equation.
(e) 3z = 12, variable z = 2
Put z = 2
⇒ 3 × 2 = 12
⇒ 6 ≠ 12
∴ 2 = 2 is not the required solution.
(f) a -4 = 5, variable a = 1
Put a = 1
⇒ 1 – 4= 5
⇒ -3 ≠ 5
∴ a = 1 is not the required solution.
(g) a + 4 = 5, variable a = -1
Put a = -1
⇒ -1 + 4 = 5
⇒ 3 ≠ 5
∴ a = -1 is not the required solution.
(h) 2x + 1 = 7, variable x = 3
Put x = 3,
2 × 3 + 1 = 7
⇒ 6 + 1 = 7
⇒ 7 = 7
∴x = 3 is the required solution.
(i) 2x – 1 = -9, variable x = 4
Put x = 4, 2 × 4 – 1 = -9
⇒ 8 – 1 = -9
⇒ 7 ≠-9
∴ x = 4 is not the required solution.
(j) 3x = -15, variable x = -5
Put x = -5
⇒ 3x – 5 = -15
⇒ -15 = -15
∴ x = —5 is the required solution. | 805 | 1,685 | {"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} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.668126 |
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# Multiplication Worksheets For Grade Extramath Pinterest Math 3rd And Division
By Peggy R. Rios on May 16 2018 05:25:14
Use mnemonics and silly stories to help you remember. A mnemonic is a special technique or learning device that helps you remember something. Stories like Times Tales can help you memorize your multiplication facts by associating the numbers with silly characters and stories. Phrases like 5 6 7 8, 56 equals 7 times 8 are also useful. There are many ways to memorize things, you just need to find the way that works best for you.
When it comes to multiplication, finger tricks are very useful tools, and they are fun! This trick is for multiplication tables of 6, 7, 8, and 9. When you need to find an answer quickly, finger tricks can save the day.
Who said learning multiplication has to be boring? By turning math into a game, you are more likely to remember what you are doing.
Fortunately for me, a wonderful 3rd grade teacher gave us the solution. It helped that he was male, and my 8-year-old daughter had a major crush on him. She ate up every word that man said, so when he taught her to skip count to some familiar tunes she came home singing her math just for fun. | 288 | 1,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-34 | latest | en | 0.970269 |
https://www.physicsforums.com/threads/why-are-dimensions-always-at-right-angles.849696/ | 1,560,906,763,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998879.63/warc/CC-MAIN-20190619003600-20190619025600-00474.warc.gz | 860,293,931 | 25,538 | # Why are dimensions always at right angles?
I asked my teacher , i didnt get any satisfactory answers , can u tell me why dimensions are always at right angles .
#### Giant
Not true at all. Dimensions aren't always at right angles. They can be whatever they want.
Not true at all. Dimensions aren't always at right angles. They can be whatever they want.
We *choose* 90 degrees because a lot of problems are avoided by doing so. For example, assume that 'a' is the angle which the axes make, then the distance formula changed to d = square_root((x2 - x1)^2 + (y2 - y1)^2 + 2*(x2 - x1)*(y2 - y1) * cos(a)); Now by choosing a=90 degrees we eliminate the cos term and get a beautiful formula, instead of an overcomplicated one. Similarly a LOT of trigonometric terms are avoided, sin(90) becomes 1 as well. If complications don't arise and instead stuff gets simplified, mathematicians do indeed take axes which are inclined at an angle.
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Gold Member
2018 Award
#### HallsofIvy
Homework Helper
I will also point out that "dimension" is the wrong word here. "Dimension" is a number; what you are calling "dimensions" are the coordinate axes. As others say here, they are not necessarily at right angles. They are often chosen to be at right angles because that simplifies the calculations.
#### Hornbein
I asked my teacher , i didnt get any satisfactory answers , can u tell me why dimensions are always at right angles .
There all all sorts of spaces. In some of those spaces, dimensions are not at right angles. But the great majority of people are not interested in these, because in our Universe the physical dimensions ARE at right angles. This is true of any space with the Euclidian metric of a^2 + b^2 + c^2 = d^2, where d is the distance between two points and a, b, and c are at right angles to one another.
Some responders are trying to tell you that the right angles are just a convention and convenience. It is possible to use coordinate axes that are not at right angles, but as far as I know no one actually does this in the real world. The real world really does have this right angle character.
#### FactChecker
Gold Member
2018 Award
The real world really does have this right angle character.
I'm afraid I have to disagree. The real world does not measure things in coordinate systems. People do. The advantage of measuring coordinates in right angles is so great that there are few exceptions. One example is magnetic North/South versus polar North/South. Magnetic North does not line up with Polar North. So if you go in the magnetic North direction, you are (usually) also moving in the East/West direction. Because of this, Latitude / Longitude measurements are not based on magnetic North even though that is the easiest thing to determine in the "real world". But humans made that decision, not the "real world".
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#### Hornbein
I'm afraid I have to disagree. The real world does not measure things in coordinate systems. People do. The advantage of measuring coordinates in right angles is so great that there are few exceptions. One example is magnetic North/South versus polar North/South. Magnetic North does not line up with Polar North. So if you go in the magnetic North direction, you are (usually) also moving in the East/West direction. Because of this, Latitude / Longitude measurements are not based on magnetic North even though that is the easiest thing to determine in the "real world". But humans made that decision, not the "real world".
#### micromass
Well, dimensions are just numbers. So dimensions being at right angles makes no sense really.
#### Hornbein
Well, dimensions are just numbers. So dimensions being at right angles makes no sense really.
A mathematician can say that dimensions are just numbers related in arbitrary ways. That's valid, but physicists deal with real objects and measurements inside of a real universe that is overwhelmingly preferential to 3 dimensions of space and one of time, all at right angles to one another. In the context of physicsforums such may be assumed.
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#### pwsnafu
A mathematician can say that dimensions are just numbers related in arbitrary ways. That's valid, but physicists deal with real objects and measurements inside of a real universe that is overwhelmingly preferential to 3 dimensions of space and one of time, all at right angles to one another. In the context of physicsforums such may be assumed.
This is the mathematics section of the website, hence people are answering in the context of mathematics. By that logic do you assume that the Earth Sciences forum or the Computing forum are also physics discussions just because the website is named Physics Forums?
#### pbuk
When dealing with "real objects", in what sense does time form an angle of 90 degrees with any of the dimensions of space?
#### Giant
When dealing with "real objects", in what sense does time form an angle of 90 degrees with any of the dimensions of space?
He's talking about relativity. It assumes a 4 dimensional space where time is another dimension. Time dilation tell us that time doesn't flow at equal rate depending on your velocity and gravitational potential. So time rate has to be measures. This is simplified by choosing a 4 dimensional space. Time does indeed flow at different rates. Experiments have been done and it's verified.
#### micromass
He's talking about relativity. It assumes a 4 dimensional space where time is another dimension. Time dilation tell us that time doesn't flow at equal rate depending on your velocity and gravitational potential. So time rate has to be measures. This is simplified by choosing a 4 dimensional space. Time does indeed flow at different rates. Experiments have been done and it's verified.
OK, how do you measure 90° angles in GR?
#### Hornbein
This is the mathematics section of the website, hence people are answering in the context of mathematics. By that logic do you assume that the Earth Sciences forum or the Computing forum are also physics discussions just because the website is named Physics Forums?
You've got a point there.
The question makes sense from the point of physics but not from the mathematics side. So I'd charitably assume that he was asking from the physics point of view.
#### Hornbein
When dealing with "real objects", in what sense does time form an angle of 90 degrees with any of the dimensions of space?
It isn't 90 degrees, but it is orthogonal. The Minkowski metric is x^2+y^2+z^2+it^2, which is pretty similar to the Euclidean metric. So one may loosely think of the angle as begin 90 degrees.
#### micromass
It isn't 90 degrees, but it is orthogonal. The Minkowski metric is x^2+y^2+z^2+it^2, which is pretty similar to the Euclidean metric. So one may loosely think of the angle as begin 90 degrees.
Are you sure the imaginary number $i$ should be in there?
And what about vectors for which this metric is zero. Should you interpret it as being orthogonal on itself?
#### Hornbein
Are you sure the imaginary number $i$ should be in there?
And what about vectors for which this metric is zero. Should you interpret it as being orthogonal on itself?
Yep. Albert himself sometimes used that notation. It can also be written as x^2+y^2+z^2-t^2. So on second thought, it should have been x^2+y^2+z^2+(it)^2 with standard operator precedence.
The lines for which the metric is zero are null lines. These are the lines traveled by light in a vacuum. As far as light is concerned, it takes zero proper time to travel anywhere in a vacuum. So it is a pseudometric.
If you are interested, you are invited to look up Minkowski spacetime. I'm sure that there are many others who can explain this better than can I. If that isn't enough, there is the Wick rotation.
#### micromass
Yep. Albert himself sometimes used that notation. It can also be written as x^2+y^2+z^2-t^2. So on second thought, it should have been x^2+y^2+z^2+(it)^2 with standard operator precedence.
The lines for which the metric is zero are null lines. These are the lines traveled by light in a vacuum. As far as light is concerned, it takes zero proper time to travel anywhere in a vacuum. So it is a pseudometric.
If you are interested, look up Minkowski spacetime. I'm sure that there are many others who can explain this better than can I.
I know Minkowski spacetime. I just disagree you can meaningfully talk about angles and orthogonality there. And when you move to GR it becomes even more problematic.
#### pbuk
It isn't 90 degrees, but it is orthogonal.
Only if you choose a particular definition for the inner product. Coordinate systems do not need to be orthogonal, but they do need to be linearly independent.
Which, getting back to the OP, is the point - it is meaningless to talk about dimensions being at right angles, however the coordinate systems we choose to describe position in any geometry are usually more convenient if their bases are orthogonal which, in the case of spatial coordinates, does imply they are at right angles.
#### Cruz Martinez
A mathematician can say that dimensions are just numbers related in arbitrary ways. That's valid, but physicists deal with real objects and measurements inside of a real universe that is overwhelmingly preferential to 3 dimensions of space and one of time, all at right angles to one another. In the context of physicsforums such may be assumed.
How does this make any sense if we are to accept mathematics as the language of physics? The dimension of spacetime (Minkowski a special case) is defined as the number of parameters needed to specify an event biunivocally in a sufficiently small neighborhood. So the dimension is 4, nothing more, how is it meaningfull to mention angles here? Also, a mathematician would not say dimensions are ''numbers related in abitrary ways''.
#### Giant
OK, how do you measure 90° angles in GR?
Ok I thought orthogonal meant 90 degrees. Sorry about that. But orthogonality test would be dot product? I'm not well read about relativity so I wont make further arguments.
#### weirdoguy
Albert himself sometimes used that notation.
But physicists don't use it for several decades now, it's problematic - it's been discussed in Relativity forum multiple times.
#### Mark Harder
Gold Member
Ok I thought orthogonal meant 90 degrees. Sorry about that. But orthogonality test would be dot product? I'm not well read about relativity so I wont make further arguments.
Yes, two vectors are orthogonal if their dot product is zero, by definition. But orthogonal and 90° need not be the same. For one thing, the angle refers to a geometric situation, where the unit vectors are directions in space. Angles have no meaning in other abstract notions of vector spaces. Secondly, one could define different types of "dot" products, for which orthogonal directions are not 90°. A physical and geometric example of this would be a crystal. As you might guess from looking at different minerals that have different shapes, quartz and salt (halite) for example, not all crystals have natural coordinate systems in which the axes meet at 90°. The periodic structures within the crystal, known as unit cells, consist of just enough atoms or molecules so that when the unit cell is translated in 3 directions by the lengths of the unit cell sides in those directions, the entire crystal is generated; and these 3 directions need not be at 90° to each other. It's analogous to periodic functions, like trig functions, that repeat themselves over every period along the axis of the independent variable.
#### Giant
Yes, two vectors are orthogonal if their dot product is zero, by definition. But orthogonal and 90° need not be the same. For one thing, the angle refers to a geometric situation, where the unit vectors are directions in space. Angles have no meaning in other abstract notions of vector spaces. Secondly, one could define different types of "dot" products, for which orthogonal directions are not 90°. A physical and geometric example of this would be a crystal. As you might guess from looking at different minerals that have different shapes, quartz and salt (halite) for example, not all crystals have natural coordinate systems in which the axes meet at 90°. The periodic structures within the crystal, known as unit cells, consist of just enough atoms or molecules so that when the unit cell is translated in 3 directions by the lengths of the unit cell sides in those directions, the entire crystal is generated; and these 3 directions need not be at 90° to each other. It's analogous to periodic functions, like trig functions, that repeat themselves over every period along the axis of the independent variable.
Wow. crystal example makes it a lot clear. Thanks thanks!
#### suremarc
I suppose that spacetime is "orthogonal" in the sense that the metric tensor has a smoothly varying orthogonal eigenbasis. But that's true for any pseudo-Riemannian manifold, so it's more a statement about a class of topological structures than the physical world.
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• Solo and co-op problem solving | 2,907 | 13,402 | {"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-2019-26 | latest | en | 0.940396 |
https://www.printablemultiplication.com/printable-multiplication-worksheets-pdf-2/ | 1,620,987,787,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00149.warc.gz | 974,037,156 | 78,618 | # Printable Multiplication Worksheets Pdf
Printable Multiplication Worksheets Pdf – Multiplication worksheets are an effective technique to help children in exercising their multiplication abilities. The multiplication tables that kids find out make up the basic base where various other superior and modern methods are taught in later on levels. Multiplication performs an extremely essential function in increasing scientific research and mathematics levels in colleges. It might be considered to be a great tool to enhance children’s capabilities at this stage, when basic being familiar with remains limited. Multiplication worksheet for children shows multiplication through a series of calculations.
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Math worksheets may also supply some practice workout routines for the child. These may help him increase his subtraction and multiplication capabilities. They may also make it possible for your youngster to build time management skills, which can give her a good edge in aggressive assessments. After they enhanced the straightforward techniques, these math worksheets may also be great to encourage young children to practice their multiplication specifics. Your youngster can have enjoyable doing these worksheets and may really feel achieved after the time.
Multiplication worksheets are really useful for moms and dads and kids likewise. They will likely use them to understand and practice diverse math specifics and fix issues every day. They may build math expertise while having fun. You will end up amazed at your kid’s changes right after with such worksheets. | 643 | 3,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-21 | longest | en | 0.950794 |
https://unimath.github.io/agda-unimath/set-theory.countable-sets.html | 1,720,942,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00367.warc.gz | 541,879,931 | 49,883 | # Countable sets
Content created by Fredrik Bakke, Egbert Rijke, Jonathan Prieto-Cubides, Elif Uskuplu and Elisabeth Stenholm.
Created on 2022-05-13.
module set-theory.countable-sets where
Imports
open import elementary-number-theory.equality-natural-numbers
open import elementary-number-theory.integers
open import elementary-number-theory.natural-numbers
open import elementary-number-theory.type-arithmetic-natural-numbers
open import foundation.action-on-identifications-functions
open import foundation.coproduct-types
open import foundation.decidable-propositions
open import foundation.decidable-subtypes
open import foundation.decidable-types
open import foundation.dependent-pair-types
open import foundation.empty-types
open import foundation.equality-coproduct-types
open import foundation.existential-quantification
open import foundation.function-types
open import foundation.functoriality-cartesian-product-types
open import foundation.functoriality-coproduct-types
open import foundation.maybe
open import foundation.negated-equality
open import foundation.negation
open import foundation.propositional-truncations
open import foundation.propositions
open import foundation.raising-universe-levels
open import foundation.sets
open import foundation.shifting-sequences
open import foundation.surjective-maps
open import foundation.unit-type
open import foundation.universe-levels
open import foundation-core.cartesian-product-types
open import foundation-core.fibers-of-maps
open import foundation-core.identity-types
open import univalent-combinatorics.standard-finite-types
## Idea
A set X is said to be countable if there is a surjective map f : ℕ → X + 1. Equivalently, a set X is countable if there is a surjective map f : type-decidable-subset P → X for some decidable subset P of X.
## Definition
### First definition of countable types
module _
{l : Level} (X : Set l)
where
enumeration : UU l
enumeration = Σ (ℕ → Maybe (type-Set X)) is-surjective
map-enumeration : enumeration → (ℕ → Maybe (type-Set X))
map-enumeration E = pr1 E
is-surjective-map-enumeration :
(E : enumeration) → is-surjective (map-enumeration E)
is-surjective-map-enumeration E = pr2 E
is-countable-Prop : Prop l
is-countable-Prop =
∃ (ℕ → Maybe (type-Set X)) (is-surjective-Prop)
is-countable : UU l
is-countable = type-Prop (is-countable-Prop)
is-prop-is-countable : is-prop is-countable
is-prop-is-countable = is-prop-type-Prop is-countable-Prop
### Second definition of countable types
module _
{l : Level} (X : Set l)
where
decidable-subprojection-ℕ : UU (lsuc l ⊔ l)
decidable-subprojection-ℕ =
Σ ( decidable-subtype l ℕ)
( λ P → type-decidable-subtype P ↠ type-Set X)
is-countable-Prop' : Prop (lsuc l ⊔ l)
is-countable-Prop' =
exists-structure-Prop
( decidable-subtype l ℕ)
( λ P → type-decidable-subtype P ↠ type-Set X)
is-countable' : UU (lsuc l ⊔ l)
is-countable' = type-Prop is-countable-Prop'
is-prop-is-countable' : is-prop is-countable'
is-prop-is-countable' = is-prop-type-Prop is-countable-Prop'
### Third definition of countable types
If a set X is inhabited, then it is countable if and only if there is a surjective map f : ℕ → X. Let us call the latter as "directly countable".
is-directly-countable-Prop : {l : Level} → Set l → Prop l
is-directly-countable-Prop X =
∃ (ℕ → type-Set X) (is-surjective-Prop)
is-directly-countable : {l : Level} → Set l → UU l
is-directly-countable X = type-Prop (is-directly-countable-Prop X)
is-prop-is-directly-countable :
{l : Level} (X : Set l) → is-prop (is-directly-countable X)
is-prop-is-directly-countable X = is-prop-type-Prop
(is-directly-countable-Prop X)
module _
{l : Level} (X : Set l) (a : type-Set X)
where
is-directly-countable-is-countable :
is-countable X → is-directly-countable X
is-directly-countable-is-countable H =
apply-universal-property-trunc-Prop H
( is-directly-countable-Prop X)
( λ P →
unit-trunc-Prop
( pair
( f ∘ (pr1 P))
( is-surjective-comp is-surjective-f (pr2 P))))
where
f : Maybe (type-Set X) → type-Set X
f (inl x) = x
f (inr star) = a
is-surjective-f : is-surjective f
is-surjective-f x = unit-trunc-Prop (pair (inl x) refl)
abstract
is-countable-is-directly-countable :
is-directly-countable X → is-countable X
is-countable-is-directly-countable H =
apply-universal-property-trunc-Prop H
( is-countable-Prop X)
( λ P →
unit-trunc-Prop
( ( λ where
zero-ℕ → inr star
(succ-ℕ n) → inl ((shift-ℕ a (pr1 P)) n)) ,
( λ where
( inl x) →
apply-universal-property-trunc-Prop (pr2 P x)
( trunc-Prop (fiber _ (inl x)))
( λ (n , p) →
unit-trunc-Prop
( succ-ℕ (succ-ℕ n) , ap inl p))
( inr star) → unit-trunc-Prop (zero-ℕ , refl))))
## Properties
### The two definitions of countability are equivalent
First, we will prove is-countable X → is-countable' X.
module _
{l : Level} (X : Set l)
where
decidable-subprojection-ℕ-enumeration :
enumeration X → decidable-subprojection-ℕ X
pr1 (pr1 (decidable-subprojection-ℕ-enumeration (f , H)) n) =
f n ≠ inr star
pr1 (pr2 (pr1 (decidable-subprojection-ℕ-enumeration (f , H)) n)) =
is-prop-neg
pr2 (pr2 (pr1 (decidable-subprojection-ℕ-enumeration (f , H)) n)) =
is-decidable-is-not-exception-Maybe (f n)
pr1 (pr2 (decidable-subprojection-ℕ-enumeration (f , H))) (n , p) =
value-is-not-exception-Maybe (f n) p
pr2 (pr2 (decidable-subprojection-ℕ-enumeration (f , H))) x =
apply-universal-property-trunc-Prop (H (inl x))
( trunc-Prop (fiber _ x))
( λ (n , p) →
unit-trunc-Prop
( ( n , is-not-exception-is-value-Maybe (f n) (x , inv p)) ,
( is-injective-inl
( ( eq-is-not-exception-Maybe
( f n)
( is-not-exception-is-value-Maybe
( f n) (x , inv p))) ∙
( p)))))
is-countable'-is-countable :
is-countable X → is-countable' X
is-countable'-is-countable H =
apply-universal-property-trunc-Prop H
( is-countable-Prop' X)
( λ E → unit-trunc-Prop (decidable-subprojection-ℕ-enumeration E))
Second, we will prove is-countable' X → is-countable X.
cases-map-decidable-subtype-ℕ :
{l : Level} (X : Set l) →
( P : decidable-subtype l ℕ) →
( f : type-decidable-subtype P → type-Set X) →
( (n : ℕ) → is-decidable (pr1 (P n)) -> Maybe (type-Set X))
cases-map-decidable-subtype-ℕ X P f n (inl x) = inl (f (n , x))
cases-map-decidable-subtype-ℕ X P f n (inr x) = inr star
module _
{l : Level} (X : Set l)
( P : decidable-subtype l ℕ)
( f : type-decidable-subtype P → type-Set X)
where
shift-decidable-subtype-ℕ : decidable-subtype l ℕ
shift-decidable-subtype-ℕ zero-ℕ =
( raise-empty l) ,
( is-prop-raise-empty ,
( inr (is-empty-raise-empty)))
shift-decidable-subtype-ℕ (succ-ℕ n) = P n
map-shift-decidable-subtype-ℕ :
type-decidable-subtype shift-decidable-subtype-ℕ → type-Set X
map-shift-decidable-subtype-ℕ (zero-ℕ , map-raise ())
map-shift-decidable-subtype-ℕ (succ-ℕ n , p) = f (n , p)
map-enumeration-decidable-subprojection-ℕ : ℕ → Maybe (type-Set X)
map-enumeration-decidable-subprojection-ℕ n =
cases-map-decidable-subtype-ℕ
( X)
( shift-decidable-subtype-ℕ)
( map-shift-decidable-subtype-ℕ)
( n)
(pr2 (pr2 (shift-decidable-subtype-ℕ n)))
abstract
is-surjective-map-enumeration-decidable-subprojection-ℕ :
( is-surjective f) →
( is-surjective map-enumeration-decidable-subprojection-ℕ)
is-surjective-map-enumeration-decidable-subprojection-ℕ H (inl x) =
( apply-universal-property-trunc-Prop (H x)
( trunc-Prop (fiber map-enumeration-decidable-subprojection-ℕ (inl x)))
( λ where
( ( n , s) , refl) →
unit-trunc-Prop
( ( succ-ℕ n) ,
( ap
( cases-map-decidable-subtype-ℕ X
( shift-decidable-subtype-ℕ)
( map-shift-decidable-subtype-ℕ)
(succ-ℕ n))
( pr1
( is-prop-is-decidable (pr1 (pr2 (P n)))
( pr2 (pr2 (P n)))
( inl s)))))))
is-surjective-map-enumeration-decidable-subprojection-ℕ H (inr star) =
( unit-trunc-Prop (0 , refl))
module _
{l : Level} (X : Set l)
where
enumeration-decidable-subprojection-ℕ :
decidable-subprojection-ℕ X → enumeration X
enumeration-decidable-subprojection-ℕ (P , (f , H)) =
( map-enumeration-decidable-subprojection-ℕ X P f) ,
( is-surjective-map-enumeration-decidable-subprojection-ℕ X P f H)
is-countable-is-countable' :
is-countable' X → is-countable X
is-countable-is-countable' H =
apply-universal-property-trunc-Prop H
( is-countable-Prop X)
( λ D →
( unit-trunc-Prop (enumeration-decidable-subprojection-ℕ D)))
## Useful Lemmas
There is a surjection from (Maybe A + Maybe B) to Maybe (A + B).
module _
{l1 l2 : Level} {A : UU l1} {B : UU l2}
where
map-maybe-coproduct : (Maybe A + Maybe B) → Maybe (A + B)
map-maybe-coproduct (inl (inl x)) = inl (inl x)
map-maybe-coproduct (inl (inr star)) = inr star
map-maybe-coproduct (inr (inl x)) = inl (inr x)
map-maybe-coproduct (inr (inr star)) = inr star
is-surjective-map-maybe-coproduct : is-surjective map-maybe-coproduct
is-surjective-map-maybe-coproduct (inl (inl x)) =
unit-trunc-Prop ((inl (inl x)) , refl)
is-surjective-map-maybe-coproduct (inl (inr x)) =
unit-trunc-Prop ((inr (inl x)) , refl)
is-surjective-map-maybe-coproduct (inr star) =
unit-trunc-Prop ((inl (inr star)) , refl)
There is a surjection from (Maybe A × Maybe B) to Maybe (A × B).
module _
{l1 l2 : Level} {A : UU l1} {B : UU l2}
where
map-maybe-product : (Maybe A × Maybe B) → Maybe (A × B)
map-maybe-product (inl a , inl b) = inl (a , b)
map-maybe-product (inl a , inr star) = inr star
map-maybe-product (inr star , inl b) = inr star
map-maybe-product (inr star , inr star) = inr star
is-surjective-map-maybe-product : is-surjective map-maybe-product
is-surjective-map-maybe-product (inl (a , b)) =
unit-trunc-Prop ((inl a , inl b) , refl)
is-surjective-map-maybe-product (inr star) =
unit-trunc-Prop ((inr star , inr star) , refl)
## Examples
The set of natural numbers ℕ is itself countable.
abstract
is-countable-ℕ : is-countable ℕ-Set
is-countable-ℕ =
unit-trunc-Prop
( ( λ where
zero-ℕ → inr star
(succ-ℕ n) → inl n) ,
( λ where
( inl n) → unit-trunc-Prop (succ-ℕ n , refl)
( inr star) → unit-trunc-Prop (zero-ℕ , refl)))
The empty set is countable.
is-countable-empty : is-countable empty-Set
is-countable-empty =
is-countable-is-countable'
( empty-Set)
( unit-trunc-Prop ((λ _ → empty-Decidable-Prop) , (λ ()) , (λ ())))
The unit set is countable.
abstract
is-countable-unit : is-countable unit-Set
is-countable-unit =
unit-trunc-Prop
( ( λ where
zero-ℕ → inl star
(succ-ℕ x) → inr star) ,
( λ where
( inl star) → unit-trunc-Prop (0 , refl)
( inr star) → unit-trunc-Prop (1 , refl)))
If X and Y are countable sets, then so is their coproduct X + Y.
module _
{l1 l2 : Level} (X : Set l1) (Y : Set l2)
where
is-countable-coproduct :
is-countable X → is-countable Y → is-countable (coproduct-Set X Y)
is-countable-coproduct H H' =
apply-twice-universal-property-trunc-Prop H' H
( is-countable-Prop (coproduct-Set X Y))
( λ h' h →
( unit-trunc-Prop
( pair
( map-maybe-coproduct ∘
( map-coproduct (pr1 h) (pr1 h') ∘ map-ℕ-to-ℕ+ℕ))
( is-surjective-comp
( is-surjective-map-maybe-coproduct)
( is-surjective-comp
( is-surjective-map-coproduct (pr2 h) (pr2 h'))
( is-surjective-is-equiv (is-equiv-map-ℕ-to-ℕ+ℕ)))))))
If X and Y are countable sets, then so is their coproduct X × Y.
module _
{l1 l2 : Level} (X : Set l1) (Y : Set l2)
where
is-countable-product :
is-countable X → is-countable Y → is-countable (product-Set X Y)
is-countable-product H H' =
apply-twice-universal-property-trunc-Prop H' H
( is-countable-Prop (product-Set X Y))
( λ h' h →
( unit-trunc-Prop
( pair
( map-maybe-product ∘
( map-product (pr1 h) (pr1 h') ∘ map-ℕ-to-ℕ×ℕ))
( is-surjective-comp
( is-surjective-map-maybe-product)
( is-surjective-comp
( is-surjective-map-product (pr2 h) (pr2 h'))
( is-surjective-is-equiv (is-equiv-map-ℕ-to-ℕ×ℕ)))))))
In particular, the sets ℕ + ℕ, ℕ × ℕ, and ℤ are countable.
is-countable-ℕ+ℕ : is-countable (coproduct-Set ℕ-Set ℕ-Set)
is-countable-ℕ+ℕ =
is-countable-coproduct ℕ-Set ℕ-Set is-countable-ℕ is-countable-ℕ
is-countable-ℕ×ℕ : is-countable (product-Set ℕ-Set ℕ-Set)
is-countable-ℕ×ℕ =
is-countable-product ℕ-Set ℕ-Set is-countable-ℕ is-countable-ℕ
is-countable-ℤ : is-countable (ℤ-Set)
is-countable-ℤ =
is-countable-coproduct (ℕ-Set) (coproduct-Set unit-Set ℕ-Set)
( is-countable-ℕ)
( is-countable-coproduct (unit-Set) (ℕ-Set)
( is-countable-unit) (is-countable-ℕ))
All standart finite sets are countable.
is-countable-Fin-Set : (n : ℕ) → is-countable (Fin-Set n)
is-countable-Fin-Set zero-ℕ = is-countable-empty
is-countable-Fin-Set (succ-ℕ n) =
is-countable-coproduct (Fin-Set n) (unit-Set)
( is-countable-Fin-Set n) (is-countable-unit) | 4,138 | 12,419 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.570522 |
https://www.khanacademy.org/math/basic-geo/basic-geometry-pythagorean-theorem/pythag-theorem/v/pythagorean-theorem | 1,477,292,480,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719542.42/warc/CC-MAIN-20161020183839-00362-ip-10-171-6-4.ec2.internal.warc.gz | 940,334,420 | 65,271 | # Intro to the Pythagorean theorem 2
Sal introduces the famous and super important Pythagorean theorem! Created by Sal Khan and CK-12 Foundation.
Video transcript
Let's now talk about what is easily one of the most famous theorems in all of mathematics. And that's the Pythagorean theorem. And it deals with right triangles. So a right triangle is a triangle that has a 90 degree angle in it. So the way I drew it right here, this is our 90 degree angle. If you've never seen a 90 degree angle before, the way to think about it is, if this side goes straight left to right, this side goes straight up and down. These sides are perpendicular, or the angle between them is 90 degrees, or it is a right angle. And the Pythagorean theorem tells us that if we're dealing with a right triangle-- let me write that down-- if we're dealing with a right triangle-- not a wrong triangle-- if we're dealing with a right triangle, which is a triangle that has a right angle, or a 90 degree angle in it, then the relationship between their sides is this. So this side is a, this side is b, and this side is c. And remember, the c that we're dealing with right here is the side opposite the 90 degree angle. It's important to keep track of which side is which. The Pythagorean theorem tells us that if and only if this is a right triangle, then a squared plus b squared is going to be equal to c squared. And we can use this information. If we know two of these, we can then use this theorem, this formula to solve for the third. And I'll give you one more piece of terminology here. This long side, the side that is the longest side of our right triangle, the side that is opposite of our right angle, this right here-- it's c in this example-- this is called a hypotenuse. A very fancy word for a very simple idea. The longest side of a right triangle, the side that is opposite the 90 degree angle, is called the hypotentuse. Now that we know the Pythagorean theorem, let's actually use it. Because it's one thing to know something, but it's a lot more fun to use it. So let's say I have the following right triangle. Let me draw it a little bit neater than that. It's a right triangle. This side over here has length 9. This side over here has length 7. And my question is, what is this side over here? Maybe we can call that-- we'll call that c. Well, c, in this case, once again, it is the hypotenuse. It is the longest side. So we know that the sum of the squares of the other side is going to be equal to c squared. So by the Pythagorean theorem, 9 squared plus 7 squared is going to be equal to c squared. 9 squared is 81, plus 7 squared is 49. 80 plus 40 is 120. Then we're going to have the 1 plus the 9, that's another 10, so this is going to be equal to 130. So let me write it this way. The left-hand side is going to be equal to 130, and that is equal to c squared. So what's c going to be equal to? Let me rewrite it over here. c squared is equal to 130, or we could say that c is equal to the square root of 130. And notice, I'm only taking the principal root here, because c has to be positive. We're dealing with a distance, so we can't take the negative square root. So we'll only take the principal square root right here. And if we want to simplify this a little bit, we know how to simplify our radicals. 130 is 2 times 65, which is 5 times 13. Well, these are all prime numbers, so that's about as simple as I can get. c is equal to the square root of 130. Let's do another one of these. Maybe I want to keep this Pythagorean theorem right there, just so we always remember what we're referring to. So let's say I have a triangle that looks like this. Let's see. Let's say it looks like that. And this is the right angle, up here. Let's say that this side, I'm going to call it a. The side, it's going to have length 21. And this side right here is going to be of length 35. So your instinct to solve for a, might say, hey, 21 squared plus 35 squared is going to be equal to a squared. But notice, in this situation, 35 is a hypotenuse. 35 is our c. It's the longest side of our right triangle. So what the Pythagorean theorem tells us is that a squared plus the other non-longest side-- the other non-hypotenuse squared-- so a squared plus 21 squared is going to be equal to 35 squared. You always have to remember, the c squared right here, the c that we're talking about, is always going to be the longest side of your right triangle. The side that is opposite of our right angle. This is the side that's opposite of the right angle. So a squared plus 21 squared is equal to 35 squared. And what do we have here? So 21 squared-- I'm tempted to use a calculator, but I won't. So 21 times 21: 1 times 21 is 21, 2 times 21 is 42. It is 441. 35 squared. Once again, I'm tempted to use a calculator, but I won't. 35 times 35: 5 times 5 is 25. Carry the 2. 5 times 3 is 15, plus 2 is 17. Put a 0 here, get rid of that thing. 3 times 5 is 15. 3 times 3 is 9, plus 1 is 10. So it is 11-- let me do it in order-- 5 plus 0 is 5, 7 plus 5 is 12, 1 plus 1 is 2, bring down the 1. 1225. So this tells us that a squared plus 441 is going to be equal to 35 squared, which is 1225. Now, we could subtract 441 from both sides of this equation. The left-hand side just becomes a squared. The right-hand side, what do we get? We get 5 minus 1 is 4. We want to-- let me write this a little bit neater here. Minus 441. So the left-hand side, once again, they cancel out. a squared is equal to-- and then on the right-hand side, what do we have to do? That's larger than that, but 2 is not larger than 4, so we're going to have to borrow. So that becomes a 12, or regrouped, depending on how you want to view it. That becomes a 1. 1 is not greater than 4, so we're going to have to borrow again. Get rid of that. And then this becomes an 11. 5 minus 1 is 4. 12 minus 4 is 8. 11 minus 4 is 7. So a squared is equal to 784. And we could write, then, that a is equal to the square root of 784. And once again, I'm very tempted to use a calculator, but let's, well, let's not. Let's not use it. So this is 2 times, what? 392. And then this-- 390 times 2 is 78, yeah. And then this is 2 times, what? This is 2 times 196. That's right. 190 times 2 is-- yeah, that's 2 times 196. 196 is 2 times-- I want to make sure I don't make a careless mistake. 196 is 2 times 98. Let's keep going down here. 98 is 2 times 49. And, of course, we know what that is. So notice, we have 2 times 2, times 2, times 2. So this is 2 to the fourth power. So it's 16 times 49. So a is equal to the square root of 16 times 49. I picked those numbers because they're both perfect squares. So this is equal to the square root of 16 is 4, times the square root of 49 is 7. It's equal to 28. So this side right here is going to be equal to 28, by the Pythagorean theorem. Let's do one more of these. Can never get enough practice. So let's say I have another triangle. I'll draw this one big. There you go. That's my triangle. That is the right angle. This side is 24. This side is 12. We'll call this side right here b. Now, once again, always identify the hypotenuse. That's the longest side, the side opposite the 90 degree angle. You might say, hey, I don't know that's the longest side. I don't know what b is yet. How do I know this is longest? And there, in that situation, you say, well, it's the side opposite the 90 degree angle. So if that's the hypotenuse, then this squared plus that squared is going to be equal to 24 squared. So the Pythagorean theorem-- b squared plus 12 squared is equal to 24 squared. Or we could subtract 12 squared from both sides. We say, b squared is equal to 24 squared minus 12 squared, which we know is 144, and that b is equal to the square root of 24 squared minus 12 squared. Now I'm tempted to use a calculator, and I'll give into the temptation. So let's do it. The last one was so painful, I'm still recovering. So 24 squared minus 12 squared is equal to 24.78. So this actually turns into-- let me do it without a-- well, I'll do it halfway. 24 squared minus 12 squared is equal to 432. So b is equal to the square root of 432. And let's factor this again. We saw what the answer is, but maybe we can write it in kind of a simplified radical form. So this is 2 times 216. 216, I believe, is a-- let me see. I believe that's a perfect square. So let me take the square root of 216. Nope, not a perfect square. So 216, let's just keep going. 216 is 2 times 108. 108 is, we could say, 4 times what? 25 plus another 2-- 4 times 27, which is 9 times 3. So what do we have here? We have 2 times 2, times 4, so this right here is a 16. 16 times 9 times 3. Is that right? I'm using a different calculator. 16 times 9 times 3 is equal to 432. So this is going to be equal to-- b is equal to the square root of 16 times 9, times 3, which is equal to the square root of 16, which is 4 times the square root of 9, which is 3, times the square root of 3, which is equal to 12 roots of 3. So b is 12 times the square root of 3. Hopefully you found that useful. | 2,506 | 9,055 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2016-44 | longest | en | 0.950164 |
https://www.geeksforgeeks.org/class-12-rd-sharma-solutions-chapter-21-areas-of-bounded-regions-exercise-21-4/ | 1,708,866,698,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00410.warc.gz | 799,302,953 | 61,773 | Class 12 RD Sharma Solutions- Chapter 21 Areas of Bounded Regions – Exercise 21.4
Question 1. Find the area of the region between the parabola x = 4y − y2 and the line x = 2y − 3.
Solution:
Area of the bounded region
Question 2. Find the area bounded by the parabola x = 8 + 2y − y2; the y-axis and the lines y = −1 and y = 3.
Solution:
Area of the bounded region
(ii) By using vertical strips
Solution:
Area of the bounded region
Question 4. Find the area of the region bounded the parabola y2 = 2x and straight line x − y = 4.
Solution:
Area of the bounded region
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Next | 182 | 602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.858736 |
https://kidsworksheetfun.com/comparing-fractions-worksheet-5th-grade/ | 1,708,851,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474594.56/warc/CC-MAIN-20240225071740-20240225101740-00232.warc.gz | 358,332,006 | 25,336 | # Comparing Fractions Worksheet 5th Grade
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Comparing Fractions Worksheets Fractions Worksheets Math Fractions Worksheets Math Fractions | 908 | 5,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-10 | latest | en | 0.90648 |
https://www.mathway.com/examples/finite-math/equations-and-inequalities/solving-by-factoring?id=31 | 1,725,750,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00634.warc.gz | 839,216,501 | 21,925 | # Finite Math Examples
Step 1
Rewrite as .
Step 2
Since both terms are perfect squares, factor using the difference of squares formula, where and .
Step 3
If any individual factor on the left side of the equation is equal to , the entire expression will be equal to .
Step 4
Set equal to and solve for .
Step 4.1
Set equal to .
Step 4.2
Subtract from both sides of the equation.
Step 5
Set equal to and solve for .
Step 5.1
Set equal to .
Step 5.2
Add to both sides of the equation.
Step 6
The final solution is all the values that make true. | 149 | 543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.885935 |
https://physics.stackexchange.com/questions/168357/reflected-and-refracted-light-have-same-frequency-as-that-of-the-incident-light | 1,720,908,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00515.warc.gz | 403,638,789 | 40,801 | # Reflected and refracted light have same frequency as that of the incident light frequency. Why?
My text book says-
When a monochromatic light is incident on a surface separating two media, the refracted and reflected light both have the same frequency as the incident frequency.
Can anyone explain why? I think we must look into the behavior of the atoms of the oscillator to understand the above statement.
• I'd consider it a conservation of energy issue. If the frequencies were different, we'd have difficulty keeping the same energy flux without changing the amplitude. Commented Mar 4, 2015 at 10:10
In refraction and reflection the incoming electromagnetic wave causes the electron density of the refracting material to oscillate. This happens because at any point in space the wave produces an oscillating electric field (and magnetic field, though that isn't relevant here) so any material that has a non-zero polarisability will respond by developing an oscillating dipole. This oscillating dipole then emits EM radiation, as any oscillating dipole will do. However the emitted wave will have a phase shift relative to the incoming wave, and this causes the velocity of the EM wave in the solid to be different from the speed in the vacuum. Hence the refractive index is different from 1 and we get refraction and reflection. A search of this site will find several questions that go into this process in more detail.
The point of all this is that the oscillations of the electron density in the material are at the same frequency of the incoming wave because they are driven by it. Therefore the frequency of the reradiated light is also the same frequency as the incoming wave. The process cannot change the frequency of the light.
The energy of a photon doesn't change when moving from one medium to another as pointed out by Andrew in a comment.
Considering that $E = h\nu$, $\nu$ being the frequency of the photon and $h$ Planck's constant, we see that the frequency has to stay the same when going from one medium to another. Since the frequency is the same, then the wavelength of the refracted photon will change in the process.
• Another perspective is to drop the notion of pure photons in matter and adopt composite quasiparticles involving excitations within the material itself. These excitations are massive, resulting in propagating v<c giving rise to the refractive index as well as the dispersion relation based on the phonon spectrum of the material. Commented Mar 4, 2015 at 10:34
It's actually important to recall that this is true only by definition, because it's part and parcel of what refraction means. But refraction is not the only thing that can happen when light passes between two mediums, even though it is probably by far the most common thing that happens in this situation.
There are solutions of Maxwell's equations where light passes between two mediums and the interaction of electromagnetic field and medium is linear, and, experimentally, we observe these interactions often and in keeping with Maxwellian theory. In such a case, the physics is as described by John Rennie's Answer.
But intense beams incident on a frequency doubling or other nonlinear material, where the physics is no longer linear, do not conserve the frequency. Hence my somewhat pedantic point about being true by definition. One can't prove that frequency is conserved in general, as shown by the nonlinear counterexamples.
There are very good intuitive answers already, I would like to provide a mathematical version.
The propagation of electromagnetic fields is governed by the Maxwell equations. A self-consistent solution of these equations has to fulfil certain conditions at the interface of a material.
These conditions have to be fulfilled at all times.
Say the conditions match at a certain time. You know that wave solutions oscillate with $$e^{i\omega t}$$. So if $$\omega$$ is different in the two media, then you clearly can't fulfill the interface conditions at subsequent times. Therefore the frequencies have to be equal.
Bottom line:
Consistency with Maxwell's equations requires that the frequencies in the media are the same.
Light reflects and refracts due to the interaction of incident light with the atoms of the medium. These atoms always take up the frequency of the incident light which forces them to vibrate and emit light of same frequency. Hence, the frequency remains the same. | 890 | 4,449 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.949496 |
https://www.michaelglove.com/post/the-platonic-solids-are-intelligent | 1,695,794,230,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00790.warc.gz | 981,124,785 | 211,482 | top of page
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# The Platonic Solids are Intelligent!
Updated: Jul 11
The Platonic solids are more than just shapes we learned about in math class. We can draw Metatron’s cube in three-point perspective on a piece of paper, but we know that it is more than that. We can create three dimensional models of Metatron’s cube that turn squares into cubes, but we know there is more to it than that. We know that Metatron's cube is multi-dimensional. We know that that the Platonic solids within Metatron’s cube rotate and create energy. The more we learn about our universe, the more we appreciate the fullness and diversity of life that abounds within it.
“You wish to work with the Platonic Solids, the shapes. Treat them as intelligent life for that is what they are. They respond to the signals you send through your mind, through your thoughts and vision.”
One of my favorite quotes from our “Team” is, “In order for intelligence to act, there must be intelligence to be acted upon”. The first reference to intelligence refers to our intelligence. The second reference is to the intelligence throughout the universe. The intelligence that is found in energy itself. You may not believe that energy has intelligence and that is fine. There is much about energy that our scientists still do not agree on. Unfortunately, it is beyond the scope of this exercise to debate whether energy has intelligence. I am just assuming that it does. I also believe that we can direct that energy based upon our thoughts.
Platonic Solids are Energy
I hope you had the opportunity to work with the models of the Platonic solids in the previous exercise. Hopefully you were able to feel something in your hands for each one. It is not necessary that you felt something for each one in order to move on to this next exercise.
“You have spent time creating physical models representing the platonic solids. Be aware that physical models are all approximations to ideal geometric objects and are never perfect. If a perfect dodecahedron exists, it does so only in the realm of thought. Hence the only perfect model exists within the mind.”
When you read the above quote, do not get hung up on the concept of the “perfect model”. We can build all the physical models that we want, but they will never work the same as a mental model. When we create a mental model of the Platonic solids within Metatron’s cube in our minds, we can direct the outcome that we are seeking through our directed thought. We use our directed thought to connect with the consciousness of the Platonic solids.
Platonic Solids Rotating
In this exercise, we want to start seeing the Platonic solids as three-dimensional shapes that rotate. I have videos that show each one of the Platonic solids rotating in space. The goal is to be able to visualize each shape rotating in your mind. We are going to follow the same procedure for all five of the solids.
For each shape, place a three-dimensional model of the shape in your hand. Then watch the short video. They are each about 30 seconds long. Then close your eyes and imagine the image using your inner vision. If you are able to visualize the rotating Platonic solid after viewing the short video for the first time, that is great. If not repeat the video and the visualization. Stop after three attempts to visualize the shape and move on to the next one.
Like all the exercises we will be doing, make sure you can devote your full attention to this. Do not attempt this while doing anything else. I would start out doing two or three of these together. When you have practiced them some, then you might want to try to visualize all of them in one session. Take it slow for the first couple of times. Let’s begin.
Rotating Tetrahedron
If you have a three-dimensional model of a tetrahedron, then place it in your hand. Take a deep, gentle breath and see if you can feel the energy of the tetrahedron. Click on the link below to play the video.
When the video is done playing, close your eyes and visualize the rotating tetrahedron in your mind’s eye. Give yourself about a minute to do the visualization. If you were able to visualize the rotating tetrahedron great. Then move on to the next shape. If not, play the video over again. And then try the visualization again. If you are still having trouble visualizing the shape, try it one more time. Don’t worry if you don’t get it after the third try. Move on to the next shape.
Rotating Hexahedron
If you have a three-dimensional model of a hexahedron, then place it in your hand. Take a deep, gentle breath and see if you can feel the energy of the hexahedron. Click on the link below to play the video.
When the video is done playing, close your eyes and visualize the rotating hexahedron in your mind’s eye. Give yourself about a minute to do the visualization. If you were able to visualize the rotating hexahedron great. Then move onto the next shape. If not, play the video over again. And then try the visualization again. If you are still having trouble visualizing the shape, try it one more time. Don’t worry if you don’t get it after the third try. Move on to the next shape.
Rotating Octahedron
If you have a three-dimensional model of an octahedron, place it in your hand. Take a deep, gentle breath and see if you can feel the energy of the octahedron. Click on the link below to play the video.
When the video is done playing, close your eyes and visualize the rotating octahedron in your mind’s eye. Give yourself about a minute to do the visualization. If you were able to visualize the rotating octahedron great. Then move on to the next shape. If not, play the video over again. And then try the visualization again. If you are still having trouble visualizing the shape, try it one more time. Don’t worry if you don’t get it after the third try. If this is your first time doing the visualization, I suggest you stop here. Otherwise, continue on to the next shape.
Rotating Dodecahedron
If you don’t have a model of a dodecahedron then just play the next video. Place a model of a dodecahedron in your hands. Take a deep, gentle breath in. What do you feel? When you are ready, click on the video below to see the dodecahedron moving in space.
When the video is done playing, close your eyes and visualize the shape of the dodecahedron rotating in your mind’s eye. Replay the video if you don’t see a clear image of the rotating dodecahedron. If you are still having trouble after the second replay of the video, then try it one more time. Then move on to the next visualization.
Rotating Icosahedron
If you don’t have a model of a icosahedron then just the next video. Place a model of a icosahedron in your hands. Take a deep, gentle breath in. What do you feel? When you are ready, click on the video below to see the icosahedron moving in space.
When the video is done playing, close your eyes and visualize the shape of the icosahedron rotating in your mind’s eye. Replay the video if you don’t see a clear image of the rotating icosahedron. If you are still having trouble after the second replay of the video, then try it one more time.
Work with this visualization exercise until you can easily create the image of a three-dimensional Metatron’s cube with all the Platonic solids within your mind’s eye. Once you feel comfortable doing | 1,640 | 7,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.972726 |
https://www.mrexcel.com/board/threads/arrays-in-excel.22262/ | 1,721,287,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00573.warc.gz | 799,721,495 | 21,955 | # Array's In Excel
#### skay4242
##### New Member
I am attempting to sort 3 numbers and return the lowest non zero number. The following formula works: {=MIN(IF((D5:D7)>0,(D5:D7)))}
However the location of the numbers D5:D7 are scattered and not next to each other. Is there a way to use arrays with multiple references that are separated? ie {D5,D7,E2}
### Excel Facts
Save Often
If you start asking yourself if now is a good time to save your Excel workbook, the answer is Yes
Two options...
A) Apply the aary formula to a range of consecutive cells to which you can link the non-consecutive cells;
(B) array-enter:
=MIN(IF(SETV(EVAL(CHAR(123)&D5&","&D7&","&E2&CHAR(125))),GETV()))
SETV, GETV, and EVAL are part of the morefunc.xll add-in, downloadable from:
http://longre.free.fr/english/index.html
This message was edited by Aladin Akyurek on 2002-09-23 11:16
Thanks for the assistance. As it turns out I went back to the original 3 cells and altered the formulas to give only one result. The first solution also does not work due to the fact that i really have three matrices of data with which I need to compare. Putting each sell next to each other slows down the processing. I am running 10000 monte carlo simulations around the data sets so speed is crucial. Thanks again.
On 2002-09-23 11:34, skay4242 wrote:
Thanks for the assistance. As it turns out I went back to the original 3 cells and altered the formulas to give only one result. The first solution also does not work due to the fact that i really have three matrices of data with which I need to compare. Putting each sell next to each other slows down the processing. I am running 10000 monte carlo simulations around the data sets so speed is crucial. Thanks again.
Try B to see whether that works.
Hi,
Actually another options (at least it worked for me ) is available:
=IF(MIN(Arrays)=0,SMALL(Arrays,2),MIN(Arrays))
Where the name Array refer to the range.
This is an array-formula which means that it´s entered with Ctrl´+SHift+Enter and XL will confirm by using brackets {} around the formula,
{=IF(MIN(Arrays)=0,SMALL(Arrays,2),MIN(Arrays))}
Kind regards,
Dennis
On 2002-09-23 14:14, XL-Dennis wrote:
Hi,
Actually another options (at least it worked for me ) is available:
=IF(MIN(Arrays)=0,SMALL(Arrays,2),MIN(Arrays))
Where the name Array refer to the range.
This is an array-formula which means that it´s entered with Ctrl´+SHift+Enter and XL will confirm by using brackets {} around the formula,
{=IF(MIN(Arrays)=0,SMALL(Arrays,2),MIN(Arrays))}
Kind regards,
Dennis
Try when Arrays refer to the nonconsec range with values...
D5=0
D7=3
E2=0
Aladin,
You may laugh but back to the OP´s question (only 3 cells....) and in view of Your finding we can expand the formula to cover all possibilities with nested IF-statements, right?
See below answer from me
Kind regards,
Dennis
This message was edited by XL-Dennis on 2002-09-23 15:15
The formula was not correct.
This message was edited by Dave Patton on 2002-09-23 15:36
Dave,
I can´t get Your solution to work?
Nevertheless, Aladin take a look on this solution and see if You beat this "horrible" formula :wink:
Diverse1.xls
ABCDEFGH
1
200
3
41
5
61
7
8Array=Blad1!\$E\$2;Blad1!\$C\$2;Blad1!\$E\$4
9
10
Blad1
Kind regards,
Dennis
[...] You may laugh
I've been punished... My Netscape has now that grinning face as background I can't get rid off. Did you do that to me by any chance?
but back to the OP´s question (only 3 cells....) and in view of Your finding we can expand the formula to cover all possibilities with nested IF-statements, right?
My question is to You:
Is it possible or not?
Yeah, right. If it's just a few cells, it's no doubt a viable option. I have no idea though what the performance will be as part of the solver model the OP appears to use.
Aladin
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Go back | 1,363 | 5,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.929817 |
https://www.slideshare.net/ColleenFarrelly/machine-learning-by-analogy-59094152 | 1,657,009,295,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00193.warc.gz | 1,038,387,233 | 74,362 | Successfully reported this slideshow.
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# Machine Learning by Analogy
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#### Description
A start guide to the concepts and algorithms in machine learning, including regression frameworks, ensemble methods, clustering, optimization, and more. Mathematical knowledge is not assumed, and pictures/analogies demonstrate the key concepts behind popular and cutting-edge methods in data analysis.
Updated to include newer algorithms, such as XGBoost, and more geometrically/topologically-based algorithms. Also includes a short overview of time series analysis
#### Transcript
1. 1. Colleen M. Farrelly
2. 2. Many machine learning methods exist in the literature and in industry. ◦ What works well for one problem may not work well for the next problem. ◦ In addition to poor model fit, an incorrect application of methods can lead to incorrect inference. Implications for data-driven business decisions. Low future confidence in data science and its results. Lower quality software products. Understanding the intuition and mathematics behind these methods can ameliorate these problems. ◦ This talk focuses on building intuition. ◦ Links to theoretical papers underlying each method.
3. 3. Total variance of a normally- distributed outcome as cookie jar. Error as empty space. Predictors accounting for pieces of the total variance as cookies. ◦ Based on relationship to predictor and to each other. Cookies accounting for the same piece of variance as those smooshed together. Many statistical assumptions need to be met. www.zelcoviacookies.com
4. 4. Extends multiple regression. ◦ Many types of outcome distributions. ◦ Transform an outcome variable to create a linear relationship with predictors. Sort of like silly putty stretching the outcome variable in the data space. Does not work on high- dimensional data where predictors > observations. Only certain outcome distributions. ◦ Exponential family as example. tomstock.photoshelter.com
5. 5. Impose size and/or variable overlap constraints (penalties) on generalized linear models. ◦ Elastic net as hybrid of these constraints. ◦ Can handle large numbers of predictors. Reduce the number of predictors. ◦ Shrink some predictor estimates to 0. ◦ Examine sets of similar predictors. Similar to eating irrelevant cookies in the regression cookie jar or a cowboy at the origin roping coefficients that get too close
6. 6. Homotopy arrow example ◦ Red and blue arrows can be deformed into each other by wiggling and stretching the line path with anchors at start and finish of line ◦ Yellow arrow crosses holes and would need to backtrack or break to the surface to freely wiggle into the blue or red line Homotopy method in LASSO/LARS wiggles an easy regression path into an optimal regression path ◦ Avoids obstacles that can trap other regression estimators (peaks, valleys, saddles…) Homotopy as path equivalence ◦ Intrinsic property of topological spaces (such as data manifolds)
7. 7. Instead of fitting model to data, fit model to tangent space (what isn’t the data). ◦ Deals with collinearity, as parallel vectors share a tangent space (only one selected of collinear group). ◦ LASSO and LARS extensions. ◦ Rao scoring for selection. Effect estimates (angles). Model selection criteria. Information criteria. Deviance scoring.
8. 8. Fit all possible models and figure out likelihood of each given observed data. ◦ Based on Bayes’ Theorem and conditional probability. ◦ Instead of giving likelihood that data came from a specific parameterized population (univariate), figure out likelihood of set of data coming from sets of population (multivariate). ◦ Can select naïve prior (no assumptions on model or population) or make an informed guess (assumptions about population or important factors in model). ◦ Combine multiple models according to their likelihoods into a blended model given data.
9. 9. Semi-Parametric Extensions of Regression
10. 10. Extends a generalized linear models by estimating unknown (nonlinear) functions of an outcome on intervals of a predictor. Use of “knots” to break function into intervals. Similar to a downhill skier. ◦ Slope as a function of outcome. ◦ Skier as spline. ◦ Flags as knots anchoring the skier as he travels down the slope. www.dearsportsfan.com
11. 11. Multivariate adaptive regression splines as an extension of spline models to multivariate data ◦ Knots placed according to multivariate structure and splines fit between knots ◦ Much like fixing a rope to the peaks and valleys of a mountain range, where the rope has enough slack to hug the terrain between its fixed points
12. 12. Extends spline models to the relationships of many predictors to an outcome. ◦ Also allows for a “silly- putty” transformation of the outcome variable. Like multiple skiers on many courses and mountains to estimate many relationships in the dataset. www.filigreeinn.com
13. 13. Chop data into partitions and then fit multiple regression models to each partition. Divide-and-conquer approach. Examples: ◦ Multivariate adaptive regression splines ◦ Regression trees ◦ Morse-Smale regression www.pinterest.com
14. 14. Based on a branch of math called topology. ◦ Study of changes in function behavior on shapes. ◦ Used to classify similarities/ differences between shapes. ◦ Data clouds turned into discrete shape combinations (simplices). Use these principles to partition data and fit elastic net models to each piece. ◦ Break data into multiple toy sets. ◦ Analyze sets for underlying properties of each toy. Useful visual output for additional data mining.
15. 15. Linear or non-linear support beams used to separate group data for classification or map data for kernel- based regression. Much like scaffolding and support beams separating and holding up parts of a high rise. http://en.wikipedia.org/wiki/Support_vector_machine leadertom.en.ec21.com
16. 16. Based on processing complex, nonlinear information the way the human brain does via a series of feature mappings. colah.github.io www.alz.org Arrows denote mapping functions, which take one topological space to another
17. 17. Type of shallow, wide neural network. Reduces framework to a penalized linear algebra problem, rather than iterative training (much faster to solve). Based on random mappings. Shown to converge to correct classification/regression (universal approximation property—may require unreasonably wide networks). Semi-supervised learning extensions.
18. 18. Added layers in neural network to solve width problem in single-layer networks for universal approximation. More effective in learning features of the data. Like sifting data with multiple sifters to distill finer and finer pieces of the data. Computationally intensive and requires architecture design and optimization.
19. 19. Recent extension of deep learning framework from spreadsheet data to multiple simultaneous spreadsheets ◦ Spreadsheets may be of the same dimension or different dimensions ◦ Could process multiple or hierarchical networks via adjacency matrices Like sifting through data with multiple inputs of varying sizes and textures
20. 20. CHAPTER 3:
21. 21. Classifies data according to optimal partitioning of data (visualized as a high- dimensional cube). Slices data into squares, cubes, and hypercubes (4+ dimensional cubes). ◦ Like finding the best place to cut through the data with a sword. Option to prune tree for better model (glue some of the pieces back together). Notoriously unstable. Optimization possible to arrive at best possible trees (genetic algorithms). tvtropes.org
22. 22. Optimization technique. ◦ Minimize a loss function in regression. Accomplished by choosing a variable per step that achieves largest minimization. Climber trying to descend a mountain. ◦ Minimize height above sea level per step. Directions (N, S, E, W) as a collection of variables. Climber rappels down cliff according to these rules. www.chinatravelca.com
23. 23. Optimization helpers. ◦ Find global maximum or minimum by searching many areas simultaneously. ◦ Can impose restrictions. Teams of mountain climbers trying to find the summit. Restrictions as climber supplies, hours of daylight left to find summit… Genetic algorithm as population of mountain climbers each climbing on his/her own. wallpapers.brothersoft.com
24. 24. One climber exploring multiple routes simultaneously. Individual optimized through a series of gates at each update until superposition converges on one state ◦ Say wave 2 is the optimal combination of predictors ◦ Resultant combination slowly flattens to wave 2, with the states of wave 1disappearing http://philschatz.com/physics -book/contents/m42249.html
25. 25. Combine multiple models of the same type (ex. trees) for better prediction. ◦ Single models unstable. Equally-good estimates from different models. ◦ Use bootstrapping. Multiple “marble draws” followed by model creation. ◦ Creates diversity of features. ◦ Creates models with different biases and error.
26. 26. Uses gradient descent algorithm to model an outcome based on predictors (linear, tree, spline…). ◦ Proceeds in steps, adding variables and adjusting weights according to the algorithm. Like putting together a puzzle. ◦ Algorithm first focuses on most important parts of the picture (such as the Mona Lisa’s eyes). ◦ Then adds nuances that help identify other missing pieces (hands, landscape…). play.google.com
27. 27. Adds a penalty term to boosted regression ◦ Can be formulated as LASSO/ridge regression/elastic net with constraints ◦ Also leverages hardware to further speed up boosting ensemble play.google.com
28. 28. Single tree models poor predictors. Grow a forest of them for a strong predictor (another ensemble method). Algorithm: ◦ Takes random sample of data. ◦ Builds one tree. ◦ Repeats until forest grown. ◦ Averages across forest to identify strong predictors.
29. 29. CHAPTER 4:
30. 30. Principle Component Analysis (PCA) ◦ Variance partitioning to combine pieces of variables into new linear subspace. ◦ Smashing cookies by kind and combining into a flat hybrid cookie in previous regression model. Manifold Learning ◦ PCA-like algorithms that combine pieces into a new nonlinear subspace. ◦ Non-flat cookie combining. Useful as pre-processing step for prediction models. ◦ Reduce dimension. ◦ Obtain uncorrelated, non- overlapping variables (bases). marmaladeandmileposts. com
31. 31. Balanced sampling for low-frequency predictors. ◦ Stratified samples (i.e. sample from bag of mostly white marbles and few red marbles with constraint that 1/5th of draws must be red marbles). Dimension reduction/mapping pre-processing ◦ Principle component, manifold learning… ◦ Hybrid of neural network methods and tree models.
32. 32. Aggregation of multiple types of models. ◦ Like a small town election. ◦ Different people have different views of the politics and care about different issues. Different modeling methods capture different pieces of the data and vote in different pieces. ◦ Leverage strengths, minimize weaknesses ◦ Diversity of methods to better explore underlying data geometry Avoids multiple testing issues.
33. 33. Subsembles ◦ Partition data into training sets Randomly selected or through partition algorithm ◦ Train a model on each data partition ◦ Combine into final weighted prediction model This is similar to national elections. ◦ Each elector in the electoral college learns for whom his constituents voted. ◦ The final electoral college pools these individual votes. Full Training Dataset Model Training Datasets 1, 2, 3, 4 Final Subsemble Model
34. 34. Combines superlearning and subsembles for better prediction ◦ Diversity improves subsemble method (better able to explore data geometry) ◦ Bootstrapping improves superlearner pieces (more diversity within each method) ◦ Preliminary empirical evidence shows efficacy of combination.
35. 35. CHAPTER 5:
36. 36. Classification of a data point based on the classification of its nearest neighboring points. Like a junior high lunchroom and student clicks. ◦ Students at the same table tend to be more similar to each other than to a distant table. www.wrestlecrap.com
37. 37. Iteratively separating groups within a dataset based on similarity. Like untangling tangled balls of yarn into multiple, single- colored balls (ideally). ◦ Each step untangles the mess a little further. ◦ Few stopping guidelines (when it looks separated). krazymommakreations.blogspot.com
38. 38. Hybrid of supervised and unsupervised learning (groupings and prediction). Uses graphical representation of data and its relationships. Algorithms with connections to topology, differential geometry, and Markov chains. Useful in combination with other machine learning methods to provide extra insight (ex. spectral clustering).
39. 39. K-means algorithm with weighting and dimension reduction components of similarity measure. ◦ Simplify balls of string to warm colors and cool colors before untangling. Can be reformulated as a graph clustering problem. ◦ Partition subcomponents of a graph based on flow equations. www.simplepastimes.com
40. 40. Multivariate technique similar to mode or density clustering. ◦ Find peaks and valleys in data according to an input function on the data (level set slices)— much like a watershed on mountains. ◦ Separate data based on shared peaks and valleys across slices (shared multivariate density/gradient). ◦ Many nice theoretical developments on validity and convergence.
41. 41. Topological clustering. ◦ Define distance metric. ◦ Slice multidimensional dataset with Morse function. ◦ Examine function behavior across slice. ◦ Cluster function behavior. ◦ Iterate through multiple slices to obtain hierarchy of function behavior. Much like examining the behavior of multiple objects across a flip book. ◦ Nesting ◦ Cluster overlap Filtered functions then used to create various resolutions of a modified Reeb graph summary of topology.
42. 42. Time Series Forecasting
43. 43. Similar to decomposing superposed states ◦ Seasonal trends ◦ Yearly trends ◦ Trend averages ◦ Dependencies on previous time point Knit individual forecasted pieces into a complete forecast by superposing these individual forecasts Several extensions to neural networks, time- lagged machine learning models…
44. 44. A time-series method incorporating predictors ◦ Constant predictors at initial time point ◦ Varying predictors at multiple time points Creates a sort of correlation web between predictors and time points ◦ Can handle multiple time lags and multivariate outcomes ◦ Can handle any GLM outcome links Related to partial differential equations of dynamic systems
45. 45. Data-based mining for SEM relationships/time-lag components ◦ Leverages conditional probability between predictors to find dependencies ◦ Does not require a priori model formulation like SEM Peeking at data to create a dependency web over time or predictors/outcome Can be validated by a follow-up SEM based on network structure Time 1 Outcome Predictor 2 Time 2 Outcome Time 3 Outcome
46. 46. Technically: ◦ Matrix decomposition (similar to PCA/manifold learning) ◦ Followed by spectral methods ◦ Cleaning of time-lagged covariance matrix ◦ Reconstruction with simple forecast Kind of like deconstructing, cleaning, a rebuilding a car engine
47. 47. Combines k-nearest neighbors-based clustering with memoryless state changes converging to a transition distribution (weighted directed graph) ◦ Reduce an observation to a pattern ◦ Remember patterns seen (across time or space) ◦ Match new observations to this set of patterns ◦ Computationally more feasible than k-means clustering Like remembering classes of chess board configurations across games played
48. 48. Many machine learning methods exist today, and many more are being developed every day. Methods come with certain assumptions that must be met. ◦ Breaking assumptions can lead to poor model fit or incorrect inference. ◦ Matching a method to a problem not only can help with better prediction and inference; it can also lead to faster computation times and better presentations to clients. Development depends on problem-matching and deep understanding of the mathematics behind the methods used.
49. 49. Parametric Regression: ◦ Draper, N. R., Smith, H., & Pownell, E. (1966). Applied regression analysis (Vol. 3). New York: Wiley. ◦ McCullagh, P. (1984). Generalized linear models. European Journal of Operational Research, 16(3), 285-292. ◦ Zou, H., & Hastie, T. (2005). Regularization and variable selection via the elastic net. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 67(2), 301-320. ◦ Augugliaro, L., Mineo, A. M., & Wit, E. C. (2013). Differential geometric least angle regression: a differential geometric approach to sparse generalized linear models. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 75(3), 471-498. ◦ Raftery, A. E., Madigan, D., & Hoeting, J. A. (1997). Bayesian model averaging for linear regression models. Journal of the American Statistical Association, 92(437), 179-191. ◦ Osborne, M. R., & Turlach, B. A. (2011). A homotopy algorithm for the quantile regression lasso and related piecewise linear problems. Journal of Computational and Graphical Statistics, 20(4), 972-987. ◦ Drori, I., & Donoho, D. L. (2006, May). Solution of l 1 minimization problems by LARS/homotopy methods. In Acoustics, Speech and Signal Processing, 2006. ICASSP 2006 Proceedings. 2006 IEEE International Conference on (Vol. 3, pp. III-III). IEEE.
50. 50. Semi-Parametric Regression: ◦ Marsh, L. C., & Cormier, D. R. (2001). Spline regression models (Vol. 137). Sage. ◦ Hastie, T., & Tibshirani, R. (1986). Generalized additive models. Statistical science, 297-310. ◦ McZgee, V. E., & Carleton, W. T. (1970). Piecewise regression. Journal of the American Statistical Association, 65(331), 1109-1124. ◦ Gerber, S., Rübel, O., Bremer, P. T., Pascucci, V., & Whitaker, R. T. (2013). Morse– Smale Regression. Journal of Computational and Graphical Statistics, 22(1), 193- 214. ◦ Hearst, M. A., Dumais, S. T., Osman, E., Platt, J., & Scholkopf, B. (1998). Support vector machines. Intelligent Systems and their Applications, IEEE, 13(4), 18-28. ◦ Hornik, K., Stinchcombe, M., & White, H. (1989). Multilayer feedforward networks are universal approximators. Neural networks, 2(5), 359-366. ◦ Krizhevsky, A., Sutskever, I., & Hinton, G. E. (2012). Imagenet classification with deep convolutional neural networks. In Advances in neural information processing systems (pp. 1097-1105). ◦ Huang, G. B., Wang, D. H., & Lan, Y. (2011). Extreme learning machines: a survey. International Journal of Machine Learning and Cybernetics, 2(2), 107-122. ◦ Friedman, J. H. (1991). Multivariate adaptive regression splines. The annals of statistics, 1-67. ◦ Abadi, M., Agarwal, A., Barham, P., Brevdo, E., Chen, Z., Citro, C., ... & Ghemawat, S. (2016). Tensorflow: Large-scale machine learning on heterogeneous distributed systems. arXiv preprint arXiv:1603.04467.
51. 51. Nonparametric Regression: ◦ Buntine, W. (1992). Learning classification trees. Statistics and computing, 2(2), 63-73. ◦ Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378. ◦ Bäck, T. (1996). Evolutionary algorithms in theory and practice: evolution strategies, evolutionary programming, genetic algorithms. Oxford university press. ◦ Zhang, G. (2011). Quantum-inspired evolutionary algorithms: a survey and empirical study. Journal of Heuristics, 17(3), 303-351. ◦ Dietterich, T. G. (2000). Ensemble methods in machine learning. In Multiple classifier systems (pp. 1-15). Springer Berlin Heidelberg. ◦ Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378. ◦ Breiman, L. (2001). Random forests. Machine learning, 45(1), 5-32. ◦ Chen, T., & Guestrin, C. (2016, August). Xgboost: A scalable tree boosting system. In Proceedings of the 22nd acm sigkdd international conference on knowledge discovery and data mining (pp. 785-794). ACM.
52. 52. Supervised with Unsupervised Methods: ◦ van der Maaten, L. J., Postma, E. O., & van den Herik, H. J. (2009). Dimensionality reduction: A comparative review. Journal of Machine Learning Research, 10(1-41), 66-71. ◦ Kuncheva, L. I., & Rodríguez, J. J. (2007). An experimental study on rotation forest ensembles. In Multiple Classifier Systems (pp. 459-468). Springer Berlin Heidelberg. ◦ van der Laan, M. J., Polley, E. C., & Hubbard, A. E. (2007). Super learner. Statistical applications in genetics and molecular biology, 6(1). ◦ Sapp, S. K. (2014). Subsemble: A Flexible Subset Ensemble Prediction Method (Doctoral dissertation, University of California, Berkeley).
53. 53. Unsupervised Methods: ◦ Fukunaga, K., & Narendra, P. M. (1975). A branch and bound algorithm for computing k-nearest neighbors. Computers, IEEE Transactions on, 100(7), 750-753. ◦ MacQueen, J. (1967, June). Some methods for classification and analysis of multivariate observations. In Proceedings of the fifth Berkeley symposium on mathematical statistics and probability (Vol. 1, No. 14, pp. 281-297). ◦ Chartrand, G., & Oellermann, O. R. (1993). Applied and algorithmic graph theory. ◦ Ng, A. Y., Jordan, M. I., & Weiss, Y. (2002). On spectral clustering: Analysis and an algorithm. Advances in neural information processing systems, 2, 849-856. ◦ Singh, G., Mémoli, F., & Carlsson, G. E. (2007, September). Topological Methods for the Analysis of High Dimensional Data Sets and 3D Object Recognition. In SPBG (pp. 91- 100). ◦ Chen, Y. C., Genovese, C. R., & Wasserman, L. (2016). A comprehensive approach to mode clustering. Electronic Journal of Statistics, 10(1), 210-241.
54. 54. Time Series ◦ Wu, J. P., & Wei, S. (1989). Time series analysis. Hunan Science and Technology Press, ChangSha. ◦ Vautard, R., Yiou, P., & Ghil, M. (1992). Singular-spectrum analysis: A toolkit for short, noisy chaotic signals. Physica D: Nonlinear Phenomena, 58(1), 95-126. ◦ Dunham, M. H., Meng, Y., & Huang, J. (2004, November). Extensible markov model. In Data Mining, 2004. ICDM'04. Fourth IEEE International Conference on (pp. 371-374). IEEE. ◦ Ullman, J. B., & Bentler, P. M. (2003). Structural equation modeling. John Wiley & Sons, Inc.. ◦ Heckerman, D., Geiger, D., & Chickering, D. M. (1994, July). Learning Bayesian networks: The combination of knowledge and statistical data. In Proceedings of the Tenth international conference on Uncertainty in artificial intelligence (pp. 293-301). Morgan Kaufmann Publishers Inc..
#### Editor's Notes
• Assumes predictors and observations are independent.
Assumes errors are normally distributed and independent.
Assumes linear relationship between predictors and outcome.
Assumes normally-distributed outcome.
Draper, N. R., Smith, H., & Pownell, E. (1966). Applied regression analysis (Vol. 3). New York: Wiley.
• Same assumptions as multiple regression, minus outcome’s normal distribution (link function extends to non-normal distributions).
McCullagh, P. (1984). Generalized linear models. European Journal of Operational Research, 16(3), 285-292.
• Relaxes predictor independence requirement and adds penalty term.
Adds a penalty to reduce generalized linear model’s model size.
Zou, H., & Hastie, T. (2005). Regularization and variable selection via the elastic net. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 67(2), 301-320.
• Exists for several types of models, including survival, binomial, and Poisson regression models
Augugliaro, L., Mineo, A. M., & Wit, E. C. (2013). Differential geometric least angle regression: a differential geometric approach to sparse generalized linear models. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 75(3), 471-498.
Augugliaro, L., & Mineo, A. M. (2015). Using the dglars Package to Estimate a Sparse Generalized Linear Model. In Advances in Statistical Models for Data Analysis (pp. 1-8). Springer International Publishing.
• Joint model likelihood given multivariate data vs. parameter likelihood given univariate data
Raftery, A. E., Madigan, D., & Hoeting, J. A. (1997). Bayesian model averaging for linear regression models. Journal of the American Statistical Association, 92(437), 179-191.
Raftery, A. E., Gneiting, T., Balabdaoui, F., & Polakowski, M. (2005). Using Bayesian model averaging to calibrate forecast ensembles. Monthly Weather Review, 133(5), 1155-1174.
Madigan, D., Raftery, A. E., Volinsky, C., & Hoeting, J. (1996). Bayesian model averaging. In Proceedings of the AAAI Workshop on Integrating Multiple Learned Models, Portland, OR (pp. 77-83).
• Relaxes linear relationship of predictors to outcome requirement.
Marsh, L. C., & Cormier, D. R. (2001). Spline regression models (Vol. 137). Sage.
• Generally flexible and able to handle high-dimensional data compared to other spline methods.
Friedman, J. H. (1991). Multivariate adaptive regression splines. The annals of statistics, 1-67.
• Relaxes linear relationship of predictors to outcome requirement and outcome normality assumption.
Hastie, T., & Tibshirani, R. (1986). Generalized additive models. Statistical science, 297-310.
• Regression by partition.
McZgee, V. E., & Carleton, W. T. (1970). Piecewise regression. Journal of the American Statistical Association, 65(331), 1109-1124.
• Regression by data partitions defined by multivariate function behavior.
Gerber, S., Rübel, O., Bremer, P. T., Pascucci, V., & Whitaker, R. T. (2013). Morse–Smale Regression. Journal of Computational and Graphical Statistics, 22(1), 193-214.
• Classification tool by combined data splits.
Singularity issues.
Computationally taxing, particularly as number of groups increases.
Can be used in semi-parametric regression as nonlinear mapping of regression kernels.
Hearst, M. A., Dumais, S. T., Osman, E., Platt, J., & Scholkopf, B. (1998). Support vector machines. Intelligent Systems and their Applications, IEEE, 13(4), 18-28.
• Computationally expensive in traditional algorithms and rooted in topological maps.
Cannot handle lots of variables compared to number of observations.
Cannot handle non-independent data.
Hornik, K., Stinchcombe, M., & White, H. (1989). Multilayer feedforward networks are universal approximators. Neural networks, 2(5), 359-366.
• Random mappings to reduce MLP to linear system of equations.
Huang, G. B., Wang, D. H., & Lan, Y. (2011). Extreme learning machines: a survey. International Journal of Machine Learning and Cybernetics, 2(2), 107-122.
• Computationally expensive neural network extension.
Still suffers from singularities which hinder performance.
Krizhevsky, A., Sutskever, I., & Hinton, G. E. (2012). Imagenet classification with deep convolutional neural networks. In Advances in neural information processing systems (pp. 1097-1105).
• More flexible than deep learning architectures (different types of simultaneous input data—graphs, spreadsheets, images…).
Abadi, M., Agarwal, A., Barham, P., Brevdo, E., Chen, Z., Citro, C., ... & Ghemawat, S. (2016). Tensorflow: Large-scale machine learning on heterogeneous distributed systems. arXiv preprint arXiv:1603.04467.
• Does not assume anything about data and regression parameters.
Can be unstable (many optimal solutions).
Can be used to classify or regress.
Buntine, W. (1992). Learning classification trees. Statistics and computing, 2(2), 63-73.
• A type of optimization that works by finding the quickest route to a solution.
Can be very accurate, but takes an inordinate amount of time for some problems (big data, neural network training…).
Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378.
• Optimization technique.
Computationally expensive but guaranteed to converge.
Bäck, T. (1996). Evolutionary algorithms in theory and practice: evolution strategies, evolutionary programming, genetic algorithms. Oxford university press.
• Generally much faster convergence than genetic algorithms.
Zhang, G. (2011). Quantum-inspired evolutionary algorithms: a survey and empirical study. Journal of Heuristics, 17(3), 303-351.
• Exploit weak learners.
Dietterich, T. G. (2000). Ensemble methods in machine learning. In Multiple classifier systems (pp. 1-15). Springer Berlin Heidelberg.
• Useful when the previous types of regression fail, when the data is “noisy” with little known about which predictors are important, and when outliers are likely to exist in the data (piece-wise modeling method of sorts).
Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378.
• Speeds up boosting but requires extensive tuning and does not give interpretable model (only prediction). Has been used a lot in recent Kaggle competitions, particle physics datasets.
Chen, T., & Guestrin, C. (2016, August). Xgboost: A scalable tree boosting system. In Proceedings of the 22nd acm sigkdd international conference on knowledge discovery and data mining (pp. 785-794). ACM.
• Useful in modeling complex, nonlinear regression/classification problems.
Also distance-based metric. Handles noisy data well.
Breiman, L. (2001). Random forests. Machine learning, 45(1), 5-32.
• Variance combined.
van der Maaten, L. J., Postma, E. O., & van den Herik, H. J. (2009). Dimensionality reduction: A comparative review. Journal of Machine Learning Research, 10(1-41), 66-71.
• Deep connections to sampling theory and topology.
Kuncheva, L. I., & Rodríguez, J. J. (2007). An experimental study on rotation forest ensembles. In Multiple Classifier Systems (pp. 459-468). Springer Berlin Heidelberg.
• Bagging of different base models (same bootstrap or different bootstrap).
van der Laan, M. J., Polley, E. C., & Hubbard, A. E. (2007). Super learner. Statistical applications in genetics and molecular biology, 6(1).
• Good prediction with ability to parallelize for distributed computing. Tends to be faster than superlearners (smaller training samples).
Sapp, S. K. (2014). Subsemble: A Flexible Subset Ensemble Prediction Method (Doctoral dissertation, University of California, Berkeley).
• New formulation by author (Colleen M. Farrelly). Requires a minimum sample size of ~30 per bootstrap within method.
• Local classification.
Forms basis of many other machine learning algorithms (manifold learning, clustering, certain Markov chains…).
Fukunaga, K., & Narendra, P. M. (1975). A branch and bound algorithm for computing k-nearest neighbors. Computers, IEEE Transactions on, 100(7), 750-753.
• Creates groupings of like objects in the dataset based on different linear/nonlinear metrics.
Assumes groups are separable (no overlap).
Issues with computation as number of observations increases.
MacQueen, J. (1967, June). Some methods for classification and analysis of multivariate observations. In Proceedings of the fifth Berkeley symposium on mathematical statistics and probability (Vol. 1, No. 14, pp. 281-297).
• Data by pictures.
Chartrand, G., & Oellermann, O. R. (1993). Applied and algorithmic graph theory.
• Weighted kernel k-means (weighted subcomponent clustering).
Ng, A. Y., Jordan, M. I., & Weiss, Y. (2002). On spectral clustering: Analysis and an algorithm. Advances in neural information processing systems, 2, 849-856.
• Related to density clustering but relies on Morse critical points and filter functions to define complexes (clusters)
Gerber, S., Bremer, P. T., Pascucci, V., & Whitaker, R. (2010). Visual exploration of high dimensional scalar functions. IEEE Transactions on Visualization and Computer Graphics, 16(6), 1271-1280.
Gerber, S., & Potter, K. (2011). Morse-Smale approximation, regression and visualization.
Chen, Y. C., Genovese, C. R., & Wasserman, L. (2016). A comprehensive approach to mode clustering. Electronic Journal of Statistics, 10(1), 210-241.
• Clustering based on multiple levels of function behavior—recent Nature paper and analytics company.
Singh, G., Mémoli, F., & Carlsson, G. E. (2007, September). Topological Methods for the Analysis of High Dimensional Data Sets and 3D Object Recognition. In SPBG (pp. 91-100).
• Typical method, requires stationarity assumptions, constant error over time, and no predictors/confounders.
Wu, J. P., & Wei, S. (1989). Time series analysis. Hunan Science and Technology Press, ChangSha.
• Stationarity and error assumptions; assumptions about outcome distribution (must be GLM in current software).
Ullman, J. B., & Bentler, P. M. (2003). Structural equation modeling. John Wiley & Sons, Inc..
• Some restrictions on predictor/outcome probability distributions (binomial/multinomial).
Heckerman, D., Geiger, D., & Chickering, D. M. (1994, July). Learning Bayesian networks: The combination of knowledge and statistical data. In Proceedings of the Tenth international conference on Uncertainty in artificial intelligence (pp. 293-301). Morgan Kaufmann Publishers Inc..
• Can handle multivariate time series. Based on nonlinear dynamics decomposition/spectral analysis methods.
Vautard, R., Yiou, P., & Ghil, M. (1992). Singular-spectrum analysis: A toolkit for short, noisy chaotic signals. Physica D: Nonlinear Phenomena, 58(1), 95-126.
• Clustering memories.
Dunham, M. H., Meng, Y., & Huang, J. (2004, November). Extensible markov model. In Data Mining, 2004. ICDM'04. Fourth IEEE International Conference on (pp. 371-374). IEEE.
• #### Description
A start guide to the concepts and algorithms in machine learning, including regression frameworks, ensemble methods, clustering, optimization, and more. Mathematical knowledge is not assumed, and pictures/analogies demonstrate the key concepts behind popular and cutting-edge methods in data analysis.
Updated to include newer algorithms, such as XGBoost, and more geometrically/topologically-based algorithms. Also includes a short overview of time series analysis
#### Transcript
1. 1. Colleen M. Farrelly
2. 2. Many machine learning methods exist in the literature and in industry. ◦ What works well for one problem may not work well for the next problem. ◦ In addition to poor model fit, an incorrect application of methods can lead to incorrect inference. Implications for data-driven business decisions. Low future confidence in data science and its results. Lower quality software products. Understanding the intuition and mathematics behind these methods can ameliorate these problems. ◦ This talk focuses on building intuition. ◦ Links to theoretical papers underlying each method.
3. 3. Total variance of a normally- distributed outcome as cookie jar. Error as empty space. Predictors accounting for pieces of the total variance as cookies. ◦ Based on relationship to predictor and to each other. Cookies accounting for the same piece of variance as those smooshed together. Many statistical assumptions need to be met. www.zelcoviacookies.com
4. 4. Extends multiple regression. ◦ Many types of outcome distributions. ◦ Transform an outcome variable to create a linear relationship with predictors. Sort of like silly putty stretching the outcome variable in the data space. Does not work on high- dimensional data where predictors > observations. Only certain outcome distributions. ◦ Exponential family as example. tomstock.photoshelter.com
5. 5. Impose size and/or variable overlap constraints (penalties) on generalized linear models. ◦ Elastic net as hybrid of these constraints. ◦ Can handle large numbers of predictors. Reduce the number of predictors. ◦ Shrink some predictor estimates to 0. ◦ Examine sets of similar predictors. Similar to eating irrelevant cookies in the regression cookie jar or a cowboy at the origin roping coefficients that get too close
6. 6. Homotopy arrow example ◦ Red and blue arrows can be deformed into each other by wiggling and stretching the line path with anchors at start and finish of line ◦ Yellow arrow crosses holes and would need to backtrack or break to the surface to freely wiggle into the blue or red line Homotopy method in LASSO/LARS wiggles an easy regression path into an optimal regression path ◦ Avoids obstacles that can trap other regression estimators (peaks, valleys, saddles…) Homotopy as path equivalence ◦ Intrinsic property of topological spaces (such as data manifolds)
7. 7. Instead of fitting model to data, fit model to tangent space (what isn’t the data). ◦ Deals with collinearity, as parallel vectors share a tangent space (only one selected of collinear group). ◦ LASSO and LARS extensions. ◦ Rao scoring for selection. Effect estimates (angles). Model selection criteria. Information criteria. Deviance scoring.
8. 8. Fit all possible models and figure out likelihood of each given observed data. ◦ Based on Bayes’ Theorem and conditional probability. ◦ Instead of giving likelihood that data came from a specific parameterized population (univariate), figure out likelihood of set of data coming from sets of population (multivariate). ◦ Can select naïve prior (no assumptions on model or population) or make an informed guess (assumptions about population or important factors in model). ◦ Combine multiple models according to their likelihoods into a blended model given data.
9. 9. Semi-Parametric Extensions of Regression
10. 10. Extends a generalized linear models by estimating unknown (nonlinear) functions of an outcome on intervals of a predictor. Use of “knots” to break function into intervals. Similar to a downhill skier. ◦ Slope as a function of outcome. ◦ Skier as spline. ◦ Flags as knots anchoring the skier as he travels down the slope. www.dearsportsfan.com
11. 11. Multivariate adaptive regression splines as an extension of spline models to multivariate data ◦ Knots placed according to multivariate structure and splines fit between knots ◦ Much like fixing a rope to the peaks and valleys of a mountain range, where the rope has enough slack to hug the terrain between its fixed points
12. 12. Extends spline models to the relationships of many predictors to an outcome. ◦ Also allows for a “silly- putty” transformation of the outcome variable. Like multiple skiers on many courses and mountains to estimate many relationships in the dataset. www.filigreeinn.com
13. 13. Chop data into partitions and then fit multiple regression models to each partition. Divide-and-conquer approach. Examples: ◦ Multivariate adaptive regression splines ◦ Regression trees ◦ Morse-Smale regression www.pinterest.com
14. 14. Based on a branch of math called topology. ◦ Study of changes in function behavior on shapes. ◦ Used to classify similarities/ differences between shapes. ◦ Data clouds turned into discrete shape combinations (simplices). Use these principles to partition data and fit elastic net models to each piece. ◦ Break data into multiple toy sets. ◦ Analyze sets for underlying properties of each toy. Useful visual output for additional data mining.
15. 15. Linear or non-linear support beams used to separate group data for classification or map data for kernel- based regression. Much like scaffolding and support beams separating and holding up parts of a high rise. http://en.wikipedia.org/wiki/Support_vector_machine leadertom.en.ec21.com
16. 16. Based on processing complex, nonlinear information the way the human brain does via a series of feature mappings. colah.github.io www.alz.org Arrows denote mapping functions, which take one topological space to another
17. 17. Type of shallow, wide neural network. Reduces framework to a penalized linear algebra problem, rather than iterative training (much faster to solve). Based on random mappings. Shown to converge to correct classification/regression (universal approximation property—may require unreasonably wide networks). Semi-supervised learning extensions.
18. 18. Added layers in neural network to solve width problem in single-layer networks for universal approximation. More effective in learning features of the data. Like sifting data with multiple sifters to distill finer and finer pieces of the data. Computationally intensive and requires architecture design and optimization.
19. 19. Recent extension of deep learning framework from spreadsheet data to multiple simultaneous spreadsheets ◦ Spreadsheets may be of the same dimension or different dimensions ◦ Could process multiple or hierarchical networks via adjacency matrices Like sifting through data with multiple inputs of varying sizes and textures
20. 20. CHAPTER 3:
21. 21. Classifies data according to optimal partitioning of data (visualized as a high- dimensional cube). Slices data into squares, cubes, and hypercubes (4+ dimensional cubes). ◦ Like finding the best place to cut through the data with a sword. Option to prune tree for better model (glue some of the pieces back together). Notoriously unstable. Optimization possible to arrive at best possible trees (genetic algorithms). tvtropes.org
22. 22. Optimization technique. ◦ Minimize a loss function in regression. Accomplished by choosing a variable per step that achieves largest minimization. Climber trying to descend a mountain. ◦ Minimize height above sea level per step. Directions (N, S, E, W) as a collection of variables. Climber rappels down cliff according to these rules. www.chinatravelca.com
23. 23. Optimization helpers. ◦ Find global maximum or minimum by searching many areas simultaneously. ◦ Can impose restrictions. Teams of mountain climbers trying to find the summit. Restrictions as climber supplies, hours of daylight left to find summit… Genetic algorithm as population of mountain climbers each climbing on his/her own. wallpapers.brothersoft.com
24. 24. One climber exploring multiple routes simultaneously. Individual optimized through a series of gates at each update until superposition converges on one state ◦ Say wave 2 is the optimal combination of predictors ◦ Resultant combination slowly flattens to wave 2, with the states of wave 1disappearing http://philschatz.com/physics -book/contents/m42249.html
25. 25. Combine multiple models of the same type (ex. trees) for better prediction. ◦ Single models unstable. Equally-good estimates from different models. ◦ Use bootstrapping. Multiple “marble draws” followed by model creation. ◦ Creates diversity of features. ◦ Creates models with different biases and error.
26. 26. Uses gradient descent algorithm to model an outcome based on predictors (linear, tree, spline…). ◦ Proceeds in steps, adding variables and adjusting weights according to the algorithm. Like putting together a puzzle. ◦ Algorithm first focuses on most important parts of the picture (such as the Mona Lisa’s eyes). ◦ Then adds nuances that help identify other missing pieces (hands, landscape…). play.google.com
27. 27. Adds a penalty term to boosted regression ◦ Can be formulated as LASSO/ridge regression/elastic net with constraints ◦ Also leverages hardware to further speed up boosting ensemble play.google.com
28. 28. Single tree models poor predictors. Grow a forest of them for a strong predictor (another ensemble method). Algorithm: ◦ Takes random sample of data. ◦ Builds one tree. ◦ Repeats until forest grown. ◦ Averages across forest to identify strong predictors.
29. 29. CHAPTER 4:
30. 30. Principle Component Analysis (PCA) ◦ Variance partitioning to combine pieces of variables into new linear subspace. ◦ Smashing cookies by kind and combining into a flat hybrid cookie in previous regression model. Manifold Learning ◦ PCA-like algorithms that combine pieces into a new nonlinear subspace. ◦ Non-flat cookie combining. Useful as pre-processing step for prediction models. ◦ Reduce dimension. ◦ Obtain uncorrelated, non- overlapping variables (bases). marmaladeandmileposts. com
31. 31. Balanced sampling for low-frequency predictors. ◦ Stratified samples (i.e. sample from bag of mostly white marbles and few red marbles with constraint that 1/5th of draws must be red marbles). Dimension reduction/mapping pre-processing ◦ Principle component, manifold learning… ◦ Hybrid of neural network methods and tree models.
32. 32. Aggregation of multiple types of models. ◦ Like a small town election. ◦ Different people have different views of the politics and care about different issues. Different modeling methods capture different pieces of the data and vote in different pieces. ◦ Leverage strengths, minimize weaknesses ◦ Diversity of methods to better explore underlying data geometry Avoids multiple testing issues.
33. 33. Subsembles ◦ Partition data into training sets Randomly selected or through partition algorithm ◦ Train a model on each data partition ◦ Combine into final weighted prediction model This is similar to national elections. ◦ Each elector in the electoral college learns for whom his constituents voted. ◦ The final electoral college pools these individual votes. Full Training Dataset Model Training Datasets 1, 2, 3, 4 Final Subsemble Model
34. 34. Combines superlearning and subsembles for better prediction ◦ Diversity improves subsemble method (better able to explore data geometry) ◦ Bootstrapping improves superlearner pieces (more diversity within each method) ◦ Preliminary empirical evidence shows efficacy of combination.
35. 35. CHAPTER 5:
36. 36. Classification of a data point based on the classification of its nearest neighboring points. Like a junior high lunchroom and student clicks. ◦ Students at the same table tend to be more similar to each other than to a distant table. www.wrestlecrap.com
37. 37. Iteratively separating groups within a dataset based on similarity. Like untangling tangled balls of yarn into multiple, single- colored balls (ideally). ◦ Each step untangles the mess a little further. ◦ Few stopping guidelines (when it looks separated). krazymommakreations.blogspot.com
38. 38. Hybrid of supervised and unsupervised learning (groupings and prediction). Uses graphical representation of data and its relationships. Algorithms with connections to topology, differential geometry, and Markov chains. Useful in combination with other machine learning methods to provide extra insight (ex. spectral clustering).
39. 39. K-means algorithm with weighting and dimension reduction components of similarity measure. ◦ Simplify balls of string to warm colors and cool colors before untangling. Can be reformulated as a graph clustering problem. ◦ Partition subcomponents of a graph based on flow equations. www.simplepastimes.com
40. 40. Multivariate technique similar to mode or density clustering. ◦ Find peaks and valleys in data according to an input function on the data (level set slices)— much like a watershed on mountains. ◦ Separate data based on shared peaks and valleys across slices (shared multivariate density/gradient). ◦ Many nice theoretical developments on validity and convergence.
41. 41. Topological clustering. ◦ Define distance metric. ◦ Slice multidimensional dataset with Morse function. ◦ Examine function behavior across slice. ◦ Cluster function behavior. ◦ Iterate through multiple slices to obtain hierarchy of function behavior. Much like examining the behavior of multiple objects across a flip book. ◦ Nesting ◦ Cluster overlap Filtered functions then used to create various resolutions of a modified Reeb graph summary of topology.
42. 42. Time Series Forecasting
43. 43. Similar to decomposing superposed states ◦ Seasonal trends ◦ Yearly trends ◦ Trend averages ◦ Dependencies on previous time point Knit individual forecasted pieces into a complete forecast by superposing these individual forecasts Several extensions to neural networks, time- lagged machine learning models…
44. 44. A time-series method incorporating predictors ◦ Constant predictors at initial time point ◦ Varying predictors at multiple time points Creates a sort of correlation web between predictors and time points ◦ Can handle multiple time lags and multivariate outcomes ◦ Can handle any GLM outcome links Related to partial differential equations of dynamic systems
45. 45. Data-based mining for SEM relationships/time-lag components ◦ Leverages conditional probability between predictors to find dependencies ◦ Does not require a priori model formulation like SEM Peeking at data to create a dependency web over time or predictors/outcome Can be validated by a follow-up SEM based on network structure Time 1 Outcome Predictor 2 Time 2 Outcome Time 3 Outcome
46. 46. Technically: ◦ Matrix decomposition (similar to PCA/manifold learning) ◦ Followed by spectral methods ◦ Cleaning of time-lagged covariance matrix ◦ Reconstruction with simple forecast Kind of like deconstructing, cleaning, a rebuilding a car engine
47. 47. Combines k-nearest neighbors-based clustering with memoryless state changes converging to a transition distribution (weighted directed graph) ◦ Reduce an observation to a pattern ◦ Remember patterns seen (across time or space) ◦ Match new observations to this set of patterns ◦ Computationally more feasible than k-means clustering Like remembering classes of chess board configurations across games played
48. 48. Many machine learning methods exist today, and many more are being developed every day. Methods come with certain assumptions that must be met. ◦ Breaking assumptions can lead to poor model fit or incorrect inference. ◦ Matching a method to a problem not only can help with better prediction and inference; it can also lead to faster computation times and better presentations to clients. Development depends on problem-matching and deep understanding of the mathematics behind the methods used.
49. 49. Parametric Regression: ◦ Draper, N. R., Smith, H., & Pownell, E. (1966). Applied regression analysis (Vol. 3). New York: Wiley. ◦ McCullagh, P. (1984). Generalized linear models. European Journal of Operational Research, 16(3), 285-292. ◦ Zou, H., & Hastie, T. (2005). Regularization and variable selection via the elastic net. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 67(2), 301-320. ◦ Augugliaro, L., Mineo, A. M., & Wit, E. C. (2013). Differential geometric least angle regression: a differential geometric approach to sparse generalized linear models. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 75(3), 471-498. ◦ Raftery, A. E., Madigan, D., & Hoeting, J. A. (1997). Bayesian model averaging for linear regression models. Journal of the American Statistical Association, 92(437), 179-191. ◦ Osborne, M. R., & Turlach, B. A. (2011). A homotopy algorithm for the quantile regression lasso and related piecewise linear problems. Journal of Computational and Graphical Statistics, 20(4), 972-987. ◦ Drori, I., & Donoho, D. L. (2006, May). Solution of l 1 minimization problems by LARS/homotopy methods. In Acoustics, Speech and Signal Processing, 2006. ICASSP 2006 Proceedings. 2006 IEEE International Conference on (Vol. 3, pp. III-III). IEEE.
50. 50. Semi-Parametric Regression: ◦ Marsh, L. C., & Cormier, D. R. (2001). Spline regression models (Vol. 137). Sage. ◦ Hastie, T., & Tibshirani, R. (1986). Generalized additive models. Statistical science, 297-310. ◦ McZgee, V. E., & Carleton, W. T. (1970). Piecewise regression. Journal of the American Statistical Association, 65(331), 1109-1124. ◦ Gerber, S., Rübel, O., Bremer, P. T., Pascucci, V., & Whitaker, R. T. (2013). Morse– Smale Regression. Journal of Computational and Graphical Statistics, 22(1), 193- 214. ◦ Hearst, M. A., Dumais, S. T., Osman, E., Platt, J., & Scholkopf, B. (1998). Support vector machines. Intelligent Systems and their Applications, IEEE, 13(4), 18-28. ◦ Hornik, K., Stinchcombe, M., & White, H. (1989). Multilayer feedforward networks are universal approximators. Neural networks, 2(5), 359-366. ◦ Krizhevsky, A., Sutskever, I., & Hinton, G. E. (2012). Imagenet classification with deep convolutional neural networks. In Advances in neural information processing systems (pp. 1097-1105). ◦ Huang, G. B., Wang, D. H., & Lan, Y. (2011). Extreme learning machines: a survey. International Journal of Machine Learning and Cybernetics, 2(2), 107-122. ◦ Friedman, J. H. (1991). Multivariate adaptive regression splines. The annals of statistics, 1-67. ◦ Abadi, M., Agarwal, A., Barham, P., Brevdo, E., Chen, Z., Citro, C., ... & Ghemawat, S. (2016). Tensorflow: Large-scale machine learning on heterogeneous distributed systems. arXiv preprint arXiv:1603.04467.
51. 51. Nonparametric Regression: ◦ Buntine, W. (1992). Learning classification trees. Statistics and computing, 2(2), 63-73. ◦ Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378. ◦ Bäck, T. (1996). Evolutionary algorithms in theory and practice: evolution strategies, evolutionary programming, genetic algorithms. Oxford university press. ◦ Zhang, G. (2011). Quantum-inspired evolutionary algorithms: a survey and empirical study. Journal of Heuristics, 17(3), 303-351. ◦ Dietterich, T. G. (2000). Ensemble methods in machine learning. In Multiple classifier systems (pp. 1-15). Springer Berlin Heidelberg. ◦ Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378. ◦ Breiman, L. (2001). Random forests. Machine learning, 45(1), 5-32. ◦ Chen, T., & Guestrin, C. (2016, August). Xgboost: A scalable tree boosting system. In Proceedings of the 22nd acm sigkdd international conference on knowledge discovery and data mining (pp. 785-794). ACM.
52. 52. Supervised with Unsupervised Methods: ◦ van der Maaten, L. J., Postma, E. O., & van den Herik, H. J. (2009). Dimensionality reduction: A comparative review. Journal of Machine Learning Research, 10(1-41), 66-71. ◦ Kuncheva, L. I., & Rodríguez, J. J. (2007). An experimental study on rotation forest ensembles. In Multiple Classifier Systems (pp. 459-468). Springer Berlin Heidelberg. ◦ van der Laan, M. J., Polley, E. C., & Hubbard, A. E. (2007). Super learner. Statistical applications in genetics and molecular biology, 6(1). ◦ Sapp, S. K. (2014). Subsemble: A Flexible Subset Ensemble Prediction Method (Doctoral dissertation, University of California, Berkeley).
53. 53. Unsupervised Methods: ◦ Fukunaga, K., & Narendra, P. M. (1975). A branch and bound algorithm for computing k-nearest neighbors. Computers, IEEE Transactions on, 100(7), 750-753. ◦ MacQueen, J. (1967, June). Some methods for classification and analysis of multivariate observations. In Proceedings of the fifth Berkeley symposium on mathematical statistics and probability (Vol. 1, No. 14, pp. 281-297). ◦ Chartrand, G., & Oellermann, O. R. (1993). Applied and algorithmic graph theory. ◦ Ng, A. Y., Jordan, M. I., & Weiss, Y. (2002). On spectral clustering: Analysis and an algorithm. Advances in neural information processing systems, 2, 849-856. ◦ Singh, G., Mémoli, F., & Carlsson, G. E. (2007, September). Topological Methods for the Analysis of High Dimensional Data Sets and 3D Object Recognition. In SPBG (pp. 91- 100). ◦ Chen, Y. C., Genovese, C. R., & Wasserman, L. (2016). A comprehensive approach to mode clustering. Electronic Journal of Statistics, 10(1), 210-241.
54. 54. Time Series ◦ Wu, J. P., & Wei, S. (1989). Time series analysis. Hunan Science and Technology Press, ChangSha. ◦ Vautard, R., Yiou, P., & Ghil, M. (1992). Singular-spectrum analysis: A toolkit for short, noisy chaotic signals. Physica D: Nonlinear Phenomena, 58(1), 95-126. ◦ Dunham, M. H., Meng, Y., & Huang, J. (2004, November). Extensible markov model. In Data Mining, 2004. ICDM'04. Fourth IEEE International Conference on (pp. 371-374). IEEE. ◦ Ullman, J. B., & Bentler, P. M. (2003). Structural equation modeling. John Wiley & Sons, Inc.. ◦ Heckerman, D., Geiger, D., & Chickering, D. M. (1994, July). Learning Bayesian networks: The combination of knowledge and statistical data. In Proceedings of the Tenth international conference on Uncertainty in artificial intelligence (pp. 293-301). Morgan Kaufmann Publishers Inc..
#### Editor's Notes
• Assumes predictors and observations are independent.
Assumes errors are normally distributed and independent.
Assumes linear relationship between predictors and outcome.
Assumes normally-distributed outcome.
Draper, N. R., Smith, H., & Pownell, E. (1966). Applied regression analysis (Vol. 3). New York: Wiley.
• Same assumptions as multiple regression, minus outcome’s normal distribution (link function extends to non-normal distributions).
McCullagh, P. (1984). Generalized linear models. European Journal of Operational Research, 16(3), 285-292.
• Relaxes predictor independence requirement and adds penalty term.
Adds a penalty to reduce generalized linear model’s model size.
Zou, H., & Hastie, T. (2005). Regularization and variable selection via the elastic net. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 67(2), 301-320.
• Exists for several types of models, including survival, binomial, and Poisson regression models
Augugliaro, L., Mineo, A. M., & Wit, E. C. (2013). Differential geometric least angle regression: a differential geometric approach to sparse generalized linear models. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 75(3), 471-498.
Augugliaro, L., & Mineo, A. M. (2015). Using the dglars Package to Estimate a Sparse Generalized Linear Model. In Advances in Statistical Models for Data Analysis (pp. 1-8). Springer International Publishing.
• Joint model likelihood given multivariate data vs. parameter likelihood given univariate data
Raftery, A. E., Madigan, D., & Hoeting, J. A. (1997). Bayesian model averaging for linear regression models. Journal of the American Statistical Association, 92(437), 179-191.
Raftery, A. E., Gneiting, T., Balabdaoui, F., & Polakowski, M. (2005). Using Bayesian model averaging to calibrate forecast ensembles. Monthly Weather Review, 133(5), 1155-1174.
Madigan, D., Raftery, A. E., Volinsky, C., & Hoeting, J. (1996). Bayesian model averaging. In Proceedings of the AAAI Workshop on Integrating Multiple Learned Models, Portland, OR (pp. 77-83).
• Relaxes linear relationship of predictors to outcome requirement.
Marsh, L. C., & Cormier, D. R. (2001). Spline regression models (Vol. 137). Sage.
• Generally flexible and able to handle high-dimensional data compared to other spline methods.
Friedman, J. H. (1991). Multivariate adaptive regression splines. The annals of statistics, 1-67.
• Relaxes linear relationship of predictors to outcome requirement and outcome normality assumption.
Hastie, T., & Tibshirani, R. (1986). Generalized additive models. Statistical science, 297-310.
• Regression by partition.
McZgee, V. E., & Carleton, W. T. (1970). Piecewise regression. Journal of the American Statistical Association, 65(331), 1109-1124.
• Regression by data partitions defined by multivariate function behavior.
Gerber, S., Rübel, O., Bremer, P. T., Pascucci, V., & Whitaker, R. T. (2013). Morse–Smale Regression. Journal of Computational and Graphical Statistics, 22(1), 193-214.
• Classification tool by combined data splits.
Singularity issues.
Computationally taxing, particularly as number of groups increases.
Can be used in semi-parametric regression as nonlinear mapping of regression kernels.
Hearst, M. A., Dumais, S. T., Osman, E., Platt, J., & Scholkopf, B. (1998). Support vector machines. Intelligent Systems and their Applications, IEEE, 13(4), 18-28.
• Computationally expensive in traditional algorithms and rooted in topological maps.
Cannot handle lots of variables compared to number of observations.
Cannot handle non-independent data.
Hornik, K., Stinchcombe, M., & White, H. (1989). Multilayer feedforward networks are universal approximators. Neural networks, 2(5), 359-366.
• Random mappings to reduce MLP to linear system of equations.
Huang, G. B., Wang, D. H., & Lan, Y. (2011). Extreme learning machines: a survey. International Journal of Machine Learning and Cybernetics, 2(2), 107-122.
• Computationally expensive neural network extension.
Still suffers from singularities which hinder performance.
Krizhevsky, A., Sutskever, I., & Hinton, G. E. (2012). Imagenet classification with deep convolutional neural networks. In Advances in neural information processing systems (pp. 1097-1105).
• More flexible than deep learning architectures (different types of simultaneous input data—graphs, spreadsheets, images…).
Abadi, M., Agarwal, A., Barham, P., Brevdo, E., Chen, Z., Citro, C., ... & Ghemawat, S. (2016). Tensorflow: Large-scale machine learning on heterogeneous distributed systems. arXiv preprint arXiv:1603.04467.
• Does not assume anything about data and regression parameters.
Can be unstable (many optimal solutions).
Can be used to classify or regress.
Buntine, W. (1992). Learning classification trees. Statistics and computing, 2(2), 63-73.
• A type of optimization that works by finding the quickest route to a solution.
Can be very accurate, but takes an inordinate amount of time for some problems (big data, neural network training…).
Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378.
• Optimization technique.
Computationally expensive but guaranteed to converge.
Bäck, T. (1996). Evolutionary algorithms in theory and practice: evolution strategies, evolutionary programming, genetic algorithms. Oxford university press.
• Generally much faster convergence than genetic algorithms.
Zhang, G. (2011). Quantum-inspired evolutionary algorithms: a survey and empirical study. Journal of Heuristics, 17(3), 303-351.
• Exploit weak learners.
Dietterich, T. G. (2000). Ensemble methods in machine learning. In Multiple classifier systems (pp. 1-15). Springer Berlin Heidelberg.
• Useful when the previous types of regression fail, when the data is “noisy” with little known about which predictors are important, and when outliers are likely to exist in the data (piece-wise modeling method of sorts).
Friedman, J. H. (2002). Stochastic gradient boosting. Computational Statistics & Data Analysis, 38(4), 367-378.
• Speeds up boosting but requires extensive tuning and does not give interpretable model (only prediction). Has been used a lot in recent Kaggle competitions, particle physics datasets.
Chen, T., & Guestrin, C. (2016, August). Xgboost: A scalable tree boosting system. In Proceedings of the 22nd acm sigkdd international conference on knowledge discovery and data mining (pp. 785-794). ACM.
• Useful in modeling complex, nonlinear regression/classification problems.
Also distance-based metric. Handles noisy data well.
Breiman, L. (2001). Random forests. Machine learning, 45(1), 5-32.
• Variance combined.
van der Maaten, L. J., Postma, E. O., & van den Herik, H. J. (2009). Dimensionality reduction: A comparative review. Journal of Machine Learning Research, 10(1-41), 66-71.
• Deep connections to sampling theory and topology.
Kuncheva, L. I., & Rodríguez, J. J. (2007). An experimental study on rotation forest ensembles. In Multiple Classifier Systems (pp. 459-468). Springer Berlin Heidelberg.
• Bagging of different base models (same bootstrap or different bootstrap).
van der Laan, M. J., Polley, E. C., & Hubbard, A. E. (2007). Super learner. Statistical applications in genetics and molecular biology, 6(1).
• Good prediction with ability to parallelize for distributed computing. Tends to be faster than superlearners (smaller training samples).
Sapp, S. K. (2014). Subsemble: A Flexible Subset Ensemble Prediction Method (Doctoral dissertation, University of California, Berkeley).
• New formulation by author (Colleen M. Farrelly). Requires a minimum sample size of ~30 per bootstrap within method.
• Local classification.
Forms basis of many other machine learning algorithms (manifold learning, clustering, certain Markov chains…).
Fukunaga, K., & Narendra, P. M. (1975). A branch and bound algorithm for computing k-nearest neighbors. Computers, IEEE Transactions on, 100(7), 750-753.
• Creates groupings of like objects in the dataset based on different linear/nonlinear metrics.
Assumes groups are separable (no overlap).
Issues with computation as number of observations increases.
MacQueen, J. (1967, June). Some methods for classification and analysis of multivariate observations. In Proceedings of the fifth Berkeley symposium on mathematical statistics and probability (Vol. 1, No. 14, pp. 281-297).
• Data by pictures.
Chartrand, G., & Oellermann, O. R. (1993). Applied and algorithmic graph theory.
• Weighted kernel k-means (weighted subcomponent clustering).
Ng, A. Y., Jordan, M. I., & Weiss, Y. (2002). On spectral clustering: Analysis and an algorithm. Advances in neural information processing systems, 2, 849-856.
• Related to density clustering but relies on Morse critical points and filter functions to define complexes (clusters)
Gerber, S., Bremer, P. T., Pascucci, V., & Whitaker, R. (2010). Visual exploration of high dimensional scalar functions. IEEE Transactions on Visualization and Computer Graphics, 16(6), 1271-1280.
Gerber, S., & Potter, K. (2011). Morse-Smale approximation, regression and visualization.
Chen, Y. C., Genovese, C. R., & Wasserman, L. (2016). A comprehensive approach to mode clustering. Electronic Journal of Statistics, 10(1), 210-241.
• Clustering based on multiple levels of function behavior—recent Nature paper and analytics company.
Singh, G., Mémoli, F., & Carlsson, G. E. (2007, September). Topological Methods for the Analysis of High Dimensional Data Sets and 3D Object Recognition. In SPBG (pp. 91-100).
• Typical method, requires stationarity assumptions, constant error over time, and no predictors/confounders.
Wu, J. P., & Wei, S. (1989). Time series analysis. Hunan Science and Technology Press, ChangSha.
• Stationarity and error assumptions; assumptions about outcome distribution (must be GLM in current software).
Ullman, J. B., & Bentler, P. M. (2003). Structural equation modeling. John Wiley & Sons, Inc..
• Some restrictions on predictor/outcome probability distributions (binomial/multinomial).
Heckerman, D., Geiger, D., & Chickering, D. M. (1994, July). Learning Bayesian networks: The combination of knowledge and statistical data. In Proceedings of the Tenth international conference on Uncertainty in artificial intelligence (pp. 293-301). Morgan Kaufmann Publishers Inc..
• Can handle multivariate time series. Based on nonlinear dynamics decomposition/spectral analysis methods.
Vautard, R., Yiou, P., & Ghil, M. (1992). Singular-spectrum analysis: A toolkit for short, noisy chaotic signals. Physica D: Nonlinear Phenomena, 58(1), 95-126.
• Clustering memories.
Dunham, M. H., Meng, Y., & Huang, J. (2004, November). Extensible markov model. In Data Mining, 2004. ICDM'04. Fourth IEEE International Conference on (pp. 371-374). IEEE.
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See all | 16,821 | 67,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.896438 |
https://parametrichouse.com/1-02/ | 1,701,188,843,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00069.warc.gz | 503,495,150 | 20,005 | #### 1- Single Point Attractor
In this Grasshopper tutorial, you will learn how to use a single point attractor in your design. By understanding the concepts you can get ready for more complicated attractors and how to control them in next tutorials.
#### 2- Multiple Attractor Points
In this Grasshopper tutorial, we will understand what happens when we use two or more points as an attractor and how it affects our grid of points.
#### 4-Single Curve Attractor
In this grasshopper tutorial i will use a curve as an attractor and use it to deform a surface. You will learn how to define the distance between a set of point and a single curve.
#### 5- Parametric Curve Attractor
In this Grasshopper tutorial I will explain how you can use the u,v coordinate system to produce a parametric curve on a surface and use it as an attractor to affect the frames of triangular panels
#### 6-Multiple Curves
In this grasshopper tutorial, we will learn how to use a set of curves for attraction. You will learn how to use advanced grasshopper techniques to analyze the distance between the point grids and the attractors.
#### 7- Image Attractor
In this grasshopper tutorial, I will model a facade louver system based on scaling. First I will divide the facade into louvers and then I will extract the lines and scale them based on the height of the surface. then we will remodel the louvers with another technique and use point attractor and image attractor to affect the facade.
#### 8- Based on height
In this grasshopper tutorial, I will model a facade louver system based on scaling. First I will divide the facade into louvers and then I will extract the lines and scale them based on the height of the surface. then we will remodel the louvers with another technique and use point attractor and image attractor to affect the facade.
#### 9- Based on colour
In this tutorial, I will explain how you can use Gradient to color a set of panels based on their height. First we will produce the panels by isotrim and then we will extract the panels center and use the z component to define the height of the panels. Finally we will use....
#### 10 -Pt Attarctor triangles
In this Paracourse lessons you can use a series of point attractors to change the scale of a square grid and then connect the corners to the middle of the scaled border.
#### 11- Paneling Attractors
In this Paracourse Lesson, we are going to use different attractors in Paneling tools to model a parametric grid that deforms with Point/Curve/Random and surface curvature. Finally, we are going to smooth the result.
#### 12- Parametric Hexagons
In this Paracourse Lesson, we are going to model a series of hexagons which scale based on point attractors. We are also going to study how to fix the cells on the closed Nurbs surface seam.
#### 13- Rotating Panels
In this Paracourse lesson, we will model a series of rotating panels which extrude until they reach their neighboring surface. We will use three different techniques
#### 14- Paneling Tools (List)
In this Paracourse lesson, we will use the Paneling Tools plugin to cover a Nurbs surface with a series of modules. We will use point attractors to distribute the modules and at the end, we will count each module and export them in different layers.
#### 3- Parametric Point
In this Grasshopper tutorial, I will teach you how to define a parametric point in grasshopper and use it to design a framed facade. First we will find the distance between the attractor and the points and then we will define the extrusion | 794 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-50 | longest | en | 0.910194 |
http://us.metamath.org/mpeuni/uvcval.html | 1,653,386,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00422.warc.gz | 60,304,772 | 5,318 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > uvcval Structured version Visualization version GIF version
Theorem uvcval 20172
Description: Value of a single unit vector in a free module. (Contributed by Stefan O'Rear, 3-Feb-2015.)
Hypotheses
Ref Expression
uvcfval.u 𝑈 = (𝑅 unitVec 𝐼)
uvcfval.o 1 = (1r𝑅)
uvcfval.z 0 = (0g𝑅)
Assertion
Ref Expression
uvcval ((𝑅𝑉𝐼𝑊𝐽𝐼) → (𝑈𝐽) = (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )))
Distinct variable groups: 1 ,𝑘 𝑅,𝑘 𝑘,𝐼 0 ,𝑘 𝑘,𝐽
Allowed substitution hints: 𝑈(𝑘) 𝑉(𝑘) 𝑊(𝑘)
Proof of Theorem uvcval
Dummy variable 𝑗 is distinct from all other variables.
StepHypRef Expression
1 uvcfval.u . . . . 5 𝑈 = (𝑅 unitVec 𝐼)
2 uvcfval.o . . . . 5 1 = (1r𝑅)
3 uvcfval.z . . . . 5 0 = (0g𝑅)
41, 2, 3uvcfval 20171 . . . 4 ((𝑅𝑉𝐼𝑊) → 𝑈 = (𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 ))))
54fveq1d 6231 . . 3 ((𝑅𝑉𝐼𝑊) → (𝑈𝐽) = ((𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 )))‘𝐽))
653adant3 1101 . 2 ((𝑅𝑉𝐼𝑊𝐽𝐼) → (𝑈𝐽) = ((𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 )))‘𝐽))
7 simp3 1083 . . 3 ((𝑅𝑉𝐼𝑊𝐽𝐼) → 𝐽𝐼)
8 mptexg 6525 . . . 4 (𝐼𝑊 → (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )) ∈ V)
983ad2ant2 1103 . . 3 ((𝑅𝑉𝐼𝑊𝐽𝐼) → (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )) ∈ V)
10 eqeq2 2662 . . . . . 6 (𝑗 = 𝐽 → (𝑘 = 𝑗𝑘 = 𝐽))
1110ifbid 4141 . . . . 5 (𝑗 = 𝐽 → if(𝑘 = 𝑗, 1 , 0 ) = if(𝑘 = 𝐽, 1 , 0 ))
1211mpteq2dv 4778 . . . 4 (𝑗 = 𝐽 → (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 )) = (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )))
13 eqid 2651 . . . 4 (𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 ))) = (𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 )))
1412, 13fvmptg 6319 . . 3 ((𝐽𝐼 ∧ (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )) ∈ V) → ((𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 )))‘𝐽) = (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )))
157, 9, 14syl2anc 694 . 2 ((𝑅𝑉𝐼𝑊𝐽𝐼) → ((𝑗𝐼 ↦ (𝑘𝐼 ↦ if(𝑘 = 𝑗, 1 , 0 )))‘𝐽) = (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )))
166, 15eqtrd 2685 1 ((𝑅𝑉𝐼𝑊𝐽𝐼) → (𝑈𝐽) = (𝑘𝐼 ↦ if(𝑘 = 𝐽, 1 , 0 )))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1054 = wceq 1523 ∈ wcel 2030 Vcvv 3231 ifcif 4119 ↦ cmpt 4762 ‘cfv 5926 (class class class)co 6690 0gc0g 16147 1rcur 18547 unitVec cuvc 20169 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pr 4936 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-ral 2946 df-rex 2947 df-reu 2948 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-nul 3949 df-if 4120 df-sn 4211 df-pr 4213 df-op 4217 df-uni 4469 df-iun 4554 df-br 4686 df-opab 4746 df-mpt 4763 df-id 5053 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-uvc 20170 This theorem is referenced by: uvcvval 20173
Copyright terms: Public domain W3C validator | 2,040 | 3,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-21 | latest | en | 0.238974 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/construct-a-triangle-abc-in-which-bc-8-cm-b-45-and-ab-ac-3-5/constructions/6295177 | 1,670,240,050,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00097.warc.gz | 106,782,300 | 8,167 | # Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB − AC = 3.5 cm.
The below given steps will be followed to draw the required triangle.
Step I: Draw the line segment BC = 8 cm and at point B, make an angle of 45°, say ∠XBC.
Step II: Cut the line segment BD = 3.5 cm (equal to AB − AC) on ray BX.
Step III: Join DC and draw the perpendicular bisector PQ of DC.
Step IV: Let it intersect BX at point A. Join AC. ΔABC is the required triangle.
• 26
What are you looking for? | 147 | 489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-49 | latest | en | 0.848347 |
https://www.openvdb.org/documentation/doxygen/classnanovdb_1_1Stats_3_01ValueT_00_011_01_4.html | 1,669,976,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00642.warc.gz | 965,510,504 | 4,826 | OpenVDB 10.0.1
Stats< ValueT, 1 > Class Template Reference
This class computes statistics (minimum value, maximum value, mean, variance and standard deviation) of a population of floating-point values. More...
`#include <nanovdb/util/GridStats.h>`
Inherits Extrema< ValueT, 1 >.
## Public Types
using ValueType = ValueT
## Public Member Functions
Stats ()
Statsadd (const ValueT &val, uint64_t n)
Add n samples with constant value val. More...
Add the samples from the other Stats instance. More...
size_t size () const
double avg () const
Return the arithmetic mean, i.e. average, value. More...
double mean () const
Return the arithmetic mean, i.e. average, value. More...
double var () const
Return the population variance. More...
double variance () const
Return the population variance. More...
double std () const
Return the standard deviation (=Sqrt(variance)) as defined from the (biased) population variance. More...
double stdDev () const
Return the standard deviation (=Sqrt(variance)) as defined from the (biased) population variance. More...
## Static Public Member Functions
static constexpr bool hasMinMax ()
static constexpr bool hasAverage ()
static constexpr bool hasStdDeviation ()
## Protected Types
using BaseT = Extrema< ValueT, 1 >
using RealT = double
size_t mSize
double mAvg
double mAux
## Detailed Description
### template<typename ValueT> class nanovdb::Stats< ValueT, 1 >
This class computes statistics (minimum value, maximum value, mean, variance and standard deviation) of a population of floating-point values.
variance = Mean[ (X-Mean[X])^2 ] = Mean[X^2] - Mean[X]^2, standard deviation = sqrt(variance)
Note
This class employs incremental computation and double precision.
## Member Typedef Documentation
using BaseT = Extrema
protected
using RealT = double
protected
using ValueType = ValueT
## Constructor & Destructor Documentation
Stats ( )
inline
## Member Function Documentation
Stats& add ( const ValueT & val )
inline
Stats& add ( const ValueT & val, uint64_t n )
inline
Add n samples with constant value val.
Stats& add ( const Stats< ValueT, 1 > & other )
inline
Add the samples from the other Stats instance.
double avg ( ) const
inline
Return the arithmetic mean, i.e. average, value.
static constexpr bool hasAverage ( )
inlinestatic
static constexpr bool hasMinMax ( )
inlinestatic
static constexpr bool hasStdDeviation ( )
inlinestatic
double mean ( ) const
inline
Return the arithmetic mean, i.e. average, value.
size_t size ( ) const
inline
double std ( ) const
inline
Return the standard deviation (=Sqrt(variance)) as defined from the (biased) population variance.
double stdDev ( ) const
inline
Return the standard deviation (=Sqrt(variance)) as defined from the (biased) population variance.
double var ( ) const
inline
Return the population variance.
Note
The unbiased sample variance = population variance * num/(num-1)
double variance ( ) const
inline
Return the population variance.
Note
The unbiased sample variance = population variance * num/(num-1)
## Member Data Documentation
double mAux
protected
double mAvg
protected
size_t mSize
protected | 755 | 3,185 | {"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} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.513137 |
https://www.physicsforums.com/threads/about-current-from-a-battery-going-through-our-body.883408/ | 1,532,195,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00577.warc.gz | 964,131,747 | 17,868 | # About current from a battery going through our body
1. Aug 28, 2016
### KFC
Hi all,
I am reading some materials on electrical parameters of a AA battery which give 1.5V and about 2000mAh. As my understanding, the 2000mAh means it will produce 2A if we keep taking current from the battery for an hour. 2A is huge current for human being. I wonder why it won't hurt the body. Is it because it take so slow for the electron to accumulate so to get 2A and once some electrons arrive the body, they move to the ground immediately? So what happens if we isolate the body from the ground, will it cause any injure by a AA battery by holding two terminals for even 30 minutes?
2. Aug 28, 2016
### davenn
through the appropriate resistance
No ... it's because the body's skin resistance is sufficiently high enough that the 1.5V of the battery cannot breakdown that resistance and allow correct to flow
no, doesn't matter if you are grounded or not
Dave
3. Aug 28, 2016
### Aaron Crowl
You've misunderstood the units. mAh is a unit of energy (sort of). 2000mAh means that the battery could put out 200 Amps for 6 minutes, 2 Amps for an hour, 1 Amp for 2 hours, or 1mA for 2000 hours. It depends on the circuit that the battery is in.
There are more parameters for batteries.
A battery can only put out a limited current. This is usually represented with a "C" rating where ((Amp output)=C * (storage rating)) If your battery had a "C" of 1 then we drop the h from mAh and say that the battery can only safely put out 2000mA without blowing up. The "C" of your AA battery is probably much lower than 1. If "C=0.1" then the battery can only output 200mA.
batteries drop voltage as they discharge. The profile of this depends on the chemistry and number of cells.
4. Aug 28, 2016
### davenn
yes, this is correct
but from the OP's point of view about why there is no electric shock, it's not the relevant answer
Dave
5. Sep 6, 2016
### KFC
thanks for your explanation. So if for regular battery, C=0.1, you mean the maximum current is just 200mA by assuming the load is not zero ohm, is that right? I am curious where do we find the parameter of the battery for that C and mAh.
6. Sep 6, 2016
### Aaron Crowl
That's right. An AA battery has 1.5V so if our max current is 200mA then the lowest load resistance we could have is 7.5Ohm before the battery is in an unsafe region of operation. I just did a little Googling. It looks like the "C" rating is more common for NiMh, NiCad, and lithium batteries but I can't find anything like a chart for regular alkaline AA batteries.
7. Sep 7, 2016
### CWatters
It won't hurt because the current will not be 2A....
The resistance between my left hand and right hand is currently about 1,000,000 Ohms (I've just measured it). Most of that is the resistance of my skin. If I hold an AA battery with a thumb on each end the current flowing through me will (according to Ohms law) be..
I = V/R
= 1.5/1,000,000
= 1.5*10-6 Amps
That's quite a lot less than 2A and not enough to cause an electric shock. It's generally considered that currents of 10mA will be painful and that currents of 100 to 200mA can kill you.
However the resistance of the human body can vary quite a lot. Google says it can fall as low as 500 Ohms under certain conditions. This means the maximum current that an AA battery could send through me is around..
I = V/R
= 1.5/500
= 3 *10-3 Amps
= 3mA
As I recall most international safety standards limit accessible voltages to around 36-42V. Why? Because that would generate a worse case current of..
42/500 = 84mA and that is quite close to the 100mA considered capable of killing someone.
Aside: Batteries have some internal resistance which will also limit the maximum current they can deliver BUT usually it will be the external resistance that dominates the calculation. For example the internal resistance of a NiMH cell can be of order 10-3 Ohms which is negligible compared to the 500 to 1,000,000 Ohms of the human body.
8. Sep 8, 2016
### CraigHB
There's also something called power density which indicates how much power a battery can deliver per unit weight, can also be expressed in terms of volume. That's usually considered along with a battery's energy density which is the capacity (amp-hours) per unit weight, can also be expressed in terms of volume. This is important in considering how much power a battery can actually deliver efficiently.
Some power is always lost inside a battery as a function of its IR (internal resistance). The IR of a battery has to be considered in a circuit so even if you short a battery with a wire, current flow will be a function of the battery's IR according to Ohm's law. An alkaline AA battery has fairly high IR compared to other battery chemistries, about 200mΩ. So a short circuit across its terminals would result in a current flow about 6 Amps. With a capacity of 2000mAh, it could to that for about 20 minutes. On the other hand, a high drain or high C rate lithium-ion battery may have IR of only a few mΩ. In that case massive currents can flow when shorted, enough to cause extreme heating and fire. So you can see the power density of a high C rate Li-Ion battery is much greater than that of an alkaline battery even though energy density is not hugely different.
The 1C rate of an AA Alkaline battery as you mentioned (2 Amps) would get into an area of poor efficiency. In a circuit there would be the power loss of that 2A across the battery's IR which would be close to a Watt. Using rough numbers that's compared to about 2 Watts for the load. You can see a significant portion of the total power consumed is lost inside the battery. For a load of 2 Amps you'd want a battery with higher power density (lower IR) to provide better efficiency. Typically alkalines are used to provide only lower currents because they are primary (can not be recharged) and don't deliver higher currents efficiently. For higher power applications you would go with a secondary battery (rechargeable) having higher power density like Li-Ion or NiMH.
Last edited: Sep 8, 2016 | 1,522 | 6,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-30 | latest | en | 0.94159 |
https://netlib.sandia.gov/slatec/src/bskin.f | 1,716,014,529,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00628.warc.gz | 380,558,034 | 5,026 | *DECK BSKIN SUBROUTINE BSKIN (X, N, KODE, M, Y, NZ, IERR) C***BEGIN PROLOGUE BSKIN C***PURPOSE Compute repeated integrals of the K-zero Bessel function. C***LIBRARY SLATEC C***CATEGORY C10F C***TYPE SINGLE PRECISION (BSKIN-S, DBSKIN-D) C***KEYWORDS BICKLEY FUNCTIONS, EXPONENTIAL INTEGRAL, C INTEGRALS OF BESSEL FUNCTIONS, K-ZERO BESSEL FUNCTION C***AUTHOR Amos, D. E., (SNLA) C***DESCRIPTION C C The following definitions are used in BSKIN: C C Definition 1 C KI(0,X) = K-zero Bessel function. C C Definition 2 C KI(N,X) = Bickley Function C = integral from X to infinity of KI(N-1,t)dt C for X .ge. 0 and N = 1,2,... C ____________________________________________________________________ C BSKIN computes sequences of Bickley functions (repeated integrals C of the K0 Bessel function); i.e. for fixed X and N and K=1,..., C BSKIN computes the M-member sequence C C Y(K) = KI(N+K-1,X) for KODE=1 C or C Y(K) = EXP(X)*KI(N+K-1,X) for KODE=2, C C for N.ge.0 and X.ge.0 (N and X cannot be zero simultaneously). C C INPUT C X - Argument, X .ge. 0.0E0 C N - Order of first member of the sequence N .ge. 0 C KODE - Selection parameter C KODE = 1 returns Y(K)= KI(N+K-1,X), K=1,M C = 2 returns Y(K)=EXP(X)*KI(N+K-1,X), K=1,M C M - Number of members in the sequence, M.ge.1 C C OUTPUT C Y - A vector of dimension at least M containing the C sequence selected by KODE. C NZ - Underflow flag C NZ = 0 means computation completed C = M means an exponential underflow occurred on C KODE=1. Y(K)=0.0E0, K=1,...,M is returned C IERR - Error flag C IERR = 0, Normal return, computation completed. C = 1, Input error, no computation. C = 2, Error, no computation. The C termination condition was not met. C C The nominal computational accuracy is the maximum of unit C roundoff (=R1MACH(4)) and 1.0e-18 since critical constants C are given to only 18 digits. C C DBSKIN is the double precision version of BSKIN. C C *Long Description: C C Numerical recurrence on C C (L-1)*KI(L,X) = X(KI(L-3,X) - KI(L-1,X)) + (L-2)*KI(L-2,X) C C is stable where recurrence is carried forward or backward C away from INT(X+0.5). The power series for indices 0,1 and 2 C on 0.le.X.le. 2 starts a stable recurrence for indices C greater than 2. If N is sufficiently large (N.gt.NLIM), the C uniform asymptotic expansion for N to INFINITY is more C economical. On X.gt.2 the recursion is started by evaluating C the uniform expansion for the three members whose indices are C closest to INT(X+0.5) within the set N,...,N+M-1. Forward C recurrence, backward recurrence or both, complete the C sequence depending on the relation of INT(X+0.5) to the C indices N,...,N+M-1. C C***REFERENCES D. E. Amos, Uniform asymptotic expansions for C exponential integrals E(N,X) and Bickley functions C KI(N,X), ACM Transactions on Mathematical Software, C 1983. C D. E. Amos, A portable Fortran subroutine for the C Bickley functions KI(N,X), Algorithm 609, ACM C Transactions on Mathematical Software, 1983. C***ROUTINES CALLED BKIAS, BKISR, EXINT, GAMRN, I1MACH, R1MACH C***REVISION HISTORY (YYMMDD) C 820601 DATE WRITTEN C 890531 Changed all specific intrinsics to generic. (WRB) C 891009 Removed unreferenced statement label. (WRB) C 891009 REVISION DATE from Version 3.2 C 891214 Prologue converted to Version 4.0 format. (BAB) C 920501 Reformatted the REFERENCES section. (WRB) C***END PROLOGUE BSKIN INTEGER I, ICASE, IERR, IL, I1M, K, KK, KODE, KTRMS, M, * M3, N, NE, NFLG, NL, NLIM, NN, NP, NS, NT, NZ INTEGER I1MACH REAL A, ENLIM, EXI, FN, GR, H, HN, HRTPI, SS, TOL, T1, T2, W, X, * XLIM, XNLIM, XP, Y, YS, YSS REAL GAMRN, R1MACH DIMENSION EXI(102), A(50), YS(3), YSS(3), H(31), Y(*) SAVE A, HRTPI C----------------------------------------------------------------------- C COEFFICIENTS IN SERIES OF EXPONENTIAL INTEGRALS C----------------------------------------------------------------------- DATA A(1), A(2), A(3), A(4), A(5), A(6), A(7), A(8), A(9), A(10), * A(11), A(12), A(13), A(14), A(15), A(16), A(17), A(18), A(19), * A(20), A(21), A(22), A(23), A(24) /1.00000000000000000E+00, * 5.00000000000000000E-01,3.75000000000000000E-01, * 3.12500000000000000E-01,2.73437500000000000E-01, * 2.46093750000000000E-01,2.25585937500000000E-01, * 2.09472656250000000E-01,1.96380615234375000E-01, * 1.85470581054687500E-01,1.76197052001953125E-01, * 1.68188095092773438E-01,1.61180257797241211E-01, * 1.54981017112731934E-01,1.49445980787277222E-01, * 1.44464448094367981E-01,1.39949934091418982E-01, * 1.35833759559318423E-01,1.32060599571559578E-01, * 1.28585320635465905E-01,1.25370687619579257E-01, * 1.22385671247684513E-01,1.19604178719328047E-01, * 1.17004087877603524E-01/ DATA A(25), A(26), A(27), A(28), A(29), A(30), A(31), A(32), * A(33), A(34), A(35), A(36), A(37), A(38), A(39), A(40), A(41), * A(42), A(43), A(44), A(45), A(46), A(47), A(48) * /1.14566502713486784E-01,1.12275172659217048E-01, * 1.10116034723462874E-01,1.08076848895250599E-01, * 1.06146905164978267E-01,1.04316786110409676E-01, * 1.02578173008569515E-01,1.00923686347140974E-01, * 9.93467537479668965E-02,9.78414999033007314E-02, * 9.64026543164874854E-02,9.50254735405376642E-02, * 9.37056752969190855E-02,9.24393823875012600E-02, * 9.12230747245078224E-02,9.00535481254756708E-02, * 8.89278787739072249E-02,8.78433924473961612E-02, * 8.67976377754033498E-02,8.57883629175498224E-02, * 8.48134951571231199E-02,8.38711229887106408E-02, * 8.29594803475290034E-02,8.20769326842574183E-02/ DATA A(49), A(50) /8.12219646354630702E-02,8.03931690779583449E-02 * / C----------------------------------------------------------------------- C SQRT(PI)/2 C----------------------------------------------------------------------- DATA HRTPI /8.86226925452758014E-01/ C C***FIRST EXECUTABLE STATEMENT BSKIN IERR = 0 NZ=0 IF (X.LT.0.0E0) IERR=1 IF (N.LT.0) IERR=1 IF (KODE.LT.1 .OR. KODE.GT.2) IERR=1 IF (M.LT.1) IERR=1 IF (X.EQ.0.0E0 .AND. N.EQ.0) IERR=1 IF (IERR.NE.0) RETURN IF (X.EQ.0.0E0) GO TO 300 I1M = -I1MACH(12) T1 = 2.3026E0*R1MACH(5)*I1M XLIM = T1 - 3.228086E0 T2 = T1 + N + M - 1 IF (T2.GT.1000.0E0) XLIM = T1 - 0.5E0*(LOG(T2)-0.451583E0) IF (X.GT.XLIM .AND. KODE.EQ.1) GO TO 320 TOL = MAX(R1MACH(4),1.0E-18) I1M = I1MACH(11) C----------------------------------------------------------------------- C LN(NLIM) = 0.125*LN(EPS), NLIM = 2*KTRMS+N C----------------------------------------------------------------------- XNLIM = 0.287823E0*(I1M-1)*R1MACH(5) ENLIM = EXP(XNLIM) NLIM = INT(ENLIM) + 2 NLIM = MIN(100,NLIM) NLIM = MAX(20,NLIM) M3 = MIN(M,3) NL = N + M - 1 IF (X.GT.2.0E0) GO TO 130 IF (N.GT.NLIM) GO TO 280 C----------------------------------------------------------------------- C COMPUTATION BY SERIES FOR 0.LE.X.LE.2 C----------------------------------------------------------------------- NFLG = 0 NN = N IF (NL.LE.2) GO TO 60 M3 = 3 NN = 0 NFLG = 1 60 CONTINUE XP = 1.0E0 IF (KODE.EQ.2) XP = EXP(X) DO 80 I=1,M3 CALL BKISR(X, NN, W, IERR) IF(IERR.NE.0) RETURN W = W*XP IF (NN.LT.N) GO TO 70 KK = NN - N + 1 Y(KK) = W 70 CONTINUE YS(I) = W NN = NN + 1 80 CONTINUE IF (NFLG.EQ.0) RETURN NS = NN XP = 1.0E0 90 CONTINUE C----------------------------------------------------------------------- C FORWARD RECURSION SCALED BY EXP(X) ON ICASE=0,1,2 C----------------------------------------------------------------------- FN = NS - 1 IL = NL - NS + 1 IF (IL.LE.0) RETURN DO 110 I=1,IL T1 = YS(2) T2 = YS(3) YS(3) = (X*(YS(1)-YS(3))+(FN-1.0E0)*YS(2))/FN YS(2) = T2 YS(1) = T1 FN = FN + 1.0E0 IF (NS.LT.N) GO TO 100 KK = NS - N + 1 Y(KK) = YS(3)*XP 100 CONTINUE NS = NS + 1 110 CONTINUE RETURN C----------------------------------------------------------------------- C COMPUTATION BY ASYMPTOTIC EXPANSION FOR X.GT.2 C----------------------------------------------------------------------- 130 CONTINUE W = X + 0.5E0 NT = INT(W) IF (NL.GT.NT) GO TO 270 C----------------------------------------------------------------------- C CASE NL.LE.NT, ICASE=0 C----------------------------------------------------------------------- ICASE = 0 NN = NL NFLG = MIN(M-M3,1) 140 CONTINUE KK = (NLIM-NN)/2 KTRMS = MAX(0,KK) NS = NN + 1 NP = NN - M3 + 1 XP = 1.0E0 IF (KODE.EQ.1) XP = EXP(-X) DO 150 I=1,M3 KK = I CALL BKIAS(X, NP, KTRMS, A, W, KK, NE, GR, H, IERR) IF(IERR.NE.0) RETURN YS(I) = W NP = NP + 1 150 CONTINUE C----------------------------------------------------------------------- C SUM SERIES OF EXPONENTIAL INTEGRALS BACKWARD C----------------------------------------------------------------------- IF (KTRMS.EQ.0) GO TO 160 NE = KTRMS + KTRMS + 1 NP = NN - M3 + 2 CALL EXINT(X, NP, 2, NE, TOL, EXI, NZ, IERR) IF(NZ.NE.0) GO TO 320 IF(IERR.EQ.2) RETURN 160 CONTINUE DO 190 I=1,M3 SS = 0.0E0 IF (KTRMS.EQ.0) GO TO 180 KK = I + KTRMS + KTRMS - 2 IL = KTRMS DO 170 K=1,KTRMS SS = SS + A(IL)*EXI(KK) KK = KK - 2 IL = IL - 1 170 CONTINUE 180 CONTINUE YS(I) = YS(I) + SS 190 CONTINUE IF (ICASE.EQ.1) GO TO 200 IF (NFLG.NE.0) GO TO 220 200 CONTINUE DO 210 I=1,M3 Y(I) = YS(I)*XP 210 CONTINUE IF (ICASE.EQ.1 .AND. NFLG.EQ.1) GO TO 90 RETURN 220 CONTINUE C----------------------------------------------------------------------- C BACKWARD RECURSION SCALED BY EXP(X) ICASE=0,2 C----------------------------------------------------------------------- KK = NN - N + 1 K = M3 DO 230 I=1,M3 Y(KK) = YS(K)*XP YSS(I) = YS(I) KK = KK - 1 K = K - 1 230 CONTINUE IL = KK IF (IL.LE.0) GO TO 250 FN = NN - 3 DO 240 I=1,IL T1 = YS(2) T2 = YS(1) YS(1) = YS(2) + ((FN+2.0E0)*YS(3)-(FN+1.0E0)*YS(1))/X YS(2) = T2 YS(3) = T1 Y(KK) = YS(1)*XP KK = KK - 1 FN = FN - 1.0E0 240 CONTINUE 250 CONTINUE IF (ICASE.NE.2) RETURN DO 260 I=1,M3 YS(I) = YSS(I) 260 CONTINUE GO TO 90 270 CONTINUE IF (N.LT.NT) GO TO 290 C----------------------------------------------------------------------- C ICASE=1, NT.LE.N.LE.NL WITH FORWARD RECURSION C----------------------------------------------------------------------- 280 CONTINUE NN = N + M3 - 1 NFLG = MIN(M-M3,1) ICASE = 1 GO TO 140 C----------------------------------------------------------------------- C ICASE=2, N.LT.NT.LT.NL WITH BOTH FORWARD AND BACKWARD RECURSION C----------------------------------------------------------------------- 290 CONTINUE NN = NT + 1 NFLG = MIN(M-M3,1) ICASE = 2 GO TO 140 C----------------------------------------------------------------------- C X=0 CASE C----------------------------------------------------------------------- 300 CONTINUE FN = N HN = 0.5E0*FN GR = GAMRN(HN) Y(1) = HRTPI*GR IF (M.EQ.1) RETURN Y(2) = HRTPI/(HN*GR) IF (M.EQ.2) RETURN DO 310 K=3,M Y(K) = FN*Y(K-2)/(FN+1.0E0) FN = FN + 1.0E0 310 CONTINUE RETURN C----------------------------------------------------------------------- C UNDERFLOW ON KODE=1, X.GT.XLIM C----------------------------------------------------------------------- 320 CONTINUE NZ=M DO 330 I=1,M Y(I) = 0.0E0 330 CONTINUE RETURN END | 3,921 | 10,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.719012 |
http://www.energyvanguard.com/blog-building-science-HERS-BPI/bid/55629/Why-Is-Air-Conditioner-Capacity-Measured-in-Tons | 1,371,720,970,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711240143/warc/CC-MAIN-20130516133400-00032-ip-10-60-113-184.ec2.internal.warc.gz | 423,240,003 | 19,615 | ### Need a Speaker?
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# Energy Vanguard Blog
### Why Is Air Conditioner Capacity Measured in Tons?
A few years ago, a student of mine told a funny story in a home energy rater class. He was an HVAC contractor and said he was installing a new air conditioner for an elderly woman. As he was explaining things to her, he mentioned that they would be installing a 4 ton unit. "Oh, my," she said. "How are you going to get something so big into my back yard?"
The confusion here is completely natural. HVAC and home energy pros find this story funny because when you say an air conditioner is 4 tons, we know it's not weight. It's a number that tells how much heat the air conditioner can remove from the house in an hour. (Let's ignore the issues of nominal vs. actual capacity and AHRI de-rating.) A 4 ton air conditioner is one that can remove 48,000 BTUs of heat per hour from the house. For most people, though, 4 tons means 8000 pounds. (A BTU is a British Thermal Unit, approximately the amount of heat you get from burning one kitchen match all the way down.)
Most pros also know how such a common term as 'ton' turned into a bit of HVAC jargon. Before Willis Carrier invented the modern air conditioner, people used to cool buildings in the summertime with ice harvested from rivers and lakes in the wintertime. A Green Homes America article quotes ice production figures from the 19th century Ice and Refrigeration journal, indicating that the 1890 crop from the Hudson River was about 4 million tons.
OK, so people used to cool and refrigerate with ice. How does that equate to air conditioning capacity in BTUs per hour, you ask? Well, let's get quantitative and find out.
When ice is below freezing and it absorbs heat, the temperature increases. When ice is at its melting point, 32° F, and it absorbs heat, its temperature doesn't change. Instead, it melts. If you've had a physics or chemistry class, you may recall that the amount of heat needed to melt ice is called the latent heat of fusion. In Imperial units, that number is 143 BTUs per pound.
That's actually a lot of heat to pump into a pound frozen water. Once the ice is melted into liquid water, it takes only 1 BTU per pound to raise the temperature 1 degree. So if you've got a pound of ice at 32° F, you put 143 BTUs into it to melt it completely. Then it takes only 180 more BTUs to raise the temperature of that pound of water from 32° F to 212° F, the boiling point.
Anyway, getting back to our main discussion, if you have a ton of ice, it takes (143 BTU/lb) x (2000 lbs) = 286,000 BTUs to melt it completely. You could do that in one hour or 10 hours or a year, depending on how quickly you pump heat into it. Somewhere along the line, though, someone decided to use 1 day—24 hours—as the standard time reference here. If the ice melts uniformly over the 24 hours, it absorbs heat at the rate of 286,000 / 24 hrs = 11,917 BTU/hr.
Rounding that number up makes it a nice, round 12,000 BTU/hr. In air conditioning jargon, then, a ton of AC capacity is equal to 12,000 BTU/hr. There it is.
If you're wondering how this term got institutionalized, it was probably the usual way. People in the industry start using it, and then the professional organizations make it official. An architecture website has a quote from 1912 that claims the American Society of Mechanical Engineers standardized it. It sounds likely, but their numbers don't work out, so I'm gonna go with Honest Abe on this one and remain skeptical (until someone in the comments shows me what's wrong with my thinking anyway).
For the fearless: If you want to read some funny HVAC banter on this topic, check out this thread in the HVAC-Talk forum. And if you figure out what 'heat of zaporization' is, let me know!
Related Articles
It's Called an Air Conditioner — Not an Air Cooler!
Photo of blue ice by ezioman from flickr.com, used under a Creative Commons license. Photo of ice harvest by Robert N. Dennis, in the public domain, from Wikimedia Commons. Abe Lincoln photo and quote from The Canterbury Tales by Chaucer.
Bless you, Allison for bringing unanticipated joy to my day! I laughed out loud about Abe's quote and the credits! Canterbury Tales, Indeed - and I thought it was from Homer's Iliad!
Posted @ Friday, September 21, 2012 9:31 AM by Rob George
Rob G.: And bless you, too! I thought not a single person would notice Chaucer in there. I also laughed out loud.
Posted @ Friday, September 21, 2012 9:34 AM by Allison Bailes
Hmm... and just when I thought you were going to demystify nominal capacity vs. AHRI capacity vs. design capacity.
BTW, the website you linked to with the bit about the ASME had a typo (228,000 vs. 288,000). Here's the actual source:
http://bit.ly/Pwiwuo
Posted @ Saturday, September 22, 2012 3:10 AM by David Butler
David B.: Hey, I did mention that issue here! It's on my list to cover.
Yes, that's exactly what I was referring to when I said, "their numbers don't work out." Thanks for that link to an original source from 1920. The discussion of whether to use 143.5 or 144 BTU/lb for the latent heat of fusion of ice is pretty nerdy stuff.
Posted @ Saturday, September 22, 2012 6:18 AM by Allison Bailes
Very Interesting! Thanks. What I am having a hard time understanding...and maybe it's because I didn't read closely enough or I just don't have the brain capacity to figure it out...but why if the whole thing came from a ton of ice melting in a 24 hour period...which would mean 12,000 BTUs per hour, do we refer to a ton of HVAC capacity as only 12,000 BTUs and not 188,000 BTUs which is where the ton of ice is?
Posted @ Saturday, September 22, 2012 4:08 PM by grusso
grusso: You ask a great question, and it's not because of lack of brain power on your part that this doesn't make sense. It's because of laziness on the part of a lot people who talk. It happens for the same reason that builders and other folks who work with areas all the time say a house is '2500 feet' when they really mean it's 2500 square feet. A ton of AC capacity, as you (almost) correctly state, is 288,000 BTUs, sort of. But it's a rate, not a quantity, and a ton of AC capacity has been defined as 288,000 BTUs melting over 24 hours, or 12,000 BTUs per hour. Without the 'per hour' in there, it looks like a quantity, but it's really a rate.
Posted @ Saturday, September 22, 2012 4:22 PM by Allison Bailes
Expanding a bit further on grusso's question... when folks had to physically handle their energy source around (e.g., ice), a daily metric seemed appropriate, e.g, tons of ice a day. So in that sense, tons of ice was also a rate. But when automatic energy conversion systems like compressors entered the scene, engineers standardized on hourly rates, pretty much across the board.
Yo, Alison. Yeah, you did mention nominal capacity, etc... just to say "Let's ignore..." I'll be looking for your blog on that subject.
Posted @ Saturday, September 22, 2012 8:44 PM by David Butler
What is ton in AC..plz explin details..
Posted @ Monday, May 13, 2013 3:49 AM by anupam
Post Comment Name * Email * Website (optional) Comment * Allowed tags: link, bold, italics Receive email when someone replies. Subscribe to this blog by email. | 1,823 | 7,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-20 | longest | en | 0.957446 |
https://forum.allaboutcircuits.com/threads/remaining-battery-charge-percentage.194217/ | 1,695,301,449,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00862.warc.gz | 292,086,528 | 24,615 | # remaining battery charge percentage
#### Kittu20
Joined Oct 12, 2022
355
I'm currently working on a project that involves monitoring the remaining battery charge percentage of a device.
specifications of the Forte ER34615 battery:
1. Chemistry: Lithium-thionyl chloride (Li-SOCl2)
2. Nominal Voltage: 3.6V
3. Capacity: Typically around 19,000 mAh (milliampere-hours)
Remaining charge percentage = (Current capacity / Maximum capacity) * 100
the typical maximum capacity for the Forte ER34615 battery is around 19,000 mAh (milliampere-hours).
I am not sure I am using correct formula and i don't understand how can we get current capacity.
#### Ian0
Joined Aug 7, 2020
8,361
I'm currently working on a project that involves monitoring the remaining battery charge percentage of a device.
specifications of the Forte ER34615 battery:
1. Chemistry: Lithium-thionyl chloride (Li-SOCl2)
2. Nominal Voltage: 3.6V
3. Capacity: Typically around 19,000 mAh (milliampere-hours)
Remaining charge percentage = (Current capacity / Maximum capacity) * 100
the typical maximum capacity for the Forte ER34615 battery is around 19,000 mAh (milliampere-hours).
I am not sure I am using correct formula and i don't understand how can we get current capacity.
You need to use a technique known as a Coulomb counter. It is a current-controller oscillator which counts down from full capacity. That's what is in most of the "fuel-gauge" ICs
#### Audioguru again
Joined Oct 21, 2019
6,151
The datasheet for the expensive Chinese battery has no details except its capacity is rated at a very low current of only 2mA and when its 3.6V has dropped down to only 2V. Its maximum continuous current is rated at only 200mA.
#### Kittu20
Joined Oct 12, 2022
355
You need to use a technique known as a Coulomb counter. It is a current-controller oscillator which counts down from full capacity. That's what is in most of the "fuel-gauge" ICs
First of all I want to know whether the formula which I am using is correct or not. ? @Audioguru again
If it's correct than i am interested to know how can we get current capacity.?
#### Alec_t
Joined Sep 17, 2013
13,766
There may be some confusion here over the term 'current capacity' It is a reference to the remaining charge available; not to the flow of charge (electrical current).
#### Ian0
Joined Aug 7, 2020
8,361
There may be some confusion here over the term 'current capacity' It is a reference to the remaining charge available; not to the flow of charge (electrical current).
Would "present capacity" be a better choice of phrase?
#### Audioguru again
Joined Oct 21, 2019
6,151
The rated capacity occurs only with a tiny 2mA load and when the battery voltage has dropped down to 2V.
Is 2mA and 2V good enough for powering your circuit?
#### crutschow
Joined Mar 14, 2008
32,839
To perhaps clarify a "coulomb counter" measures the current into or out of the battery and multiplies by the time, to estimate the mAh left in the battery.
It does not look at the voltage to make the estimate.
#### Ya’akov
Joined Jan 27, 2019
8,145
Hmm... what you seem to want to know when you say "current capacity" is the SoC (State of Charge).
Different cell chemistries allow different methods for determining SoC. Some allow a pretty good estimate based on one terminal voltage. Oddball gaseous hydrogen cells sometimes used in satellites are particularly easy because the internal pressure is directly proportional to SoC.
But you have a lithium cell. It is impossible to reliably know SoC for a lithium cell without accounting for all the charge that went in and how much has come out. That's what a coulomb counter is all about.
A coulomb (C), is the SI unit for quantity of charge. It is defined as the amount of charge that moves past a point in a conductor in 1 second when the current is 1A.
A coulomb counter is the name for the device at the heart of a battery fuel gauge like the one in a laptop. The coulomb counter watches the charging process and counts the coulombs of charge that go into the battery. When the battery begins to discharge, it subtracts the number of coulombs leaving the battery from the total charge it saw go in. This is a close approximation of SoC.
It is not prefect due to losses and unknown charge/discharge. This is why for a laptop fuel gauge to be as accurate as possible, you must fully discharge the battery, then fully charge it. At that point the coulomb counter has a good accounting of the charge in the battery and can add and deduct as charge-discharge cycles happen.
So, to get the "current capacity" (say SoC, instead) you will need to use a coulomb counter. Fortunately options are abundant and some are very easy to use. You can get breakouts of counters from places like Adafruit and Sparkfun for learning and evaluation, or clones of those breakouts from the usual sources.
#### nsaspook
Joined Aug 27, 2009
11,738
The OP is lucky to have picked a very linear (energy delivered per coulomb circulated) primary battery to monitor SoC where a simple a coulomb counter is effective. The details for tracking Flooded Lead Acid batteries SoC (where some coulombs are more equal than others) , with any type of precision, with dynamical load and charging profiles, between full charge reset points is a good way to explore exactly how batteries are not electronic devices with lots of nth degree deviations from linearity.
https://github.com/nsaspook/mbmc_k42/blob/master/mbmc_k42.X/bsoc.h
https://github.com/nsaspook/mbmc_k42/blob/master/mbmc_k42.X/bsoc.c
Last edited:
#### Kittu20
Joined Oct 12, 2022
355
I am providing some values and interested to know how the battery performs within the voltage range of 3.6V to 3.3V.
Battery provide maximum 3.6 V.
Load require minimum 3.3 V and maximum 3.6V.
Battery discharge when it drop voltage from 3.6 to 2V.
how the battery performs within the voltage range of 3.6V to 3.3V.
as the battery voltage decreases 3.6V to 3.3V during , does current delivered by the battery may also decrease? Does battery provide constant of 24mA?
#### Ian0
Joined Aug 7, 2020
8,361
how the battery performs within the voltage range of 3.6V to 3.3V.
as the battery voltage decreases 3.6V to 3.3V during , does current delivered by the battery may also decrease? Does battery provide constant of 24mA?
That depends entirely on the load.
If the load is resistive, the current will decrease proportional to the voltage.
If the load takes a constant power (for instance a LED with a switched-mode current regulator) the current will increase as the battery voltage decreases.
#### nsaspook
Joined Aug 27, 2009
11,738
I am providing some values and interested to know how the battery performs within the voltage range of 3.6V to 3.3V.
Battery provide maximum 3.6 V.
Load require minimum 3.3 V and maximum 3.6V.
Battery discharge when it drop voltage from 3.6 to 2V.
how the battery performs within the voltage range of 3.6V to 3.3V.
as the battery voltage decreases 3.6V to 3.3V during , does current delivered by the battery may also decrease? Does battery provide constant of 24mA?
Do the experiment and let us know what the results are.
#### Audioguru again
Joined Oct 21, 2019
6,151
The Chinese battery you found has a datasheet without detailed specs. Maybe because it has poor performance.
I found a German battery of the same size and chemistry that has detailed specs. Its voltage does not drop during its discharging time but its capacity at your high load of 24mA kills it after about 600 hours.
#### Attachments
• 102.8 KB Views: 4 | 1,892 | 7,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.937194 |
https://discourse.looker.com/t/how-to-split-a-number-with-dots-in-a-column/13575 | 1,566,645,791,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320734.85/warc/CC-MAIN-20190824105853-20190824131853-00378.warc.gz | 438,545,485 | 5,248 | # How to split a number with dots in a column
Hi!
I have this column in my table which I can’t change the format.
17.298.3
17.46.1
17.4.7
17.298.12
etc…
Actually, this should be an ordinal number, that’s what I am looking for to do.
The 1st part (17) is useless.
After that, I need to use to 2nd and the 3rd number and apply them a formula in order to find out the ordinal number.
I.e.:
298.3 is (298-3)*15+20
46.1 is (46-3)*15+20
etc…
My biggest problem here is how to split the column in order to get 3 “new” columns.
Thanks a lot!!!
Guiche
It begs for some regular expressions functions in Looker
Is the first “useless” number always 2 digits?
You can try split function in your dimension, this works for snowflake DWH.
Blockquote select split(‘127.0.0.1’, ‘.’)[0];
select split(‘127.0.0.1’, ‘.’)[1];
I’d say yes
I’ve found a work around guys, do not spend more time in my case please xD | 278 | 904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.860593 |
https://lscsoft.docs.ligo.org/lalsuite/lal/group___unit_raise__c.html | 1,656,658,412,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00180.warc.gz | 406,490,570 | 4,382 | LAL 7.1.7.1-15d842a
Module UnitRaise.c
## Detailed Description
Raises an LALUnit structure to a specified rational power.
This function raises the LALUnit structure *input to the rational power *power. In this way, units such as $$\mathrm{s}^{1/2}$$ and $$\mathrm{m}^{-1}$$ can be created using existing units.
### Algorithm
The function first multiplies the overall power of ten input->powerOfTen by the rational number *power, checking to make sure that the resulting power is still an integer. It then multiplies each of the rational powers in *input by *power by naïve multiplication of rational numbers
$\left(\frac{N_1}{1+D_1}\right)\left( \frac{N_2}{1+D_2} \right) = \frac{N_1 N_2}{1 + (1+D_1)(1+D_2)-1}$
and then calls XLALUnitNormalize() to bring the result into standard form.
## Prototypes
LALUnitXLALUnitRaiseRAT4 (LALUnit *output, const LALUnit *input, const RAT4 *power)
Raises a LALUnit structure to a rational power given by the RAT4 structure power. More...
LALUnitXLALUnitRaiseINT2 (LALUnit *output, const LALUnit *input, INT2 power)
Raises a LALUnit structure to an integer power power. More...
LALUnitXLALUnitSquare (LALUnit *output, const LALUnit *input)
Produces the square of a LALUnit structure. More...
LALUnitXLALUnitSqrt (LALUnit *output, const LALUnit *input)
Produces the square-root of a LALUnit structure. More...
LALUnitXLALUnitInvert (LALUnit *output, const LALUnit *input)
UNDOCUMENTED. More...
## ◆ XLALUnitRaiseRAT4()
LALUnit* XLALUnitRaiseRAT4 ( LALUnit * output, const LALUnit * input, const RAT4 * power )
Raises a LALUnit structure to a rational power given by the RAT4 structure power.
Definition at line 56 of file UnitRaise.c.
## ◆ XLALUnitRaiseINT2()
LALUnit* XLALUnitRaiseINT2 ( LALUnit * output, const LALUnit * input, INT2 power )
Raises a LALUnit structure to an integer power power.
Definition at line 106 of file UnitRaise.c.
## ◆ XLALUnitSquare()
LALUnit* XLALUnitSquare ( LALUnit * output, const LALUnit * input )
Produces the square of a LALUnit structure.
Definition at line 120 of file UnitRaise.c.
## ◆ XLALUnitSqrt()
LALUnit* XLALUnitSqrt ( LALUnit * output, const LALUnit * input )
Produces the square-root of a LALUnit structure.
Definition at line 133 of file UnitRaise.c.
## ◆ XLALUnitInvert()
LALUnit* XLALUnitInvert ( LALUnit * output, const LALUnit * input )
UNDOCUMENTED.
Definition at line 144 of file UnitRaise.c. | 708 | 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": 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} | 3.0625 | 3 | CC-MAIN-2022-27 | latest | en | 0.541548 |
https://www.mcas.k12.in.us/Page/19629 | 1,590,923,791,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00046.warc.gz | 802,334,779 | 67,067 | # E-Learning Day 10 - 3/16/2020
• Welcome to our E-Learning Day #10!
Today’s date is Monday, March 16, 2020. Below you will find your assignments for the day!
*I can be reached by email if you have any questions. kpryor02@mcas.k12.in.us
* Please label each day's assignment AND PAPER CLIP/STAPLE all assignments.
* Please pay attention to which math book you will be working out of EACH day. It will either be PRACTICE AND PROBLEM SOLVING or INSTRUCTION BOOK. There is a difference!
**MAKE SURE TO COMPLETE THE EXIT TICKET FOR EACH DAY! THIS WILL COUNT FOR ATTENDANCE!!**
I CAN…….write the missing digraphs in words.
Steps of the Lesson: (30 minutes)
1. Watch this lesson on phonics sounds.
1. Copy the following words below.
2. Sort the words into the appropriate columns - ie, igh, kn, and wr
Sort the Words!
Word List ie igh kn wr lie high right knock pie knee bright wrong tie might night wrap tight
Work to be turned in: Chart with ie, igh, kn, and wr words.
Writing
I CAN….. write sentences about a topic and use correct punctuation and capitalization.
Steps of the Lesson: (30 minutes)
1. Click on the video The Cat in the Hat, if you have not read it.
1. Choose ONE topic to write about below.
• Write 5 or more sentences about how you would spend the day with The Cat in the Hat if he came to visit you at your house.
• Write 5 or more sentences about how you would get The Cat in the Hat out of your house if he came to visit you.
3. Check your sentences for capitals, end marks and complete thoughts.
Work to be turned in: 5 sentences
Math
I CAN…...... count from any number on the 120 chart.
Steps of the Lesson: (30minutes)
1. Watch the “120 Chart” Day 1 Video
1. Complete pages 117-118 in your iReady Math INSTRUCTION BOOK. NOT the Practice and Problem Solving Book.
2. Answer these questions: How many numbers are across the first row of the 120 chart? How do you know? Do you think there are 10 numbers in the next row? Why? How do the numbers change in each row? In each column?
Work to be turned in: INSTRUCTION BOOK PAGES 117-118 and answers to the questions
EXIT TICKET | 543 | 2,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-24 | latest | en | 0.889583 |
https://www.physicsforums.com/threads/please-help-me-with-this-series-test-thank-you.810539/ | 1,531,801,238,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00304.warc.gz | 931,267,146 | 13,641 | 1. Apr 25, 2015
### sun1234
2. Apr 25, 2015
### micromass
Just because the $n$th term converges to $0$, doesn't mean the series converges. Unless you can explain better why it might imply this in this case.
3. Apr 25, 2015
### micromass
And the $p$-test only works for positive series (a series whose terms are positive).
4. Apr 25, 2015
### sun1234
That's what I think of. Also how do you know when to test for absolute converges and conditional converges? Thank you for trying to help.
5. Apr 25, 2015
### Staff: Mentor
Instead of answering that question, I think it would be a good idea for you to step back and take a closer look at the two tests you used, the p-series test and what you call the "nth term test."
As already stated, the p-series applies only to series consisting of positive terms. You also misused the other test that you used. What exactly does that test say?
6. Apr 26, 2015
### HallsofIvy
If a series "converges absolutely" then there is no point in asking if it converges conditionally. So it would seem to make sense to first try to show that a series converges absolutely and only if it doesn't try to show that it converges conditionally.
One test you do not mention is the "alternating sequence test": if, for $a_n> 0$, $\lim_{n\to 0} a_n= 0$ then $\sum (-1)^n a_n$ converges. | 368 | 1,323 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-30 | latest | en | 0.931743 |
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